Applied Mathematics
The Cauchy problem for
pseudohyperbolic equations
Gennadi Demidenko
1?Sobolev Institute of Mathematics, SB RAS, Novosibirsk, Russia e-mail:demidenk@math.nsc.ru
Received: 15.02.2000
Summary.
In this paper we study the Cauchy problem for a class of partial dierential equations not solved with respect to the high-est derivative. The members of this class are calledpseudohyperbolicequations. We establish the solvability conditions, obtain an \energy" inequality for solutions, and prove the well-posedness of the Cauchy problem in the weighted Sobolev spacesW2lr.
Key words:
equations not solved with respect to the highest deri-vative, the weighted Sobolev spaces, the Cauchy problem, pseudohy-perbolic equationsMathematics Subject Classication (1991): 35M05, 35M20
1. Introduction
In the present paper we consider a class of equations not solved with respect to the highest derivative
(1:1) L0(Dx)Dltu+l
;1 X
k=0Ll;k(Dx)Dktu=f(tx)
whereL0(Dx) and Ll;k(Dx) are linear dierential operators.
? The research was nancially supported by the Russian Foundation for Basic
Research (No. 98-01-00553, No. 99-01-00533) and by the Ministry of general and professional education of the Russian Federation.
To the best of our knowledge, equations not solved with respect to the highest derivative were rst studied by H. Poincare 1]. Subse-quently, they were considered in some articles by various mathemati-cians. This was connected with the research of certain hydrodynam-ics problems. The most intense interest to equations of the form (1.1) arose in connection with the study of the problem on small oscilla-tions of rotating uids by S. L. Sobolev 2]. His investigation was the rst profound research of equations not solved with respect to the highest derivative and originated the systematic study of such equa-tions. Therefore equations of the form (1.1) are often called equations of the Sobolev type.
At present there are a great number of theoretical and applied works devoted to the study of equations of the Sobolev type (see the bibliographies in 3{6]). The authors of 6] dened some class-es of equations and systems not solved with rclass-espect to the highclass-est derivatives and studied boundary value problems for the classes. In the present paper we shall consider the Cauchy problem for one of the classes. The class was called pseudohyperbolic in 6].
2. Denitions and main results
In this paper, we consider linear dierential operators of the form
L(DtDx) =L0(Dx)Dlt+l
;1 X
k=0Ll;k(Dx)Dkt x= (x1:::xn):
Let = (01:::n) be a vector with positive real entries and
1=i is an integer. We assume that
1.
The symbolL(ii),= (1:::n), ofL(DtDx) is an-homo-geneous function i.e.,
L(c0ici) =cL(ii)
8c >0
where = (1:::n).
2.
L0(Dx) is a quasi-elliptic operator i.e.,L0(i) = 0, 2Rn, if andonly if= 0.
Denition 1.
Equation (1.1) is called pseudohyperbolic if conditions 1 and 2 hold forL(DtDx) and the equation(2:1) (i)l+l;1 X
k=0
Ll;k(i)
L0(i) (i)k= 0 2Rnnf0g
has only real roots 1()l(). If the roots of (2.1) are distinct
The class of pseudohyperbolic equations contains quasi-hyperbolic equations with 0 = 1 = ::: = n = 1=(2m). Quasi-hyperbolic
equations were introduced by S. A. Gal'pern, see 7].
Example 1.Let Pn k=1ak4k>0 for8 2Rnnf0g. Then Dtu+iXn k=1akD4xku=f(tx) and D2tu ; n X k=1akD4xku=f(tx)
are pseudohyperbolic equations. For the rst equation the symbol of
L(DtDx) has the form
L(ii) =;jj
2i+iXn
k=1ak4k
where 0 = 1=2,1 = :::=n = 1=4. For the second equation, we
have L(ii) =jj 22 ; n X k=1ak4k where0=1 =:::=n= 1=4.
Now we turn to the Cauchy problem for strongly pseudohyperbolic equations
(2:2) L(DtDx)u=f(tx) t >0 x2Rn
Dktujt=0= 0 k= 0l;1:
Let G Rn+1. Then we use the symbol W
lr
2(G) to denote the
Sobolev space with the weighted functione;t. The norm inWlr
2(G) is dened as follows ku(tx)W lr 2(G)k=ke ;tu(tx)Wlr 2 (G)k:
We solve (2.2) inW2lr(R+n+1), wherer= (1=1:::1=n) and >0.
We also assume thatDktu(tx) is a member ofW0(1;k 0)r 2 (R+n+1) for 0kl. Lethi2 = n P i=12= i i ,jj= n P
i=1i, andmin= min(1n). We
denote the Fourier transform ofu(tx) inxby ^u(t). Let ~u() be
the Fourier transform ofu(tx) =e;tu(tx) in (tx).
Theorem 1.
Let u(tx) be in W2lr(Rn+1) > 0 and let Dktu(tx) be inW0(1;k 0)r 2 (Rn+1) for k= 0:::l. Then (2:3) khi (1;l 0)( jj++hi 0)l ;1u~ ()L2(Rn+1)k ckL(DtDx)u(tx)L2(Rn+1)kwhere the positive constant c is independent ofu(tx).
Theorem 2.
Letjj=2+l0>1 s= (0=1:::0=n) >0 andf(tx)2W
0s
2(R+n+1)\L2(R+1L1(Rn)). Then there exists a unique
solution u(tx) to (2.2) such that u(tx)2W
lr 2(R+n+1) and (2:4) ku(tx)W lr 2(R+n+1)k+ l X k=0 kDktu(tx)W 0(1;k 0)r 2 (R+n+1)k c kf(tx)W 0s 2(R+n+1)k+ kf(tx)L1(Rn)kL2(R+1)
where the positive constant c does not depend onf(tx).
Theorem 3.
LetN be a natural number jj=2 +l01 >0 jj=2 +l0+Nmin>1jj=2 +l0+ (N;1)min f(tx)2W 0s 2(R+n+1) (1 +hxi)N jjf(tx) 2L2(R+1L1(Rn)) and (2:5) Z Rn xf(tx)dx= 0 j j= 0N;1:Then there exists a unique solution u(tx) to (2.2) such that u(tx)
is a member of W2lr(R+n+1) and (2:6) kuW lr 2(R+n+1)k+ l X k=0 kDktuW 0(1;k 0)r 2 (R+n+1)k c kfW 0s 2(R+n+1)k+ k(1 +hxi)N jjf(tx)L 1(Rn)kL2(R+1) :
Here c does not depend on f(tx), c >0.
Estimate (2.3) is an analog of an energy inequality for strongly hy-perbolic operators, see 8, 9].
3. An \energy" inequality and uniqueness of the solution
to the Cauchy problem
First, for the symbols of strongly pseudohyperbolic operators we es-tablish properties which are similar to the properties of the symbols of strongly hyperbolic operators.
Lemma 1.
There exist constantsc1 >0 and c2>0 such that(3:1) M(i+ i) =;Im L(i+ i)DL(i+ i) c1 hi 2(1;l 0)( ji+ j+hi 0)2l ;2 and (3:2) jL(i+ i)jc2 hi(1 ;l 0)( jj+ +hi 0)l ;1 where 0, ()2Rn+1.
Proof. By denition (3.1),M(i+ i) is equal to
;
1 2i
L(i+ i)DL(i+ i);L(i+ i)DL(i+ i) =; 1 2i(L0(i))2 l Y p=1(i+ ;ip()) l X k=1( ;i) Y j6=k (i+ ;ij()) ; l Y p=1(i+ ;ip()) l X k=1i Y j6=k (i+ ;ij() = (L0(i))2Xl k=1 Y j6=k ji+ ;ij()j2:
Since the roots of (2.1) are the (1n)-homogeneous functions
with the homogeneity index 0, it follows that
m(i+ i) =Xl k=1 Y j6=k ji+ ;ij()j 2
is the (01n)-homogeneous function with the homogeneity
index 20(l;1). Hence, there exists a positive constantasuch that
(3:3) m(i+ i)a(ji+ j+hi 0)2(l
;1) ( )
2Rn+2:
By the homogeneity, with the inequality
(3.3) is immediate. Let us establish (3.4).
Let 6= 0. Since the roots j() are real numbers, (3.4) holds. If
= 0, then (3.4) follows from the relations
j()6=k() j=6 k 2Rnnf0g j(0) = 0 j= 1:::l:
The identityM(i+ i) (L0(i))2m(i+ i),together with
the estimates a1hi1 ;l 0 jL0(i)ja2hi1 ;l 0 0< a 1 a2 2Rn and (3.3), yields (3.1).
By easy calculations, we derive from the denition ofM() that jM(i+ i)jjL(i+ i)j l X k=1 jL0(i)j Y j6=k ji+ ;ij()j cjL(i+ i)jhi (1;l 0)( ji+ j+hi 0)l ;1:
Whence and from (3.1) we infer (3.2). ut
Proof of Theorem 1. By Parseval's equality we have
kL(DtDx)u(tx)L2(Rn+1)k=kL(Dt+Dx)u(tx)L2(Rn+1)k
=kL(i+i)~u()L2(Rn+1)k:
This equality, together with (3.2), yields (2.3). ut
Corollary 1.
The Cauchy problem (2.2) has a unique solution in the introduced class of functions.Proof. Let u(tx) be a solution to (2.2)f(tx) = 0 and let u(tx) be the extension ofu(tx) by zero fort <0. Thenu(tx) satises the conditions of Theorem 1. Whence and from (2.3) we have
khi(1 ;l 0)( jj++hi 0)l ;1u~ ()L2(Rn+1)k= 0 i:e: u(tx) = 0:
4. Approximate solutions of the Cauchy problem
If we apply formally the Fourier transform in x to (2.2), then we obtain the following Cauchy problem
(4:1) L(Dti)v= ^f(t) t >0
Dktvjt=0= 0 k= 0:::l;1
with 2Rn. The solution to (4.1) can be written as follows
(4:2) v(t) =Zt 0 J(t ;)^f()d 2Rnnf0g where (4:3) J(t) = 12 Z ;( ) eit L(ii)d
and ;() is a contour embraced all roots of (2.1). Formula (4.2) can be easily veried by using the following
Lemma 2.
The integralJ(t) is a solution of the Cauchy problemL(Dti)J = 0
DjtJjt=0= 0 j= 0:::l;2 Dl ;1
t Jjt=0=L0(i) ;1:
We rewrite the integral dened by (4.3) as follows.
Lemma 3.
Let 2Rnnf0g. Then(4:4) J(t) =i1;l(L 0(i));1 l X k=1ak()e itk( ) whereak() = Q j6=k (k();j()) ;1 forl >1 anda 1() = 1for l= 1.
Proof. If l= 1, then (4.4) is obvious. By the denition, we have
J(t) = 12 Z ;( ) eit L0(i) Ql j=1i(;j()) d l >1:
Since the roots 1()l() are distinct real numbers, it follows that J(t) =i1;l(L 0(i));1 l X k=1ak() 12i Z ;( ) eit ;k() d:
With this equality established, (4.4) is immediate. ut
Since the rootsk() are real numbers, from Lemma 3 we conclude
that the integrals
1 Z
0 e
;(i+)tDktJ(t)dt k 0
are dened for8 >0 and 2Rnnf0g. Therefore, by Lemma 2 it is
not hard to prove the following
Lemma 4.
Let >0 and2Rnnf0g. Then 1 Z 0e ;(i+)tDktJ(t)dt (i+)k(L(i+i)) ;1 k= 0:::l ;1 1 Z 0e ;(i+)tDltJ(t)dt (i+)l(L(i+i)) ;1 ;(L0(i)) ;1:Let m() = 1 for hi > 1=m and m() = 0 for hi < 1=m.
We consider the sequence fvm(t)g, vm(t) = m()v(t) with
the function v(t) dened by (4.2). By (4.2){(4.4), vm(t) has no
singularities. Consequently, if the right-hand sidef(tx) of the equa-tion in (2.2) satises the condiequa-tions of Theorem 2, then we can apply the inverse Fourier transform in to vm(t). In section 5 we
es-timate um(tx) = F;1vm](tx) in the Wlr
2(R+n+1)-norm and prove
thatfum(tx)gconverges to a solution of (2.2).
5. The estimates of approximate solutions
Lemma 5.
Lets= (00=1:::0=n). Then(5:1) X 1;l 0 1 kDxu m L2(R+n+1) kckf W 0s 2(R+n+1)k
where the positive constant c does not depend uponm and f(tx). If
q 1 and m!1, then (5:2) X 1;l 0 1 kDxum+q(tx);Dxum(tx) L2(R+n+1)k!0:
Proof. By Parseval's equality, the normkDxum(tx) L2(R+n+1)k is equal to k(i) m()Zt 0 J(t ;)^f()d L2(R+n+1)k:
Let (t) be Heaviside's function. By the properties of the Fourier transform, we can write the normkDxum L2(R+n+1)kas follows
kDxum L2(R+n+1)k =k(i)m() 1 Z ;1 e;(t; )(t ;)J(t;)()^f()dL2(Rn+1)k =k(i)m()( 1 Z 0 e ;(i+)tJ(t)dt) ~f () L2(Rn+1)k:
Whence and from Lemma 4, we infer
kDxum(tx) L2(R+n+1)k=k(i)m() ~ f() L(i+i) L2(Rn+1)k: By (3.2), we have (5:3) kDxumL2(R+n+1)k c hil 0 ;1m() (jj++hi 0)l ;1 ~ f() L2(Rn+1) : Consequently, kDxum(tx) L2(R+n+1)k c1 kf(tx) W 0s 2(R+n+1)k for = 1. If = 1;l0, then kDxum(tx) L2(R+n+1)k c lkf(tx) L2(R+n+1)k:
With these estimates established, (5.1) is immediate.
Let q 1 and 1;l0 1. Arguing as above, it is easy to
show that kDxum+q(tx);Dxum(tx) L2(R+n+1)k c (m+q();m())hil 0 ;1 (jj++hi 0)l ;1 ~ f() L2(Rn+1) :
The norm in the left side of this inequality tends to zero asm!1. u
Lemma 6.
Letjj=2 +l0>1. Then (5:4) X 0<1;l 0 kDxu m(tx) L2(R+n+1) k c kf(tx) L2(R+n+1)k+ kf(tx) L1(Rn)k L2(R+1) :Here the positive constantcis independent of mand f(tx). Ifq1
and m!1, then
(5:5) X
0<1;l 0
kDxum+q(tx);Dxum(tx) L2(R+n+1)k!0:
Proof. Estimate (5.3) with = 0 yields
kumL2(R+n+1)k c k hil 0;1m() (jj++hi 0)l ;1 ~ f() L2(Rn+1)k c lkhil 0;1m() ~f () L2(Rn+1)k c lkhil 0 ;1f^(t) L 2(R+1 fhi>1g)k + clkhil 0 ;1m() ^f(t)L 2(R+1 fhi<1g)k:
Taking into account the inequalityjj=2>1;l0 >0, using
Parse-val's equality and the Riemann{Lebesgue theorem, we have
kum(tx) L2(R+n+1)k c lkf(tx) L2(R+n+1)k +c1l kf(tx) L1(Rn)k L2(R+1) :
Here the positive constantc1 depends only on n,l, and .
By similar calculations, we derive that
kum+q(tx);um(tx) L2(R+n+1)k c lkhil 0;1(m+q() ;m()) ^f(t) L2(R+1 fhi<1g)k c l Z h i<1=m hi2(l 0 ;1)d 1=2 kf(tx) L1(Rn)k L2(R+1) :
Since jj=2>1;l0, it follows that
lim
m!1
kum+q(tx);um(tx) L2(R+n+1)k= 0:
By the same way, we obtain (5.4) and (5.5) for every such that 0< <1;l0. ut
Lemma 7.
Letjj=2 +l0>1. Then (5:6) kDltum(tx) L2(R+n+1)kc kf(tx) L2(R+n+1)k +kkf(tx) L1(Rn)k L2(R+1)k :Here the positive constantc does not depend on m and f(tx). If q1, then
(5:7) mlim
!1
kDltum+q(tx);Dltum(tx) L2(R+n+1)k= 0:
Proof. By Parseval's equality, we have
kDltu
m(tx) L2(R+n+1)
k=kDltu^
m(t) L2(R+n+1)
k:
Whence and from Lemma 2 it follows that
kDltu m(tx) L2(R+n+1) kk m() ^f(t) L0(i) L2(R+n+1)k +km() t Z 0 DltJ(t ;)^f()d L2(R+n+1)k=I1m+I2m:
Consider in turn both quantities on the right side of this equality. The inequalities jL0(i)j ahi1
;l
0, a > 0, and
jj=2 +l0 > 1
hold. Reasoning as in the proof of (5.4), we have (5:8) I1mc1 kf L2(R+n+1)k+ kf(tx) L1(Rn)k L2(R+1) :
Here the positive constantc1 does not depend on mand f(tx).
Let us estimateI2m. By (4.4), we have
I2m l X k=1 k m()ak() L0(i) t Z 0 e i(t; ) k( )( k())lf^()d L2(R+n+1)k:
Since k(), k= 1:::l, are the (12:::n)-homogeneous
func-tions with the homogeneity index 0, it follows that
jk()j=hi 0 jk( 0) j 0== hi jak()j= Y j6=k hi ; 0 jk( 0) ;j( 0) jchi ; 0(l ;1)
I2m ckm()(L0(i)) ;1 hi 0 t Z 0 ^ f()d L2(R+n+1)k ca ;1 km()hi(l+1) 0 ;1 t Z 0 ^ f()d L2(R+n+1)k:
Applying Jung's inequality to the norm in the right side, we obtain
I2m ca ;1 1 Z 0 e ;tdt k m() hi (l+1)0 ;1f^() L 2(R+n+1)k:
Similarly, taking into account the conditionjj=2 +l0 >1, we can
obtain the inequality
I2m c2 kf(tx) L2(R+n+1)k+ kf(tx) L1(Rn)k L2(R+1)
where the positive constant c2 does not depends on m and f(tx).
Whence and from (5.8) we infer (5.6). The proof of (5.7) is carried out by the same scheme. ut
Lemma 8.
Lets be equal to (00=1:::0=n) jj=2 +l0 >1.Then (5:9) kDktum(tx) W 0(1;k 0) 2 (R+n+1)kc kf(tx) W 0s 2(R+n+1)k + kf(tx) L1(Rn)k L2(R+1) k= 1:::l
where the positive constantc does not depend on m and f. If q 1,
then (5:10) limm !1 kDktum+q(tx);Dktum(tx) W 0(1;k 0) 2 (R+n+1)k= 0:
Proof. Let kbe an integer 1kl 0< 1;k0 and
Amk=kDktDxum(tx) L2(R+n+1)k:
Ifk=l, then we can establish the following estimates
Amlk (i) L0(i)m() ^f(t) L2(R+n+1)k +k(i)m() t Z 0 DltJ(t ;)^f()d L2(R+n+1)k
ckm()hi+l 0 ;1f^(t) L 2(R+n+1)k +ckm()hi+(l+1) 0 ;1f^(t) L 2(R+n+1)k
in the same way as in the proof of Lemma 7. Whence and from the inequalityjj=2>1;l0 it follows that
Amlc1 kf(tx) W 0s 2(R+n+1)k+kkf(tx) L1(Rn)k L2(R+1)k
where the positive constantc1 is independent of mand f(tx).
Let k < l. By the properties of the Fourier transform, Lemmas 2 and 4, in the same way as in the proof of Lemma 5 we obtain
Amk =km()(i) t Z 0 DktJ(t ;)^f()d L2(R+n+1)k km()(i)(i+)k(L(i+i)) ;1f~ () L2(Rn+1)k:
Whence and from Lemma 1, we infer
Amk c km() hi+l 0 ;1( jj+)k (jj++hi 0)l ;1 ~ f() L2(Rn+1)k:
Since jj=2>1;l0 and 1;k0 , it follows that
Amkc2 kf(tx) W 0s 2(R+n+1)k+ kf(tx) L1(Rn)k L2(R+1)
where the positive constantc2 is independent of mand f(tx).
By above results, (5.9) is readily veried. We can establish (5.10) in a similar way as in the proof of (5.9). ut
6. Existence of the solution to the Cauchy problem
In section 2 we established that (2.2) has a unique solution. Now we prove that this solution exists.
Proof of Theorem 2. We assume that the conditions of Theorem 2 are satised. Since jj=2 +l0 > 1, from Lemma 5, Lemma 6, and
Lemma 7 it follows that fum(tx)g is a fundamental sequence in
W2lr(R+n+1). Moreover, kumW lr 2(R+n+1)kc kf W 0s 2(R+n+1)k+ kfL1(Rn)kL2(R+1)
where the positive constant c is independent of m and f(tx). By completeness ofW2lr(R+n+1), it follows that there exists a limit
func-tionu(tx) such thatu(tx)2W
lr
2(R+n+1).
By Lemma 8 and the properties of generalized dierentiation it follows that generalized derivatives
DxDktu(tx) k0+ 1 0kl
exist and lim
m!1
kDxDktum(tx);DxDktu(tx) L2(R+n+1)k= 0:
Furthermore, the inequality
kDxDktuL2(R+n+1)kc kfW 0s 2(R+n+1)k + kfL1(Rn)kL2(R+1) :
holds. By the denition of um(tx), we have
kL(DtDx)u m(tx) ;f(tx) L2(R+n+1)k =kL(Dti)^um(t);f^(t) L2(R+n+1)k =km()L(Dti)v(t);f^(t) L2(R+n+1)k =k(m();1) ^f(t) L2(R+n+1)k:
The last norm tends to zero as m !1. Hence u(tx) is a solution
to (2.2) and (2.4) holds. ut
We essentially used thatjj=2+l0>1 in the proofs of Lemma 6,
Lemma 7, and Lemma 8. Now we establish analogs of these lemmas in the case ofjj=2 +l0 1.
Lemma 9.
Let the conditions of Theorem 3 be satised. Then(6:1) X 0<1;l 0 kDxumL2(R+n+1)k+ l X k=1 kDktumW 0(1;k 0) 2 (R+n+1)k c kf W 0s 2(R+n+1)k+ k(1 +hxi)N jjf L 1(Rn)k L2(R+1) :
Here the positive constantc is independent of m and f(tx). If q1, then (6:2) limm !1 n X 0<1;l 0 kDxum+q(tx);Dxum(tx) L2(R+n+1)k +Xl k=1 kDktum+q(tx);Dktum(tx) W 0(1;k 0) 2 (R+n+1)k o = 0:
Proof. We consider the norm kum(tx) L2(R+n+1)k and use
ortho-gonality conditions (2.5) in order to obtain (6.1) and (6.2). By the rst estimate in the proof of Lemma 6, we have (6:3) kumL2(R+n+1)k c lkhil 0 ;1f^(t) L 2(R+1 fjj>1g)k +clkhil 0;1m() ^f(t) L 2(R+1 fjj<1g)k:
Ifjj=2+l0>1, then (6.3) yields (5.4) otherwise, we can not bound
the second summand in (6.3) by a constant which is independent of
m. However, since (2.5) holds, it follows that (6:4) ^ f(t) = 1(2)n=2Z1 0::: 1 Z 0 ( Z Rn e;i 1::: Ny ( ;iy)Nf(ty)dy) N;1 Y j=1 N;j j d
where = (12:::N). We emphasize that (iy)N arises as a
factor in the right side of (6.4). Hence, hil 0
;1f^(t) is a summable
function in a neighborhood of = 0 and
kum(tx) L2(R+n+1)k c lkf(tx) L2(R+n+1)k +c1l k(1 +hxi) Njjf(tx) L 1(Rn)k L2(R+1)
where the positive constants c and c1 are independent of m and
f(tx). By the same way it is easy to show lim
m!1
kum+q(tx);um(tx) L2(R+n+1)k= 0:
Arguing as in the proofs of Lemmas 6|8, using (6.4), we can esti-mate the rest of summands in the left-hand side of (6.1). ut
Proof of Theorem 3. First we note that Lemma 5 holds for every value of jj=2 +l0. Whence and from Lemma 9 it follows in the
same way as in the proof of Theorem 2 thatfum(tx)g converges to
the solution of (2.2) and estimate (2.6) holds. ut
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