Solvability conditions of the Cauchy problem for the system of internal waves

Selcuk Journal of

Applied Mathematics

Solvability conditions of the Cauchy problem

for the system of internal waves

Inessa Matveeva

?

Sobolev Institute of Mathematics, SB RAS, 630090 Novosibirsk, Russia e-mail:[email protected]

Received: February 15, 2000

Summary.

The author studies the Cauchy problem for a system of integrodifferential equations describing internal waves for the Boussi-nesq approximation. The necessary and sucient conditions of solv-ability in weighted Sobolev spaces are established.

Key words:

integrodifferential equations, Cauchy problem, condi-tions of solvability, weighted Sobolev spaces

Mathematics Subject Classication (1991): 35M05, 35Q35

Sel cuk J. Appl. Math. Vol. 2, No. 1, pp. 69{82, 2001

1. Introduction

In the present paper we study the Cauchy problem for the following system of integrodifferential equations

(1) D t u 1+ D x 1 p= 0 D t u 2+ D x2 p= 0 D t u 3+ ! 2 t R 0 u 3 ds+D x 3 p= 0 divu= 0

where !>0 is a real constant. The system describes internal waves

for the Boussinesq approximation.

? The research was nancially supported by the Russian Foundation for

Ba-sic Research (No. 99-01-00533) and by the Ministry of education of the Russian Federation.

System (1) is not a system of the Cauchy { Kovalevsky type. At present there are a great number of works devoted to the research of equations and systems not solved with respect to the highest deriva-tive (for example, see the bibliography in the book 1]). These inves-tigations show that the theory of boundary value problems for such equations and systems has certain peculiarities in comparison with the theory of boundary value problems for classical equations and systems.

2. Denitions and main results

Consider the Cauchy problem for (1) in the half-spaceR + 4 = f(tx) : t>0 x2R 3 g (2) D t u 1+ D x 1 p= 0 t>0 x2R 3  D t u 2+ D x 2 p= 0 D t u 3+ ! 2 t R 0 u 3 ds+D x3 p= 0 divu= 0 u j j t=0= u 0 j( x) j = 123 x2R 3 : Suppose that r 1 and r 2 are integers, 1 <q<1,>0. Introduce

the following weighted functional spaces. By L

q  we denote the space of functions

v(tx) with the norm kv(tx)L q ( R + 4) k=ke ;t v(tx)L q( R + 4) k: ByW r1r2 q  ( R +

4) we denote the weighted Sobolev space of functions v(tx) with the norm

kv(tx)W r 1 r 2 q  ( R + 4) k=kD r 1 t v(tx)L q ( R + 4) k + X jjr 2 kD  x v(tx)L q ( R + 4) k: By L 1( R

3) we denote the space of functions

v(x) with the norm kv(x)L 1( R 3) k=k(1 +jxj) ; v(x)L 1( R 3) k:

Results obtained in 2] give the following assertion for the Cauchy problem (2).

Theorem 1.

Let u 0 j( x) 2 W 1 q( R 3), j = 12, u 0 3( x) 2 W 1 q( R 3) \ L 1( R

3), and let the compatibility condition

(3) divu

0( x) = 0

be satised. If q > 3=2 then the Cauchy problem (2) has a unique solution (4) u j( tx)2W 11 q ( R + 4)  j= 123 p(tx)2W 02 q ( R + 4)  >0 and the following estimate is valid:

3 X j=1 ku j( tx)W 11 q ( R + 4) k+kp(tx)W 02 q ( R + 4) k c  3 X j=1 ku 0 j( x)W 1 q( R 3) k+ku 0 3( x)L 1( R 3) k  

where the constant c>0 is independent of u 0(

x).

The question is whether the restriction (q >3=2) on the summand

exponent is essential or not. We answer to the question in the present paper.

Theorem 2.

Let u 0 j( x) 2 W 1 q( R 3), j = 12, u 0 3( x) 2 W 1 q( R 3) \ L 1;1( R

3), and let condition (3) be satised. The Cauchy problem (2) has a unique solution (4) for 1<q 3=2if and only if

(5) Z R3 u 0 3( x)dx= 0:

Note that, for the rst time, the necessity of additional require-ments on data was discovered by S. A. Gal'pern 3] while construct-ing theL

2-theory of the Cauchy problem for a class of systems not of

Cauchy { Kovalevsky type. Analogous peculiarities for mixed bound-ary value problems were observed by G. V. Demidenko 4].

In section 3, we construct a sequence of approximate solutions of the Cauchy problem (2). In section 4, we establish L

q-estimates for

the sequence and prove Theorem 2. Deriving the estimates we follow the scheme proposed by G. V. Demidenko in 4] and described in detail in 1].

3. Approximate solutions

Consider the following algebraic system with the parameters = +i2C,>0,2R 3 nf0g (6) v 1+ i 1 v 4 = f 1( ) v 2+ i 2 v 4 = f 2( ) v 3+ ! 2 ;1 v 3+ i 3 v 4 = f 3( ) i 1 v 1+ i 2 v 2+ i 3 v 3 = 0 :

Suppose that the functionsf j(

),j= 123, satisfy the condition

(7) i 1 f 1( ) +i 2 f 2( ) +i 3 f 3( ) = 0:

System (6) and condition (7) result from formal application of the Fourier transform inxand the Laplace transform intto the Cauchy

problem (2) and condition (3). It is obvious that a solution of (6) can be written as follows v 1( ) =; 1 i 1 v 4( )+ 1 f 1( ) (8) v 2( ) =; 1 i 2 v 4( )+ 1 f 2( ) v 3( ) =; 2+ ! 2 i 3 v 4( )+ 2+ ! 2 f 3( ) (9) v 4( ) = i! 2  3 2 jj 2+ ! 2(  2 1+  2 2) f 3( ): Since v j(

),j= 1:::4, are analytic and bounded functions of ,

Re =>0, by Theorem 5.2 of 1, Chapter 1] the functions w j( t) = (2) ;1=2 1 Z ;1 e (i +)t v j( i+)d j= 1:::4 do not depend on>0.

Now we construct an approximate solution of the Cauchy prob-lem (2). We could obtain a formal solution of the probprob-lem by ap-plying the inverse Fourier transform in x to the functions w

j( t), j = 1:::4. However, the function w

4(

t) has nonintegrable

Fourier operator. For this purpose we use the integral representation proposed by S. V. Uspenski 5] for functions f(x)2L

q( R n): (10) f(x) = lim h!0 (2) ;n h ;1 Z h v ;1 Z Rn Z Rn e i(x;y ) G(v)f(y)ddydv whereG() = 2Njj 2Nexp( ;jj 2N),

N is an arbitrary natural number

(see obtaining the representation in 1, Chapter 1]). Denote the Fourier transform of u

0 3( x) by ^u 0 3( ). Let f 3( ) = (2) ;1=2^ u 0 3(

). Dene the functions

(11) p h( tx) = (2) ;3=2 h ;1 Z h v ;1 Z R 3 e i x G(v)w 4( t)ddv = (2) ;2 h ;1 Z h v ;1 Z R3 Z R1 e (i +)t+i x G(v)  i! 2  3 2 jj 2+ ! 2(  2 1+  2 2)^ u 0 3( )dddv u 1h( tx) =u 0 1( x); t Z 0 D x 1 p h( sx)ds (12) u 2h( tx) =u 0 2( x); t Z 0 D x 2 p h( sx)ds u 3h( tx) = cos(!t)u 0 3( x); t Z 0 cos(!(t;s))D x 3 p h( sx)ds:

It is easy to prove that

D t u 1h+ D x1 p h 0 D t u 2h+ D x 2 p h 0 D t u 3h+ ! 2 t R 0 u 3h ds+D x 3 p h 0 u jh j t=0= u 0 j( x) j= 123 and, by representation (10), kD x 1 u 1h+ D x 2 u 2h+ D x 3 u 3h L q( R 3) k!0 h!0:

Therefore, we can consider the vector-function (u h(

tx)p h(

tx)) as

4. Necessary and su cient conditions of solvability

In this section we prove that condition (5) is necessary and sucient for the Cauchy problem (2) to be solvable in the weighted Sobolev space W r 1 r 2 q  ( R + 4) for 1 <q3=2.

First, we establish estimates of the approximate solution (u h( tx) p h( tx)):

Lemma 1.

Let u 0 j( x) 2 W 1 q( R 3),

j = 123, and let compatibility condition (3) be satised. Then,

(13) 3 X j=1 ku jh( tx)W 11 q ( R + 4) k+ 3 X k =1 kD x k p h( tx)W 01 q ( R + 4) k c 3 X j=1 ku 0 j( x)W 1 q( R 3) k q>1 where the constant c>0 is independent of u

0( x). Furthermore, (14) 3 X j=1 ku jh 1( tx);u jh 2( tx)W 11 q ( R + 4) k + 3 X k =1 kD x k p h1( tx);D x k p h2( tx)W 01 q ( R + 4) k!0 h 1  h 2 !0: Proof. Now we prove that

(15) 3 X k =1 kD x k p h( tx)L q ( R + 4) kcku 0 3( x)L q( R 3) k: From (11) we obtain e ;t D x k p h( tx) =;(2) ;2 h ;1 Z h v ;1 Z R 3 e i x G(v)  Z R1 e i t ! 2  3  k 2 jj 2+ ! 2(  2 1+  2 2)^ u 0 3( )dddv:

Using properties of the Fourier transform, we have (16) e ;t D x k p h( tx) =c h ;1 Z h v ;1 Z R Z R e i(x;y )s G(sv)

 8 < : Z R 3 Z R 1 e i y +i t ! 2  3  k 2 jj 2+ ! 2(  2 1+  2 2)^ u 0 3( )dd 9 = dsdydv:

Denote the expression in the curly braces by P(ty). Establish the

following estimate (17) kP(ty)L q( R + 4) kcku 0 3( x)L q( R 3) k: Since 1 Z 0 e ; d = 1=  =+i >0

the functionP(ty) can be written as follows P(ty) = Z R 4 e i y +i t () Z R 4 e ;i ;i x e ; ( )u 0 3( x)d dxdd where () = ! 2  3  k 2 jj 2+ ! 2(  2 1 +  2 2) 

( ) is the Heaviside function. It is not hard to verify that the function () satises the conditions of Lizorkin's theorem on multipliers6].

Hence, kP(ty)L q( R 4) kcke ; ( )u 0 3( x)L q( R 4) kcku 0 3( x)L q( R 3) k

and (17) is proved. By integral representation (10) and estimate (17), from (16) we obtain (15).

Similarly, it is easily shown that (18) 3 X k =1 3 X j=1 kD 2 x k xj p h( tx)L q ( R + 4) kc 3 X l=1 kD x l u 0 3( x)L q( R 3) k: Estimatethe functionsu jh(

tx),j= 123.Consider the following

function v(tx) = t Z 0 g(t;s)w(sx)ds

where jg(t)jc fort 0, the function w(tx) belongs toL q (

R + 4).

Now we prove thatv(tx) belongs to the spaceL q (

R +

4) too. Rewrite

the functionv(tx) as follows e ;t v(tx) = t Z e ;(t;s) g(t;s)e ;s w(sx)ds

= 1 Z ;1 e ;(t;s) (t;s)g(t;s)e ;s (s)w(sx)ds=f 1( t)f 2( tx) where f 1( t) =e ;t (t)g(t) f 2( tx) =e ;t (t)w(tx):

From this representation we have

kv(tx)L q ( R + 4) kkv(tx)L q ( R 4) k =kkv(tx)L q ( R 1) kL q( R 3) k =kkf 1( t)f 2( tx)L q( R 1) kL q( R 3) k:

Applying Young's inequality, we obtain

kv(tx)L q ( R + 4) kkkf 1( t)L 1( R 1) kkf 2( tx)L q( R 1) kL q( R 3) k =ke ;t g(t)L 1( R + 1) kkw(tx)L q ( R + 4) k c  kw(tx)L q ( R + 4) k:

Thus, taking into account (15) and (18), from (12) we obtain

ku k h( tx)W 11 q ( R + 4) kku 0 k( x)W 11 q ( R + 4) k +kD x k p h( tx)L q ( R + 4) k+k t Z 0 D x k p h( sx)dsL q ( R + 4) k + 3 X j=1 k t Z 0 D xjx k p h( sx)dsL q ( R + 4) kc 3 X j=1 ku 0 j( x)W 1 q( R 3) k k= 12 ku 3h( tx)W 11 q ( R + 4) kkcos(!t)u 0 3( x)W 11 q ( R + 4) k +kD x 3 p h( tx)L q ( R + 4) k +!k t Z 0 sin(!(t;s))D x3 p h( sx)dsL q ( R + 4) k +k t Z 0 cos(!(t;s))D x 3 p h( sx)dsL q ( R + 4) k + 3 X j=1 k t Z cos(!(t;s))D x j x 3 p h( sx)dsL q ( R + 4) k

c 3 X j=1 ku 0 j( x)W 1 q( R 3) k:

Estimate (13) is proved. The proof of (14) is carried out by the same scheme. ut

Lemma 2.

Letu 0 3( x)2L q( R 3) \L 1;1( R

3) and let condition (5) be satised. Then, (19) kp h( tx)L q ( R + 4) k c  ku 0 3( x)L q( R 3) k+ku 0 3( x)L 1;1( R 3) k   where the constant c>0 is independent of u

0 3( x). Furthermore, (20) kp h 1( tx);p h 2( tx)L q ( R + 4) k!0 h 1  h 2 !0: Proof. By (11) we have e ;t p h( tx) = (2) ;2 1 Z h v ;1 Z R 4 e i t+i x G(v)  i! 2  3 2 jj 2+ ! 2(  2 1 +  2 2)^ u 0 3( )dddv+ (2) ;2 h ;1 Z 1 v ;1 Z R 4 e i t+i x G(v)  i! 2  3 2 jj 2+ ! 2(  2 1+  2 2)^ u 0 3( )dddv=P 1( tx) +P 2( tx):

Consider the rst summand. Rewrite it as follows

P 1( tx) = (2) ;2 1 Z h v ;1 Z R 4 e i t+i x G(v)  i! 2  3 2 jj 2+ ! 2(  2 1 +  2 2) Z R 4 e ;i ;i y e ; ( )u 0 3( y)d dydddv:

It is not hard to verify that the function

jj i! 2  3 2 jj 2+ ! 2(  2 1+  2 2)

satises the conditions of Lizorkin's theorem on multipliers. Conse-quently, kP 1( tx)L q( R + 4) kck 1 Z h v ;1 Z R3 e i x G(v)jj ;1^ u 0 3( )ddvL q( R 3) k:

Taking into account a formula for the inverse Fourier transform of a product, by Minkowski's inequality we obtain

kP 1( tx)L q( R + 4) k c 1 Z h v ;1 k Z R 3 Z R 3 e i(x;y ) G(v)jj ;1 u 0 3( y)ddyL q( R 3) kdv:

According to Young's inequality, we nd that

kP 1( tx)L q( R + 4) k c 1 Z h v ;1 k Z R 3 e i x G(v)jj ;1 dL 1( R 3) kdvku 0 3( x)L q( R 3) k:

Using the change of variables

s j =  j v j= 123 y l= x l =v l= 123 we have kP 1( tx)L q( R + 4) k c 1 Z h dvk Z R 3 e isy G(s)jsj ;1 ds L 1( R 3) kku 0 3( x)L q( R 3) k:

Choosing suciently largeN in the denition ofG(s), we obtain

(21) kP 1( tx)L q( R + 4) kcku 0 3( x)L q( R 3) k:

Estimate the function P 2(

tx). From (5) it follows that

^ u 0 3( x) =;(2) ;3=2 1 Z 0 Z R 3 e ;i y( iy)u 0 3( y)dyd:

Arguing as above, it is easy to show that

kP 2( tx)L q( R + 4) kc 3 X k =1 h ;1 Z v ;1

k Z R 3 e i x G(v)  k jj dL q( R 3) kdvkx k u 0 3( x)L 1( R 3) k:

Using the change of variables

s j =  j v j= 123 y l= x l =v l= 123 we have kP 2( tx)L q( R + 4) kc 3 X k =1 h ;1 Z 1 v ;1;3+3=q dv k Z R 3 e isy G(s) s k jsj dsL q( R 3) kku 0 3( x)L 1;1( R 3) k:

Since the integral 1 R

1 v

;4+3=q

dv converges for q > 1, choosing

su-ciently large N in the denition ofG(s), we obtain

(22) kP 2( tx)L q( R + 4) kcku 0 3( x)L 1;1( R 3) k:

Estimate (19) follows from (21) and (22). In analogous way, one can prove (20). ut

Taking into account (14) and (20), by completeness of the Sobolev space W r 1 r 2 q  ( R +

4), there exist functions u j( tx) 2 W 11 q ( R + 4), j = 123,p(tx)2W 02 q ( R + 4) such that ku jh( tx);u j( tx)W 11 q ( R + 4) k!0 kp h( tx);p(tx)W 02 q ( R + 4) k!0

ash!0. Moreover, according to (13) and (19), it holds that 3 X j=1 ku j( tx)W 11 q ( R + 4) k+kp(tx)W 02 q ( R + 4) k c  3 X j=1 ku 0 j( x)W 1 q( R 3) k+ku 0 3( x)L 1;1( R 3) k  :

It is easily shown that the vector-function (u(tx)p(tx))is a solution

of the Cauchy problem (2).

Thus, we proved the rst part of the assertion of Theorem 2 i.e., we showed that condition (5) is sucient for the Cauchy problem (2) to have solution (4). As follows from Lemma 1,

u(tx)2W 11 q ( R + 4)  rp(tx)2W 01 q ( R + 4)

forq >1. Therefore, to obtain the second part of assertion of Theorem

2 we have to prove that condition (5) is necessary for the function

p(tx) to belong toL q ( R + 4) for q3=2.

Lemma 3.

Letu 0 3( x)2L q( R 3) \L 1;1( R

3). For the function p(tx) to belong to L q ( R + 4) for q 3=2 it is necessary to have Z R 3 u 0 3( x)dx= 0: Proof. Suppose that Z

R 3 u 0 3( x)dx6= 0 and p(tx)2L q ( R + 4) for q 3=2. Since kp(tx)L q ( R + 4) kc<1

then, by the Hausdor{Young inequality, we have

kp~( )L q 0(R 4) kc<1 1 q + 1 q 0 = 1  where ~ p( ) = (2) ;2 1 Z 0 Z R 3 e ; t;i x p(tx)dtdx =i+: Hence, sup ">0 1 Z ;1 Z "<j j<1 jp~( )j q 0 ddc:

Obviously, the functions ~ u j( ) = (2) ;2 1 Z 0 Z R 3 e ; t;i x u j( tx)dtdx j= 123

and the function ~p( ) satisfy system (6) for f j( ) = (2) ;2 Z R e ;i x u 0 j( x)dx j = 123

and are given by (8), (9). By Hadamard's lemma, the function ^u 0 3(

)

can be represented as follows ^ u 0 3( ) = ^u 0 3(0) + 3 X j=1  j U j( ) where U j( ) = (2) ;3=2 1 Z 0 Z R 3 e ;i x( ;ix j) u 0 3( x)dxd: Consequently, ~ p( ) = (2) ;1=2 i! 2  3 2 jj 2+ ! 2(  2 1+  2 2)^ u 0 3(0) +(2) ;1=2 i! 2  3 2 jj 2+ ! 2(  2 1 +  2 2) 3 X j=1  j U j( ) =J 1+ J 2 : It is obvious that sup ">0 1 Z ;1 Z "<j j<1 J 2 q 0 ddcku 0 3( x)L 1;1( R 3) k q 0 c: Hence, sup ">0 1 Z ;1 Z "<j j<1 jJ 1 j q 0 ddc: By denition, jJ 1 j= (2) ;1=2 j 3 j jj 2 ! 2 2+ ! 2(  2 1+  2 2) =jj 2 ju^ 0 3(0) j (2) ;1=2 j 3 j jj 2 ! 2  2+ 2+ ! 2 ju^ 0 3(0) j: Then, sup ">0 Z  j 3 j jj 2  q 0 ju^ 0 3(0) j q 0 dc:

Sinceq 3=2 it follows that the function (j 3 j=jj 2)q 0 is not integrable in a neighborhood of= 0. Therefore, if condition (5) is not fullled,

then ^u 0 3(0) 6 = 0 and lim "!0 Z "<j j<1  j 3 j jj 2  q 0 ju^ 0 3(0) j q 0 d=1:

This contradiction concludes the proof. ut

Thus, we proved that condition (5) is necessary and sucient for the Cauchy problem (2) to have solution (4). We omit the uniqueness proof. It is carried out by the scheme described in detail in 1].

Theorem 2 is proved.

References

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