AAAAbstract and Applied Analysis1687-04091085-3375Hindawi Publishing Corporation18237110.1155/2009/182371182371Review ArticleWell-Posedness of the Cauchy Problem for Hyperbolic Equations with Non-Lipschitz CoefficientsAlievAkbar B.ShukurovaGulnara D.SobolevskiiPavelBaku State UniversityAcademic Zahid Xalilov str., 23AZ 1148 BakuAzerbaijanbsu.edu.az200916062009200912032009160520092009Copyright © 2009This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We consider hyperbolic equations with anisotropic elliptic part and some non-Lipschitz coefficients. We prove well-posedness of the corresponding Cauchy problem in some functional spaces. These functional spaces have finite smoothness with respect to variables corresponding to regular coefficients and infinite smoothness with respect to variables corresponding to singular coefficients.

1. Introduction

Let us consider the Cauchy problem for a second-order hyperbolic equation: ü-i,j=1naij(t)uxixj+j=1nbj(t)uxj+c(t)u=0,(t,x)[0,T]×Rn,u(0,x)=u0(x),ut(0,x)=u1(x),xRn, where the matrix (aij(t)) is real and symmetric for all t(0,T], ü=utt.

Suppose that (1.1) is strictly hyperbolic, that is, there exists λ0>0 such that a(t,ξ)i,j=1naij(t)ξiξj|ξ|2λ0>0, for all (t,ξ)(0,T]×Rn{0}.

It is known that if a(t,ξ) satisfies the Lipschitz condition and bj(t),c(t)L(0,T),j=1,2,,n, then for any u0Hs(Rn),u1Hs-1(Rn) the problem (1.1), (1.2) has a unique solution u(·)C([0,T];Hs(Rn))C1([0,T],Hs-1(Rn)), where s1 (see [1, Chapter 5] and [2, Chapter 3]).

If we reject the Lipschitz condition, this result, generally speaking, stops to be valid (see ).

In the paper  it is proved that if a(t,ξ)LLω(0,T), that is, if a(t,ξ) satisfies the logarithmic Lipschitz condition: |a(t+τ,ξ)-a(t,ξ)|c|τ|·|log|τ||·ω(|τ|), where ω(|τ|) monotonically decreasing tends to zero, and log|τ|·ω(|τ|) tends to infinity, then there exists δ>0 such that, for all u0Hs(Rn), u1Hs-1(Rn) the problem (1.1), (1.2) has a unique solution uC([0,T];Hs-δ(Rn))C1([0,T],Hs-1-δ(Rn)) (this behavior goes under the name of loss of derivatives).

In the paper  it is considered the case when ai,j(t)=0,ij, a part of coefficients belongs to the class LLω(0,T), and another part of coefficients satisfies the Lipschitz condition. It is proved that the loss of derivatives occurs in those variables xk for which appropriate coefficient akk(t) belongs to the class LLω(0,T).

It is interesting to investigate the Cauchy problem for (1.1), with singular coefficients. Many interesting results have been obtained in this direction. For example, in the paper  it is supposed that for each ξRn{0}a(t,ξ)C1(0,T] and tq|ȧ(t,ξ)|c,(t,ξ)(0,T]×Rn{0}, where q1, c>0. It is proved that if q=1, the problem (1.1), (1.2) is well-posed in C(Rn). If q>1 and tp|a(t,ξ)|c,(t,ξ)(0,T]×Rn{0}, where p[0,1)[0,q-1), then the problem (1.1), (1.2) is well-posed in the Geverey class γ(s)(Rn), s<(q-p)/(q-1) (see ). If the coefficients aij(t) satisfy only Holder conditions of order α<1 then in  it is established that the problem (1.1), (1.2) is γ(s) well-posed for all s<1/(1-α). In this direction see also the results obtained in the papers .

In this paper we consider the Cauchy problem for a higher-order hyperbolic equation with anisotropic elliptic part: ü+k=1n(-1)lkak(t)Dxk2lku+|α:l|1bα(t)Dxαu=0,(t,x)[0,T]×Rn,u(0,x)=u0(x),ut(0,x)=u1(x),xRn, where lkN,{1,2,,},αkN{0},k=1,2,,n,|α:l|=α1/l1++αn/ln.

Here the coefficients ak(t) satisfy different conditions of type (1.6) and (1.7), so that qk and pk corresponding to different k are different. The smoothness of the solution depending on smoothness on initial data with respect to each variable xk depends not only on lk but also on qk and pk.

2. Statement of the Problem and Results

We considered the Cauchy problem (1.8). Suppose that ak(t) and bα(t) satisfy the following conditions: ak(t)a>0,t[0,T],k=1,2,,n,tqk|ȧk(t)|c,t(0,T],k=1,2,,n,bα(t)L(0,T),|α:l|1.

In order to formulate the basic results we introduce some denotation. Let H be some Hilbert space. By W2λ,L(Rm,H) we will denote a functional space with the norm uW2λ,L(Rm,H)=[Rm(1+k=1mηk2Lk)λû(η)H2dη]1/2, where L=(L1,,Lm),LjN,j=1,2,,m,λ0, and û(η)=Fx[u](η);Fx is a Fourier transformation with respect to variable xRn.

For s1 by γβs,L(Rm,H) we will denote a functional space with the norm uγβs,L(Rm,H)=[exp{β|k=1mηk2|1/s}û(η)H2dη]1/2.

Denote W2λ,L(Rm,R)=W2λ,L(Rm),γβs,L(Rm,R)=γβs,L(Rm), C(Rm;H)=λ0W2λ,L(Rm;H),γ(s)(Rm;H)=β0γβ(s)(Rm;H).

If L=(1,,1) then W2λ,L(Rm,H)=Hλ(Rm;H), γβs,L(Rm,H)=γβs(Rm,H), and γβs,L(Rm,R)=γβ(s), where γβ(s) is the Geverey space of order s (see [12, 13]). If λH then W2λ,L(Rm,H) is Hilbert-valued anisotropic Sobolev space W2(λL1,,λLm)(Rm;H). For the read valued functions the anisotropic Sobolev spaces are stated in . The basic results led in  are also valid for abstract-valued functions.

We introduce also the following denotation: x=(x1,,xn1),x′′=(xn1+1,,xn),ξ=(ξ1,,ξn1),ξ′′=(ξn1+1,,ξn),l=(l1,,ln1),l′′=(ln1+1,,ln),|ξ|=k=1nξk2lk,|ξ|l=k=1n1ξk2lk,|ξ′′|l′′=k=n1+1nξk2lk,n2=n-n1.

The main results are the following theorems.

Theorem 2.1.

Let the conditions (2.1)–(2.3) be satisfied, where qk[0,1),fork=1,2,,n1,qk=1,fork=n1+1,,n.

Then for any λ0,λ′′0 the energy estimates E(t,λ,λ′′)ME(0,λ,λ′′+λ0), hold, where M and λ0 are some constants indepent of t[0,T], E(t,λ,λ′′+λ)=Rn(1+|ξ|l)λ(1+|ξ′′|l′′)λ′′+λ[|v̇(t,ξ)|2+(1+|ξ|l)|v(t,ξ)|2]dξ,λ0,v̇(t,ξ)=v(t,ξ)t.

Theorem 2.2.

Let the conditions (2.1)–(2.3) be satisfied, where qk[0,1),fork=1,2,,n1,qk=q>1,fork=n1+1,,n. Additionally, let the conditions tp|ak(t)|c,t[0,T],fork=n1+1,,n. be satisfied, where p[0,1)[0,q-1). Then for any β>0,λ0, and 1s<(q-p)/(q-1) the energy estimates, (t,β,s,λ)M(0,β+δ,s,λ), hold, where M and δ are some constants independent of t[0,T], (t,β,s,λ)=Rnexp{β|ξ′′|l1/s}(1+|ξ|l)λ[|v̇(t,ξ)|2+(1+|ξ|l)|v(t,ξ)|2]dξ.

Remark 2.3.

It is clear by our notation that E(t,λ,λ′′)u̇(t,·)W2λ′′,l′′(Rx′′n2;W2λ+1,l(Rxn1))+u(t,·)W2λ′′,l′′(Rx′′n2;W2λ+1,l(Rxn1))+u(t,·)W2λ′′+1,l′′(Rx′′n2;W2λ,l(Rxn1))2E(λ,λ′′,t), and we can write (t,β,s,λ)=u(t,·)γβs,l′′(Rx′′n2;W2λ,l(Rxn1)).

Remark 2.4.

It is possible to replace the conditions a1(t),,an1(t)C1(0,T] and (2.8) or (2.12) by Lipschitz conditions.

The following theorems are obtained from Theorems 2.1 and 2.2.

Theorem 2.5.

Let condition (2.1)–(2.9) be satisfied. Then for any s0,u0C(Rxn2;W2s+1,l(Rxn1)), u1C(Rxn2;W2s,l(Rxn1)) the problem (1.1), (1.2) admits a unique solution uC([0,T];C(Rx′′n2;W2s+1,l(Rxn1)))C1([0,T];C(Rx′′n2;W2s,l(Rxn1))).

Theorem 2.6.

Let conditions (2.1)–(2.3) and (2.12)–(2.14) be satisfied. Then for any s0,1s<(q-p)/(q-1),u0γs(Rxn2;W2s+1,l(Rxn1)), u1γs(Rxn2;W2s,l(Rxn1)) the problem (1.1), (1.2) admits a unique solution uC([0,T]  ;γs(Rxn2;W2s+1,l(Rxn1)))C1([0,T]  ;γs(Rxn2;W2s,l(Rxn1))).

In particular it follows from Theorem 2.1 that if the conditions (2.1)–(2.3) are satisfied, then the problem (1.1), (1.2) is well-posed in C(Rn), and if the conditions (2.1)–(2.3) and (2.12)–(2.14) are satisfied then the problem (1.1), (1.2) is well-posed in the Geverey class γ(s).

3. Proof of Theorems

At first we reduce some auxiliary statements.

We denote v(t,ξ)=Fx[u](t,ξ) and define the weighted energetic function in the following way: Φ(t)=Φ(t,ξ,λ,λ′′,β,r)=[|v̇(t,ξ)|2+(1+|ξ|l+d(t,ξ′′))|v(t,ξ)|2]·H(t,ξ), where H(t,ξ)=H(t,ξ,λ,λ′′,β,r)=(1+|ξ|l)λ(1+|ξ′′|l′′)λ′′×exp[-0tα(τ,ξ′′)dτ+β|ξ′′|l′′(q-1)/r],λ0,λ′′0,β>0,r={s(q-1), forq>1,1, forq=1,d(t,ξ′′){k=n1+1nak(T)ξk2lk,forTr|ξ′′|l′′1,k=n1+1nak(|ξ′′|l′′-1/r)ξk2lk,forTr|ξ′′|l′′>1,tr|ξ′′|l′′1,k=n1+1nak(t)ξk2lk,fortr|ξ′′|l′′>1,α(t,ξ′′){|d(t,ξ′′)-k=n1+1nak(t)ξk2lk|,fortr|ξ′′|l′′1,|k=n1+1nȧk(t)ξk2lk|k=n1+1nak(t)ξk2lk,fortr|ξ′′|l′′>1.

The following auxiliary lemmas are proved similar to the paper . The proofs of the lemmas are in appendix.

Lemma 3.1.

If qk=1,k=n1+1,,n, then there exits such c1>0,c2>0, that a|ξ′′|l′′d(t,ξ′′)[c1+c2ln(1+|ξ′′|l′′)  ]|ξ′′|l′′.

If qk>1,k=n1+1,,n, then there exits such c1>0,c2>0, that a|ξ′′|l′′d(t,ξ′′)[c1+c2|ξ′′|l′′p/r]|ξ′′|l′′.

Lemma 3.2.

If qk=1,k=1,2,,n1, then there exits such constant c3>0,γ>0, that 0tα(τ,ξ)dτc3+c4ln(1+|ξ|l).

If qk>1,k=1,2,,n1 then there exits such c3>0,c4>0, that 0tα(τ,ξ)dξc3+c4|ξ′′|l′′(q-1)/r.

By the definition of Φ(t)=Φ(t,ξ,λ,λ′′,β,r) we have dΦ(t)dt=2Re[v̈(t,ξ)v̇(t,ξ)¯+(1+|ξ|+d(t,ξ))v(t,ξ)v̇(t,ξ)¯]H(t,ξ)+ḋ(t,ξ)|v(t,ξ)|2H(t,ξ)-α(t,ξ)Φ(t).

On the other hand from (1.8) we have v̈(t,ξ)+k=1nak(t)  ξk2kv(t,ξ)+|α:|1bα(t)(iξ)αv(t,ξ)=0,v(0,ξ)=v0(ξ),v̇(0,ξ)=v1(ξ), where v0(ξ)=F[u0](ξ),v1(ξ)=F[u1](ξ),v̈(t,ξ)=2v(t,ξ)/t2.

From (3.6) and (3.7) we obtain dΦ(t)dt=2Re[-k=1n1ak(t)  ξk2k+(1+|ξ|)+(d(t,ξ′′)-k=n1+1nak(t)  ξk2k)]×v(t,ξ)v̇(t,ξ)¯H(t,ξ)-2Re|α:|1bα(t)(iξ)αv(t,ξ)v̇(t,ξ)¯H(t,ξ)+ḋ(t,ξ′′)|v(t,ξ)|2H(t,ξ)-α(t,ξ)Φ(t).

If tr|ξ′′|<1, then by definition of d(t,ξ) and α(t,ξ′′) we have dΦ(t)dt=2Re[-k=1n1ak(t)  ξk2k+(1+|ξ|)+α(t,ξ)]v(t,ξ)v̇(t,ξ)¯H(t,ξ)-2Re|α:|1bα(t)(iξ)αv(t,ξ)v̇(t,ξ)¯H(t,ξ)-α(t,ξ)Φ(t).

By our supposition qk<1 for  k=1,2,,n1. Therefore we can easily see that aak(t)aT,k=1,2,,n1 with some constant aT>a.

Using the Cauchy inequality, definition of α(t,ξ), H(t,ξ), and φ(t) we have

2Reα(t,ξ)v(t,ξ)v̇(t,ξ)¯H(t,ξ)-α(t,ξ)Φ(t)0,2Re|α:|1bα(t)(iξ)αv(t,ξ)v̇(t,ξ)¯H(t,ξ)2bT|α:l|1|ξα||v(t,ξ)|·|v̇(t,ξ)|·H(t,ξ)2bTc5[(1+k=1n|ξk|2lk)|v(t,ξ)|2+|v̇(t,ξ)|2]H(t,ξ), where bT=sup|α:l|1bα(t)L(0,T),c5=supξRn(|α:|1|ξα|)2/(k=1n|ξk|2lk+1).

From (3.10)–(3.13) we get that when tr|ξ′′|l′′<1, then there exists such a constant M1>0, that dΦ(t)dtM1Φ(t). If tr|ξ′′|l′′1 then by definition of d(t,ξ) and α(t,ξ′′) from (3.9) we have that dΦ(t)dt=2Re[-k=1n1ak(t)ξk2k-|α:|1bα(t)(iξ)αv(t,ξ)v̇(t,ξ)¯]H(t,ξ)+k=n1+1nȧk(t)  ξk2k|v(t,ξ)|2H(t,ξ)-|k=n1+1nȧk(t)ξk2k|k=n1+1nak(t)ξk2kΦ(t). On the other hand k=n1+1nȧk(t)  ξk2k|v(t,ξ)|2H(t,ξ)-|k=n1+1nȧk(t)ξk2k|k=n1+1nak(t)ξk2kΦ(t)=k=n1+1nȧk(t)ξk2k|v(t,ξ)|2H(t,ξ)-|k=n1+1nȧk(t)ξk2k|k=n1+1nak(t)ξk2k×[|v̇(t,ξ)|2+(1+|ξ|l2+k=n1+1nak(t)ξk2k)|v(t,ξ)|2]H(t,ξ)0. From (3.13), (3.15), and (3.16) we again get inequality (3.14).

It follows from (3.14) that Φ(t)MΦ(0),t[0,T], where M=M1eT.

Proof of Theorem <xref ref-type="statement" rid="thm1">2.1</xref>.

Let qk=q=1,k=n1+1,,n, then r=1. From Lemmas 3.1 and 3.2 we have RnΦ(t,ξ,λ,λ′′,0,1)dξRn(1+|ξ|)λ(1+|ξ′′|′′)λ′′×[|v̇(t,ξ)|2+(1+|ξ|l2+|ξ|l′′(c1+c2ln(1+|ξ′′|l′′)))]|v(t,ξ)|2×exp(c3+c4ln(1+|ξ′′|l′′))dξc6Rn(1+|ξ|)λ(1+|ξ′′|′′)λ′′+c2[|v̇(t,ξ)|2+(1+|ξ|)|v(t,ξ)|2]dξ=c6E(t,λ,λ′′+λ0), where λ0=c4+1, c6=max{1,c1ec3,c2ec3}.

Thus, RnΦ(t,ξ,λ,λ′′,0,1)dξc6E(t,λ,λ′′+λ0).

On the other hand from the definition of Φ and E we have RnΦ(t,ξ,λ,λ′′,0,1)dξc7E(t,λ,λ′′). It follows from (3.17)–(3.20) that E(t,λ,λ′′)c8E(0,λ,λ′′+d).

Proof of Theorem <xref ref-type="statement" rid="thm2">2.2</xref>.

Let qk=q>1,k=n1+1,,n, then r=(q-1)s. Taking into account Lemmas 3.1 and 3.2 and Theorem 2.5 we have RnΦ(t,ξ,λ,0,β,r)dξRn(1+|ξ|)λ[|v(t,ξ)|2+(1+|ξ|l+c1|ξ′′|l+c2|ξ′′|p/2+1)]×|v(t,ξ)|2exp(c3+(β+c4)|ξ|(q-1)/2)dξ.

Further using the inequality ηp/2+1c9exp(cη1/s) we obtain RnΦ(t,ξ,λ,0,β,r)dξc10(t,λ,s,β+δ), where δ=c4+c.

On the other hand from the definition of ϕ and we have RnΦ(t,ξ,λ,0,β,r)dξc11(t,λ,s,β).

From inequalities (3.17), (3.23), and (3.24) it follows that (t,λ,s,β)c12(0,λ,s,β+d).

Proof of Theorem <xref ref-type="statement" rid="thm3">2.5</xref>.

For any ξRn the problem (3.7), (3.8) has a unique solution v(t,ξ)C1[0,T] (see [15, Chapter I]).

Let u0C(Rx′′n2;W2λ+1,l(Rxn1)), u1C(Rx′′n2;W2λ,l(Rxn1)), then for any s0,λ0, E(0,λ,s+λ0)cs,λ, where cs,λ>0 is some constant dependent on s0 and λ0.

Taking into account Theorem 2.1 it follows from (3.20) that E(t,λ,s)Mcλ,s,t[0,T], that is, u̇(t,·)W2s,l′′(Rx′′n2;W2λ,l(Rxn1))+u(t,·)W2s+1,l′′(Rx′′n2;W2λ,l(Rxn1))+u(t,·)W2s,l′′(Rx′′n2;W2λ+1,l(Rxn1))Mcλ,s,t[0,T]. It follows from (3.28) that uC([0,T];C(Rx′′n2;W2λ+1,l(Rxn1))),u̇C([0,T];C(Rx′′n2;W2λ,l(Rxn1))).

By the expression of u(t,x) it follows that the function u(t,x) is the solution of problem (1.8).

The uniqueness of the solution is proved by standard method.

The proof of Theorem 2.6 is carried out in the similar way.

AppendicesA. Proof of LemmasProof of Lemma <xref ref-type="statement" rid="lem1">3.1</xref>.

Let qk=1,k=n1+1,,n. Then from (2.2) we have ak(t)ak(T)+|ak(t)-ak(T)|ak(T)+tT|ȧk(s)|dsak(T)+clnTtc1+c2ln(1+1t). It follows from (2.1) and (2.14) that a|ξ′′|l′′d(t,ξ′′).

By definition of d(t,ξ′′) for T|ξ′′|l′′1 we have d(t,ξ′′)=k=n1nak(T)ξk2lkc1|ξ′′|l′′. If T|ξ′′|l′′>1, and t|ξ′′|l′′<1, then from (A.1) we have d(t,ξ′′)=k=n1nak(|ξ′′|l′′-1)ξk2lk[c1+c2ln(1+1|ξ′′|l′′-1)]k=n1nξk2lk=(c1+c2ln(1+|ξ′′|l′′))|ξ′′|l′′. If t|ξ′′|l′′>1, then using (A.1) we get d(t,ξ′′)=k=n1+1nak(t)ξk2lk[c1+c2ln(1+1t)]|ξ′′|l′′[c1+c2ln(1+|ξ′′|l′′)]|ξ′′|l′′. Consequently if q=1, the statement of the lemma follows from (A.2)–(A.5).

Let qk>1,k=n1+1,,n. By definition of d(t,ξ′′) for Tr|ξ′′|l′′<1 we have d(t,ξ′′)c1|ξ′′|l′′.

If Tr|ξ′′|l′′>1 and tr|ξ′′|<1 then d(t,ξ′′)=k=n1+1nak(|ξ′′|l′′-1/r)ξk2lkk=n1+1nM(|ξ′′|l′′-1/r)pξk2lk=M|ξ′′|l′′1+p/r.

If tr|ξ′′|>1 then d(t,ξ′′)=k=n1+1nak(t)ξk2lkk=n1+1nMtpξk2lk=M|ξ′′|l′′·|ξ′′|p/r=M|ξ′′|1+p/r.

Thus if qk>1,k=n1+1,,n then the statement of the lemma follows from (A.2), (A.6), and (A.8).

The lemma is proved.

Proof of Lemma <xref ref-type="statement" rid="lem2">3.2</xref>.

At first we consider the case when qk=1,k=n1+1,,n. If T|ξ′′|′′1, then 0tα(τ,ξ′′)dτ0Tα(τ,ξ′′)dτ0T|k=n1nak(T)ξk2k-k=n1nak(τ)ξk2k|dτk=n1nξk2k0T|ak(T)-ak(τ)|dτT·maxk=n1+1,,nak(T)|ξ′′|′′+|ξk′′|′′0Tak(τ)dτaT, where aT=maxk=n1+1,,nak(τ)+(1/T)maxk=n1+1,,n0Tak(τ)dτ<+.

If T|ξ′′|>1, then 0tα(τ,ξ)ds0|ξ′′|l′′-1α(s,ξ)dτ+|ξ′′|′′-1Tα(s,ξ)ds0|ξ′′|l′′-1|d(τ,ξ)-k=n1nak(τ)ξk2lk|dτ+|ξ′′|′′-1T|k=n1nȧk(τ)ξk2k|k=n1nak(τ)ξk2kdτ0|ξ′′|l′′-1d(τ,ξ)dτ+k=n1nξk2lk·0|ξ′′|l′′-1αk(τ)dτ+cak=n1nξk2k|ξ′′|′′-1Tdττ0|ξ′′|′′-1[c1+c2n(1+|ξ′′|′′)]|ξ′′|′′dτ+k=n1nξk2k·c0|ξ′′|′′-1nTτdτ+cak=n1n|ξ′′|′′-1Tdττ=c1+c2n(1+|ξ′′|′′-1)+c|ξ′′|′′·0|ξ′′|′′-1nTτdτ+ca|ξ′′|′′-1Tdττc3+c4n(1+|ξ′′|′′).

Now let us consider the case qk>1,k=n1+1,,n. In this case r=(q-1)s. If Tr|ξ′′|′′1, then 0tα(τ,ξ′′)dτ0Tα(τ,ξ′′)dτk=n1+1n0T|ak(T)-ak(τ)|ξk2kdτmaxk=n1+1,,nak(T)T|ξ′′|′′+0Tcτ-pdτ|ξ′′|′′aT·T1-r+c·11-pT1-p|ξ′′|′′aTT1-r+c1-pT1-p-r. If Tr|ξ′′|'>1, then 0tα(τ,ξ′′)dτ0|ξ′′|-1/rα(τ,ξ)dτ+|ξ′′|′′-1/rTα(τ,ξ′′)dτ0|ξ′′|′′-1/r|d(τ,ξ)-k=n1+1nak(τ)ξk2lk|dτ+|ξ′′|′′-1/rTα(τ,ξ′′)dτk=n1+1nak(|ξ′′|′′-1/r)ξk2k0|ξ′′|l′′-1/rdτ+k=n1+1nξk2k0|ξ′′|′′-1/rak(τ)dτ+|ξ′′|′′-1/rT|k=n1+1nȧk(τ)ξk2k|k=n1+1nak(τ)ξk2kdτc(|ξ′′|′′-1/r)p·|ξ′′|′′·0|ξ′′|′′-1/rdτ+c|ξ′′|′′·0|ξ′′|-1/rdττpdτ+ca|ξ′′|-1/rTdττqc|ξ′′|′′p/r+1·|ξ′′|-1/r+c|ξ′′|′′·11-p(|ξ′′|-1/r)1-p+ca11-q(T1-q-(|ξ′′|-1/r)1-q)<c|ξ′′|′′1-((1-p)/r)+c1-p|ξ′′|′′1-((1-p)/r)+ca(q-1)|ξ′′|(q-1)/r.

As r=(q-1)s, and s<(q-p)/(q-1), it follows that 1-(1-p)/s<1/s and (q-1)/r=1/s. Then according to the Young inequality there exists such δ>0 that |ξ′′|1-((1-p)/r)c1δ+δ1|ξ′′|1/s. Thus, by (A.9)–(A.13) the following inequality is valid: 0tα(τ,ξ′′)dτδ|ξ|1/s+cδ, where δ=δ1a(2+p)/(1-p)+(c/a(q-1))cδ=c1δc(2+δ)/(1-p).

B. Example

Let us consider the Cauchy problem in [0,T)×R2: utt-(1+t2)uxx-(1+t23)uyy=0,u(0,x,y)=φ1(x)ψ1(y),ut(o,x,y)=φ2(x)ψ2(y), where φ1(x),φ2(x)C(R)=s0W2s(R),ψ1(y)W22(R),ψ2(y)W21(R),u=u(t,x,y).

It follows from Theorem 2.5 that the problem (B.1) has a unique solution uC([0,T];C(R;W22(R)))C1([0,T];C(R;W21(R))).

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