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LIBRWARY
Michigan State
University

 

 

 

This is to certify that the

dissertation entitled

Solutions to a Nonlinear Heat Equation
with Critical Exponent

presented by

Canan Celik

has been accepted towards fulfillment
of the requirements for

 

Ph . D . degree in Mathematics

Wigwfl

\ -
K7Mauor professor

 

Date {/flé/‘QDD I

MSU is an Affirmative Action/Equal Opportunity Institution 0-12771

PLACE IN RETURN Box to remove this checkout from your record.
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6/01 cJCIRCIDateDue.p65-p. 15

 

Solutions to a Nonlinear Heat Equation with Critical

Exponent

By

Canon Celik

A DISSERTATION

Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of

DOCTOR OF PHILOSOPHY

Department of Mathematics

2001

 

ABSTRACT

Solutions to a Nonlinear Heat Equation with Critical Exponent
By

Canan Celik

In this study we consider the nonlinear heat equation it, = Au+ Iulp‘lu on Q x IR+
with some boundary conditions and the initial condition u(:1:,0) = u0(2:) 6 L162),
where Q C R" and p > 1. For the one dimensional case, it is well known for p < 3
that this problem has a local solution for any initial condition no 6 L162). But the
existence and uniqueness of a local solution in L1 for the critical exponent p = 3
was widely open and this work is to answer this open question. First, we prove for
the Cauchy problem that there is no local solution in L1 for some no 6 L1. Then
using the nonlocal existence of the Cauchy problem by a cutoff function argument,
we prove the nonlocal existence of a solution for the Dirichlet problem which answers
this open question. Moreover we generalize the nonlocal existence result for the n-
dimensional case with the critical exponent p = l + 3;. More general nonlinearity is
also considered for Dirichlet boundary value problems. And also we prove the same
result for the mixed boundary condition with the same initial data no. Finally, we
establish the global existence in L”6 with ||uo||1+.E sufficiently small and e > 0, for
the critical exponent p = 3 and n = 1. By using some numerical methods we also

give some numerical simulations to these results.

 

 

To My Parents

iii

ACKNOWLEDGMENTS

I would like to express my gratitude and thanks to my advisor, Professor
Zhengfang Zhou for his constant help, encouragement, and excellent advice. I would
also like to thank Professors Dennis Dunninger, Sheldon N ewhouse, Michael Frazier,
Baisheng Yan and Clifford Weil for their time and for their valuable suggestions.

My warm thanks would go to my family for their continual inspiration and en-
couragement.

Finally, I would like to thank Devrim Karaaslanli for his constant support during

my study.

iv

TABLE OF CONTENTS

Introduction

6

7

0.1 An Overview ..............................
0.2 Statement of the Problem ......................

N onlocal Existence for the Cauchy Problem

Nonlocal Existence of a Solution for the Dirichlet Problem
2.1 Blow up of Solutions for the Dirichlet Problem .........
2.2 Nonlocal Existence of Solutions for the Dirichlet Problem . .

Nonlocal Existence for the n-dimensional Dirichlet Problem
3.1 Nonlocal Existence of the n—dimensional Cauchy Problem . .
3.2 Nonlocal Existence for the n-dimensional Dirichlet Problem

Nonlocal Existence of Solutions for Mixed Boundary Conditions

4.1 Blow up of Solutions for the Mixed Boundary Conditions . .

4.2 Nonlocal Existence of Solutions for the Mixed Boundary Con-
ditions ..................................

Global Existence for Small Initial Data

Numerical Results

Summary and Conclusion

BIBLIOGRAPHY

13
13
26

35
36
40

52
53

61

64

70

82

85

 

Introduction

0.1 An Overview

In this study we consider solutions to the nonlinear heat equation

ut(.’c,t) = Au(:t:,t) + |u(:r,t)|”‘1u(:t,t) in Q x (0,T),
u(:r,t) = 0 on an x (0,T), (1)
u(:z:, 0) = u0(a:) in Q,

where Q C IR" is a smooth bounded domain and p > 1.

Throughout this work we will use the following definitions.

Definition 0.1 A function u = u(:c, t) is a solution of the initial and boundary value
problem (1) in Q x [0, T], T being a positive number, if and only if u E C’ ( [0, T], 01(9))

satisfies the integral equation

Wat) = LK(m,y,t)uO(i/)dy+ [0 [Q K($,y,t-S)IU(y, s)l”‘1u(y, S)dyds

where uo 6 L962) and K denotes the heat kernel for the linear heat equation with

corresponding boundary condition.

Definition 0.2 A solution of (1) on Q x (0,T) (where Q C R" or Q = R") for some
T < +00 is called a local (in time) solution. The supremum of all such T’s for which

a solution exists is called the maximal time of eaistence Tm. When Tm” = 00 the

1

solution is called global. When Tm“ < +00, the solution of (1 ) is not global or the

solution blows up in finite time.

Before stating the main problem, we will give some related results to this initial
and boundary value problem.

This initial-boundary value problem was studied by many authors. First H. Eijita
[1], [2] studied the initial value problem

ut(:r,t) = Au(:c,t) + up(a:,t) in R” x (0,T),

u(:r, t) = u0(:r) on IR",

where uo(:r) Z 0, for the following two cases.
Case (i) If n(p — 1) / 2 < 1, he proved that no non-negative global solution exists for
any non-trivial initial data uo 6 L1. That is, every positive solution to this initial

value problem blows up in a finite time.

Case (ii) If n(p — 1) / 2 > 1, global solutions do exist for small initial data uo 6 L1.
To be precise, for any It > 0 we can choose 6 > 0 such that initial value problem has
a global solution whenever 0 S u0(:c) S 6e‘k'z'2. To prove the global existence result,
he used the comparison argument with the solution of the same initial value problem
with the initial data u0(:r) = (Se-WI”.

The critical case n(p— l) / 2 = 1 was studied by K. Hayakawa [3] for n = 1, 2 and by
K. Kobayashi, T. Sinao and H. Tanaka [4] for general n and they proved that no non-
negative global solution exists for any non-trivial initial data. For n = 1,2, Hayakawa
first obtained the local solution u(:r, t) for sufficiently small T and he found to > 0,
co > 0 and [30 > 0, so that they satisfy u(z, to) Z c0e30'z'2. Next by using an iteration
argument, he obtained the non global existence to the same initial value problem

replacing the initial data by u0(:r) = c0e30'z'2. Then by a comparison argument he

proved that the initial value problem has no global solution.

Also F. B. Weissler [7] obtained that for n(p — 1)/2 g 1 with 1 __<_ p < 00, non-
negative LP solutions to the Cauchy problem must blow up in L” norm in finite time.
And for the case n(p — 1)/ 2 > 1, he gave some sufficient conditions on the initial data
no which guarantee the existence of a global solution.

The initial and boundary value problem (1) was also studied by many authors.
H. Brezis and T. Cazenave [13] studied the question of local existence for this initial
and boundary value problem. It is well known that if uo E L°°(Sl), then there is
a unique solution defined on a maximal interval [0,Tm). The solution u here is a
classical solution (u is C1 in t E (0,Tm) and 0'2 in a: E Q) of (1) on O x (0,Tm)
and u e L°°((0,T) X 9) for all T < Tm“. They were concerned with the question
of what happens if uo ¢ L°°(Q). More precisely, they assumed that no 6 Lq(Q) for
some 1 S q < 00.

This type of problem was first considered by F. B. Weissler [8], [9] who obtained
some local existence and uniqueness results in C ([0,T], L4 ((2)) In particular he
proved the existence result for the case q > n(p — 1) / 2 and q > p. The uniqueness for
the same case was proved by P. Baras [5] by standard contraction mapping argument
by using regularizing properties of —A. But for the critical case p = q = n(p — 1) / 2;
i.e., for p = q = n/(n — 2) with n 2 3, W. M. Ni and P. Sacks [11] proved that
uniqueness fails.

Some related non uniqueness results were also proved in A. Haraux and F. B.
Weissler [6] and further results were obtained by Y. Giga [10], who constructed a
unique solution to the initial and boundary value problem in LP1(O,T; L“). In his
work, the main relation between p1 and p2 was 1 = 3(1 — —1—), p1 > 7', provided that

E r P1
the initial data no G L" with r = n(p —1)/2 > 1.

0.2 Statement of the Problem

The main problem considered in this study is motivated by H. Brezis and T. Cazenave
[13] where they obtained the following local existence results where no 6 L962),

1 s q < 00 for two cases.

Case (i) Ifq> "(i212 (resp. qzzw—z—il)andq2 1 (resp. q> 1) withnz 1, they
obtained the existence and uniqueness of a local solution in C ([0, T], Lq(Q)) which is

a classical solution of the initial boundary value problem on 5-2 x (0, T).

Case (ii) If q < "(p—fl}, they showed that there exists no local solution in any

reasonable sense for some initial data uo 6 L962).

To prove the existence part of Case (i) for q > 33:32:12 and q 2 1, they first
established the existence of a solution u 6 L°°((0, T), Lq(fl)) fl L°° ((0, T), Lp"(fl)) by

Zoe

using a contraction mapping argument and then they completed the proof showing
that u E C([0, T], Lq(Q)) flLfifc((0,T), L°°(Q)). The proof of existence for q = 532:9
and q > 1 was shown by a similar strateg with some minor technical differences.
They proved the uniqueness part of Case (i) by using the solution that was constructed
in the existence part by the contraction mapping argument.

So the value q = n(p —- 1) / 2 plays a critical role in the existence and uniqueness
of local solution for this initial and boundary value problem.
Remark. To see that q = ELIE) is the critical exponent, we observe the following

2

dilation argument:

if v is the solution of

vt 2 ”$2: + Up)

We 0) = 9(27).

then u(a:, t) = k‘v(k:c, k2t) is the solution of

at = “xx + up,

u(:o, 0) = k’g(ka:).
By using the equation ut = u;m + u’D with u(:c, t) = k’i)(ka:, k2t), we get
kiwi), = ks+2um + 193%"

which gives the condition 3 = 51—.

H

Also note that,

llu(0)lqu = fuq($:0)d$ = / trauma -——— “mom."

when 8 = 1;.
So, combining these two conditions for s, we have q = 392—12, which is the critical
exponent.
Note that if u blows up in Lq at t = T, then so does it at t = ET; even if they have
initial data with the same L" norm.

The value q = n(p — 1) / 2 plays a critical role for the existence and uniqueness of
a local solution and the existence and uniqueness of a local solution for the doubly
critical case which is q = n(p — 1) /2 and q = l was widely open. Therefore we

attacked the following open problem proposed by H. Brezis and T. Cazenave [13]. For

simplicity we first consider the one-dimensional case, n = 1, (p = 3) with S2 = (-1, 1)
and then we study the general n-dimensional case with S2 = (—1, 1) x - -- x (--1, 1).

So we have the following problem in one-dimension,

”Is there some no 6 L1(—1, l) for which there is no local solution of (1)?”
where the initial-boundary value problem (1) takes the following form for the one-

dimensional case.

ut(a:,t) = um(:1:,t) + u3(:t:,t) in (—1, 1) X (0,T),
u(:i:1,t) = 0 in (0, T), (3)
u(a:,0) = no in (—1,1).

This study is organized as follows.

In Chapter 1 we consider the Cauchy problem and we prove that, for some initial
data no 6 L1, there is no local solution to the Cauchy problem in L1.

In Chapter 2, we study this initial and boundary value problem and in the first
section, by using a delicate dilation argument with some technical lemmas, we prove
the blow up of the solution in a finite time. Physically blow up of solution means
the total heat would go to infinity in finite time. In the second section, by using the
nonlocal existence result that we obtained in Chapter 1 for the Cauchy problem and
using a cutoff function argument, we prove the nonlocal existence of solutions for the
Dirichlet problem for some initial data no 6 L1.

In Chapter 3, we first generalize the nonlocal existence result for the n—dimensional
Cauchy problem and by using this result we proved the nonlocal existence result for
n-dimensional Dirichlet problem with the similar arguments that we used to prove
the one-dimensional case.

In addition, we prove the nonlocal existence result by generalizing the nonlinearity

in the Dirichlet Problem.

Since we prove all of these nonlocal existence results for the cubic domain 9, for
general it i.e., for {2 = (—c, c) x x (—c, c), where c is a positive constant, our goal
in this chapter is to generalize these results to an arbitrary domain 9 C R“ by using
comparison argument of solutions.

In Chapter 4, we consider the same problem with the mixed boundary conditions;
that means having u and the normal derivative of u in the boundary condition. In the
first section, using a dilation argument and proving similar technical lemmas as we
used for the Dirichlet problem, we prove that a solution to the mixed boundary value
problem also blows up in finite time. Moreover, by using the comparison principle
with the solution of Dirichlet problem, we prove the nonlocal existence result for the
mixed boundary condition with the same initial data uo.

In Chapter 5, we give the sufficient conditions for q and the initial data uo to
guarantee the global existence of the solution for the Dirichlet boundary condition.
Mainly we proved that for any fixed 6 > 0, if the initial condition uo 6 L1“, then the
Dirichlet problem has a global solution in L1+€ for ||u0||1+€ sufficiently small.

In Chapter 6, we showed some numerical simulations and graphs for our results
for the Dirichlet problem by using finite difference method.

Finally in Chapter 7 we give the summary and conclusion to our study.

Notation: Throughout this study uc, K 6 denote the solution to the Cauchy problem
and the kernel of the Cauchy problem, ud, K 4 denote the solution to the Dirichlet

problem and the kernel of the Dirichlet problem.

CHAPTER 1

Nonlocal Existence for the Cauchy

Problem

For the motivation to prove the nonlocal existence of L1 solution for the Dirichlet
problem, we first prove the nonlocal existence of an L1 solution for the Cauchy prob-

lem

114(3)” = u$2($1t)+ ’U. ($,t) in R X (0,T), (1.1)

u(:1:, t) = uo(:r) on R,

for some no 6 LICK).

First we choose a particular initial data uo(.r) = Ice-’9‘”2 6 EUR) and observe the
following result. After this observation, we prove in Theorem 1.1 that the Cauchy
problem has no local Ll solution for initial data of the form uo = Z ckke"k%c2 where

k.
ck 2 0 will be determined later.

Observation. For the Cauchy problem (1.1) with the initial data u0(a:) = Ice-kin:2 E

L1(IR), by using the fact that

2 2

1" "'01

/ e“ 4: ”at/2d el+4at
e = —,
IR V 47rt y \/ 1 + 4at

 

 

The linear solution is

L k -—k23:2
uC (x,t) = ————e 1+4k’t.

\/1+ 4k2t

And the solution to the Cauchy problem (1.1) is

to, t) = uses) + f f Kc($ay,t — s)(u.(y, s>>3dyds,

where the Cauchy kernel K C(so, y, t) is given by

 

 

|z_ I2
8— 4t
KC 2:, ,t = .
( y ) Tm:
Note that
k3 _3k2 2
L 3 _ —5;
(uc (y,s)) — (1+4k28)3/2€1+4k

and / Kc(:c,y,t — s)d:r = 1.
1R
Since ucL(:I:, t) > 0 and uc(:r, t) > ucL(:t:, t), we have

uc(:r,t) > /0 /RKc(:r,y,t — s)(uCL(y,s))3dyds.

Then for any 6 > 0,

6
Manama» = [0 Airman

> [06/ft/Kc(a:,y,t— s)(ucL(y,s))3dydsda:dt

[ll%W% /mH~

31:2 2

k3e'ifiiyj

= [06/0/( 1+41:28)”?dydsdt
= — ———d

\/;/0 f0 (1+4k2s) Sdt

6

7r 1
= — —ln1 4k2tdt.

,/;/0 4 (+ >

 

s)d:rdydsdt

So we can use the above estimate to establish the following nonlocal existence theorem

for the Cauchy problem.

Theorem 1.1 For some initial data no 6 L1(IR) the Cauchy problem (1.1) has no

local solution in C([0, 6], L1(IR)).

Proof of Theorem 1.1 Let uo = 20.196421? where ck 2 0 will be determined

k
later. The linear solution to the Cauchy problem is

Ckk 4:212
UCL($, t) = E —e 1+4k7t .

k \/1 + 4th

So the solution of the problem is

aw) > / f Kauai/J - s)(u.L<y.s)>3dyds,

10

Note that

l+4kas .

 

Then we have

__8____— :- ,__, Ck3k3 _3k2 2

—5:
R:——_/47r( t - 3) 1+ 4k2s)3/2e 1+“ dy.

 

[Kc(:r,y,t—s )(uc( y,s ))3dy >
IR

So for any 6 > 0,

6
IIUcIIL1(Rx(o,6)) = / fuc() (,)ztdxdt
0 1R
6
> / f/t /KC( SC ,y,t 3)(UcL(y,S))3dydsd$dt
0 R
6
= f/A(ucL(y,3))3/Kc( a: ,y,t s)d:rdydsdt
0 o
6 3 3 2 2
Ck k —-3k E
> ftf0/Z(1+4k2s)3/28W'ddedt
= 20/ / —C"3—:—dsdt
k 0 0 (1+4k S)
‘1
= :C/ ch3ln(1+4k2t)dt
o

k

 

6
Z C;A/2c21n(l+ 4k2t)dt Z ;6021n(1+ 2’626).

Now we choose ck such that

i) Zcid ln(1 + 2k26) = 00; and
1:

ii) ZC" < 00 for uo 6 L1.
1:

We will find a sequence {[9]-Bil such that C}, 52$ 0 only when k = kj. For k big enough,

11

we have

c261n(1 + 21926) 2 ck3 ln 1:.

So if we set ck3v1nk = Unit)”4 for some 19 = k,- —+ 00 as j —> 00; that is,

ck = W, for k = kj,j = 1,2, - -- Condition (i) will be satisfied.

.. 1 1 . . . '24 .
For (11), we set W = :73 Wthh 1mpl1es that k = e3 = 3,] =1,2,---. Then
for

_ 00 161-24 _(82124)$2
“0‘27 ‘3 ,
i=1 3

the Cauchy problem (1.1) has no local solution in L1.

12

CHAPTER 2

Nonlocal Existence of a Solution

for the Dirichlet Problem

In this chapter, first we will prove for the one-dimensional Dirichlet problem with the

explicit initial data,

ut(:r,t) = um(a:,t)+u3(:r,t) in (—1,1)X(O,T),
u(:l:1,t) = 0, (2-1)
u(z,0) = eke-W" in (—1,1),

that the solution blows up in finite time for some constant c in the initial data. In
Section 2.2 we will establish much stronger result; that is, the nonlocal existence of

L1 solutions to the same problem for some initial data uo G L1.

2.1 Blow up of Solutions for the Dirichlet Problem

In this section we will use the dilation argument to reduce the Dirichlet problem (2.1)

to another problem as follows. Now let v be the solution of

13

v¢(x,t) = vm(a:,t) + v3(:r,t) in (——k, k) x (0,T),
v(ik.t) = 0. (2.2)
v(a:,0) = ce“"’2 in (—k, k).

Then it is clear by the dilation argument that u(r, t) = ku(ka:, k2t) is the solution of

the original initial and boundary value problem

ut(a:,t) = um(:c,t) + u3(a:,t) in (—1, 1) x (0,T),
u(ztl,t) = 0, (2-3)
u(x, 0) = eke—kg“:2 in (—1,1).

Note that the dilation argument preserves the L1 norm of the solution; i.e.,
H u ”Ll(_1’1)=“ u ”Ll(_k,k). As it was discussed in the introduction, if it can be shown

that v blows up in finite time T, then u will also blow up in finite time k—Z.

Now we will prove that the solution 2) to problem (2.2) blows up in finite time.

By Fourier series, the linear solution to problem (2.2) is

k
vL($at) 2] Kd($,y,t)C€_y2dy
—k

where the Dirichlet kernel Kd(:c, y, t) is given by

 

)_°°1inj7rk+y.j_7rk+r
a(,trvy. )—;ke_J—T —(—2— ——)Slnk( 2 ).

l4

Then the solution of (2.2) is

k: t 1:
watt) = (Katine-via + / /.K"("”’y’t — s)lv(y. s)l3dyds

and the L1 norm of v in (—k, k) is

k k k
[u(x,t)dr = f / Kd(:r,y,t)ce’y2dydr
—k -Ic —k
k t k
+ f j / Kd(:r.y.t-s)lv(y.s)l3dydsdrc
—k 0 -k

t k k
2 / / Iv(y.s)l3/ Kd($,y,t-3)d$dyd8- (2.4)
0 —k —k

where

k k oo . ,
l_'2«2 __ , 371' k+:1: , j7l' k+y

t_ d —— (t 3) _ _. _ _ .

/_de(2:,3;, s) a: [4: :1 e 1:7 sm k( 2 )sm k( 2 )dx

Prl

Since

00 '232 . k
E |e-fi:r<t-s>sin%(%)| < oo,

1:1

we can integrate the Dirichlet kernel with respect to :r and get

 

is co . k ,
1 '21?” it k+ 7r k—l—a:
_ .. z- —Lr(t—s>- 9_ __£ - J_
[k Kd(:r,y,t s)d:t: ke k Sin k( 2 )1— sm k( 2 )da:

i=1 ’°

00 2 _ _ 7T 19—]-
= _ _ _ J La—(tS) J __31

Elia“ ( 1) 1e .. ,( 2 >

15

Now let us denote the x-integral of the Dirichlet kernel by H k, i.e.,

OO

2 -—'-’-vr'~’ _8, '7r k+
H.(t—s,y)=§j3;[1-(—1rletr~ <‘ wife-211).

i=1

Our goal is to show that Hk(t — s,y) 2 C for y E [‘7", g] and 0 < s < t < T, for some

constant C', so that we will get by (2.4),

k t k
/ v(:r,t)da: _>_ 01/ / [u(y, s)|3dyds
—k 0 —k

for some constant Cl. By using this estimate for the L1 norm of v and some technical
lemmas, we will prove the finite time blow up of solution in L1.
Note that H k(t — s, y) = l for the Cauchy problem for which we have already shown

the nonexistence of the local solution.

Now we will give some technical lemmas which will be used to prove our main
blow up result in a bounded domain. In the first lemma we will prove that v is a

decreasing fimcion of a: for all a: E [0, k] and t 2 0.
Lemma 2.1 vx(a:,t) S 0 for all a: E [0, k] andt 2 0.

Proof of Lemma 2.1 By the initial condition u(cc,0) = ce‘xz, we have
vx(a:,0) = —2cze‘m2 g 0 for all a: 6 [0,19]. Since the solution is symmetric;
i.e., u(zr,t) = v(—:1:,t), we get vm(0,t) = 0. By using the maximum principle for the
heat equation, we have v(:z:,t) 2 0 for all :1: E [0, k] and t 2 0 which implies that
vz(k,t) S 0 by the boundary condition v(k,t) = 0. Since vx(0,t) = 0, o3(:r,0) S 0
and vz(k, t) S 0, again by the maximum principle we conclude that 121(33, t) S 0 for

alle [0,19] andtZO.

16

In the following lemma it will be proven that Hk(t — s,y) 2% for y E [—’c 5, 5,
0 < s < t < T, for k big enough, which is the crucial estimate to prove the blow up

result for this Dirichlet problem.

Lemma 2.2 For any fired T > 0, 3kg such that Hk(t — s,y) 2 % for k 2 kg and
y€[’7k,§],0<s<tST.
Proof of Lemma 2.2
_ _oo___1-j_"_’3i£#<t—s)
H.(t s,y)—£1341 ( )lsm k 2 )e .

Let aj = %[1— (—1)j]. Then

Hk(,)=t—sy Zajsin—(l—C-g— —-——ye)—k22_7"(t 3).

00

Note that Z a]- sin j 7ru is the Fourier Sine series of 1 on [0, 1]. Then standard Fourier
j=1

series theory claims

00

gaj sinjiru = 1 on [i g].
If we replace u by— -1-(k—-2— + y), then
°° . , jrr k+y _ —k k
Zajsmk( 2 )—1 on [2,2].

i=1

Also note that since {aj} is monotone decreasing and partial sums of sin jiru are

17

bounded, 2a] sinjiru converges uniformly on [3, 21' Now we choose 6 — -. Then

3': —1

[2a,- sinjrru] <8 —

j>s

3M s.t

Vs 2 ManduE [i2].

_ 2,2
Let Elj = a]- sinjtru, T8 = :67. and bj = e k2 (t—s). Then,

128

Hk(t — 87 y)

(X)
= :6.ij
alibi-'l Z c~labil-

:Mzii

Z
= j=M+l
00
First we will estimatel Z djbj|,
j=M+l
00
[251%]: IZCT-Tml
j=M+1 j=M+1
= | 2 73b.- :3 "obj—1|
j=M+1 j=M+2
oo
= ITM+1bM+1+ Z: Tj(b3_bj-1)l
j=M+2
S ITM+1|bM+1+ Z llelbj-bj—il
j=M+2
00
= ITM+1|bM+1+ Z |Tj|(bj_1—b,-)
j=M+2
1 1 l
S ng+1+§bM+1 =4bM+1
1
S 4 since bM+1§1, (2.5)
M
and next we will estimate 2 Zijbj
j=1

18

M a
up!
“0"

n
K9.

+
[V] :
um
Co?

I

J=l J=1 3:1
1 M
2 1—g—Za,(1—b,)
j=1
1 M
2 1 — g — 240- b,) since la,| g 4
j=1
1 _MQWQT
2 1— g — 4M(1 — e—s— ). (2.6)
Now combining (2.5) and (2.6),
M 00
Hie“ ‘ 8.31) Z Zéjbj —| Z éjbjl
i=1 j=M+1
1 —M27r2T 1
Hk(t—S,y) Z 1—§_4M(1—6T)—Z

1 —M27r2 _

= 1—3g—4M(1—e_7_k T).

Since T and M are fixed (T will be determined after Lermna 2.3), we choose It: so that

 

_ 2 2‘
k2 2 M 7r1T
111(1— m)
which implies that
_ 2,2 _
4M(1— e—a—‘Z 7‘) g %.

Consequently we have

1 _ 21:2
Hk(t—s,y)21—3§—4M(1—e—v—"Z T1) 21—4

OOIF-i
OOIH
NIH

19

So Lemma 2.2 will follow.
Since Hk(t — s ,y) >1-2on [—k 2, S] by Lemma 2. 2 and since the solution 2) is decreasing

on [0, k] by Lemma 2.1, we get

Ic t k

/ v(:t:,t)da: 2 fo/k. |u(y,s)|3Hk(t—s,y)dyds

—k g
[0 [90(3) 3)]3 )(Hk t— s ,y)dyds
_ 3
2 / /__Iv<y,s>| dyds
1 t k
— f / lv(y,s)l3dyds- (2.7)
4 0 —k

Before proving that the maximal time of existence for v is finite; i.e., v blows up in a

IV

IV

IV

finite time, we observe the following.

For each It > 0, let Tk be the life span for S2,, = (—k, k). It is by the maximum
principle that {T1,} is a decreasing sequence, since the solution on the domain Qk+1
with the existence time Tk+1 is bigger than the solution on T2,, with the existence
time T1,. If we can also show that {Tk} is bounded, then {Tk} has a limit. In the
following lemma, it will be proven that the maximal existence time for v is finite for
k big enough; i.e., {Tk} is bounded, and the L2 norm of 1) goes to infinity in finite
time. So by using H61der’s inequality it will follow that the L1 norm will also go to

infinity at that time since I: is fixed.

Lemma 2.3 Tk < 00 for k big enough.

Proof of Lemma 2.3 To prove Lemma 2.3, we will use the following abstract theorem

by H. A. Levine [12].

20

Theorem. (Levine) Let u : [0, T) ——> D (D is a dense domain in a real Hilbert space

H) be a strongly differentiable solution in the D norm to the problem

13%:- = —A(t)u +.7:(u(t)), t E [0,T), u(0) = uo, no 6 D

where f(0) = 0. Suppose that, for each t, A(t) and P are symmetric and

i) ($,Px) > 0 for all a: E D,:z: 74 0;

ii) (2:, A(t):z:) Z 0 for all a: E D;

iii) (an, A(t)a:) g 0 for all a: e D;

iv) condition F1 is satisfied where

(F1) .7 : D —> H has a symmetric Frechet derivative .75; at each a: E D, a: i——> f} is
strongly continuous, the scalar valued function 9 : D —+ IR defined by

mm) = /1(;(p.),..)dp
(the potential associated with .7) satisfies

2(a + 1)g(a:) S (x,.7:'(:r))
forallccEDandsomea>0.
Finally let uo satisfy

9(u0) > —(u0, A(0)u0).

21

Then the interval [0, T) of existence of u is bounded, and in fact,

T < (2a +1)(uo,Pu0)
_ 02(2a + 2)(g(u0) — -;-(u0,A(0)u0))

 

while also

hm f <u<n),Pu(n)>dn = +oo

t—+T-

and consequently
lim sup(u(t), Pu(t)) = +00.

t—iT-

For the Dirichlet problem (2.2), we have P = I, A(t)v = —vn, f(v) = v3 and the

domain 52;. = (—k, Is). So the conditions of Levine’s Theorem are satisfied as follows.

It is obvious that A(t) and P are symmetric for each t and conditions (i) and (ii) are

trivial. Since A(t) = 0, (iii) is also satisfied. Condition (iv) holds for a g 1 since
9(33) = fol(.7:(pa:),r)dp = fol p3r4dp = i154 satisfies
2(a +1)g(r) S ($6,.7:(:1:)) when Oz 3 1.
Also 17(0) = 0 and the initial data no = ce’m2 satisfies the condition

1
9(120) > 2(00’ A(0)u0) when G = 4

22

as follows,

NIH

which implies that

 

Since

k
k k :7"
4 / (e432 — 232e'2z2)dx 4 / e—hzda: 4 2 \/2e“4$2d:1:
—k

 

 

 

I. S ;’° 3 7' . = m;
/ 8—412da? / e’4‘E2da: 75 3‘4“:de
-k —k 3%

_32

it is enough to choose c2 > 6, in particular we can choose c = 4; i.e., v0 = 4e

Then by the theorem above the interval of existence [0, Tk) for v is bounded. In fact

 

Tk < (201 + 1)(1’0, ’00)
- a2<2a + woo) — two. A(0)v0)

while

t

lim (?I(n).v(n))dn = +00-
t—iTk— 0

23

By Hélder’s inequality, we have

2/3

t k t k
lim [Of—k(u(y,s),u(y,s))dyds£ lim 0 (1k |v(y,s)|3dy) (2k)1/3ds

t—+Tk— t—iTk-

which implies that

t k
lim / / [u(y, s)|3dyds = +00.
t—m: o -k

Now we recall that

k 1 t k
/ v(.r,t)d:c Z —/ / [u(y,s)|3dyds
—k 4 o —k

and we get

k t 1:
lim v(:r,t)da: 2 lim %/ / |v(y,s)|3dyds = +00.
0 —k

t—+Tk_ _k t-iTk-

Hence,

1:

lim v(.r,t)d:r = +00
t—*Tk- —k

which means that the solution 1) of problem (2.2) blows up in finite time in L1.

Next, we will make sure that Tk g T from Lemma 2.2. Since we choose the initial

_ 2 . . . .
data as 120 = 4e 3 , the max1mal ex13tence time for u 18

24

 

 

T < (1,0,50) _ ffk16e“2“2dz < ffoo16e-2xzdx _ 16%
k _ 0250 _ 0230 _ 02160 — 01250

where

 

(2a +2>(g<vo) — i<vo,A(o>vo>)_

fl” = (2a + 1)
Since a = 1 in our case,
4 1
160 = §[g(1’0) - §(’Uo, A(Ol’lioll-
So,
1 1 k 4 —42:2 1 k —21'2 3 2 —2:1.-2
lg(v0) — 5(UO,A(O)’UO)] = 4- &4 e do: — 5 k(32e -— 4 a: e )da:

1: 2 k 2 k 2
= 64 f e‘“ da: — 16 / 12‘2"” +32 / 626—2z dr
—k _k -k

k k k
7' 7" 7'
Z 64 2 e'432da‘ — 16x/2/ 2 e‘4x2dm + 32\/2 2 2:62e—4x2da:
it it at
$5 2 —4 2
= (64 — 16¢§ + 64\/§:L' )e 3 do:
5%
1
2 f (64 -— 16\/2 + 64\/2x2)e—4z2d:r
—1
2 (64 — 16\/§)e_42 = 1.5155 2 1,

which implies that ,80 Z 1;.

25

Therefore

16 I! ,, _
Tkg 2<163f=150397gl6=T

 

 

So we can take T = 16 in Lemma 2.2.

By combining Lemma 2.1, Lemma 2.2 and Lemma 2.3, we prove that the solution 1)
of the problem (2.2) blows up in finite time Tk which implies that the solution it of
the Dirichlet problem (2.3) also blows up at £5» by the dilation argument. The next

theorem follows from the above.

Theorem 2.1 The solution to the Dirichlet problem

ut(a:,t) = um(a:,t) + u3(a:,t) in (-—1, 1) x (0, T),
u(ztl, t) = 0, (2-8)
u(x,0) = u0(a:) in (—1,1).

blows up in finite time for no = 4ke‘k232 E L1(—1, 1).

2.2 Nonlocal Existence of Solutions for the Dirich-

let Problem

In the first section, we proved that for some initial data no 6 L1(—1, 1) the solution
to the Dirichlet problem blows up in finite time. Now we will prove a stronger result;
namely, there is no local L1 solution to the same Dirichlet problem for some initial
data no 6 L1. First by using a cutoff and lower solution argument we prove some

estimates for a family of particular initial data and then we will construct the initial

26

data no E L1(—1, 1) for which there is no local L1 solution to the Dirichlet problem.

Throughout this section we denote the solution of the Dirichlet problem by ud.

udt(xit) : udzx($7t) + ud3($at) in (—1)1) X (0,T)
ud(:l:1,t) = 0, (2.9)
ud(x, O) = ke‘kz‘”2 in (—1,1).

By Fourier series, the linear solution to the problem (2.9) is

1
udL(a:,t)=/ Kd(x,y,t)ke_k2y2dy
-1

where the Dirichlet kernel is

1 1
+y)sinj7r( +3

 

 

oo
Kd(a:, y, t) = Z (3.2"!2t sinj7r( ).

3:1

Then the solution of (2.9) is

t 1
ud(z,t)=udb(a=,t>+/ f Kd<x,y,t—s>ud3(y,s>dyds.
O -—1

Difficulty. The difficulty for the Dirichlet problem is that we do not have a good
explicit formula for the solution in a bounded domain as we have for the Cauchy
problem. In the proof of the Cauchy problem, f3 Kc(a:,y,t — s)da: = 1 is used to

simplify the argument. For the Dirichlet problem we do not have that but we know

27

that fa Kd(:1:, y,t)da: exponentially decays for t > 0.

Idea. The idea to prove the nonlocal existence for the Dirichlet problem is to use
the nonlocal existence for the Cauchy problem (which is proven in Chapter 1). To
do this, we first construct a cutoff function and estimate the Dirichlet kernel Kd in
terms of Kc. Then using this estimate we find a lower bound for the linear solution
of the Dirichlet problem it} in terms of the linear solution of the Cauchy problem
1161’. Combining the estimate for K4 and it}, we estimate Hudll L1 and obtain the

nonlocal existence for the Dirichlet problem.

In the following lemma we estimate the Dirichlet kernel Kd in terms of the Cauchy

kernel K 6.

Lemma 2.4

l l
K x, ,t >e‘12‘ —
d( y )— (1+(as—y)2 1+;

 

 

”(C(37) y, t)

when la: —y| S Jfi and |x| 3 %.

Proof of Lemma 2.4 The idea to prove this lemma is to construct a cutoff function

9 so that the fimction

10(23,y,t)= ch($,y,t) _ Kd($1yat)

will satisfy the following conditions:
(i) L(w) _<_ 0 where L(w) = wt — w”,

(ii)w§00ntheboundary|x—y|=-‘}——2.with|x|Siandatt20.

28

Since L(Kd) = 0, we want 11) to satisfy

L(’UJ) = L(ch —' Kd) = L(ch) : L(g)Kc _' 2VKc-Vg S 0.

 

 

-2113 2 -2113
4!. — — 4g _ _
Kc(x,y,t)=e\/m implies that we: (it 3’) 61/1771: =EE—3QKC.

So 9 must satisfy

wa=umxel$;wKansm

 

that is,

 

$30.

$—
gt_gm+( ty)

To satisfy this inequality and to have 0 on the boundary (y = x :1: i), we choose the

cutoff function as

 

1 1
1)

:13, ,t =e‘at —

where a will be determined later. So we want

(z—y) 1 1 6x-y2—2
g$=—a( 2— 1 ( )
t 1+(113—y) 1+5

 

 

 

 

gt — 93:2: +

 

 

(HI-y)
t

IfIx-yl 2 ‘y—g,thengt—9xx+ gar SOfor anyaZO. If lx—yl < %,weneed

to find a so that this expression will be negative.

29

2—6Sx—y)
(1+(I-y)

2
)
(1-1)

 

So, if we choose a 2 ; i.e., if a _>_ 12, then we have

 

(iv-y)
t

gt_g:r:r+ 91:30-

Set a = 12. Then the cutoff function is

1 1

9(x1yat) 28—12% —

1+(:z:—y)2 1+§)'

 

 

So L(w) S 0 and (i) is satisfied.

Now we will show that condition (ii) will be satisfied by w with this cutoff function;
i.e., we will show that w 5 O on the boundary and at t = 0. Since g(:r, a: :i: i, t) = 0,
w s 0 on the boundary, In: — yl = 71? and |a:| S 31-. To show that w 3 O at t = 0, it is

enough to show that

14-715
ling w(a:,y,t)h(y)dy§0 for every hZO.
—¢ z_71§
Since
. “T15 “75
$1333 1 w(x,y.t)h(y)dy = f 1 (9(x,y,t)Kc(x,y,t)-Ka(x,y,t))h(y)dy
11—75 I—72-

s Ch(a=) — h(r)

whereC§1,wehavewSOatt=0.

Hence, (i) and (ii) implies by maximum principle that, w S 0 inside the domain which

30

means

I l
K :13, ,t >e"12‘ —

 

 

”(C(23) y, t)

when lx—yl S 71—; and |x| S i.
In the following Lemma we use the estimate for Kd to get an estimate for udL which

is the linear solution of the Dirichlet problem.

Lemma 2.5

udL(:L',t) 2 c1e_12‘ucL(:c, t)

l

4, with ud(:c, O) = kekzxz, where Cl is a uniform constant independent of

when |:z:| S

11:.

Proof of Lemma 2.5 To show this estimate for the linear solution of the Dirichlet
problem, the main idea is to use the crucial estimate for the Dirichlet kernel that we

proved in Lemma 2.4.

The linear solution for the Dirichlet problem is

l
udL(-'c,t)= / Kd(z,y,t)ke-"2y2dy
—l

31

By using the estimate of Lemma 2.4 for Kd(:c, y, t), we have

1
ud"(:v,t) = Katine-”12y
-—1

$+-‘/——2 k2 2
2 / Kd(a:,y, t)ke_ y dy since lac — ylS
3-715 :72

 

 

 

 

:n+ 1
7' l l
2 “(a-12% , — ,)K.(a:,y,t>ke-’°2y2dy
#715 1+ (1: - y) 1+ 5
l
3 l 1 1
> ce‘nth :17, ,t he ’k2 3’2 d :1: — > c
_ /_ (:1) 1111-4 (1+(I—y)2 1,_%)_
% -1==_-11L2
—12t e 4‘ —k2 2
= ce ——ke y dy
s «4
I:
—12t__ 37 2 —(i7—‘“‘ ‘)s 2
= ce e2—H «c t ds where 3: kg
V47rt :_'=
_+ 4H2 2 k
= ce—me T‘ W" ‘ e—(ir‘.:'°f‘><s———r..:: Nd,
v41rt :45

22

—k z
_12t61+4'=" V4k2t ”2 _U2 1 + 416% km
= 66 e du, u = —(s — ———),
\/47rt \/1+ 4k2t u, 4k2t 1 + 418::

u _ /1+4k2t(_§_ km ) u 1_+4k2t(k_ k2: )
‘— 416% 4 1+4k2t’ 2: 4m 1+4k2t
_k22

—12t k__el+4k2t 112 82—” . _ .1.
ce fi_1+4k 0 du 1f 4S:1:<O

 

IV

_kz 2

_ 144th 0 _2 .
ce12tJ'ie———l———Z+4k uleudu 1f OSxS

l
4

V7?

c1e_12‘uc[‘(z, t),

IV

IV

. 0 _ 2 _ 2
where c1 = 6% and c’ = nun(c1',02’) where ful e " du 2 c1’ and f0"? e " du 2 c2’.

32

Now by using the estimates in Lemma 2.4 and 2.5, we will prove that there is no

local solution to the Dirichlet problem in L1.

Lemma 2-6 “udHL1((-1,1)x(0,6)) = 00 for any filled <5 > 0 with no = 2:ckke"‘2$2

where the ch ’3 will be determined later.

Proof of Lemma 2.6
For any fixed 6 > O,

6 1
”udHL1((—l,1)x(0,6)) = f / ud(x, t)d:cdt
o —1

2 f 6 [1 jot [.11 Kd(a:,y,t—s)(udL(y,s))3dydsdxdt
Z [06 [:fo :Kd( 1‘ y,t —3L)(Ud (y, 5))3dyd3d17dt
2 f / [W ys, ))3 _1 Km, y,t —s)dxdydsdt

l
4

“HA

By using Lemma 2.4, we get

A

i 4
Kd(:t:, y,t — s)da: Z / ce—12(t—‘)Kc($, y, t — s)d:r
-1

T
l _%_12;d
4 6—4———t-a

lulu-

—12(t—s)

> ce

_ :41\/47r( (t — s)
e—-12(t s) “:36 a: _ y

= ‘ where u = _—

73%.;6 4(t — s)

e--12(t -8) 7346—81“

2 —:—/

2 Ce fa“ 5') where C = constant.

33

By using the fact that

k22

36 £732:
8))3>Z(13+4k23) 3/2’

 

4 Kd(:1:,y,t — s)d:rdydsdt

llud||L1((—1,1)x(0.6)) Z [Gt/01‘1“ 1
Z

2 f f (u, (,)ys)3(Ce“l2(‘"‘))dydsdt

12
lb

N" MHA

M“ NH

2 /6/:; 41Ce"12(‘—“3)(ole—12314611“3))3dydsdt by Lemma 2.5,

= /060 Col e"12“24"/_:1 (2161‘ (y, 8))3dydsdt

31:2 2

“313381-4413:
12t— 243
> [Daft—Cele [cf—:26 (1+4k23) 3/2 dydsdt

2 20/06 [010:3 +4k2s -——d3dt where C’ =constant

: SCI/6Ck31n(1+4k2tt)d

 

1
Now as we discussed for the Cauchy problem in Chapter 1, we choose ck = W
which will implies that k— - e3“ — j, j- — 1, 2, .So for
00 148—(821242)$
11,0 = Z2 —8j2
i=1 jz
Lemma 2.4, 2.5 and 2.6 prove the following theorem.
0° 1 . 1.2
Theorem 2.2 For 110 = ZFeflde—(e2 4W, Dirichlet problem (2.9) has no local
j=1

solution in L1(—1,1).

34

CHAPTER 3

Nonlocal Existence for the

n-dimensional Dirichlet Problem

In Chapter 2, for the one-dimensional Dirichlet problem, we proved that for some
no 6 L1 there is no local solution in L1. In this chapter we generalize this result to
the n-dimensional case by using the similar argument as in the one-dimensional case.

We also prove the nonlocal existence of solution by generalizing the nonlinearity
in the n—dimensional Dirichlet Problem

Moreover, since we prove all of these nonlocal existence results for the cubic
domain {2, for general n i.e., for Q = (—c,c) x x (—c,c), where c is a positive
constant, we generalize all these results to an arbitrary domain 9 C R" by comparing

the solution for cubic domain.

Throughout this chapter we denote, a: = (2:1, . . . ,zrn), d2: = dxl . . .dzn and

|:t:|2 = 2:12 +---+xn2.

35

The Dirichlet problem in n—dimensions has the form

u,(z,t) = Au(:r,t)+|u(a:,t)|¥21u(x,t) in n x (0,T),
amt) = 0 on an x (0,T), (3.1)

u(x,0) = u0(a:) in S2,
wherefl= (—1,1) x x (—l,1) CR" forn> 1.
First we will prove the result for the cubic domain; i.e., S2 = (—-1, 1) x- - -x (—-1, 1) C IR"
and then we generalize the result for an arbitrary domain 9 C R" in Section 3.2.

To repeat the similar argument as in the one-dimensional case, we first prove the

nonlocal existence of the Cauchy problem for the n—dimensional case with
u0(:c) = Z ckkne’kzzz.
k
3.1 N onlocal Existence of the n—dimensional

Cauchy Problem

The Cauchy problem in n-dimensions has the form

ut(a:,t) = A143,” + |u(z,t)|;u(x,t) in IR" X (0,T): (3.2)

u(x,0) = u0(a:) on R",

where uo(:z:) = chk"e—k2$2. The linear solution to this n-dimensional Cauchy
k

problem is

ucL(a:,t) = Kc"($,y,t)u0(y)dy,
Rn

36

where the kernel K c" for the n-dimensional Cauchy problem is

-|r-;z|2
e 4:

Kc "(33 3M): W

So the linear solution is

_k2|1!2
ckkneu+4k t)
L _
uc (2:,t) — E
k

 

(1 + 4k2t)n/2'

Then the solution to (3.2) is

t
mu) = uc"(:v,t) + / 1mm» — s>(uc(y,s))%+1dyds
0 IR"

2
> / Ken”, lat - 8)(ucL(y, 8))%+1dyd3
0 R"

 

where
-+1 (%+1)_(2+1)k21212
n " (1+4): c)
< 2 z k k e
k (1 + 4k2t)%(:+1)
For any 6 > 0,

5
lluCl|L1(R"x(0,6)) = [o/uc(a:,t)dxdt

6 t
> f f / Kcn<a2,y,t—s)(ucL(y,s>>%+ldydsdxdt
fl Rn
/O:/:(/ ucL (y,s ))%+1/ KC"(:c, y,t —3)d:1:dydsdt

(influx?
> ZCk§+1kn(— n‘+1)e W
n(l)+4k2t%(:+1)

Kc"(:z:, y,t — s)d:cdydsdt.
Rn

V

 

37

As in the one-dimensional case, the integral of the Cauchy kernel on IR" is also equal

to 1 for the n-dimensional case, i.e.,

2

_I—

Kc"(a:,y,t — s)d:z: =/ id]: = 1.
Rn

Rn (47Tt)n/2

So,

(2+1)k2y2

ck %+lk2+fle- (1+-41:25)
”ucllL1(IRx(o, 5)) > 2:1: / [uh (I +4k2s)1+'3 dydsdt.

 

%+1ky \/1+4k23
——=u, and dy=——
Vl+4k28 '/Z+1k

fi+1k2+n 1+4}:2 3 n _u2
lludllLl(Rnx(0,6)) > 23/0: [0: fuck (1+4k2s) 1+3 (\/2_+1k) e | | dudsdt

1 2
= :06 Cki+( k 6""IQdudsdt
kfi+1)]:2 7r 9.
= Cu Eff/0t: +504“ ———dsdt where C',,=(%+1)2

By substituting, du, we get

 

 

 

2 0 23/6: —ck%+11n( 1+4th tt)d
2 On 2/24 —ckn+11n (1+4k2t)dt
2

Cn:6ck%+11n(1+ 41:26).
I:

38

Now we choose ck such that

i) Zak-3+161n(1+ 211926) = 00; and
k

ii) Zak < 00 for no E L1.
1:

Similarly, we will find a sequence {kj}J°-:1 such that ch 714 0 only when k = kj. For k

 

big enough,
6ck%+1 ln(l + 2k26) 2 ck%+1\/1n k.
If we set cg“ Ink = (lnlc)1/4 for some 19 = k,- —2 00 as j —+ 00, we get
Ck: -——1——-, fork: kj,j=1, 2,
(lnk)8+4n

So (i ) is satisfied.

.. l l _61-_8_n
For (11), we set ck = ——I —which implies that k— — e] " and
(111 k)— 8+4n :j2

00 | n 16
FEE—6'” 6 8 6‘” " “I" e L1(IR")

2
i=1 ‘7

for which there is no local solution for the n—dimensional case. So the next theorem

follows.

00
1 , n . .
Theorem 3.1 For u()(cr =28 —28" e(e2 “1'2, the n-dzmenszonal Cauchy

problem (3. 2) has no local solution in LICK").

39

3.2 Nonlocal Existence for the n-dimensional

Dirichlet Problem

In this section, we will prove the nonlocal existence of solutions for the n-dimensional
Dirichlet problem. The idea is to use an argument similar to that used in the
one-dimensional case by using the nonlocal existence result of the Cauchy problem

in n-dimensional case which was proven in the previous section.

The n-dimensional Dirichlet problem is

u.(x,t) = Au(:c,t) + |u(x,t)|%u(x,t) in a x (0,T),
u(x,t) = 0 on 09 X (0,T), (3.3)
u(m,0) = 2c]..,llc"e"‘72"‘2 in Q,

I:

wherefl=(—l,1)x---x(—1,1)CR"forn>1.

The linear solution to the n-dimensional Dirichlet problem is

mat) = /_ - - - /_ K."(x,y,t)uo(y>dy

where n—dimensional Dirichlet kernel Kd" is

 

 

. . 1+ . . 1+ n
"(aw/t.) _Z Ze'(j12++j"’r2tsmj17r( le)...smjn7.( 25")

171:1 31:1

)---sm]n7r( .

 

 

X sin j17l'(

4O

To generalize the nonlocal existence to the n—dimensional case, we prove the
following lemmas which are generalizations of the lemmas proven in Chapter 2 for

the one-dimensional Dirichlet problem with some technical differences.

To get an estimate for the n-dimensional Dirichlet kernel K4" in terms of the
n—dimensional Cauchy kernel K c", by a similar argument that we used for the
one—dimensional case, we choose the cutoff fimction as a product of cutoff fimctions

with a = 12n in the following lemma.

Lemma 3.1

n - n n 1 1
Kd($1yat)zel2tH(l+($._y.)2—l+l
1 z 2

 

 

i=1

when |a:,--y,-| 3 % and |$,| Sifori=1,...,n.

Proof of Lemma 3.1 To prove this estimate for the n—dimensional Dirichlet kernel,
we use the cutoff function argument as in the one-dimensional case. First we denote
the n-dimensional cutoff function by g... The idea is to construct gn as the product

of the cutoff functions in each dimension. The function 21),, which is given by

“171(3), ya t) = gann(1L‘,y,t) _ Kdn($, ya t)

must satisfy the following conditions:
(i) L(wn) S 0 where L(wn) = 11)... — Aw",
(ii) wn S O on the boundary lac.- — yil = 3‘5 with |x,~| g % for'i = 1,--- ,n and at

t=0.

41

Since L(Kd") = O, we want 11),, to satisfy

L(wn) = L(gann _ Kdn) = L(gnK on)— — Lg( anc _ 2VIQ" vgn S 0

-1211:
e 4:

where Kc"(,:1: y, t—) — W and implies that

So gn must satisfy

L(g,,)Kc" + E;—y)Kcn ' Vgn < 0;

that is,
x _
g,,t—Agn+£t—y)-Vgn 30.

To satisfy this inequality and to have 0 on the bmmdary (y,- = 1:.- i —\}——2-), we choose

the n—dimensional cutoff function as

 

 

e—at
gnhxy’)_1:1(1+(x.l)2—1+

42

So gn will satisfy

 

 

 

 

(:3 — y) —-at n 1 1
9n _ A91: + Vgn = —ae I I _ _.
Zn 6(13,‘ - yi)2 — H" 1 l
e (M (1+(501- — yil2l3 k—1(1+($k — 31k)2 1+ £5»

k¥i

_eat n 2(a):- _. yi)2
(12:; t(1 — (2:,- - 3102)?) S 0

 

 

(av-y)
t

Iflxi—yilZ:j—gforz':l,...,n,thengnt—Agn+ -Vgn5_0foranya20.

If lat,- — yil < —‘}-—§, we need to find a so that this inequality will be satisfied.

71 2-6(:r.--—y.-)2

——T.—T5
If a Z 2 (1Tb. 3") ) 1 ; i.e., if a Z 12n, then we have
.-=1 (———:.+(.,_.,,.) ‘ fig)

 

gut — Agra +

(_$_;_y_)..Vgn-SO

So, it is enough to choose or = 12n. Then the cutoff function is

 

 

" 1
971(xay7t) = 8-12,“ H( 2 _—
i=1 1+(ZL'i—y3‘) 1+

By the choice of gn, we have L(wn) _<_ O and (i) is satisfied.

Now we will show that condition (ii) is satisfied for the cutoff function; i.e., we will

show that wn S O on the boundary and at t = 0.

Since gn(:c,:r,- :t ‘—1r§,t) = 0 when lac,- — 3),] = i and |a:,-| S l1 11),, S 0 on the boundary.

43

To prove that wn S O at t = 0, it is enough to show that

2n+ 1 x1+ 1
7' 7'
lirré 2 2 wn(x,y,t)hn(y)dy S 0 for every hn Z 0.
urn-715 m1—715
Since
311+ 1 31+
7 7"
hm 2 ' 2 wn($ayvt)hn(y)dy
t—oO x _ 1 11- 1
" W 75
. 3n+§5 31+ 12
= 111% 1 ' (gn($,y,t)Kcn(.’L‘,y,t) _ Kdn(xi y’t))hn(y)dy
$n"$ 31—75

3 Chn(:r) — hn(a:)

wheregnSCS1,wehavewngOatt=O.

Hence by the maximum principle, wn < 0 inside the domain which means

12 — n n 1 l
Kd (33,y,t)28 12 tl-I(1+($._y.)2 _1+

 

 

)Kc"(:v. at)

NIH

i=1

when Imi—yilgfiand |a:,~|$%fori=1,...,n.

Next we use this estimate for K4" to get an estimate for udL for the n—dimensional

case by the following lemma.

Lemma 3.2
udL(a:, t) 2 cue"12"‘ucl’(:c, t)

when lay-I g i- fori = 1,...,n and ud(a:,0) = k"e"k2'x|2, where on is a uniform

constant independent of k.

44

Proof of Lemma 3.2 By recalling the linear solution for the n-dimensional Dirichlet

problem

1 1
udL(a:,t) 2/ ”f Kd(m,y,t)k”e_k2'y|2dy
-—1 -—1

and using the estimate for Kd"(z, y, t) from Lemma 3.1, we get

1 1
udL(a:,t) =/ .../ Kd"(:c,y,t)k"e—k2'y'2dy
—1 —1

 

 

 

 

 

$n+7 31+7 1
2 f 2 Kd($,y,t)k"e_k2'y|2dy since lac,- - 3),] g — Vi
...-715 x.— 71- x/i
373+ 1 31+
7' 72' 1
2 f 2" '1/ 8—12nt1:1(1($+( )2 — 1+% —TC)K "(53 yatlkne_k2|y|2dy
_ 1 i "' yi
Zn 7; 1'1— 71-
_-|:: ~y|2
371+")? “1 T —12nt n 1 1 8- n e-k2lyl2
= 1 ... e H(1+($--—y)2— 1+l2)(47rt):1/2ke dy
Inn—75 21—712- i=1 2 1
_l‘t '2
> f % —12nt e— 2? kn -k2|y|2d
_ —%. . . _% Ene —(4nt)"/2 e y
" 1 1
since :23,- — > 6, where 6 = constant
ll’4:> :_l>_‘[(1+(:z:.-—y.-)2 1+9—

_(1 -y-2)

= e—l2nt 1:1[/—:%e___\/______ l lite—kzyizdyi].
I

Now by using the following inequality from the proof of Lemma 2.4,

421.2

'2
1 x..._ .
Z e‘L‘TgyL)‘ —k2 2 kg l+4két .
——ke y‘dy,> C—— for i=1,...,n,

— \/1 + 4k2t

"_1_L)_'—y2 -k2(112+---+$n2)

4; Ice-’9 12 ~ kne (1+4k2t) _ ~ L
l:[L/L y dyil 2 C" (1 + 4k2t)n/2 — Cnuc (mat)

 

45

where C" = H Ci. Hence it follows that

i=1

udL (:r, t) Z cue—12ntucL(a:, t),

where en: (31.5,, is a constant independent of k.

Similarly, by using the estimates in Lemma 3.1 and 3.2, we will prove in Lemma 3.3
that for some no 6 L1 there is no local L1 solution for the n-dimensional Dirichlet

problem.

Lemma 3.3 For any fixedé > O, ||ud|]L1((_1,1)x(0,5)) = 00 with no = chkne—k2'xl2

where the ck ’s will be determined later.

Proof of Lemma 3.3 For any 6 > O, by the solution of the n-dimensional Dirichlet

problem,

6 1 1
“udllLl(((—1,l)X---X(—l,1))X(0,6)) =/ [1.../1ud(xat)d$dt
f: [_11" 1/1 [0 1.11”].11Kdn((x’ty’)(“dL(3/a3))%+ldydsdxdt
f: [0 ff” ff /_i ink.“ (miyit—SXWLW, 3))%+1d$dydsdt.
l l l _

IV

IV

NH

46

By using Lemma 3.1 to estimate Kd",

 

l\
AIH N'"

l
3 1
K?! t'— S)d > e(—-—12nt—s)
d (“A 5” fj" [:8 fi(1+(a:.1- 1+

1
1 _
2 i=1 2

1
~ —12n(t—s) 4 n _
cue / .H/ Kc (11:, y,t s)d:z:

 

Z
_1 :1
4 4
1 1 _ 4’?” 2
4 4 e —s
...... ~ —12n(t—s) d
— e ... (E
c" _/—_l /;1_ (41r t — 3))"/2
4 4
~ —l2n(t—s) n _yi _. .
e -a I
= 27 [ _ 4:5 54.31%] where ”1.: _<_yl
7r 4:1 4(t_y:) 4(t — 8)
-12n(t— a) fi fit:- 2
2C1; W 8-1“ duz]
ewn_—/_—2

Z Cue—12'4“” where Cn = constant.

Also by using the fact that

<2+uk21x12

( ))2+1 > Z c %+1kn<% +1>e 42TH—
uCL ) 3
(8y (1 + 4k2t)%(%+1>

 

it follows that

47

_1_
||ud||L1 2 f:/:/i Cn—e—una ”(ML (y, 3))n 3+1dydsdt

= /:6/t—Cncn8—12nt—24s‘/Jm I (ucL(y, S))%+1dyd8dt
= /6/tCnC118—12nt—248/ m/ (U CL(y, S))%+ldyd8dt
0 0 _ _

(2+uk21yl2

6 t fi+1kn(% +1) W
> / f0 Cncne_12m_24’ /... 26k e dydsdt
-1 (1+4k2t)2 %(%+1>
4
ck2+1k2
Cu '2 [6 jock (1+4k2s) ————dsdt where Cn' =const

Cn’Z/o 101$“an + 4k2t)dt
Cu '2]; —ckn+1 1n 1(+4k2t)dt

Cn’ :6ck%+11n(1 +4k26).
k

N“ M»
I
M“ an-

IV

Cne"12"(t‘s)(cue-12nsucl‘(y, s))%+1dydsdt by Lemma 3.2,

blu—
I
bla—

N'" A1.—
5"" NH

N" NH

 

M.-

IV

IV

IV

IV

1

Following the same argument as in the Cauchy problem, we choose ck = (—l_k)—’_L_
n 8+4n

_16i8n

n

which implies that k = e3 and

Hafiz

16 Ba .
u()(=a:) :32 —e "j " e(82] W2 E LICK"),

for which there is no local solution for the n-dimensional case. So the next theorem

follows.

48

00 1 . 16 l 87: 2j 16in8n 2
Theorem 3.2 For uo(:r) = Zj—2e’” " e‘(e )lxl E L1 , the n-dimensional
i=1

Dirichlet problem has no local solution in L1((—1, 1) x x (—1,1)).

Remark Note that we can repeat the proof of the nonlocal existence result for

the n—dimensional case with the cubic domain (20 = (—c, c) x- - -x (—c, c) where c > 1.

In the first corollary below, we will prove that the nonlinear term |u|%+1 in the
Dirichlet problem can be generalized for the nonlocal existence result and in the
second corollary we will show that the cubic domain can be generalized to an arbitrary

domain in R" for the same result.

Corollary 3.1 For the Dirichlet problem,

ut(:z:,t) = Au(a:,t) + f(u) in 525 x (0,T),

u(a‘,t) = 0 on 6520 x (0, T), (3.4)
u(x,0) = u0(:1:) in OD ,
°° 1 .m ,mgi“
where u()(sc) = 21—6” " {(82 ”1'2 E L1 and DD = (—c,c) x x (—c,c) C

2
i=1 3

R", c is a positive constant, if the nonlinear term f (u) 2 |u|%+1, there is no local

solution in L1 for problem (3.4).

Proof of Corollary 3.1 For the proof, it is enough to follow the proof of the
Dirichlet problem in the n—dimensional case (proof of Theorem 3.2) by using the
comparison f (u) 2 |u|§+1, since we already proved that there is no local solution to

problem (3.4) for f (u) = |u|%+1.

49

Corollary 3.2 The Dirichlet problem

ut(:z:,t) = Au(:t,t) + f(u) in Q x (0,T),
u(x,t) = 0 on 60 x (0,T), (3-5)
u(x, O) = u0(:c) in Q,

where Q C R" is any smooth bounded domain and f (u) is as in Corollary 3.1, has no
_16i8n ”ii——

00 811
1
local L1 solution for uo(a:) = Z 358“] " e‘“ ”3'2 6 L1.
i=1
Proof of Corollary 3.2 To prove the nonlocal existence result for an arbitrary
domain, we will use the nonlocal existence result for the cubic domain to compare

the solutions.

Now let a be the solution of

at(a:,t) =Aa(a:,t)+f(a) in 9:. x (0,T),
t) = 0 on 690 x (0,T), (3.6)
0) = u0(a:) in $20 ,

:1

(x,
(at,

C1

where $20 = (—c,c) x x (—c,c) C Q for some constant c > 0.

Then on QB problem (3.5) takes the form

ut(:c,t) = Au(:z:,t) + f(u) in DD x (0,T),
u(as,t) 2 0 on 690 x (0,T), (3.7)

u(m,0) = u()(at) in QC] .

If we take the difference of the equation (3.6) and (3.7), we get

50

(u-fi)t=A(u-fi)+f(U)-f(fi) in 9m X (0,T),
(u—fi) 20 on 89.3 x (0,T), (3.8)

(21—21): at t=0,in QB.

Now we multiply and divide the nonlinear term f (u) — f (11) by u — a, and get

1
f(u) — f(u) = g(u - a) where g = [0 f'(ru + (l — r)21)dr.
Let w = u — a. Then (3.8) takes the form,

w, = Aw+gw in 0D x (0,T),
w 2 0 on 6525 x (0,T), (3.9)

w=0 at t=0,in (20.

Hence by maximum principle we have w(:c, t) Z O for every (:5, t) E [0, T ] x $20 ;
i.e., u(z,t) Z u(x,t) for every (:12,t) E [0,T] x 90 which proves Corollary 3.2 since

we already proved that there is no local solution for problem (3.6).

51

CHAPTER 4

Nonlocal Existence of Solutions for

Mixed Boundary Conditions

In this chapter, we study the same nonlinear equation with the mixed boundary
conditions; i.e., considering the boundary conditions in terms of u and the normal
derivative of u, which is denoted by %.

Let 21 be the solution of the one-dimensional mixed bmmdary value problem,

11:03:13) = fizz(x,t) +u3(a:,t) in (—1, 1) x (0,T),
$0M) + (Want) = 0 m = 2:1 in (0,T), (4.1)
i2(113,0) = 710 in (—1,1)

where B Z 0 is a constant and 110 6 L1.

As we proved for the Dirichlet problem, in the first section we prove the blow up of
the solution for this mixed boundary value problem with the initial data of the form

120(3) = ce“"'2 6 L1(—1, 1) and in the second section we prove the nonlocal existence

n

Of solution for 110(3) = 2 ,—e"3 e_(e2j n

2 )(32), first for the one-dimensional case

i=1
and then for the n-dimensional case.

52

4.1 Blow up of Solutions for the Mixed Boundary
Conditions

In this section we consider the one-dimensional mixed boundary value problem,

a,(x,t) = a..(x,t) + fl3(m,t) in (—1, 1) x (0,T),
g(u) + fifl(x,t) = 0 a: = :tl in (0,T), (42)

$2

i2(:1:,0) = ce‘ in (—-1,1).

Let 11L denote the linear solution to the mixed boundary value problem (4.2). Then,

1
flL(w.t)= f K($.y,t)ce"”2dy
—1

where the kernel f((z, y, t) for the mixed boundary condition consists of the following

two sets of eigenfunctions with corresponding eigenvalues An and p",

i) cos x/Xna: with fin tan fin = ,6
ii) sin flux with fl tan x/Xn = — x/Xn
such that

,. °° NS
K 2:, ,t = e'Agt n cos finmcos fin
( y ) f2 2¢Xn+sin2fin 3’

 

 

+ i 6”"it 2W" sin n a: sin V}? y
”:1 2W7. — sin zfin " n

Then the solution of (4.2) is

53

u(a: ,t)=/:( K(a: ,y,t) ce Jydy+A/:( K(:c ,ty, s)|ii(y, s)|3dyds.

Hence the L1 norm of ii in (—1, 1) is

[:fi(x,t)da: = fl [1 K(:z:, y,t)ce ydyda:
+1.11/0 [1 Kc” 9’ 3)IU(y, 8)|3dydsdx
[0]; lu(y13)|3/:—R(,$y,— s)d:cdyds, (4.3)

By an argument similar to that used in the Dirichlet problem, to simplify the inequal-

IV

ity (4.3), we first take the (rt-integral of the kernel K; that is,

 

1 ~ 1 0° zfi
K :13, ,t — 3 d3: = / 6421“”) n cos x/Xna: cos fin dz:
/—1 ( y ) —1 "2:; 2A,, + SinZfin y

 

+ / e”1n "3 _" sin ans: sin pnydx.
_1 ”2:; Zfin — sm 2W1: f

So,

 

1 0° 1
2A,,
K a: ,y,t s)—da: - 64%“ 3) cos Any / cos Anardzz:
[.1 ( n; 2f” +sin2\/_n —1

 

+ie‘“a(t—s) zfi" sinfi y/1 sin ,u xda:
”:1 2W7. — sin 2\/'fin " _1 n

54

1
Since / sin pnxda: = 0, we get
—1

 

2x0: 23in\/_n
K( (a: , s)d:c= €_’\'2‘(t 3) " ncosfny
/_l1yt 21:1 2f" +sin 2f" \/—,,

And we denote this sum by H, i.e.,

~ °° 4 sin x/X
H t — s, = 6—)‘?‘(t_3) " cos x/Xn .
( y) 2 2A,. + sin 2x5n y

 

71:]

In the following lemmas, we will prove similar results as in the Dirichlet problem and
get the same astimate below for the L1 norm of 11 as we obtained for the solution of

the Dirichlet problem,

1 t 1
/ fl(z,t)d$ _>_ C1] / |fi(y, s)|3dyds
—1 0 —1

for some constant C1. In the first lemma, we prove that the solution 17. is a decreasing

function of a: for all a: 6 [0,1] and t Z 0.
Lemma 4.1 For all a: 6 [0,1] andt Z 0 fl$(x,t) g 0 .

Proof of Lemma 4.1

Since 11(1, t) > O by the maximum principle and fl > O, the mixed boundary condition
gn—fi(l,t) + fi&(l,t) = 0 implies that 5—3fi(l,t) = fi$(1,t) S 0 for t Z 0. By using the
initial data 1‘2(:z:,0) = ce‘xz, 11$(a2,0) = —2c:z:e“‘”2 S 0 for all a: 6 [0,1] and since the
solution is symmetric, we get 111(0, t) = 0. So, by using maximum principle, we get

11$(x,t) g 0 for all :1: 6 [0,1] and t Z 0.

55

In the next lemma, we prove the same estimate as in Lemma 2.2 for H.
Lemma 4.2 For any fixed T1 > 0, H(t — s,y) 2 % fory E [—%,-;-], 0 < s < t _<_ T].

Proof of Lemma 4.2

~ °° 43in A
H t -- s, = e”\%(‘”) n
( y) 2 2x/Xn + sin 2A,,

71:]

cos \fA—ny.

 

4 sin «in

Let a7, = . Then

2\/Xn + SlIl 2\/Xn

 

H(t — s, y) = Z (3“’\?'(t“')c1n cos x/Xny.

n=1

00

To prove the lemma we first observe that 2 an cos x/Xny convergas uniformly on
11:]

[- 1-

00
Remark. Note that 2 an cos flay converges uniformly on [—

lob-0

,

NIH

1 I

5, 5] proving the

11:1
following two claims.

Claim 1. {an} is monotonically decreasing with an —i 0 as n —> 00.

Proof of Claim 1.
To prove that {an} is monotonically decreasing, it is enough to show that an 2 an+1

when An S An“. So we need to get

4 sin A, > 4 sin fin“
2A,. + sin 2x/Xn _ 2x/Xn+1 + sin 2”,,“

 

 

When An S Aft-+1-

Multiplying and dividing this inequality by cos fin, we get

56

sin A" cos x/Xn > sin fin“ cos fin+1
cos ”(flax/3,, + 2 sin \/:\-,, cos x/Xn) — cos fin+1(2\/X,,+1 + 2 sin \/:\-,,+1 cos fin“),

 

 

i.e.,

tan fin > tan \/—n+1

mik+sinfin “ ;%+sinfn+l

 

 

Since x/Xn tan fin = 6 for An, we get

 

 

_L _.a__
fin > fin+1
\fi ' _ f 1
5.7%: + 3‘“ ‘5" 3.7%: + Si“ “3+1

or

1 > 1
arm”: + sin «1.) “ s/X..(COL+— + sin A...»

 

 

ByusingtanJ—n: A i,weget\/—,, =mr+6,, where6,, —->Oasn——>oo.
So,

 

 

fl 6 B 6,, 62
t 6,,= =—————=—1——— " —....
an(mr + ) mr + 6n n1r(1 + 7%) mr mr + (mt)2 ]
On the other hand tan(n7r + 6n) = tan 6,, = 6,, — . . . , which implies that
2
5n = E. _ ’8 +

 

Since 8602 An = tan2 fin + 1 = £72: + 1, which impiles that cos x/Xn = i4!—

\/An+132,

57

we get

 

 

1 1
> .
fin( V A'n + H2 + Sin fin) — fin+l(v /\n+1 + 52 + SIII fin+1)

And sin fin = sin(n7r + 6n) = (~1)"-f-— + - - - implies that
7r

x/X, sin i/X, = (mr + (ma—mi + ...) = (—1)"5 + 0(1).

”71'

So, it is enough to show that

fins/A. + [32 + sin WA.) 3 Vials/Ann + 52 + sin fin“)

where

fin+l( V An+1 + fl2 + Sin An+1) = (fin + fin+1 _ fin“ V An + .82

+ \/).,,+1 + 52 — s/An + W + sin x/Xw)

= finN/An + fl2 + finH + x/anin AM]
n+1

+(x/Xn+1 — x/Xn)(\//\,,+1 + [32 + sin fin“). (4.4)

 

Since («AMI—fin) = (n+1)7r+6,,+1—mr—6,, = 7r and x/Xn sin fin = (-—1)"fi+0(1),
from (4.4) it follows that

(—1)"B + 0(1) 3 7r(n7r + 0(1)) + bounded terms

which satisfies the inequality an“ s a,, for every n.

58

Hence, {an} is monotonically decreasing.

M
Claim 2. IZcos x/Xnyl is bounded.

n=1

Proof of Claim 2.

Since x/Xn = mr + 6,,, we have

M M
I: cos Vinyl = I Z cos(n7r + 6,,)y|
”=1 n=1

M
= | 2 cos mry cos 6,,y — sin mry sin 6,,yl

n=l

M
= Izcosmryh — 6.2g? + ...] — sin mrylény + 5,3313 + ---l|

n=l

M M M
= | 2 cos mry -— 2 cos mry6,,2y2 — 2 sin n7ry6,,y — ...|

By the proof of Claim 1, 6,, is monotonically decreasing and goes to zero as n ——> 00

and also partial sums of cos mry and sin mry are bounded. Hence we conclude that

M
| 2 cos Vinyl is also bounded. So the proof of remark is completed.

n=l
00
By Fourier series theory and the remark above, we conclude that Z a,, cos Any = 1
n=1

on [-—1,1]. Using this result and repeating the same argument as in the proof of

Lemma 2.2, we get

59

which completes the proof of Lemma 4.2.

Since H (t — s ,y)> — 2on —§, 5] by Lemma 4.2 and since 11 is decreasing by Lemma

4.1, we get

1
fil(:1:,t)da: Z /0/:( |u( y,s s)|3H(t—s,y)dyds
—1
Z /0/%I11(y s)I3H(,t—sy)dyds
il
1 ~ 3
2 — é|u(y,~i)| dyds
2 0 —TI
1 t 1
2 —/ / |i1(y,s)l3dyds-
4 o —1

Again by using the abstract theorem of H. A. Levine [12] which is stated in Section 2.1
we can also get the blow up result for the mixed boundary value problem. Similarly
to satisfy conditions of the theorem, we choose the initial data as 110(3) 2 46-12.

Then the maximal time of existence for 11 is

 

T“ S a2<2a + 2)(g<in >— uoanommo)

where a g 1 and 9(3) = i334, and also

t

in (n<n),n<n>>dn = +so.
t—vTo 0

60

This implies that

1 t 1
lim u(m,t)da: 2 lim —l-/ / |u(y,s)|3dyds = +00,
0 —1

t—+TO' _1 t—sTo“
1

lim u(x,t)da: = +00.
t—+T0- _1

Hence we have the following theorem for the mixed boundary value problem (4.1).

Theorem 4.1 For 110(3)) = 4e32, the solution to the mixed boundary value problem

{4.1) blows up in finite time.

4.2 Nonlocal Existence of Solutions for the Mixed
Boundary Conditions

In this section, first we prove the nonlocal existence of solutions for the one-
dimensional mixed boundary value problem (4.1) and then generalize this result to
the n-dimensional case.

The idea is to compare the solution of the mixed boundary value problem with

the solution of the Dirichlet problem for which we already proved that there is no

1 neg-1848 (82,1?
local solution in L1 for initial data u0(2: )=Ze j—Qen )zz.
Recall that the one-dimensional Dirichlet problem is
nest) = n..<s,t) + u3($,t) in (—1,1) x (0,T),
u(i1,t) = 0, (4-5)

u(at,0) = u()(rr) in (—1,1).

00
]_ , .24
where 110(1)) = Z 383248—(e23 )22.

i=1 3

61

By taking the difference of equations (4.1) and (4.5), we get

(u — u), = (i1 -— u)“ +213 — u3 in (—1,1)X(O,T),

—a%—(x,t)+fl(u—u)(x,t)=—a—nZO, :1: =41 te(0,T), (4-6)

 

k(i'I.—u)(:1:,0)=0 a: 6 (—1,1).

Note that —%n'1 2 0 by Lemma 2.1.
Let c = ii -— u. Since 17.3 — u3 = 392(11 — u), for some 9, problem (4.6) takes the

following form,

e,(:c,t) = 63$(x,t) + 3926(x,t) in (—1, 1) x (0,T),

%($,t)+fie(x,t)=—%:-ZO :1: =11 te(0,T), (4-7)
e(:c,0)=0 a: 6 (—1,1).

Hence, by the maximum principle 6 Z 0; i.e., a 2 u, which proves the following

theorem for the mixed boundary value problem.

00

1 . J.24 . -
Theorem 4.2 For {10(3) = E 356246032 >32, there is no local L1 solution to the
i=1

one-dimensional mixed boundary value problem (4 .1)

Remark. By using the nonlocal existence result for the n-dimensional Dirichlet
problem, we can generalize the same result for the n-dimensional mixed boundary

value problem with the same comparison argument above.

62

Theorem 4.3 For the n-dimensional mixed boundary value problem,

71:06 t)— Au(fl?15)+IU(:1=t)|%u(ac,it) in 9 x (0,T),
%“.:(:v t)+fiu(z t)- — 0 on an x (0,T), (4.8)
($10) = 110(3) in 9,

Q1

where Q C IR" is any smooth bounded domain and

m 16 871
~ 1 in”? _(e2i+)|z|2 1 - - . 1
uo(2:) = 2 ,—2e e E L , there is no local solution in L .
- .7
1:1

Remark. Note that we can also prove the finite time blow up of solutions for the
mixed boundary value problem, which was proven in Theorem 4.1, by comparison

argument above.

63

CHAPTER 5

Global Existence for Small Initial

Data

In this chapter we will give some sufiicient conditions on q and uo for the existence

of global solution to the Dirichlet problem

u,(a‘,t) = ux$($,t) + u3(m,t) in (——1, 1) x (0,T),
u(:l:1,t) = 0, (5.1)
u(as, 0) = uo in (—1,1).

where uo 6 L9. Mainly we will prove that for any fixed 6 > 0, if the initial condition
uo 6 L1“, then the Dirichlet problem (5,1) has a global solution in LI‘LE for ||u0||1+e

sufficiently small.

For notational simplicity, we will use the following form of the solution throughout

this chapter.

t
u(t) = e‘Auo +/ e(t_8)Au3(s)ds
0

where e‘Auo denotes the linear solution of problem (5.1). Before proving the global

64

existence result, we give some useful inequalities for the linear solution of (5.1).

Remark. Let 1 < p < 2. Then it is easy to see that for all t 2 0, e‘A : L1 —+ L1 with

norm M1 3 1; i.e.,

wenhswms

and e‘A : L2 —> L2 with norm M2 = em where '71 is the first (negative) eigenvalue of

the Laplacian with vanishing Dirichlet boundary condition; i.e.,

kmflhSéWWM-

Now we recall the following interpolation theorem in [14] by L. Nirenberg,

Theorem. (Nirenberg) Let T(t) be a continuous mapping of LP into L” with
norm M1 and L9 into Lq with norm M2. Then T(t) is a continuous mapping of

L" into L' with the norm M S M1’\M21_)‘ where 1 g p S r S q 3 co and% = %+1—‘q—A—.

We conclude by this theorem that e‘A : D” —+ LP with the norm M S MIAle—A with

A = 2—;2. Hence
Met/inn. s e‘ilg’F—annp for 1 < p < 2.

For p 2 2, we can get a similar estimate by the following interpolation argument.
e‘A : L2 —-> L2 with norm M1 = em as above. Note that for all t Z O, e‘A : L°° —> L°°

with norm M2 3 1 by the maximum principle; i.e.,

kmMmSHNm-

65

Then again by the interpolation theorem [14] of L. Nirenberg, we have e‘A : L” —-+ L”

with the norm M g M,*M21-* with ,\ = g; i.e.,
g
IIe‘AnIIP s ample”. VP 6 [2,os).

First by the following lemma, we get the estimate on ||u(s)||oo in terms of ]]u0||1+,,

which is a crucial estimate to prove the global existence theorem.

Lemma 5.1 ||u(s)]]°o g G(M, T)||u0||1+,s'2<1_1+¢7, where C(M, T) : 12+ x R+ _. 12+

is an increasing function of M and T respectively when ||u(s)]]1+6 S M + l on [0,T].
Proof of Lemma 5.1 We prove Lemma 5.1 by using the following local existence

theorem by H. Brezis and T. Cazenave, (Theorem 1 in [13]).

Theorem. (Brezis,Cazenave) Assume q > n(p — 1) / 2 (resp. q = n(p — 1) / 2 ) and
q 2 1 (resp. q > 1), n 2 1. Given any uo 6 L97, there exist a time T = T(uo) > 0 and
a unique function u E C([O, T], Lq(Q)) with u(O) = uo, which is a classical solution of

(5.1). Moreover, we have smoothing efi’ect and continuous dependence; namely
llu - ’UllLs + tn/2qll’u - ’UllLoo S 0|on — vollm-

for all t E (0,T] where T = min(T(uo),T(vo)) and C' can be estimated in terms of

”wills: and llvollns.

By replacing q = 1 + e and p = 3 in this theorem, we get an L°° estimate for the

solution u of (5.1),

66

IIU(s)lloo s C(M,T)||u0|]1+€s'flll—+s5

6 66
where G(M, T) = echJrUfiTfifiz and c1 = c1(e).

Now we prove the global existence theorem.

Theorem 5.1 For any fixed 6 > 0, there is a 6 > 0 such that ||u0||1+6 S 6 implies

that u(t) globally exists and
||u(t)||1+€ S 26 for O S t S 00.

Proof of Theorem 5.1 First, let us choose 6 > 0 such that 1—‘:—‘(G(26, 1)6)2 < —7

1-:,L‘(G'(25.1)5)2

and e < 2 where 7 = 7112—4156. By the local existence theorem that is

stated in Lemma 5.1, 3T1 > 0 such that u(t) exists on [0,T1] and ||u(t)|]1+, S 26 for

t E [0, T1]. Now we will prove the following claims to prove the global existence.
Claim 1. ||u(t)||1+, < 26 for 0 S t S 1.
Claim 2. ]|u(1)|]1+e S 6.

Proof of Claim 1. Suppose Claim 1 is not true. Then set
if = min{t > 0|||u(t)||,+, 2 26}.

We have ||u(T)||1+, = 26 and “haul... < 25 for all 0 g t < T with o < T g 1.

Consequently, for any t E [0,T], the remark in this section indicates that there is a

67

’7 < 0 ('y = 711272) such that

t
IIU(t)|Ii+s : €7tHU0H1+e+f lle(“")Au3(S)lli+sds
0
t
s €7t||u0H1+e+f si<‘-i>lln(s)1h+.nu<s)Hints
0

t
~ __1__
s erunonnfi/ e”“'i>llu<s>lh+.(a(2nTiiinnouhs ......d
0

Note that the last inequality follows by Lemma 5.1.
Let B = (G'(26, Cl"))2||u0||'f’+6 for notational simplicity. Then

t
e-i‘nnmnn. s llu0]]1+e + B / e-run(s)1h..s-r‘sds. (5.2)
0
Now let the right hand side of (5.2) be H(t), i.e.,
t I
H(t) = ll'uo||1+s + 8/ e—qs]]u(s)]]1+es_md3-
0
Then

H’(t) = Be-r||n(t)||,+.t-fi-s g BH(t)t"1‘3s‘e.

H ’(t)

H (t) S Bt‘llfi. Integrating this inequality from O to t yields

 

Hence

H(t) 1+6
nH—(O—) S e

 

 

Btiis and H(t)gH(0)e1—?B“i7,

which implies by (5.2) that

IIU(t)l|1+e< «21th s [Insulinei‘sl-f-‘B‘fi‘. (5.3)

68

Note that for t E [0,T] C [0,1] we get

1 ‘ 1 c 1 2 2
e'yte—fiBtm _<_ e—‘f—B S e—ji(G(26,1)) 6 < 2,

by the choice of 6, which contradicts to the fact that ||u(T)||1+, = 26 (by using (5.3)).

So the proof of Claim 1 is completed.

Proof of Claim 2. After proving Claim 1, the inequality (5.3) immediately proves

Claim 2 since
IIU(1)||1+. S 667+1+£B S 6

by the choice of 6.

Finally, the global existence follows by a simple induction argument. By Claim
1 and 2, we can easily conclude that ||u(j)||1+6 S 6 for each j = 0,1,.... Using
u(j) as new initial data, we can solve the equation for t E [j, j + 1], and we get

|]u(t)||1+, < 26 for all t E [j,j +1] by Claim 1.

As a consequence, we can state the generalized global existence result as follows.

Corollary 5.1 For q > 1, Problem (5.1) has a global solution in L" for all uo E L9

with |]u0||q sufficiently small.

69

CHAPTER 6

Numerical Results

In this chapter, by using finite difference method, we give some numerical simulations

for the Dirichlet problem

u,(x,t) = max) + ]u(x,t)|%+1u(x,t) in a x (0,T),
u(x,t) = O on 052 x (0,T), (6-1)
u(x, 0) = u0(x) in Q,

for n=1 and n=2 with Q = (0,1) and Q = (0,1) x (0.1) respectively. We
take the initial data as an = eke—k2($"%)2 for the one—dimensional case and

-k2((x—

uo = ck2e %)2+(y—%)2) for the two dimensional case where c will be determined

later.

Let U be the numerical solution by using finite difference method and U (j, k) denote

U(x,-,tk) where x, = x0 + ij and ti, = to + kAt.

1

 

for the stability condition.

Note that we choose

uh I

t
(ll-’17)2

7O

For one-dimensional case, the approximations are

At

 

UtUi k) =

U(j +1,k) — 2U(j,k) + U(j —1,k)

any, k) = (my .

 

By substituting these approximations in the equation (6.1), we get

U(j, k + 1) = U(j, k) + U(j + 1, k) — 2U(j, k) + U(j — 1, k)) + U3(j, k).

—Dt—(
(D116)2
Similarly adding the approximation of um, to the formula above and replacing U 3 by
U 2, we get the numerical results for the two dimensional case.

To apply the finite difference method and graph the numerical solutions we used
a matlab code.

As a result of these numerical simulations for n=1 and n=2, we can see that the

L1 norm of U, which is calculated as Z AxU (j, k) = B(k), is infinity in a finite time
for both cases. j

The rest of this chapter is devoted to show the graphs of the numerical solutions
by using finite difference method for different Ax, Ay, time steps, k and to determine
constant c in the initial data to get the blow up of solutions for both cases. We also

give the tables for B(k) which is the L1 norm of U to see the exact steps of the blow

up results.

71

2.5

1.5

0.5

GRAPHS FOR ONE-DIMENSIONAL CASE

-13

 

 

—
-
_
r—
i—

 

Figure 6.1. Graph of U for Ax = 1/10, t =

72

1
100 .—
0 4(10)?’

k =10, C: 0.7.

 

10

L4

12

08

06

GA

02

 

 

 

 

73

n _
0 5 10 15 20 25 30 35 40 45
X
1
Figure 6.2. Graph of U for Ax = 1/50, t = 1000475635, k = 10, c = 0.7.

50

 

 

 

 

3 l I 1 1 n r 1 1 1
2.5 n 1
2 r -
3 1.5 - a
1 ~ _
0.5 ~ _
0 1 1 1 1 1 1 1 1 1
O 10 20 30 40 {:21 60 70 80 90 100

Figure 6.3. Graph of U for Ax = 1/100, t = IOOO'ZUIEIF’ K = 10, c = 0.7.

For the one-dimensional case, we can observe from the graphs in Figure 6.1, 6.2
and 6.3 that if c = 0.7, then we have smooth numerical solutions. We obtained this
constant c after some experiments by fixing k and c and running the program with
smaller Ax, Ay and bigger time t. So, after running the program for c = 0.8, 0.9
with different step sizes, we see that L1 norm of U, which is B(k), is infinity if c > 0.9
in a finite time;

i.e., we have finite time blow up of the numerical solution.

74

In Table 6.1 and 6.2 below, we give the values of the L1 norm of U with c = 0.7
and c = 0.9 to see the difference when the solution blows up.
For c = 0.7, Table 6.1 shows that, the values of L1 norm of U (B(k)) is getting smaller
and smaller even if we increase time. However for c = 0.9, it is getting bigger and

l
bigger and at time step 779, B(779) = 00; i.e., the solution blows up at t = 779——

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

75

4(50)2'
Table 6.1. L1 norm of U for Ax = —1— t = 1000-—1—— K = 10 c = 0 7
50’ 4(50)2’ ’ '
B(100)= 1.4201
B(300)= 1.4344
B(400)= 1.3710
B(500)= 1.2924
B(600)= 1.1219
B(700)= 1.1219
B(800)= 1.0034
B(900)= 0.9550
B(1000)= 0.8765
1
1 _ _ _ _
Table 6.2. L norm of U for Ax — E, t — 1000M, K — 10, c — 0.9
B(100)= 2.3652
B(300)= 2.5889
B(400)= 2.8179
B(500)= 3.1277
B(600)= 3.6672
B(700)= 5.1234
B(779): inf
B(900)= NaN
B(1000)= NaN

GRAPHS FOR TWO DIMENSIONAL CASE

 

 

Figure 6.4. Graph of U for Ax = 1/20, Ay = 1/20, t = 80R2107, k = 10, c = 0.5.

76

.........................................

 

.. ...... ~_ ..... m. ...... we ..... .1 ..... w» ...... Au“

x10”

 

10).._-1"'
8x
6\.~-"
4s
2s

 

O

k = 10, C: 2.5.

1
20)2’

Figure 6.5. Graph of U for Ax = 1/20, Ay = 1/20, t = 80.4(

77

20\._,..._.

 

 

Figure 6.6. Graph of U for Ax = 1/50, Ay = 1/50, t = 80.4—(516)—2-, k = 10, c = 0.5.

78

x10

 

 

 

 

 

 

Figure 6.7. Graph of U for Ax = 1/50, Ay = 1/50, t = 8021—51072, k = 10, c = 2.5.

As we can see from the graphs for the two dimensional case in Figure 6.4 and 6.6,
if c = 0.5 with the specified time step, we have smooth solutions, however if we take
c = 0.6, 0.7, 0.8 - -- with smaller Ax, Ag and bigger time t, we see that the value
of the L1 norm of U is getting bigger and bigger. But for c = 3, we can see that the
solution blows up in finite time for different space discretizations. Note that in Figure
6.5 and 6.7, to see the graph before the blow up, we show the graphs for c = 2.5 but

not for c = 3 since we could not get the graphs for c = 3 with the blow up.

79

Also note that for all two dimensional graphs above, we showed the results for

1
t— — 80— 7—420)o rt = 80712-51??- even though we run the program to show the tables
1

below for bigger time such as t— - 150. ——21——4(0)20 rt = 150W. The reason is the
same as above, we can not get the graphs after the blow up time.
- 1
For example, we can see in Table‘6.3 that, if we take Ax = 220’ Ay = 20’
1
t— — 150-———— 4k(2O)2’ = 10, c = 0.5, we do not have blow up and the values of B(k)’s are

decreasing after sometime. But if we increase c to c = 3, in Table 6.4, we see that
the values of B (k) are increasing very fast and at time step 32 we get the blow up of

the solution since the L1 norm of U is infinity.

Table 6.3. L1 norm of U for Ax = Ay = %, t = 1504—(210—)§, k = 10, c = 0.5

 

B(5+i) B(10+i) B(15+i) B(20+i) B(25+i) B(30+i)
1: 1.6551 1.7149 1.7497 1.7604 1.7478 1.7155
1:30 1.6681 1.6099 1.5448 1.4755 1.4043 1.3328
i=60 1.2620 1.1929 1.1259 1.0613 0.9995 0.9405
i=90 0.8843 0.8310 0.7805 0.7326 0.6875 0.6449
1:120 0.6047 0.5668 0.5312 0.4977 0.4662 0.4365

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1 _ _ 1 _ 1 _ _

Table 6.4. L norm of U for Ax — Ay — 20, t — 1504(20)2, k — 10, c — 3
B(5+i) B(10+i) B(15+i) B(20+i) B(25+i) B(30+i)
i=0 13.5557 20.6655 34.4732 92.1197 4.1872e+008 3.6071e+238
i-30 B(32) =inf NaN NaN NaN NaN NaN
i=60 NaN NaN NaN NaN NaN NaN
i=90 NaN NaN NaN NaN NaN NaN
i=120 NaN NaN NaN NaN NaN NaN

 

 

 

 

 

 

 

 

 

80

To obtain more accurate results, in Table 6.5 and Table 6.6 below we take

1
Ax = Ay = 50 (smaller step size) with the same It and c and observe that the results

are the same as above.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1 _ _ _1__ _ 1 _ _
Table 6.5. L norm of U for Ax — Ay — 50, t — 1504(50)2, k —- 10, c — 0.5

B(5+i) B(10+i) B(15+i) B(20+i) B(25+i) B(30+i)

i=0 1.5864 1.6033 1.6183 1.6316 1.6436 1.6546

i=30 1.6648 1.6742 1.6829 1.6911 1.6987 1.7057

i=60 1.7123 1.7183 1.7238 1.7289 1.7334 1.7373

i=90 1.7408 1.7437 1.7461 1.7479 1.7493 1.7501

i=120 1.7503 1.7501 1.7493 1.7481 1.7463 1.7441

Table 6.6. L1 norm of U for Ax = Ay = —1- t = 150—-1—— k = 10 c = 3
50’ 4(50)2’ ’

B(5+i) B(10+i) B(15+i) B(20+i) B(25+i) B(30+i)
i=0 10.0145 10.7594 11.5212 12.3074 13.1256 13.9833
i=30 14.8889 15.8514 16.8813 17.9910 19.1954 20.5129
i=60 21.9665 23.5860 25.4102 27.4916 29.9032 32.7496
i=90 36.1884 40.4692 46.0182 53.6394 65.0922 85.3100
i=120 137.5040 1.6476e+003 3.8270e+052 B(138)= inf NaN NaN

 

 

 

 

 

 

 

 

 

Hence we conclude that, if c is big enough, then we have finite time blow up of

solutions for the Dirichlet problem as we proved theoretically in the previous chapters.

81

CHAPTER 7

Summary and Conclusion

As a summary, in this study we considered solutions to the nonlinear heat equation
of the form u,(x, t) = Au(x, t) + |u(x, t)|1"’1u(x, t) on Q x IR+ for the critical exponent
p = 1 + £- with some bmmdary conditions and the initial condition u(x, 0) = uo(x) E
L1(Q) where Q C IR".

The main problem in this study, which was motivated by H. Brezis and T.
Cazenave [13], was the local existence question of the following Dirichlet problem

for the doubly critical case q = @ and q = 1, i.e., for p = 1 + :2".

u,(x,t) = Au(x,t) + |u(x,t)|"'1u(x, t) in Q X (0,T),
u(x,t) = 0 on 69 x (0,T), (7~1)
u(x, 0) = u0(x) in Q,

where no 6 L1(Q) and Q C R" is any smooth bounded domain.
Before the nonlocal existence results, we obtained the blow up results for some initial
condition uo E L1 for both vanishing Dirichlet boundary condition and the mixed

boundary condition.

We first studied one dimensional cases for simplicity with the critical exponent

82

p = 3 and then generalized the results for n-dimensions with the critical exponent
p=l+§

For the motivation, before proving the nonlocal existence result for the one—
dimensional Dirichlet problem, we first considered the one-dimensional Cauchy prob-
lem and proved that for some 110 6 L1, there is no local solution to the Cauchy
problem in L1.

For the one-dimensional Dirichlet problem, first we take 9 = (—1,1). The dif-
ficulty for the Dirichlet problem was that there was not a good explicit formula for
the solution. By using some technical lemmas, we first proved that for some uo 6 L1;

—lcz:1:2

namely for ac = 4ke , the solution to the Dirichlet problem blows up in finite
time. The main idea in this proof was to use a delicate dilation argument with the
uniform convergence on the integral of the Dirichlet kernel.

By using the nonlocal existence result of the Cauchy problem, we proved the
same result for the Dirichlet problem. The crucial step in this proof was to get
an estimate for the Dirichlet kernel in terms of the Cauchy kernel by using a lower
solution argument. By this estimate, first we compared the linear solutions of these
problems and showed that the linear solution of Dirichlet problem can be estimated
by that of the Cauchy problem. Finally with these two estimates, we proved that
“ad” [,1 = 00 for some initial data uo 6 L1; i.e., for this initial data no 6 L1, there is
no local solution for the Dirichlet problem in L1.

We also generalized the nonlocal existence result for the n-dimensional Dirichlet
problem with the cubic domain 0 = (—1,1) x x (—1, 1). To prove the nonlocal
existence result for the Dirichlet problem, we first generalized the nonlocal existence
result for the n—dimensional Cauchy problem. Then we used a new cutoff function
for the n-dimensional case to estimate the n-dimensional Dirichlet kernel in terms of

the Cauchy kernel and by generalizing the same steps as in the one-dimensional case,

we showed the nonlocal existence result for the n-dimensional case.

83

Since we already proved the nonlocal existence result for a cubic domain, again
by a comparison argument of solutions, we proved the same result for an arbitrary
domain 52 C R".

Moreover, we obtained the nonlocal existence result for the general nonlinear term;
i.e., we showed that, if we replace the nonlinaer term |u|P‘1u by f (u) in problem (7.1)
satisfying f (u) 2 lulp‘lu, the nonlocal existence result follows by comparing the
solutions since we already proved the nonlocal existence result for lulp‘lu.

By replacing the Dirichlet boundary condition with the mixed boundary condition;
i.e., describing the boundary condition in terms of u and 332-, we also considered local
existence question for the mixed boundary value problem. We proved both blow up
and nonlocal existence results for this new problem. To prove the blow up result,
we showed the uniform convergence of the integral of the mixed boundary kernel
as we proved for the integral of the Dirichlet kernel. By comparing the solution of
the mixed boundary value problem with that of the Dirichlet problem for which we
already showed the nonlocal existence, we proved that the solution of the Dirichlet
problem is dominated by the solution of the mixed boundary value problem which
leads to the nonlocal existence of solution for the mixed boundary value problem.

In addition, we studied the global existence of solutions for the one—dimensional
Dirichlet problem under some sufficent conditions on q and the initial data an. Mainly
we proved that for any fixed 6 > 0, if the initial condition no 6 L1“, then the Dirichlet
problem has a global solution in L1+6 for ||u0||1+, sufficiently small. To prove this we
used the local existence result by H. Brezis and T. Cazenave [13] for q 2 1 with some
useful interpolation inequalities for the linear solution and an induction argument.

Finally, we showed some numerical simulations to our results for the Dirichlet
problem by using finite difference method and these numerical results confirmed our

theoretical results from the previous chapters.

84

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