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This is to certify that the
dissertation entitled

Global Existence Of Solutions To Nonlinear Wave Equations
By Weighted Strichartz Inequalities

presented by

Nimet Alpay

has been accepted towards fulfillment
of the requirements for the

Ph.D. degree in Mathematics

 

 

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Date

MSU is an Affirmative Action/Equal Opportunity Institution

 

 

 

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DATE DUE

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— ‘
2/05 c:/ClRC/DatoDuo.hdd—p.15

 

Global Existence of Solutions to Nonlinear Wave Equations

By Weighted Strichartz Inequalities

By

Nimet Alpay

A DISSERTATION

Submitted to
Michigan State University
in partial fulfillment of the requirements

for the degree of

DOCTOR OF PHILOSOPHY

Department of Mathematics

2004

ABSTRACT

Global Existence of Solutions to Nonlinear Wave Equations

By Weighted Strichartz Inequalities

By

Nimet Alpay

We study the global existence of the solution to the nonlinear wave equation
with small data under the assumption of spherical symmetry. More specifically, we
consider the equation Du = Fl(u, Du) with small and radially symmetric data in
the Minkowski space M = IR x IR" where F satisfies a certain condition for exis-
tence. First, using a conformal transformation, the Minkowski space M = IR x IR“
is compactly embedded in a compact subset of the Einstein universe E = R x S".
This method is called the “conformal compactification of Minkowski space.” Then
a weighted version of the Strichartz estimates on the Einstein universe is proved to
show the global existence of a solution. This estimate is proved for the linear inhomo-
geneous wave equation with zero Cauchy data. Then a standard iteration argument
will bring the result of global existence. This work basically simplifies and extends

some of the existing results.

ACKNOWLEDGMENTS

I would like to express my gratitude and sincere thanks my advisor, Professor
Zhengfang Zhou for his warm and constant help, guidance and encouragement. He
made some difficult times during my study go smoothly and my years at Michigan
State University became more memorable because of him.

I would also like to appreciate my committee members, Professor Peter Bates,
Professor Michael Frazier, Professor Clifford Wei] and Professor Baisheng Yan for
their time and advice.

Special gratitude goes to my mother in 'Ihrkey, whose constant love, support and
prayer helped me survive the long Ph.D. years.

I would also like to express my warmest thanks to my dear husband Zafer Alpay
for his continual love, support, help and encouragement. I could not have graduated
without him. I should also mention my son Aytekin and my daughter Dilara for all
the joy they have put in my life.

Finally, I would like to thank God for blessing me to accomplish all these things.

iii

TABLE OF CONTENTS

Introduction 1
0.1 An Overview ............................... 1
0.2 The Statement of Our Work ....................... 4

1 Compactification of Minkowski Space 8
1.1 Definition and Properties of the Conformal

Transform ................................. 8
1.2 Transforming the d’Alembertian ..................... 12

2 A Weighted Strichartz Inequality on the Einstein Universe 17
2.1 Weighted Strichartz Estimates for n = 3 ................ 17
2.2 Weighted Strichartz Estimates for Higher

Dimensions: ................................ 31

3 Global Existence Theorems for Semilinear Wave Equation 53
3.1 Global Existence Theorem for n=3. ................... 53
3.2 Global Existence Theorem for Higher Dimensions: ........... 60

4 Weighted Strichartz Inequality for the Quasilinear Case 65
4.1 Weighted Strichartz Estimates for Qasilinear Equation when n=3 . . 65
4.2 Global Existence Theorem for Quasilinear Equation when n=3 . . . . 76

iv

Introduction

0. 1 An Overview

In this dissertation, we study the global existence of the solution to the following

nonlinear Cauchy problem

Utt - Au = FNMDU),
u(0,x) = f(x), (I)

lit“), X) = 9(X),

where (t,x) 6 IR X IR", and the initial data and the nonlinear term are spherically
symmetric.

The space M = IR x IR" is called the Minkowski Space. Since (1) appears in many
different areas of applied mathematics, physics and engineering such as sound waves,
electromagnetic waves, vibrating strings and drum heads, the Cauchy problem has
been studied by many authors extensively with considerable success.

When Fl(u, Du) = |u|l with l > 1 Equation (1) becomes the semilinear wave
equation,

u“ — Au = lull,

u(0, x) = f(x), (2)

at“): X) = 90‘):

where (t,x) 6 IR x IR".

Now let us give some historical background on this problem. Equation (2) was
first studied by F. John [15]. In 1979, he showed for n = 3 ifl > 1 + \/2 then (2)
always has global classical solutions provided the initial data are smooth, compactly
supported and are sufficiently small. In this work [15] John also proved that when
1 < l < 1 + V2, there is no global solutions of (2) for any smooth, non-trivial data
with compact support. His proof is based on an iteration argument involving weighted
L°° estimates for certain solutions of the linear inhomogeneous wave equation with

zero Cauchy data. Specifically, he proved an inequality which is equivalent to the

following:
llt(t — Ixi>’—2wn,,..(,,i+3, s Czllt’(t — le>‘“‘2)Fan(Rr3,,
where
F(t,x) = O for t — |x| 3 1, (3)
1+v5<z<3

and w is a solution of

wtt -' Aw = F(t,X),

w(0,x) = wt(0,x) = 0 for (t,x) E IRE”

The number 3 in condition (3) is the “conformal power” when n = 3. For higher

dimensions the conformal power for problem (1) is

n+3

lconf = n-

The name “conformal power” arises from the fact that l = Iconf is the unique power

so that the equations of the form Utt - Au = |u|l are preserved under conformal trans-
formations for the Minkowski metric. Using variants of the Strichartz inequality [32],
optimal results had been known for the case I > Iconf. These results were obtained
by H. Lindblad and C. D. Sogge [20]. However, the results for l = lconf are due to
Strauss [31]. The case where l < Icon; is also obtained by V. Georgiev, H. Lindblad,
and C. D. Sogge [12].

The strange number 1+ J2 in condition (3) appeared first in Strauss’ work [30] on
scattering of small amplitude semilinear Schrodinger equations in three dimensions.
In fact, he conjectured that for n _>_ 2 there should always be global solutions of (1) if
I is greater than a critical power (C(n), where [C(n) is the positive root of the quadratic
equation

(n—l)l2-(n+1)l-—2=O.

More specifically,

 

n+1+Jn2+10n-7
[C(n): 2(71-1)

 

This conjecture was shortly verified by Glassey [13] for n = 2. He showed that

3 + J1_7
2
data is sufficiently small. John’s results, which we mentioned above for n = 3, also

 

the global existence for solutions of (2) exists when I > as long as the initial
supported the conjecture since he proved the global existence for the equation (2)
when l > [6(3) = 1+ J2.

Later, Schaeffer [24] showed that there is no global solutions of (2) in three dimen-

3+Jfi
2

 

sions when l = 1+ J2 and also in two dimensions when l = . In 1981, John’s
results led Sideris [27] to show that for all n, there can be blow up for arbitrarily
small data if I < (C(n).

In 1996, Lindblad and Sogge [27] showed that for n S 8, one has global solutions
to (2) if lc(n) < l < Iconf. They also showed that under the assumption of spherical

symmetry, for n 2 3 global solutions of (1) always exist if [C(n) < l < Iconf. For odd

spatial dimensions, the last result was obtained independently by Kubo [18].
When Fl(u, Du) = lutll or lurll with l > 1, Equation (1) becomes a quasilinear

wave equation,

utt - An = [Utlli lurll:
u(0,x) = f(x), (4)
WW") = 90‘),
where (t,x) 6 IR x IR".
Kleinerman and Ponce [17] proved that Equation (4) has small global solutions
if the initial data are small and 1 satisfies (n — 1)(l — 1)2 — 21 > 0. However, it was
conjectured that the critical value for the quasilinear case is lo(n) = n + 1

n — 1
Toward verifying this conjecture for 10(2) 2 3 finite-time blow-up has been shown

for n _>_ 2.

 

by John [16] for nonlinearity lurl3. The case 10(3) = 2 was verified by Sideris [32],
who proved global existence of small radially symmetric solutions for nonlinearity
lutllJ > 2. Schacffer [25], for n = 5, established global existence of small radially
symmetric solutions for l > 3. For higher dimensions the problem of global existence

is still open.

0.2 The Statement of Our Work

In this dissertation we will study the global existence of the solution of problem (1)
with subconformal powers under the assumption of spherical symmetry .

The method we are going to use is a conformal compactification. In this, we are
motivated by the works of Christodoulou [5]; Baez, Segal and Zhou [2]; and Belchev,
Kepka and Zhou [3] on nonlinear wave equations, as well as the recent results in the
rigorous theory of nonlinear quantum fields.

The method of conformal compactification is based on the idea by Penrose [23]

dating back to 1963. The key idea is to map the Minkowski space M 2 IR x IR” to a
compact subset of the Einstein universe E = IR x S" by conformally embedding. To
be more specific, let (t,x1,x2, . . . ,xn) be the usual coordinates in M = IR x IR" and
let 9 = dt2 — def - ---— tie-3, be the Minkowski metric on M and let 5; = (151"? — dS2
be the metric on Einstein universe where (152 is the canonical metric on the n-
dimensional unit sphere 5". Any point in E = IR x S" will be represented by
(T, yl, . . . ,yn+1) where T 6 IR is the Einstein time and y? + - - - + y121+1=1.

The map c : M -> E, which is known as the Penrose transform [23], is a conformal

map (i.e. preserves angles) and is given by

C(t,.’L‘1, ' ' ' 91:71) = (Taylt ' ' ° ail/n+1),

 

 

 

where
t2-—x2
sinT = p(x) t, cosT =p(x) (1— ), T E (-7r,1r),
yj =p5rj, j=1,...,n,
t2-X2
yn+1=p' (1+ 4 )
1
2 ’2
t2—x2
p(x) 2 [9+ (1— 4 ) ] (5)
fl
x2433.
i=1

In terms of spherical coordinates on S", the map 0 can be written as
sin T = p(x) t, sinp = p(x) 7‘, cosp + cosT = 2p(x),

where

. X _
yn+1=cosp, (y1,...,yn)=smpw andw=;ES" 1, pE[0,1r).

When t = 0, the transformation is exactly stereographic transformation from

IR" to S". It is easy to see that the map 0 is well-defined.

The map c is a conformal map from the Minkowski space M to a compact subset
of the Einstein universe E, with a conformal factor p. That is, c*§ = 1229. Therefore,
it preserves the angular variable, while if r is the radial variable and t is the time
variable in Minkowski space, then under the map 0 the corresponding variables in E
are related as follows:

p = arctan(t + r) — arctan(t — r),

T = arctan(t + r) + arctan(t — 7‘).

It can also easily be seen that the image of M under this map is

c(M)={cosT+cosp=2p(x)>0}={p—7r<T<7r—p}.

This set is the so-called “Einstein diamond.” Here p is actually the distance on S"
from the north pole measured in the standard metric.

We make the observation here that C(M) is bounded in E since IT] < 7r. The

boundary of C(M) has two parts C- and 0+. Namely,

C—={p-T=7r},

C+={p+T=7r}.

These are the two lightcones of the “diamond” shape image set. On this boundary
p(x) = (cosT + cos p) /2 is zero.

Let CI = 63 — A and CI = 8% — A31: be the d’Alembertians relative to met-
rics g and g respectively, where A 3n is the Laplace—Beltrami Operator on S". Since
the operators Cl and Dc = CI + ("—2—_1)2 are conformally covariant, Dc is called the
conformal d’Alembertian. Therefore, the solutions of the wave equation on M and
the conformal wave equation on E are in one-to-one correspondence via the relation

u —» P(”_l)/2 Ule(M)'

After we embed the Minkowski space M into a compact subset of the Einstein
universe under this conformal map, we can work with the conformal wave equation
on the Einstein universe, where the Einstein time is finite. In order to get the global
existence of solutions to the conformal wave equation, we will prove a weighted version
of Strichartz estimate on C(M) for the solution of the linear inhomogeneous wave

equation
Clcw(T,/)) = F(T, p),
(6)
“((03 p) : wT(0ap) : 0?
where T 6 (—7r, 1r), p 6 [0, 1r) and F is spherically symmetric.
This idea was initiated by Georgiev [11].

The regular Strichartz inequality for equation (1) is

llwllL2(n+l)/(n—l)(R_1+_+fl) S C llFllL2(n+l)/(n+3)(IR-1:41)-

We are going to prove a weighted version of this inequality for the solution to
the corresponding conformal wave equation on the Einstein universe. The weights
we are going to use will be some suitable powers of p = %(cos T + cos p), which is the
spherical coordinates form of p(x) given in (5). Since these weights will behave well
with respect to the iterations, we will easily get the global existence of solutions of
(2) when [C(n) < l < Iconf.

For the quasilinear equation, we prove a similar weighted Strichartz inequality on
C(M) for the solution of Equation (6). Since the right hand side of Equation (6) will be
different for the quasilinear equation, we will prove the inequality for the derivatives
of the solution of Equation (6). Of course the powers of weights in this case are going
to be different than those of in semilinear case. Then, a standard iteration argument

will give us the global existence.

CHAPTER 1

Compactification of Minkowski

Space

To study infinity in Minkowski Space M = IR x IR", Penrose gave an interesting idea
[23]. He suggested that the Minkowski space M can be conformally embedded into
a compact subset of the Einstein universe E = IR x S". The map c: M -—v E he
proposed would do this job, and it is called the Penrose Transform. Since this map
preserves the angles, it is a conformal map. In this chapter we define the map 6 and

state some of its basic properties.

1.1 Definition and Properties of the Conformal

Transform

We consider the Minkowski Space M = R x IR" endowed with the metric

12
g = dt2 —dx2 = dt2 - Ede?

i=1

We also consider the Einstein universe E = IR x S" endowed with the metric

g = dT2 - (152,

where (182 is the canonical metric on the unit sphere S".

Any point on M is of the form (t,xl,:r2, . . . ,3") where t 6 IR, (1:1, . . . ,x,,) 6 IR"

and any point on E is of the form (T, yl, . . . ,y,,+1) where y? + - . - + y12,+1=1. We

will call T the Einstein time.
Definition 1.1. The map c ; M —» E is defined by

c(t,x) = c(t,r1,x2,...,$n) = (T,y1,...,y,,+1),

where

 

sinT = p(x) t,cosT = p(x) (1—

yj=p(x)-:rj,j=1,...,n,

t2—x2
yn+1=p(X)~ (1+ 4 ),

 

1

2 2 2‘2
2 t - X

11
gig.
i=1

 

p(X)

x2

It is easy to see that this map c is well-defined. In fact,

 

22 t2“X22
cos2T+sin2T=pt +p2(1— 4 ) =1,

which implies that T is well-defined. Also C(t, x) e E since 23:11 y]?

=1.

It is more convenient to define the map c in spherical coordinates. To do that

it
let a point (t,x) E M be represented by (t,r,w) where r = [xl and w = El- 6 3""1.
Also let a point (T, yl, . . . ,yn+1) E E be represented by (T, p, w) where p E [0,1r) is

the distance on S" from the north pole; thus p and w are defined by the embedding

[0,7r) x 51H __. S",

(PM) —-’ (y1,.--,yn+1),
where (311,. . .,y,,) = sinp - w and yn+1= cosp.
Definition 1.2. In spherical coordinates the map c: M —+ E is defined by
c(t,r,w) = (T, p, to) where

t2—7‘2

4

 

sinT=pt, cosT=p-(1— ), TE(—1r,7r),
(1.1)

2 2

t—r
4

 

sinp=p1‘, COS/)=p-(I+ ), pE[0,7r)

with w E S"-1 unchanged and

1
2 ‘2
p: t2+ 1_t2—r2 .
4

We can also rewrite the metrics g on M and g on E using spherical coordinates as

 

g = dt2 — dr2 — 7‘2dw2, (1.2)

where dw2 is the canonical metric on Sn'l, and

Q = dT2 — 61,02 — sin2p dw2 = dT2 — dp2 - p27‘2 (is? (1.3)

This “physical metric” is just the pushforward of the standard Minkowski metric

10

defined by (1.2) and they turn out to be conformally equivalent. The following propo-

sition will show the conformality of 0.

Proposition 1.1. The map c is a conformal transformation with a conformal factor

p. More precisely, c*g = p29.

Proof. Let 8t, Br, 61 and 8,, be the respective partial derivative operators. If one
considers 6t and ar as the sections of the tangent bundle TM of M, one will see
that at and 6,- induce vector fields on compactifited Minkowski space C(M), which

are defined by 5t = as, and Er = ctfip. So we have the following:

—_6T s. -_6_T. at

9
and then differentiating equations (1.1), we can calculate, 2911—, %, g—g, g—f. Thus
we get the following expressions for 5t and 37-:
— I I . . .
at = -(1+ cosTcosp)3T — —s1nTsmpdp,
2 2 (1.4)

— l 1
6r = —§sinTsinp8T + 2 (1+ cosTcosp)8p.

If we compare the metrics (1.2) and (1.3) and observe the fact that TM decom-

poses orthogonally into S"-1 and span of {8b 67-}, it will suffice to show that

g(5ti5t) = 192, g(51‘351‘)= _pzr g(3ti37‘) = 0' (1'5)

But using the equations (1.4) and the fact that 0p and 6,, are orthonormal in the

metric g, the verification of (1.5) comes straightforward. Cl

One can easily see from the equations (1.1) that in spherical coordinates p has the
following form

p = (cos T + cos p)

[\DIH

11

which shows that p can be extended on the whole Einstein universe E. As we men-
tioned earlier the lightcones Ci = {:ET + p = 7r} are the boundary of the compacti-

fied Minkowski space c( M) Therefore on this boundary p vanishes.

1.2 Transforming the d’Alembertian

We consider the d’Alembertian operator CI relative to the metric g in M and denote
by [j the d’Alembertian operator relative to the metric g in E. Then we will prove
that Cl and Dc are conformally covariant operators, where Dc = I: + (15.1)2 is the
“curved d’Alembertian” on E.

We use the following notation for the remainder of this chapter. We denote by
u a function on the Minkowski space, and by v, a function on the Einstein universe
A(

1

related to u by u = p v o c) where A = 35—.

 

Lemma 1.1. Forp =

t2 —x2 2 2
t2 + (I — 4 ) , the following is true:

 

3 __
Up)‘ = ”pa;— , where A = 22—1.

Proof. Since on M the operator C] is simply CI = 8,2 — A, where A is the Laplacian

in x, by direct evaluation we note that

 

 

_ 3 2_ 2
maps-" 3.1)“ x), (1.6)
2 4
Therefore,
n—1 n—3 "-5 n—1 n—3
E12)": 2 - 2 12—1".q(dp,dp)+——2 pTDp- (1-7)
4 2

But since g(dp, dp) = %, we plug this and (1.6) into (1.7) to get the result easily.

[3

12

Lemma 1.2. Let 1‘) = v o 0. Then the following formula is true:

(C1200 c = p-2 [317 + (n —1)p—3g(dp,d'D).

Proof. Let {61, . . .,6n+1} be a basis for TM. Denote by 5,- : c.a,- the vector field
induced by 8,- on C(M). Then {31, . . . , 31,“) forms a local basis for TE on C(M).
Now let us introduce some notation. We let gij = 9(6),, 63-) and 57,-]- : §(3,-,5j). Let
(gij ) be the inverse matrix of (gij) and similarly, let (git ) be the inverse matrix of (gij).
Also let D be |det(g,~j)|, and D be |dct(g,-j)|. Then analogously to the Riemannian

case, we have the following in terms of local coordinates
_ 1 _ -- _ _
Clv=—_0- '23 @634). 1.8
m l (g 2 ) ( )

Because of the conformality of the map c, we have gij o c = p2 gij and therefore
g” o c = p‘2 9”. Using these we also get that D o c = p2("+1) D. Then from (1.8)

we get

 

(Dv)°c=pn 0MP Zg'Jp"+1\/50j v)

HJIDZ
After performing the derivative above, we get

(Dv)oc= 1(”J—6v-l-(n—I)z%gij(8ip)(8jt7)

J—E
2017+

gt ”.6le

+(n - 1):?‘3 9(dp, 6117)- 0

Proposition 1.2. Let u be a 1function in M and v be a function in E related to u by

u = p’\(v o c), where A— - .Then

n 3
(Clcv)oc=p_+Clu.

13

Proof. Since u = p’\(v o c) and Dc v = If] v + A2 v by Lemma 1.2 we get

3 -3 n 3
19%(06 v) o c = p’\ Eli”) + (n — 1);)12— g(dp,di7) +p+ A217 (1.9)

where ’D = v o c as before.

By applying the product rule for the d’Alembertian operator we get that
Cl(p’\ c) = pA E] 27 + (Cl phi-t + 29(dp, dc).
Then applying Lemma1.1 for E] p’\ we obtain
El(p)‘ t7) = p’\ [317 + A2p%3 i7 + (n — ”pi? g(dp,d17). (1-10)
Since the right hand side of (1.9) is just (1.10), combining these we get

p%3(l:]c v) o c = Cl(p)‘ 17).

Proposition 1.3. The equation

Du = Utt — Au = F)(u, Du) in IR x IR",
u(o.x> = f(X) . x e R", (1.11)

ut(0,X) = 9(X) , x 6 IR".

where initial data and F] are spherically symmetric corresponds to the equation

Elev = (CI + A2)v = F)(T, p, v, Du),

v(o,p) = (P’Af) o c—l, (1.12)

—)\—1 1

vr(0, p) = (p 9) 0 C‘

14

in the Einstein universe with

pklcf ifF[(u, Du) = |u|l

Fz(T,p, v. Dv) = pk|T(u)|l ifF)(u, Du) = (utll (1.13)

pklR(v)I’ ira1u,Du>= lurl’.

 

 

where T E (—7r,7r), p E [0,7r) and the notation A,k and T(v) are as: A = n g 1,
k: l(n— 1)- (n+3)
2 t
1 . . .
T(v) = 2 (—AsmTcospv — s1nTsmp up +(1+ cosTcosp )vT) ,
(1.14)
R(v) = % (—AcosTsin pv +(1+ cosTcosp)vp — sinTsinp UT).

Proof. By Proposition 1.2, we have that

(Elev) o c = p—EP - F)(u, Du).

_ n+3
In the case where F)(u, Du) = lull, we see that F1(T, p, 1), Do) = p_£_T2|p)‘v|l since
11 _
2
F)(u, Du) = Iutll, we differentiate the composition (p’\v) o c with respect to t and get

1 _
u = (p’\v) o c. Since A = we get F1(T,p,v, Dv) = pklvll. In the case where

 

0t(pAv 0 C) = APA—lptv + PAU’TTt + vppt), (1-15)

where pt, T; and Pt denote the derivatives with respect to t. One can easily verify the

expressions for these derivatives using the equations from Definitions 1.1 and 1.2, i. e.,

1
pt = —§psinTcosp,
1
Tt = 5(1 + cosTcosp),

1
pt = -§sinTsinp.

15

Putting these derivatives in (1.15) and simplifying it, we get exactly the second
case of (1.13). In the case where Fl(u, Du) = lurll, we will repeat a similar calculation.

If we differentiate (pAv) o c with respect to r we get
311va 0 C) = Ari—1:19:41 + pA(vT - Tr + Up ' Pr)-
Using the derivative expressions

1
pr = —§p - cosTsinp,

1
Tr = —5 sin Tsinp,

H

p,— = —(1 + cosTcosp),

to

we get the third case of (1.13).
In order to see how the Cauchy data are transformed we need to recall the one-

to-one correspondence between the functions in M and functions in E as discussed

earlier.
So letting u = (pAv) o c we get u(0,-) = p’\v(0,.) o c, which gives

v(0, .) = (p—Af) o c'l. Also
em. .) = p0T(p*v) o c = pt+1vT(o,.>o c,

which gives vT(0, .) = (p'A‘lg) o c’l. Therefore equation (1.11) transforms to the

equation (1.12) in E. C]

16

CHAPTER 2

A Weighted Strichartz Inequality

on the Einstein Universe

In this chapter we are going to prove a weighted version of the inequality of Strichartz,
on the Einstein universe. Strichartz stated his estimate in Minkowski space for the

equation
Clu(t,x) = F(t,x), (t,x) 6 IR x IR",

u(0,x) = ut(0,x) = 0
in [32] as
lIuIIL2(n+l)/(n—l)(leRn) S CIIFIIL2(n+1)/(n+3)(Ran)'

We will prove a weighted version of this estimate on the Einstein Universe.

2.1 Weighted Strichartz Estimates for n = 3

We will consider the transformed equation (1.12) for n = 3, when F)(u, Du) = lull.

Then ( 1.12) will take the form

(CI +1)v = p’"3|v|l. (2.1)

17

Here we are going to make the assumption that the function is spherically sym-
metric. Let us call the right hand side of (2.1) to be H (T, p). We do not put the
angular variable to here, since we have radial symmetry.

The following theorem states the weighted Strichartz inequality for the linear
inhomogeneous wave equation when n = 3. Here let us remember that the image of

the Minkowski space under the conformal map c is

C(M) = {cosT+cosp = 2p(x) > 0}.

Theorem 2.1. Assume that H is a spherically symmetric function compactly sup-
ported on {(T, p) : T — p 2 —(5 for some positive small 6} on C(M). If w solves the

following inhomogeneous wave equation on c( M) with zero Cauchy data

(13 +1)w(T.p) = H(T,/9),

(2.2)
w(0.10) = wr(0ip) = 0
and if2 < q < 4, then we have
a -13
”(COST + cosp) w|| ”(d 114)) g 0, ”(COST + cos p) HHLq,(C(M)) (2.3)
whereq’=—g—, 7:1—2, a+fi>l—E, anda>—l.
q - 1 q q q

Here recalling that p = %(cosT + cos p), we see that the weights in (2.3) are some
powers of p. Since the powers of the weights behave well with respect to iteration,
we can easily get the global solutions of conformal wave equation on E when n = 3
and 1 + J2 < l S 3. In what follows C will denote a generic constant which may not
be the same at each occurrence.

The following theorem will state the Hardy-Littlewood Inequality for Ffactional

Integrals, whose proof will be omitted here and can be found at [28]. We will use this

18

theorem to prove the following two lemmas.

Theorem 2.2. (Hardy-Littlewood Inequality for Fractional Integrals )

Fix 0 < it < 1 and exponents l < p < q < oo satisfying

1 1
1—(-—-—)=K..
P q

Then, if

 

1mm = f ”if?” ds,

there is a constant C = C(K,p, q) so that

II1(f)IILQ(]R) S C llflleaR)

Lemma 2.1. If
6

~ _ k(u)
f“) ‘0/ IuI-aitl-fiiu — €I" d“

 

then

“film 5 Cllklqu/

where a,fi,7,q and q’ are as in Theorem 2.1, and k E LqI.

(2.4)

Proof. To prove this lemma we are going to use an iterative argument and Hardy-

Littlewood Inequality for fractional integrals as a tool. We got the motivation for this

from the work of Georgiev, Lindblad and Sogge [12]. First, we break the integral for

Q:

f (5 ) into two pieces as

 

 

W k( ) 6 k( )
.. u U
0 6/2

19

Kl 1 W

 

 

 

 

In the domain of the first integral above, we have u — E > — ,so < —-—,
2 ' ' 2 I -€l7 1511
since 7 = 1 — a > 0 by the assumption on q. So we get
~ C 2k( _u()
f(é) S d’u + 6/ du
KIM" WI “ pram (mu-fll
Let
1 €/2 k( )
~ It
f1(€)= —/ _ du and
KP” IUI 0‘
0
mm

 

u
2 IuI-Olt’I-f’lu—slr

We will show (2.4) for f1 and f2 separately.
Let us do it for f2 first, since it is the easier one to estimate. We fix a number 9

4 4
by9= a —l+a+fl. Sinceoz-I-fi > 1 - a, by assumption we havefl >0. Then we

rewrite the integral for f2 as

 

9 du. 2.5
=IE' flu I‘alél kfi+)‘9lu-€l' ( )

In the domain of the integration we have g < u < 5, so u behaves like é. Therefore

we have |u|_°K ~ lfil-a, and then

Isl—”+0 lul‘” ~ |€|’”’fl+0-

4
By the choice of 0 we have —a — B + 0 = -q- — 1 which is positive since q < 4.

20

Therefore we get

Iii-W lul‘“ ~ lél‘0’fl” 2 In — 51-0—5”.

Then we can bound f2 in (2.5) as

 

5
~ It
f2(€) _<_ 1610/ IU_ “ELM du. (2.6)
6/2

Since 0 > 0, |§|9 is bounded by a constant and again by choice of 9, it is obvious that
'y—a—fl+0= 2. Thusweget
°° k< )
~ u
< c —— du,
12(5) _ f In _ 5'2),

—00

and we can use Theorem 2.2 to get the estimate for f2.

 

In our case p = q’ = E 1, so the number K. in Theorem 2.2 will be

(q—I I) 2
5:1— ——— :—,
q q q

which is the exponent of the denominator in (2.6). Therefore the part for f2 is quite
easily done by the use of Theorem 2.2.
~ 4
In order to prove (2.4) for f1({), we are going to fix the same 6 = 9_1+a+fl > 0

and rewrite f1(§) as

6/2

_ [file (C(11)
0

Since 9 > 0, ltfil‘9 is bounded by a constant. Then we have f1(§) S Cf1(£), where

21

€/2
MWW/flia

2/4 5/2
f1(€)SCf1(€)=E£m0/%% du+——l€|7€fi+0€£I—$J5 du

5/2
_ ._ A k
= C2 (7 fi+0>f1(g) + C [6'7-582lul-a

6/4

 

6

In the domain of the second term above, which is i < u < 2’ u and 5 will behave
alike. So we get lul'“ ~ Iél'a- Then lél‘i‘fi” lul‘” ~ law-[3+9 = 1512/4 2

2
In - {[2/‘1 since 7 — a — [3 + 19 = II by definition of (9 and '7. Placing this bound in

f1(€) we get

- , . °° k
11(tISC2-<i-fi+9>fi(§)+ / fidu-

Then taking the Lq norm of both sides and using the Minkowski’s inequality we get

That the second term on the right hand side above is bounded by “k“ qu comes from

 

 

 

the Hardy-Littlewood Inequality for Fractional Integrals. Thus we arrive at

l/\

. , 1/(1
”month ( / 2*‘1-fi+">02<h<s»q ds) +Cllk||ch

1- _ a .
= 22 (I 13+)”flqu+C||klqu/.

22

2 1
By definition of 6 and by assumptions 7 = 1 — — and a > —- we have
4
1
E-(y-Bwta) <0.

Since fl is a constant multiple of f1, we get [|f1(£)||Lq S llklqu/

If we combine the bounds for f1 and f2, we get (2.4). C]

Lemma 2.2. If

 

9(6)
u)=/: |secu|”| secélfll sin(u — {)l7 d6
then
llfIILq(—5.§) S C ||9l|Lq/(_,,‘;zt)

7r
where a, )3, ’7, q and q’ are as in Theorem 2.1, and 6 is a constant such that 0 < 6 < 2'

Proof. Since 0 < 6 < %, in the interval [—6, g], sin(u — 5) is going to behave like

(u —- 5). So we can replace |sin(u — €)|7 by |u — ([7 in the integral above and we get

 

u (I B
f(u) s / 'C°S|:'_'§|‘f" mods.

Since we need the L9 norm of f, we will multiply f (u) by a test function k(u) in

qu(—6, %) and integrate over [—6, g]. So we get

n/2 ”/2 u

/) f(u ud)u < / [cosulal cosflfi 9(5) k(u) d5 du. (2.7)

|u--€|7

 

If we change the order of integration in (2.7), we can write the right hand side of (2.7)

23

1r/2 71/2
/ |cosu|al cosélfl

In _ £11 91o Mu) dude.

Then we change the variables introducing

 

u=-72E—uand§~=-72-r—£.
The integral becomes
we -B
sinuasin ~~~._ ~ ~
/ f ' -' '. 5' g<t>k<u>dudt (2.8)
0 0 lu—El"

In the domain of the integration we can again replace sing by E and sinu by 11 since
they behave alike. For the sake of simplicity we are just going to use the variables u

and E for a and 5. So (2.8) is bounded by

 

which can be written as

6 §
u"r 3
9(5)] ' ' '5' k(u)dud£.
0

 

|u--€|7

o\t~f

If we call

6
Inlalélfi
Iu — £11
0

fit) =

 

k(u) du,

24

then the left hand side of (2.7) would be bounded by

(ed: 5 191th Ilflqu.

l
“\M21
:2
EL
it
E,
Q.
Q
|/\
c:\'
to
A
L")
‘51

Then since by Lemma 2.1 we have ||f||Lq S C “k“LqI and since we have k 6 L9,, we

get the result. 0

Proof of Theorem 2.1: We will use the solution formula of Problem (2.2). To get

the solution of Equation (2.2), we will recall the operator CI in the Einstein universe:
[:1 = aTT — A53,

where A 33 = app + 2cot p19,, is the Laplace-Beltrami operator on S3. So we get the

following equation
([3 +1)w = wTT — wpp — 2cotpwp + w = H(T,p).
Multiplying this by sin p, and arranging the terms we obtain

(sin p u’lTT — (sin p u1)pp = sin p H(T, p).

This is the one dimensional inhomogeneous linear wave equation whose solution for-

25

mula by Duhamel’s principle is

T p+T—s
/ sin P0 H (8, PO) dpo d8,
Olp- T+d

Sin p u) (T p) =

lollt-t

and from here we get the solution of (2.2) as

 

T p+T—s
1
w(T, p) = 2sinp / / sin/)0 H (.9, p0) dpo ds.
0 Ip-T+8|

In order to get the norm in (2.3) we are going to multiply this solution by
1
(cosT+cosp)’0
the following bound.

 

K (T, p) where K (T, p) 6 qu(c(M)) and integrate. Then we get

 

. 1 .
rr- -TTP+T SIK( Tip)||H(s.po)llsmpol' 'lsmzpl

fl
/0// 1 |coss+cos1p0| dpo ds dpdT.
0...... 218mm-

 

 

[COS T'l' COS PIG [cos s+ cos p0]fl

Cancelling sinp in the integrand, and using the half angle formulas for cosine we

rewrite the above integral as

 

2
T, Hs, sin 9’ sin 3’
”mm— . K1 pm (Po)|| pol [ccsv ',,1 pl
Hf / 1 ,1“ 1,... e.,..W
'7 [’7
OOIP-T+8l 5 | 31an |sinp0
[COST—2'2 TECOSZQ‘BIO [coss—zmcos—rfnl

 

 

(2.9)

2
where '7 = 1 — -, q' = —q—1. Since q > 2 by assumption, we have 7 > 0.
q q '

26

Our aim is to show (2.9) is bounded by

[lKlqu/ ”(cosp + cosT)—B H'qup

For this, we introduce new variables. Let

 

 

u_T-+-p €_s+p0
_T2 ’ ‘ 2 ’
_ —p =s-m
”— 2 ’7’ 2

Then, p0 = 5 — 17 and p = u — v. With the new variables let

mesmsmmwi
|cos5 cosnlfl

 

GKmF=

and

a2w=uanmhmwwf

Claim 1: With the new variables the integration domain will be —6 S r] S v S 5 S u,
where 6 E (0, g).
Proof of Claim 1:

_s—p0<s—|p—T+s| <s—p+T—s=T—p=

 

 

 

 

 

 

 

 

 

0)n- 2 - 2 — 2 2 ”'
00v = T—p : T—p+ausr+T—s = s—e-aus» S
T 2 2 2
s+[p- +S|Sm=6
2 2
s+po s+p+T—s T+p
: < = — =
(111) 5 2 _ 2 2 u
(iv) 17 = S _2'00 > —6, where 6 is a constant number in (0, g).

27

The integral in (2.9) will be rewritten as:

Jluav)CN£;n)
[fff lsecu|0| SGCUIQI sec€|fi| secnlfil sin(u — v)l7| sin(€ — fill“7 dn dv d6 du.

—65n3v5£su

 

(2.10)

Claim 2: In the domain of the integral (2.10), we have

sin({ — 7)) sin(u — v) > sin(u — 5) sin(v — 7)).

Proof of Claim 2: In the domain of the integral (2.10) we have 0 < 5 — v < 1r
and 0 < u - 1) < 1r. Therefore we get sin(E — v) sin(u -- 7]) > 0. Using trigonometric

identities we obtain

(sin 5 cos 1) — sin 12 cos {)(sin 11 cos n — sin n cos 1:) > 0.

Multiplying these factors and arranging the terms we get

sin(cosvsinucosn + sinvcosésinncosu > sinécosvsinncosu +sinvcos§sinucosn.

Now we add
—sin€cosncosusinn — cosésinnsinucosv

to both sides:
sinécosvsinucosn + sinvcosésinncosu — sin(cosncosusinn — cosEsinnsinucosv

> sinécosvsinncosu+ sinvcosEsinucosn —- sinfcosncosusinn - cosfsinnsinucosv.

Factoring the expressions on both sides, we get

(sin 6 cos n — cos 6 sin n)(sin 11 cos 1) — cos it sin 1)) >

(sin 11 cos 5 — cos it sin {)(sin 1) cos n — cos 1) sin 7))

28

which is

sin(é — 7)) sin(u — v) > sin(u — é) sin(v — 7)).
Since 7 > 0, we get

1 1
|sin(5 — n) sin(u — v)l" < |sin(u — o sin(v — um"

 

 

Putting this into the integral (2.10) and changing the order of integration we obtain

 

 

5 v
M”) / C(ém)
d d d .
{[64lseculnlsecglfiISinW—élll . IseC’UlaIsecnlfllsinw—TIW 7) vdé u
’ ‘ ‘ —o

(2.11)

~ G(€,n)
L t , l =
e G“ L) falsecvla lsecnlfiISinW ’ 77W

 

d7]. Then (2.11) becomes

u:

u

// ISBCUIO‘|sec§|fi3|sin(u_§)l, ,/J(U1U)G(€,v) dvdédu.

—6—6

 

Applying Holders inequality to the last integral above we get

 

1'5 u
{llseculabecfllfilsinw _g)p“J(“’”)“Lg'IIG(€:U)I|L3 dédu. (2.12)

By Lemma 2.2, ||C~}'(£,v)||Lg _<_ C||G(€,n)||Lq/. So (2.12) is bounded by

 

1;
5 u ”0(5. any,
J , d 1' ,.
f” (u UHILg’ |secu|0 |sec§|fl | sin(u — {)P 0 u
—6 _‘

29

Applying Holders inequality to this integral, we will see that the above is bounded by

 

d6

L2

“ IIG(€,n)||LqI
||J(u, 12)”qu '7

an 5 lseCUI" |sec{|3lsin(u “0|“7

The application of Lemma 2.2 to the second norm above will give us the following

bound:

“Jo. 22>”ng IIG(£. any).
€71
Putting back J and G here and tracing the estimates back to (2.8) we get what we

aimed to prove the theorem. C]

30

2.2 Weighted Strichartz Estimates for Higher
Dimensions:

The argument for the case n = 3 is going to be a model for the argument of higher
dimensions. In order to get the weighted Strichartz Estimate for n = 3, we actually
used the solution formula for the equation. Here, for higher dimensions, we will again
need the explicit solution formula for the wave equation. For this, we consider the

transformed equation (1.11) for n 2 3, which is
_ 3
Elev = (U + A2)v = p-%Fl(p’\v), (2-13)

where A = 9-5—1 and [j is the d’Alembertian operator on the Einstein universe. We
recall that Cl is defined as [:1 = 6% —— Asa where A512 is the Laplace~Beltrami operator
on S". Here we want to remind our assumption on the spherical symmetry of F. As
before let us call the right hand side of (2.13) be H (T, p). Since the angular variable
does not appear explicitly, we do not put it here as a variable.

The following theorem states the weighted Strichartz inequality for the linear

inhomogeneous wave equation for higher dimensions.

Theorem 2.3. Assume that H is a spherically symmetric function compactly sup-
ported on {(T, p) : T — p Z —6 for some positive small 6} on c(M). Iffor n 2 3, odd
w solves the following inhomogeneous wave equation in c(M) with zero Cauchy data

(D + 12mm p) = H(T,p).
(2.14)

10(0. l’) = wr(0,/)) = 0»

31

 

(n+1)

and if 2 < q < 2 , then we have

 

 

n
a . -fi
||(cosT+ cosp) wlqu(c(M)) 3 C3 “(cosT+ cosp) H”L¢'(c(M)) (2.15)
whereq’=—q—, 7=(n—1)(l—l), a+fl>n_1—n+1, anda>—-—.
q-1 2 q 2 (1

In (2.15), the weights are again some suitable powers of the conformal factor
p = %(cosT + cos p), which will be very useful to get the existence results.
In order to prove this theorem, we first need the explicit solution formula for the

equation (2.14). The following theorem will provide us with the solution.

Theorem 2.4. The solution of (2.14) for n 2 3 odd is given by the formula

T p+T—s

1 n—l
map): , (_1 2 f / Pm(u)H<s.po)(sinpo)Tdpods (2.16)
2(smp)" )/ 0| T l
p- +s

 

where
_ cos(T — s) — cosp cospo
_ sinp sinpo

 

(2.17)
n—3

and Pm is the Legendre polynomial of degree m = T'

For the proof of this theorem we will need to make some observations and straight-

forward calculations. These are to be done in the next propositions.
Proposition 2.1.
(i) In the domain of the integral (2.16), —1 S u S 1.

(ii) p = 1 at the endpoints where p0 = p+T- s or p0 = |p— T+s|.

32

Proof.

(1) Let |p—T+s| S p0 S p+T-s. From the lower bound of fig, we get T—s S p0+p.
Since the variables lie in the interval [0, 7r) and since cosine function is decreasing

on this interval we get
cos(po + p) S cos(T — 3).
Then using the trigonometric formula for cos(po + p) we get
— sinpo sinp S cos(T — s) — cospo cosp.

Since the sine function is positive on the interval [0, 7r), dividing both sides by

sinpo sinp we get

_1 < cos(T — s) — cospo cosp =

 

sinpo sinp

In order to get p S 1, we will use the upper bound for p0. From that we have
[)0 - p S T — 3. By the same reasons as above, we have cos(po — p) 2 cos(T — s)
and then

sin/)0 sinp 2 cos(T — s) — COS/)0 COS/I.
Dividing both sides by sinpo sinp we get

cos(T — s) — cospo cosp _

 

1 Z . .
s1np0 Slnp

(ii) In the formula (2.17), we plug p = p + T — 3. Then using the trigonometric

33

formula for cos(p + (T — s)), we obtain

_ cos(T —- s) — coszp cos(T - s) + cosp sinp sin(T — s)

sinzp cos(T - s) + sinp cosp sin(T — s)

from which it is easy to see that p = 1. For the other end point we do the same

thing.

For later use, we will need the first and second derivatives of u with respect to

the variables T and p. We will denote them by pT, pp, #TT, ppp.

Proposition 2.2.

 

 

 

 

, sin(T — s
2) #T = --.——.-——)
smp s1np0
T _
ii) #77 = --————C.OS( . s)
smp s1np0
cospo —cosp cos T—s
222) ”p = . 2 . ( )
sm ps1np0
, (l + cos2p ) cos(T — s) — 2c03p cospo
21)) ”pp = . 3 .
sm psmpo
2 2"
v — = —
#TT pr sin2p
1 — [12
- 2 2
vi 1 — 1 =
) IT I p sin2p

Proof. Using the formula (2.17), it is very easy to see i) and ii).

iii) By direct differentiation of (2.17) with respect to p, and arranging the terms we

get

l =
I p sinzp sinzpo

sin2p cospo sinpo — cosp sinpo cos(T - s) + 0032p cospo sinpo

Factoring out cospo sinpo from the first and the last terms in the numerator and

using cos2p + sinzp = 1 we get

 

_ cospo sinpo — cosp sinpo cos(T — s)
”p sin2p sin2p0 '

34

Cancelling out sinpo from the denominator and the numerator we get iii).
iv) Differentiating iii) with respect to p and cancelling out the common factors from

the numerator and the denominator we get

sin2p cos(T -— s) — 2cosp cospo + 2 coszp cos(T — s)

 

“pp — sin3 psinpo

Factoring out cos(T — s) from the first and last terms in the numerator and using
that coszp + sinzp = 1, we get iv).

v) Replacing the factor (1+ cos2p) in “PP by 2 — sin2p and doing the straight forward
calculation by using ii) and iv), we easily get v).

vi) Squaring i) and iii), and subtracting them we get

Iii—#2:
sin2(T—s)—c032p sin2(T—s)—cos2p0 —coszp cos2(T—s)+2 cospo cosp cos(T—s)

 

sin4p sinzpo

Factoring out — coszp from the second and the fourth terms in the numerator, and

replacing the first term in the numerator by 1 — cosz(T — s) we get

1 - cosz(T — s) — coszp — coszpo + 2 cosp cospo cos(T — s)

 

2 2
1 — 1 =
I T I p sin4p sin2p0

Then we add and subtract coszp 003ng to the numerator above and then rewrite the

whole expression as

_ (0032(T — s) + coszp coszpo — 2cosp cospo cos(T — s))

 

2 _ 2 =
”T Hp sin4p sinzpo

1 —— 0032/10 + 0082p 008po — cos2p

 

sin4p sinzpo

 

 

= _ ”2 + (1" COSZPO) " 0082p (1 — coszpo)
Sin?!) sin4p sin2p0

35

____ u +(1_cos2p0)(1_coszp)
sin4p sin2p0

 

 

2 ~ 2 ' 2
$111 SlIl
# + 90 p

 

sin2p sin4p sin2p0
2 2
1 1 1— 1
I + = I

 

 

— sinzp sin2p sin2p '
We will also have to recall the Legendre Polynomials and some of their basic
properties. The next definition and the following proposition will serve as reminders
for the Legendre Polynomials. A detailed discussion and the proofs can be found at

[19].

Definition 2.1. The Legendre Polynomials are defined by Rodrigue’s formula -

1 d’"
P1n(.’17)= m ax—m($2 —1)m,m = 0,1,2,. . .

for arbitrary real or complex values of the variable :17. The general expression for the

Legendre Polynomial of degree m is obtained from the above formula as

[Tn/2] k
_ (—1) (2m — 2k)! m—2k
Pm($) — [c220 2mk!(m _ k)!(m — 2k)! :1: ,

 

where the symbol [v] denotes the largest integer less than or equal to v.
Proposition 2.3.

(i) The Legendre Polynomial y = Pm(:v) of degree m satisfies the differential equation

(1 — :12)y” — 2233/ + m(m +1)y = 0.

(ii) Pm(1) =1 form=0,1,2,... .
(1'11) |Pm(:L‘)| s 1 for —1 g :1: g 1.

36

(iv) P,’,,(1)= mm = 0,1,2,...

The proofs will be omitted here and can be found at [19].
Observation: When n = 3, (2.16) agrees with the solution that we derived earlier
for the proof of Theorem 2.1. This is because for n = 3, The Legendre polynomial in
(2.16) will be of degree 0, and by Definition (2.1), P001) =

Now we are ready to prove Theorem 2.4.

Proof of Theorem 2.4. We will show that the formula (2.16) satisfies (2.14). For

this, let us first write down the equation (2.14) explicitly
([3 + A2)w = wTT — Asn’w + A210,

where A511 is the Laplace-Beltrami operator on S 7‘. Since we make the assumption of

radial symmetry, there will be no angular derivatives. Therefore, the equation will be

_ 2
(Cl + A2)w = wTT — wpp — (n — 1) cotp wp + (231) w = H(T, p), (2.18)
For simplicity we rewrite (2.16) as
w(T, p) %(sinp) n2 10/TR( (T p, 3) ds, (2.19)
where
p+T—s . ”_1
R<T.p.s) = [I H le(u)H(S,Po)(Smpo >poo. (2.20)
— 3

and then differentiate it.

T
wT(T, p): s—(inp) LTZR [IN (T p,T )+/RT(T,p,s)ds:l .
0

37

Since H(T, p, T) = 0 by definition, we get
1 1 T
n-
wT(T.p> = 5(sinp)‘LIJ / RT(T.p.s>ds.
0

Differentiating this one more time with respect to T, we get

I f 12 T
wTT(T, p) = §(sinp)— RT(T, p, T) + [R77(T,p,s) ds . (2.21)
0

Now we need to find the derivatives of H(T, p, s). For this, we differentiate (2.20)

with respect to T.

n—l
RT(T:P13) = Pm(u(po = p + T - S)H(s.p + T - S)(Sin(p + T - 8))T

+Pm<1(po = In — T+ sums, lp — T+ s|)(sin(p — T + snail

p+T—s
I . n—l
+ / Pm(u)uTH(s.po)(smpo>Tdpo.
Ip—T+8|

By part (ii) of Proposition 2.1 and part (ii) of Proposition 2.3, we obtain,

1-1 _1
RT(T, p, s) =H(s,p + T - s)(sin(p + T — swif— + H(s, |p — T + s|)(sin(p — T + 3))ET

p+T—s
I _ n—l
+ / Pm(l‘)l‘TH(Sa Po)(SmP0 )poo- (222)
lp-T+SI

Therefore,
n—l
RT(T, p, T) =2H(T, p)(Sinp )T- (223)

We will differentiate (2.22) with respect to T again. In what follows 02H will denote

38

the derivative of H with respect to the second variable.

n—l

RTT(T, p, s) = 82H(s,p + T — s)(sin(p + T—s))T

 

+H(s,p+T—s)(n )(sin(p+T—s))£2'§cos(p+T—s)

+ 62H(s,p — T + s)(—1)(sin(p — T+s))1:’2-_1

+ H(s,p — T + s)(n _1)(sin(p — T+s))fl2§ cos(p — T + s)(—1)

2
I . n—l
+ Pm(u(po = p + T - 8)(ur(po = p + T - 8)H(s.p + T - S)(Sm(p + T-S))T

 

n—l
+ P1201010 =1) - T + 8)(/tr(fio =1) - T + S)H(s, p - T + 8)(Sin(fl - T+8))T

p+T—s
n I . n—l
+ / (P17100112 + Pm(u)ur)H(s. po)(smpo )T dpo.
lp-T+8|

We plug p0 = p+T— s and p0 = p— T+ s in HT whose expression is in part (i) of
Proposition 2.2. We also use part (ii) of Proposition 2.1 and part (iv) of Proposition

2.3 to simplify the expression for RTT» and we obtain

12mm.) = ‘9sz + T — s)(sin(p+ was?
Tl - 1

+( 2

_ 62mm) — T + S)(Sin(p — T+s))E2—1

— (n;1)H(s,p— T+ s)(sin(p — T+s))fl§'§ cos(p — T+ s)

m(m + 1) sin(T — s) . "_1
" - — T
2 Sinpsin(p+T—3)H(8’p+T s)(sm(p+T 3))

m(m + 1) sin(T — s) . "_1
— H - T — T T
2 Sinpsin(p _ T + s) (31p + s)(sm(p +s))

 

)H(s,p + T — s)(sin(p + T—s))&5_3' cos(p + T — s)

 

 

 

p+T—s
II I _ n~1
+ / (BA/011% + Pm(#)l‘TT)H(Sa po)(81npo )7 (190- (2-24)
lp-T+8|

39

We then plug (2.23) and (2.24) in (2.21) to get wTT-

wTT(T, p) = H(T, p) + é—(sinprmll'l‘2 [82H(s,p + T — s)(sin(p + T—s))£§_1 ds

 

+%(n;1)(3inp)—EE—Q/H(s’p + T —- s)(sin(p+ T—s))fl§§ cos(p+ T - 3) ds

T
_ $(sinprfl‘2‘1‘2b/82Hbm — T + s)(sin(p — T+s))1§_l d8

 

_n2%( 1)(sinp) £12120]ku T+s)(sin(p— T+s))Tcos(p— T+s)ds

 

_1 1 n 1 ‘1
m(m2+ )(Sinp) LZJb/sinpssiiill((l)+T— S)H(31P+T—s)(Sin(/’+T"3))E2— d3

 

1 _ —l
— m(m____1_)(2+ (sinp) £12.20qu Sing S) ”(8, p - T + S)(Sin(P — T+8))ET d3

2 sinp sin (p T + s)
1Tp-’r-T-—s
+(1 n—
+§(Sii1p)2 / f(P m(m HT + Pm(l‘)l‘TT)H(9 po)(sinpo) 1’ dpo d3
0 Ip-T+3l

(2.25)

For the p—derivatives of w, we differentiate (2.19) with respect to p and get

 

wp(T, p)= _1(n 21)(sinp)m2_z cosp 0/TR( (T p, s) ds+ (sinp) )QYHO/IRP(T,) p,s

OI‘

 

T
wp(T,p) = —1(n—1)(sinp) 1nglcotp/H(T,/1,s)ds+-;-(sinp)"'21O/TRP(T,p,s)ds.

0
(2. 26)

40

Now we need an expression for RP- So we differentiate (2.20) with respect to p

n—l
Rp(T,p. 8) =Pm(#(po = p + T - S)H(s,p + T - S)(Sin(p + T—S))T

— Pmmpo = p — T + s>H(s.p — T + s)(sin(p — Tenn?

p'l'T—3 I . n—l
+ f Pm<1>1pH<.-.poem/)0) (1m-
Ip-T+8|

By parts (ii) of both Proposition 2.1 and Proposition 2.3 we get

Rp(T,p,8) =H(s,p+ T — s)(sin(p+T—3))1§—l _ H(s,p _ T+ s)(sin(p _ T+s))£2'l

p+T-s
I . 11—1
+ / Pm(#)#pH(8,/90)(SIDP0) 2 (We (2-27)
lp-T+d

Then we plug (2.20) and (2.27) in (2.26) to get explicit formula for wp

 

T p+T-s
_. (r-l) (n—l)
wp(T, p) = — n 4 l(sinp)— l cotp / Pm(/1)H(s,po)(sinpo) dpo ds
0 lp-T+SI
1 f 12 T
n— _1
+ ~2-(sinp)_ [H(s,p+T— s)(sin(p+T—s))97_ ds
0

T
_ %(Sinp)‘m§‘u [H(s,p - T + s)(sin(p — T+s))1T1' ds
0

T p+T—s
1 . _ n—l I . (n-l)
+ 5(SlnP) LTZ/ / Pm(”)l‘pH(31p0)(SmPO) de ds- (2-28)
0 lp-T+SI

We also need the second derivative of w with respect to p. For this, we differentiate

(2.26) with respect to p again.

41

n _ 2
wpp(T»P) =%( 1)

 

T
(sinp)g—2_z cosp cotp /( R( T, p,s
0

n—l "-1

)(sinp)_ (sin/2)’2/R(T, p, 3) d3

 

+<

 

T
— n-l
1)(sinp)_ 2 cotp/RP(T,p,s)

 

)(sinp) n2 cosp [RA (T p, 5) ds

NlH

+%( (Sin/))_ "2 IZRPP( (T, I), 5') d3.

Rewriting the factors in each term we get

 

T

1 _ 1 2 n+3

wpp(T,p) =5“ 4 ) (sinprLrlcoép [H(T,/mas
0

 

T
— )(sinp)_m2fl/R(T,p,s)ds
0

 

T
_ n—l
_ (n 2 1) (Silva—LT) cotp [RP(T’ p, 8) d8
0

T
1 _(n—l)
+ §(sinp) [Rpp(T, p, 3) d3
0

Replacing the factor cos2p in the first term above by l — sin2p and combining the

like terms, we end up with

T
2- +3
wpp<T,p>=%(" 4 1)(sinm—Lrl" [H(T,p.s>ds
0

 

42

 

T

_ 2 (n-l)

+ $01 41) (sinp)_ fR(T,p,s)ds
0

T
_ n—l
_ (n 2 1)(Slnp)—L2—2 cotp [Rp(T, p, 8) d8

0

 

T
1 (n—l)
+ §(sinp)_ /Rpp(T, p, 3) ds. (2.29)
0

Now we need the first and second derivatives of R with respect to p. We already
found the first derivative in (2.27). To get the second derivative, we differentiate
(2.27) with respect to p again. As before, here 62H denotes the derivative of the

function H with respect to its second variable.

RMT’ p’ 3) = 62H(3v P + T - 8)(Sin(p + T—s))£21

 

 

+ (n ; 1)H(s,p + T - s)(sin(p + T—s))fl2é cos(p + T — 3)
— 82H(s,p — T + s)(sin(p - T+3))E§_l
_ (n;1)H(S’p — T + s)(sin(p - T+s))£§§ 003(1) — T + 3)

I . 71—]
+ Pm(u(po = p + T - S)(up(po = p + T - 8)H(s,p + T - 8)(Sm(p + T-8))T

_ an(#(p0 = p " T + 3)(#p(P0 = p - T + 3)H(s,p — T + s)(sin(p — T+3))E§'l

p+T—s 1

II I . n—
+ / [Pmmmfiwmmmpp]H(s,poxsmpof—Hdpo.
lp-T+sl

By part (ii) of Proposition 2.1 and part (iv) of Proposition 2.3 we replace the
m(m + 1)
2
using part (iii) of Proposition 2.2 we can evaluate up at the end points p0 = p + T —

factors Pg,(p(p0 = p + T -- s) and P,’,,(p(p0 = p — T + s) by . Moreover,

sandpo=p—T+s.
sin(T—s)

— and
sinp sin(p + T — s)

 

Atp0=p+T-s pp:

43

sin(T — s)
sinp sin(p — T + s)

 

atp0=p—T+s up:—

Then we get Rpp as

WT, Pa 3) =32H<S» p + T - s)(sin(p + 21:3»?

(72-1)
2

—3
+ H(s,p+T—-s)(sin(p+T—s))£2_cos(p+T—s)

 

— 82H(s, p — T + s)(sin(p — T+s))fl2l

(n21)H(s,p - T + s)(Sin(P — T+3llié;3 cos(p ‘ T + S)

— m(m+1) sin(T— S) H(s,p+T— s)(sin(p+T—s))£2_

 

H

 

2 sinp sin(p + T - s)
_ m(m +1) sin(T — s)

71—]
H - T " — T T
2 Sinp sin(p _ T + s) (Sip + s)(sin(p +8»

 

p+T—s
II I . n—l
+ / [Pm(#)u§+Pm(u)#ple(s,po)(Smpo) 2 dpo- (2-30)
Ip-T+8|

N ow we plug back (2.20),(2.27) and (2.30) in to (2.29) to get the explicit expression

 

 

 

 

for w,,,,.
3 T p+T- 3
11+
wpp(T,/)) = $0124— )(Sin/)) 2 / / Pm (MW 8 p0) (sin/)0) n12 d/Hls
0 Ip- T+8l
1 1)21T—p+T s n 1
._ 71——
_§(n4) (sinp) 2 / /Pm H(s ,p0)(sinp0) 2 dpds
0 l0- T+S|
_1 ( —1) n—l
— (n 2 )(sinp)- n cotp /[H(.s-,p+T— s)(sin(p+T—s))—2_
-1
— H(s,p — T + s)(sin(p — T+s))12_] d3
( 1) 1 l T p+T-S n l
n — . _ 7 _ ’ . _
— 2 (sum) cotp / / Pm(#)upH(8,p0)(Smpo)L2_l dpods
0 lp-T+s|

44

+%( (sinp) Lle/[62H( (8 P+T—3)(Sin(P+T— Snai—

— 3211(3, p — :r + s)(sin(p — T+s))"T‘1]ds

 

 

 

 

T
_ 1 (71—1) —3
+ 4 )(sinp)_ fH(s,p+T— s)(sin(p+T—s))fl2_cos(p+T—s)ds
0
1 fl 12 T
_ 11-- 11—3
- (n 4 )(sinp)— /H(s,p — T + s)(sin(p — T+3))_2_ cos(p — T + 3) d3
0
T
_(mm (sin "21 / sin(T H(s + T — s)(sin( + T-s))£2_l d3
4+ p) s1npsmp+T-s) ,p p
0
T
m(m + 1) rflrll sin(T , n—l
— — T
——4— (sinp) O/Sinp sin(p: T) + S)H(s,p T + s)(sm(p T+s)) ds
1 l T p+T— s n l
. _(71- 2 II I . g - 2
+ §(Slnp) / / [PM/1M?) + Pm(u)/tple(s, po)(smpo) dpo ds.
0 lp-T+8l

(2.31)

Now we are going to plug these derivatives (2.31), (2.25), (2.28) and the function
(2.16) into the equation (2.18), to see if (2.18) holds.

2

 

wTT—wpp—(n—l)cotpwp+(n— =H(T, p)

ls1np)$22—110/T[62H(s,p+T—s)(sin(p+T— s))12_

IQ:

— 32mg, 1) — T + s)(sin(p — T+s))12_l] ds (2.32)

45

+(n

 

) (sinp)_&2__12 [H(s,p + T — s)(sin(p + T—s))fl2_3 cos(p + T - .9) ds

(2.33)

 

_ (11:1- 1)(Sinp)_m2_Q/H(s,p — T + s)(sin(p — T+8))12§ cos(p — T + 8) d3

(2.34)

 

 

m(m + 1) )_§P_,z_1_2 sin(T — s) , n—l
— — 2 d
4 (sinp) / s1np sin ()0 + T _ S)H(s, p + T s)(s1n(p + T 3)) s

 

 

 

 

 

 

(2.35)
_m(m4+1(sinp)£12_1_2/ sin(T H(s,p — T + s)(sin(p — T+s))fl2_lds
sinp sin (p T + s)
(2.36)
+(1 n 1 T p+T— s ”—1
+5 (sinp) )LTZ / [[13 u)u§r+Pm ((1)/1n] H(s,po)(sinpo)L’_2 dpods
0 |p— T+s|
(2.37)
Tp—+T s
—;(nz4_1)(sinp) m/ [P m(,p)H(s p0)( sinpo) )g_2_12 dpds (2.38)
0 lp-T+sl
1( n l T p+T— s
+- 2 7114 2(sinp) )L'Z—l/ /P m,p()H(s p0)( sinpo) )g—T2 dpds (2.39)
0 Ip- T+8|
T
+ (n _ )(sinp)—m2_llcotp f[H(s,p+T— s)(sin(p+T—s))n_2__1
0
— H(s, p — T + s)(sin(p — T+s))12l] ds (2.40)
(n — 1) 71—1 T [HT—8 I (151)
+ (sinpr cocp / / Pm<u>upH(s,po)(sinpo> dpods (2.41)
0 lp-T+8|

46

T
— %(sinp)”m2_12 /[62H(s,p+ T— s)(sin(p+ T—s))£2‘
0

— 6211(3, p — :r + s)(sin(p — T+s))n—2__l]ds (2.42)

 

1) (sinp)_gn_2_l‘2 [H(s,p + T — s)(sin(p + T—s))&2':2 cos(p + T - 3) ds

 

 

 

 

 

(2.43)
T
( — 1) , _ "‘1 . n—3
+ 4 (s1np) [H(s,p - T + s)(s1n(p — T+s))_2_ cos(p — T + 5) d3
0
(2.44)
n— 1 _ n—l
+ m(m 1)(sinp) 2 j sin(T S) H(s, p + T — s)(sin(p + T—s))—2_ d3
sinp sin (p+T—s)
(2.45)
m( m + 1) )_£22_1) sin(T , n—l
— — T
+ 4 (sinp) /sinp( sin (—p T) + S)H(s,p T + s)(sm(p T+s)) ds
(2.46)
T p+T— s
452 £252
S-(irip) /[ Pm(u) up + P m( )upp] H (8 po)(sinpo) dpo d8
0 IP- T+8|
(2.47)
2 1 T p+T—s 1
— n— n—
+ (12—41(sinpr cot2p/ / Pm(p)H(s,p0)(sinp0) 2 dpds (2.48)
0 lp-T+8|
(71 - 1)

 

T
(sinp)_m2_12 cotp f[H(s, p + T — s)(sin(p + T—s))fl2-_l
O

— H(s, p — T + s)(sin(p — T+s))£2_1]ds (2.49)

47

p+T— s

T
_ 1 (71—1))(11—1)
)(sinp)- cotp/ / Pg,() )ppH(s, p0)( sinpo) dpods (2. 50)

 

 

0 |p-— T+s|
1 1 T p+T- s 1
11 n.—
+ 2(7141 2(sinp) )_$—2_l/ /Pm H(S ,p0)(sinp0) 2 dpds. (2.51)
0 lp— T+S|

Now, if we look at the numbered terms above, we will see that some of them will
cancel each other: (2.32) and (2.42), (2.33) and (2.43), (2.34) and (2.44), (2.35) and
(2.45), (2.36) and (2.46), (2.40) and (2.49), (2.41) and (2.50) will cancel each other

and we are left with

11—1)2

 

wTT-wpp—(n—1)cotpwp+( =H(T,p)

+ (2.37) + (2.33) + (2.39) + (2.47) + (2.48) + (2.51).

Combining (2.37) with (2.47), also (2.39) with (2.51) and replacing the factor

1
cot2p in (2. 48) with 2 — 1, we end up with

 

 

 

 

 

sinp
_ )2
wTT —— wpp — (n —1)cotpwp + = H(T, p)
1 T p+T- s l
n n-
+- §(sinp) LT) / [[P )(u%~— u.)+P’ (uxm— w] H(8.P0)(SinP0)LTldPOdS
0 IP- T+8l
(2.52)
2 1 T p+T- s
_ Tl.“
+0141) (sinp) 2 / [Pm 21,)H(s p0)( sinpo) n21 dpds (2.53)
0 Ip- T+8|
1( 2 1) 3 T p+T—s 1
n ._ . _ 11+ ' n—
— -2- 4 (sum) LT) / / Pm(#)H(8~pol(SmPO)£—Tldpds (2.54)
0 lp- T+8l
1 3 T p+T— 3
71+
+ 2(n412(sinp) 2 f / P(m p,H)(s p0)( sinpo) n21 dpds (2.55)
0 Ip- T+S|

48

 

T p+T—s
__ 1 2 _ n—l . "—1
— (" 2 ) (sinp) L24 / / Pm(M)H(S,po)(Sm/)0)Lr2deS- (2.56)
0 |p—T+8I

Then (2.53) cancels with (2.56) and if we combine (2.52), (2.54) and (2.55) we

 

obtain
(D + 12mm) = H(T,p) + gem/2)- "'1
T p+T—s 1 3 1 _1
/ [P1,7;(l‘)(#%‘_u3)+P1,n(#)(/1TT-upp) (n _ if" _ ) Singp Pm(u)lH(s, po)(sinpo)LT2 dpm
0lp—T+8|

By parts (v) and (vi) of Proposition 2.2, we write the expression for p31 — pg and

“TT — ppp and we obtain

(Cl + A2)w(T,p) = H(T, p) + $(sinprmifl.

 

T p+T—s
/ / [(1-u2)P:.(u)-2#P1’n(u)+(n-1);”_3)Pm(u)]H(8.p0)(Sinpo)fl?fldpodS-
0lp-T+8l

Since Pm(/1) is the Legendre Polynomial of degree m = ”—2—, it satisfies the
differential equation given in part (i) of Proposition 2.3, which corresponds to the
integrand above. Therefore the integral in the expression above vanishes and we end
up with (Cl + A2)w(T, p) = H (T, p). It is clear that (2.16) also satisfies the initial

conditions in (2.14). Therefore (2.16) is a solution of (2.14). D

49

Proof of Theorem 2.3. In order to get the norm in (2.15), we are going to multi-

1
(COST + cos ’0)... K(T.p) where K (T. p) e Lq’(c(M)) and

integrate on the c(M). Since the Jacobian on the sphere is (sinp )"‘1dpdT, we get

ply the solution (2.16) by

 

1

 

ln-l

7' “'T T "“7"‘3|K(T,10)|ll'flb‘nooll|«‘3ir1/)o|("‘1)/2
|coss +cosp0

/ 1 1
2 ' (Tl—1V?
0 0 0 |p—T+8I Ismpl |cosT + COS pl“ I coss + COS POIB

lfi | sinp
dpo d8 dp dT.

 

n—1)/2

After we cancel out |sinp |( in the integrand and use the half angle formulas

for cosine we rewrite the above integral as

 

 

. I 1 .
|K(T.p)l |H(s.po)| IsmpoP/q film W
7r 1r T p+T—s 3+ 3-
lcos—QEQ cos —230|
[ff ,/ a 5 ’ dpo d8 dpdT
0 0 0 Ip—T+sl #0825130“ T‘ l lcmfirp‘lcoss-Bflfll |sinp l7lsinpo I”
(2.57)
l 1

where7=(n—1)(§—E), (1’: qzl.

Our aim is to show that (2.57) is bounded by
-H
||K||qu H(COST+COSp) HUM.

For this, we introduce the new variables. Let

 

 

_T+p s+p0
1“ 2 ’ €— 2 ’
_T—l’ _S-Po
”‘ 2 ’ "_ 2

Then we have p = u — v and p0 = 6 — 17. With these new variables let

|H(po, s)l |sinpol2/q'

G(€,77) : |cos£cos 17|fl

 

J(u, v) = 1mm»): IsianW.

50

As before, with the new variables the integration domain will be —6 _<_ 17 S v S 5 S u

where 6 E (0, g)' Therefore the integral (2.57) can be rewritten as

 

J(u,v) C(67))
ff/f ISBCUIQI sec’Ulo‘l sec{l’gl sec nlfil sin(u — v)|“r| sin(g _ mp d" dvdédu'
-5<n<v<§<u
(2.58)

As in the proof of Theorem 2.1 we have

1 1
|sin(u — v) sin(f — 17)|7 < |sin(u — f) sin(v — 17)]7'

 

 

This is because 7 > 0 by the assumption q > 2. Putting this into the integral (2.58)

and changing the order of integration we obtain

 

 

 

 

 

2 . .. G
///6 J(“ v) . (6,17). ,dndvdfidu.
6 Isecula |seC€|5 | sin(u — 5)]? 6 Isecula lsecélfi I Sln(u _ OI"
(2.59)
Let
” 005.77)
C ’ = d .
(6 v) Isec'ul" Isecélfilsinm — 5))? 77
-—6
Then (2.59) becomes
2 .. 5
J . C? . d d .
// |secu|0 |S€C€|1fi|sin(u— Op] (11 v) (5.0) vd€ u
-6 —6 _6
Applying Holders inequality to the last integral above we get
32' u 1
J G d 2.60
f/lsecu|a|secglfilsin(u _€)|7“ (u ’U)“ Lq’“ ({ U)||qu{ u. ( )
—6—0

At this point, we can again use Lemma 2.1 and Lemma 2.2 since all conditions

51

are satisfied. Therefore by Lemma 2.2

||G(€,v)||Lg s Uncanny)
’7

So (2.60) is bounded by

7r

||G(n.€)lquI
{Wyn ’ Iseculalsecafilsinm— on

 

d6 du.

If we apply the Holder’s inequality to the whole integral above, we will see that it is
bounded by

HG( 77 0|qu!

6|secul" | secél/i | sin(u — {)P

IIJWUII d5

L3

 

The application of Lemma 2.2 to the second norm above will give us the bound

J , . G , .
II (“"212 n (mum;

If we put back J and G here, we obtain what we aimed to prove in the theorem. Cl

52

CHAPTER 3

Global Existence Theorems for

Semilinear Wave Equation

In this chapter we are going to state and prove some global existence theorems for
the radially symmetric wave equation. For the proof of existence theorems we will
use the Weighted Strichartz Estimates that we proved in the previous chapter. We
will prove the existence of a solution up to the critical power. As we mentioned in
the introduction the critical power for the semilinear case is the positive root of the
quadratic equation (n — 1)!2 — (n + 1)l — 2 = 0 for n 2 2.

We first state and prove the existence theorem for the case n = 3, since it will be

a model for higher dimensions.

3.1 Global Existence Theorem for n=3.

We consider the following problem on M = R x R3:

Clu = Utt - Au = I'ull in R x R3,
u(0,x) = f(x),x 6 R3, (3-1)

ut(0,x) = g(x),x E R3,

53

where l > 1 and the initial data are spherically symmetric on M .'
As we mentioned earlier in the introduction, we are interested in the subconformal
case, where l < 3. Here, 3 is the conformal power for the three dimensional case.

Now we are ready to state the theorem.

Theorem 3.1. Assume that the nonlinear term and the initial data in (3.1) are
spherically symmetric and 1+ x/i < l < 3. Then (3.1) has a unique (weak) global

solution u in M = IR x 1R3, if the initial data are sufliciently small.

In order to prove this theorem we will use the transformed equation in the Ein-
stein Universe. By Proposition 1.3 we have had the equation (3.1) transformed, via
Penrose’s conformal map, into the Einstein universe corresponding to the following

equation.

Elev = (on — A53 +1)v(T.p) = pl_3|u|l, T e (—1r, 7r),p e [0, 1r),
v(0.p)

vT(0,/)) = §(/)) .1) E [0, 7r).

f(p) 2p 6 [0, 7r). (3.2)

where f(p) =(p‘1f)° 0’1 and W) = (13—29) 0 c‘1 by (1-12)-

Since a function u in the Minkowski space is in one-to—one correspondence with
a function u in the Einstein universe via the relation u —i v = p‘lu, once we prove
the global existence of a solution to (3.2), then using this one-to-one relation we get

the global existence for (3.1). Therefore we should first prove the global existence of

a solution to (3.2) in the Einstein universe. This is done in the following theorem.

Theorem 3.2. Assume that the initial data satisfying (3.2) are spherically symmetric
and l + \f2- < l < 3. Then (3.2) has a unique global (weak) solution 1) on E, if the

initial data are sufficiently small.

54

Lemma 3.1. Let 11.1 E 0, and for m = 0,1,2,... let um be defined recursively by
requiring
Ucvm = Pl_3lvm-1|l,
vm(0, p) = f(p),
3rvm(0, p) = 900),
where fig are as before. Then ifl + x/i < l < 3 and iffor fired a > %3——:lé2 we set

Qm = ”(COST + COS p)a vmllLl+l(c(1w)) and

Rm = ||(cosT + cospla (vm - “Um—1)“L1+1(c(M)>

Then we have form = 0,1,2...

,. 1
21.) R4n+1 S §Rm.

Proof. The proof will be by induction. For m = 0, (i) is trivial and we will only
need to prove (ii). i. e.,we will show R1 S é—RO. By the recursive definition v0(T, p)

is the solution of the homogenous equation

Dcvo = 0,

~

00(0,p) = f,87~v0(0,p) = g

and v1(T, p) is the solution of the homogeneous equation

mm = p’-3Ivol’,

‘-

“0(0ap) = f v 31"U()(0,p) = f]

55

By the Duhamel’s principle, the solution to above equation is

T p+T—s
1

lnp .

 

v1(T, p) = v0(T, p) +

Then

T p+T—s
1 _
111-710: 2 . f / sinpopl 3l’UolldpoalS
smp
0 Ip-T+SI

 

is the solution of the inhomogeneous equation with zero Cauchy data. By Theorem

2.1 we have

121 = “cos?“ + cos p)”(v1 — vollle+1(c(M)) s cn<cosT + cwp)‘fi+l_3lvollll 1+1
L (c
, 1
Since 1) = §(cosT + cosp) and q = l + 1.
Now, choosing [i = -E, and evaluating the exponent —/3 + l — 3, we get

1

HI = llp"(v1- vo)|le+1(C(M)) s cupa‘lool’anrl

(6(

That is,

Is,
. l
Rlsc / pa<’+”loo|’+1 =Cllpavolle+1(c(M))
(M)

= Cllpavolle+1(c(M)) ' “pavoulj‘l”(c(Mll'

l-l

L’+1(c(M)) can be made to be

Since the initial data are sufficiently small, C llpavoll

1 1
less than 5. Therefore we get R1 S 5120.

56

In order to complete the induction argument, let us first notice that for j,m 2 0

vm+1 — vj+1 satisfies the following inhomogeneous equation with vanishing Cauchy

data.
C] _ , _ l—3 l _ , l
C(vm-H Uri-1) ‘ P (lvml lvjl )v

(Um-H _ vj-l-l) IT=0 = 0,

a11 (Um-H - vj+l)lT=0 = 0-

We can write right hand side of this equation as
p” (loml’ - ml) 5 Cp’-3(loml + lojl>’-1(om — ojl,

where C is a positive constant. Now, we will use Theorem 2.1 with q = l + 1, and

fl = —Ta. With this choice of q and 5, all conditions of Theorem 2.1 will be satisfied:

i) Condition on q is 2 < q < 4:

Since1+\/§<l<3, weget2+\/§<l+1=q<4.

ii) Condition on (),/3 is a + [3 > 1— —:
q

l(3—l) B=_g=l-3 weget

SinCGQ>—1—_-l-§- , l 1-l2,
l(3—l) l—3 l—3 4 4
“+fl>T_—l‘2‘+1—_l‘2—m—1‘m-1—a'

iii) Condition on a is a > —— :
q

COM»;
1—12 1+1“

This will imply that l2 — 21— 1 > 0, which holds as long as l > 1+ «5. Thus, by

Theorem 2.1 we have the estimate

57

”(COST + cosp)" (Uni+1 - vj+1)(”Ll+1 S

CII °°sT+coso> P'3(lvm|+lvjl)”1 -% II +

(cosT + cosp ), we rewrite this as

NHH

Since p =

$+l—3(

IIPG(Um+1—vj+1)IIL’+l(c(M))SCIIp lvaHUJI)’ (Um—v3) II L+(c( (M))

f;- + l — 3, it will be equal to al. So the left hand side

(LHS) of the above inequality is bounded by

If we evaluate the exponent

pf-f
LHS SC / pa)(l+1 (vm _"Uj)+ l(IUmI + Iva) _WI deT
c(M)

Rewrite this as

l
m

 

a(l+l) (l+1)(l—l) l 1 gl—1)Sl+1)
LHS SC / P I P0 I (Um _ vj)+(Iva ‘I' Iva) deT

c(A/I)

 

Now, we use Holder’s inequality with conjugate numbers I and l 1.
LHS < C ”1)an —Uj )IILHI “POUUmI +IUjI)IILl+1

So we end up with

Ilpa(vm+1— oj+1>llu+l s c ”pawn; — ojllle (om + ca)“. (3.3)

Ifj = —1, Qj = 0. Hence we get

IIpa(v1n+1 “ ”OlIILHl S C IIPavaILHl Q5714,

which gives

NIH

1 . _
QnH—l S EQm 'I' Q0 Slnce CQin l <

by the assumption of sufficiently small data. Then by the induction hypothesis we
get Qm+1 S 2Q0, which is part (i) of the lemma.

To see part (ii), let j = m — 1 in (3.3). Then

IIPa(vm+1 _ Um)“ 1+1 S C IIPOWm “‘ I’m—1)” [+1 (Qm + Qm_1)l_1.
L L

1 1
Therefore Rm+1 S 53m if C (Qm + 62m“)!-1 < 5 and this comes from induction

argument. C]

Now we are ready to prove Theorem 3.2.

Proof of Theorem 3.2. The existence part comes easily from Lemma 3.1. let um

be defined as above. Then by part (ii) of Lemma 3.1 for m = 0, 1,2,. . . we have
1 1 _m
Rm+1£ ER": S WRO = 0(2 )
This bound yields
Ilp’-3<Iom+ll’ — Ivmll)llLu+1)/l s Rm 3 0(2-m)

by repeating the proof of the lemma.

Therefore lumll ——+ lull in LU‘Hl/l as m —+ 00. Also we have by the part (i) of

59

Lemma 3.1 that
Ilpl‘3lvml’ll 1+1 3 c at. s 0 626
L
by repeating the proof of Lemma 3.1 with j = 1. Thus |vm|l is also uniformly bounded
in L(l+1)/l'
Therefore we conclude that um must converge to a weak solution of (3.2). Since the

proof of the bound for Rm“ yields the uniqueness part, this completes the argument.

Cl

Proof of Theorem 3.1. By theorem 3.2 we know that (3.2) has a global (weak)
solution 1) in E. As we said earlier, a function u in E is one-to-one correspondence
with a function u in M via the relation 1) —s u = p‘U. Hence, letting u = pi) we obtain

that Clu == H(u) as we discussed earlier. Also

u(O. .) = pow.) =pf' = f,

otlo. .) = olpcmlo, .) = p2vT(0, .) = 12% = 9.

Hence the problem (3.1) has a unique global (weak) solution as well. [:1

3.2 Global Existence Theorem for Higher Dimen-
sions:

The rest of this chapter will be devoted to the existence theorem for radially symmetric
semilinear wave equation when n 2 3. In this, we will take the theorems for the 3

dimensional case into consideration as our models.

60

As before, we state the problem on Minkowski Space first.

[In = utt — Au = |u|I in IR x R",
u(0,x) = f (x),x 6 IR", (34)

ut(0,x) = g(x),x 6 IR",
where l > 1 and the initial data are spherically symmetric.

Theorem 3.3. Let n 2 3 be odd and assume that the nonlinear term and the initial
data in (3.4) are spherically symmetric with lc(n) < l < Z—i: where lc(n) is the
positive root of (n — 1)l2 — (n + 1)l — 2 = 0. Then (3.4) has a unique (weak) global

solution u in M = IR x IR", if the initial data are sufficiently small.

As before we will again use the transformed equation in the Einstein universe. By

Proposition 1.12, equation (3.4) is transformed in to the following equation

(D + as = pklvl’,
c(o, p) = fol, (3.5)

”TIOa p) = 57(1)).

whereTE (—7r,7r), p6 [0,1r) ,A= n;1 andk=l(n-1);(n+3).

 

 

Theorem 3.4. Let n 2 3 be odd and assume that the nonlinear term and the initial
data in (3.5) are spherically symmetric with lc(n) < l < Z—g, where lc(n) is the
positive root of (n — 1)l2 — (n + 1)l — 2 = 0. Then (3.5) has a unique (weak) global

solution v in E, if the initial data are sufficiently small.

As we did in the three dimensional case, we will prove the theorem by a standard

iteration argument. This will be done in the next lemma.

61

Lemma 3.2. Let v-1 5 0 and for m = 0,1,2,. .. . Let vm be defined recursively by

requiring
Dcvm = kavm—l It:

~

'Um(0,p) = f(pli
5rvm(0.P) = g(p),

n+3

and iffor fired a > 1_—k:§ we set

 

where fig are as before. Iflc(n) < l < n 1

Qm = “(cosTcosp)“ vaILl+1(C(M)) and

Rm = “(COSTCOS mu (”m - Urn—lllILl+1(C(M)) 1
then we have for m = 0,1,2,...

i) Qm S 2Q0a
.. 1
ll) Rm+1 S 2R7”.

l(n—1)——(n+3)

2 and lc(n) is the positive root of(n—1)l2-(n+1)l—2 = 0.

Herek=

 

Proof: The proof will be by induction just like the proof of Lemma 3.1. Actually,
we will just repeat the proof of Lemma 3.1 with q = l+ 1,01 > —1—kl—2 and ,6 = —%.

Here we will use Theorem 2.3 with this choice of q and ,3. Let’s check if all the

conditions of Theorem 2.3 are satisfied.

1
i) Condition on q is 2 < q < 2:+1:

 

lc(n) is the positive root of (n - 1)l2 — (n + 1)l — 2 = 0.i.e.,

 

n+1+\/n2+10n—7

 

3 . 1
Since1<lc(n)<l<7—7:%,weget2<lc(n)+1<l+1=q<2::—:1.

62

ii)

iii)

. . 2
Condition on 0,3718 a + ,3 > 7 — -:
q

 

 

1 1 l a k
licroo=(n—1><§—E>.o>—;-7w”='7=1—l2
. _ _(‘n-llU-l)
Slnceq—l+1,7— 2 1+1
So

a+B>— kl + k
‘ 1—12 1-12

l(n—1)—(n+3)

 

When we plug in k = here , and simplify the expression we

 

 

 

2
end up with
l(l—1)—(n+3)_ 2
(1+6 > 2(l+1) — '7 q.
Condition on a is a > —2 :
kl (n+3)l-(n—1)l2 1
a > — = __.
1—12 2(1—12) 1+1

This will imply that (n — 1)l2 — (n + 1)l — 2 > 0, which holds as long as

n+1+\/n2+10n—7
2(n—1)

 

 

l > lc(n) =

Thus by Theorem 2.3 we will have the estimate

H(cosT +cosp>° (vm+1 - vj+1)||Ll+i s

a _
c I|<oosT + cospflpklloml + IojI)‘ 1(,,m_ ”leILL'i—l’

OI'

°+k l—l
IIPa(Um+1 - Uj+1lIILl+l S C IIl9T (I’UmI + IUjI) ('Um _ 'UjllILtyl- (3-6)

63

If we evaluate the exponent % + k, it will be equal to

 

 

 

3 k _l _12
l _ _

Hence the left hand side (LHS) of (3.6) is bounded by

l 1 l “41’
l 1 ~ +1
(mesa / p"<’+1><om—oy>+uoml+lojllui—zdodr

c(M)

The rest of the proof will be just a repetition of the proof of Lemma 3.1 and

therefore we will just refer to that proof and not repeat it here. D

The proof of Theorem 3.4 is just the repeat of the proof Theorem 3.2 and therefore

we will skip it.

64

CHAPTER 4

Weighted Strichartz Inequality for

the Quasilinear Case

This chapter will discuss the inequality of Strichartz for the quasilinear wave equa-
tion on the Einstein universe. As before, we will prove the estimate first in the three
dimensions and take that argument as a model to prove the inequality for higher di-
mensions. We will first deal with a quasilinear equation of this form Utt — Au = lutll

for some 1 > 1.

4.1 Weighted Strichartz Estimates for Qasilinear
Equation when n=3

Recalling Proposition 1.3, we will see that when Fl(u, Du) = lutll, equation (1.11)

will be transformed to equation ( 1.12) with H(T, p, v, Dv) = pk |T(v)|l where T(v) is

l(n— 1)— (n+3)
2

(1.12) will involve derivatives of v with respect to T and p, and so we will prove the

 

as in (1.14) and k = . Therefore the right hand side of equation

estimates for the derivatives of v.
The following theorem states the weighted estimate of Strichartz for the

T —derivative of the solution to inhomogeneous wave equation in three dimensions.

65

Theorem 4.1. Assume that H is a spherically symmetric function on c(M) supported

on {(T, p) : T — p 2 -6 for some positive small 6} and w satisfies the equation

Dow = (l:1 +1)w(T.p) = H(T,p).

(4.1)
w(o. p) = wimp) = o.
If3 < q < 4, then we have
H (cos T + cos p)0 wT IILQ(c(M)) 507 H (cos T + cos p)_5 HIILQ’(c(M)) (4.2)

2 3
whereq’=-i-,7=1—-, a+fi>—1 and a>1——.
q-1 q q

Proof: From the proof of Theorem 2.1, we know that the solution of equation (4.1) is

 

T p+T—s
1 .
w(T,/J) = 2sinp / / smpo H (s.po)dpo d3-
0 Ip-T+3|

The derivative of w with respect to T is

wT(T, p) = T1(T, p) + T2(T, p) with

 

T
T1(T,p) = 2Silnp/sin(p+T—s)H(s,p+T—s)ds, and
O
T2(1)(T, p) if T < p
T2030) =

where
Tom”, p) + T§3>(T, o) if T > p,

 

 

2sinp
T-p
T(2)(T p) - _2si1np / sin(T— p— s)H(s,T— p— s)ds,
0

66

1
2sinp

 

T
T2(3)(T, p) = sin(p — T + s)H(s. p — T + 3) ds.

Tp

We will prove (4.2) for each of these integrals and we start with T1 (T, p).

Letp=p+T-sin T1(T,p). Thens=p+T—pand

p+T
/ sin ofllp + T — o. 5) do.
p

 

T1(T’ p) = 2sinp

In order to get (4.2), we will use a duality argument. For this, multiply T1 (T, p) above

by (cosp + cos T)“ - K(T, p), where K(T, p) E Lq’(c(M)) and integrate over c(M).
1
7‘ N-TP'I'T K T, sin ” H +T— ’, ~
I < p)|| pll (p p p)| |cos(p+T_fi) + cowl

l 1
|cos(p+T—p) + cos pl” |cosp + cos Tl“

 

[3 ISin2 pI

 

dp dp dT.

 

0 0 p 2|sinp|

Using the half angle formulas for Cosine we rewrite this as

2
|sinp I‘l—I

1
)I T+ T+ 2~

.72 _g_2- r3
1 ICOSI COS I dpdpdT,

0 O p |cos—T-3'-3cos—%—ET+ '2)” |cosz;£cosZ§-3|

 

2
1r rr—Tp+T|K(T,p)| |sin (3|? |H(p+T-p,f)

 

|sinp I7 |Sin {3|7
C!

(4.3)

2
where '7 = 1 — a and q’ = q—qi. Since q > 3 by assumption, we have 7 > 0.

Our aim is to show (4.3) is bounded by

||K||qu ”(oosr + cos 10‘” 11“”, .

67

For this we introduce the new variables. Let

T+p T—p =T+p—2p=u -

 

Then p = u — v and f) = u — 6. With the new variables, let

2

- 1 . s 1
G(€.u)=H(p+T—p,p)lsmp|57———B.
|cosucos€|

2
Mo. o) = Ker, p) |sinp li’.

The new integration domain will be —6 S E S v S u, where 6 6 (019° We can see

this easily in the following way:

(i) v S u is trivial.

T
(ii) SincepZ/),{=u—p$u—p= ——:—p—p=v.
(p + T)
7r 2
6 E (0, 2)’ by the assumption on the domain.

T+p

(iii)£=u—p= 2 —(p+T)+s=— +s>—T+sZ—6forsome

Therefore we get —6 S E S v S u. Using the new variables we can rewrite the integral

(4.3) as

 

III G(§,u) J(u,v) 7 dédvdu.

—6<€<v<u lseculfilsecilfilseculalsecvlalsinm - elilsinlo -- v)|

Next we will separate the integrals and obtain

 

 

zzr u u
J(u,v) G(£,u)

d d . 4.4

{4 ISCCu|fi|secv|0|sin(u — ”)II “U 5 ISBCEIBIsec uIQISin(u - 5W d6 u ( )

68

Now write a and [3 as followsa = 01 + 02 ,fi = 51 + 62, where
1 2 1 4
01>——,02>1—-,l31>—-andfl2>-—2. (4.5)
q q q q

This choice of (11,02, [31,62 is also consistent with the assumption that a + [3 > —1

3
and 0: >1— - since
0

1 2
a=al+a2>-—+1——=l-§and
q q q
l 2 l 4
a+fi=al+ag+fi1+fig>—-+1———-+——2=—1.
q q q q

Then we write a and ,3 in this way in (4.4) and get

rr/2

 

 

/ 1 J(u, v) ldv/ G(,€ u) dédu
6 |secu|a2+52 6 Iseculfillsecv|a|sin(u—v) Isecélfilseculallsinw— {)I7 '

 

 

 

 

 

 

(4.6)
Now we let
u)=Z .](u v) v and
lseculfil lsecvlalsin(u - v)|7
~ _ 0(6) 1!.)
GI“) “ / lscccllsccollsinlo — 0|" di
in (4.6) and then apply the Hélder’s inequality
rr/2
02+52 ~ ~ 0:24-62 " "
/ |cosu| J(u)G(u) du g I‘lcosul LT J(u)“Lq |lG(u)HLq, (4.7)
—6

1 2
where ; + E = 1, which implies that r = a—q—2. Note that since q > 3 we have r > 1.

69

In (4.7), for the first norm to be bounded we need
, . 2
((12 + (5)7 > —1,l.e., 02 “‘I' ,82 > a — 1.

By our choice of 02,132, this is satisfied, so the first norm ill (4.7) is bounded.
4
For the other two norms in (4.7) we will use Lemma 2.2. Since 0 +61 > 1 — a, 6 +

4 l 1
(11 > 1 — a, (11 > —E and 61 > —— by our choice, we can use that lemma and get
‘1

I|J(a)HLq g C “Johan“, and

||é<o>||Lq s C ||G(€.u)lqu/.

Writing these bounds in (4.7) and putting back the definitions of G and J we get
that (4.3) is bounded by ”KHLqI ”(cos p + cos T)_3 H ”up This completes the proof
of (4.2) for T1 (T, p).

Now with a similar argument, we will do the proof for T2“) (T, p).

 

T2(1)( T, p)

T

[Sin n(—p T+S)H(s,p—T+S)ds.
=2si1np
0

As before let )5 = p - T + s. Then

 

2smp

p
1 1 .- - - -
Ti)(Tip)= . [SIHpH(p-p+T.p)dp-
p—T

We multiply this with our weight function (cos p + cos T)OK (T, p), where K (T, p) E

qu(c(M)) and integrate over c(M). Here, we will use the half angle formulas for

70

cosine as before.

2 2
”_T p lK(T.p)IIsian<7IH(p—p+T) p)I - Isinpli’ dpdpdtr

hos—Ef—cos —73|5

lsin pillsin (3|7

 

1

0 0 p—T |cosgzl'if—il‘coszfglfi|coslw§2cosZ§BVll

 

(4.8)

 

2
where q’= q 1 and7=1—Easin T1(T,p).

T T—
Letu=——;—£,v=2 —p,=€

With the new variables, we define

2p—p+T

2 =p+v. Thenp=u—v,andp=§—v.

2
- - . - 1
G(€.v) = H(p- p+T.p)Ismp|3'——-—fl,
|cos£cosv|
2
J(u,v) = K(T,p)|sin ply.
The new domain will be —6 S v S 5 5 u, where 6 E (0, g), since

(i) v s u is trivial,

(ii) Sincep>Twe havep>0,also€=p+v,so§2v,

(iii) Sincep<p, we have£=p+v£p+-—;— =u,

T -
(iv) v = —2—8 2 —6 by the assumption on compactly supported H.

Then (4.8) becomes

 

1} u
/// “I ”Hm” ‘awm
|sec§|5|secv|5|secu|0|secv|a|sin(u — v)|7 |sin(é —v)|'7
-5-5 -5

71

Here we will change the order of integration to separate the integrals and get

ml:

U.

u
u “) cm)
at did .
ff Isecvlfilseculalsinm—v)llf lsecélfilsccvlalsiolé—o)ll 5 L "

_5_5

 

 

7r
Since u S 2’ we can replace the upper limit in the last integral above by 32:.

Now let a = 01 + (12, [i = [31 + [32, where al,ag,[31, [32 are as in (4.6) and get

«1:1

u

 

 

/ j J’(“ v) (5 ) dgdv du.
6 6 |secv|10‘2+'32 |secv|5l |secu|a|31n(u — v)|7v Isecélfllsecvlflsmfi — v)|7

(4.9)

u:

 

We call C(v )=/ G“ v) , d5 and then change the order of

|sec§|l3|sec vlo‘l |sin(é — v)|7

integration in( 4.9) to get

TI

2 2

fleesv|a2+32é(v)/ J(u,v) dudv.
6 |secv|31|secu|a|sin(u — v)|7

_ U

 

If we denote the inner integral above to be j (v) and then apply Hélder’s inequality

to the above integral, we get
i
[|coso102+326(o)i<o) s Illcosvlazw’zIILr||G(v)||La ||j(v)||La.

1 2
where ; + a = 1, which implies that r = —q—.

q —
The rest of the proof will be the same with the proof for the term T1(T, p) and

will not be repeated here.

72

For Té2)(T, p), we let ,3 = T — p — 3. Then T2(2)(T, p) becomes

T—p
/ sianlT — p — pp) dp.

 

m __1
T2 (T,p)— 2sinp

As before, we multiply this by (cosp + cos T)OK (T, p), where K (T, p) E Lq,(c(M))

and integrate over c(M).

2
T_ _, Elsinplg'
)llcos—2--‘2 cos—-‘§—-3| dpdpdT,

l 1

° ° ° Icosizficos—i—Elfilcosiific was)“

 

7r 1r—TT- PIK(T» P)I ISinPIy IH(T— ‘13— Pop

 

kaWfll

am

2
where q’ = i, 7 = 1 — - as before.
q-1 q

T T— T— —2"
Letu=——2i£, v= ——p,£= __p2_p

We can define G(€, v) and J (u, v) as before with the new variables.

= v—p. Thenp= u—v, p=v—{.

Then the new domain will be -6 g g S v S u, where 6 E (0, g) since
(i) v S u is trivial,

(ii) SincefiZOandv=€+p,wegethé,

 

 

T—
(iii) Since p S T—p, £- — v— p>

assumption on the domain.

Then (4.10) becomes

ml:

 

/// “TWWW , ham
6 |secu|a|secvlalsecvlfilsecélfilsmm — v)|7 |sm(v — §)|'7

73

If we change the order of integration and separate the integrals we get

177?

2
// J(u, v) Idu/ G(€, u) dgdv
_ |secu|°|secv|3|sin( (u — v) |sec§|f3|secv|a|sin(v — €)|7 I

_5 v

 

 

The rest of the proof for T52) (T, p) is the same as previous ones.

For T53)(T, p), we again change the variable as ,5 = p — T + 3. Then

712(3)(T,) p) =

 

p
sinp/Sin /3II( [3 — p + T, p) (1,6.
0
The proof for this term will be just like the proof for T2(1)(T, p) and hence will not be

repeated here. D

Theorem 4.2. Assume that H is a spherically symmetric function on c(M) supported
on {(T, p) : T - p _>_ -6 for some positive small 6} and w satisfies the equation (4 .1).

If3 < q < 4, then we have
' -/3
”(cosTcos p)” sm WPIILq(c(M)) 5 C7 ”(cosTcos p) HUM/(c(M)) (4.10)

where (1,13, 7,q, q’ are as in Theorem 4.1.

Proof. Since the solution to equation (4.1) is

 

T p+T—s
1 .
w(T,/1) = 2sinp] / SlnpoH(s,/)0) dpo ds,
0 lp—T+8|

74

it is easy to calculate that

T p+T—s
1
wp(T, p) = —§(sin /))-2 cos p/ / sin pOH(s,p0) dpo ds
0 lp-T+SI

sin(p+T—s)H(s,p+T—s)ds

+ (sin p)—1

NIH

1
— §(sinp)_1 sin(lp — T + s|)H(s, |p — T + s|)ds.

O\HO\~]

Multiplying this by sin p we get

 

 

 

1 T p+T—s
. < . .
sm pwplr. p) _ m p f / smpoH(s, po) dpodc (R1)
0 lp-T+8|
T
+2sinp/sm(p+T—s)H(s,p+T—s)ds (R2)
T
+ 2sinp fsm(|p—T+s|)H(s,|p—T+s|)ds. (R3)
0

In order to prove the theorem, we will prove (4.10) for each term (R1), (R2), (R3)
of sinp wp(T, p). But proving (4.10) for (R1) is going to be same as Theorem 2.1
since the conditions on a and fl here will be covered by the conditions on a and 6 in
Theorem 2.1. Also, proving (4.10) for (R2) and (R3) is going to be same as Theorem
4.1. This finishes the proof of Theorem 4.2. D

75

4.2 Global Existence Theorem for Quasilinear
Equation when n=3

We consider the following problem on M = IR x 1R3:

Cl’u = utt - Au = l’utll in IR X R3,
u(0,x) = f(x) , x E R3, (4-11)

ut(0,x) = g(x) , x E R3,

where l > 1 and initial data are spherically symmetric and sufficiently small on M.

As before, we are interested in the subconformal case where l < 3.

Theorem 4.3. Assume that the nonlinear term and the initial data in (4.11 ) are
spherically symmetric and 2 < l < 3. Then (4.11) has a unique (weak) global solution

u in M, if the initial data are sufficiently small.

By Proposition 1.3 we have had equation (4.11) transformed, via Penrose’s con-

formal map, into the Einstein universe corresponding to the following equation:

Dev = (611T — A33 +1)’U(T,/)) = pl—3IT(U)Ili T E (-77, 7r)rp E [0) 7T),

~

'U(0,p) = f(p) 1 P 6 I01"),

vr(0.p) = g(p) . 06 PM).
(4.12)

where f(p) = (P—lf) e c"1 and W?) = (p‘29) ° 6‘1 by (1-12) and
T(v) = -21-(-sinTcos;m — sinTsinpvp + (1 + COSTCOSp)’UT).

Now we will prove the existence of the global solution to equation (4.12) in E.

Theorem 4.4. Assume that the initial data satisfying (4 .12) are spherically symmet-
ric and 2 < l < 3. Then (4.12) has a unique (weak) global solution v on E, if the

76

initial data are sufficiently small.

Lemma 4.1. Let v_1 E 0, and for m = 0,1,2,... let vm be defined recursively by

requiring
Dc'Um = Pl—BIT(Um—1)Ilr

’Um(0, P) = f(p))
Brvm(0i p) = 900).

- l— 2
where f,gT,p are as before. If2 < l < 3 and iffor fixed a > l—_—1- we set

Qm = H(cosT + cos p)0 T(vmlIIL‘+1(c(M)) and

Rm = “(COST + COS p)a (T(vm) — T(vm—lllIILl+l(c(M)) -

Then for m = 0,1,2,... we have
.. 1
22) R171+1 S ERm.

Proof. We will argue by induction. For m = 0, it is easy to see (i) and (ii).
For j,m Z 0, vm+1 — vj+1 satisfies the inhomogeneous equation with vanishing

Cauchy data

who...“ — v.41) = p"3(lT(vm)|’ — |T(vj)|’),
(’Um+1 — ‘Uj+1)(0sP) = 0, (4'13)

5T(vm+1 - vj+1)(0iP) = 0.

We can write the right hand side of this equation as

p’-3(IT<om)l’ — |T(vj)l’) = Cp’-3(IT<om)I + |T(vj)l)"1(T(vm) — T(oj».

77

where G is a positive constant. We let q = l+1 and 6 > —a—1 and apply Theorem 4.1
to equation (4.13) with this choice of 6 and q. We have to make sure that conditions
of Theorem 4.1 are satisfied for this 6 and q.

(i) Condition on q is 3 < q < 4:

Since2<l<3,weget3<l+l=q<4.

(ii) Condition on a,fl is a + (3 > —1:

 

 

 

— 3-2l l—2 3—2l
' 3 — — =— =—1.
Sincea>l_1and[> a 1 [_1,wegeta+fi>l_1+l_l
(iii) Condition onaisa>1—ZI-:
l—2 3 l—2 .
a>m>l—l+1—l+lh01dSSlDC€l>2.

Thus, by Theorem 4.1 we have the following estimate:

II (COS TCOS p)a(vm+1 - vj+1)TIILl+1 (c(M))

_ l-l
so (coszrcosp)"+1pl 3 IN) )1 + (T(v)) (T(o )- T(v)) i i .
II ( m J ) m J “LT—(c(M))
. 1 . .
Slnce p = §(cosT + cos p), we can rewrite thls as

“pa(vm+1 _ vj+llIILl+l(c(M)) S

C a+1+l—3 T m T . l—lT m —T , .
no (I (o )|+| (o.)l) < (o ) <o.))IILaFi(C(M))

The exponent a + 1 + l - 3 is equal to al. Therefore, the left hand side (LHS) of the

above inequality is bounded by

l
m
(LHS)sc / p"(l+1)(T(vm)-T(vj))Lll(lT(vm)l+lT(vj)I)fli2’Qfl dpdT
(M)

78

 

Now, we use Hélder’s inequality with conjugate numbers l and and get

l — 1
(LHS) S C IIP"(T(vm) - T(vj))llLl+1 IIPO(IT(Um)I + IT(vj)I)IIlL—Hl»1 -

So we end up with

IIPa(vm+1 — Uj+1lTIILl+1 S C IIPa(T(’Um) — T(vjllIILHl (Qm + Qj)l_1- (4-14)

Ifj = —1,Qj = 0. Hence we get

1

1 . _
IIP0(Um+1 - volTIILHI S 2Qm’ s1nce CQinl < -

2. (4.15)

by the assumption of sufficiently small data. With a similar argument, using Theorem

4.2, we also get
a - 1 - l-l 1
II? smp (vm+1 _ ”OlplIILl+l(c(M)) S EQm smce CQm < '2' (4-16)
We also have from earlier argument (from the proof of Lemma 3.1) that
a 1 . (_1 1
IIP (Um-F1 _ ”OlIILl+1(C(A,I)) S §Qm Slnce CQm < 5 (4°17)

When we add up (4.15), (4.16) and (4.17) we get

NIH

Qm + Q0 since 0625,11 <

Nah--l

IIPOT(vm+1)IILl+1(C(M)) S

Then by the induction hypothesis we get Qm+1 S 2Q0, which is part (i) of the lemma.

In order to prove part (ii) we let j = m — 1 in (4.14). Then

IIPa(vm+1 — UranIILlH S CIIPO(T(vm) - T(vm—lllIILHl (Qm + Qm—1)l_1 (4-18)

79

With a similar argument using Theorem 4.2, we also get

“pa sinp (vm+1”vm)pIILl+l(c(M)) S CIIpa(T('Um) —T(vm-1))IILI+1(C(M)) (Qm‘I’Qm—lll—1-

(4.19)

We also have by the proof of Lemma 3.1, that

”pew...“ - pony... s alpaca...) — T(om..i))uLai (em + easy-1. (4.20)
If we add (4.18), (4.19), and (4.20) together, we get

IIP“(T(vm+1) - T(vm))llL1+1 S Cllp"‘(T(vm) - T(vm—l))llL1+1(Qm + Qm—1)l_1-

Therefore we get Rm+1 S éRm since C(Qm + C,2,n_1)l—1 < % and this finishes the

proof. C]

Proof of Theorem 4.4: The proof will follow the same steps the proof of Theorem

3.2 using Lemma 4.1 and therefore we will just refer to the proof of Theorem 3.2. D

80

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