# Well-Posedness of the Cauchy Problem for  Hyperbolic Equations with Non-Lipschitz Coefficients

Authors: Akbar B. Aliev; Gulnara D. Shukurova
Journal: Abstract and Applied Analysis
 (2009)

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## Abstract

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.


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## Body

## 1. Introduction

Let us consider the Cauchy problem for a second-order hyperbolic equation:(1.1)ü-∑i,j=1naij(t)uxixj+∑j=1nbj(t)uxj+c(t)u=0,(t,x)∈[0,T]×Rn,(1.2)u(0,x)=u0(x),ut(0,x)=u1(x),x∈Rn,
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
(1.3)a(t,ξ)≡∑i,j=1naij(t)ξiξj|ξ|2≥λ0>0,
for all (t,ξ)∈(0,T]×Rn∖{0}.It is known that ifa(t,ξ) satisfies the Lipschitz condition and bj(t),c(t)∈L∞(0,T),j=1,2,…,n, then for any u0∈Hs(Rn),u1∈Hs-1(Rn) the problem (1.1), (1.2) has a unique solution
(1.4)u(·)∈C([0,T];Hs(Rn))∩C1([0,T],Hs-1(Rn)),
where s≥1 (see [1, Chapter 5] and [2, Chapter 3]).If we reject the Lipschitz condition, this result, generally speaking, stops to be valid (see [3]).In the paper [4] it is proved that if a(t,ξ)∈LLω(0,T), that is, if a(t,ξ) satisfies the logarithmic Lipschitz condition:
(1.5)|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 u0∈Hs(Rn), u1∈Hs-1(Rn) the problem (1.1), (1.2) has a unique solution u∈C([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 [5] it is considered the case when ai,j(t)=0,i≠j, 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 [6] it is supposed that for each ξ∈Rn∖{0}a(t,ξ)∈C1(0,T] and
(1.6)tq|ȧ(t,ξ)|≤c,(t,ξ)∈(0,T]×Rn∖{0},
where q≥1, 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
(1.7)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 [6]). If the coefficients aij(t) satisfy only Holder conditions of order α<1 then in [3] 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 [6–13].In this paper we consider the Cauchy problem for a higher-order hyperbolic equation with anisotropic elliptic part:(1.8)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),x∈Rn,
where lk∈N,{1,2,…,},αk∈N∪{0},k=1,2,…,n,|α:l|=α1/l1+⋯+αn/ln.Here the coefficientsak(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:
(2.1)ak(t)≥a>0,t∈[0,T],k=1,2,…,n,(2.2)tqk|ȧk(t)|≤c,t∈(0,T],k=1,2,…,n,(2.3)bα(t)∈L∞(0,T),|α:l|≤1.In order to formulate the basic results we introduce some denotation. LetH be some Hilbert space. By W2λ,L(Rm,H) we will denote a functional space with the norm
(2.4)∥u∥W2λ,L(Rm,H)=[∫Rm(1+∑k=1mηk2Lk)λ∥û(η)∥H2dη]1/2,
where L=(L1,…,Lm),Lj∈N,j=1,2,…,m,λ≥0, and û(η)=Fx[u](η);Fx is a Fourier transformation with respect to variable x∈Rn.Fors≥1 by γβs,L(Rm,H) we will denote a functional space with the norm
(2.5)∥u∥γβs,L(Rm,H)=[∫exp{β|∑k=1mηk2|1/s}∥û(η)∥H2dη]1/2.DenoteW2λ,L(Rm,R)=W2λ,L(Rm),γβs,L(Rm,R)=γβs,L(Rm),
(2.6)C∞(Rm;H)=⋂λ≥0W2λ,L(Rm;H),γ(s)(Rm;H)=⋂β≥0γβ(s)(Rm;H).IfL=(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 λ∈H 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 [14]. The basic results led in [14] are also valid for abstract-valued functions.We introduce also the following denotation:(2.7)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
(2.8)qk∈[0,1),fork=1,2,…,n1,(2.9)qk=1,fork=n1+1,…,n.
Then for anyλ′≥0,λ′′≥0 the energy estimates
(2.10)E(t,λ′,λ′′)≤ME(0,λ′,λ′′+λ0),
hold, where M and λ0 are some constants indepent of t∈[0,T],
(2.11)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,λ≥0,v̇(t,ξ)=∂v(t,ξ)∂t.Theorem 2.2.
Let the conditions (2.1)–(2.3) be satisfied, where
(2.12)qk∈[0,1),fork=1,2,…,n1,(2.13)qk=q>1,fork=n1+1,…,n.
Additionally, let the conditions
(2.14)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,λ′≥0, and 1≤s<(q-p)/(q-1) the energy estimates,
(2.15)ℰ(t,β,s,λ′)≤Mℰ(0,β+δ,s,λ′),
hold, where M and δ are some constants independent of t∈[0,T],
(2.16)ℰ(t,β,s,λ′)=∫Rnexp{β|ξ′′|l′′1/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(2.17)E(t,λ′,λ′′)≤∥u̇(t,·)∥W2λ′′,l′′(Rx′′n2;W2λ′+1,l′(Rx′n1))+∥u(t,·)∥W2λ′′,l′′(Rx′′n2;W2λ′+1,l′(Rx′n1))+∥u(t,·)∥W2λ′′+1,l′′(Rx′′n2;W2λ′,l′(Rx′n1))≤2E(λ′,λ′′,t),
and we can write
(2.18)ℰ(t,β,s,λ′)=∥u(t,·)∥γβs,l′′(Rx′′n2;W2λ′,l′(Rx′n1)).Remark 2.4.
It is possible to replace the conditionsa1(t),…,an1(t)∈C1(0,T] and (2.8) or (2.12) by Lipschitz conditions.The following theorems are obtained from Theorems2.1 and 2.2.Theorem 2.5.
Let condition (2.1)–(2.9) be satisfied. Then for any s≥0,u0∈C∞(Rx′′n2;W2s+1,l′(Rx′n1)), u1∈C∞(Rx′′n2;W2s,l′(Rx′n1)) the problem (1.1), (1.2) admits a unique solution
(2.19)u∈C([0,T];C∞(Rx′′n2;W2s+1,l′(Rx′n1)))∩C1([0,T];C∞(Rx′′n2;W2s,l′(Rx′n1))).Theorem 2.6.
Let conditions (2.1)–(2.3) and (2.12)–(2.14) be satisfied. Then for any s′≥0,1≤s′′<(q-p)/(q-1),u0∈γs′′(Rx′′n2;W2s′+1,l′(Rx′n1)), u1∈γs′′(Rx′′n2;W2s′,l′(Rx′n1)) the problem (1.1), (1.2) admits a unique solution
(2.20)u∈C([0,T];γs′′(Rx′′n2;W2s′+1,l′(Rx′n1)))∩C1([0,T];γs′′(Rx′′n2;W2s′,l′(Rx′n1))).In particular it follows from Theorem2.1 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 denotev(t,ξ)=Fx[u](t,ξ) and define the weighted energetic function in the following way:
(3.1)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
(3.2)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],λ′≥0,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 [6]. The proofs of the lemmas are in appendix.Lemma 3.1.
Ifqk=1,k=n1+1,…,n, then there exits such c1>0,c2>0, that
(3.3)a|ξ′′|l′′≤d(t,ξ′′)≤[c1+c2ln(1+|ξ′′|l′′)]|ξ′′|l′′.
Ifqk>1,k=n1+1,…,n, then there exits such c1>0,c2>0, that
(3.4)a|ξ′′|l′′≤d(t,ξ′′)≤[c1+c2|ξ′′|l′′p/r]|ξ′′|l′′.Lemma 3.2.
Ifqk=1,k=1,2,…,n1, then there exits such constant c3>0,γ>0, that ∫0tα(τ,ξ)dτ≤c3+c4ln(1+|ξ′′|l′′).
Ifqk>1,k=1,2,…,n1 then there exits such c3>0,c4>0, that
(3.5)∫0tα(τ,ξ)dξ≤c3+c4|ξ′′|l′′(q-1)/r.By the definition ofΦ(t)=Φ(t,ξ,λ′,λ′′,β,r) we have
(3.6)dΦ(t)dt=2Re[v̈(t,ξ)v̇(t,ξ)¯+(1+|ξ′|ℓ′+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
(3.7)v̈(t,ξ)+∑k=1nak(t)ξk2ℓkv(t,ξ)+∑|α:ℓ|≤1bα(t)(iξ)αv(t,ξ)=0,(3.8)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
(3.9)dΦ(t)dt=2Re[-∑k=1n1ak(t)ξk2ℓk+(1+|ξ′|ℓ′)+(d(t,ξ′′)-∑k=n1+1nak(t)ξk2ℓk)]×v(t,ξ)v̇(t,ξ)¯H(t,ξ)-2Re∑|α:ℓ|≤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).Iftr|ξ′′|<1, then by definition of d(t,ξ) and α(t,ξ′′) we have
(3.10)dΦ(t)dt=2Re[-∑k=1n1ak(t)ξk2ℓk+(1+|ξ′|ℓ′)+α(t,ξ)]v(t,ξ)v̇(t,ξ)¯H(t,ξ)-2Re∑|α:ℓ|≤1bα(t)(iξ)αv(t,ξ)v̇(t,ξ)¯H(t,ξ)-α(t,ξ)Φ(t).By our suppositionqk<1 fork=1,2,…,n1. Therefore we can easily see that
(3.11)a≤ak(t)≤aT,k=1,2,…,n1
with some constant aT>a.Using the Cauchy inequality, definition ofα(t,ξ), H(t,ξ), and φ(t) we have(3.12)2Reα(t,ξ)v(t,ξ)v̇(t,ξ)¯H(t,ξ)-α(t,ξ)Φ(t)≤0,(3.13)2Re∑|α:ℓ|≤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|≤1∥bα(t)∥L∞(0,T),c5=supξ∈Rn(∑|α:ℓ|≤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
(3.14)dΦ(t)dt≤M1Φ(t).
If tr|ξ′′|l′′≥1 then by definition of d(t,ξ) and α(t,ξ′′) from (3.9) we have that
(3.15)dΦ(t)dt=2Re[-∑k=1n1ak(t)ξk2ℓk-∑|α:ℓ|≤1bα(t)(iξ)αv(t,ξ)v̇(t,ξ)¯]H(t,ξ)+∑k=n1+1nȧk(t)ξk2ℓk|v(t,ξ)|2H(t,ξ)-|∑k=n1+1nȧk(t)ξk2ℓk|∑k=n1+1nak(t)ξk2ℓkΦ(t).
On the other hand
(3.16)∑k=n1+1nȧk(t)ξk2ℓk|v(t,ξ)|2H(t,ξ)-|∑k=n1+1nȧk(t)ξk2ℓk|∑k=n1+1nak(t)ξk2ℓkΦ(t)=∑k=n1+1nȧk(t)ξk2ℓk|v(t,ξ)|2H(t,ξ)-|∑k=n1+1nȧk(t)ξk2ℓk|∑k=n1+1nak(t)ξk2ℓk×[|v̇(t,ξ)|2+(1+|ξ′|l′2+∑k=n1+1nak(t)ξk2ℓk)|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
(3.17)Φ(t)≤MΦ(0),t∈[0,T],
where M=M1eT.Proof of Theorem2.1.
Letqk=q=1,k=n1+1,…,n, then r=1. From Lemmas 3.1 and 3.2 we have
(3.18)∫RnΦ(t,ξ,λ′,λ′′,0,1)dξ≤∫Rn(1+|ξ′|ℓ′)λ′(1+|ξ′′|ℓ′′)λ′′×[|v̇(t,ξ)|2+(1+|ξ′|l′2+|ξ′′|l′′(c1+c2ln(1+|ξ′′|l′′)))]|v(t,ξ)|2×exp(c3+c4ln(1+|ξ′′|l′′))dξ≤c6∫Rn(1+|ξ′|ℓ′)λ′(1+|ξ′′|ℓ′′)λ′′+c2[|v̇(t,ξ)|2+(1+|ξ|ℓ)|v(t,ξ)|2]dξ=c6E(t,λ′,λ′′+λ0),
where λ0=c4+1, c6=max{1,c1ec3,c2ec3}.
Thus,(3.19)∫RnΦ(t,ξ,λ′,λ′′,0,1)dξ≤c6E(t,λ′,λ′′+λ0).
On the other hand from the definition ofΦ and E we have
(3.20)∫RnΦ(t,ξ,λ′,λ′′,0,1)dξ≥c7E(t,λ′,λ′′).
It follows from (3.17)–(3.20) that
(3.21)E(t,λ′,λ′′)≤c8E(0,λ′,λ′′+d).Proof of Theorem2.2.
Letqk=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
(3.22)∫RnΦ(t,ξ,λ′,0,β,r)dξ≤∫Rn(1+|ξ′|ℓ)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 inequalityηp/2+1≤c9exp(cη1/s) we obtain
(3.23)∫RnΦ(t,ξ,λ′,0,β,r)dξ≤c10ℰ(t,λ′,s,β+δ),
where δ=c4+c.
On the other hand from the definition ofϕ and ℰ we have
(3.24)∫RnΦ(t,ξ,λ′,0,β,r)dξ≥c11ℰ(t,λ′,s,β).
From inequalities (3.17), (3.23), and (3.24) it follows that
(3.25)ℰ(t,λ′,s,β)≤c12ℰ(0,λ′,s,β+d).Proof of Theorem2.5.
For anyξ∈Rn the problem (3.7), (3.8) has a unique solution v(t,ξ)∈C1[0,T] (see [15, Chapter I]).
Letu0∈C∞(Rx′′n2;W2λ′+1,l′(Rx′n1)), u1∈C∞(Rx′′n2;W2λ′,l′(Rx′n1)), then for any s≥0,λ′≥0,
(3.26)E(0,λ′,s+λ0)≤cs,λ′,
where cs,λ>0 is some constant dependent on s≥0 and λ′≥0.
Taking into account Theorem2.1 it follows from (3.20) that
(3.27)E(t,λ′,s)≤Mcλ′,s,t∈[0,T],
that is,
(3.28)∥u̇(t,·)∥W2s,l′′(Rx′′n2;W2λ′,l′(Rx′n1))+∥u(t,·)∥W2s+1,l′′(Rx′′n2;W2λ′,l′(Rx′n1))+∥u(t,·)∥W2s,l′′(Rx′′n2;W2λ′+1,l′(Rx′n1))≤Mcλ′,s,t∈[0,T].
It follows from (3.28) that
(3.29)u∈C([0,T];C∞(Rx′′n2;W2λ′+1,l′(Rx′n1))),u̇∈C([0,T];C∞(Rx′′n2;W2λ′,l′(Rx′n1))).
By the expression ofu(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 Theorem2.6 is carried out in the similar way.



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*Source: 182371-2009-08-26.xml*