BLOW-UP PROBLEMS FOR THE HEAT EQUATION WITH LOCAL
NONLINEAR NEUMANN BOUNDARY CONDITIONS
By
Xin Yang

A DISSERTATION
Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of
Mathematics — Doctor of Philosophy
2017

ABSTRACT
BLOW-UP PROBLEMS FOR THE HEAT EQUATION WITH LOCAL
NONLINEAR NEUMANN BOUNDARY CONDITIONS
By
Xin Yang
This thesis studies the blow-up problem for the heat equation ut = ∆u in a C 2 bounded
open subset Ω of Rn (n ≥ 2) with positive initial data u0 and a local nonlinear Neumann
∂u = uq on partial boundary Γ ⊆ ∂Ω for some q > 1 and ∂u = 0 on
boundary condition: ∂n
1
∂n

the rest of the boundary. The motivation of the study is the partial damage to the insulation
on the surface of space shuttles caused by high speed flying subjects.
First, we establish the local existence and uniqueness of the classical solution for such a
problem. Secondly, we show the finite-time blowup of the solution and estimate both upper
and lower bounds of the blow-up time T ∗ . In addition, the asymptotic behaviour of T ∗ on
q, M0 (the maximum of the initial data) and |Γ1 | (the surface area of Γ1 ) are studied.
• As q

1, the order of T ∗ is exactly (q − 1)−1 .
0, the order of T ∗ is at least ln(M0−1 ); if the region near Γ1 is convex, then

• As M0

−(q−1)

the order of T ∗ is at least M0
−(q−1)

least M0

/ ln(M0−1 ); if Ω is convex, then the order of T ∗ is at

. On the other hand, if the initial data u0 does not oscillate too much,
−(q−1)

then the order of T ∗ is at most M0
• As |Γ1 |

.

0, the order of T ∗ is at least ln(|Γ1 |−1 ) and at most |Γ1 |−1 . If the region
1

−
near Γ1 is convex, then the order of T ∗ is at least |Γ1 | n−1

and |Γ1 |−1

ln |Γ1 |−1

2

ln |Γ1 |−1 for n ≥ 3

for n = 2. If Ω is convex, then the order of T ∗ is at least

1

−
|Γ1 | n−1 for n ≥ 3 and |Γ1 |−1 ln |Γ1 |−1 for n = 2.

Finally, we provide two strategies from engineering point of view (which means by changing
the setup of the original problem) to prevent the finite-time blowup. Moreover, if the region
near Γ1 is convex, then one of the strategies is applied to bound the solution from above by
M1 for any M1 > M0 .
For the space shuttle mentioned in the motivation of this thesis, Γ1 is on its left wing
of the shuttle, so the region near Γ1 is indeed convex. In addition, the relation between T ∗
and small surface area |Γ1 | is of particular interest for this problem. As an application of
the above estimates to this problem, let n = 3 and |Γ1 |
1

|Γ1 |− 2

0, then the order of T ∗ is between

ln |Γ1 |−1 and |Γ1 |−1 . On the other hand, one of the strategies can be applied to

prevent the temperature from being too high.
This thesis seems to be the first to systematically study the heat equation with piecewise
continuous Neumann boundary conditions. It also seems to be the first to investigate the
relation between T ∗ and |Γ1 |, especially when |Γ1 |

0. The key innovative part of this

thesis is Chapter 4. First, the new method developed in Chapter 4 is able to derive a
lower bound for T ∗ without the convexity assumption of the domain which was a common
requirement in the historical works. Secondly, even for the convex domains, the lower bound
estimate obtained by this new method improves the previous results significantly. Thirdly,
this method does not involve any differential inequality argument which was an essential
technique in the past on the blow-up time estimate.

ACKNOWLEDGMENTS

I would like to express my deepest gratitude to my thesis adviser, Dr. Zhengfang Zhou,
for his inspiring guidance and tremendous support during my graduate study. He not only
introduces such an interesting thesis problem to me, but also provides technical suggestions
on crucial steps. Moreover, his consistent encouragement gives me confidence to conquer all
the difficulties.
I would like to thank Dr. Keith Promislow, Dr. Jeffrey Schenker, Dr. Willie Wong and
Dr. Baisheng Yan for serving as my doctoral committee members and providing invaluable
suggestions.
I am grateful to Dr. Gabriel Nagy, Dr. Keith Promislow, Dr. Benjamin Schmidt and
Dr. Baisheng Yan for supporting my postdoctoral applications.
I also would like to thank the professors who supported my research in one way or another:
Dr. Keith Promislow, for discussions on the math model of this thesis; Dr. Benjamin
Schmidt, for discussions on the geometric properties of the boundary; Dr. Baisheng Yan,
for suggestions on my research from time to time. In addition, I appreciate the research
assistantships provided by Dr. Patricia Lamm and Dr. Zhengfang Zhou which facilitate my
research progress.
During these years’ graduate study, I have broaden my view by taking courses in varies
areas. Special thanks go to Dr. Gabor Francsics, Dr. Jun Kitagawa, Dr. Keith Promislow,
Dr. Willie Wong and Dr. Zhengfang Zhou on PDEs; Dr. Jeffrey Schenker, Dr. Ignacio
Uriarte-Tuero and Dr. Alexander Volberg on Analysis; Dr. Yingda Cheng, Dr. Di Liu and
Dr. Jianliang Qian on Numerical analysis; Dr. Thomas Parker and Dr. Benjamin Schmidt

iv

on Geometric analysis.
I also want to take this opportunity to thank all my friends for their help and sharing of
challenges and happiness together.
Last but not least, I owe a debt of gratitude to my family for their whole-hearted love,
understanding and support.

v

TABLE OF CONTENTS

LIST OF FIGURES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viii
Chapter 1 Introduction . . . . . . . . . . . . . . . . . . . .
1.1 Motivation and mathematical model . . . . . . . . . . .
1.2 Historical works . . . . . . . . . . . . . . . . . . . . . . .
1.2.1 Blow-up phenomenon for Cauchy problems . . . .
1.2.2 Parabolic blow-up problems in bounded domains
1.3 Difficulties and main ideas . . . . . . . . . . . . . . . . .
1.4 Notations . . . . . . . . . . . . . . . . . . . . . . . . . .
1.5 Main results . . . . . . . . . . . . . . . . . . . . . . . . .

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Chapter 2 Existence and Uniqueness . . . . . . . . . . . . . . . .
2.1 Main theorem and outline of the approach . . . . . . . . . . . .
2.2 Jump relation . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.1 Heat potentials with continuous density . . . . . . . . . .
2.2.2 Heat potentials with piecewise continuous density . . . .
2.3 Linear case . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3.1 Definition of the local solution . . . . . . . . . . . . . . .
2.3.2 Auxilliary lemmas . . . . . . . . . . . . . . . . . . . . . .
2.3.3 Existence . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3.4 Comparison principles, uniqueness and global solution . .
2.4 Nonlinear case . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4.1 Definition of the local solution . . . . . . . . . . . . . . .
2.4.2 Comparison principles and uniqueness . . . . . . . . . .
2.4.3 Existence . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4.4 Maximal solution and application to the target problem .

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Chapter 3 Upper Bound Estimate of the Blow-up Time
3.1 Main theorem and outline of the approach . . . . . . . .
3.2 Approximation for C 2 domain from inside . . . . . . . .
3.3 Approximation for the solution by smoother functions . .
3.4 Upper bound on life span: case of C 2 domain . . . . . .

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Chapter 4 Lower Bound Estimate of the Blow-up
4.1 Main theorems and outline of the approach . . . .
4.2 Weak solution and representation formula . . . .
4.2.1 Weak solution . . . . . . . . . . . . . . . .
4.2.2 Representation formula . . . . . . . . . . .
4.2.3 Time-shifted representation formula . . . .
4.3 Lower bound on life span: case of C 2 domain . .

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vi

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Chapter 5 Prevention of Blowup . . . . . . . . . . . . . . . . . . . . . . . . .
5.1 Main theorems and outline of the approach . . . . . . . . . . . . . . . . . . .
5.2 Repairing the broken part . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2.1 Prevention of finite-time blowup . . . . . . . . . . . . . . . . . . . . .
5.2.2 Control of the solution under a given value for convex domains . . . .
5.2.3 Control of the solution under a given value for the domain with local
convexity near Γ1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.3 Adding a pump . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

148
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4.4

4.5

4.6

4.3.1 A traditional way by Gronwall-type technique . . . . . . . . . .
4.3.2 Better estimate by a new method . . . . . . . . . . . . . . . . .
Lower bound on life span: case of convex C 2 domain . . . . . . . . . .
4.4.1 Main idea . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4.2 Auxiliary lemmas . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4.3 Proof of Theorem 4.1.4 . . . . . . . . . . . . . . . . . . . . . . .
Lower bound on life span: case of C 2 domain with local convexity near
4.5.1 Estimates for boundary integrals . . . . . . . . . . . . . . . . .
4.5.2 Proof of Theorem 4.1.3 . . . . . . . . . . . . . . . . . . . . . .
Comparison with previous works . . . . . . . . . . . . . . . . . . . . . .

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Γ1
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161
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BIBLIOGRAPHY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171

vii

LIST OF FIGURES

Figure 1.1: Mathematical Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2

Figure 3.1: Γ1,j . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

75

Figure 5.1: Model with a pump . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151

viii

Chapter 1
Introduction

1.1

Motivation and mathematical model

This thesis is partially motivated by the Space Shuttle Columbia disaster in 2003. When
the space shuttle was launched, a piece of foam broke off from its external tank and struck
the left wing damaging the insulation there. As a result, the shuttle disintegrated during
its reentry to the atmosphere due to the enormous heat generated near the damaged part.
Actually, such damages on the wings were also found in previous shuttles too. But the
engineers suspected that the previous damages were so small that the shuttle managed to
land safely before the temperature became too high. Motivated by this, this thesis intends
to study the relation between the blow-up time of the temperature inside the shuttle and
the area of the broken part on the left wing. The goal is to rigorously verify the engineers’
conjecture from a mathematical perspective. In addition, some strategies that could prevent
the blowup will also be explored.
In Figure 1.1, let u be the inside temperature of the space shuttle. During the re-entry
of the shuttle to the atmosphere, the air was compressed at a very high speed. Then many
chemical reactions happened and produced enormous radiative heat flux, which was the main
source of the heat. In physics, the radiative heat flux is proportional to the fourth power of
the temperature. Due to this nonlinear effect, we consider a simplified model as follows. On

1

Γ2 :

∂u
∂n

Ω

=0

ut − ∆u = 0

Γ1 :

∂u
∂n

= uq

Figure 1.1: Mathematical Model
∂u = H(u) ∼ uq for some q > 1; on the other part Γ , ∂u = 0, since
the broken part Γ1 , ∂n
2 ∂n

the insulation there is intact. Finally, the inside temperature of the shuttle is supposed to
satisfy the heat equation. Thus, the following math model is adopted (see more descriptions
in Section 1.4).




ut (x, t) = ∆u(x, t) in Ω × (0, T ],








 ∂u(x,t)
= uq (x, t) on Γ1 × (0, T ],
∂n(x)


∂u(x,t)

=0


∂n(x)






 u(x, 0) = u0 (x)

(1.1.1)

on Γ2 × (0, T ],
in

Ω,

where
q > 1, Γ1 = ∅, u0 ∈ C 1 (Ω), u0 (x) ≥ 0, u0 (x) ≡ 0.

2

(1.1.2)

1.2
1.2.1

Historical works
Blow-up phenomenon for Cauchy problems

In the seminal work [9], Fujita studied the Cauchy problem



 ut (x, t) − ∆u(x, t) = up (x, t) in Rn × (0, ∞),

(1.2.1)

in Rn ,



 u(x, 0) = ψ(x)

where n ≥ 1, ψ ∈ C 2 (Rn ) is nonnegative and ψ, Di ψ, Dij ψ are all bounded on Rn . It
is shown that if 1 < p < 1 + n2 , then the only nonnegative global solution is u ≡ 0, and
if p > 1 + n2 , then there exist positive global solutions for positive and sufficiently small
ψ. Since then, there is vast literature studying the nonlinear blow-up phenomenons. The
number 1 + n2 is called the critical power in the sense that when p < 1 + n2 , any positive
solution blows up in finite time; when p > 1 + n2 , there exist positive global solutions. The
existence of such a critical power is a feature of this kind of blow-up problem. The study
of the borderline case is more involved and usually obtained separately after the subcritical
and supercritical are established. In the works [14] and [19], it is shown that the critical
power p = 1 + n2 case belongs to the blow-up regime.
Similar questions were also asked for the nonlinear wave equations and the situation there
is more complicated.



utt (x, t) − ∆u(x, t) = |u|p (x, t) in Rn × (0, ∞),




u(x, 0) = ψ0 (x)
in Rn ,





 u (x, 0) = ψ (x)
in Rn ,
t
1

3

(1.2.2)

where n ≥ 1, ψ0 and ψ1 are nonnegative, compactly supported and let either of them be
positive somewhere. In the pioneering work [17], John showed that when n = 3, the critical
√
√
power for (1.2.2) is p = 1 + 2. Again this means if 1 < p < 1 + 2, then any solution
√
blows up in finite time; if p > 1 + 2, then there exist global solutions for suitably small
initial data. For general dimensions, the problem is also called the Strauss conjecture and
the critical power is conjectured to be the positive root of the quadratic equation below for
n ≥ 2 and infinity for n = 1.

(n − 1)p2 − (n + 1)p − 2 = 0.

Now this guess has been confirmed after several decades’ work. The subcritical cases can be
found in [13, 39]. The supercritical cases are dealt with in [10, 12, 26]. Finally, the borderline
cases are also proved to be in the blow-up regime, see [37,46]. The one dimensional case was
discussed in [13] and [18].
It is also interesting to notice a problem which combines both nonlinear heat and wave
equations.



ut (x, t) + utt (x, t) − ∆u(x, t) = |u|p (x, t) in Rn × (0, ∞),




u(x, 0) = ψ0 (x)
in Rn ,





 u (x, 0) = ψ (x)
in Rn ,
t
1
where ψ0 and ψ1 are compactly supported. See [42, 47] for more details.

4

1.2.2

Parabolic blow-up problems in bounded domains

Now let us focus on the parabolic type of nonlinear equations. Besides the Cauchy problems,
people also study the boundary value problems including both Dirichlet and Neumann types.



ut (x, t) − ∆u(x, t) = f x, t, u(x, t)






in

on ∂Ω × (0, T ],

F x, t, u(x, t) = 0






 u(x, 0) = ψ(x)
or

Ω × (0, T ],




ut (x, t) − ∆u(x, t) = f x, t, u(x, t)





∂u(x,t)

in

Ω,

in

Ω × (0, T ],

on ∂Ω × (0, T ],

= F x, t, u(x, t)
∂n(x)






 u(x, 0) = ψ(x)

in

(1.2.3)

(1.2.4)

Ω.

For detailed discussions on the history, we refer the readers to the surveys [5, 24] and the
books [7, 15, 35].
The typical examples are



ut (x, t) − ∆u(x, t) = up (x, t) in






on ∂Ω × (0, T ],

u(x, t) = 0






 u(x, 0) = ψ(x)
and

Ω × (0, T ],

in




ut (x, t) − ∆u(x, t) = 0 in





∂u(x,t)

= uq (x, t)
∂n(x)






 u(x, 0) = ψ(x)

Ω,

Ω × (0, T ],

on ∂Ω × (0, T ],
in

(1.2.5)

(1.2.6)

Ω,

where p > 1, q > 1 and the initial data is positive. But the blow-up properties of (1.2.5) and
5

(1.2.6) are quite different from (1.2.1). More precisely, for the problem (1.2.5), there exists
some positive global solution, see [28, 38]. On the other hand, for the problem (1.2.6), any
solution blows up in finite time, see [16, 36, 44].
For the more general problems (1.2.3) and (1.2.4), the research topics include the local
and global existence and uniqueness of the solutions [1–4, 21, 27, 44]; nonexistence of global
solutions and upper bound for the blow-up time [16, 21–23, 25, 27, 31, 36, 44]; lower bound for
the blow-up time [31–34, 43]; blow-up sets, blow-up rate and the asymptotic behaviour of
the solutions near the blow-up time [8, 11, 16, 21, 27, 29, 36, 45].
When considering the bounds of the blow-up time, the upper bound is usually related to
the nonexistence of the global solutions and this area has brought extensive attention over
several decades. Various methods on this issue have been developed, such as the comparison
method, the concavity method, the Green’s function method, the energy method, and the
unbounded Fourier coefficient method (see [23]). Most methods have in common that they
first consider some nonlinear functional of the solution and try to establish a first order
differential inequality for that functional, then it is shown that such a differential inequality
can not hold beyond some finite time T which serves as an upper bound.
The lower bound was not studied as much in the past but was paid more attention in
recent years. However, the lower bound can be argued to be more useful in practice, since
it provides an estimate of the safe time. In contrast to the upper bound case, not many
methods have been developed to deal with the lower bound. But the existing methods again
have in common that they first consider some nonlinear functional of the solution and try to
establish a first order differential inequality for that functional, then it is shown that such
differential inequality will hold for at least some time T , which serves as a lower bound.

6

1.3

Difficulties and main ideas

Generally speaking, there are two main difficulties. First, Although there has been vast
literature on the blow-up problem of the parabolic type, few of them deal with discontinuous
Neumann boundary conditions. Second, the existing works on the estimate of the lower
bound of the blow-up time only work for convex domains. But by examining the graph
1.1, the domain is clearly not convex. In the following, we discuss these difficulties and the
corresponding strategies in more details.
First, for the theory on existence and uniqueness of the solution to (1.2.4), the key tool
is the jump relation of the single-layer potentials (see Theorem 2.2.1). In order to generalize the theory to the linear problem (2.3.1) with piecewise continuous Neumann boundary
conditions, we establish an adapted version of the jump relation in Theorem 2.2.6. Taking
advantage of this adapted relation, a classical solution can be constructed to satisfy (2.3.1)
pointwise. In addition, such a solution also fits the condition (2.3.2). By imposing the condition (2.3.2) into the definition of the solution (see Definition 2.3.1), the uniqueness follows
from the maximum principle and the Hopf lemma. After the theory is established for the
linear problem, the existence and uniqueness for the nonlinear case (2.4.1) will be justified
by applying the iterative arguments and fixed point theorems.
Secondly, for the estimate of the upper bound for the blow-up time, we adopt the idea
in [36] which introduces a suitable energy function and shows the finite blowup of this energy
function. But due to the discontinuity of the normal derivative, the original argument does
not carry through directly. So we need to additionally introduce a sequence of approximated
solutions {vj }j≥1 to u, see (3.3.5), to justify the argument.
Thirdly, for the estimate of the lower bound for the blow-up time, our method is new

7

in order to deal with more general domains. In Subsection 4.3.2, the lower bound of T ∗ is
derived without any convexity assumption of the domain which is a common requirement in
the previous works. Let M (t) denote the supremum of the solution on Ω × [0, t]. The idea
is to chop the range of M (t) into suitable pieces and find a lower bound for the time spent
in each piece by analysing the representation formula of the solution. Then adding all these
lower bounds together yields a lower bound for T ∗ . This strategy does not introduce any
differential inequalities which often appeared in the historical works on the blow-up time
estimate. The proofs in Subsection 4.4.3 and Subsection 4.5.2 adopt the similar idea but
with some convexity assumptions on the domain. These assumptions make it possible to
chop the range of M (t) in a more delicate way to improve the estimate.
Finally, for the strategies to prevent the finite-time blowup, the ideas are similar to those
in Chapter 4.

1.4

Notations

In this thesis, unless stated otherwise,
• Ω represents a bounded open subset in Rn (n ≥ 2) with C 2 boundary ∂Ω.
• Γ1 and Γ2 denote two disjoint relatively open subsets of ∂Ω. ∂Γ1 = ∂Γ2

Γ is C 1 .

Moreover, Γ1 = ∅ and ∂Ω = Γ1 ∪ Γ ∪ Γ2 .
• |Ω| and |Γ1 | represents the volume of Ω and the surface area of Γ1 respectively. That
is,
|Ω| =

|Γ1 | =

dx,
Ω

8

dS(x).
Γ1

The normal derivative in (1.1.1) is understood in the following way: for any (x, t) ∈ ∂Ω ×
(0, T ],
∂u(x, t)
∂n(x)

−
lim (Du)(xh , t) · →
n (x),
+

(1.4.1)

h→0

−
where →
n (x) denotes the exterior unit normal vector at x and xh

−
x − h→
n (x) for x ∈ ∂Ω.

Since ∂Ω is C 2 , xh belongs to Ω when h is positive and sufficiently small.
In the above notations, Γ1 is not allowed to be empty, since otherwise the blowup will not
occur. On the other hand, Γ2 is allowed to be empty and in that case, problem (1.1.1) has
been studied extensively in the past. In Section 4.6, the results obtained in this thesis will be
compared with the previous results when Γ2 is empty. Finally, in some extreme situations, Γ
may also be empty. For example, let Ω = {x ∈ Rn : 12 < |x| < 1}, Γ1 = {x ∈ Rn : |x| = 21 }
and Γ2 = {x ∈ Rn : |x| = 1}. It is worth mentioning that all the results in this thesis also
apply to this situation.
For any function f : A → R, we follow the convention to denote the supremum norm of
f to be
||f ||∞,A = sup |f (x)|.
x∈A

When there is no ambiguity, ||f ||∞ will be used short for ||f ||∞,A . For any T > 0, define

AT = C 2,1 (Ω × (0, T ]) ∩ C(Ω × [0, T ])

and

BT = {g : (Γ1 ∪ Γ2 ) × (0, T ] → R for i = 1 or 2, g|Γ ×(0,T ] is uniformly continuous}.
i

9

For any g ∈ BT , the restriction function g|Γ ×(0,T ] (i = 1 or 2) has a unique continuous
i
extension to Γi × [0, T ] which will be denoted as gi . But one should notice that g may not
be able to extend to a continuous function on ∂Ω × (0, T ], since it may have a jump between
Γ1 and Γ2 . We endow BT with the supremum norm:

||g||∞, (Γ ∪Γ )×(0,T ] =
1 2

sup

|g(x, t)|.

(x,t)∈(Γ1 ∪Γ2 )×(0,T ]

It is readily seen that BT is a Banach space.
We write
M0 = max u0 (x)

(1.4.2)

x∈Ω

and denote M (t) to be the supremum of the solution u to (1.1.1) on Ω × [0, t]:

M (t) =

sup

u(x, τ ).

(1.4.3)

(x,τ )∈Ω×[0,t]

Φ always refers to the fundamental solution to the heat equation:

Φ(x, t) =

1
(4πt)n/2

exp −

|x|2
,
4t

∀ (x, t) ∈ Rn × (0, ∞).

(1.4.4)

The surface integral with respect to the variable x will be denoted as dS(x). In addition,
C = C(a, b . . . ) and Ci = Ci (a, b . . . ) represent positive constants which only depend on the
parameters a, b . . . . One should also note that C and Ci may stand for different constants
from line to line.

10

1.5

Main results

The solution to (1.1.1) is understood in the following way.
Definition 1.5.1. For any T > 0, a solution to (1.1.1) on Ω × [0, T ] means a function u ∈
C 2,1 Ω×(0, T ]
∂u(x,t)
∂n(x)

C Ω×[0, T ] that satisfies (1.1.1) pointwise and for any (x, t) ∈ Γ×(0, T ],

exists and
1
∂u(x, t)
= uq (x, t).
∂n(x)
2

(1.5.1)

Definition 1.5.2. The maximal existence time T ∗ for (1.1.1) is defined as

T ∗ = sup{T ≥ 0 : there exists a solution to (1.1.1) on Ω × [0, T ]}.

When T ∗ > 0, a function u ∈ C 2,1 Ω×(0, T ∗ )

C Ω×[0, T ∗ ) is called a maximal solution

if u|Ω×[0,T ] is a solution to (1.1.1) on Ω × [0, T ] for any T ∈ (0, T ∗ ).
Based on these two definitions, the existence and uniqueness theory of the maximal
solution u is established in Theorem 2.1.1. In addition, Theorem 2.1.1 also claims the
positivity of u at any positive time. For convenience, we will just call u to be solution
instead of maximal solution. Actually, we deal with the existence and uniqueness theory in
much more general settings. First, the key tool, jump relation of the single-layer potential
with piecewise continuous density, is set up in Subsection 2.2.2. Secondly, the theory for
linear case is built in Section 2.3. Finally, the theory for the nonlinear case in a very general
form is framed in Section 2.4. The targeted problem (1.1.1) is just a special case in Section
2.4.
Theorem 3.1.1 concludes that the solution u does not exist globally. Moreover, it is the
supremum norm M (t) that blows up first. As a convention, we just call T ∗ to be the blow-up
11

time. If the initial data u0 is strictly positive, then an explicit formula for an upper bound
of T ∗ is given in Theorem 3.1.1.
In Theorem 4.1.1, a lower bound for T ∗ is obtained for any C 2 domain. If it is locally
convex near Γ1 , then Theorem 4.1.3 improves the lower bound estimate significantly. Combining these estimates, the asymptotic behaviour of T ∗ on q, M0 and |Γ1 | are understood
quite well. The following is a summary of the conclusions.
1, the order of T ∗ is exactly (q − 1)−1 .

• As q

0, the order of T ∗ is at least ln(M0−1 ); if the region near Γ1 is convex, then

• As M0

−(q−1)

the order of T ∗ is at least M0
−(q−1)

least M0

/ ln(M0−1 ); if Ω is convex, then the order of T ∗ is at

. On the other hand, if the initial data u0 does not oscillate too much,
−(q−1)

then the order of T ∗ is at most M0
• As |Γ1 |

.

0, the order of T ∗ is at least ln(|Γ1 |−1 ) and at most |Γ1 |−1 . If the region
1

−
near Γ1 is convex, then the order of T ∗ is at least |Γ1 | n−1

and |Γ1 |−1

ln |Γ1 |−1

2

ln |Γ1 |−1 for n ≥ 3

for n = 2. If Ω is convex, then the order of T ∗ is at least

1

−
|Γ1 | n−1 for n ≥ 3 and |Γ1 |−1 ln |Γ1 |−1 for n = 2.

In order to compare with the previous results on the lower bound estimate of T ∗ for the
convex domains and Γ1 = ∂Ω, we also derive a result for the convex domains in Section 4.4,
see Theorem 4.1.4. Since the results will be compared under the assumption that Γ1 = ∂Ω,
the order of the lower bound on |Γ1 | as |Γ1 |
the lower bound on M0 as M0

0 is not of interest. Instead, the order of

0 or M0 → ∞ is more important. So the lower bound in

Theorem 4.1.4 has a better order on M0 , no matter M0

0 or M0 → ∞, than those in

Theorem 4.1.3 and Remark 4.5.11, but it loses order on |Γ1 |−1 as |Γ1 |

12

0.

Finally, we investigate two strategies to prevent the blowup: repairing the broken part
and adding a pump. For the method of repairing the broken part, it is shown in Theorem
5.1.2 that if the area of the broken part decreases at an exponential rate, then the solution
will not blow up in finite time. Furthermore, let the region near Γ1 be convex and M0 denote
the maximum of the initial data. Then Theorem 5.1.4 claims that for any M1 > M0 , the
solution will be bounded by M1 (M1 > M0 ) if the area of the broken part decreases at a
super exponential rate. For the method of adding a pump, it is shown in Theorem 5.1.5 that
by adding a suitable pump near the broken part, the solution will not blow up in finite time.

13

Chapter 2
Existence and Uniqueness

2.1

Main theorem and outline of the approach

The goal of this chapter is to establish the existence and uniqueness of the targeted problem
(1.1.1). The main theorem is as follows.
Theorem 2.1.1. The maximal existence time T ∗ for (1.1.1) is positive and there exists
a unique maximal solution u ∈ C 2,1 Ω × (0, T ∗ ) ∩ C Ω × [0, T ∗ ) to (1.1.1). Moreover,
u(x, t) > 0 for any (x, t) ∈ Ω × (0, T ∗ ).
Since (1.1.1) is a nonlinear problem, the strategy is to first deal with the linear case and
then build the nonlinear case by fixed point theorems. The arguments for the nonlinear
problem is standard, the main difficulty lies in the linear case. The key technique is a
generalization of the jump relation for the heat potentials. Previously, the jump relation of
the heat potentials with continuous density is well-known, but due to the discontinuity of
the normal derivative in (1.1.1), we need to extend the result to the heat potentials with
piecewise continuous density.
The organization of this chapter is as follows. Section 2.2 discusses the jump relation of
the heat potentials with piecewise continuous density. In Section 2.3, we study the linear
problem (2.3.1) and prove the global existence and uniqueness of the classical solution. In
Section 2.4, we first set up the local existence and uniqueness for the nonlinear problem
14

(2.4.1) and then discuss the maximal solution. As a corollary, Theorem 2.1.1 is justified.

2.2

Jump relation

The key tool in the proof of the existence of the solution to the parabolic equations with
Neumann boundary conditions is the jump relation of the single-layer and double-layer potentials. Historically, people study these potentials with continuous density, but in order
to adapt to the current problem, we generalize the results for the potentials with piecewise
continuous density in this section.

2.2.1

Heat potentials with continuous density

Let g ∈ C ∂Ω × [0, T ] . The single-layer heat potential with density g is given by
t

Φ(x − y, t − τ )g(y, τ ) dS(y) dτ,

U (x, t) =
0

∀ (x, t) ∈ Ω × (0, T ],

(2.2.1)

∂Ω

where Φ is defined as in (1.4.4) and dS means the surface integral. The double-layer heat
potential is given by
t

V (x, t) =
0

∂Φ(x − y, t − τ )
g(y, τ ) dS(y) dτ,
∂n(y)
∂Ω

∀ (x, t) ∈ Ω × (0, T ],

(2.2.2)

−
where →
n (y) denotes the exterior unit normal vector at y. The following Theorem 2.2.1 and
2.2.2 are two fundamental properties of the jump relations.
Theorem 2.2.1. Let g ∈ C ∂Ω × [0, T ] and U be the single-layer heat potential defined in

15

(2.2.1). Then for any (x, t) ∈ ∂Ω × (0, T ],

−
lim DU (xh , t) · →
n (x) =

h→0+

t
0

1
∂Φ(x − y, t − τ )
g(y, τ ) dS(y) dτ + g(x, t),
∂n(x)
2
∂Ω

(2.2.3)

−
where xh = x − h→
n (x). Moreover, the convergence in the above limit is uniform on ∂Ω ×
[τ0 , T ] for any τ0 > 0.
Proof. The pointwise convergence of (2.2.3) can be found in Theorem 1, page 137 in Section
−
2, Chapter 5 of [7], note that the author in [7] uses →
ν (x) to denote the interior unit normal
direction at x, so the jump is −g(x, t)/2 which differs a sign with (2.2.3). Actually in
that theorem, it proves the pointwise convergence in much more general cases: arbitrary
parabolic kernels; C 1,α domains; and convergence from interior cones. As a result, the
uniform convergence is not clear in that setting. But if restricted to the heat kernel Φ, C 2
domain case and convergence from the normal direction, then the uniform convergence of
(2.2.3) can be established through the same proof.
Theorem 2.2.2. Let g ∈ C ∂Ω × [0, T ] and V be the double-layer heat potential defined in
(2.2.2). Then

lim V (xh , t) = V (x, t) −

h→0+

1
g(x, t)
2

∀ (x, t) ∈ ∂Ω × (0, T ],

(2.2.4)

−
where xh = x − h→
n (x). Moreover, the convergence in the above limit is uniform on ∂Ω ×
[τ0 , T ] for any τ0 > 0. In particular, V can be continuously extended from Ω × (0, T ] into
Ω × (0, T ].
Proof. We refer the readers to the proof of Theorem 9.5, page 176, Section 2, Chapter 9
−
−
of [20], note that the author in [20] uses →
ν (y), rather than →
n (y), to denote exterior unit
16

normal direction at y. In that proof, it only deals with the dimension n = 2 or 3, but the
proof for arbitrary dimensions can be carried out almost identically.
Proposition 2.2.3. Theorem 2.2.1 and Theorem 2.2.2 are equivalent.
Proof. By the explicit formulas (2.2.1) and (2.2.2), it suffices to show the uniform convergence
on ∂Ω × (0, T ] of the following limit.
t

lim

h→0+ 0
t

∂Ω

−
−
DΦ xh − y, t − τ · →
n (y) − →
n (x) g(y, τ ) dS(y) dτ

−
−
DΦ(x − y, t − τ ) · →
n (y) − →
n (x) g(y, τ ) dS(y) dτ.

=
0

(2.2.5)

∂Ω

By Corollary 3.2.2, there exists σ > 0 such that for any h < σ and for any x ∈ ∂Ω, there
exists an interior ball touching x with radius h. As a result, |xh − x| ≤ |xh − y| for any
y ∈ ∂Ω. Thus,
|x − y| ≤ |x − xh | + |xh − y| ≤ 2 |xh − y|.
Now using the fact ∂Ω ∈ C 2 again, we get

→
−
−
n (y) − →
n (x) ≤ C|y − x| ≤ C |xh − y|.

Consequently,
C |xh − y|2
|x − y|2
−
−
DΦ xh − y, t − τ · →
n (y) − →
n (x) g(y, τ ) ≤
exp − h
4(t − τ )
(t − τ )n/2+1
≤
≤

17

|x − y|2
exp − h
8(t − τ )
(t − τ )n/2
C
C
(t − τ )n/2

exp −

|x − y|2
.
32(t − τ )

Since the right hand side of the above inequality is integrable on ∂Ω × [0, t], if we split the
integral as following
t

t

=
0

∂Ω

+
0

∂Ω

= I + II,
∂Ω

then the integral I is uniformly small and the uniform convergence on II is clear. As a
result, the uniform convergence of (2.2.5) is established.

2.2.2

Heat potentials with piecewise continuous density

The previous subsection introduces jump relation of the single-layer and double-layer potentials with continuous density. But Theorem 2.2.1 and Theorem 2.2.2 are still not enough
for our problems. For example, in order to show the existence of the solution to (2.3.1),
the boundary functions β and g are only assumed in BT , not in C ∂Ω × [0, T ] . Thus we
need to adapt this jump relation to our case. So in this section, we will establish a similar
jump relation for the single-layer heat potential with piecewise continuous density as that
in Theorem 2.2.1. This jump relation for the single-layer heat potential with discontinuous
density in Theorem 2.2.6 will be mainly applied to show the existence and uniqueness of the
solution to (2.3.1) and (2.4.1).
First, we extend the definition of the single-layer heat potential to include the function
space BT in which the function is only piecewise continuous on the boundary. For any
ϕ ∈ BT , define the single-layer heat potential with density ϕ to be
t

Φ(x − y, t − τ )ϕ(y, τ ) dS(y) dτ,

U (x, t) =
0

∂Ω

18

∀ (x, t) ∈ Ω × (0, T ].

(2.2.6)

Then we introduce a similar definition in which the integration is on Γi (i=1 or 2):
t

Ui (x, t) =

Φ(x − y, t − τ )ϕ(y, τ ) dS(y) dτ,
0

∀ (x, t) ∈ Ω × (0, T ].

(2.2.7)

Γi

Recalling that the interface between Γ1 and Γ2 is defined to be Γ = Γ1 ∩ Γ2 . Thus, the
surface measure of Γ is 0, which implies that

U (x, t) = U1 (x, t) + U2 (x, t).

Recalling also the notations in Section 1.4: for i = 1 or 2, ϕ|Γ ×(0,T ] is uniformly coni
tinuous, thus we can extend this function to a continuous function ϕi on Γi × [0, T ]. In the
following, when x ∈ Γ and t > 0, Lemma 2.2.5 will establish the jump relation for Ui (i = 1
or 2) at (x, t) with the jump ϕi (x, t)/4; when x ∈ Γi (i = 1 or 2) and t > 0, Lemma 2.2.4
state the jump relation for Ui at (x, t) with the jump ϕi (x, t)/2 (here ϕi (x, t)/2=ϕ(x, t)/2,
since x ∈ Γi ).
The essence of the proof is the same as that of Theorem 2.2.1. Before the proof, we
introduce some notations that will be used in the proof and later argument. We write 0 and
˜
0 to be the origins in Rn and Rn−1 respectively and en denotes the point (0, 0, · · · , 0, 1) in
Rn . For any point y = (y1 , y2 , · · · , yn ) ∈ Rn , we write

y˜ = (y1 , y2 , · · · , yn−1 ).

For any r > 0,
Br

B(0, r)

19

means the ball in Rn with radius r and

B(˜
0, r)

Br

represents the ball in Rn−1 with radius r. Γ is used to denote the Gamma function, i.e.
Γ : (0, ∞) → R defined by
∞

Γ(a) =

ta−1 e−t dt.

0

It should not be confused with the partial boundaries Γ1 , Γ2 and Γ.
Lemma 2.2.4. Assume ϕ ∈ BT and i = 1 or 2. Define Ui as in (2.2.7). Then for any
x ∈ Γi and t ∈ (0, T ],

−
lim DUi (xh , t) · →
n (x) =
+

h→0

t
0

1
∂Φ(x − y, t − τ )
ϕ(y, τ ) dS(y) dτ + ϕ(x, t),
∂n(x)
2
Γi

(2.2.8)

−
where xh = x − h→
n (x).
Proof. The proof is almost the same as that of Theorem 2.2.1, since Γi is a relatively open
subset of ∂Ω and ϕ is continuous on Γi × (0, T ].
But the situation when x ∈ Γ is different from that in Lemma 2.2.4, since Γ is the
boundary of Γi . The following lemma claims that in this situation, the jump is only half of
that in Lemma 2.2.4. The proof for Lemma 2.2.5 is also similar to that of Theorem 2.2.1,
but for completeness, we include a detailed proof.
Lemma 2.2.5. Assume ϕ ∈ BT and i = 1 or 2. Define Ui as in (2.2.7). Then for any

20

x∈Γ

Γ1 ∩ Γ2 and t ∈ (0, T ],

−
lim DUi (xh , t) · →
n (x) =
+

h→0

t

1
∂Φ(x − y, t − τ )
ϕ(y, τ ) dS(y) dτ + ϕi (x, t),
∂n(x)
4
Γi

0

(2.2.9)

−
where xh = x−h→
n (x) and ϕi represents the continuous extension of ϕ|Γ ×(0,T ] on Γi ×[0, T ].
i
Proof. We assume i = 1 (The case i = 2 is similar). By (1.4.4), (2.2.9) becomes
−
(xh − y) · →
n (x)

t

lim −

h→0+
t

=−
0

exp

−

|xh − y|2
4(t − τ )

(t − τ )n/2+1
−
(x − y) · →
n (x)
|x − y|2
exp −
4(t − τ )
Γ1 (t − τ )n/2+1
0

Γ1

ϕ1 (y, τ ) dS(y) dτ

ϕ1 (y, τ ) dS(y) dτ +

(4π)n/2
ϕ1 (x, t).
2
(2.2.10)

Without loss of generality, we assume x = 0, otherwise we can do a translation. After this,
−
we further assume →
n (0) = −en , otherwise we can do a rotation which preserves dot product
−
and the distance. By these two simplifications, we have x = 0 and →
n (x) = −en , therefore
xh = hen and (2.2.10) is reduced to
t

h − yn

lim

h→0+ 0
t

=−
0

Γ1

Γ1

exp

(t − τ )n/2+1
yn

(t − τ )n/2+1

exp

−

|hen − y|2
−
4(t − τ )
|y|2
4(t − τ )

ϕ1 (y, τ ) dS(y) dτ

ϕ1 (y, τ ) dS(y) dτ +

(4π)n/2
ϕ1 (0, t).
2

By a change of variable in τ , it is equivalent to
t

h − yn

lim

h→0+ 0
t

=−
0

Γ1

Γ1

τ n/2+1

yn
τ n/2+1

exp

exp
−

−
|y|2
4τ

|hen − y|2
4τ

ϕ1 (y, t − τ ) dS(y) dτ

ϕ1 (y, t − τ ) dS(y) dτ +

21

(4π)n/2
ϕ1 (0, t).
2

(2.2.11)

Because ∂Ω ∈ C 2 and Γ = ∂Γ1 ∈ C 1 , we can straighten the boundary. More specifically,
after relabeling the coordinates, there exist φ1 ∈ C 2 : Rn−1 → R, φ2 ∈ C 1 : Rn−2 → R,
η0 > 0 and a neighborhood Sη0 ⊂ ∂Ω of 0 such that Sη0 can be parametrized as

Sη0 = { y˜, φ1 (˜
y ) : y˜ ∈ Bη0 }

and for any y ∈ Γ ∩ Sη0 , we not only have yn = φ1 (˜
y ), but also yn−1 = φ2 (y1 , y2 , · · · , yn−2 ).
Fixing η0 and then for any η < η0 , we define

Sη = { y˜, φ1 (˜
y ) : y˜ ∈ Bη },

which is a subset of Sη0 and a small neighborhood of 0. Then we denote

Sη,1 = Sη ∩ Γ1 ,

Bη,1 = {˜
y ∈ Bη : y˜, φ1 (˜
y ) ∈ Sη,1 },

Sη = Sη ∩ Γ,

Pη = {˜
y ∈ Bη : y˜, φ1 (˜
y ) ∈ Sη }.

After these preparations, we begin the technical proof. Given any

> 0, we want to find

δ = δ( ) > 0 such that for any 0 < h < δ, the difference between the two sides of (2.2.11) is
within Cε for some constant C.
For any η ∈ (0, η0 ) which will be determined later, we split the integral over Γ1 in (2.2.11)
into two parts: Γ = S + Γ \S . Since Γ1 \ Sη,1 is away from 0, it is easy to see there
1
η,1
1 η,1

22

exists δ1 = δ1 (η, ) such that when 0 < h < δ1 , then
t

h − yn
τ n/2+1

Γ1 \Sη,1

0
t

exp

yn

+

τ n/2+1

Γ1 \Sη,1

0

−

|hen − y|2
4τ

(2.2.12)

|y|2
−
4τ

exp

ϕ1 (y, t − τ ) dS(y) dτ

ϕ1 (y, t − τ ) dS(y) dτ < .

−
˜ = 0.
˜ As a result, for any y ∈ Sη,1 ,
Next since →
n (0) = −en , then Dφ1 (0)

˜ = |Dφ1 (θ˜
|yn | = |φ1 (˜
y )| = |φ1 (˜
y ) − φ1 (0)|
y ) · y˜|
(2.2.13)
˜ |˜
≤|Dφ1 (θ˜
y ) − Dφ1 (0)|
y| ≤

C |˜
y |2 ,

where by the mean value theorem θ is some number between 0 and 1. By (2.2.13), together
with the fact |hen − y| ≥ |˜
y |, we attain

|yn | exp

−

|hen − y|2
4τ

≤ C |˜
y |2 exp

−

|˜
y |2
.
4τ

Noticing
|˜
y |2

t
0

Sη,1

τ n/2+1

exp

−

|˜
y |2
4τ

dS(y) dτ < ∞,

then it follows from Lebesgue’s dominated convergence theorem that
t

yn

lim

h→0+ 0
t

τ n/2+1

yn

=
0

Sη,1

Sη,1

τ n/2+1

exp

exp
−

|y|2
4τ

−

|hen − y|2
4τ

ϕ1 (y, t − τ ) dS(y) dτ

ϕ1 (y, t − τ ) dS(y) dτ.

23

As a result, there exists δ2 = δ2 (η, ) such that when 0 < h < δ2 , then
t

yn

0

Sη,1

τ n/2+1

t

yn

−
0

Sη,1

−

exp

τ n/2+1

|hen − y|2
4τ

|y|2
−
4τ

exp

ϕ1 (y, t − τ ) dS(y) dτ
(2.2.14)

ϕ1 (y, t − τ ) dS(y) dτ < .

Now it suffices to verify that |Iη (h, t) − 21 (4π)n/2 ϕ1 (0, t)| < Cε, where
t

h

Iη (h, t)
0

Sη,1

τ n/2+1

exp

−

|hen − y|2
4τ

ϕ1 (y, t − τ ) dS(y) dτ.

(2.2.15)

Recalling that yn = φ(˜
y ), (2.2.15) can be rewritten as
t

h

Iη (h, t) =
0

Bη,1
t

h

=
0

τ n/2+1

Bη,1

τ n/2+1

exp

−

|hen − y|2
4τ

exp

−

|˜
y |2 + |h − yn |2
4τ

ϕ1 (y, t − τ )

1 + |Dφ1 (˜
y )|2 d˜
y dτ

ϕ1 (y, t − τ )

1 + |Dφ1 (˜
y )|2 d˜
y dτ,
(2.2.16)

where y = y˜, φ1 (˜
y ) . Iη is hard to compute, so we approximate it by a simpler function.
We define I˜η (h, t) as following
t

I˜η (h, t) =
0

h
Bη,1

t

h

=
0

τ n/2+1

Bη,1

τ n/2+1

|hen − (˜
y , 0)|2
4τ

exp

−

exp

|˜
y |2 + h2
−
4τ

ϕ1 (0, t − τ ) d˜
y dτ
(2.2.17)

ϕ1 (0, t − τ ) d˜
y dτ.

Our strategy is to show that I˜η (h, t) is close to both 21 (4π)n/2 ϕ1 (0, t) and Iη (h, t).
Based on (2.2.17) and noticing h is strictly positive, so the integrand is absolutely integrable. Thus, we can reverse the order of integration. Then using the change of variable

24

τ → σ = |˜
y |2 + h2 /(4τ ), we attain
t

I˜η (h, t) =

h

−

exp

0 τ n/2+1
∞
4n/2 σ n/2−1 h

Bη,1

=

y |2 +h2
Bη,1 |˜
4t

(|˜
y |2 + h2 )n/2

|˜
y |2 + h2
4τ

ϕ1 (0, t − τ ) dτ d˜
y

e−σ ϕ1 0, t −

|˜
y |2 + h2
dσ d˜
y
4σ

∞

=

h

4n/2

(|˜
y |2 + h2 )n/2

Bη,1

σ n/2−1 e−σ ϕ1 0, t −
|˜
y |2 +h2
4t

h

= 4n/2
Bη,1

(|˜
y |2

|˜
y |2 + h2
dσ d˜
y
4σ

+ h2 )n/2

H |˜
y |2 + h2 , t d˜
y,

(2.2.18)

where
∞

H(λ, t)

λ
4t

σ n/2−1 e−σ ϕ1 0, t −

λ
dσ.
4σ

It is readily to see that
lim H(λ, t) = Γ

λ→0

n
ϕ1 (0, t).
2

(2.2.19)

Consequently, there exists δ3 = δ3 ( ) such that when η < δ3 and 0 < h < δ3 , then

H |˜
y |2 + h2 , t − Γ

n
ϕ (0, t) < ,
2 1

∀ y˜ ∈ Bη,1 .

(2.2.20)

After taking care of the H term in (2.2.18), let’s consider the following integration

h
(|˜
y |2 + h2 )n/2

d˜
y,

Bη,1

where the integrand h |˜
y |2 + h2

−n/2

is radial in y˜ ∈ Rn−1 and positive when h > 0. Since

25

Γ = ∂Γ1 ∈ C 1 , it ensures that Pη almost bisects Bη when η is small, which means Bη,1 is
close to a hemisphere and
|Bη,1 |

1
= .
2
η→0 |Bη |
lim

This limit is the essential reason why the jump is 21 ϕ(x, t) in (2.2.8). As a result, we can
find δ4 = δ4 ( ) such that for any η < δ4 ,

Bη,1

1− < 1
2

Bη

h
2
(|˜
y | +h2 )n/2

d˜
y

h
(|˜
y |2 +h2 )n/2

d˜
y

<1+ ,

i.e.

h
(|˜
y |2 + h2 )n/2

d˜
y−

h

1
2

(|˜
y |2 + h2 )n/2

d˜
y <

Bη

Bη,1

Next, we will estimate
Bη

h
(|˜
y |2 + h2 )n/2

2

d˜
y . Making the change of variable y˜ → z˜

h
(|˜
y |2 + h2 )n/2

1

d˜
y=

Bη

(|˜
z |2 + 1)n/2

d˜
z.

Bη/h

On one hand,
(|˜
z |2 + 1)n/2
Bη/h

(2.2.21)

Bη

h
(|˜
y |2 +h2 )n/2

1

d˜
y.

1

d˜
z≤
Rn−1

26

(|˜
z |2 + 1)n/2

d˜
z=

π n/2
,
Γ n2

y˜/h,

while on the other hand,

lim

1

h→0
Bη/h

(|˜
z |2 + 1)n/2

1

d˜
z=
Rn−1

(|˜
z |2 + 1)n/2

d˜
z=

π n/2
.
Γ n2

Thus, there exists δ5 = δ5 (η, ) such that for any 0 < h < δ5 ,
h
(|˜
y |2 + h2 )n/2

d˜
y−

π n/2
<
Γ n2

(2.2.22)

Bη

and
h
(|˜
y |2 + h2 )n/2

d˜
y−

π n/2
2Γ n2

< Cε

(2.2.23)

Bη,1

by noticing (2.2.21). It then follows from (2.2.18), (2.2.20) and (2.2.23) that
(4π)n/2
ϕ1 (0, t) < Cε.
I˜η (h, t) −
2

(2.2.24)

Now it suffices to show that I˜η (h, t) is close to Iη (h, t). Firstly, because of (2.2.13),
|˜
y |2 + |h − yn |2 is comparable to |˜
y |2 + h2 . More precisely, there exist positive constants
m1 < 1 and M1 > 1 such that

m1 |˜
y |2 + h2 ≤ |˜
y |2 + |h − yn |2 ≤ M1 |˜
y |2 + h2 .

(2.2.25)

We can equivalently write it to be

m1 |hen − (˜
y , 0)|2 ≤ |hen − y|2 ≤ M1 |hen − (˜
y , 0)|2 .

27

(2.2.26)

Consequently, it follows from (2.2.16) and (2.2.17) that

|Iη (h, t) − I˜η (h, t)|
t

≤
0

h
τ n/2+1 Bη,1
t

+
0

e−

h
τ n/2+1 Bη,1

e

|hen −y|2
4τ

−

− e−

|hen −(˜
y ,0)|2
4τ

|hen −(˜
y ,0)|2
4τ

ϕ1 (y, t − τ )

ϕ1 (y, t − τ )

1 + |Dφ1 (˜
y )|2 − ϕ1 (0, t − τ ) d˜
y dτ

1 + |Dφ1 (˜
y )|2 d˜
y dτ

I + II.

(2.2.27)

For II, since ϕ1 ∈ C Γ1 × [0, T ] , then there exists δ6 = δ6 ( ) such that when η < δ6 ,

ϕ1 (y, t − τ )

1 + |Dφ1 (˜
y )|2 − ϕ1 (0, t − τ ) < ,

∀ y ∈ Bη,1 , τ ∈ [0, t].

As a result,
t

II ≤

h
τ n/2+1 Bη,1

0
t

=
0

h
τ n/2+1 Bη,1
t

=

h
Bη,1

0

h
Bη,1

0

|hen −(˜
y ,0)|2
4τ

−

|˜
y |2 +h2
4τ

e

e

1
τ n/2+1

∞

≤

−

e

−

|˜
y |2

n/2
+ h2

h

=C
Bη,1

|˜
y |2 + h2

d˜
y dτ

|˜
y |2 +h2
4τ

4n/2

n/2

d˜
y dτ

dτ d˜
y

σ n/2−1 e−σ dσ d˜
y

d˜
y,

where the second inequality is due to the change of variable τ → σ

28

|˜
y |2 +h2
4τ .

Now by

another change of variabel y˜ → z˜

II ≤ C

y˜/h, we get

1
Rn−1

(|˜
z |2 + 1)n/2

d˜
z = Cε.

(2.2.28)

To estimate I, firstly it is easy to see that for any h > 0 and y ∈ Bη,1 ,

h ≤ |hen − (˜
y , 0)|.

(2.2.29)

Then by (2.2.13),

|hen − y|2 − |hen − (˜
y , 0)|2
= (h − yn )2 − h2
= |yn | |2h − yn |
≤ C |˜
y |2 2h + |˜
y |2
≤ C |hen − (˜
y , 0)|3 .

Now it follows from the mean value theorem and (2.2.26) that

e

−

|hen −y|2
4τ

−

−e

|hen −(˜
y ,0)|2
4τ

2

1 − m1 |hen −(˜y,0)|
4τ
≤
|hen − y|2 − |hen − (˜
y , 0)|2
e
4τ
2
|hen − (˜
y , 0)|3 − m1 |hen −(˜y,0)|
4τ
≤C
e
(2.2.30)
τ

Thus, based on (2.2.29) and (2.2.30), we attain
m |hen −(˜
y ,0)|2
− 1
−n/2−2
4τ
τ
e

t

I≤
0

Bη,1

29

|hen − (˜
y , 0)|4 d˜
y dτ.

Again, by reversing the order of integration and the change of variable τ → σ

|hen −(˜
y ,0)|2
,
4τ

we get

I≤

|hen

t

− (˜
y , 0)|4

Bη,1

m |hen −(˜
y ,0)|2
− 1
−n/2−2
4τ
τ
e

dτ d˜
y

0

∞
1
σ n/2 e−m1 σ dσ d˜
y
n−2
y , 0)|
0
Bη,1 |hen − (˜
1
≤C
d˜
y.
y |n−2
Bη,1 |˜

≤C

Hence, there exists δ7 = δ7 ( ) such that when η < δ7 , then

I< .

(2.2.31)

Combining (2.2.31) and (2.2.28), we get

|Iη (h, t) − I˜η (h, t)| < Cε.

Therefore, we finish the proof.
In summary, for any > 0, we firstly determine δ3 ( ), δ4 ( ), δ6 ( ), δ7 ( ) and choose
η < min{η0 , δ3 , δ4 , δ6 , δ7 }. Then we determine δ1 (η, ), δ2 (η, ), δ5 (η, ) and choose δ <
min{δ1 , δ2 , δ3 , δ5 }.
Theorem 2.2.6. Let ϕ ∈ BT and define U to be the single-layer heat potential with density
ϕ as (2.2.6).

30

(1) If x ∈ Γ1 ∪ Γ2 and t ∈ (0, T ], then

−
lim DU (xh , t)·→
n (x) =

h→0+

t
0

1
∂Φ(x − y, t − τ )
ϕ(y, τ ) dS(y)dτ + ϕ(x, t). (2.2.32)
∂n(x)
2
∂Ω

(2) If x ∈ Γ and t ∈ (0, T ], then
t

−
lim DU (xh , t) · →
n (x) =

h→0+

∂Φ(x − y, t − τ )
ϕ(y, τ ) dS(y) dτ
∂n(x)
∂Ω

0

+

1
ϕ (x, t) + ϕ2 (x, t) ,
4 1

(2.2.33)

where ϕi (i = 1 or 2) represents the continuous extension of ϕ|Γ ×(0,T ] on Γi × [0, T ].
i
Proof.

(1) Without loss of generality, we suppose x ∈ Γ1 , then by Lemma 2.2.4,

−
lim DU1 (xh , t) · →
n (x) =

h→0+

t
0

1
∂Φ(x − y, t − τ )
ϕ(y, τ ) dS(y) dτ + ϕ(x, t).
∂n(x)
2
Γ1

In addition, since the distance between x and Γ2 is positive, then it is easy to see that

−
lim DU2 (xh , t) · →
n (x) =

h→0+

t
0

∂Φ(x − y, t − τ )
ϕ(y, τ ) dS(y) dτ.
∂n(x)
Γ2

Adding these two equations together, (2.2.32) follows.
(2) (2.2.33) is directly implied by Lemma 2.2.5.

31

2.3
2.3.1

Linear case
Definition of the local solution

In this section, we will show the existence and uniqueness of the solution to the following
linear initial-boundary value problem:



ut (x, t) − ∆u(x, t) = f (x, t)






in

Ω × (0, T ],

∂u(x,t)

+ β(x, t)u(x, t) = g(x, t) on (Γ1 ∪ Γ2 ) × (0, T ],
∂n(x)






 u(x, 0) = ψ(x)
on Ω,

(2.3.1)

where f ∈ C α,α/2 Ω × [0, T ] , β, g ∈ BT , ψ ∈ C 1 (Ω). We will first show the existence and
then apply the existence result to derive the uniqueness.
Definition 2.3.1. For any T > 0, a solution to (2.3.1) on Ω × [0, T ] means a function u in
∂u(x,t)

AT that satisfies (2.3.1) pointwise and for any (x, t) ∈ Γ × (0, T ], ∂n(x) exists and
1
∂u(x, t) 1
+ β1 (x, t) + β2 (x, t) u(x, t) =
g (x, t) + g2 (x, t) ,
∂n(x)
2
2 1

(2.3.2)

where βi and gi denote the continuous extensions of β|Γ ×(0,T ] and g|Γ ×(0,T ] to Γi × [0, T ]
i
i
(i = 1 or 2).

2.3.2

Auxilliary lemmas

Before showing the existence, we state some preliminary results.
Lemma 2.3.2. Suppose Ω is an open bounded subset in Rn with ∂Ω ∈ C 2 , then there exists

32

a constant C > 0 such that for any x, y ∈ ∂Ω,

−
|(x − y) · →
n (x)| ≤ C |x − y|2 .

Proof. This is a standard fact, we omit the proof.
Lemma 2.3.3. Suppose Ω is an open, bounded set in Rn with ∂Ω ∈ C 2 , 0 ≤ a < n − 1,
0 ≤ b < n − 1, then there exists a constant C = C(a, b, n, Ω) such that for any x, z ∈ ∂Ω,


 C |x − z|n−1−a−b if a + b > n − 1,

dS(y)
≤
a
b

∂Ω |x − y| |y − z|
 C

if a + b < n − 1.

Proof. We refer the readers to ( [7], Lemma 1, Sec. 2, Chap. 5).
The following Lemma is mentioned in ( [7], Theorem 2, Sec. 2, Chap. 5), but it does not
explicitly give the estimate (2.3.6), which will be used in some other places of this thesis.
So for the convenience of the readers, we decide to include a complete proof for it. For any
T > 0, define
DΩ,T = {(x, t; y, τ ) x, y ∈ Ω, x = y, 0 ≤ τ < t ≤ T }

(2.3.3)

to be the domain of {Kj }j≥0 mentioned in Lemma 2.3.4. The sequence of the functions
{Kj }j≥0 will be utilized to find the explicit formula for ϕ in (2.3.26), where K ∗ is defined
in (2.3.24). K ∗ is evidently bounded by K in Lemma 2.3.4.
Lemma 2.3.4. Given K0 : DT,Ω → R. Let C be a positive constant such that for any
(x, t; y, τ ) ∈ DT,Ω ,

|K0 (x, t; y, τ )| ≤ C (t − τ )−3/4 |x − y|−(n−3/2) .
33

(2.3.4)

For any j ≥ 1, define Kj : DT,Ω → R by
t

Kj (x, t; y, τ )

τ

∂Ω

K0 (x, t; z, σ) Kj−1 (z, σ; y, τ ) dS(z) dσ.

Then all the Kj (j ≥ 1) are well-defined and the series

∞
j=0 |Kj |

(2.3.5)

converges uniformly to

some function K on DT,Ω . Moreover, there exists some constant C ∗ = C ∗ (n, Ω, T ) such
that for any (x, t; y, τ ) ∈ DT,Ω ,

K(x, t; y, τ ) ≤ C ∗ (t − τ )−3/4 |x − y|−(n−3/2) .

(2.3.6)

Proof. We first justify that all the Kj (j ≥ 1) are well-defined. In fact, by (2.3.4) and (2.3.5),
one has
t

|K1 (x, t; y, τ )| ≤ C

τ

(t − σ)−3/4 |x − z|−(n−3/2) (σ − τ )−3/4 |z − y|−(n−3/2) dS(z) dσ
∂Ω

t

=C

|x − z|−(n−3/2) |z − y|−(n−3/2) dS(z).

(t − σ)−3/4 (σ − τ )−3/4 dσ
∂Ω

τ

Making the change of variable ρ

σ−τ
t−τ

|K1 (x, t; y, τ )| ≤ C (t − τ )−1/2

for σ, then we obtain

|x − z|−(n−3/2) |z − y|−(n−3/2) dS(z).
∂Ω

Now there are two cases:
• If n > 2, then by Lemma 2.3.3,

|x − z|−(n−3/2) |z − y|−(n−3/2) dS(z) ≤ C |x − y|−(n−2) .
∂Ω

34

Therefore
|K1 (x, t; y, τ )| ≤ C (t − τ )−1/2 |x − y|−(n−2) .

(2.3.7)

• If n = 2, then we can not use Lemma 2.3.3 directly since (n − 3/2) + (n − 3/2) = n − 1
is on the borderline. However, since Ω is bounded, we have |z − y|−(n−3/2) ≤ C |z −
y|−(n−3/2) |z − y|−1/4 . Then Lemma 2.3.3 can be applied to get

|x − z|−(n−3/2) |z − y|−(n−3/2) dS(z)
∂Ω

|x − z|−(n−3/2) |z − y|−(n−5/4) dS(z)

≤C
∂Ω

≤ C |x − y|−1/4 .

Consequently,

|K1 (x, t; y, τ )| ≤ C (t − τ )−1/2 |x − y|−1/4 = C (t − τ )−1/2 |x − y|−(n−7/4) . (2.3.8)

Comparing (2.3.4) with (2.3.7) and (2.3.8), the exponent of (t − τ ) term is added by 1/4 and
the exponent of (x − y) term increases by at least 1/4. In other words, the singularity of K1
in both time and space variables is weaker than K0 by a certain number 1/4. Thus, after
finite steps, we can find j0 (only depending on n) such that

Kj0 (x, t; y, τ ) ≤ C,

∀ (x, t; y, τ ) ∈ DΩ,T ,

for some constant C. Moreover, for any j > j0 , Kj is also well-defined and bounded.

35

(2.3.9)

From (2.3.4) and Lemma 2.3.3, there exists a constant C1 such that

|K0 (x, t; y, τ )| ≤ C1 (t − τ )−3/4 |x − y|−(n−3/2) ,

∀ (x, t; y, τ ) ∈ DΩ,T

(2.3.10)

and
dS(y)
∂Ω

≤ C1 ,

|x − y|n−3/2

∀ x ∈ ∂Ω.

For the rest of the proof, C1 and C will be fixed. We start from (2.3.9) to prove by induction
that for any m ≥ 0 and for any (x, t; y, τ ) ∈ DΩ,T ,

|Kj0 +m (x, t; y, τ )| ≤

where Q

C Qm (t − τ )m/4
,
Γ 1 + m/4

(2.3.11)

C12 Γ(1/4).

When m = 0, (2.3.11) is just (2.3.9). Now we suppose (2.3.11) is true for m = i and try to
verify it for m = i + 1. Applying (2.3.11) with m = i and (2.3.4), we obtain
t

|Kj0 +i+1 (x, t; y, τ )| ≤
≤
≤
=

τ

∂Ω

|K0 (x, t; z, σ) Kj0 +i (z, σ; y, τ )| dS(z) dσ

C1 C Qi
Γ 1 + i/4

t
τ

C12 C Qi
Γ 1 + i/4

|x − z|−(n−3/2) dS(z)

(t − σ)−3/4 (σ − τ )i/4 dσ
∂Ω

t

(t − σ)−3/4 (σ − τ )i/4 dσ

τ

C Qi+1
Γ(1/4) Γ 1 + i/4

t

(t − σ)−3/4 (σ − τ )i/4 dσ,

τ

where the third inequality and the fourth equality are due to the definitions of C1 and Q

36

respectively. By the change of variable ρ

|Kj0 +i+1 (x, t; y, τ )| ≤
=

σ−τ
t−τ

for σ, we attain

C Qi+1 (t − τ )(i+1)/4 1
(1 − ρ)−3/4 ρi/4 dρ
Γ(1/4) Γ 1 + i/4
0
C Qi+1 (t − τ )(i+1)/4
.
Γ 1 + (i + 1)/4

Consequently, (2.3.11) is true for m = i + 1 and therefore it is true for any m ≥ 0.
By (2.3.11), we get that for any m ≥ 0 and for any (x, t; y, τ ) ∈ DΩ,T ,

|Kj0 +m (x, t; y, τ )| ≤

C Qm T m/4
.
Γ(1 + m/4)

(2.3.12)

This estimate is significant because the right hand side of (2.3.12) is independent of x, y, t,
τ . By applying the ratio test, we can prove
∞
m=0

Qm T m/4
< ∞.
Γ(1 + m/4)

(2.3.13)

Next, due to (2.3.4) and the fact that the singularity of K0 is stronger than any other
Kj (j ≥ 1), we can find a constant C2 = C2 (n, Ω, T ) such that for any (x, t; y, τ ) ∈ DΩ,T ,
j0

|Kj (x, t; y, τ )| ≤ C2 (t − τ )−3/4 |x − y|−(n−3/2) .

(2.3.14)

j=0
∞

|Kj (x, t; y, τ )|

Combining (2.3.14), (2.3.12) and (2.3.13) together, it is readily to see that
j=0

converges to K(x, t; y, τ ) uniformly on DΩ,T and moreover, there exists some constant C ∗ ,

37

only depending on n, Ω and T such that for any (x, t; y, τ ) ∈ DT,Ω ,

K(x, t; y, τ ) ≤ C ∗ (t − τ )−3/4 |x − y|−(n−3/2) .

.
Lemma 2.3.5. If f ∈ C α,α/2 Ω × [0, T ] and
t

Φ(x − y, t − τ ) f (y, τ ) dS(y) dτ,

W (x, t)
0

∀ (x, t) ∈ Ω × [0, T ],

Ω

then W ∈ C 2,1 Ω × (0, T ] and

(Wt − ∆W )(x, t) = f (x, t),

∀ (x, t) ∈ Ω × (0, T ].

Proof. This is a standard argument, we refer the readers to ( [7], Theorem 9, Sec. 5, Chap.
1).

2.3.3

Existence

The idea of the proof for the following Theorem 2.3.6 is analogous to ( [7], Theorem 2, Sec.
3, Chap. 5).
Theorem 2.3.6. For any T > 0, there exists a solution u ∈ AT to (2.3.1) on Ω × [0, T ].
Proof. We will construct a solution u to (2.3.1). Firstly, since ψ ∈ C 1 (Ω) and ∂Ω ∈ C 2 ,
one can extend ψ to a larger domain. More precisely, there exists an open set Ω1 ⊃ Ω and
ψ1 ∈ C 1 (Ω1 ) such that ψ1 agrees with ψ on Ω. In the rest of the proof, for convenience, we

38

just write ψ for ψ1 . We are looking for a solution u in the following form:
t

Φ(x − y, t) ψ(y) dy +

u(x, t) =
Ω1

Φ(x − y, t − τ ) f (y, τ ) dy dτ
0

Ω

(2.3.15)

t

Φ(x − y, t − τ ) ϕ(y, τ ) dS(y) dτ,

+
0

∀ (x, t) ∈ Ω × [0, T ],

∂Ω

where ϕ ∈ BT will be determined later.
Due to Lemma 2.3.5, it is easy to see that the function u defined in (2.3.15) belongs to
AT and satisfies the first and the third equations in (2.3.1), so in order to verify u to be the
solution, the only things left to check are

∂u(x, t)
+ β(x, t)u(x, t) = g(x, t),
∂n(x)

∀ (x, t) ∈ (Γ1 ∪ Γ2 ) × (0, T ]

(2.3.16)

and

∂u(x, t) 1
1
+ β1 (x, t)+β2 (x, t) u(x, t) =
g (x, t)+g2 (x, t) ,
∂n(x) 2
2 1

∀ (x, t) ∈ Γ×(0, T ]. (2.3.17)

The plan is to firstly find a function ϕ ∈ BT such that u defined in (2.3.15) satisfies (2.3.16),
then we will prove this u satisfies (2.3.17) as well.
By (1.4.1) and (2.3.15), for any (x, t) ∈ (Γ1 ∪ Γ2 ) × (0, T ], it follows from the Lebesgue’s
dominated convergence theorem and Theorem 2.2.6 that
t
∂u(x, t)
∂Φ(x − y, t)
∂Φ(x − y, t − τ )
=
ψ(y) dy +
f (y, τ ) dy dτ
∂n(x)
∂n(x)
∂n(x)
Ω1
0 Ω
t
∂Φ(x − y, t − τ )
1
+
ϕ(y, τ ) dS(y) dτ + ϕ(x, t).
(2.3.18)
∂n(x)
2
0 ∂Ω

39

Therefore, (2.3.16) is reduced to for any (x, t) ∈ (Γ1 ∪ Γ2 ) × (0, T ],
t

ϕ(x, t) =

K(x, t; y, τ ) ϕ(y, τ ) dS(y) dτ + H(x, t),
0

(2.3.19)

∂Ω

where
K(x, t; y, τ ) = −2

∂Φ(x − y, t − τ )
+ β(x, t) Φ(x − y, t − τ )
∂n(x)

and

∂Φ(x − y, t)
+ β(x, t) Φ(x − y, t) ψ(y) dy
∂n(x)
Ω1
t
∂Φ(x − y, t − τ )
−2
+ β(x, t) Φ(x − y, t − τ ) f (y, τ ) dy dτ
∂n(x)
0 Ω

H(x, t) = − 2

+ 2 g(x, t).

In other words, the proof of (2.3.16) becomes the search for a fixed point ϕ ∈ BT of (2.3.19).
In the following, we will construct a fixed point of (2.3.19) in BT . Noticing

1
(t − τ )n/2

−

e

|x−y|2
4(t−τ )

|x−y|2
−
|x − y|2 2n−3
4
e 4(t−τ ) (t − τ )−3/4 |x − y|−(n−3/2)
=
t−τ

≤ C(t − τ )−3/4 |x − y|−(n−3/2) ,

so by the similar argument, we have
|x − y|2
(t − τ )n/2+1

e

−

|x−y|2
4(t−τ )

≤ C(t − τ )−3/4 |x − y|−(n−3/2) .

40

Then it follows from Lemma 2.3.2 that for any x, y ∈ ∂Ω, 0 ≤ τ < t ≤ T ,

1

|K(x, t; y, τ )| ≤ C

(t − τ )n/2

+

|x − y|2
(t − τ )n/2+1

−

e

|x−y|2
4(t−τ )

≤ C (t − τ )−3/4 |x − y|−(n−3/2) .

(2.3.20)

Then using the fact ψ ∈ C 1 (Ω1 ) and the integration by parts, we obtain
∂Φ(x − y, t)
ψ(y) dy
∂n(x)
Ω1
−
Dy Φ(x − y, t) · →
n (x) ψ(y) dy

=−
Ω1

−
Φ(x − y, t) Dψ(y) · →
n (x) dy.

−
−
Φ(x − y, t) ψ(y) →
n (y) · →
n (x) dy +

=−

(2.3.21)

Ω1

∂Ω1

Consequently, as a function in (x, t),

∂Φ(x − y, t)
ψ(y) dy ∈ C ∂Ω × [0, T ] ⊂ BT .
∂n(x)
Ω1
Then it is readily to check that H ∈ BT .
Next, two sequences of functions {Kj }j≥0 and {ϕj }j≥0 will be constructed as following.
First, define K0 = K on DΩ,T and ϕ0 = H on (Γ1 ∪ Γ2 ) × (0, T ]. Then for any j ≥ 1, define
t

Kj (x, t; y, τ ) =

τ

K0 (x, t; z, σ) Kj−1 (z, σ; y, τ ) dS(z) dσ

(2.3.22)

Ki (x, t; y, τ ) H(y, τ ) dS(y) dτ + H(x, t)

(2.3.23)

∂Ω

on DΩ,T and
j−1

ϕj (x, t) =

i=0 0

t
∂Ω

41

on (Γ1 ∪ Γ2 ) × (0, T ]. Because of (2.3.4), we can prove by induction that for any j ≥ 0, ϕj
is well defined and belongs to BT . Our goal is to show that ϕj (x, t) uniformly converges to
some function ϕ(x, t) on (Γ1 ∪ Γ2 ) × (0, T ] as j → ∞, which makes ϕ to be the fixed point
of (2.3.19) in BT . Writing
∞

K ∗ (x, t; y, τ )

Kj (x, t; y, τ ),

(2.3.24)

j=0

by Lemma 2.3.4, we know K ∗ is well-defined and

∞
j=0 Kj

converges uniformly to K ∗ on

DT,Ω . Moreover, there exists a constant C ∗ = C ∗ (n, Ω, T ) such that for any (x, t; y, τ ) ∈
DΩ,T ,
|K ∗ (x, t; y, τ )| ≤ C ∗ (t − τ )−3/4 |x − y|−(n−3/2) .

(2.3.25)

Consequently, it follows from (2.3.23) and (2.3.24) that ϕj converges uniformly to the function ϕ on (Γ1 ∪ Γ2 ) × (0, T ], where
t

K ∗ (x, t; y, τ ) H(y, τ ) dS(y) dτ + H(x, t).

ϕ(x, t)
0

(2.3.26)

∂Ω

Thus, ϕ is a fixed point of (2.3.19) in BT and therefore the function u defined in (2.3.15)
satisfies (2.3.16).
Now as our plan, it only left to show this function u satisfies (2.3.17) as well. Making use
∂u(x,t)

of (2.3.15), (1.4.1) and Theorem 2.2.6, we get for any x ∈ Γ, 0 < t ≤ T , ∂n(x) exists and
t
∂u(x, t)
∂Φ(x − y, t)
∂Φ(x − y, t − τ )
=
ψ(y) dy +
f (y, τ ) dy dτ
∂n(x)
∂n(x)
∂n(x)
Ω1
0 Ω
t

+
0

∂Φ(x − y, t − τ )
1
1
ϕ(y, τ ) dS(y) dτ + ϕ1 (x, t) + ϕ2 (x, t).
∂n(x)
4
4
∂Ω

42

(2.3.27)

Then we choose two sequences of points {ξk }k≥1 ⊂ Γ1 and {zj }j≥1 ⊂ Γ2 which converge to
x, it follows from (2.3.18) that
t
∂Φ(ξk − y, t)
∂Φ(ξk − y, t − τ )
∂u(ξk , t)
=
ψ(y) dy +
f (y, τ ) dy dτ
∂n(ξk )
∂n(ξk )
∂n(ξk )
Ω1
0 Ω
t

+
0

∂Φ(ξk − y, t − τ )
1
ϕ(y, τ ) dS(y) dτ + ϕ(ξk , t)
∂n(ξk )
2
∂Ω

and
t
∂u(zj , t)
∂Φ(zj − y, t)
∂Φ(zj − y, t − τ )
=
ψ(y) dy +
f (y, τ ) dy dτ
∂n(zj )
∂n(zj )
∂n(zj )
Ω1
0 Ω
t
∂Φ(zj − y, t − τ )
1
+
ϕ(y, τ ) dS(y) dτ + ϕ(zj , t).
∂n(zj )
2
0 ∂Ω

Taking k → ∞ and j → ∞, we obtain
t
∂Φ(x − y, t)
∂Φ(x − y, t − τ )
∂u(ξk , t)
=
ψ(y) dy +
f (y, τ ) dy dτ
∂n(x)
∂n(x)
k→∞ ∂n(ξk )
Ω1
0 Ω
(2.3.28)
t
∂Φ(x − y, t − τ )
1
+
ϕ(y, τ ) dS(y) dτ + ϕ1 (x, t)
∂n(x)
2
0 ∂Ω

lim

and
t
∂u(zj , t)
∂Φ(x − y, t)
∂Φ(x − y, t − τ )
=
ψ(y) dy +
f (y, τ ) dy dτ
∂n(x)
∂n(x)
j→∞ ∂n(zj )
Ω1
0 Ω
(2.3.29)
t
∂Φ(x − y, t − τ )
1
+
ϕ(y, τ ) dS(y) dτ + ϕ2 (x, t).
∂n(x)
2
0 ∂Ω

lim

Adding (2.3.28) and (2.3.29) together and noticing (2.3.27), we attain

∂u(zj , t)
∂u(x, t)
1
∂u(ξk , t)
=
lim
+ lim
.
∂n(x)
2 k→∞ ∂n(ξk )
j→∞ ∂n(zj )

43

(2.3.30)

Moreover, since u satisfies (2.3.16), we have

∂u(ξk , t)
+ β(ξk , t) u(ξk , t) = g(ξk , t),
∂n(ξk )
∂u(zj , t)
+ β(zj , t) u(zj , t) = g(zj , t).
∂n(zj )
Sending k → ∞ and j → ∞, we obtain

∂u(ξk , t)
= g1 (x, t) − β1 (x, t) u(x, t)
k→∞ ∂n(ξk )
∂u(zj , t)
lim
= g2 (x, t) − β2 (x, t) u(x, t)
j→∞ ∂n(zj )
lim

(2.3.31)
(2.3.32)

Combining (2.3.30), (2.3.31) and (2.3.32) together, (2.3.17) follows.

2.3.4

Comparison principles, uniqueness and global solution

Next, we will prove the comparison principle and the uniqueness of the solution by applying
Theorem 2.3.6. But before that, let’s prove the following easier comparison result.
Lemma 2.3.7. Suppose in (2.3.1), f ≥ 0 on Ω × [0, T ], ψ > 0 on Ω and

inf

(Γ1 ∪Γ2 )×[0,T ]

g > 0,

then the solution u > 0 on Ω × [0, T ].
Proof. Since ψ > 0 on Ω, we have m

min ψ > 0. Now we claim u > 0 on Ω × [0, T ]. If
Ω

not, then there will exist x0 ∈ Ω and t0 ∈ (0, T ] such that

u(x0 , t0 ) = 0 =
44

min u.
Ω×[0,t0 ]

By the strong maximum principle, x0 ∈ ∂Ω. If x0 ∈ Γ1 ∪ Γ2 , then

0 < g(x0 , t0 ) =

∂u(x0 , t0 )
∂u(x0 , t0 )
+ β(x0 , t0 ) u(x0 , t0 ) =
≤ 0,
∂n(x0 )
∂n(x0 )

which is impossible. If x0 ∈ Γ, then
1
g (x , t ) + g2 (x0 , t0 )
2 1 0 0
∂u(x0 , t0 ) 1
+ β1 (x0 , t0 ) + β2 (x0 , t0 ) u(x0 , t0 )
=
∂n(x0 )
2

0<

=

∂u(x0 , t0 )
≤ 0,
∂n(x0 )

which is also a contradiction. Thus, the Lemma follows.
Corollary 2.3.8. Suppose in (2.3.1), f ≥ 0 on Ω × [0, T ], ψ ≥ 0 on Ω and g ≥ 0 on
(Γ1 ∪ Γ2 ) × (0, T ], then the solution u ≥ 0 on Ω × [0, T ]. In particular, the solution to (2.3.1)
on Ω × [0, T ] is unique. If further assuming that ψ ≡ 0 on Ω, then u(x, t) > 0 for any
(x, t) ∈ Ω × (0, T ].
Proof. Due to Theorem 2.3.6, there exists a solution v ∈ AT to the following problem:



vt (x, t) − ∆v(x, t) = 1






in

∂v(x,t)

Ω × (0, T ],

+ β(x, t)v(x, t) = 1 on (Γ1 ∪ Γ2 ) × (0, T ],
∂n(x)






 v(x, 0) = 1
on Ω.

45

Now for any > 0, define w = u + v, then w satisfies



(w )t (x, t) − ∆w (x, t) = f + ≥ ε






in

Ω × (0, T ],

∂w (x,t)

+ β(x, t)w (x, t) = g + ≥ ε on (Γ1 ∪ Γ2 ) × (0, T ],
∂n(x)






 w (x, 0) = ψ + ≥ ε
on Ω.

By invoking Lemma 2.3.7, w ≥ 0 on Ω × [0, T ]. Taking → 0, we get u ≥ 0 on Ω × [0, T ].
Now we further assume that ψ is not identical 0 on Ω. If there exists some (x0 , t0 ) ∈
Ω × (0, T ] such that u(x0 , t0 ) = 0, then

u(x0 , t0 ) =

min u.
Ω×[0,t0 ]

By strong maximum principle and ψ ≡ 0, x0 has to be on the boundary ∂Ω. In addition, it
follows from the boundary condition

∂u(x0 , t0 )
+ β(x0 , t0 )u(x0 , t0 ) = g(x0 , t0 )
∂n(x0 )
that

∂u(x0 ,t0 )
∂n(x0 )

≥ 0. But on the other hand, the Hopf lemma implies that

∂u(x0 ,t0 )
∂n(x0 )

< 0,

which is a contradiction. So u(x, t) > 0 for any (x, t) ∈ Ω × (0, T ].
Corollary 2.3.9. Let u be a solution to (2.3.1). Then for any T > 0, there exist constants
C = C(Ω, β, T ) such that for any x ∈ Ω and 0 ≤ t ≤ T ,

|u(x, t)| ≤ C

sup |f | +
Ω×(0,T ]

sup

|g| + sup |ψ| .

(Γ1 ∪Γ2 )×(0,T ]

Ω

Proof. Similar to the proof of Corollary 2.3.8, there exists a solution v ∈ AT to the following
46

problem:



vt (x, t) − ∆v(x, t) = 1






in

Ω × (0, T ],

∂v(x,t)

+ β(x, t)v(x, t) = 1 on (Γ1 ∪ Γ2 ) × (0, T ],
∂n(x)






 v(x, 0) = 1
on Ω.
Let K = supΩ×(0,T ] |f | + sup(Γ ∪Γ )×(0,T ] |g| + supΩ |ψ|. Then it follows from Corollary
1 2
2.3.8 that |u| ≤ Kv. Since v is bounded on Ω × [0, T ], the conclusion follows.
By combining Theorem 2.3.6 and Corollary 2.3.8, the existence and uniqueness of the
global solutions is established.
Theorem 2.3.10. The linear problem (2.3.1) has a unique global solution u ∈ C 2,1 Ω ×
(0, ∞) ∩ C Ω × [0, ∞) , where the global solution means a function that solves (2.3.1) for
any T > 0.

2.4
2.4.1

Nonlinear case
Definition of the local solution

This section is devoted to the existence and uniqueness of the solution to the following
nonlinear initial-boundary value problem:




ut (x, t) − ∆u(x, t) = f (x, t) in Ω × (0, T ],








 ∂u(x,t)
= F x, u(x, t)
on Γ1 × (0, T ],
∂n(x)


∂u(x,t)

=0


∂n(x)






 u(x, 0) = ψ(x)

on Γ2 × (0, T ],
on Ω,

47

(2.4.1)

where f ∈ C α,α/2 Ω × [0, T ] , F ∈ C 1 (Γ1 × R) and ψ ∈ C 1 (Ω). For any (x, σ) ∈ Γ1 × R,
we will use Fxi (x, σ) (1 ≤ i ≤ n) to denote the ith spatial derivative of F and use Fσ (x, σ)
to denote the derivative of its (n + 1)th variable.
Definition 2.4.1. For any T > 0, a solution to (2.4.1) on Ω × [0, T ] means a function u in
∂u(x,t)

AT that satisfies (2.4.1) pointwise and for any (x, t) ∈ Γ × (0, T ], ∂n(x) exists and
∂u(x, t)
1
= F x, u(x, t) .
∂n(x)
2

2.4.2

(2.4.2)

Comparison principles and uniqueness

This time, we will first show some comparison principles and then discuss the existence of
the solution.
Theorem 2.4.2. Suppose ui ∈ AT (i = 1, 2) is the solution to (2.4.1) on Ω × [0, T ] with
right hand side fi , Fi and ψi . If f1 ≥ f2 on Ω × [0, T ], F1 ≥ F2 on Γ1 × R and ψ1 ≥ ψ2 on
Ω, then u1 ≥ u2 on Ω × [0, T ]. As a consequence, the solution to (2.4.1) is unique.
Proof. Let w = u1 − u2 , f = f1 − f2 and ψ = ψ1 − ψ2 . Then f ≥ 0 on Ω × [0, T ] and ψ ≥ 0
on Ω. In addition, since F1 ≥ F2 on Γ1 × R, we have

F1 x, u1 (x, t) − F2 x, u2 (x, t)
≥ F1 x, u1 (x, t) − F1 x, u2 (x, t)
= β(x, t)w(x, t),

where
1

β(x, t) =
0

(F1 )σ x, su1 (x, t) + (1 − s)u2 (x, t) ds.
48

Thus, w satisfies the equations below




wt (x, t) − ∆w(x, t) = f (x, t) ≥ 0 in Ω × (0, T ],








 ∂w(x,t)
− β(x, t)w(x, t) ≥ 0
on Γ1 × (0, T ],
∂n(x)


∂w(x,t)

=0


∂n(x)






 w(x, 0) = ψ(x) ≥ 0

on Γ2 × (0, T ],
on Ω.

Now it follows from Corollary 2.3.8 that w ≥ 0.
Theorem 2.4.3. Suppose u ∈ AT is the solution to (2.4.1) with f ≥ 0 on Ω × [0, T ],
F (·, 0) ≥ 0 on Γ1 and ψ ≥ 0 on Ω, then u ≥ 0 on Ω × [0, T ]. If further assuming ψ ≡ 0,
then u(x, t) > 0, ∀ x ∈ Ω, 0 < t ≤ T .
Proof. To prove the first statement, we write

F x, u(x, t) = F x, u(x, t) − F (x, 0) + F (x, 0) = β(x, t)u(x, t) + F (x, 0),

where
1

β(x, t) =

Fσ x, su(x, t) ds.
0

Hence, u satisfies














ut (x, t) − ∆u(x, t) = f (x, t) ≥ 0
∂u(x,t)
∂n(x)

in

Ω × (0, T ],

− β(x, t) u(x, t) = F (x, 0) ≥ 0 on Γ1 × (0, T ],



∂u(x,t)

=0


∂n(x)






 u(x, 0) = ψ(x) ≥ 0

on Γ2 × (0, T ],
on Ω.

49

It then follows from Corollary 2.3.8 that u ≥ 0.
Now suppose additionally that ψ ≡ 0, then similar to the proof of Corollary 2.3.8, by
applying the strong maximum principle and the Hopf lemma, we get u(x, t) > 0, ∀ x ∈
Ω, 0 < t ≤ T .

2.4.3

Existence

Now we turn to the existence of the solution. As a common process to deal with the nonlinear
problem, we will take advantage of the theories for the linear problems and some fixed point
theorems. For any T > 0 and R > 0, define XT = C Ω × [0, T ] and XT,R = {v ∈ XT :
||v|| ≤ R}, where the norm in XT is the supremum norm as following

||u|| =

|u(x, t)|,

max
(x,t)∈Ω×[0,T ]

∀ u ∈ XT .

Then XT is a Banach space and XT,R is a closed and thus complete subset of XT . For
any v ∈ XT,R , it follows from Theorem 2.3.6 and Corollary 2.3.8 that there exists a unique
solution u ∈ AT to the following problem




ut (x, t) − ∆u(x, t) = f (x, t) in Ω × (0, T ],








 ∂u(x,t)
= F x, v(x, t)
on Γ1 × (0, T ],
∂n(x)


∂u(x,t)

=0


∂n(x)






 u(x, 0) = ψ(x)

(2.4.3)

on Γ2 × (0, T ],
on Ω.

Thus, it determines a mapping ΨT : XT,R → AT . Our strategy is to pick up suitable T
and R such that ΨT has a fixed point in XT,R , which turns out to be the unique solution to

50

(2.4.1).
In the proof of Theorem 2.4.5, we will utilize the Schauder fixed point theorem, which
requires to verify the following three things:
(i) ΨT maps XT,R to AT ∩ XT,R for some suitably T and R;
(ii) ΨT : XT,R → XT,R is continuous;
(iii) ΨT (XT,R ) is precompact in XT,R .
The requirement (iii) is the most technical part, so in order to make the argument more
readable, we state the following Lemma 2.4.4 separately which will be used in the proof of
(iii). Actually, Lemma 2.4.4 is a fact mentioned without justification in the proof of ( [7],
Theorem 13, Sec. 5, Chap. 7), but here for the convenience of the readers, we would like to
give a complete proof.
Lemma 2.4.4. Given T > 0 and {ϕj }j≥1 ⊂ L∞ (Γ1 ∪ Γ2 ) × (0, T ] , we define
t

wj (x, t) =

0

∂Ω

Φ(x − y, t − τ ) ϕj (y, τ ) dS(y) dτ,

∀ (x, t) ∈ Ω × [0, T ].

(2.4.4)

If {ϕj }j≥1 are uniformly bounded on (Γ1 ∪ Γ2 ) × (0, T ], then {wj }j≥1 are uniformly bounded
and equicontinuous on Ω × [0, T ].
Proof. From the assumption, there is a constant C ∗ such that for any j ≥ 1 and (x, t) ∈
(Γ1 ∪ Γ2 ) × (0, T ],
|ϕj (x, t)| ≤ C ∗ .

(2.4.5)

In the rest of the proof, we will use C to denote a constant which only depends on n, Ω, T

51

and C ∗ . Noticing that

|Φ(x − y, t − τ )| ≤

|x−y|2
−
C (t − τ )−n/2 e 4(t−τ )

≤ C (t − τ )−3/4 |x − y|−(n−3/2) ,

(2.4.6)

it then follows from (2.4.5) and Lemma 2.3.3 that
t

|wj (x, t)| ≤ C

0

∂Ω
t

≤C

(t − τ )−3/4 |x − y|−(n−3/2) dS(y) dτ
(t − τ )−3/4 dτ

(2.4.7)

0

≤ C t1/4 ≤ C T 1/4 = C.
This showed the uniform boundedness. Next, we will prove the equicontinuity, which means
for any > 0, there exists δ > 0 such that the following (1) and (2) are satisfied.
(1) For any j ≥ 1, t ∈ [0, T ] and x1 , x2 in Ω with |x1 − x2 | < δ, we have

|wj (x1 , t) − wj (x2 , t)| ≤ Cε.

(2) For any j ≥ 1, x ∈ Ω, 0 ≤ t1 < t2 ≤ T with t2 − t1 < δ, we have

|wj (x, t1 ) − wj (x, t2 )| ≤ Cε.

By a change of variable in τ , we attain
t

wj (x, t) =

0

∂Ω

Φ(x − y, τ ) ϕj (y, t − τ ) dS(y) dτ,

52

∀ (x, t) ∈ Ω × [0, T ].

(2.4.8)

Sometimes it is easier to use (2.4.8) to do the estimate.
To prove (1), we split the interval [0, t] into [0, ] and [ , t] and take advantage of the
uniform boundedness of {ϕj }j≥1 , it follows from (2.4.8) that
|x −y|2
− 14τ
−n/2
τ
e

t

|wj (x1 , t) − wj (x2 , t)| ≤C

|x −y|2
− 24τ
−e

dS(y) dτ

∂Ω

+C
0

dS(y) dτ

|x −y|2
− 24τ
−n/2
τ
e

dS(y) dτ

∂Ω

+C
0

|x −y|2
− 14τ
−n/2
τ
e

∂Ω

I + II + III.

Analogous to the derivation of (2.4.7), we get II ≤ C 1/4 and III ≤ C 1/4 . To estimate I,
we exploit the mean value theorem and find for any τ ∈ (0, T ], y ∈ ∂Ω,
|x −y|2
− 14τ
e

|x −y|2
− 24τ
−e

≤ C τ −1/2 |x1 − x2 |.

As a result,
t

I≤C
∂Ω

τ −(n+1)/2 |x1 − x2 | dS(y) dτ

≤ C −(n+1)/2 δ.

As long as we take δ ≤ (n+3)/2 , then I ≤ Cε. Thus, we finish the proof of (1).
In order to prove (2), we consider two cases.

53

• t1 ≤ 2ε. In this case, we choose δ ≤ ε, then again by (2.4.7),

|wj (x, t2 ) − wj (x, t1 )| ≤ |wj (x, t2 )| + |wj (x, t1 )|
1/4

≤ C t2

1/4

+ C t1

≤ C (2ε + δ)1/4 + C (2ε)1/4 ≤ C 1/4 .

• t1 > 2ε. In this case, using (2.4.4), we get

|wj (x, t2 ) − wj (x, t1 )|
t1 −

−

1

≤
0 ∂Ω
t2

+
t1 −
t1

+
t1 −

e

(t2 − τ )n/2
1
(t2 − τ )n/2
1
(t1 − τ )n/2

|x−y|2
4(t2 −τ )

−

1
(t1 − τ )n/2

−

e

−

|x−y|2
4(t2 −τ ) |ϕ (y, τ )| dS(y) dτ
j

−

|x−y|2
4(t1 −τ ) |ϕ (y, τ )| dS(y) dτ
j

e

e

|x−y|2
4(t1 −τ )

|ϕj (y, τ )| dS(y) dτ

I + II + III.

If we choose δ ≤ ε, then similar to (2.4.7) and using the mean value theorem,

1/4

II ≤ C t2
≤

− (t1 − )1/4

C
(t1 − )3/4

(t2 − t1 + )

≤ C −3/4 (δ + ) ≤ C 1/4 .

54

(2.4.9)

Similarly,
1/4

III ≤ C t1

− (t1 − )1/4 ≤ C 1/4 .

(2.4.10)

Employing the mean value theorem again, there exists a constant C such that for any
τ ∈ [0, t1 ), x ∈ Ω, y ∈ ∂Ω, we have

1
(t2

− τ )n/2

−

e

|x−y|2
4(t2 −τ )

1

−

(t1

− τ )n/2

e

−

|x−y|2
4(t1 −τ )

≤ C (t1 − τ )−n/2−1 |t2 − t1 |
≤ C (t1 − τ )−n/2−1 δ.

If we choose δ ≤ εn/2+3 , then
t1 −

I≤C
0

≤C

∂Ω

−n/2−1 δ

(t1 − τ )−n/2−1 dS(y) dτ
(2.4.11)

≤C .

Combining (2.4.11), (2.4.9), (2.4.10), we proved (2) and therefore the Lemma follows.

Now similar to the arguments in ( [7], Theorem 13, Sec. 5, Chap. 7) and ( [27], Theorem
1.3), we conclude the following theorem on the existence of the solution.
Theorem 2.4.5. For the nonlinear equation (2.4.1) with f, F, ψ described there.
(1) There exists T0 > 0 such that for any 0 < T ≤ T0 , there exists a unique solution
u ∈ AT to (2.4.1) on Ω × [0, T ].
(2) If F is bounded on Γ1 × R, then for any T > 0, there exists a unique solution u ∈ AT
to (2.4.1) on Ω × [0, T ].

55

Proof. Just as the heuristic idea before Lemma 2.4.4, in order to prove the existence of a
solution, we will use Schauder fixed point theorem to prove ΨT has a fixed point in XT,R
for suitable T and R. Thus, we need to verify the following three requirements:
(i) ΨT maps XT,R to AT ∩ XT,R ;
(ii) ΨT : XT,R → XT,R is continuous;
(iii) ΨT (XT,R ) is precompact in XT,R .
In the following, we will prove (1) and (2) in Theorem 2.4.5 together. Actually, the proofs
of requirements (ii) and (iii) for (1) and (2) are identically the same, only the proofs of
requirement (i) has slightly difference.
Firstly, given T > 0, let’s recall how we construct u

ΨT (v) for any v ∈ XT,R . We

will use the same notations as those in the proof of Theorem 2.3.6 with β = 0 and g(x, t) =
F x, v(x, t) ✶Γ1 (x), where

✶Γ1 (x)



 1, x ∈ Γ1 ,

(2.4.12)


 0, x ∈
/ Γ1 ,

is the indicator function. Thus u has the following expression: for any (x, t) ∈ Ω × [0, T ],
t

Φ(x − y, t) ψ(y) dy +

u(x, t)
Ω1

Φ(x − y, t − τ ) f (y, τ ) dy dτ
0

Ω

t

Φ(x − y, t − τ ) ϕ(y, τ ) dS(y) dτ.

+
0

∂Ω

Here ϕ ∈ BT satisfies for any (x, t) ∈ (Γ1 ∪ Γ2 ) × (0, T ],
t

ϕ(x, t) =

K(x, t; y, τ ) ϕ(y, τ ) dS(y) dτ + H(x, t),
0

∂Ω

56

(2.4.13)

where
K(x, t; y, τ ) = −2

∂Φ(x − y, t − τ )
∂n(x)

(2.4.14)

and

H(x, t) = − 2

∂Φ(x − y, t)
ψ(y) dy
∂n(x)
Ω1
t

−2
0

∂Φ(x − y, t − τ )
f (y, τ ) dy dτ
∂n(x)
Ω

(2.4.15)

+ 2F x, v(x, t) ✶Γ1 (x).
Since this K also satisfies (2.3.4), we can follow the same way as the derivations of (2.3.26),
(2.3.25) to get
t

K ∗ (x, t; y, τ ) H(y, τ ) dS(y) dτ + H(x, t)

ϕ(x, t) =
0

(2.4.16)

∂Ω

for some function K ∗ . Moreover, there exists a constant C ∗ = C ∗ (n, Ω, T ), such that

|K ∗ (x, t; y, τ )| ≤ C ∗ (t − τ )−3/4 |x − y|−(n−3/2) .

(2.4.17)

Next, we will first assume requirement (i) and prove requirements (ii) and (iii), then we
will confirm requirement (i) for the Cases (1) and (2) in Theorem 2.4.5 respectively.
Given T > 0, we assume there exists R > 0 such that ΨT : XT,R → AT ∩ XT,R . Define
MF , MFσ : [0, ∞) → R by
MF (r) =

sup
Γ1 ×[−r,r]

57

|F (x, σ)|

and
MFσ (r) =

sup

|Fσ (x, σ)|.

Γ1 ×[−r,r]

Then for any fixed r ≥ 0, both MF (r) and MF (r) are finite number since F ∈ C 1 (R). In
the following, for any v ∈ XT,R , we define u, ϕ and H as in (2.4.13), (2.4.14) and (2.4.15)
respectively. For any vj ∈ XT,R (j ≥ 1), we define uj , ϕj and Hj analogously.
• Proof of Requirement (ii). Given {vj }j≥1 ⊂ XT,R and vj → v in XT,R , we want
to show ΨT (vj ) → ΨT (v) in XT,R . Since v and all the vj (j ≥ 1) belong to XT,R ,
then for any (x, t) ∈ Ω × [0, T ], |v(x, t)| ≤ R and |vj (x, t)| ≤ R. Thus, by the mean
value theorem and the fact MF (R) < ∞, it follows from (2.4.15) that Hj ⇒ H on
(Γ1 ∪ Γ2 ) × (0, T ] (here “ ⇒ ” means uniform convergence). Then by (2.4.16) and
(2.4.17), ϕj ⇒ ϕ on (Γ1 ∪ Γ2 ) × (0, T ]. Finally, due to the expression (2.4.13), we have
uj ⇒ u on Ω × [0, T ], which is equivalent to say ΨT (vj ) → ΨT (v) in XT,R .
• Proof of Requirement (iii). In this proof, we will use C to denote a constant which
is independent of j, x and t, but may depend on n, Ω, Ω1 , T , R, MF (R), sup |f |,
sup |ψ| and sup |Dψ|. C may be different from line to line. Given any sequence
{vj }j≥1 ⊂ XT,R , we want to show {ΨT (vj )}j≥1 has a subsequence which converges
to some function u in XT,R . Since vj ∈ XT,R for any j ≥ 1, then for any j ≥ 1 and
for any (x, t) ∈ Ω × [0, T ], |vj (x, t)| ≤ R. Now recalling (2.3.21), we know
∂Φ(x − y, t)
ψ(y) dy
∂n(x)
Ω1
is bounded by some constant C. Consequently by (2.4.15), these exists some constant

58

C such that for any j ≥ 1 and for any (x, t) ∈ Ω × [0, T ],

|Hj (x, t)| ≤ C.

Then due to (2.4.16) and (2.4.17), there exists some constant C such that for any j ≥ 1
and for any (x, t) ∈ Ω × [0, T ],
|ϕj (x, t)| ≤ C.
Now using (2.4.13) and Lemma 2.4.4, we find {uj }j≥1 is uniformly bounded and
equicontinuous on Ω × [0, T ]. Hence it follows from the Arzela-Ascoli theorem that
{uj }j≥1 has a subsequence {ujk }k≥1 which converges uniformly to some function u
on Ω × [0, T ]. Since ujk ∈ XT,R , it is readily to see that u is also in XT,R . Thus,
ΨT (XT,R ) is precompact in XT,R .
Now we turn to verify Requirement (i).
• Proof of Requirement (i) for (1). We will find 0 < T0 ≤ 1 such that for any 0 < T ≤ T0 ,
there exists R > 0 such that ΨT maps XT,R to AT ∩XT,R . In this proof, C will denote
a constant which is independent of x, t, R and T , but may depend on n, Ω, Ω1 , sup |f |,
sup |ψ| and sup |Dψ|. C may be different from line to line. For the first term of (2.4.15),
we recall (2.3.21) again to get for any (x, t) ∈ (Γ1 ∪ Γ2 ) × (0, T ],
∂Φ(x − y, t)
ψ(y) dy
∂n(x)
Ω1
−
−
Φ(x − y, t) ψ(y) →
n (y) · →
n (x) dy +

≤
∂Ω1

Ω1

|x − y|−n dy + C ≤ C.

≤ C

−
Φ(x − y, t) Dψ(y) · →
n (x) dy
(2.4.18)

∂Ω1

59

For the second term of (2.4.15), we have for any (x, t) ∈ (Γ1 ∪ Γ2 ) × (0, T ],
t
0

∂Φ(x − y, t − τ )
f (y, τ ) dy dτ
∂n(x)
Ω
t

(t − τ )−3/4 |x − y|−(n−1/2) dS(y) dτ

≤ C sup |f |
0

Ω
1/4

≤ C t1/4 ≤ C T 1/4 ≤ C T0

≤ C,

(2.4.19)

where the last inequality is due to the assumption T0 ≤ 1. Then it follows from
(2.4.18), (2.4.19) and (2.4.15) that for any (x, t) ∈ (Γ1 ∪ Γ2 ) × (0, T ],

|H(x, t)| ≤ C + C MF (R).

(2.4.20)

Although the constant C ∗ in (2.4.17) depends on T , it is readily to see that C ∗ is an
increasing function in T . As a result, when T is bounded by 1, C ∗ will also be bounded
by some constant C, which only depends on n and Ω. Based on this observation and
(2.4.16), we get for any (x, t) ∈ (Γ1 ∪ Γ2 ) × (0, T ],

|ϕ(x, t)| ≤ C ∗ [C + C MF (R)]

t
0

(t − τ )−3/4 |x − y|−(n−3/2) dS(y) dτ
∂Ω

+ C + C MF (R)
≤ [C + C MF (R)] T 1/4 + C + C MF (R)
≤ C + C MF (R).

(2.4.21)

60

Now by (2.4.13) and (2.4.21), we obtain for any (x, t) ∈ Ω × [0, T ],
t

(t − τ )−3/4 |x − y|−(n−3/2) dS(y) dτ

|u(x, t)| ≤ sup |ψ| + t sup |f | + sup |ϕ|
0

∂Ω

≤ C + C + C MF (R) C T 1/4
1/4

≤ C + C MF (R) T0

1/4

C1 + C1 MF (R) T0

.

(2.4.22)

Hence, if we choose R and T0 ≤ 1 such that

R = 2C1

1/4

and T0

then we have ||u|| ≤ R and therefore u

MF (2C1 ) < 1,

(2.4.23)

ΨT (v) ∈ XT,R .

• Proof of Requirement (i) for (2). We will prove for any T > 0, there exists R > 0 such
that ΨT maps XT,R to AT ∩ XT,R . In this proof, F is a bounded function on Γ1 × R,
so sup |F | < ∞. In the following, we will use C to denote a constant just like that in
Γ1 ×R

the proof for (1) but allowing C to depend on T and sup |F |. As the same derivations
Γ1 ×R

as (2.4.18), (2.4.19) and (2.4.20), we attain for any (x, t) ∈ (Γ1 ∪ Γ2 ) × (0, T ],

|H(x, t)| ≤ C + C MF (R)
≤ C + C sup |F | = C.
R

61

(2.4.24)

Then based on (2.4.16) and (2.4.24), we get for any (x, t) ∈ (Γ1 ∪ Γ2 ) × (0, T ],
t

|ϕ(x, t)| ≤ C ∗ C
0

(t − τ )−3/4 |x − y|−(n−3/2) dS(y) dτ + C
∂Ω

≤ C T 1/4 + C
= C.

Therefore, by (2.4.13) and (2.4.21) again, we obtain for any (x, t) ∈ Ω × [0, T ],
t

(t − τ )−3/4 |x − y|−(n−3/2) dS(y) dτ

|u(x, t)| ≤ sup |ψ| + t sup |f | + sup |ϕ|
0

∂Ω

≤ C + C T 1/4
C2 .

Thus, as long as we take R > C2 , we have ||u|| ≤ R and consequently u

ΨT (v) ∈

XT,R .

2.4.4

Maximal solution and application to the target problem

As we can see from Theorem 2.4.2 that the solution to (2.4.1) is only proved to be global
when the function F is bounded on R. Thus, we need to study the maximal solution when
F is unbounded.
Definition 2.4.6. We call

T∗

sup{T ≥ 0 : there exsits a solution to (2.4.1) on Ω × [0, T ]}

62

the maximal existence time for (2.4.1). Moreover, a function u ∈ C 2,1 Ω × (0, T ∗ ) ∩ C Ω ×
[0, T ∗ ) is called a maximal solution if it solves (2.4.1) on Ω × [0, T ] for any T ∈ (0, T ∗ ).
Remark 2.4.7. It follows from Theorem 2.4.5 and Theorem 2.4.2 that the maximal solution
exists and is unique.
Just like ( [27], Corollary 1.1), when T ∗ is finite, it coincides with the blow-up time of
the L∞ norm of u. Thus in order to estimate T ∗ , one only needs to focus on the L∞ norm
of the solution. We state it more precisely in the following theorem.
Theorem 2.4.8. Let T ∗ be the maximal existence time for (2.4.1) and u be the maximal
solution. If T ∗ < ∞, then

lim
t

sup

T ∗ (x,τ )∈Ω×[0,t]

|u(x, τ )| = ∞.

Proof. We prove by contradiciton. Assume

lim
t

sup

T ∗ (x,τ )∈Ω×[0,t]

|u(x, τ )| < ∞.

Then we denote K = supΩ×[0,T ∗ ) |u| < ∞. The strategy below is to find a solution that
exists beyond T ∗ , which contradicts to the definition of T ∗ .
To do this, we firstly construct a bounded and C 1 function F : Γ1 × R → R such that
F (x, σ) = F (x, σ) for any x ∈ Γ1 and |σ| ≤ K + 1. Then by Theorem 2.4.5, for any T > 0,

63

the problem














u˜t (x, t) − ∆˜
u(x, t) = f (x, t) in
∂u
˜(x,t)
∂n(x)

Ω × (0, T ],

on Γ1 × (0, T ],

= F x, u˜(x, t)

(2.4.25)



∂u
˜(x,t)

=0


∂n(x)






 u˜(x, 0) = ψ(x)

on Γ2 × (0, T ],
on Ω,

has a unique solution u˜. Especially, when T = T ∗ + 1, there exists a solution u˜ ∈ AT ∗ +1 to
(2.4.25).
Since u also solves (2.4.25) for T < T ∗ , then by uniqueness, u˜ = u on Ω × [0, T ∗ ).
Therefore
|˜
u| = K.

sup
Ω×[0,T ∗ )

Now by continuity, there exists

> 0 such that

sup

|˜
u| ≤ K + 1.

Ω×[0,T ∗ + ]

Recalling that F (x, σ) and F (x, σ) coincide when |σ| ≤ K + 1, so u˜ is the solution of (2.4.1)
on Ω × [0, T ∗ + ]. This contradicts to the definition of T ∗ .
As a particular application of the theories established in this section, one can apply
Theorem 2.4.5, Theorem 2.4.2, Theorem 2.4.3 with f = 0, F (x, σ) = σ q and ψ = u0 to our
targeted problem (1.1.1) to obtain Theorem 2.1.1

64

Chapter 3
Upper Bound Estimate of the
Blow-up Time

3.1

Main theorem and outline of the approach

The goal of this chapter is to show the finite-time blowup of the solution to (1.1.1) and find
an upper bound for the blow-up time. In this chapter, M (t) is defined as in (1.4.3). The
next theorem is the main result of this chapter.
Theorem 3.1.1. Let T ∗ be the maximal existence time for (1.1.1). Then T ∗ < ∞ and

lim M (t) = ∞,

t T∗

where M (t) is defined as in (1.4.3). In addition, if min u0 (x) > 0, then
x∈Ω

T∗ ≤

1
1−q
u
(x) dx.
(q − 1)|Γ1 | Ω 0

Inspired by [36], the idea of this proof is very simple, define

u1−q (x, t) dx.

h(t) =
Ω

65

(3.1.1)

Then by formal computations (assuming u is smooth up to the boundary),

h (t) = (1 − q)
Ω

u−q ut dx
u−q ∆u dx

= (1 − q)
Ω

∇ · (u−q Du) dx

≤ (1 − q)
Ω

u−q

= (1 − q)
∂Ω

∂u
dS
∂n

= (1 − q)|Γ1 |.

(3.1.2)

(3.1.2) implies that h(t) decreases at a speed which is at least (q − 1)|Γ1 | every unit time. As
a result, (3.1.1) is justified. Although the solution u is smooth inside the domain, it is not C 1
up to the boundary. So we need to consider the integral of u1−q over interior domains of Ω
and then take limit as the interior domains approach Ω. In addition, in the limiting process,
the uniform limit of the normal derivative is required, so we also have to approximate u by
a sequence of smoother functions whose normal derivative is uniformly continuous up to the
boundary after any positive time and strictly less than the blow-up time.
The organization of this chapter is as follows. In Section 3.2, we explore some geometric
properties of the C 2 domains which make it possible to approximate from inside. All the
conclusions in this section should be standard, but for the sake of completeness, we also
include the detailed proofs. Section 3.3 discusses how to approximate the solution by some
functions with desired regularity of the normal derivative. Finally, Section 3.4 carries out
the rigorous proof for the main theorem.

66

3.2

Approximation for C 2 domain from inside

For any bounded domain Ω ⊂ Rn and for any h > 0, we define

Ωh = {x ∈ Ω : dist(x, ∂Ω) > h}

(3.2.1)

Ωch = {x ∈
/ Ω : dist(x, ∂Ω) > h}.

(3.2.2)

and

Lemma 3.2.1. Let Ω be a bounded, open subset of Rn with ∂Ω ∈ C 2 . Then for any x ∈ ∂Ω,
there exist positive constants r and σ which only depend on x and Ω such that for any
y ∈ B(x, r) ∩ ∂Ω, there is an interior ball touching y with radius σ. Namely,

−
B σ (y − σ →
n (y)) ∩ Ωc = {y}.

(3.2.3)

Analogously, there is also an exterior ball at y with radius σ. Namely,

−
B σ (y + σ →
n (y)) ∩ Ω = {y}.

(3.2.4)

Proof. We will only prove (3.2.3), since the proof of (3.2.4) is almost identical. Fix any
−
˜ −1),
x ∈ ∂Ω, by translation and rotation, we can assume x to be the origin and →
n (x) = (0,
˜ denotes the origin in Rn−1 .
where 0
Since ∂Ω ∈ C 2 , there exist a C 2 function φ : Rn−1 → R and r ∈ (0, 1] (depending on φ
and Ω) such that

67

(1) For any y ∈ B4r ∩ ∂Ω,

→
−
n (y) · en < 0;

yn = φ(˜
y );

|Dφ(˜
y )| ≤ 1/4.

(3.2.5)

(2) B4r ∩ Ω = B4r ∩ {y : yn > φ(˜
y )}.
c

(3) B4r ∩ Ω = B4r ∩ {y : yn < φ(˜
y )}.
Now for any y ∈ Br ∩ ∂Ω, writing y = (˜
y , φ(˜
y )), then
(Dφ(˜
y ), −1)
→
−
,
n (y) =
< Dφ(˜
y) >
1 + |x|2 . We will show that there exists a positive

where < · > is defined as < x >=

constant σ, which only depends on φ and Ω such that (3.2.3) is satisfied.
Let σ ∈ (0, r] be determined later. Define

−
y0 = y − σ→
n (y).

Then
y 0 = y˜ −

σDφ(˜
y)
< Dφ(˜
y) >

and yn0 = φ(˜
y) +

σ
,
< Dφ(˜
y) >

where y 0 denotes the first n − 1 components of y 0 and yn0 denotes the last component of y 0 .
For any z = (˜
z , zn ) ∈ B σ (y 0 ), we have |z − y 0 | ≤ σ and

|z| ≤ |z − y 0 | + |y 0 | ≤ σ + (|y| + σ) ≤ 3r.

So in order to show (3.2.3), it suffices to prove zn > φ(˜
z ) for any z ∈ B σ (y 0 ) \ {y}. Since
68

|z − y 0 | ≤ σ, then
zn ≥ yn0 −

σ 2 − |˜
z − y 0 |2 .

(3.2.6)

Next, the goal is to verify
yn0 −

σ 2 − |˜
z − y 0 |2 ≥ φ(˜
z ).

(3.2.7)

Namely,
φ(˜
z ) − φ(˜
y) ≤

σ
−
< Dφ(˜
y) >

σ 2 − |˜
z − y 0 |.

By Taylor expansion, there exists some η between y˜ and z˜ such that

1
φ(˜
z ) − φ(˜
y ) = Dφ(˜
y ) · (˜
z − y˜) + (˜
z − y˜)T D2 φ(η)(˜
z − y˜).
2
Let λ = max|ξ|≤3r ||D2 φ(ξ)||2 , where || · ||2 denotes the matrix norm. Then
1
φ(˜
z ) − φ(˜
y ) ≤ Dφ(˜
y ) · (˜
z − y˜) + λ|˜
z − y˜|2 .
2

So it suffices to show

1
σ
z − y˜|2 ≤
−
Dφ(˜
y ) · (˜
z − y˜) + λ|˜
2
< Dφ(˜
y) >

σ 2 − |˜
z − y 0 |.

Noticing

σ 2 − |˜
z − y 0 | = σ 2 − |˜
z − y˜ +
=

σDφ(˜
y) 2
|
< Dφ(˜
y) >

σ2
σ
2−2
−
|˜
z
−
y
˜
|
Dφ(˜
y ) · (˜
z − y˜).
< Dφ(˜
y) >
< Dφ(˜
y ) >2

69

(3.2.8)

Let
A = Dφ(˜
y ) · (˜
z − y˜);

B=

σ
;
< Dφ(˜
y) >

E = |˜
z − y˜|2 .

Then (3.2.8) boils down to

1
A + λE ≤ B −
2

B 2 − E − 2AB,

that is
1
B 2 − E − 2AB ≤ B − A − λE.
2

(3.2.9)

Since

|˜
z − y˜| ≤ |˜
z − y 0 | + |y 0 − y˜|
≤ σ + σ = 2σ,

then |A| ≤ |˜
z − y˜|/4 ≤ σ/2 and E ≤ 4σ 2 . On the other hand, |Dφ(˜
y )| ≤ 1/4 implies that
B ≥ 2σ/3. As a result, as long as we take

σ

min{r,

1
},
12λ

B − A − 12 λE ≥ 0. Hence (3.2.9) reduces to
1
2
B 2 − E − 2AB ≤ B − A − λE .
2

70

Simplifying the above inequality, we obtain an equivalent form

1
λE(B − A) ≤ E + A2 + λ2 E 2 ,
4

which is always true because of the fact that

|λE(B − A)| ≤ λE(σ/2 + σ) ≤ E/8 ≤ E.

(3.2.10)

Thus, we proved (3.2.7). Then it follows from (3.2.6) that zn ≥ φ(˜
z ), which means z ∈ Ω.
In addition, if zn = φ(˜
z ), then (3.2.10) should take “equal sign”. This implies E = 0, which
means z˜ = y˜. Moreover, since both (3.2.6) and (3.2.7) should also take “equal sign”, we have
zn = φ(˜
z ) = φ(˜
y ) = yn , which implies z = y. As a result, (3.2.3) is justified.
Corollary 3.2.2. Let Ω be a bounded, open subset of Rn with ∂Ω ∈ C 2 . Then there exists
a positive constant σ = σ(Ω) such that for any x ∈ ∂Ω, there exists an interior ball and also
an exterior ball at x with radius σ.
Proof. It follows from (3.2.1) and the compactness of ∂Ω.
Corollary 3.2.3. Let Ω be a bounded, open subset of Rn with ∂Ω ∈ C 2 . Then there exists
a positive constant σ = σ(Ω) such that
−
(1) for any h ∈ (0, σ), the map Ψh : ∂Ω → ∂Ωh defined by Ψh (x) = x − h→
n (x) is a C 1
diffeomorphism.
−
(2) for any h ∈ (0, σ), the map Ψh : ∂Ω → ∂Ωch defined by Ψh (x) = x + h→
n (x) is a C 1
diffeomorphism.
Proof. Again, we will only prove for (1), since the proof for (2) is analogous.
71

Define σ1 as that in Corollary 3.2.2. Then for any h ∈ (0, σ1 ) and for any x ∈ ∂Ω, there
−
n (x)) ∩ Ωc = {x}. As a
exists an interior ball touching x with radius h. Namely, B h (x − h→
result, dist(Ψh (x), ∂Ω) = h. So Ψh is well-defined. Moreover, this also implies that Ψh is an
injection. To show Ψh is surjective, take any y ∈ Ω such that dist(y, ∂Ω) = h. Then there
−
exists x ∈ ∂Ω such that |y − x| = h. As a result, y = x − h→
n (x) = Ψh (x).
−
Now Ψh has been proved to be a bijection. Next, since ∂Ω is C 2 , then →
n is C 1 on ∂Ω.
Hence, by the inverse function theorem, there exists σ2 > 0 such that for any h ∈ (0, σ2 ),
Ψh is a C 1 diffeomorphism. Finally, by taking σ = min{σ1 , σ2 }, (1) is justified.
The following corollary is a simple version of the tubular neighborhood theorem.
Corollary 3.2.4. Let Ω be a bounded, open subset of Rn with ∂Ω ∈ C 2 . Then there exists a
positive constant σ = σ(Ω) such that the map Ψ : ∂Ω×(−σ, σ) → {x ∈ Rn : dist (x, ∂Ω) < σ}
−
defined by Ψ(x, h) = x − h→
n (x) is a C 1 diffeomorphism.
Proof. First, by Corollary 3.2.3, there exists σ > 0, only depending on Ω, such that Ψ :
−
∂Ω × (−σ, σ) → {x ∈ Rn : dist (x, ∂Ω) < σ} defined by Ψ(x, h) = x − h→
n (x) is a bijection.
Then again by the inverse function theorem, by choosing σ smaller enough, Ψ is a C 1
diffeomorphism.
Lemma 3.2.5. Let Ω be a bounded, open subset of Rn with ∂Ω ∈ C 2 . Then there exists a
positive constant σ = σ(Ω) such that for any h ∈ (0, σ) and for any x ∈ ∂Ω,

−
→
−
n→
h Ψh (x) = n (x),

(3.2.11)

where Ψh is defined as in Corollary 3.2.3 and −
n→
h denotes the exterior unit normal vector
with respect to ∂Ωh .
72

Proof. By Corollary 3.2.3, there exists σ = σ(Ω) such that for any h ∈ (0, σ), Ψh is a
C 1 diffeomorphism between ∂Ω to ∂Ωh . Next fix h ∈ (0, σ) and x ∈ ∂Ω, we will show
−
→
−
n→
h Ψh (x) = n (x).
Just like the proof for Theorem 2.2.6, by translating and rotating the coordinates, we
−
can assume that x to be the origin 0 and →
n (0) = −en = (0, 0, · · · , 0, −1). As a result,
Ψh (x) = hen and then it is equivalent to prove that −
n→
h (hen ) = −en . Suppose the point
near 0 can be parametrized by

y = (˜
y , yn ) = (˜
y , φ(˜
y )).

So when y is near 0, the exterior unit normal vector at y is

(Dφ(˜
y ), −1)
→
−
.
n (y) =
< Dφ(˜
y) >

Then the point near hen can be parametrized by

z = Ψh (y) = (˜
y , φ(˜
y )) − h

(Dφ(˜
y ), −1)
< Dφ(˜
y) >

F (˜
y ).

˜ n−1 .
As a result, the tangent plane at hen with respect to ∂Ωh is spanned by {Di F (0)}
i=1
Thus in order to prove −
n→
h (hen ) = −en , it suffices to show en is perpendicular to each
˜ or equivalently the nth component of Di F (0)
˜ is 0. For each 1 ≤ i ≤ n − 1, by direct
Di F (0),
˜ = 0, we find the nth component of Di F (0)
˜ is indeed 0.
calculations and noticing Di φ(0)

73

3.3

Approximation for the solution by smoother functions

By Theorem 2.1.1, there exists a unique nonnegative maximal solution u ∈ C 2,1 Ω ×
(0, T ∗ ) ∩ C Ω × [0, T ∗ ) to (1.1.1) such that




ut (x, t) = ∆u(x, t) in Ω × (0, T ∗ ),








 ∂u(x,t)
= uq (x, t) on Γ1 × (0, T ∗ ),
∂n(x)


∂u(x,t)

=0


∂n(x)






 u(x, 0) = u0 (x)

on Γ2

(3.3.1)

× (0, T ∗ ),

on Ω,

where T ∗ is the maximal existence time to (1.1.1). It is readily seen that the normal derivative
of this solution is not continuous along the boundary. But in many discussions, it really
requires the continuity of the normal derivative. Thus, the purpose of this section is to
construct a sequence of functions that has normal derivatives of better behavior hear he
boundary.
Lemma 3.3.1. Assume g is continuous on ∂Ω × [0, T ] and ψ ∈ C 1 (Ω). Let v be the solution
to




vt (x, t) = ∆v(x, t) in





∂v(x,t)

Ω × (0, T ],

on ∂Ω × (0, T ],

= g(x, t)
∂n(x)






 v(x, 0) = ψ(x)

in

Ω.

Then for any τ0 > 0, the following limit converges uniformly on ∂Ω × [τ0 , T ].

lim

h→0+

∂v(xh , t)
∂v(x, t)
=
.
∂n(x)
∂n(x)
74

Proof. We will mimic the proof of Theorem 2.3.6 with f = 0, β = 0, Γ1 = ∂Ω and Γ2 = ∅.
First, we extend ψ to a C 1 function to a larger domain Ω1

Ω. Then from that proof, the

solution v can be written as
t

Φ(x − y, t)ψ(y) dy +

v(x, t) =
Ω1

Φ(x − y, t − τ )ϕ(y, τ ) dS(y) dτ
0

(3.3.2)

∂Ω

v1 (x, t) + v2 (x, t),

(3.3.3)

where ϕ is a continuous function on ∂Ω × [0, T ] that is related to g and ψ.
Now fix any τ0 ∈ (0, T ), it is readily seen that the following limit converges uniformly on
∂Ω × [τ0 , T ].
∂v (x, t)
∂v1 (xh , t)
= 1
.
∂n(x)
h→0+ ∂n(x)
lim

On the other hand, Theorem 2.2.1 implies the uniform convergence of

∂v2 (xh , t)
∂v (x, t)
= 2
∂n(x)
h→0+ ∂n(x)
lim

on ∂Ω × [τ0 , T ].
Next, we take a sequence of boundary pieces {Γ1,j }j≥1 such that Γ1,j ⊂ Γ1 and Γ1,j

Γ1

(see Figure 3.1). Then a sequence of C ∞ cut-off functions {ηj }∞
j=1 are chosen so that for

Γ1,j

Γ1
Figure 3.1: Γ1,j

75

each j ≥ 1,

ηj (x)




= 1,






x ∈ Γ1,j ,

∈ [0, 1],






 = 0,

x ∈ Γ1 \ Γ1,j ,

(3.3.4)

x ∈ ∂Ω \ Γ1 .

In addition, we require that ηj+1 (x) ≥ ηj (x), ∀ j ≥ 1, x ∈ ∂Ω.
For any T ∈ (0, T ∗ ) and for each j ≥ 1, we consider



(vj )t (x, t) = ∆vj (x, t)






in

Ω × (0, T ],

∂vj (x,t)
∂n(x)

= ηj (x) uq (x, t) on ∂Ω × (0, T ],






 v (x, 0) = u (x)
in Ω,
0
j

(3.3.5)

∂v (x,t)

j
= ηj (x) uq (x, t) is just the solution
where the function u in the boundary condition ∂n(x)

to (3.3.1). Since u is continuous and (3.3.5) is linear in vj , it follows from Theorem 2.3.6
that there exists a solution vj ∈ C 2,1 (Ω × 0, T ] ∩ C Ω × [0, T ] to (3.3.5). In addition, by
(3.3.4) and Corollary 2.3.8, vj ≤ u on Ω × [0, T ].
Lemma 3.3.2. For any (x, t) ∈ Ω × (0, T ], lim vj (x, t) = u(x, t).
j→∞

Proof. Define the indicator function ✶Γ1 as (2.4.12) and let wj = u − vj , then




(wj )t (x, t) = ∆wj (x, t)





∂wj (x,t)
= gj (x, t)
∂n(x)







 wj (x, 0) = 0
where gj (x, t) = [✶Γ1 (x) − ηj (x)]uq (x, t).

76

in

Ω × (0, T ],

on

(Γ1 ∪ Γ2 ) × (0, T ],

on

Ω,

(3.3.6)

Then similar to the proof of Theorem 2.3.6, for any (x, t) ∈ Ω × (0, T ],
t

wj (x, t) =

0

∂Ω

Φ(x − y, t − τ ) ϕj (y, τ ) dS(y) dτ,

(3.3.7)

where ϕj ∈ BT satisfies for any x ∈ Γ1 ∪ Γ2 and t ∈ (0, T ],
t

ϕj (x, t) =

0

∂Ω

K(x, t; y, τ ) ϕj (y, τ ) dS(y) dτ + 2gj (x, t)

(3.3.8)

with
K(x, t; y, τ ) = −2

∂Φ(x − y, t − τ )
.
∂n(x)

Since K satisfies (2.3.4), we can follow the same way as the derivations of (2.3.25) and
(2.3.26) to obtain

|K ∗ (x, t; y, τ )| ≤ C ∗ (t − τ )−3/4 |x − y|−(n−3/2) .

(3.3.9)

for some constant C ∗ = C ∗ (n, Ω, T ). Moreover,
t

ϕj (x, t) =

0

∂Ω

K ∗ (x, t; y, τ ) Hj (y, τ ) dS(y) dτ + 2gj (x, t).

(3.3.10)

Due to the fact that u is bounded on Ω × [0, T ] and the choice of {ηj }j≥1 , we know Hj
is also bounded on Ω × [0, T ] and

lim Hj (x, t) = 0,

j→∞

∀ x ∈ Γ1 ∪ Γ2 , t ∈ (0, T ].

Then it follows from (3.3.9), (3.3.10) and the Lebesgue’s dominated convergence theorem

77

that
∀ x ∈ Γ1 ∪ Γ2 , t ∈ (0, T ].

lim ϕj (x, t) = 0,

j→∞

In addition, the boundedness of Hj implies the boundedness of ϕj , hence by (3.3.7) and the
Lebesgue’s dominated convergence theorem, we obtain

lim wj (x, t) = 0,

j→∞

∀ (x, t) ∈ Ω × (0, T ].

Lemma 3.3.3. For any j ≥ 1 and T ∈ (0, T ∗ ), let vj be the solution to (3.3.5) on Ω×[0, T ] .
Then for any φ ∈ C Ω × [0, T ] and for any 0 < t1 < t2 ≤ T ,

lim

t2

k→∞ t1

φ(x, t)
∂Ω1/k

t2
∂vj (x, t)
∂vj (x, t)
φ(x, t)
dS(x) dt =
dS(x) dt,
∂nk (x)
∂n(x)
∂Ω
t1

where −
n→
k denotes the normal derivative with respect to ∂Ω1/k .
Proof. By Corollary 3.2.3 and Lemma 3.2.5, for sufficiently large k, the function Ψk : ∂Ω →
∂Ω1/k defined by
Ψk (ξ) = ξ −

1→
−
n (ξ),
k

∀ ξ ∈ ∂Ω,

is a C 1 bijection such that

−
→
−
n→
k (Ψk (ξ)) = n (ξ),

78

∀ ξ ∈ ∂Ω.

As a result, by the change of variable x = Ψk (ξ) for x and denoting dS(x) = Fk (ξ)dS(ξ),

∂Ω1/k

=
∂Ω

=
∂Ω

φ(x, t)Dvj (x, t) · −
n→
k (x) dS(x)

φ Ψk (ξ), t Dvj (Ψk (ξ), t) · −
n→
k Ψk (ξ) Fk (ξ) dS(ξ)
−
φ Ψk (ξ), t Dvj (Ψk (ξ), t) · →
n (ξ)Fk (ξ) dS(ξ).

Integrating t from t1 to t2 ,
t2
t1

=

∂Ω1/k

t2
t1

∂Ω

φ(x, t)Dvj (x, t) · −
n→
k (x) dS(x) dt

−
φ Ψk (ξ), t Dvj (Ψk (ξ), t) · →
n (ξ)Fk (ξ) dS(ξ).

(3.3.11)

It is readily seen that φ Ψk (ξ), t converges uniformly to φ(ξ, t) and Fk (ξ) converges to
1 uniformly on ∂Ω × [t1 , t2 ] as k → ∞. In addition, it follows from Lemma 3.3.3 that
−
−
Dvj (Ψk (ξ), t) · →
n (ξ) converges uniformly to Dvj (ξ, t) · →
n (ξ). Thus the limit can be taken
inside the integral to justify the conclusion.

3.4

Upper bound on life span: case of C 2 domain

The goal of this section is to prove the unique solution u of (1.1.1) always blows up (i.e.
L∞ norm of u goes to ∞) in finite time. In addition, we will derive an upper bound for the
blow-up time in terms of |Γ1 |, the nonlinearity q and the initial data u0 .
The usual way to prove the blowup of a solution is to introduce a suitable energy function
and then derive a differential inequality to show the energy function blows up. This process
usually involves integration by parts and therefore requires some continuity of the spatial
79

derivative Du near the boundary. However, u is not smooth, since the normal derivative
of u is not continuous along Γ. Thus, some approximations are needed to get through this
process.
For any k, let Ω1/k be the same as in (3.2.1). For any x ∈ ∂Ω1/k , we use −
n→
k (x) to denote
−
the exterior unit normal vector at x with respect to ∂Ω1/k . For any x ∈ ∂Ω, →
n (x) represents
the exterior unit normal vector at x with respect to ∂Ω.
Proof of Theorem 3.1.1. Let u be the maximal solution to (1.1.1) as in Theorem 2.1.1.
Fix any 0 < τ0 < T < T ∗ . For any j ≥ 1, let vj be the solution to (3.3.5). Then define

0

=

min
(x,t)∈Ω×[τ0 ,T ]

v1 (x, t).

It follows from Corollary 2.3.8 that 0 > 0. Noticing that {vj }j≥1 is an increasing sequence
of functions that converges to u, so

0

∀ j ≥ 1, ∀ (x, t) ∈ Ω × [τ0 , T ].

≤ vj (x, t) ≤ u(x, t) ≤ M (T ),

(3.4.1)

Mimicking the idea in [36], for any j ≥ 1 and k ≥ 1, define hj,k : (0, T ] → R and
hj : (0, T ] → R by
hj,k (t) =

1−q

Ω1/k

vj

(x, t) dx

and
hj (t) =

1−q

Ω

vj

80

(x, t) dx.

Then

hj,k (t) = (1 − q)

−q

Ω1/k

vj (vj )t dx
−q

= (1 − q)
Ω1/k

vj ∆vj dx
−q

= (1 − q)
Ω1/k

−q−1

∇ · (vj Dvj ) + q vj

|Dvj |2 dx

−q

≤ (1 − q)
Ω1/k

∇ · (vj Dvj ) dx
−q

= (1 − q)
∂Ω1/k

vj

∂vj
dS.
∂nk

(3.4.2)

Integrating (3.4.2) for t from τ0 to T ,
T

hj,k (T ) − hj,k (τ0 ) ≤ (1 − q)

τ0

−q

∂Ω1/k

vj (x, t)

∂vj (x, t)
dS(x) dt.
∂nk (x)

Taking k → ∞, by Lebesgue’s dominated convergence theorem and Lemma 3.3.3,
T

hj (T ) − hj (τ0 ) ≤ (1 − q)

−q

τ0
T

∂Ω

τ0

∂Ω

vj (x, t)
−q

= (1 − q)

∂vj (x, t)
dS(x) dt
∂n(x)

vj (x, t)ηj (x)uq (x, t) dS(x) dt.

(3.4.3)

When j → ∞, it follows from Lemma 3.3.2 that vj goes to u pointwise. Moreover, ηj
converges almost everywhere to ✶Γ1 on ∂Ω. Thus by taking j → ∞ in (3.4.3) and noticing
the bound (3.4.1), it follows from the Lebesgue’s dominated convergence theorem that

u1−q (x, T ) dx −
Ω

Ω

u1−q (x, τ0 ) dx ≤ (1 − q)

81

T
τ0

|Γ1 | dt = (1 − q)(T − τ0 )|Γ1 |.

So

(q − 1)(T − τ0 )|Γ1 | ≤

Ω

≤
Ω

u1−q (x, τ0 ) dx −

u1−q (x, T ) dx
Ω

u1−q (x, τ0 ) dx.

Namely
1
u1−q (x, τ0 ) dx.
T ≤ τ0 +
(q − 1)|Γ1 | Ω
Noticing the right hand side of the above inequality is independent of T , so we can send T
to T ∗ , then
T ∗ ≤ τ0 +

1
u1−q (x, τ0 ) dx.
(q − 1)|Γ1 | Ω

(3.4.4)

Hence, T ∗ is finite. Then it follows from Theorem 2.4.8 and the positivity of u that

lim M (t) = ∞.

t T∗

Now if min u0 (x) > 0, then u has a strictly positive lower bound on Ω × [0, T ∗ ). As a
x∈Ω

result, by taking τ0 → 0 in (3.4.4), (3.1.1) follows.

82

Chapter 4
Lower Bound Estimate of the
Blow-up Time

4.1

Main theorems and outline of the approach

The goal of this section is to obtain lower bounds for the blow-up time. Again in this chapter,
M (t) is defined as in (1.4.3). The main results consist of three parts.
Part I: C 2 domain Ω.
Theorem 4.1.1. Assume (1.1.2). Let T ∗ be the maximal existence time for (1.1.1). Then
there exists a constant C = C(n, Ω) such that

T∗ ≥

C
− 2
ln 1 + (3M0 )−4(q−1) |Γ1 | n−1 ,
q−1

(4.1.1)

where M0 is given by (1.4.2).
From this theorem, we can study the asymptotic behaviour of T ∗ .
As q

1, T ∗ is at least of order (q − 1)−1 . Combining with the upper bound in

Theorem 3.1.1, the order of T ∗ is exactly (q − 1)−1 .
As M0

0, T ∗ is at least of order ln(M0−1 ). As M0 → ∞, T ∗ is at least of order
83

−4(q−1)

M0

.
0, T ∗ is at least of order ln |Γ1 |−1 .

As |Γ1 |

Part II: C 2 domain Ω with local convexity near Γ1 .
Definition 4.1.2. Let Ω be a bounded, open subset in Rn . Then for any Γ ⊆ ∂Ω and d > 0,
the boundary part of Ω near Γ within distance d is defined to be

[Γ]d

{x ∈ ∂Ω : dist (x, Γ) < d}.

(4.1.2)

In the following, as a standard notation, for any set A ⊆ Rn , Conv (A) denotes the convex
hull of the set A.
Theorem 4.1.3. Assume (1.1.2). Let T ∗ be the maximal existence time for (1.1.1) and
M0 be defined as in (1.4.2). Assume Conv [Γ1 ]d ⊆ Ω for some d > 0. Then there exist
constants Y0 = Y0 (n, Ω, d) and C = C(n, Ω, d) such that the following statements hold.
• Case 1: n ≥ 3. Denote
q−1

Y = M0

|Γ1 |1/(n−1) .

Y

If Y ≤ q0 , then
T∗ ≥

C
.
(q − 1)Y | ln Y |

(4.1.3)

• Case 2: n = 2. Denote
q−1

Y = M0

|Γ1 | ln

1
+1 .
|Γ1 |

Y

If Y ≤ q0 , then
T∗ ≥

C
.
(q − 1)Y | ln Y |
84

(4.1.4)

It is readily seen that the asymptotic behaviour of T ∗ has been improved a lot. More
precisely,
−(q−1)

1

−
0, the order of T ∗ is at least |Γ1 | n−1

As |Γ1 |
|Γ1 |−1

ln M0−1 .

0, the order of T ∗ is at least M0

As M0

ln |Γ1 |−1

2

ln |Γ1 |−1

for n ≥ 3 and

for n = 2.

Recalling the upper bound in Theorem 3.1.1, if u0 does not oscillate too much, which
1−q

means Ω u0
−(q−1)

M0

1−q

dx is comparable to M0

0, the order of T ∗ is at most

, then as M0

. So the lower bound is almost optimal on M0 . In addition, the order of T ∗ is at

most |Γ1 |−1 , so when n = 2, the lower bound is also almost sharp.
Part III: Convex C 2 domain Ω.
1 , there exists
Theorem 4.1.4. Assume (1.1.2). Let Ω be convex. Then for any α ∈ 0, n−1

C = C(n, Ω, α) such that

T∗ ≥

C
q−1

(q − 1)M0

|Γ1 |α

1+(n−1)α
1−(n−1)α

1

min 1,

q−1

qM0

|Γ1 |α

,

(4.1.5)

where T ∗ is the maximal existence time for (1.1.1) and M0 is given by (1.4.2). In particular,
if α is chosen to be 0 in (4.1.5), then

C1

T∗ ≥
(q

q−1
− 1)M0

min 1,

1
q−1

,

(4.1.6)

qM0

for some C1 = C1 (n, Ω).
Remark 4.1.5. As |Γ1 |

0, the lower bound (4.1.5) is not the best that we can get. By
1

−
Remark 4.5.11, the order of T ∗ is at least |Γ1 | n−1 for n ≥ 3 and |Γ1 |−1 ln |Γ1 |−1 for

85

n = 2. Noticing that when |Γ1 | is not sufficiently small, Theorem 4.1.3 and Remark 4.5.11
are not applicable. So the advantage of (4.1.5) is its validity for any Γ1 . This advantage is
important when comparing with the previous results for Γ1 = ∂Ω.
In addition, Theorem 4.1.4 concluded better asymptotic behavior of T ∗ as M0

0 than

Theorem 4.1.3 and Remark 4.5.11.
As M0

−(q−1)

0, (4.1.6) implies that T ∗ is at least of order M0

. So by the same

discussions after Theorem 4.1.3, if the initial data u0 does not oscillate too much, then
−(q−1)

the order of T ∗ is exactly M0

.
−2(q−1)

As M0 → ∞, (4.1.6) implies that T ∗ is at least of order M0

.

To obtain a lower bound for T ∗ , we first exploit the common Gronwall’s technique in
Subsection 4.3.1. But this estimate is too rough, in order to get better results, we need
to be more careful. The idea is to chop the range of M (t) into suitable pieces and find a
lower bound for the time spent in each piece by analysing the representation formula. Then
adding all these lower bounds yields a lower bound for T ∗ . In the proof for Theorem 4.1.1,
the pieces are just chosen to be [3k−1 M0 , 3k M0 ] for k ≥ 1. In the proofs of Theorem 4.1.3
and Theorem 4.1.4, due to the convexity assumptions and more delicate divisions of the
range, the results can be improved significantly. More precisely, we will chop the range to be
[Mk−1 , Mk ] for k ≥ 1, where the sequence {Mk }k≥0 satisfies a nonlinear iterative relation.
The organization of this chapter is as follows. In Section 4.2, we first prove that the
classical solution u is also the weak solution which implies the representation formula of u.
This representation formula is the fundamental tool in the later sections. In Section 4.3, two
methods are presented to obtain the lower bound of T ∗ for any C 2 domains Ω. The first
method is simpler and uses the Gronwall’s inequality. The second method performs more
86

delicate estimate and yield Theorem 4.1.1. In Section 4.4, the lower bound is derived for
any convex domains Ω and Theorem 4.1.4 is verified. We also explain the main idea in the
proof which is also used in the next section. In Section 4.5, we deal with C 2 domains Ω with
local convexity near Γ1 and justify Theorem 4.1.3. We also mention a similar result for the
convex domains in Remark 4.5.11. Finally, Section 4.6 compares the results in this chapter
with the historical works.

4.2

Weak solution and representation formula

By Theorem 2.1.1, there exists a unique nonnegative maximal solution u ∈ C 2,1 Ω ×
(0, T ∗ ) ∩ C Ω × [0, T ∗ ) to (1.1.1), where T ∗ is the maximal existence time to (1.1.1).
In this section, we will first verify that the solution u to (1.1.1) is also a weak solution (See
Definition 4.2.1) and then derive representation formulas (4.2.2) and (4.2.3) for u.

4.2.1

Weak solution

Definition 4.2.1. Let T ∗ be the maximal existence time for (1.1.1). Then a function u ∈
C Ω × [0, T ∗ ) is called a weak solution of (1.1.1) if for any t ∈ (0, T ∗ ) and for any φ ∈
C 2 Ω × [0, t] ,
t

u(y, τ )(φτ + ∆φ)(y, τ ) dy dτ =
0

Ω
t

uq (y, τ )φ(y, τ ) dS(y) dτ

−
0

Γ1

Ω
t

+
0

u(y, t)φ(y, t) − u0 (y)φ(y, 0) dy
∂φ(y, τ )
dS(y) dτ.
u(y, τ )
∂n(y)
∂Ω

(4.2.1)

In order to prove u satisfies (4.2.1), we will again take advantage of vj which is the
solution to (3.3.5).

87

Theorem 4.2.2. The maximal solution u to (1.1.1) is also a weak solution as in Definition
4.2.1.
Proof. Fix any t ∈ (0, T ∗ ). Let {vj }j≥1 be the solution to (3.3.5). For any k ≥ 1, define
Ω1/k as in (3.2.1).
Then for any 0 < τ0 < t, k ≥ 1, j ≥ 1 and φ ∈ C 2 Ω × [0, t] , we have
t

t

τ0 Ω1/k

(vj )t (y, τ ) φ(y, τ ) dy dτ =

τ0 Ω1/k

∆vj (y, τ ) φ(y, τ ) dy dτ.

Integrating by parts and arranging the terms,
t
τ0 Ω1/k

vj (y, τ ) (φτ + ∆φ)(y, τ ) dy dτ
t

=
Ω1/k

vj (y, t)φ(y, t) − vj (y, τ0 )φ(y, τ0 ) dy −

t

+
τ0 ∂Ω1/k

vj (y, τ )

τ0 ∂Ω1/k

∂vj (y, τ )
φ(y, τ ) dS(y) dτ
∂nk (y)

∂φ(y, τ )
dS(y) dτ,
∂nk (y)

where −
n→
k denotes the exterior unit normal vector with respect to ∂Ω1/k .
Sending k → ∞, by Lebesgue’s dominated convergence theorem and Lemma 3.3.3, we

88

obtain
t
τ0 Ω

vj (y, τ ) (φτ + ∆φ)(y, τ ) dy dτ
t

=
Ω

vj (y, t)φ(y, t) − vj (y, τ0 )φ(y, τ0 ) dy −
t

+
τ0 ∂Ω

vj (y, τ )

∂vj (y, τ )
φ(y, τ ) dS(y) dτ
τ0 ∂Ω ∂n(y)

∂φ(y, τ )
dS(y) dτ
∂n(y)
t

=
Ω

vj (y, t)φ(y, t) − vj (y, τ0 )φ(y, τ0 ) dy −
t

+
τ0 ∂Ω

vj (y, τ )

τ0 ∂Ω

ηj (y)uq (y, t) φ(y, τ ) dS(y) dτ

∂φ(y, τ )
dS(y) dτ.
∂n(y)

Taking j → ∞, then it follows from Lemma 3.3.2 and the Lebesgue’s dominated convergence
theorem that
t

u(y, τ ) (φτ + ∆φ)(y, τ ) dy dτ
τ0 Ω
t

=
Ω

u(y, t)φ(y, t) − u(y, τ0 )φ(y, τ0 ) dy −
t

+

u(y, τ )
τ0 ∂Ω

uq (y, τ )φ(y, τ ) dS(y) dτ

τ0 Γ1

∂φ(y, τ )
dS(y) dτ.
∂n(y)

Finally by sending τ0 → 0, (4.2.1) holds.

4.2.2

Representation formula

Next by (4.2.1) and some standard steps, we are able to attain the representation formula
of u for inside points. Note that this representation formula is different from (2.3.15) which
is used in the proof of the existence of the solution.
Lemma 4.2.3. For the maximal solution u to (1.1.1), it has the representation formula for
89

the inside points (x, t) ∈ Ω × [0, T ∗ ),
t

u(x, t) =
Ω

Φ(x − y, t) u0 (y) dy +
t

−
0

Φ(x − y, t − τ ) uq (y, τ ) dS(y) dτ

0

Γ1

(4.2.2)

∂Φ(x − y, t − τ )
u(y, τ ) dS(y) dτ.
∂n(y)
∂Ω

Proof. Fix x ∈ Ω and t ∈ (0, T ∗ ). Define φ : Ω × [0, t) → R by

φ(y, τ ) = Φ(x − y, t − τ ) =

1

1

(4π)n/2 (t − τ )n/2

e

−

|x−y|2
4(t−τ ) .

> 0, define φ : Ω × [0, t] → R by

For any

φ (y, τ ) = Φ(x − y, t + − τ ) =

1

1

(4π)n/2 (t + − τ )n/2

e

−

|x−y|2
4(t+ −τ ) .

From these, one can see that φ is smooth in its domain and

(φ )τ (y, τ ) + ∆y (φ )(y, τ ) = 0,

∀ (y, τ ) ∈ Ω × [0, t].

Applying (4.2.1) for φ = φ ,
t

φ (y, t)u(y, t) dy =
Ω

Ω

φ (y, 0)u0 (y) dy +
t

−
0

0

φ (y, τ ) uq (y, τ ) dS(y) dτ
Γ1

∂φ (y, τ )
u(y, τ ) dS(y) dτ.
∂Ω ∂n(y)

90

Sending → 0, it follows from the Lebesgue’s dominated convergence theorem that
t

u(x, t) =
Ω

φ(y, 0) u0 (y) dy +
t

−
0

φ(y, τ ) uq (y, τ ) dS(y) dτ

0

Γ1

∂φ(y, τ )
u(y, τ ) dS(y) dτ
∂Ω ∂n(y)
t

=
Ω

Φ(x − y, t) u0 (y) dy +
t

−
0

0

Φ(x − y, t − τ ) uq (y, τ ) dS(y) dτ
Γ1

∂Φ(x − y, t − τ )
u(y, τ ) dS(y) dτ.
∂n(y)
∂Ω

The last equality is because φ(y, τ ) = Φ(x − y, t − τ ). Now we have proved (4.2.2) for
(x, t) ∈ Ω × (0, T ∗ ), next it is obvious to see that (4.2.2) holds for any point (x, t) ∈ Ω × {0},
thus the Theorem follows.
Lemma 4.2.3 only gives the formula for the inside points, but we still need the formula
for the boundary points (x, t) ∈ ∂Ω × [0, T ∗ ) . In order to get that, we combine Lemma 4.2.3
and Theorem 2.2.2.
Corollary 4.2.4. For the maximal solution u to (1.1.1), it has the representation formula
for the boundary points (x, t) ∈ ∂Ω × [0, T ∗ ),
t

u(x, t) =2
Ω

Φ(x − y, t) u0 (y) dy + 2
t

−2
0

0

Φ(x − y, t − τ ) uq (y, τ ) dS(y) dτ
Γ1

(4.2.3)

∂Φ(x − y, t − τ )
u(y, τ ) dS(y) dτ.
∂n(y)
∂Ω

−
Proof. For any h > 0, we write xh = x − h→
n (x) for x ∈ ∂Ω. As shown in the proof of
Theorem 2.2.2, when h is sufficiently small, xh ∈ Ω for any x ∈ ∂Ω. Consequently we can

91

apply Lemma 4.2.3 to conclude that
t

u(xh , t) =

Ω

Φ(xh − y, t) u0 (y) dy +
t

−
0

0

Γ1

Φ(xh − y, t − τ ) uq (y, τ ) dS(y) dτ

∂Φ(xh − y, t − τ )
u(y, τ ) dS(y) dτ.
∂n(y)
∂Ω

Taking h → 0+ , then it follows from Theorem 2.2.2 that
t

u(x, t) =
Ω

Φ(x − y, t) u0 (y) dy +
t

−
0

0

Φ(x − y, t − τ ) uq (y, τ ) dS(y) dτ
Γ1

∂Φ(x − y, t − τ )
1
u(y, τ ) dS(y) dτ + u(x, t),
∂n(y)
2
∂Ω

which implies (4.2.3). Now we have proved (4.2.3) for (x, t) ∈ ∂Ω × (0, T ∗ ), next it is obvious
to see that (4.2.3) holds for any point (x, t) ∈ ∂Ω × {0}, hence the Theorem follows.

4.2.3

Time-shifted representation formula

In Corollary 4.2.4, it derived the representation formula (4.2.3), where the initial time is 0
and the initial data is u(·, 0) = u0 (·) ∈ C 1 (Ω). Now for any T ∈ (0, T ∗ ), we are asking
that if regarding T to be the initial time and u(·, T ) to be the initial data, then is there a
representation formula similar to (4.2.3)? It seems trivial, but we should be careful, since u0
is in C 1 (Ω) but u(·, T ) does not. The next lemma claims that as long as u is the solution to
(1.1.1) with the assumption (1.1.2), then for any T ∈ [0, T ∗ ), there also holds a representation
formula which is similar to (4.2.3) but with “initial data u(·, T )”.
Lemma 4.2.5. Assume (1.1.2). Let T ∗ be the maximal existence time and u be the maximal

92

solution to (1.1.1). Then for any x ∈ ∂Ω, T ∈ [0, T ∗ ) and t ∈ [0, T ∗ − T ),
t

Φ(x − y, t − τ ) uq (y, T + τ ) dS(y) dτ

Φ(x − y, t) u(y, T ) dy + 2

u(x, T + t) = 2
Ω

0
t

−2
0

Γ1

∂Φ(x − y, t − τ )
u(y, T + τ ) dS(y) dτ
∂n(y)
∂Ω

(4.2.4)

Proof. When T = 0, (4.2.4) is just the representation formula (4.2.3) which has been proven.
Next let T > 0. We intend to verify (4.2.4) which can be regarded as a representation formula
with initial time T and initial data u(·, T ).
Define v : Ω × [0, T ∗ − T ) → R by v(x, t) = u(x, T + t). Then v ∈ C 2,1 Ω × (0, T ∗ −
T)

C Ω × [0, T ∗ − T ) and satisfies














vt (x, t) = ∆v(x, t)
∂v(x,t)
∂n(x)

in

Ω × (0, T ∗ − T ),

= uq (x, T + t) on Γ1 × (0, T ∗ − T ),
(4.2.5)



∂v(x,t)

=0


∂n(x)






 v(x, 0) = u(x, T )

on Γ2 × (0, T ∗ − T ),
in

Ω.

Note that (4.2.5) is a linear problem in v, since u is a fixed function.
We continuously extend u(·, T ) to Rn and still denote it to be u(·, T ). Then for any
j ≥ 1, we choose the standard mollifier {η } >0 and define

gj (x) = η j (·) ∗ u(·, T ) (x),

93

where j is chosen to be so small that

max |gj (x) − u(x, T )| ≤ 1/j.

(4.2.6)

x∈Ω

Since gj ∈ C 1 (Ω), it follows from Theorem 2.3.10 that there exists vj ∈ C 2,1 Ω × (0, T ∗ −
T)

C Ω × [0, T ∗ − T ) such that




(vj )t (x, t) = ∆vj (x, t) in Ω × (0, T ∗ − T ),






∂vj (x,t)


 ∂n(x)
= uq (x, T + t) on Γ1 × (0, T ∗ − T ),

∂vj (x,t)


=0


∂n(x)






 vj (x, 0) = gj (x)

(4.2.7)

on Γ2 × (0, T ∗ − T ),
in

Ω.

In addition, by an analogous argument as that for Subsections 4.2.1 and 4.2.2, there exists
a representation formula for (4.2.7): for any (x, t) ∈ ∂Ω × [0, T ∗ − T ),
t

vj (x, t) = 2

Ω

Φ(x − y, t) gj (y) dy + 2
t

−2
0

0

Φ(x − y, t − τ ) uq (y, T + τ ) dS(y) dτ
Γ1

∂Φ(x − y, t − τ )
vj (y, τ ) dS(y) dτ.
∂n(y)
∂Ω

Let wj = vj − v, then wj ∈ C 2,1 Ω × (0, T ∗ − T )




(wj )t (x, t) = ∆wj (x, t)






∂w


 ∂nj (x, t) = 0

C Ω × [0, T ∗ − T ) and satisfies

in

Ω × (0, T ∗ − T ),

on Γ1 × (0, T ∗ − T ),


∂wj


on Γ2 × (0, T ∗ − T ),


∂n (x, t) = 0






 wj (x, 0) = gj (x) − u(x, T ) in Ω.

94

(4.2.8)

So it follows from the maximum principle and the Hopf lemma that for any (x, t) ∈ Ω ×
[0, T ∗ − T ),
|wj (x, t)| ≤ max |gj (x) − u(x, T )| ≤ 1/j.
x∈Ω

Thus
∀ (x, t) ∈ Ω × [0, T ∗ − T ).

|vj (x, t) − v(x, t)| ≤ 1/j,

Now fixing any point (x, t) ∈ ∂Ω × [0, T ∗ − T ) and let j → ∞ in (4.2.8), then it follows
from Lebesgue’s dominated convergence theorem that
t

Φ(x − y, t − τ ) uq (y, T + τ ) dS(y) dτ

Φ(x − y, t) u(y, T ) dy + 2

v(x, t) = 2

0

Ω
t

−2
0

Γ1

∂Φ(x − y, t − τ )
v(y, τ ) dS(y) dτ.
∂n(y)
∂Ω

Finally noticing that v(x, t) = u(x, T + t) and v(y, τ ) = u(y, T + τ ), we obtain (4.2.4).

4.3

Lower bound on life span: case of C 2 domain

4.3.1

A traditional way by Gronwall-type technique

Lemma 4.3.1. Let Ω be a bounded open subset in Rn with ∂Ω ∈ C 2 . Then there exists a
constant C = C(Ω, n) such that for any x ∈ ∂Ω and t > 0,

1
t(n−1)/2 ∂Ω

−

e

|x−y|2
4t

dS(y) ≤ C.

Proof. It is easy to prove this conclusion by simply using the definition of a C 2 boundary
and the dimension of ∂Ω is n − 1, we omit the proof here.

95

Theorem 4.3.2. Assume (1.1.2). Let T ∗ be the maximal existence time for (1.1.1). Then
there exists a constant C = C(n, q, Ω) such that

T∗

≥C

2
− n+2

ln |Γ1

|−1

− (n + 2)(q − 1) ln M0 − ln(q − 1) − ln C

2
n+2

,

(4.3.1)

where M0 is defined as in (1.4.2) and C remains bounded as q → 1. As a result, no matter
|Γ1 |

0, M0

1, we have T ∗ → ∞.

0 or q

Proof. In the following, C will be used to denote a positive constant which only depends
on n, Ω, q and is bounded when q

1. Moreover, C may be different from line to line.

We prove by analyzing the representation formula (4.2.3) for u on the boundary points
(x, t) ∈ ∂Ω × [0, T ∗ ):
t

u(x, t) = 2
Ω

Φ(x − y, t) u0 (y) dy + 2
t

−2
0

0

Φ(x − y, t − τ ) uq (y, τ ) dS(y) dτ
Γ1

∂Φ(x − y, t − τ )
u(y, τ ) dS(y) dτ
∂n(y)
∂Ω

= I + II + III.

(4.3.2)

Define M , M : [0, T ∗ ) → R by

M (t) = max u(y, t)
y∈∂Ω

and
M (t) = max M (τ ).
τ ∈[0,t]

96

(4.3.3)

It is clear that M is increasing and also blows up at T ∗ . It is also easy to see that

I ≤ 2M0 ,
t

II ≤ C

|x−y|2
−
(t − τ )−n/2 e 4(t−τ )

M q (τ )

0

(4.3.4)

dS(y) dτ

Γ1
t

=C

|x−y|2
− 4τ
−n/2
τ
e

M q (t − τ )

0

(4.3.5)
dS(y) dτ

Γ1

and
|x − y|2

t

III ≤ C

M (τ )
0

∂Ω

(t − τ )n/2+1

e

−

|x−y|2
4(t−τ )

dS(y) dτ

2

|x − y|2 − |x−y|
4τ dS(y) dτ
e
∂Ω τ n/2+1
0
2
t
1 − |x−y|
8τ dS(y) dτ
M (t − τ )
≤C
e
∂Ω τ n/2
0
t

M (t − τ )

=C

t

≤C

(4.3.6)

M (t − τ ) τ −1/2 dτ.

0

In (4.3.6), the first inequality is due to Lemma 2.3.2, the last inequality is due to Lemma
4.3.1 and the second inequality is because
2

|x − y|2 − |x−y|
8τ
e
≤ sup r e−r/8 ≤ C.
τ
r≥0

1 , it follows
Now we are trying to estimate (4.3.5) and (4.3.6) further. Taking m = 1 + n+1

97

from Holder’s inequality that

2

τ

−n/2 − |x−y|
4τ

e

dS(y) ≤ τ

1/m

2

−n/2

e

Γ1

− m|x−y|
4τ

|Γ1 |(m−1)/m

dS(y)

Γ1

=τ

1/m

2

− n2 + n−1
2m

τ

− m|x−y|
− n−1
2
4τ

e

|Γ1 |(m−1)/m

dS(y)

Γ1
n

n−1

≤ C τ − 2 + 2m |Γ1 |
2n+1

m−1
m

1

= C τ − 2n+4 |Γ1 | n+2 ,

where the second inequality is because of Lemma 4.3.1. Thus (4.3.5) leads to
t

1

II ≤ C |Γ1 | n+2

2n+1

−
M q (t − τ ) τ 2n+4 dτ.

(4.3.7)

0

By Holder’s inequality again,
t

− 2n+1
M q (t − τ ) τ 2n+4

t

dτ ≤

qm
M m−1 (t − τ ) dτ

m−1
m

t

0

0

M q(n+2) (τ ) dτ

1
n+2

0

=

dτ

1
m

0
t

=

(2n+1)m
− 2n+4
τ

t

τ

2n+1
− 2n+2

dτ

n+1
n+2

0
t

1
C t 2n+4

M q(n+2) (τ ) dτ

1
n+2

.

0

Based on this, (4.3.7) becomes

II ≤

1
1
C |Γ1 | n+2 t 2n+4

t
0

98

M q(n+2) (τ ) dτ

1
n+2

.

(4.3.8)

To estimate III, it follows from Holder’s inequality that
t

III ≤ C

1
n+2

M n+2 (τ ) dτ

t

τ

0
t

n

= C t 2n+4

0
1
n+2

M n+2 (τ ) dτ

− 12 n+2
n+1

n+1
n+2

dτ

.

(4.3.9)

0

Combining (4.3.2), (4.3.4), (4.3.8), (4.3.9), we obtain

u(x, t) ≤C

t

1
1
M0 + |Γ1 | n+2 t 2n+4
t

n
+ t 2n+4

1
n+2

M q(n+2) (τ ) dτ

0
1
n+2

M n+2 (τ ) dτ

.

0

Since x is arbitrary on ∂Ω, by raising both sides to the power n + 2,
t

M n+2 (t) ≤ C M0n+2 + |Γ1 | t1/2

t

M q(n+2) (τ ) dτ + tn/2

0

M n+2 (τ ) dτ .

0

Thus, due to the definition of M ,

M n+2 (t) ≤ C M0n+2 + |Γ1 | t1/2
≤ C 1 + tn/2

t

M

q(n+2)

t

(τ ) dτ + tn/2

M

0

n+2

(τ ) dτ

0
t

M0n+2 + |Γ1 |

M

q(n+2)

t

(τ ) dτ +

0

M

n+2

(τ ) dτ .

0

As a consequence,

M

n+2

(t) ≤ C

1 + tn/2

M0n+2

t

+ |Γ1 |

M
0

q(n+2)

t

(τ ) dτ +

M
0

99

n+2

(τ ) dτ .

(4.3.10)

We define
t

E(t) = M0n+2 + |Γ1 |

M

q(n+2)

t

(τ ) dτ +

M

n+2

(τ ) dτ,

(4.3.11)

0

0

then E(0) = M0n+2 and E(t) is increasing. Now (4.3.10) becomes

M

n+2

(t) ≤ C 1 + tn/2 E(t)

and consequently

E (t) = |Γ1 | M

q(n+2)

(t) + M

(n+2)

(t)

q

≤ C |Γ1 | 1 + tn/2 E q (t) + C 1 + tn/2 E(t).

(4.3.12)

Moreover, E(t) also blows up at T ∗ , since M is increasing. (4.3.12) looks like the Bernoulli
equation, so we multiply both sides by E −q (t) and define Ψ(t)

E 1−q (t). Then Ψ(t) → 0

as t approaches to T ∗ and

q

Ψ (t) + C(q − 1) 1 + tn/2 Ψ(t) ≥ −C(q − 1)|Γ1 | 1 + tn/2 .

(4.3.13)

We introduce the integration factor µ(t) which is defined as
t

µ(t)

exp C

(q − 1) 1 + τ n/2 dτ ,

∀ t ≥ 0.

0

It is readily seen that
n

µ(t) ≤ C exp C t1+ 2 .

100

(4.3.14)

Multiplying (4.3.13) by µ(t), one gets

µ(t)Ψ(t)

q

≥ −C(q − 1)|Γ1 | 1 + tn/2 µ(t).

−(n+2)(q−1)

Integrating this inequality and noticing that µ(0)Ψ(0) = M0

−(n+2)(q−1)

µ(t)Ψ(t) ≥ M0

t

− C(q − 1)|Γ1 |

1 + τ n/2

q

, one obtains

µ(τ ) dτ.

(4.3.15)

0

It follows from (4.3.14) that
t

1 + τ n/2

q

t

µ(τ ) dτ ≤ C

0

1 + τ n/2

q

n

exp C τ 1+ 2 dτ

0

≤ C 1 + tn/2

q

n

t exp C t1+ 2
n

≤ C exp (C + 1) t1+ 2 .

Plugging in (4.3.15), we obtain

−(n+2)(q−1)

µ(t)Ψ(t) ≥ M0

n

− C(q − 1)|Γ1 | exp C t1+ 2 .

Taking t → T ∗ , one obtains

n

−(n+2)(q−1)

C(q − 1)|Γ1 | exp C (T ∗ )1+ 2 ≥ M0
n

−(n+2)(q−1)

exp C (T ∗ )1+ 2 ≥ C −1 (q − 1)−1 |Γ1 |−1 M0
n+2

C (T ∗ ) 2 ≥ ln |Γ1 |−1 − (n + 2)(q − 1) ln M0 − ln(q − 1) − ln C.

101

Hence,

T∗

4.3.2

≥C

2
− n+2

ln |Γ1

|−1

− (n + 2)(q − 1) ln M0 − ln(q − 1) − ln C

2
n+2

.

Better estimate by a new method

The aim of this subsection is the same as the last subsection, which is to find a lower bound
of T ∗ . But this subsection will provide a different method which leads to a better lower
bound as in (4.1.1).
Comparing (4.1.1) with (4.3.1), for convenience of statement, we use T1 and T2 to represent the lower bounds in (4.1.1) and (4.3.1) respectively. The advantage of (4.1.1) is in the
following aspects.
• T1 is always positive, but T2 will be negative unless |Γ1 | is sufficiently small, M0 is
sufficiently small or q is sufficiently close to 1.
• As q

1 , while the order of T is only ln 1
1, the order of T1 is q−1
2
q−1

2
n+2 .

• As |Γ1 |

0, the order of T1 is ln |Γ1 | , while the order of T2 is only ln |Γ1 |
1
1

• As M0

0, the order of T1 is ln M1 , while the order of T2 is only ln M1
0
0

2
n+2 .
2
n+2 .

The problem of the method in Subsection 4.3.1 is that the estimate for the integral terms
lose a lot when t is large. For example, in (4.3.6), the term

1
∂Ω

τ n/2

|x−y|2

e− 8τ

102

dS(y)

is bounded by C τ −1/2 . When τ is small, this estimate is okay. But when τ is large, the
2

|x−y|
1 e− 8τ dS(y) obviously decays like τ −n/2 , so the bound τ −1/2 is too rough.
term ∂Ω n/2
τ

As a consequence, the main change in this subsection is that we divide the range of M (t)
into small pieces and analyse each piece separately. Once we found the lower bound of the
time spent in each piece, adding them together yields a lower bound for T ∗ .
In the rest of this thesis, we define

n−1

2
e−|x−y| /(4τ ) dS(y).

sup sup τ − 2

B1

τ >0 x∈∂Ω

(4.3.16)

∂Ω

It is shown in Lemma 4.3.1 that B1 is a finite positive constant depending only on Ω and n.
1 , let
In addition, for convenience of notation, for any α ∈ 0, n−1

nα

1 − (n − 1)α
.
2

(4.3.17)

It is readily seen that 0 < nα ≤ 21 .
Lemma 4.3.3. Let Ω and Γ1 be the same as in (1.1.1). Then there exists C = C(n, Ω) such
1 , x ∈ ∂Ω and t > 0,
that for any α ∈ 0, n−1
t

Φ(x − y, t − τ ) dS(y) dτ ≤
0

Γ1

C
|Γ1 |α tnα .
1 − (n − 1)α

In particular, if α = 0, then
t

Φ(x − y, t − τ ) dS(y) dτ ≤ C
0

Γ1

103

√

t.

(4.3.18)

1 . We denote
Proof. Let x ∈ ∂Ω, t > 0 and α ∈ 0, n−1
t

Φ(x − y, t − τ ) dS(y) dτ.

LHS =
0

Γ1

By a change of variable in τ ,
t

Φ(x − y, τ ) dS(y) dτ

LHS =
0

=

Γ1

1
(4π)n/2 0

t

2
e−|x−y| /(4τ ) dS(y) dτ.

τ −n/2

(4.3.19)

Γ1

For any m ≥ 1, applying Holder’s inequality,

2
e−m |x−y| /(4τ )

2
e−|x−y| /(4τ ) dS(y) ≤

1/m

Γ1

Γ1

|Γ1 |(m−1)/m .

(4.3.20)

Recalling the definition of B1 in (4.3.16),

2
e−m |x−y| /(4τ ) dτ =
Γ1

2
e− |x−y| /[4(τ /m)] dτ
Γ1

≤

τ (n−1)/2
B1
m

≤ τ (n−1)/2 B1 .

Combining this inequality with (4.3.20),

Γ1

2
1/m
e−|x−y| /(4τ ) dS(y) ≤ B1 τ (n−1)/(2m) |Γ1 |(m−1)/m

≤ (B1 + 1) τ (n−1)/(2m) |Γ1 |(m−1)/m .

104

(4.3.21)

Plugging (4.3.21) into (4.3.19),

LHS ≤

B1 + 1
(4π)n/2

t

|Γ1 |(m−1)/m

n

n−1

τ − 2 + 2m dτ.

(4.3.22)

0

Let
m=

1
.
1−α

Then m ≥ 1 and (m − 1)/m = α. Therefore (4.3.22) becomes

LHS ≤
=

B1 + 1
(4π)n/2

|Γ1 |α

B1 + 1
(4π)n/2 nα

t

τ nα −1 dτ

0

|Γ1 |α tnα ,

1 .
where the last equality is due to the assumption that α < n−1

Proof of Theorem 4.1.1. In this proof, C denote the constants which only depend on n
and Ω, the values of C may be different in different places. But C ∗ will represent a fixed
value which also depends only on n and Ω. Recalling that M (t) is defined as in (1.4.3). For
any k ≥ 0, define
Mk = 3k M0

(4.3.23)

and Tk to be the first time that the function M (t) reaches Mk . Obviously T0 = 0. For any
k ≥ 1, denote
tk

Tk − Tk−1

(4.3.24)

to be the time spent in the kth step. For any k ≥ 1, we are trying to find a lower bound tk∗
for tk , then summing up all tk∗ gives a lower bound for T ∗ .

105

By the maximum principle and the Hopf lemma, there exists xk ∈ Γ1 such that

u(xk , Tk ) = Mk .

(4.3.25)

Applying the time-shifted representation formula (4.2.4) with T = Tk−1 and (x, t) = (xk , tk ),
then

u(xk , Tk ) = 2

Ω

Φ(xk − y, tk ) u(y, Tk−1 ) dy
∂Φ(xk − y, tk − τ )
u(y, Tk−1 + τ ) dS(y) dτ
∂n(y)
∂Ω

tk

−2
0

tk

+2
0

Γ1

Φ(xk − y, tk − τ ) uq (y, Tk−1 + τ ) dS(y) dτ.

(4.3.26)

Combining (4.3.25) and (4.3.26),

Mk ≤ 2Mk−1
q

+ 2Mk

Φ(xk − y, tk ) dy + 2Mk

Ω
tk
0

Γ1

tk
0

∂Φ(xk − y, tk − τ )
dS(y) dτ
∂n(y)
∂Ω

Φ(xk − y, tk − τ ) dS(y) dτ

Ik + IIk + IIIk .

(4.3.27)

Since Φ is the fundamental solution of the heat equation, it is evident that

Ik ≤ 2Mk−1 .

(4.3.28)

Secondly it is easy to check that
IIk ≤ C

106

tk Mk .

(4.3.29)

Now define
r=

1
,
2(n − 1)

then it follows from (4.3.18) that

1/4

IIIk ≤ C |Γ1 |r tk

q

Mk .

(4.3.30)

Combining (4.3.27), (4.3.28), (4.3.29) and (4.3.30), there exists a constant C = C(n, Ω) such
that
1/4

tk Mk + C |Γ1 |r tk

Mk ≤ 2Mk−1 + C

q

Mk .

Recalling that Mk = 3k M0 and Mk−1 = 3k−1 M0 , so

1/4 qk q
3 M0 .

3k M0 ≤ 2 · 3k−1 M0 + C

tk 3k M0 + C |Γ1 |r tk

Subtracting 2 · 3k−1 M0 and then dividing by 3k−1 M0 , we obtain the existence of a constant
C = C(n, Ω) such that

1≤C

1/4 (q−1)k q−1
3
M0 .

tk + C |Γ1 |r tk

(4.3.31)

Rearranging (4.3.31),

1/4 2

tk

q−1 (q−1)k 1/4
3
tk

+ |Γ1 |r M0

−

1
≥ 0.
C
1/4

Regarding the above inequality to be a quadratic inequality in tk , then it is readily seen
1/4

that tk

has to be greater than the positive root of the corresponding quadratic equation,

107

that is
1/4

tk

≥

1
2

q−1 (q−1)k
3

− |Γ1 |r M0

2(q−1) 2(q−1)k
3

|Γ1 |2r M0

+

+

4
C

.

After some algebraic simplification, we obtain

1/4

tk

1

≥

2(q−1) 2(q−1)k
3
|Γ1 |2r M0

C

.
+

4
C

Hence, there exists C ∗ = C ∗ (n, Ω) such that

tk ≥

C∗
4(q−1) 4(q−1)k
3

|Γ1 |4r M0

.
+1

After obtaining the lower bound for each tk , we can add all of them together to get a
lower bound for T ∗ . Namely,
∞

T∗

≥

C∗

1

4r 4(q−1) 34(q−1)k
k=1 |Γ1 | M0

.
+1

Therefore,
∞

T ∗ ≥ C∗
1

=

1
4(q−1) 4(q−1)x
|Γ1 |4r M0
3

dx
+1

C∗
1
ln 1 +
.
4(q−1) 4(q−1)
4(q − 1) ln(3)
|Γ1 |4r M0
3

1 , then (4.1.1) follows.
Recalling that r = 2(n−1)

108

4.4

Lower bound on life span: case of convex C 2 domain

4.4.1

Main idea

First of all, let us recall the method in Subsection 4.3.1. Define M as in (4.3.3). Then for any
t > 0, there exists x0 ∈ ∂Ω such that u(x0 , t) = M (t), so it follows from the representation
formula (4.2.3) that
t

Φ(x0 − y, t) dy + 2

M (t) ≤ 2M0

Ω
t

0

0

∂Φ(x0 − y, t − τ )
dS(y) dτ
∂n(y)
∂Ω

Φ(x0 − y, t − τ ) dS(y) dτ

M q (τ )

+2

M (τ )

Γ1

I + II + III.

(4.4.1)

I, II and III are called the constant functional, linear functional and nonlinear functional
in M (t) respectively. After estimating
∂Φ(x0 − y, t − τ )
dS(y) and
∂n(y)
∂Ω

Φ(x0 − y, t − τ ) dS(y),
Γ1

the lower bound in Theorem 4.3.2 is achieved by applying a Gronwall-type technique to
(4.4.1). However, this lower bound is only logarithm of |Γ1 |−1 as |Γ1 |

0. The obstruction

that prevents this method obtaining a polynomial order of |Γ1 |−1 for the lower bound is
explained through the following remark.

109

Remark 4.4.1. Consider the following two simple integral inequalities. First,


 φ1 (t) ≤ A + t φ1 (τ ) dτ + |Γ1 | t φq (τ ) dτ,
0
0 1

t > 0,
(4.4.2)


 φ (0) = A > 0.
1
It is easy to see by the Gronwall’s inequality that the blow-up time T1∗ satisfies

T1∗ ≥

1
1
,
ln 1 + q−1
q−1
A
|Γ1 |

which is of order ln(|Γ1 |−1 ) as |Γ1 | → 0. Second,


 φ2 (t) ≤ A + |Γ1 | t φq (τ ) dτ,
0 2

t > 0,
(4.4.3)


 φ (0) = A > 0.
2
It is easy to see by Gronwall’s inequality that the blow-up time T2∗ satisfies

T2∗ ≥

1
,
(q − 1)Aq−1 |Γ1 |

which is of order |Γ1 |−1 . From these two differential equations, the obstruction that prevents
the lower bound being a polynomial order of |Γ1 |−1 is the linear term 0t φ1 (τ ) dτ in (4.4.2).
Corresponding to (4.4.1), it is the linear term II:
t

M (τ )
0

∂Φ(x0 − y, t − τ )
dS(y) dτ.
∂n(y)
∂Ω

If the linear term II can be eliminated from (4.4.1), then the lower bound is expected to
be a polynomial order |Γ1 |−α for some α > 0. Taking advantage of the convexity of Ω, the

110

identity (4.4.8) in Lemma 4.4.2 can be used to absorb the linear term II into the constant
term I in (4.4.1). Let’s see how it works.
First, if t ∈ (0, T ∗ ) satisfies

M (t) > M0

and

max u(x, t) = M (t),

x∈∂Ω

(4.4.4)

then there exists a point x0 ∈ ∂Ω such that u(x0 , t) = max u(x, t) = M (t). Thus, it follows
x∈∂Ω

from (4.3.3) and (4.4.1) that

M (t) ≤ 2M0

t

Φ(x0 − y, t) dy + 2M (t)
Ω

0
t

+ 2M q (t)
0

∂Φ(x0 − y, t − τ )
dS(y) dτ
∂n(y)
∂Ω

Φ(x0 − y, t − τ ) dS(y) dτ.

(4.4.5)

Γ1

Invoking (4.4.8),
t
0

∂Φ(x0 − y, t − τ )
1
dS(y) dτ = −
Φ(x0 − y, t) dy.
∂n(y)
2
∂Ω
Ω

Plugging this identity into (4.4.5) and simplifying, one has

M (t)
Ω

Φ(x0 − y, t) dy ≤ M0

t

Φ(x0 − y, t) dy + M q (t)
Ω

0

Φ(x0 − y, t − τ ) dS(y) dτ.
Γ1

(4.4.6)
Now this estimate does not contain the linear term, which should enable us to get a lower
bound of polynomial order |Γ1 |−β for some β > 0.
To continue from (4.4.6), the Gronwall-type technique will not work, since (4.4.6) is
proved to be true only for t satisfying (4.4.4). Then what kind of time t satisfies (4.4.4)?
By the maximum principle, if at some t > 0, M (t) > M (t1 ) for any 0 ≤ t1 < t, then such
111

t satisfies (4.4.6). As an instance, for any λ1 > 1, if we write M1 = λ1 M0 and denote T1
to be the first time that M (t) reaches M1 , then T1 satisfies (4.4.4). Another disadvantage
of (4.4.6) is that although it gets rid of the linear functional of M (t), there is an extra
term Ω Φ(x0 − y, t) dy on the left hand side, which decays like t−n/2 when t becomes large.
Hence, to avoid the effect of this decay, λ1 should be kept small. Taking these restrictions
into consideration, we need to come up with some delicate strategies. The rough idea is as
follows. We will firstly choose a suitably small λ1 such that there is still a lower bound t∗
for T1 , where T1 is the first time for M (t) to reach λ1 M0 . Then we regard u(·, T1 ) as the
“initial data” and repeat the first step. Finally if such process can proceed for at least L0
steps, then L0 t∗ is a lower bound for T ∗ , since the time in each step has a lower bound t∗
(the choice of t∗ will be the same in each step).

4.4.2

Auxiliary lemmas

The second conclusion of the next lemma is the only place that the convexity is used in this
section.
Lemma 4.4.2. Let Φ be the heat kernel as in (1.4.4). Then
t

Φ(x − y, t) dy −
Ω

0

∂Φ(x − y, t − τ )
1
dS(y) dτ = ,
∂n(y)
2
∂Ω

∀ x ∈ ∂Ω, t > 0.

(4.4.7)

In addition, if Ω is convex, then
t

Φ(x − y, t) dy +
Ω

0

∂Φ(x − y, t − τ )
1
dS(y) dτ = ,
∂n(y)
2
∂Ω

112

∀ x ∈ ∂Ω, t > 0.

(4.4.8)

Proof. The problem



ut (x, t) = ∆u(x, t) in





∂u(x,t)

Ω × (0, ∞),

on ∂Ω × (0, ∞),

=0
∂n(x)






 u(x, 0) = 1

in

(4.4.9)

Ω,

obviously has the unique solution u ≡ 1 on Ω × [0, ∞). As a result, plugging u ≡ 1 into the
representation formula (4.2.3) (taking Γ1 = ∅), (4.4.7) follows.
∂Φ(x−y,t−τ )
∂n(y)

Now if Ω is convex, then

≤ 0 for any x, y ∈ ∂Ω. Thus, (4.4.7) implies

(4.4.8).
Lemma 4.4.3. Define F : ∂Ω × [0, 1] → R to be

F (x, t) =






Ω Φ(x − y, t) dy



 1/2

for x ∈ ∂Ω, t ∈ (0, 1],
for x ∈ ∂Ω, t = 0.

Then F is continuous on ∂Ω × [0, 1]. As a result,

b1

min F

(4.4.10)

∂Ω×[0,1]

is a positive constant depending only on Ω and the dimension n.
Proof. Since ∂Ω has been assumed to be C 2 , the proof can be carried out by standard
analysis. We can also prove it by applying (4.4.7) and noticing the uniform decay for x ∈ ∂Ω
of the following integral
t

lim

t→0 0

∂Φ(x − y, t − τ )
dS(y) dτ = 0.
∂n(y)
∂Ω

113

The details are omitted here.
Next, M0 and M (t) are still defined as in (1.4.2) and (1.4.3). In addition, we define

Eq = (q − 1)q−1 /q q ,

∀ q > 1.

(4.4.11)

By elementary calculus,

1
1
1
< Eq < min
,
3q
q (q − 1) e

< 1.

(4.4.12)

Lemma 4.4.4. For any q > 1 and m > 0, write Eq as in (4.4.11) and define g : (m, ∞) → R
by
g(λ) =

λ−m
,
λq

∀ λ > m.

(4.4.13)

Then the following two claims hold.
q
(1) For any y ∈ 0, m1−q Eq , there exists unique λ ∈ m, q−1
m such that g(λ) = y.

(2) For any y > m1−q Eq , there does not exist λ > m such that g(λ) = y.
q
Proof. Since g is strictly increasing on the interval m, q−1
m and strictly decreasing on the
q
q
interval q−1
m, ∞ , it reaches the maximum at λ = q−1
m. Noticing that

g

q
m = m1−q Eq ,
q−1

then the claims (1) and (2) follow directly.

114

4.4.3

Proof of Theorem 4.1.4

Proof. Let M (t) be defined as in (1.4.3). In the following, the first step is that for any
t∗ ∈ (0, 1], we will find a finite strictly increasing sequence {Mk }0≤k≤L such that if Tk
denotes the first time that M (t) = Mk , then Tk − Tk−1 ≥ t∗ for 1 ≤ k ≤ L. The second
step is to derive a lower bound for Lt∗ as an explicit formula in t∗ and then maximize that
lower bound for t∗ ∈ (0, 1].
Step 1. Let t∗ ∈ (0, 1] which will be determined later in Step 2. Define M0 as that in
(1.4.2) and T0 = 0. Then for k ≥ 1, suppose Mk−1 has been constructed, we are trying to
find Mk such that Tk − Tk−1 ≥ t∗ .
Denote tk = Tk − Tk−1 . We will first check what happens if tk ≤ 1. By the maximum
principle, there exists xk ∈ ∂Ω such that u(xk , Tk ) = Mk , so Tk satisfies (4.4.4). Applying
the time-shifted representation formula (4.2.4) with T = Tk−1 and (x, t) = (xk , tk ),

u(xk , Tk ) = 2

Ω

Φ(xk − y, tk ) u(y, Tk−1 ) dy
tk

−2
0

∂Φ(xk − y, tk − τ )
u(y, Tk−1 + τ ) dS(y) dτ
∂n(y)
∂Ω

tk

+2
0

Γ1

Φ(xk − y, tk − τ ) uq (y, Tk−1 + τ ) dS(y) dτ.

(4.4.14)

As a result,

Mk ≤ 2Mk−1
+ 2Mk

Φ(xk − y, tk ) dy

Ω
tk
0

tk

q

+ 2Mk

∂Φ(xk − y, tk − τ )
dS(y) dτ
∂n(y)
∂Ω

0

Γ1

Φ(xk − y, tk − τ ) dS(y) dτ.

115

(4.4.15)

Since Ω is convex, it follows from (4.4.8) that
tk
0

∂Φ(xk − y, tk − τ )
1
dS(y) dτ = −
Φ(xk − y, tk ) dy.
∂n(y)
2
∂Ω
Ω

Plugging this identity into (4.4.15) and simplifying,

(Mk − Mk−1 )

tk

q

Ω

Φ(xk − y, tk ) dy ≤ Mk

0

Γ1

Φ(xk − y, tk − τ ) dS(y) dτ.

(4.4.16)

Due to the assumption that tk ≤ 1, it follows from Lemma 4.4.3 that

Ω

Φ(xk − y, tk ) dy ≥ b1 .

(4.4.17)

In addition, Lemma 4.3.3 implies the existence of a constant C = C(n, Ω) such that
tk
0

Γ1

Φ(xk − y, tk − τ ) dS(y) dτ ≤

α
C |Γ1 |α tn
k ,
nα

(4.4.18)

where nα is defined in (4.3.17). Plugging (4.4.17) and (4.4.18) into (4.4.16),
α
C |Γ1 |α tn
Mk − Mk−1
k .
≤
q
b1 n α
Mk

(4.4.19)

In summary, this paragraph claims that if tk ≤ 1, then Mk will satisfy (4.4.19).
Based on the above observation, denote

δ1 =

α
C |Γ1 |α tn
∗
,
b1 n α

(4.4.20)

where the constant C is the same as that in (4.4.19). Then we define Mk to be the solution
116

(if it exists) to
Mk − Mk−1
= δ1 .
q
Mk

(4.4.21)

With such a choice for Mk , it is evident that

tk ≥ min{1, t∗ } = t∗ .

On the other hand, by applying Lemma 4.4.4, (4.4.21) has a solution Mk > Mk−1 if and
1−q

only if δ1 ≤ Mk−1 Eq . In addition, as long as such a solution exists, Mk can be chosen to
satisfy
Mk−1 < Mk ≤

q
M
.
q − 1 k−1

Thus, the strategy of constructing {Mk } is summarized as following. First, M0 is defined
to be the same as in (1.4.2). Next, for k ≥ 1, suppose Mk−1 has been constructed, then
based on Lemma 4.4.4, whether defining Mk depends on how large Mk−1 is.
q−1

q
If Mk−1 δ1 ≤ Eq , then we define Mk ∈ Mk−1 , q−1
Mk−1 to be the solution to

(4.4.21).
q−1

If Mk−1 δ1 > Eq , then there does not exist Mk > Mk−1 which solves (4.4.21). So we
do not define Mk and stop the construction.
Based on this construction, if {Mk }1≤k≤L0 have been defined, then for any 1 ≤ k ≤ L0 ,
Tk − Tk−1 ≥ t∗ . Therefore, Tk ≥ kt∗ for any 1 ≤ k ≤ L0 . Applying Theorem 3.1.1,
L0 ≤ T ∗ /t∗ < ∞, which means the cardinality of {Mk } has be to finite. So we can assume
the constructed sequence is {Mk }0≤k≤L for some finite L.

117

Step 2. By Lemma 4.4.5,

L>

1
1
− 3q ,
q−1
10(q − 1) M
δ
1
0

so
T ∗ ≥ Lt∗ >

1
1
− 3q t∗ .
10(q − 1) M q−1 δ1
0

(4.4.22)

Plugging (4.4.20) into (4.4.22),

T∗ ≥
=

1
b1 nα
α − 3q t
t1−n
∗
∗
q−1
α
10(q − 1) M
C
|Γ
|
1
0
3q
C1 nα
α −t ,
t1−n
∗
∗
q−1
α
10(q − 1) qM
|Γ
|
1
0

(4.4.23)

where C1 = b1 /(3C) is a constant only depending on n and Ω.
In order to maximize the right hand side of (4.4.23), let

A=

C 1 nα
q−1

qM0

|Γ1 |α

,

β = 1 − nα ∈

1
,1
2

and define
t∗

(min{1, βA})1/(1−β) .

(4.4.24)

Then it follows from Lemma 4.4.6 that t∗ maximizes the right hand side of (4.4.23) on (0, 1]
and

T∗ ≥

3q
β/(1−β)
(1 − β)A min{1, βA}
.
10(q − 1)

118

Noticing that β ≥ 1/2, so

T∗ ≥
≥

≥

β/(1−β)

A
3q
(1 − β)A min 1,
10(q − 1)
2
3C1 n2α
q−1

10(q − 1)M0
C2

q−1

(q − 1)M0

min 1,

|Γ1 |α

|Γ1 |α

1
nα −1

C 1 nα
q−1

2qM0

|Γ1 |α
1
nα −1

1

min 1,

q−1

qM0

|Γ1 |α

,

(4.4.25)

where
C2 =

3C1 n2α
10

1
nα −1

C nα
min 1, 1
2

(4.4.26)

is a constant depending on n, Ω and α.
In particular, if we choose α = 0 in (4.4.25) and (4.4.26), then it follows from nα =
1−(n−1)α
2

= 12 that
T∗ ≥

C3

min 1,
q−1

(q − 1)M0

1
q−1

,

qM0

where C3 is a positive constant only depending on n and Ω.
Lemma 4.4.5. Given q > 1, M0 > 0 and δ1 > 0, denote Eq as (4.4.11) and construct
a finite sequence {Mk }0≤k≤L inductively as follows. For k ≥ 1, suppose Mk−1 has been
constructed, then based on Lemma 4.4.4, whether defining Mk depends on how large Mk−1
is.
q−1

q
If Mk−1 δ1 ≤ Eq , then we define Mk ∈ Mk−1 , q−1
Mk−1 to be the solution to

Mk − Mk−1
= δ1 .
q
Mk
q−1

(4.4.27)

If Mk−1 δ1 > Eq , then there does not exist Mk > Mk−1 which solves (4.4.27). So we
119

do not define Mk and stop the construction.
Denote the last term of this construction to be ML , then

L>

1
1
− 3q .
10(q − 1) M q−1 δ1
0

(4.4.28)

Proof. First, we want to mention that the construction indeed stops in finite steps. In fact,
it follows from (4.4.27) that the sequence {Mk } is strictly increasing and

q−1

q

Mk = Mk−1 + Mk δ1 ≥ 1 + M0

δ1 Mk−1 .

As a result,
q−1 k

Mk ≥ 1 + δ1 M0
q−1

Thus, when k is sufficiently large, Mk

M0 .

will exceed Eq /δ1 , which forces the construction to

stop.
Next, suppose the construction stops at ML , that is to say, the constructed sequence is
{Mk }0≤k≤L , then the lower bound (4.4.28) for L will be justified based on two situations.
q−1

Case 1. M0

δ1 > Eq . In this case, it follows from (4.4.12) that
1
q−1
M0 δ1

<

1
< 3q,
Eq

so the right hand side of (4.4.28) is negative. Thus, (4.4.28) holds since L ≥ 0.
q−1

Case 2. M0

δ1 ≤ Eq . In this case, it is evident that L ≥ 1. Therefore, since the last

120

term of the sequence is ML , then

q−1

ML−1 δ1 ≤ Eq

q−1

δ1 > Eq .

q−1

δ1 .

and ML

According to the recursive relation (4.4.27),

Mk−1 = Mk 1 − Mk

Raising both sides of the above equality to the power q − 1 and multiplying by δ1 ,

q−1

q−1

Mk−1 δ1 = Mk

q−1

Let xk = Mk

q−1

δ1 1 − Mk

δ1

q−1

.

δ1 . Then

xk−1 = xk (1 − xk )q−1 ,

∀ 1 ≤ k ≤ L.

(4.4.29)

Moreover,
q−1

x0 = M0

δ1 ,

xL−1 ≤ Eq

and xL > Eq .

q
ML−1 , so
Noticing that ML ≤ q−1

xL =

q−1
ML q−1
q
1
Eq = .
xL−1 ≤
ML−1
q−1
q

Since the right hand side of (4.4.29) is a nonlinear function in xk , it is better to consider
the “reversed” relation of (4.4.29). Thus, we define a new sequence {yk }0≤k≤L in the

121

following way: y0

min{1/2, Eq } and

yk

yk−1 (1 − yk−1 )q−1 ,

∀ 1 ≤ k ≤ L.

(4.4.30)

In addition, define h : (0, 1) → R by

h(t) = t (1 − t)q−1 .

It is easy to see that h is strictly increasing on (0, 1/q] and strictly decreasing on [1/q, 1).
As a result, it follows from 0 < y0 ≤ Eq < xL ≤ 1/q that

y1 = h(y0 ) < h(xL ) = xL−1 .

Keep doing this, we get yk < xL−k for any 0 ≤ k ≤ L. In particular, when k = L,
q−1

yL < x0 = M0

δ1 .

Since {yk } is a decreasing positive sequence and y0 ≤ 1/2, then yk ≤ 1/2 for any
0 ≤ k ≤ L. As a result, it follows from (4.4.30) and the mean value theorem that for any
1 ≤ k ≤ L,
yk ≥ yk−1 1 − 2(q − 1)yk−1 .
Recalling (4.4.12) again,
yk−1 ≤ y0 ≤ Eq <

1
,
(q − 1)e

so
1 − 2(q − 1)yk−1 > 1 −

122

2
1
> .
e
5

(4.4.31)

Hence, taking the reciprocal in (4.4.31) yields

1
1
≤
yk
yk−1 1 − 2(q − 1)yk−1
=
<

1
yk−1
1
yk−1

+

2(q − 1)
1 − 2(q − 1)yk−1

+ 10(q − 1).

(4.4.32)

Summing up (4.4.32) for k from 1 to L, then

1
1
<
+ 10(q − 1)L.
yL
y0
q−1

Since yL < M0

(4.4.33)

δ1 and
y0 = min

1
, Eq
2

>

1
,
3q

it follows from (4.4.33) that

1
q−1
M0 δ1

< 3q + 10(q − 1)L.

Thus,
L>

1
1
− 3q .
q−1
10(q − 1) M
δ
1
0

Lemma 4.4.6. Given two constants A > 0 and β ∈ (0, 1), define f : (0, 1] → R by f (t) =
A tβ − t. Let
t0 = (min{1, βA})1/(1−β) .

123

Then
f (t0 ) = sup f (t) ≥ (1 − β)A (min{1, βA})β/(1−β) .

(4.4.34)

0<t≤1

Proof. For any t ∈ (0, 1], f (t) = βA tβ−1 − 1. So when 0 < t ≤ t0 , f is increasing.
• If βA ≥ 1, then
sup f (t) = f (1) = A − 1 ≥ (1 − β) A.
0<t≤1

• If 0 < βA < 1, then

sup f (t) = f (βA)1/(1−β)
0<t≤1

= A (βA)β/(1−β) − (βA)1/(1−β)
= (1 − β)A (βA)β/(1−β) .

Combining these two cases, the lemma is justified.

4.5

Lower bound on life span: case of C 2 domain with
local convexity near Γ1

The global convexity of Ω is not practical in real applications. So in this section, we try to
extend the result to locally convex case. Namely, we assume the local convexity near Γ1 as
in Definition 4.1.2.

124

4.5.1

Estimates for boundary integrals

Lemma 4.5.1. Let Ω and Γ1 be the same as in (1.1.1). Then for any d > 0, there exists
C = C(n, Ω, d) such that for any x ∈ Γ1 and t ∈ (0, 1],
t
0

∂Ω\[Γ1 ]d

∂Φ(x − y, t − τ )
d2
dS(y) dτ ≤ C exp −
.
∂n(y)
8t

(4.5.1)

Proof. In this proof, C denotes a constant which depends only on n, Ω and d. By a change
of variable in τ and the definition of Φ,
t
0

∂Ω\[Γ1 ]d

∂Φ(x − y, t − τ )
dS(y) dτ
∂n(y)

t

∂Φ(x − y, t − τ )
dS(y) dτ
∂n(y)
0 ∂Ω\[Γ1 ]d
−
t
|(x − y) · →
n (y)|
|x − y|2
≤ C
dS(y) dτ.
exp
−
n +1
4τ
0 ∂Ω\[Γ1 ]d
2
τ
=

(4.5.2)

−
Since ∂Ω is assumed to be C 2 , then |(x − y) · →
n (y)| ≤ C|x − y|2 . As a result,
−
|(x − y) · →
n (y)|
n
τ 2 +1

|x − y|2
exp −
4τ

≤

C|x − y|−n

n
|x − y|2 1+ 2
|x − y|2
exp −
τ
4τ

≤ C|x − y|−n exp −
≤ Cd−n exp −

|x − y|2
8τ

d2
.
8τ

where the last inequality is due to the fact that x ∈ Γ1 and y ∈ ∂Ω \ [Γ1 ]d .

125

(4.5.3)

Plugging (4.5.3) into (4.5.2),
t
0

≤

∂Ω\[Γ1 ]d
t
Cd−n
0

≤ Cd−n |∂Ω|

∂Φ(x − y, t − τ )
dS(y) dτ
∂n(y)
exp −

∂Ω\[Γ1 ]d
t

exp −
0

≤ Cd−n |∂Ω| t exp −

d2
8t

d2
dS(y) dτ
8τ

d2
dτ
8τ
≤ C exp −

d2
,
8t

where the last inequality is due to the assumption that t ≤ 1.
By exploiting Lemma 4.5.1, the following is a variant of the identity (4.4.8). So (4.5.4)
will play the same role in the proof of Theorem 4.1.3 as (4.4.8) did in the proof of Theorem
4.1.4.
Corollary 4.5.2. Let Ω and Γ1 be the same as in (1.1.1). Assume there exists d > 0 such
that Conv [Γ1 ]d ⊆ Ω. Then there exists C = C(n, Ω, d) such that for any x ∈ Γ1 and
t ∈ (0, 1],
t

Φ(x − y, t) dy +
0

Ω

1
d2
∂Φ(x − y, t − τ )
dS(y) dτ ≤ + C exp −
.
∂n(y)
2
8t
∂Ω

(4.5.4)

Proof. Splitting the second term on the left hand side of (4.5.4) into two parts:
t

=

∂Φ(x − y, t − τ )
dS(y) dτ
∂n(y)
0 ∂Ω
t
t
∂Φ(x − y, t − τ )
∂Φ(x − y, t − τ )
dS(y) dτ +
dS(y) dτ
∂n(y)
∂n(y)
0 [Γ1 ]d
0 ∂Ω\[Γ1 ]d

126

It follows from x ∈ Γ1 and Conv [Γ1 ]d ⊆ Ω that
∂Φ(x − y, t − τ )
≤ 0,
∂n(y)

∀ y ∈ [Γ1 ]d .

As a result,
t

∂Φ(x − y, t − τ )
dS(y)dτ
∂n(y)
0 ∂Ω
t
t
∂Φ(x − y, t − τ )
∂Φ(x − y, t − τ )
dS(y)dτ +
dS(y)dτ
=−
∂n(y)
∂n(y)
0 ∂Ω\[Γ1 ]d
0 [Γ1 ]d
t
t
∂Φ(x − y, t − τ )
∂Φ(x − y, t − τ )
dS(y)dτ + 2
dS(y)dτ. (4.5.5)
≤−
∂n(y)
∂n(y)
0 ∂Ω\[Γ1 ]
0 ∂Ω
d

Combining (4.5.5) with Lemma 4.5.1, there exists a constant C = C(n, Ω, d) such that
t

∂Φ(x − y, t − τ )
dS(y) dτ
∂n(y)
0 ∂Ω
t
∂Φ(x − y, t − τ )
d2
≤ −
dS(y) dτ + C exp −
.
∂n(y)
8t
0 ∂Ω

(4.5.6)

Hence it follows from (4.4.7) and (4.5.6) that
t

∂Φ(x − y, t − τ )
dS(y) dτ
∂n(y)
Ω
0 ∂Ω
t
∂Φ(x − y, t − τ )
d2
≤
Φ(x − y, t) dy −
dS(y) dτ + C exp −
∂n(y)
8t
Ω
0 ∂Ω
1
d2
=
+ C exp −
.
2
8t
Φ(x − y, t) dy +

Next, we introduce a simple fact which can be regarded as a rearrangement result.

127

Lemma 4.5.3. Let n ≥ 1 and f : (0, ∞) → [0, ∞) be a decreasing function. Then for any
bounded, open subset U of Rn and for any x ∈ Rn ,

f (|x − y|) dy ≤
U

f (|z|) dz

(4.5.7)

BR (0)

where R satisfies |BR (0)| = |U | (namely the volume of BR (0) equals the volume of U ).
Proof. Define
U1 = U − {x}.
Then by a change of variable z = y − x,

f (|x − y|) dy =

f (|z|) dz
U1

U

f (|z|) dz

f (|z|) dz +

=

U1 \BR (0)

U1 ∩BR (0)

I1 + I2 ,

(4.5.8)

Since f is decreasing, then

I2 ≤ f (R)|U1 \BR (0)|.

Since R is chosen such that |BR (0)| = |U | = |U1 |, then we have |BR (0)\U1 | = |U1 \BR (0)|.
As a result,
I2 ≤ f (R)|BR (0)\U1 | ≤

f (|z|) dz,

(4.5.9)

BR (0)\U1

where the last inequality is again due to the decay of f . Combining (4.5.8) and (4.5.9), we
finish the proof.
128

Definition 4.5.4. Let Ω be a bounded, open subset of Rn with C 1 boundary. Let Γ be a
relative open subset of ∂Ω. We say Γ is given by a graph if (upon relabelling and reorienting
the coordinates axes), there exists a bounded, open subset U ⊆ Rn−1 and a C 1 function
φ : Rn−1 → R such that
Γ = {(˜
y , φ(˜
y )) : y˜ ∈ U }.
In the following, for any x ∈ Rn , we will decompose it to be x = (˜
x, xn ), where x˜ denotes
the first n − 1 components of x.
Lemma 4.5.5. Let Ω be a bounded, open subset of Rn (n ≥ 3) with C 1 boundary. Let Γ be
a relatively open subset of ∂Ω that is given by a graph as defined in Definition 4.5.4. Then
there exists a constant C = C(n, ||∇φ||L∞ (U ) ), where φ and U are the same as in Definition
4.5.4, such that for any x ∈ Rn ,
1
dS(y) ≤ C|Γ|1/(n−1) ,
n−2
|x
−
y|
Γ
where |Γ|

Γ dS.

Proof. By Definition 4.5.4, without loss of generality, we can assume there exists a C 1
function φ : Rn−1 → R and a bounded, open subset U of Rn−1 such that

Γ = {(˜
y , φ(˜
y )) : y˜ ∈ U }.

129

(4.5.10)

Thus,
1 + |∇φ(˜
y )|2
1
dS(y)
=
d˜
y
n−2
x, xn ) − (˜
y , φ(˜
y ))|n−2
Γ |x − y|
U |(˜
1 + |∇φ(˜
y )|2
≤
d˜
y
|˜
x − y˜|n−2
U
≤C

1
d˜
y.
x − y˜|n−2
U |˜

Define
1
f (r) = n−2 ,
r

∀ r > 0.

Then it follows from Lemma 4.5.3 that

1
d˜
y=
f (|˜
x − y˜|) d˜
y
x − y˜|n−2
U |˜
U
f (|˜
z |) d˜
z

≤
BR (0)

= CR = C|U |1/(n−1) .

Again by the parametrization (4.5.10), it is readily seen that |U | ≤ |Γ|. Hence,

1
dS(y) ≤ C|U |1/(n−1) ≤ C|Γ|1/(n−1) .
n−2
Γ |x − y|

Corollary 4.5.6. Let Ω be a bounded, open subset of Rn (n ≥ 3) with C 1 boundary. Then
there exists a constant C = C(n, Ω) such that for any relative open subset Γ of ∂Ω and for
any x ∈ Rn ,
1
dS(y) ≤ C|Γ|1/(n−1) .
n−2
|x
−
y|
Γ
130

Proof. Since ∂Ω is C 1 , for any point x0 ∈ ∂Ω, the boundary part of Ω near x0 can be
straightened out (thus can be given by a graph as in Definition 4.5.4). As a result, we can
split ∂Ω into finite pieces:
K

∂Ω =

Ai ,

(4.5.11)

i=1

where each Ai (1 ≤ i ≤ K) is given by the graph of some C 1 function φi on some open set
Ui ⊆ Rn−1 . The number of total pieces K and ||∇φi ||L∞ (U ) only depend on Ω.
i
So for any 1 ≤ i ≤ K, Γ ∩ Ai is also a boundary part this is given by a graph. Therefore
by Lemma 4.5.5, there exists a constant C = C(n, Ω) such that for any 1 ≤ i ≤ K,

1
dS(y) ≤ C|Γ ∩ Ai |1/(n−1) .
n−2
|x
−
y|
Γ∩Ai
Hence,
K

1
1
dS(y) ≤
dS(y)
n−2
n−2
|x
−
y|
Γ∩A
Γ |x − y|
i
i=1
K

|Γ ∩ Ai |1/(n−1)

≤C
i=1

≤ CK|Γ|1/(n−1) = C|Γ|1/(n−1) .

Lemma 4.5.5 and Corollary 4.5.6 will be applied to show our desired Lemma 4.5.7 which
is an improvement of Lemma 4.3.3 when n ≥ 3.
Lemma 4.5.7. Let n ≥ 3. Let Ω and Γ1 be the same as in (1.1.1). Then there exists

131

C = C(n, Ω) such that for any x ∈ Rn and t ≥ 0,
t
0

Γ1

Φ(x − y, t − τ ) dS(y) dτ ≤ C|Γ1 |1/(n−1) .

(4.5.12)

Proof. In this proof, unless otherwise stated, C represents constants which only depend on
n and Ω. First, by the explicit formula (1.4.4) of Φ and a change of variable in τ , we have
t

t

Φ(x − y, t − τ ) dS(y) dτ = C
0

Γ1

2
τ −n/2 e−|x−y| /(4τ ) dτ dS(y).

Γ1 0

Then by the change of variable s = |x − y|2 /(4τ ) for τ ,
t

2
τ −n/2 e−|x−y| /(4τ ) dτ dS(y) ≤ C

Γ1 0

∞
n
1
s 2 −2 e−s ds dS(y).
n−2
Γ1 |x − y|
|x−y|2 /(4t)
(4.5.13)

n

Since n ≥ 3, s 2 −2 e−s is integrable on (0, ∞). As a result,
∞
∞ n
n −2 −s
1
1
2 e ds dS(y) ≤
s
s 2 −2 e−s ds dS(y)
n−2
n−2
2
|x
−
y|
|x
−
y|
Γ1
Γ1
|x−y| /(4t)
0

=C

1
dS(y).
n−2
Γ1 |x − y|

Now applying Corollary 4.5.6,

1
dS(y) ≤ C|Γ1 |1/(n−1) .
n−2
Γ1 |x − y|

The following Lemma 4.5.8, Corollary 4.5.9 and Lemma 4.5.10 are parallel results as
Lemma 4.5.5, Corollary 4.5.6 and Lemma 4.5.7, but they deal with dimension n = 2 rather
132

than n ≥ 3.
Lemma 4.5.8. Let Ω be a bounded, open subset of R2 with C 1 boundary. Let Γ be a relatively
open subset of ∂Ω that is given by a graph as defined in Definition 4.5.4. Then there exists a
constant C = C(Ω, ||∇φ||L∞ (U ) ), where φ and U are the same as in Definition 4.5.4, such
that for any x ∈ Ω,
ln
Γ

dΩ
1
dS(y) ≤ C|Γ| ln
+1 ,
|x − y|
|Γ|

where dΩ denotes the diameter of Ω, namely dΩ = sup{|u − v| : u, v ∈ Ω}.
Proof. By Definition 4.5.4, without loss of generality, we can assume there exists a C 1
function φ : R → R and a bounded, open set U ⊆ R such that

Γ = {(˜
y , φ(˜
y )) : y˜ ∈ U }.

(4.5.14)

In addition, we define
f (r) =



d

 ln rΩ , 0 < r ≤ dΩ ,


 0,

r > dΩ .

Since x = (˜
x, xn ) ∈ Ω, then for any (˜
y , φ(˜
y )) ∈ Γ,

|˜
x − y˜| ≤ |(˜
x, xn ) − (˜
y , φ(˜
y ))| ≤ dΩ .

133

(4.5.15)

As a result,

ln
Γ

dΩ
dΩ
dS(y) =
ln
|x − y|
|(˜
x, xn) − (˜
y , φ(˜
y ))|
U
≤C

ln
U

1 + |∇φ(˜
y )|2 d˜
y

dΩ
d˜
y
|˜
x − y˜|

f (|˜
x − y˜|) d˜
y.

=C

(4.5.16)

U

Now it follows from Lemma 4.5.3 that

f (|˜
x − y˜|) d˜
y≤
U

f (|˜
z |) d˜
z
BR (0)
R

f (r) dr,

=2

(4.5.17)

0

where |BR (0))| = |U |, namely 2R = |U |.
For any y˜1 , y˜2 ∈ U ,

|˜
y1 − y˜2 | ≤ |(˜
y1 , φ(˜
y1 )) − (˜
y2 , φ(˜
y2 ))| ≤ dΩ ,

which implies diam(U ) ≤ dΩ . Moreover, since U ⊆ R, then |U | ≤ diam(U ) ≤ dΩ . Thus,
R = |U |/2 ≤ dΩ /2. So it follows from (4.5.15) that
R

R

dΩ
dr
r
0
d
= R ln Ω + 1 .
R

f (r) dr =
0

ln

134

(4.5.18)

Again by the parametrization 4.5.14, it is readily seen that |U | ≤ |Γ|. Therefore,

R ≤ min

|Γ| dΩ
,
.
2 2

Define
g(r) = r ln

dΩ
+1 ,
r

∀ r > 0.

Then g is increasing when r ∈ (0, dΩ ]. In addition, (4.5.18) implies that
R

f (r) dr = g(R).
0

Next, we will estimate g(R) in the following two situations.
• |Γ| ≤ dΩ .

g(R) ≤ g(|Γ|) = |Γ| ln
= |Γ| ln
≤ C|Γ| ln

dΩ
+1
|Γ|
1
+ ln(dΩ ) + 1
|Γ|
1
+1
|Γ|

(4.5.19)

for some constant C only depending on Ω.
• |Γ| > dΩ .

g(R) ≤ g(dΩ ) = dΩ .

Define
h(r) = r ln

1
+1 ,
r

135

∀ r > 0.

(4.5.20)

Then
h (r) = −

1
< 0,
r(1 + r)2

∀ r > 0.

This implies h (r) > 0 for any r > 0, since lim h (r) = 0. Hence, h is an increasing
r→∞

function and

1
1
+ 1 = h(|Γ|) ≥ h(dΩ ) = dΩ ln
+1 .
|Γ|
dΩ

|Γ| ln

Thus, there exists a constant only depending on Ω such that

g(R) ≤ C|Γ| ln

1
+1 .
|Γ|

(4.5.21)

Combining (4.5.16) through (4.5.21), the conclusion follows.
Corollary 4.5.9. Let Ω be a bounded, open subset of R2 with C 1 boundary. Then there
exists a constant C = C(Ω) such that for any relative open subset Γ of ∂Ω and for any
x ∈ Ω,
ln
Γ

dΩ
1
dS(y) ≤ C|Γ| ln
+1 ,
|x − y|
|Γ|

where dΩ denotes the diameter of Ω, namely dΩ = sup{|u − v| : u, v ∈ Ω}.
Proof. Similar to the proof of Corollary 4.5.6, we first decompose ∂Ω as that in (4.5.11).
Then
K

dΩ
dΩ
ln
dS(y) ≤
ln
dS(y).
|x − y|
|x
−
y|
Γ
Γ∩A
i
i=1
Now since each Γ ∩ Ai is given by a graph, we can apply Lemma 4.5.8 to conclude there
136

exists a constant C = C(Ω) such that for each 1 ≤ i ≤ K,

ln
Γ∩Ai

1
dΩ
dS(y) ≤ C|Γ ∩ Ai | ln
+1 .
|x − y|
|Γ ∩ Ai |

Recalling the function h defined in (4.5.20) is an increasing function, so

|Γ ∩ Ai | ln

1
1
+ 1 ≤ |Γ| ln
+1 .
|Γ ∩ Ai |
|Γ|

As a result,
ln
Γ

1
dΩ
dS(y) ≤ C|Γ| ln
+1 .
|x − y|
|Γ|

Next, Lemma 4.5.8 and Corollary 4.5.9 will be applied to show our desired Lemma 4.5.10
which is an improvement of Lemma 4.3.3 when n = 2.
Lemma 4.5.10. Let n = 2. Let Ω and Γ1 be the same as in (1.1.1). Then there exists
C = C(Ω) such that for any x ∈ Ω and 0 < t ≤ 1,
t
0

Γ1

Φ(x − y, t − τ ) dS(y) dτ ≤ C|Γ1 | ln

1
+1 .
|Γ1 |

(4.5.22)

Proof. We proceed similarly as that in the proof of Lemma 4.5.7 until (4.5.13). Next, the situation is different since sn/2−2 e−s is not integrable near s = 0 when n = 2. For convenience,
we rewrite (4.5.13) when n = 2 as following:
t
Γ1 0

2
τ −1 e−|x−y| /(4τ ) dτ dS(y) ≤ C

∞
Γ1 |x−y|2 /(4t)

137

s−1 e−s ds dS(y).

(4.5.23)

Since t ≤ 1 and x ∈ Ω, |x − y|2 /(4t) ≥ |x − y|2 /4. Thus,
∞
|x−y|2 /(4t)

s−1 e−s ds ≤
=

≤

∞

s−1 e−s ds
2
|x−y| /4
d2Ω
∞
s−1 e−s ds
s−1 e−s ds +
2
2
|x−y| /4
dΩ
∞
d2Ω
1
−1
s ds + 2
e−s ds
2
2
dΩ d
|x−y| /4
Ω

= 2 ln

dΩ
+ C.
|x − y|

As a result,
∞
Γ1 |x−y|2 /(4t)

s−1 e−s ds dS(y) ≤ C|Γ1 | + 2

ln
Γ1

dΩ
dS(y).
|x − y|

Now applying Corollary 4.5.9,

ln
Γ1

dΩ
1
dS(y) ≤ C|Γ1 | ln
+1 .
|x − y|
|Γ1 |

Finally noticing that

|Γ1 | ≤

1

|Γ1 | ln

1 +1
ln |∂Ω|
1
= C|Γ1 | ln
+1 ,
|Γ1 |

the lemma is proved.

138

1
+1
|Γ1 |

(4.5.24)

4.5.2

Proof of Theorem 4.1.3

The idea of the proof in this subsection is similar to that in Subsection 4.4.1. The main
difference is that in the proof of Theorem 4.1.4, we treat t∗ as a variable in (0, 1] and the
choice of {Mk } depends on t∗ ; however, in the proof below, the lower bound t∗ will be a fixed
number in (0, 1] and the choice of {Mk } does not depend on t∗ . On the technical aspects,
• First, Lemma 4.5.4 will be used instead of Lemma 4.4.8 to overcome the lack of global
convexity.
• Secondly, Lemma 4.5.7 and Lemma 4.5.10 will be exploited in place of Lemma 4.3.3
to obtain higher order of |Γ1 |−1 for the lower bound of T ∗ as |Γ1 |

0. The price for

obtaining this higher order is that the results in this section only work for small |Γ1 |.
Proof of Theorem 4.1.3. We will only give the proof for the case n ≥ 3, since the proof
for n = 2 follows the same argument except applying Lemma 4.5.10 instead of Lemma 4.5.7
to estimate the last term IIIk in (4.5.27).
First, without loss of generality, we can assume d ≤ 1 for convenience. In the proof
below, Ci (1 ≤ i ≤ 3) and Cj∗ (1 ≤ j ≤ 2) denote constants which only depend on n, Ω and d.
Let M (t) be defined as in (1.4.3). The first step of the proof is to find a constant t∗ ∈ (0, 1]
and a finite strictly increasing sequence {Mk }0≤k≤L such that if Tk denotes the first time
that M (t) = Mk , then Tk − Tk−1 ≥ t∗ for 1 ≤ k ≤ L. The second step is to derive a lower
bound for Lt∗ .
Step 1. Let t∗ ∈ (0, 1] which will be determined later in this step. Define M0 as that in
(1.4.2) and T0 = 0. Then for k ≥ q, suppose Mk−1 has been constructed, we are trying to
define Mk such that Tk − Tk−1 ≥ t∗ .

139

Denote tk = Tk − Tk−1 . We will first check what happens if tk ≤ 1. By the maximum
principle and the Hopf lemma, there exists xk ∈ Γ1 such that

u(xk , Tk ) = Mk .

(4.5.25)

Applying the representation formula (4.2.4) with T = Tk−1 and (x, t) = (xk , tk ),

u(xk , Tk ) = 2

Ω

Φ(xk − y, tk ) u(y, Tk−1 ) dy
∂Φ(xk − y, t − τ )
u(y, Tk−1 + τ ) dS(y) dτ
∂n(y)
∂Ω

tk

−2
0

tk

+2
0

Γ1

Φ(xk − y, tk − τ ) uq (y, Tk−1 + τ ) dS(y) dτ.

(4.5.26)

Noticing (4.5.25), the above equality implies that

Mk ≤ 2Mk−1
q

+ 2Mk

Φ(xk − y, tk ) dy + 2Mk

Ω
tk
0

Γ1

q

+ 2Mk

Ω

Φ(xk
Ω
tk
0

Γ1

0

∂Φ(xk − y, tk − τ )
dS(y) dτ
∂n(y)
∂Ω

Φ(xk − y, tk − τ ) dS(y) dτ

= 2(Mk−1 − Mk )
+ 2Mk

tk

Φ(xk − y, tk ) dy

− y, tk ) dy +

tk
0

∂Φ(xk − y, tk − τ )
dS(y) dτ
∂n(y)
∂Ω

Φ(xk − y, tk − τ ) dS(y) dτ

Ik + IIk + IIIk .

(4.5.27)

First, since tk ≤ 1, Lemma 4.4.3 yields that

Ik ≤ 2(Mk−1 − Mk )b1 ,
140

(4.5.28)

where b1 is the same as in (4.4.10). Secondly it follows from Corollary 4.5.2 that there exists
a constant C1 such that
IIk ≤ Mk + C1 Mk exp −

d2
.
8tk

(4.5.29)

Thirdly, Lemma 4.5.7 implies the existence of a constant C2 such that

q

IIIk ≤ C2 Mk |Γ1 |1/(n−1) .

(4.5.30)

Combining (4.5.27), (4.5.28), (4.5.29) and (4.5.30),

Mk ≤ 2b1 (Mk−1 − Mk ) + Mk + C1 Mk exp −

d2
q
+ C2 Mk |Γ1 |1/(n−1) .
8tk

Subtracting Mk from both sides,

2b1 (Mk − Mk−1 ) ≤ C1 Mk exp −

d2
q
+ C2 Mk |Γ1 |1/(n−1) .
8tk

Dividing by 2b1 and rearranging the equation,
C
d2
1 − 1 exp −
2b1
8tk

q

C2 Mk |Γ1 |1/(n−1)
Mk − Mk−1 ≤
.
2b1

Define
C1∗ = max

C1
,1 ,
2b1

C2∗ =

C2
,
2b1

(4.5.31)

then
1 − C1∗ exp −

d2
8tk

q

Mk − Mk−1 ≤ C2∗ Mk |Γ1 |1/(n−1) .

141

(4.5.32)

Let us further temporarily assume tk is so small that

exp −

d2
8tk

≤

1 Mk − Mk−1
,
2C1∗
Mk

(4.5.33)

which is equivalent to

1 − C1∗ exp −

d2
8tk

1
Mk − Mk−1 ≥ (Mk − Mk−1 ).
2

Then it follows from (4.5.32) that

Mk − Mk−1
≤ 2C2∗ |Γ1 |1/(n−1) .
q
Mk

(4.5.34)

As a summary, this paragraph claims that if tk ≤ 1 and (4.5.33) holds, then Mk will satisfy
(4.5.34).
Based on this observation, denote

δ1 = 4C2∗ |Γ1 |1/(n−1) .

(4.5.35)

Then we define Mk to be the solution (if it exists) to
Mk − Mk−1
= δ1 .
q
Mk

(4.5.36)

With this choice of Mk , it is evident from (4.5.34) that either tk > 1 or tk violates (4.5.33).

142

Due to (4.5.36), tk violating (4.5.33) implies
d2
exp −
8tk

q−1

q−1
Mk δ1
M0 δ1
1 Mk − Mk−1
≥
.
>
=
2C1∗
Mk
2C1∗
2C1∗

(4.5.37)

Now if (the following requirement will be clear later)

q−1

M0

δ1 ≤

1
,
6q

(4.5.38)

then the right hand side of (4.5.37) is smaller than 1. Therefore, (4.5.37) is equivalent to
2C1∗
d2
ln
q−1
8
M0 δ1

−1

2C1∗
d2
t∗ =
ln
q−1
8
M0 δ1

−1

tk >

.

Define
.

(4.5.39)

Since d ≤ 1 and C1∗ ≥ 1, it is obvious that t∗ ∈ (0, 1]. Moreover, we can conclude that

tk ≥ min{1, t∗ } = t∗ .

On the other hand, by applying Lemma 4.4.4, (4.5.36) has a solution Mk > Mk−1 if and
1−q

only if δ1 ≤ Mk−1 Eq . In addition, as long as such a solution exists, Mk can be chosen to
satisfy
Mk−1 < Mk ≤

q
M
.
q − 1 k−1

Thus, the strategy of constructing {Mk } is summarized as following. First, M0 is defined
to be the same as in (1.4.2). Next, for k ≥ 1, suppose Mk−1 has been constructed, then

143

based on Lemma 4.4.4, whether defining Mk depends on how large Mk−1 is.
q−1

q
If Mk−1 δ1 ≤ Eq , then we define Mk ∈ Mk−1 , q−1
Mk−1 to be the solution to

(4.5.36).
q−1

If Mk−1 δ1 > Eq , then there does not exist Mk > Mk−1 which solves (4.5.36). So we
do not define Mk and stop the construction.
Based on this construction, if {Mk }1≤k≤L0 have been defined, then for any 1 ≤ k ≤ L0 ,
Tk − Tk−1 ≥ t∗ . Therefore, Tk ≥ kt∗ for any 1 ≤ k ≤ L0 . Applying Theorem 3.1.1,
L0 ≤ T ∗ /t∗ < ∞, which means the cardinality of {Mk } has be to finite. So we can assume
the constructed sequence is {Mk }0≤k≤L for some finite L.
Step 2. By Lemma 4.4.5,

L>

1
1
− 3q .
10(q − 1) M q−1 δ1
0

Taking advantage of the requirement (4.5.38),

T ∗ ≥ Lt∗ >

t∗
q−1

20(q − 1)M0

Writing
q−1

Y = M0

|Γ1 |1/(n−1) ,

then (4.5.38) reduces to
Y ≤

1
.
24C2∗ q

144

.
δ1

In addition, recalling the definition (4.5.39) of t∗ , then

T∗ ≥

C1∗
C3
ln
(q − 1)Y
2C2∗ Y

−1

for some constant C3 . Define

Y0 = min

2C2∗
1
,
,1 .
24C2∗ C1∗

Y

Then for any Y ≤ q0 ,
C1∗
ln
2C2∗ Y

≤ 2 ln

1
Y

= 2| ln Y |.

Hence,
T∗ ≥

C3
.
2(q − 1)Y | ln Y |

Remark 4.5.11. If the whole domain Ω is convex, then due to Lemma 4.4.2, (4.5.29)
becomes
IIk ≤ Mk .
Based on this change, all the exponential terms in the proof of Theorem 4.1.3 will disappear.
As a result, t∗ can be just chosen as 1 instead of the expression (4.5.39) which contains
the logarithm term in the denominator. Consequently, the logarithm term which is in the
denominator of (4.1.3) and (4.1.4) will also disappear. Namely, the lower bound in (4.1.3)
and (4.1.4) can be improved to be

T∗ ≥

C
.
(q − 1)Y
145

4.6

Comparison with previous works

As mentioned in the introduction, there are vast literature on the blow-up problems of the
parabolic type equations. But few of them deal with the lower bound estimate of the blow-up
time. A popular method dealing with the lower bound is established in [32–34]. After that,
the similar idea is also applied to some more generalized problems, see [6, 30, 40, 41, 43]. In
this section, we will compare Theorem 4.1.4 with the result in [34].
In [34], it studied the problem



ut (x, t) = ∆u(x, t)





∂u(x,t)

= F u(x, t)
∂n(x)






 u(x, 0) = u (x)
0

in

Ω × (0, T ],

on ∂Ω × (0, T ],
in

(4.6.1)

Ω,

where Ω is a convex, bounded open subset in R3 with smooth boundary and

0 ≤ F (s) ≤ ksm

(4.6.2)

for some k > 0 and m ≥ 3/2. By introducing the energy function

u4(m−1) (x, t) dx

ϕ(t) =
Ω

and adopting a Sobolev-type inequality developed in [32], they derive a first order differential
inequality for ϕ(t) and then obtain a lower bound for T ∗ :

T∗ ≥

∞

dη

ϕ(0)

K1 η + K2 η 3/2 + K3 η 3

146

,

(4.6.3)

where K1 , K2 and K3 are some positive constants depending on Ω and m.
Let us compare Theorem 4.1.4 with (4.6.3) for the problem (4.6.1) where F (s) = sq . For
convenience of statement, the lower bounds in Theorem 4.1.4 and (4.6.3) are denoted as T1
and T2 respectively.
• First, T1 works for any q > 1 and also give the exact asymptotic rate of T ∗ as q

1.

However, T2 is valid only for q ≥ 3/2, due to the restriction (4.6.2) and m ≥ 3/2 (let
k = 1 and m = q).
• Secondly, as M0

−(q−1)

0, T1 is of order M0

; however, if the initial data u0 does not

oscillate too much, that is

4(q−1)

ϕ(0) =
Ω

u0

4(q−1)

(x) dx ∼ M0

|Ω|,

then
1
∼ 4(q − 1) ln(M0−1 ),
ϕ(0)

T2 ∼ ln

which is only a logarithm order of M0−1 .
−2(q−1)

• Thirdly, as M0 → ∞, T1 is of order M0

; however, if the initial data u0 again

does not oscillate too much, then T2 grows like

−8(q−1)

T2 ∼ [ϕ(0)]−2 ∼ M0

Since M0 is large,
−2(q−1)

M0

−8(q−1)

>> M0

147

.

.

Chapter 5
Prevention of Blowup

5.1

Main theorems and outline of the approach

In this chapter, we provide two methods to prevent the finite-time blowup. The first strategy
is to repair the damaged part. The second way is by adding a suitable pump near the
1 , n is defined as in (4.3.17).
damaged part. In this chapter, for any α ∈ 0, n−1
α

Part I: Repairing the broken part
The first strategy is to repair the broken part Γ1 in the original problem (1.1.1). We are
trying to find the repairing rate at which the temperature can be prevented from blowing
up in finite time. The setup below will be studied.














ut (x, t) = ∆u(x, t) in
∂u(x,t)
∂n(x)

= uq (x, t)

Ω × (0, T ],

on Γ1,t × (0, T ],
(5.1.1)



∂u(x,t)

=0


∂n(x)






 u(x, 0) = u0 (x)

on Γ2,t × (0, T ],
in

Ω,

where Γ1,t represents the broken part at time t and Γ1,0 = Γ1 . However, since the broken
boundary in problem (5.1.1) is changing, the existence of the solution is harder to justify.
So in this part, we will assume the existence of the weak solution to (5.1.1) as in Definition

148

5.1.1 and then show that the weak solution does not blow up in finite time.
Definition 5.1.1. Given T > 0, a continuous function u on Ω × [0, T ] is called a weak
solution to (5.1.1) if for any T0 ∈ [0, T ), t ∈ (0, T − T0 ] and for any φ ∈ C 2 (Ω × [0, t]),
t
0

Ω

u(y, T0 + τ )(φτ + ∆φ)(y, τ ) dy dτ
t

=
Ω

u(y, T0 + t)φ(y, t) − u(y, T0 )φ(y, 0) dy −
t

+
0

∂Ω

u(y, T0 + τ )

0

Γ1,τ

uq (y, T0 + τ )φ(y, τ ) dS(y) dτ

∂φ(y, τ )
dS(y) dτ.
∂n(y)

(5.1.2)

It is readily seen that if u is a smooth solution, then it is a weak solution. In the rest of
this section, we will only deal with the weak solution. The first result below works for any
C 2 domain Ω and it says that as long as the area of |Γ1,t | decreases at some exponential
rate, the temperature will not blow up in finite time.
Theorem 5.1.2. Let M0 be defined as in (1.4.2). Then there exists a constant C =
C(n, Ω, q, M0 ) such that if
|Γ1,t | ≤ e−Ct |Γ1 |,
then for any weak solution u to (5.1.1) on Ω × [0, T ],

u(x, t) ≤ 3M0 eCt ,

∀ (x, t) ∈ Ω × [0, T ].

The second result deals with convex domains and it says that as long as |Γ1,t | decreases
at a suitable exponential rate, the temperature can be bounded by any data that is larger
than the initial maximum.

149

Theorem 5.1.3. Assume Ω is convex. Let M0 be defined as in (1.4.2). Then for any
B > M0 , there exists a constant C = C(n, Ω, q, M0 , B) such that if

|Γ1,t | ≤ e−Ct |Γ1 |,

then for any weak solution u to (5.1.1) on Ω × [0, T ],

u(x, t) ≤ B,

∀ x ∈ Ω, 0 ≤ t ≤ T.

Again, the global convexity may not be practical in real applications. So we want to
obtain a similar result by only assuming local convexity near Γ1 . But this time, we can only
prove the boundedness of the temperature with double-exponential decay rate of |Γ1,t |.
Theorem 5.1.4. Assume Conv [Γ1 ]d ⊆ Ω for some d > 0. Let M0 be defined as in (1.4.2).
Then for any B > M0 , there exists a constant C = C(n, Ω, d, q, M0 , B) such that if

|Γ1,t | ≤ exp 2 (n − 1) 1 − eCt

|Γ1 |,

then for any weak solution u to (5.1.1) on Ω × [0, T ],

u(x, t) ≤ B,

∀ x ∈ Ω, 0 ≤ t ≤ T.

Part II: Adding a pump In this part, we consider adding a negative source locally to

150

prevent the finite-time blowup. Now the problem becomes




ut (x, t) = ∆u(x, t) − ψ1 (x) up (x, t) in Ω × (0, T ]








 ∂u(x,t)
= uq (x, t)
on Γ1 × (0, T ]
∂n(x)


∂u(x,t)

=0


∂n(x)






 u(x, 0) = u0 (x)
where (see Figure 5.1) Ω1

Ω2

(5.1.3)

on Γ2 × (0, T ]
in

Ω, Γ1

Ω

Γ, p > 1, q > 1 and




= 1,
x ∈ Ω1 ,





ψ1 (x) = 0,
x ∈ Ω \ Ω2 ,






 ∈ (0, 1), x ∈ Ω \ Ω
2
1
is a smooth function.

Ω
Γ2
Ω2
Ω1
Γ1
Γ
Figure 5.1: Model with a pump

For problem (5.1.3), by the similar outline as that in Section 2.4, we can show it has a
local nonnegative solution and the solution can extend as long as the L∞ norm of u is finite.
151

We want to demonstrate that by choosing suitable p, the solution will not blow up, which
implies the solution is global.
Now we consider the following problem



vt (x, t) = ∆v(x, t) − ψ1 (x) v p (x, t) in





∂v(x,t)

= η(x) v q (x, t)
∂n(x)






 v(x, 0) = u (x),
0
where

Ω × (0, T ],

on ∂Ω × (0, T ],
in

(5.1.4)

Ω,




= 1,
x ∈ Γ1 ,





η(x) = 0,
x ∈ ∂Ω \ Γ,






 ∈ (0, 1), x ∈ Γ \ Γ .
1

By comparison principle, v ≥ u. So if we can prove that v is always finite, then u will not
blow up. The following conclusion is valid for any C 2 domain Ω.
Theorem 5.1.5. If p > 1, q > 1 and p > 2q − 1, then the solution v to (5.1.4) does not
blow up in finite time. As a result, the solution u to (5.1.3) exists globally.
The organization of this chapter is that in Section 5.2, we discuss how to prevent the
finite-time blowup by repairing the broken part. In Section 5.3, we provide another way to
control the solution by adding a suitable pump.

5.2

Repairing the broken part

In this section, we study the problem (5.1.1). The results are divided into three subsections,
due to different geometric properties of Ω. The Subsection 5.2.1 deals with any C 2 domains
152

but the conclusion only prevents blowup in finite time without providing any specific bound.
Next, both Subsection 5.2.2 and Subsection 5.2.3 try to control the temperature under any
value that is larger than the initial maximum. Subsection 5.2.2 assumes the global convexity
of Ω while Subsection 5.2.3 only requires the local convexity near Γ1 .

5.2.1

Prevention of finite-time blowup

The idea in this subsection is similar to that in Subsection 4.3.1. The difference is that
we will use the decay of |Γ1,t | to eliminate the nonlinear effect on the boundary. First,
we need the analogous representation formulas for the weak solution. The first lemma is
the representation formula of the solution for the inside points. Then its corollary is the
representation formula for the boundary points.
Lemma 5.2.1. Assume u is a weak solution to (5.1.1) on Ω × [0, T ] for some T > 0. Then
for any x ∈ Ω, T0 ∈ [0, T ) and t ∈ (0, T − T0 ],
t

u(x, T0 + t) =

Ω

Φ(x − y, t) u(y, T0 ) dy +
t

−
0

0

Γ1,τ

Φ(x − y, t − τ ) uq (y, T0 + τ ) dS(y) dτ

∂Φ(x − y, t − τ )
u(y, T0 + τ ) dS(y) dτ
∂n(y)
∂Ω

(5.2.1)

Proof. This proof is similar to that of Lemma 4.2.3. Fix any x ∈ Ω, T0 ∈ [0, T ) and
t ∈ (0, T − T0 ]. For

> 0, define

φ (y, τ ) = Φ(x − y, t − τ + ).

Plugging φ into (5.1.2) and sending → 0, (5.2.1) will be justified.

153

Corollary 5.2.2. Assume u is a weak solution to (5.1.1) on Ω × [0, T ] for some T > 0.
Then for any x ∈ ∂Ω, T0 ∈ [0, T ) and t ∈ (0, T − T0 ],
t

u(x, T0 + t) = 2

Ω

Φ(x − y, t)u(y, T0 ) dy + 2
t

0

Γ1,τ

Φ(x − y, t − τ )uq (y, T0 + τ )dS(y)dτ

∂Φ(x − y, t − τ )
u(y, T0 + τ )dS(y)dτ.
∂n(y)
∂Ω

−2
0

(5.2.2)

Proof. Similar to the proof for Corollary 4.2.4, by Lemma 5.2.1 and the jump relation of the
double-layer heat potential, (5.2.2) can be proved.
In the proof below as well as the proofs for Theorem 5.1.3 and Theorem 5.1.4, Corollary
5.2.2 play an essential role.
Proof of Theorem 5.1.2. Applying the representation formula (5.2.2) to (5.1.1) with T0 = 0,
we get for any x ∈ ∂Ω and t ∈ (0, T ),
t

u(x, t) = 2
Ω

Φ(x − y, t) u0 (y) dy − 2
t

0

∂Φ(x − y, t − τ )
u(y, τ ) dS(y) dτ
∂n(y)
∂Ω

Φ(x − y, t − τ ) uq (y, τ ) dS(y) dτ.

+2
0

Γ1,τ

Define M as (4.3.3). Then for any m > 1,
t

M (t) ≤ 2M0 + 2
t

+2

M (τ )
0

M q (τ )

0

Φ(x − y, t − τ ) dS(y) dτ
Γ1,τ

t

≤ 2M0 + C

∂Φ(x − y, t − τ )
dS(y) dτ
∂n(y)
∂Ω

t

1

(t − τ )− 2 M (τ ) dτ + C
0

0

154

m−1

n

n−1

M q (τ ) |Γ1,τ | m (t − τ )− 2 + 2m dτ.

In order for the integrability of the last term of the above inequality, the power − n2 + n−1
2m
should be greater than −1, which means m < n−1
n−2 . Define A : [0, T ] → R by

A(t) = |Γ1,t |.

(5.2.3)

In addition, let
α = (m − 1)/m and β = 1 − nα .
1 ), β ∈ ( 1 , 1) and
Then α ∈ (0, n−1
2
t

M (t) ≤ 2M0 + C

t

1

(t − τ )− 2 M (τ ) dτ + C

(t − τ )−β Aα (τ ) M q (τ ) dτ.

(5.2.4)

0

0

If
q−1

Aα (τ )M q−1 (τ ) ≤ 3q−1 |Γ1 |α M0

,

then
t

M (t) ≤ 2M0 + C

t

1

(t − τ )− 2 M (τ ) dτ + C

0

0

q−1

(t − τ )−β 3q−1 |Γ1 |α M0

M (τ ) dτ.

Now we are looking for a function

v(t) = 3M0 ekt

(5.2.5)

for some constant k determined later such that
t

v(t) ≥ 2M0 + C

t

1

(t − τ )− 2 v(τ ) dτ + C

0

0

155

q−1

(t − τ )−β 3q−1 |Γ1 |α M0

v(τ ) dτ.

Plugging (5.2.5) into the above inequality, we obtain an equivalent form
t

3M0 ekt ≥ 2M0 + 3CM0

0

t

q

(t − τ )−1/2 ekτ dτ + 3q CM0 |Γ1 |α

(t − τ )−β ekτ dτ.

(5.2.6)

0

Notice
t

e−kt

t

(t − τ )−β ekτ dτ =

(t − τ )−β e−k(t−τ ) dτ

0

0

t

=

τ −β e−kτ dτ

0

=

1 kt s −β −s
e ds
k 0
k

≤ k −(1−β) Γ(β)
= k −nα Γ(1 − nα ),

where Γ is the standard Gamma function. So in order to satisfy (5.2.6), it suffices to have

2
q−1
1 ≥ e−kt + Ck −1/2 + C 3q−1 M0 |Γ1 |α k −nα
3
for some constant C = C(n, Ω, α). Noticing 32 e−kt ≤ 2/3, so by taking k ≥ 36C 2 , then it
suffices to have
1
q−1
≥ C 3q−1 M0 |Γ1 |α k −nα .
6
The above inequality is equivalent to

q−1

k ≥ (C 3q−1 M0

156

|Γ1 |α )1/nα .

Therefore, it suffices to have

q−1

k = C max 1, 3q−1 M0

|∂Ω|α

1/nα

(5.2.7)

for some constant C = C(n, Ω, α).
By this choice of k, if A satisfies

q−1

Aα (τ ) v q−1 (τ ) ≤ 3q−1 M0

|Γ1 |α .

(5.2.8)

Then it follows from (5.2.6) and (5.2.8) that
t

v(t) ≥ 2M0 + C

t

1

(t − τ )− 2 v(τ ) dτ + C

0

(t − τ )−β Aα (τ )v q (τ ) dτ.

(5.2.9)

0

Since v(0) = 3M0 > M (0), then it follows from (5.2.4) and (5.2.9) that

M (t) ≤ v(t),

∀ 0 ≤ t ≤ T.

According to (5.2.5) and (5.2.8),

Aα (t) ≤ e−(q−1)kt |Γ1 |α ,

namely
A(t) ≤ exp −

(q − 1)kt
|Γ1 |.
α

1 , or equivalently by choosing m = 2n−2 , the proof is finished.
Finally, by choosing α = 2(n−1)
2n−3

157

5.2.2

Control of the solution under a given value for convex domains

In Subsection 5.2.1, we have discussed the strategy which can prevent the finite-time blowup.
But for the practical problems, it is more useful to control the temperature by a certain
value. This section provides a way to control the temperature by any value that is larger
than the maximum of the initial data under the assumption that Ω is convex. The idea of
this subsection is similar to that in Section 4.4.
Proof of Theorem 5.1.3. Define A(t) as in (5.2.3). The goal is to bound u above by B for
any B > M0 . Let

> 0 such that B = e M0 and let t∗ ∈ (0, 1] be a constant which will be

determined later. Define T0 = 0 and p0 = 1. For any k ≥ 1 and

Mk = exp (1 − 2−k ) M0 .

> 0, let

(5.2.10)

Let Tk be the first time that M (t) reaches Mk , then the maximum principle implies the
existence of xk ∈ ∂Ω such that u(xk , Tk ) = Mk . In the following, we will use induction to
show that Tk ≥ kt∗ . When k = 0, it is obviously true.
Step k (k ≥ 1): Suppose Tk−1 ≥ (k − 1)t∗ . Let tk

Tk − Tk−1 be the time spent in

the kth step. We intend to show tk ≥ t∗ , which implies Tk ≥ kt∗ . First, if tk ≥ 1, then it
has already implied that tk ≥ t∗ . So in the following, we assume tk < 1. By representation

158

formula (5.2.2), we have

u(xk , Tk ) = 2

Ω

Φ(xk − y, tk ) u(y, Tk−1 ) dy
tk

+2
0

∂Φ(xk − y, tk − τ )
u(y, Tk−1 + τ ) dS(y) dτ
∂n(y)
∂Ω

tk

+2
0

Γ1,T
k−1 +τ

Φ(xk − y, tk − τ ) uq (y, Tk−1 + τ ) dS(y) dτ.

According to the definition of Tk−1 and Tk ,

Mk ≤ 2Mk−1
q

+ 2Mk

Φ(xk − y, tk ) dy + 2Mk

Ω
tk
0

Γ1,T
k−1 +τ

tk
0

∂Φ(xk − y, tk − τ )
dS(y) dτ
∂n(y)
∂Ω

Φ(xk − y, tk − τ ) dS(y) dτ.

(5.2.11)

Since Ω is assumed to be convex, by applying Lemma 4.4.2 and Lemma 4.4.3 to the above
inequality, we obtain
tk

q

b1 (Mk − Mk−1 ) ≤ 2Mk

0
tk

q

≤ 2Mk

Γ1,T
k−1 +τ

0

Γ1,T
k−1

Φ(xk − y, tk − τ ) dS(y) dτ

Φ(xk − y, tk − τ ) dS(y) dτ.

Now Lemma 4.3.3 implies the existence of a constant C = C(n, Ω) such that

q

α
(Mk − Mk−1 )b1 ≤ CMk Aα (Tk−1 )tn
k ,

1 ). Recalling the expression (5.2.10), then
for any α ∈ (0, n−1

Mk − Mk−1 > 2−k M0
159

(5.2.12)

and
Mk < e M0 .
In addition, by the assumption that Tk−1 ≥ (k − 1)t∗ and the fact that A is a decreasing
function, we know A(Tk−1 ) ≤ A (k − 1)t∗ . Therefore, it follows from (5.2.12) that

q−1
α
2−k < Ce q M0 Aα (k − 1)t∗ tn
k .

(5.2.13)

Now if
t

−
A(t) ≤ 2 αt∗ |Γ1 |,

(5.2.14)

which implies
Aα (k − 1)t∗ ≤ |Γ1 |α 2−(k−1) ,
then it follows from (5.2.13) that

tk ≥

1/nα
1
.
2C e q M q−1 |Γ1 |α
0

Define
t∗ = min 1,

1/nα
1
.
2C e q M q−1 |Γ1 |α
0

(5.2.15)

Then tk ≥ t∗ and this finishes Step k.
In summary, if t∗ is defined as in (5.2.15) and A(t) satisfies (5.2.14), then the above
induction process is valid and can proceed forever. Thus, u will be bounded by B for all the

160

time. Noticing that

t∗ ≥ min 1,

1/nα
1
2C e q M q−1 |∂Ω|α
0

K,

where K is a constant which only depends on n, Ω, q, M0 , α and B. In order to satisfy
(5.2.14), it suffices to have
t

−
A(t) ≤ 2 αK |Γ1 |.
1 , the proof is finished.
Finally, taking α = 2(n−1)

5.2.3

Control of the solution under a given value for the domain
with local convexity near Γ1

Again, since the global convexity is not practical in real applications. In this subsection, we
try to extend the result to locally convex case. But the decreasing rate will be required to
be much faster than the convex case. The idea of this subsection is similar to the previous
subsection. But different lower bounds will be chosen for each piece, since only the local
convexity is available instead of the global convexity.
Proof of Theorem 5.1.4. In the following, unless stated otherwise, C and C1 will represent
constants which may depend on n, Ω, d, α and B. The value of C may change from line to
line. The goal is to bound u above by B. Let

> 0 such that B = e M0 and let t∗ ∈ (0, 1]

be a constant which will be determined later. Define T0 = 0 and p0 = 1. For any k ≥ 1,
define
Mk = exp (1 − 2−k ) M0 .

(5.2.16)

Let Tk be the first time that M (t) reaches Mk . Then the maximum principle implies the
161

existence of xk ∈ ∂Ω such that u(xk , Tk ) = Mk . We will use induction and choose suitable
t∗ to show that for any k ≥ 0,
k

Tk ≥
i=1
0

with the convention that
i=1

t∗
i

t∗
,
i

(5.2.17)

= 0. Since limk→∞ Tk = ∞ and Mk ≤ B, the solution u

will be bounded by B for all the time. When k = 0, (5.2.17) is obviously true.
Step k (k ≥ 1): Suppose (5.2.17) is true for k − 1, we will show it also holds for
k. Let tk

Tk − Tk−1 be the time spent in the kth step. Then it suffices to show that

tk ≥ tk∗ . If tk ≥ 1, then it has already satisfied tk ≥ tk∗ , so we assume tk < 1 below. By the
representation formula (5.2.2), we have

u(xk , Tk ) = 2

Ω

Φ(xk − y, tk ) u(y, Tk−1 ) dy
tk

−2
0

∂Φ(xk − y, tk − τ )
u(y, Tk−1 + τ ) dS(y) dτ
∂n(y)
∂Ω

tk

+2
0

Γ1,T
k−1 +τ

Φ(xk − y, tk − τ ) uq (y, Tk−1 + τ ) dS(y) dτ.

According to the definition of Tk−1 and Tk ,

Mk ≤ 2Mk−1
q

+ 2Mk

Φ(xk − y, tk ) dy + 2Mk

Ω
tk
0

Γ1,T
k−1 +τ

tk
0

∂Φ(xk − y, tk − τ )
dS(y) dτ
∂n(y)
∂Ω

Φ(xk − y, tk − τ ) dS(y) dτ.

(5.2.18)

Since Conv [Γ1 ]d ⊆ Ω, by applying Lemma 4.4.3 and Corollary 4.5.2 to the above inequality,

162

we obtain

2b1 (Mk − Mk−1 ) ≤ CMk exp −

tk
d2
q
+ 2Mk
8tk
0
Γ1,T

Φ(xk − y, tk − τ ) dS(y) dτ.

k−1 +τ

Define A(t) as in (5.2.3). Noticing |Γ1,T

k−1 +τ

| ≤ |Γ1,T

k−1

Mk − Mk−1 ≤ CMk exp −

|, Lemma 4.3.3 implies

d2
q
α
+ CMk Aα (Tk−1 )tn
k ,
8tk

1 . Again, noticing that
for some α ∈ 0, n−1

Mk − Mk−1 > 2−k M0

and
Mk < e M0 ,
then
2−k M0 ≤ Ce M0 exp −

d2
q
α
+ Ceq M0 Aα (Tk−1 )tn
k .
8tk

Dividing by M0 , then there exists a constant C1 = C1 (n, Ω, d, q, M0 , B) such that
d2
α
2−k ≤ C1 exp −
+ C1 Aα (Tk−1 )tn
k .
8tk
By induction,
k−1

Tk−1 ≥
i=1

t∗
≥ t∗ ln k.
i

163

Hence,
d2
α
2−k ≤ C1 exp −
+ C1 Aα (t∗ ln k)tn
k .
8tk

(5.2.19)

Since the right hand side of (5.2.19) is an increasing function in tk , if t∗ and A(t) are chosen
to satisfy both
C1 exp −

d2
8

t∗
k

≤ 2−k−1

(5.2.20)

and
C1 Aα (t∗ ln k)

t∗ nα
≤ 2−k−1 ,
k

(5.2.21)

then tk ≥ tk∗ and this finishes Step k.
In summary, if t∗ and A(t) are chosen to satisfy both (5.2.20) and (5.2.21), then the
above induction process is valid and can proceed forever. Thus, u will be bounded by B for
all the time. By elementary calculations, if

t∗ ≤

d2
,
8[ln C1 + 2 ln 2]

(5.2.22)

then (5.2.20) is satisfied. Next, in order to realize (5.2.21), it suffices to have

Aα (t∗ ln k) ≤ 21−k |Γ1 |α

(5.2.23)

and
C1

t∗ nα
1
≤
.
k
4|Γ1 |α

(5.2.24)

It is readily to check that if
t∗ ≤

1/nα
1
,
4C1 |∂Ω|α

164

(5.2.25)

then (5.2.24) is satisfied. According to (5.2.22) and (5.2.25), we define

t∗ = min 1,

d2
,
8[ln C1 + 2 ln 2]

1/nα
1
.
4C1 |∂Ω|α

(5.2.26)

It is readily seen that t∗ is a constant which only depends on n, Ω, d, q, M0 , B and α.
With this choice of t∗ , if A(t) satisfies

A(t) ≤ exp

ln 2
1 − et/t∗
α

|Γ1 |.

(5.2.27)

1
and requiring
Then it is readily seen that (5.2.23) holds. Finally, taking α = 2(n−1)

A(t) ≤ exp 2(n − 1) 1 − et/t∗ |Γ1 |,

then A(t) satisfies (5.2.27) automatically. Hence, the proof is finished.

5.3

Adding a pump

In this section, we provide another way to prevent the finite-time blowup. Let v be the
solution to (5.1.4). If
m≥

(p − 1)N
2

and m ≥ (q − 1)N,

then [3] shows that if the Lm norm of v does not blow up in finite time, then the L∞ norm
of v will not blow up in finite time either. Thus, it is equivalent to bound the Lm norm of
v. The idea of the proof of Theorem 5.1.5 is similar to that in [31] where the domain Ω is
assumed to be star-shaped. By using Corollary 5.3.2, we are able to generalize the proof for
any C 2 domain Ω.
165

Lemma 5.3.1. Let [a, b] be a bounded interval and f be a nonnegative C 1 function on [a, b].
Then for any r > 1 and

> 0,
b

||f ||rL∞ [a,b] ≤

b
r2 b 2r−2
1
f r (t) dt +
f
(t) dt.
b−a a
a

|f (t)|2 dt +

a

Proof. ∀ t1 , t2 ∈ [a, b],
t1

f r (t1 ) − f r (t2 ) =

rf r−1 (t)f (t) dt

t2
b

≤r

1

|f (t)|2 dt +

1

a

b

1

f 2r−2 (t) dt ,

a

where 1 is some positive constant to be determined. Integrating t2 from a to b,
b

(b − a)f r (t1 ) ≤

a

+

b

f r (t) dt + (b − a)r 1
b

(b − a)r
1

|f (t)|2 dt

a

f 2r−2 (t) dt.

a

Let 1 = /r, then

(b − a)f r (t1 ) ≤

b
a

+

b

f r (t) dt + (b − a)

|f (t)|2 dt

a

(b − a)r2

b

f 2r−2 (t) dt.

a

Since t1 is arbitrary, then

||f ||rL∞ [a,b]

b

≤

|f
a

(t)|2 dt +

b
1
r2 b 2r−2
r
f (t) dt +
f
(t) dt.
b−a a
a

166

(5.3.1)

Corollary 5.3.2. Let U = U × [a, b], where U is a bounded set in Rn−1 . Let w be a
nonnegative C 1 function on U . Then for any s ∈ [a, b],

|w(˜
x, s)|r d˜
x≤
U

U

|Dxn w(x)|2 dx +

r2
1
wr (x) dx +
w2r−2 (x) dx.
b−a U
U

Proof. For any x˜ ∈ U , define f : [a, b] → R by

f (t) = w(˜
x, t),

∀ t ∈ [a, b].

Then by Lemma 5.3.1, for any s ∈ [a, b],
b

|f (s)|r ≤

|f (t)|2 dt +

a

b
1
r2 b 2r−2
f r (t) dt +
f
(t) dt.
b−a a
a

That is,
b

|w(˜
x, s)|r ≤
a

|Dxn w(˜
x, t)|2 dt +

b
1
r2 b 2r−2
wr (˜
x, t) dt +
w
(˜
x, t) dt.
b−a a
a

Integrating x˜ on U , the conclusion follows.
Proof of Theorem 5.1.5. Let m ≥ max

(p−1)N
,2
2

and define

v m (x, t) dx.

E(t) =
Ω

167

Then

E (t) = m
Ω

=m
Ω

v m−1 vt dx
v m−1 (∆v − ψ1 v p ) dx
∇ · (v m−1 ∇v) − (m − 1) v m−2 |∇v|2 dx − m

=m
Ω

Ω

v m+q−1 dS − m(m − 1)

≤m
Γ

ψ1 v m+p−1 dx

v m−2 |∇v|2 dx − m
Ω

v m+p−1 dx.

(5.3.2)

Ω1

Next it will be shown that for any > 0, there exists a constant C = C(Γ, Ω1 , q, m) such
that

v m+q−1 dS ≤
Γ

v m−2 |∇v|2 dx + C
Ω1

v m+q−1 dx +

C

v m+2q−2 dx.

(5.3.3)

Ω1

Ω1

Noticing that v m−2 |∇v|2 = |∇(v m/2 )|2 , so by writing w = v m/2 and r = 2 + 2(q − 1)/m,
then r > 2 and (5.3.3) is equivalent to

wr dS ≤
Γ

|∇w|2 dx + C

wr dx +

Ω1

Ω1

C

w2r−2 dx.

(5.3.4)

Ω1

Fix any point x0 ∈ Γ. Since Γ is C 2 , there exists a neighborhood V of x0 in Ω1 that
can be straightened by a C 2 bijection Ψ : B1 → V . Denote Br to be the ball in Rn−1 with
radius r. Define
U = B1/2 × 0,
Then V0

1
⊂ B1 .
2

Ψ(U ) ⊂ V . Define

U0 = B1/2 × {0},

and Γ0 = Ψ(U0 ).
168

By change of variable x = Ψ(y), we obtain

wr (x) dS(x) ≤ C
Γ0

(w ◦ Ψ)r (y) dS(y)
U0

(w ◦ Ψ)r (˜
y , 0) d˜
y.

≤C
B1/2

Now applying Corollary 5.3.2 to the function w ◦ Ψ,

(w ◦ Ψ)r (˜
y , 0) d˜
y
B1/2

≤
U

|Dyn (w ◦ Ψ)(y)|2 dy + 2

(w ◦ Ψ)r (y) dy +

r2

(w ◦ Ψ)2r−2 (y) dy.
U

U

Then using the change of variable y = Ψ−1 (x),

U

|Dyn

≤C

(w ◦ Ψ)(y)|2 dy

+2

+

r2

(w ◦ Ψ)2r−2 (y) dy

U

(∇w) Ψ(y)

2

U

wr Ψ(y) dy +

dy + 2

r2

U

U
2

V0

V0

w2r−2 Ψ(y) dy
U

wr (x) dx +

∇w(x) dx + C

=C

(w ◦ Ψ)r (y) dy

C

w2r−2 (x) dx
V0

In summary, we obtain

2

wr (x) dS(x) ≤ C
Γ0

wr (x) dx +

∇w(x) dx + C
V0

C

V0

w2r−2 (x) dx

(5.3.5)

V0

Finally by a finite cover argument, (5.3.4) is justified, which also means (5.3.3) is verified.
Now combining (5.3.2) and (5.3.3) with = m − 1, we obtain that

v m+q−1 (x, t) dx + C

E (t) ≤ C
Ω1

v m+2q−2 (x, t) dx − m
Ω1

v m+p−1 (x, t) dx,
Ω1

169

for some constant C = C(Γ, Ω1 , q, m). Since p > 2q − 1 from the assumption, then

E (t) ≤ C.

This implies that E(t) will not blow up in finite time. By applying Theorem 1.1 in [3] and
the assumption that p > 2q − 1, we can conclude that u will exist globally.

170

BIBLIOGRAPHY

171

BIBLIOGRAPHY

[1] H. Amann. Quasilinear evolution equations and parabolic systems. Trans. Amer. Math.
Soc., 293(1):191–227, 1986.
[2] H. Amann. Quasilinear parabolic systems under nonlinear boundary conditions. Arch.
Rational Mech. Anal., 92(2):153–192, 1986.
[3] J. M. Arrieta, A. N. Carvalho, and A. Rodr´ıguez-Bernal. Parabolic problems with
nonlinear boundary conditions and critical nonlinearities. J. Differential Equations,
156(2):376–406, 1999.
[4] T. Cazenave, F. Dickstein, and F. B. Weissler. Global existence and blowup for
sign-changing solutions of the nonlinear heat equation. J. Differential Equations,
246(7):2669–2680, 2009.
[5] K. Deng and H. A. Levine. The role of critical exponents in blow-up theorems: the
sequel. J. Math. Anal. Appl., 243(1):85–126, 2000.
[6] C. Enache. Lower bounds for blow-up time in some non-linear parabolic problems under
Neumann boundary conditions. Glasg. Math. J., 53(3):569–575, 2011.
[7] A. Friedman. Partial differential equations of parabolic type. Prentice-Hall, Inc., Englewood Cliffs, N.J., 1964.
[8] A. Friedman and B. McLeod. Blow-up of positive solutions of semilinear heat equations.
Indiana Univ. Math. J., 34(2):425–447, 1985.
[9] H. Fujita. On the blowing up of solutions of the Cauchy problem for ut = ∆u + u1+α .
J. Fac. Sci. Univ. Tokyo Sect. I, 13:109–124 (1966), 1966.
[10] V. Georgiev, H. Lindblad, and C. D. Sogge. Weighted Strichartz estimates and global
existence for semilinear wave equations. Amer. J. Math., 119(6):1291–1319, 1997.
[11] Y. Giga and R. V. Kohn. Asymptotically self-similar blow-up of semilinear heat equations. Comm. Pure Appl. Math., 38(3):297–319, 1985.
[12] R. T. Glassey. Existence in the large for cmu = F (u) in two space dimensions. Math.
Z., 178(2):233–261, 1981.
[13] R. T. Glassey. Finite-time blow-up for solutions of nonlinear wave equations. Math. Z.,
177(3):323–340, 1981.
172

[14] K. Hayakawa. On nonexistence of global solutions of some semilinear parabolic differential equations. Proc. Japan Acad., 49:503–505, 1973.
[15] B. Hu. Blow-up theories for semilinear parabolic equations, volume 2018 of Lecture
Notes in Mathematics. Springer, Heidelberg, 2011.
[16] B. Hu and H.-M. Yin. The profile near blowup time for solution of the heat equation
with a nonlinear boundary condition. Trans. Amer. Math. Soc., 346(1):117–135, 1994.
[17] F. John. Blow-up of solutions of nonlinear wave equations in three space dimensions.
Manuscripta Math., 28(1-3):235–268, 1979.
[18] T. Kato. Blow-up of solutions of some nonlinear hyperbolic equations. Comm. Pure
Appl. Math., 33(4):501–505, 1980.
[19] K. Kobayashi, T. Sirao, and H. Tanaka. On the growing up problem for semilinear heat
equations. J. Math. Soc. Japan, 29(3):407–424, 1977.
[20] R. Kress. Linear integral equations, volume 82 of Applied Mathematical Sciences.
Springer, New York, third edition, 2014.
[21] T.-Y. Lee and W.-M. Ni. Global existence, large time behavior and life span of solutions
of a semilinear parabolic Cauchy problem. Trans. Amer. Math. Soc., 333(1):365–378,
1992.
[22] H. A. Levine. Some nonexistence and instability theorems for solutions of formally
parabolic equations of the form P ut = −Au + F (u). Arch. Rational Mech. Anal.,
51:371–386, 1973.
[23] H. A. Levine. Nonexistence of global weak solutions to some properly and improperly
posed problems of mathematical physics: the method of unbounded Fourier coefficients.
Math. Ann., 214:205–220, 1975.
[24] H. A. Levine. The role of critical exponents in blowup theorems. SIAM Rev., 32(2):262–
288, 1990.
[25] H. A. Levine and L. E. Payne. Nonexistence theorems for the heat equation with
nonlinear boundary conditions and for the porous medium equation backward in time.
J. Differential Equations, 16:319–334, 1974.
[26] H. Lindblad and C. D. Sogge. Long-time existence for small amplitude semilinear wave
equations. Amer. J. Math., 118(5):1047–1135, 1996.
[27] J. L´opez-G´omez, V. M´arquez, and N. Wolanski. Blow up results and localization of blow
up points for the heat equation with a nonlinear boundary condition. J. Differential
Equations, 92(2):384–401, 1991.
173

[28] P. Meier. On the critical exponent for reaction-diffusion equations. Arch. Rational
Mech. Anal., 109(1):63–71, 1990.
[29] C. E. Mueller and F. B. Weissler. Single point blow-up for a general semilinear heat
equation. Indiana Univ. Math. J., 34(4):881–913, 1985.
[30] L. E. Payne and G. A. Philippin. Blow-up phenomena in parabolic problems with time
dependent coefficients under Dirichlet boundary conditions. Proc. Amer. Math. Soc.,
141(7):2309–2318, 2013.
[31] L. E. Payne, G. A. Philippin, and S. Vernier Piro. Blow-up phenomena for a semilinear
heat equation with nonlinear boundary conditon, I. Z. Angew. Math. Phys., 61(6):999–
1007, 2010.
[32] L. E. Payne and P. W. Schaefer. Lower bounds for blow-up time in parabolic problems
under Neumann conditions. Appl. Anal., 85(10):1301–1311, 2006.
[33] L. E. Payne and P. W. Schaefer. Lower bounds for blow-up time in parabolic problems
under Dirichlet conditions. J. Math. Anal. Appl., 328(2):1196–1205, 2007.
[34] L. E. Payne and P. W. Schaefer. Bounds for blow-up time for the heat equation under
nonlinear boundary conditions. Proc. Roy. Soc. Edinburgh Sect. A, 139(6):1289–1296,
2009.
[35] P. Quittner and P. Souplet. Superlinear parabolic problems. Birkh¨auser Advanced Texts:
Basler Lehrb¨
ucher. [Birkh¨auser Advanced Texts: Basel Textbooks]. Birkh¨auser Verlag,
Basel, 2007. Blow-up, global existence and steady states.
[36] D. F. Rial and J. D. Rossi. Blow-up results and localization of blow-up points in an
N -dimensional smooth domain. Duke Math. J., 88(2):391–405, 1997.
[37] J. Schaeffer. The equation utt − ∆u = |u|p for the critical value of p. Proc. Roy. Soc.
Edinburgh Sect. A, 101(1-2):31–44, 1985.
[38] J. H. Shi. On the problems of preventing the blowing up of the solutions for semilinear
parabolic equations. Acta Math. Sci. (English Ed.), 4(3):291–296, 1984.
[39] T. C. Sideris. Nonexistence of global solutions to semilinear wave equations in high
dimensions. J. Differential Equations, 52(3):378–406, 1984.
[40] G. Tang, Y. Li, and X. Yang. Lower bounds for the blow-up time of the nonlinear nonlocal reaction diffusion problems in RN (N ≥ 3). Bound. Value Probl., pages 2014:265,
5, 2014.

174

[41] Y. Tao and S. Vernier Piro. Explicit lower bound of blow-up time in a fully parabolic
chemotaxis system with nonlinear cross-diffusion. J. Math. Anal. Appl., 436(1):16–28,
2016.
[42] G. Todorova and B. Yordanov. Critical exponent for a nonlinear wave equation with
damping. J. Differential Equations, 174(2):464–489, 2001.
[43] G. Viglialoro. Blow-up time of a Keller-Segel-type system with Neumann and Robin
boundary conditions. Differential Integral Equations, 29(3-4):359–376, 2016.
[44] W. Walter. On existence and nonexistence in the large of solutions of parabolic differential equations with a nonlinear boundary condition. SIAM J. Math. Anal., 6:85–90,
1975.
[45] F. B. Weissler. An L∞ blow-up estimate for a nonlinear heat equation. Comm. Pure
Appl. Math., 38(3):291–295, 1985.
[46] B. T. Yordanov and Q. S. Zhang. Finite time blow up for critical wave equations in
high dimensions. J. Funct. Anal., 231(2):361–374, 2006.
[47] Q. S. Zhang. A blow-up result for a nonlinear wave equation with damping: the critical
case. C. R. Acad. Sci. Paris S´er. I Math., 333(2):109–114, 2001.

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