initial-boundary value problems for hyperbolic systems in regions with ...

TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 176, February

1973

INITIAL-BOUNDARYVALUE PROBLEMSFOR HYPERBOLIC SYSTEMSIN REGIONSWITHCORNERS. 1(1) BY

STANLEY OSHER ABSTRACT. In recent papers Kreiss and others have shown that initialboundary value problems for strictly hyperbolic systems in regions with smooth boundaries are well-posed under uniform Lopatinskii conditions. In the present paper the author obtains new conditions which are necessary for existence and sufficient for uniqueness and for certain energy estimates to be valid for such equations in tegions with corners. The key tool is the construction of a symmetrizer which satisfies an operator valued differential equation.

I. Introduction.

estimates

for mixed

differential

symmetrizer

problem

with variable improved

[ill,

in regions

with corners.

seems

[6] has

to have

recently

in regions

real

with

[lOl,

to permit

estimate. higher

in Kreiss'

classical

and uniqueness

in the nature

energy

of the symmetrizer

was

and by Rauch

obtained

work was done by Hersh

results

estimates.

the

operator

His result

independently

Earlier

around

problems.

initial-boundary

coefficients,

[12]

partial

hyperbolic

L2 norm.

valued

Sakomoto

value

for a strictly

order equation.

obtaining

hyperbolic

type of mixed

in the

complex

existence

without

work came

boundary

is well-posed

for a single

coefficients

that a general

a smooth

energy

The work will revolve

of its own.

a semigroup

constant

for certain

a significance

results

l3J, who obtained

problems

initial-boundary

to obtain

similar

will be to obtain

for general

shown

coefficients

by Ralston

of papers

value

of a new symmetrizer

Kreiss value

series

initial-boundary

equations

introduction This

The aim of this

for equations

An important

he used

with

new idea

in order to obtain

his estimate. The symmetrizer homogeneous appear

in the Kreiss

analogous

we shall

differential

condition

Lopatinskii

use the symmetrizer

result

in a simpler

below

These

involves

for elliptic

introduced

certain

are the same

for well-posedness,

condition

paper,

Received

introduce

equation.

and indeed

systems

here to obtain

[8l.

solutions

exponential which

We shall,

special

cases

to a

solutions appear

that in the

in a later

of Kreiss'

fashion.

by the editors

December

27, 1971.

AMS(MOS)subject classifications (1970).Primary 35L50, 35L30; Secondary 78A45. trizer,

Key words and phrases. Hyperbolic energy estimate, well-posedness.

(') Research

partially

supported

equations,

under N.S.F.

141

initial

boundary

conditions,

symme-

Grant GP29-273. CopyrightC 1973, Americin Mjthemiticil Society

License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use

142

STANLEY OSHER In the regions

space

problem

give

a simple

natural

with corners

need

counterexample

and necessary

does

exist,

down the answer

has certain

a weakening

elliptic

boundary here.

The situation of Toeplitz

invertibility

invertibility Special

value

to the condition

in a natural

We then construct

for constant we shall

It should

remains

encountered operators

paper

the functional

be noted

that

that

this

sym-

calculus

in the elliptic

[4] has shown

near a corner,

an

In order to do

in a future

Kondrat'ev

for case,

that

in distinction

an

to what

section.

here appears

to be analogous

in two dimensions,

of the corner hyperbolic

such

way, and we can

coefficients.

show

is both

a new norm, with which

and hence

in the last

of the two half space

in connection

which

applies.

this

half

We shall

coefficients

properties,

problem

for each

problem.

for constant

of the norm at the corner,

We discuss

condition

a new condition

coefficients,

smoothness

Kreiss

of the corner

if a solution

at least

operators

modulo

happens

case.

is obtained,

pseudo-differential

study

in this

with variable

metrizer

below,

We introduce that

then we are led directly

estimate

problems

below.

in the sense

write

a priori

discussed

not imply well-posedness

[February

problems

e.g.

to that found in the

Douglas

is necessary,

and Howe [2l, where

but not sufficient,

for

problem.

differential

with water

waves

equations

in such regions

over sloping

wedge

problems

in optics,

e.g.

Kraus

author

[7], used

a new Weiner-Hopf

beaches,

and Levine

technique

e.g.

[5].

have

Peters

Recently,

to obtain

a closed

been

studied

[9], and in Kupka

and the

form solution

to

such a problem. II. Statement system

with

of the problem;

constant

results.

We begin

by considering

the hyperbolic

coefficients, n

(2.1)

Au + Bu + 22 C.u y

where

z = (zol3 z,,---, 4''«'

tem is hyperbolic

y=3

z ), u and

in the sense

(2.2)

'

Zi

+ Du - u = F(x, y, z, t),

F are complex r

that

for the

valued

m x m matrices

m vectors

A, B, C

and the sys-

the matrix

det [Ait; + Bit] + Cico - A] = 0

has only purely

imaginary

roots.

Here Cioj = 2"=3 Cico-, co = (d>3, • • • , ù>n), and 0. (We shall square

to solve

this

x, y, ¿>0;

(2.3)

use the convention

root of the sum of the squares

We wish

with

always

initial

problem

that

| | of a vector

of its components.) in the region:

-°° < z. < °°,

/' = 3, 4,• • •, n,

conditions

u(x, y, z, 0)= 0.


is the

+

1973]

INITIAL-BOUNDARY VALUE PROBLEMS. I We further

assume

that

0

(2.4)

143

■••

a. 0, k= p+ I,-.-,

m.

b '

conditions

u\0,

\

' / '••V

0

boundary

that

of the form

y, z, t) - SuU{0, y, z, t) = g(y, z, t),

(2.6) (b)

{Tu)m(x, n, z, t) - R{TtAlV(x, 0, z, t) = h{x, z, t),

where

u1 = («j,.. III

« 5 and rows,

/ =(«j,...,

R are constant

(m ~ p) columns

It is clear the corner

the two half

space

. with

an=

\T

,

«

IV

having

(ai+1>- • •, «m)T, / = ("p+1,---,

/ rows

«„)

and

\T

■

(m — I) columns,

and

p

respectively.

speed

of propagation

to be well-posed

considerations,

in any reasonable

that in order for

sense,

it is necessary

that

problems:

Equation

(a)

af) matrices

from finite

problem

• , tvpT,

(2.1)

boundary

in region

0 < x,

condition

(2.6(a))

t < oo, —oo < y, z. < oo,

?

and Equation

(b)

(2.1)

.

in region

0 < y, t < oo, - oo < x, z. < oo,

t

with boundary condition

^

(2.6(b))

be well-posed. We first transform

Laplace

transform

in y and all the

(2-7) [it is easily

z.

(2.1)

in time and use

variables.

We have,

ux + A~1{Bíxú2 + ¡Coi - s)u= shown

may be set equal

that the matrix to zero

D plays

(2,3).

In problem

(a), Fourier

if F = 0,

0,

only a trivial

throughout.]


x > 0. role in this problem

and

144

STANLEY OSHER It is easily

seen

that

Kreiss'

co ?, to are real, We can show linearly

(e.g.

independent

condition

[3J), that

them at x = 0, and obtain

(2.8)

this which

condition

with

ordinary decay

for both half spaces

for all

77> 0, £ real.

differential

equation

exponentially

as

has

exactly

/

x —>+ 00. Normalize

/ functions

$y(x, ú>2,ú>, s), Kreiss'

is valid

s = r¡ + iç

solutions

[February

7= 1, 2,— , /.

is, at x = 0,

(2.9)

determinant

[/, - S] [ {, ■. . , 4^] / 0

for all oj2, co, s, with Real s > 0. This formed

condition

boundary

is natural

condition

in that

if one wishes

of (2.6(a)),

to solve

the inverse

of this

(2.7) with the trans-

matrix

must

appear

in

the solution. Kreiss'

condition

for problem

(b) is completely

analogous.

We make an assumption. (2.1)

Assumption.

Kreiss'

We now proceed

by first

condition

assuming

for problems

that a solution

F = 0, h = 0, and then

seeing

(2.1')

Aux + Bu y + (Cico - s)u = 0,

(2.6')

(a)

(2.6')

(b) Let

what this

is valid

assumption

ul{0, y, co, s) - Sull(0,

We have

x, y>0. J —

y, cd, s) = g{y, to, s),

v be the

function

defined

for

y>0,

— «> 0, which

if 0 < y < oo,

if -oo < y < o.

if obeys

Aw + Bw + (Ci(ú-s)w x

= ß(3(y - 0)íXx, 0, oj, s),

y

x > 0,

(2.15)

with boundary

J

conditions

wKo, y, oj, s) - SwU(Q, y, oj, s) = g(y, oj, s)

if y > 0,

wKo, y, oj, s) - S^'HO, y, co, s) = 0

if y < 0.

We can Fourier

transform

in the analogous

solution

if we require

for x > 0, — oo < y < oo with

w\x, y, co, s) = 0

plane

0,0;, s) = 0,

0, tu, s) for x > 0.

let

w(x,

Then

145

does

indeed

in y and proceed

fashion exist

as above,

to this

to solve

finally

problem,

this

obtaining

in the right

half

w{0, y, co, s).

If a

then it must be true that

- (2.27)(b)

real,

the estimate

'dy{l+

the equation

this

method

Moreover,

we assume

in the manner we shall

discussed

derive

our first

above

main theorem

in (2.10)

to (2.18)

below.

will not be applicable

the existence

similar

((/ - TM )~ )*.

it is not too hard to obtain

of the estimates

Moreover,

oj, £.

operator

assumptions,

dy(i + \co\i+\s\2)K>U)dy,

in the case

of a symmetrizer

of variable

has an independent

coeffi-

abstract

interest.

The operators Kondrat

ev's

(y (d/dy)r

paper

[4] on elliptic

them here for technical of the matrices

K2 = 0 does

right-hand

reasons

in (2.24)

certain

However,

e.g.

Kupka

and Osher

tion

is often

cuss

this

disguised

in conical

concern

(K2

fashion

regions.

certain

if there

of eigenvalues

exist

is the constant

in

We introduced

multiplicities

We are not sure

K~ > 0 does.

matter

the solvability integral

sometimes

itself

[71. The

quite

equations

(2.6)

but

singular

to handle.

integral

equations

which

and (2.25).

not suffice,

in a somewhat

cases

in which

appearing

on the

side of (2.27)(a).)

In order to check analyze

appear

here

in a forthcoming

Definition. u(x,

Let

vector

function

which

are true for each

exact

solutions which

Thus,

with estimate

which,

algebra

nontrivial.

arising

of (2.26),

equations

of these must

(2.27),

in general,

equations

be checked

it is not surprising

often require

special

we must

are rather

can be obtained,

in the Kreiss

that

analysis.

difficult

condi-

the singular

We expect

to dis-

paper.

S(x, y, co, s) be the Fourier-Laplace

y, x, t). We shall

obtain

estimates

on/

transform Jz/(x,

of any

y, co, s)\ dxdy

co, s.

Define for fixed ru, s

(2-28)

W\2p_Q- f^1

de ¡~(l+\co\2+

\s\2)Vrdr


£

\(r £Ju(r,

6, co,

148

STANLEY OSHER

Q is some real analogous

number,

P is a nonnegative

integer,

defined

on the boundary,

e.g.

53(ex/2?, e>/2f0

(2.39)

-K,

£

(e-X/2{d/d\Yb,

e~k/2(d/dXYh)(l + |oj|2 + |s|2)

o

7=0

III. Proof of Main Theorem

the auxiliary

(3.1) with the boundary

(3.2)

I using

the symmetrizer.

We begin

problem

Adx + Bvy + {Cico-s)v=

F,

x, y>0,

conditions

?'(0, y, oj, s) = 0,

(TD) ln(y, 0, oj, s) = 0.


by considering

2.

150

STANLEY OSHER

If we inner

product

the equation

part of both sides,

(3.3)

with

v, integrate

[February

over

x, y > 0, and take

the real

we have

(i-«)B^5,9+xg|îbii2i0+ic9iîb2i2>o r¡0 > 0, with

only consider

[6] and Ralston

r¡0 any positive

U in the neighborhood

[10]

have

(4.7)

shown

that

of every

there

point

exists

constant.

Thus

we

C = Cn> T] = 0. Kreiss

a smooth

W, with

W~lM\W

K where Re /Vj- 23^;

(4.8)

for 7/ = 0, £' = £q, / > 2,

M.(¿¿, 0)

k. is purely imaginary,

(4.9)

k. /= k. if / ¡¿ /,

M.(C, r,') = M.(¿¿,0) + JN.iO + 0((r,')2)+ Ey(£').

where

£ .(£ ) has

lower

left corner

For each

purely

imaginary

entries,

of N 7{¿, ), which

block

AL, consider

Det(Alf -*')=

(4.10)

we call

£.(£»)

= 0, and the element

n s .L , has

nonzero

the characteristic

real

in the

part.

equation

Det(AI;' - k!. - (*' - k'.))

= (*'. _ k')s + i»"1,',, 7

, + '*(£' - C>

'si

0

,

s 1

+ odT,'i2+ K'-g2), Re nsl

¿0,lmmsl

Thus

the roots

= 0. are

(4.11) k! = k'.- l-P-W'st-

W

- *,+

ocivi2 + K* -¿¿I2)]1*.

Th is means that, if |£' - £'0\ < k(¿,'0)[t]']

, the roots of M, split up with l/2s,

l/2(s

part if 77 > 0. The remainders

+ 1), or l/2(s

have negative

- 1), having positive

real

real part.

Let

U. with

transformation

in the region

branch

0 < 77 < k. AC'r),

at Ç = ¿jQ, r¡ = 0,suchthat

UP

(Y-lM.ÏÏ.

(4.14)

are taken.

;

Re m['¡ > 25^, Re M^ < - 25^. We may do this

the compactness

for each

block

and for each

of the unit ball \ç \ = 1.

We may now use this

lemma

in order to obtain

transform

(4.4) with respect

to y, multiply

variables

g = Uh. Then (4.4) becomes

h +

(4.15)

A.

The result

then

follows

by

Q.E.D. by A"

'II

'12

0

'2 2

the solutions

to (4.4).

Fourier

, then make the change

of

h= 0,

hence

exp(-Lnx)ij

b=

(4.16) where

h.

is some

(m - I) vector

valued

= F"

(4.17) as a function

which

obeys

We may proceed

function

of A,,

'exp{-Ll{x)hl

U

0

(4.4).

in the same

way and obtain

"exp(-f\

V

(4.18) where

h? is some

valued

function

(m - p) vector

obeying

(4.4).

valued

(4.19)

^J_^exp(^2 ' expl-L

lly)h2

0 function

We now switch

1

■si

s, co). We define

of (,°°) independent

/>/2)/l£=F^_

that we may premultiply

of 0, then the result

7-1

/.:[?

V

(4.26) 72*(n/2)Av= FM-2\X£ We make the change

,

0)).

= /, + ;2 •

still

satisfies

At 0 = 77/2, we consider

(4.27)

rcos

now

(4.25) In this

0'

dco

N

#*,)=

of variables

5 (-In x.)

-L,

t

(oj.) exp(/oj-(A-

"exp(-0,

e"'))Av(r)¿r,

0

0

V*(co2)Av{r)dr.

We then

/S(x) = 0,

1


have,

letting

x. S2|p| 2_0,

and, if p satisfies

(4.34), then

(4-38)

(j*{0)Bv,NJ*(0)Bv) 52(7^27-22 + l).

Let

(4.52)

^jv.-^ug^^^-,

and

(4.53)

Nn = -S2.

Thus

N21 = 52(/+T*1Tn)+Re52(-T22

- 7*, 7^(7^

- l)~ 1Tn

(4 54) = 52[l-Re(T22+r*)(Tu>s-l)-An], and

^2=4l[T22-7íT^(T«..-l)"1 _12(It V. Verification

of the estimates.

(3) of Main Theorem scalar

(5.1)

operators

II. We begin

_ t*1(t We shall by noticing

_ i)-1 _ 7*). now verify

conditions

that the operator

of the form

q(co , s, co) I

exp(-A(w2,

s, co)x)x!f(x)dx,


7*

(1), (2) and involves

1973]

159

INITIAL-BOUNDARY VALUE PROBLEMS. I

where

A and

q are

except

for certain

C°° functions algebraic

of all their

branch

points

variables

where

if s = r) + iç,

multiple

r¡ > t/„ > 0,

eigenvalues

of

B~ {\ico? - s + Cico) occur.

However,

(5.2)

because

of (2.25),

\x' exp(-A(oj2,

Let

x = 1/Re

we have

s, co)x)q(co2,

s, oj)| < Kj expi-S^rj),

x > 0.

A(oj2, s, oj); we then have

(5.3)

\q((o2, s, oj)| < Kl2[Re A(oj2, s, co)]'.

If we integrate

(5.1)

by parts

j times,

we obtain

^(oj2, s, oj)

(5.4)

/ t9V j

exp(-A(oj2,

s, co)x)[—j

• (x'f{x))dx

A?(oj?, s, co We need

now to estimate

(5.5) i.e.

J~

we view this

exp(-A(oj2, s, co)x)f{x)= PxFf,

as an operator

(5.6)

acting

on the Fourier

transform

of /:

f™ exp(-ico2x)f(x)dx= Ff= f(co2)

which takes the

the operator

lower

oj2 into half

plane

- z'A(oj2, s, co). In order into

the unit

circle

to estimate

and then

use

this,

we shall

Theorem

transform

1 of Carleson

[l].

Notice

(5.7)

/.U+l)\

[//=/

is a unitary

formation

map of L2[0,

which

takes

oo] one-to-one

f(i(z

2K

VU-1)7(1-«) onto

H

of the circle.

We seek

the trans-

+ \)/{z - 1))(1/(1 - z)) into f{- ik{i{z + l)/{z-l)))(!/(!

-z)\

Let (z + 1)

M =~U-i)

(5.8) Then

the transformation

The function

A is an eigenvalue

We can show therefore,

as

for some positive

_,.

.

z + i

\z) = -r.

is

Tg = g(p-H-iA(p(z))))

(5.9)

expansion

in question

p

of ß~

p'(z)

p'(p-H-¿A(pU)))) (A/oj2 + Czoj - s) with positive

(x>2I 0 that the function

integer

A(1/oj2)

p.


real part.

has a Puiseux

160

STANLEY OSHER

LFebruary

À-,

(5-10)

A(1A) = — + T kxüL'*-1 W2

7=1

with A_j / 0. We can differentiate

this

and obtain CO

(5-11)

X'(co2)= A_, + X A.(l - ;/>X«u2)-**. 7= 1

Thus,

the function

- z'A(cu2) is one-to-one

if co2 > M(s, co), also

if &>. ) < M0(l + \s\2 + |W|2)Ql/2

for some

öi > 0.

We shall

only estimate

the operator

in (5.5) for co2 > M. This

corresponds

to

bounding

(5.13)

j\(/A(-zA(M^))))|2

-^-f-^

lAp-H-ittpiJ0))))} for e> 0j > 0. By the one-to-one

property

of A, we may make the change

of

variables

(5.14)

z = /A(AA)~ VA)Ï = ei6,

0.15)

|pU)A| -

so we must

, '^

Q0,

We may replace

f(a>2)= Af+{l-A)f, A(co2.

co, s)

The quantity wish

We next

of the form

(5.21) where

161

I

is in C°°,

A= 1

if |oj j| > M(s, co) + 1,

A= 0

if |oj,| < Al(s, oj),

involving

(1 - A) f can be easily

0 < A < 1.

estimated

appropriately.

We thus

to estimate

(5.22)

xl

We may integrate

(5.23)

I

by parts

j times,

s, co)x)qf(co2)dco2.

and obtain

f°° exp(-A(oj„ s, co)x)U—) q(co , s, co)f(co )dco . JM r 2 \cho2 A (oj2, s, co)/

We can use (5.3), modulo

different

(5.10) K,

and (5.11)

and

In order to estimate

Q22-

of {x(d/dx)Y

on them.

(5.1) with respect

of the form

Because

K,(l

+ |oj|2

to obtain

Of course,

products

differentiate bound

exp(-A(oJ2,

the same

a similar

of these

operators

of Assumption

(2.3)

3'

. Thus

as for 7'22, is valid

we must consider or Lemma

to oj2, and the effect + |s|2)

estimate

argument

(4.1),

for T...

the effect we may

will be only an increase

we are considering

in

expressions

of the form (5.24)

j"1 dx

y > 0 and

I

q is uniformly

we have

(modulo

(5.25)

Í

This (5.26)

dco exp(;'«J2y

becomes

dx I

- A(oj2, s, to)x){co2y)'q{co2,

bounded.

Integrate

by parts

/ times

s, co)y—{xTf(x))j with respect

tooj2,

constants)

exp(/oJ2y) (-t--)

a sum of terms

[w72exp(-A(oj2, s, új)x)í7(oj2,s, oj)](—]U7('

2> of the form

J "° dx J °^ exp{ico2y-)ico2,s,co)x)[cot'2xp

q\.co2, s, oj)]i^-j


{xrf(x))dx,

dx,

162

STANLEY OSHER

p = 0, 1,.. by parts

have

., /, where p times

terms

(5-27)

in x, recall

have

2

0 /)_ Next integrate

i Q(p) + |s\z) ! > co2/X(co2,

s, co). We then

* **/(*)«&.

shown

IVIp,Q