一类具有非线性积分条件的波动方程的非局部问题

A nonlocal problem for a wave equationwith nonlinear integral condition∗Yong Zhang†,Peihao Zhao‡School of Mathematics and Statistics,Lanzhou University,Lanzhou,Gansu730000,People’s Republic of China,AbstractIn this paper,we study a nonlocal problem for a wave equation with nonlinear integral condition in a cylinder.The main aim is to provethe uniqueness and existence of a generalized solution.The proofs arebased on a priori estimate and the Galerkin procedure,respectively.Keywords:Nonlocal problem,nonlinear integral condition,wave equation,generalized solution.2000Mathematics Subject Classifications:35L05,35L70, 35L991IntroductionCertain problems of modern physics and technology can be effectively de-scribed in terms of nonlocal problems for partial differential equations.These nonlocal conditions arise mainly when the data on the boundary cannot be measured directly.Thefirst paper,devoted to second-order partial differ-ential equations with nonlocal integral conditions goes back to Cannon[3]. Cannon studied a problem for a heat equation,and in most papers,devotedto nonlocal problems,parabolic and elliptic equations were studied.Mixed problems with nonlocal integral conditions for one-dimensional hyperbolic and wave equations were considered in[2,7,10,11,12].Nonlocal problemsfor a third-order one-dimensional nonlinear pseudoparabolic equation were considered in[5,6].A mixed nonlocal problem for a wave equation with linear integral con-dition was considered in[1].Sergei A.Beilin studied an n-dimensional wave equation with weighted integral nonlocal condition.∗This work was partly supported by the Fundamental Research Fund for Physics and Mathematics of Lanzhou University.†E-mail address:zyzwj309@‡E-mail address:zhaoph@1In this paper we study the uniqueness and existence of a generalized solution for the n-dimensional wave equation of the formu tt − u +c (x,t )u =f (x,t )(1.1)in a cylinder Q =Ω×(0,T ),where Ω⊂R n is a bounded domain with a smooth boundary,with Cauchy conditionsu (x,0)=φ(x ),u t (x,0)=ψ(x )(1.2)and an integral nonlocal nonlinear condition∂u∂n |S +t 0ΩK x,ξ,τ,u (ξ,τ) dξdτ=0,x ∈∂Ω,S =∂Ω×(0,T )(1.3)where φ(x ),ψ(x ),K x,ξ,τ,u (ξ,τ)are given functions.Motivation for definition of generalized solution.Define the spaceW 1,12(Q )={u (x,t ):u,∇u,u t ∈L 2(Q )}with the normu W 1,12(Q )=( u L 2(Q )+ ∇u L 2(Q )+ u t L 2(Q ))12denoteW 1,12(Q )={v (x,t ):v ∈W 1,12(Q ),v (x,T )=0}Suppose u (x,t )∈W 1,12(Q )satisfy (1.1)-(1.2)-(1.3).Multiply the equation(1.1)by the function v (x,t )∈ W 1,12(Q )and integrate it over the cylinder Q .After integrating by parts,we obtainT 0Ω(∇u ∇v −u t v t +cuv )dxdt =T 0Ωfvdxdt + T 0 ∂Ω∂u∂n vdsdt + Ωψ(x )v (x,0)dx (1.4)After substituting the condition (1.3),we haveT 0Ω(∇u ∇v −u t v t +cuv )dxdt + T 0 ∂Ωv (x,t ) t 0ΩK x,ξ,τ,u (ξ,τ)dξdτdsdt = T 0Ωfvdxdt + Ωψ(x )v (x,0)dx(1.5)Definition 1.1A function u (x,t )∈W 1,12(Q )is called a generalized solu-tion of (1.1)-(1.2)-(1.3),if it satisfies (1.5)for every v (x,t )∈ W 1,12(Q )and 2u (x,0)=φ(x ).From now on,we assume the following:(A1)c (x,t )∈C 1(Q ),c (x,t )≥c 1>0;f (x,t )∈L 2(Q );φ(x )∈W 12(Ω);ψ(x )∈L 2(Ω).(A2)K x,ξ,τ,s ∈C ∂Ω×Ω×(0,T )×R,there exist constants K =K (Q ),K 0=K 0(Q ),K 1=K 1(Q )such that|Kx,ξ,t,s 1 −K x,ξ,t,s 2 |≤K |s 1−s 2|,|K x,ξ,τ,s|≤K 0|s |,|∂K (x,ξ,t,s )∂x i|≤K 1|s |,(i =1,...,n ).The main result of this paper can be formulated as the following theo-rem.Theorem 1.2If (A1)-(A2)hold,then there exist a unique generalized so-lution to the problem (1.1)-(1.2)-(1.3).In this paper,we will use the following lemmas.It is easy to show the following lemmas,for the details of the proof,we refer the reader to [4].Lemma 1.3Let X be a Banach space,f ∈C ([0,T ]×X ;X )andf (t,x )−f (t,y ) ≤C x −y∀(x,y )∈X ×X,∀t ∈[0,T ],(1.6)then there exist 0<T ∗≤T ,such that the Cauchy problemdx (t )dt=f t,x (t ) ;x (0)=x 0;(t,x )∈[0,T ]×X(1.7)has a unique solution x(t)for t ∈[0,T ∗].Lemma 1.4Assume in addition to (1.6)the growth conditionf (t,x ) ≤a +b x ,∀(t,x )∈[0,T ]×X,(1.8)where a,b >0are constants.Then [0,T ]is the maximal existence interval for the Cauchy problem (1.7).The following two sections provide the proof of the above theorem.32UniquenessLet u 1=u 2be solutions of (1.1)-(1.2)-(1.3),then u =u 1−u 2is a solution of the same problem with f =φ=ψ=0,so u (x,t )satisfiesT 0Ω(∇u ∇v −u t v t +cuv )dxdt +T 0∂Ωv (x,t ) t 0ΩKx,ξ,τ,u 1 −K x,ξ,τ,u 2 dξdτdsdt =0(2.1)Consider the functionv (x,t )=τtu (x,η)dη,0≤t ≤τ,0,τ≤t ≤T.(2.2)Note that v (x,t )∈ W 1,12(Q ),and v t (x,t )=−u (x,t ),∀t ∈[0,τ].Integration by parts in(2.1)will give us:12 Ω |∇v (x,0)|2+u 2(x,τ) dxdt = T 0 ∂Ωv (x,t )t 0 ΩK x,ξ,τ,u 1 −K x,ξ,τ,u 2dξdτdsdt + T 0Ωcuvdxdt.With the help of elementary inequalities,we obtain the following estimates:12 Ω|∇v (x,0)|2+u 2(x,τ) dxdt ≤T 0∂Ω|v (x,t )| t 0 Ω K x,ξ,τ,u 1 −K x,ξ,τ,u 2 dξdτdsdt + T 0 Ω|cuv |dxdt.≤K T 0 ∂Ω|v (x,t )| t 0 Ω|u 1−u 2|dξdτdsdt +C 0( T 0 Ωu 2+v 2dxdt ).≤K T 0 ∂Ω |v (x,t )|2+T |Ω| t 0Ωu 2(ξ,τ)dξdτ dsdt+C 0(T 0Ω(u 2+v 2)dxdt ).(2.2)By using an inequality in[9,p.77], ∂Ωv 2ds ≤Ω|∇v |2+C ( )v 2dx(2.3)4and denoting L =T 2|Ω||∂Ω|,we obtainΩ|∇v (x,0)|2+u 2(x,τ) dx ≤Kετ0Ω|∇v |2dxdt +KC (ε)+C 0 τ0 Ωv 2dxdt +KL +C 0τ0Ωu 2dxdt.It’s trivial thatv 2=τtu (x,η)dη 2≤ττu 2(x,η)dη,andτ0Ωv 2dxdt ≤τ0Ωττu 2(x,η)dηdxdt ≤τ2τ0Ωu 2(x,t )dxdt.Using this result and denoting C 1=T 2K ·max {ε,C (ε),L }+C 0,we obtain the following inequality:Ω|∇v (x,0)|2+u 2(x,τ) dx ≤C 1 τ0Ω∇v (x,t )|2+u 2(x,t ) dxdt.(2.4)Consider the functionw (x,t )=tu (x,η)dη.Thenv (x,t )=w (x,τ)−w (x,t ),∇v (x,0)=∇w (x,τ),and also|∇v |2=|∇w (x,τ)−∇w (x,t )|2≤2|∇w (x,τ)|2+2|∇w (x,t )|2,so we haveτ0Ω|∇v |2dxdt ≤2τ Ω|∇w (x,τ)|2dx +2 τ0Ω|∇w |2(x,t )dxdt.Thus,the inequality(2.4)takes the formΩ|∇w (x,τ)|2+u 2(x,τ) dx ≤2C 1τ Ω|∇w (x,τ)|2dx +2C 1τ0Ω(|∇w |2+u 2)dxdt.(2.5)5Because of arbitrariness of τ,let τsatisfy 2C 1τ≤1.Then it follows from (2.5)that:(1−2C 1τ) Ω|∇w (x,τ)|2+u 2(x,τ) dx≤2C 1τ0Ω(|∇w |2+u 2)dxdt.(2.6)Applying the Gronwall’s inequality,we conclude thatΩ |∇w (x,τ)|2+u 2(x,τ) dx ≤0,∀τ∈[0,12C 1],whence it follows thatu (x,τ)=0,∀τ∈[0,12C 1].Obtaining the same inequality for the ”slices”τ∈[k −12C 1,k 2C 1]to cover the [0,T ],we ensure thatu (x,τ)=0,∀τ∈[0,T ].Thus,the uniqueness is proved.3ExistenceTo prove the existence,we will apply the Galerkin’s method.Let smooth function w k =w k (x )(k =1,...)such that{w k }∞k =1is an orthogonal basis of W 12(Ω)and{w k }∞k =1is an orthonormal basis of L 2(Ω).We will try to find approximate solutions in the formu m(x,t )=m k =1d k (t )w k (x ),(3.1)where coefficients d k (t )are to be determined.From (1.1)we haveΩ(u m tt w l +∇u m ∇w l +cu mw l )dx +∂Ωw l (x ) t 0 ΩK x,ξ,τ,u m (ξ,τ)dξdτds = Ωfw l dx (l =1,...,m ),(3.2)6d k(0)=αk=(φ,w k)L2(Ω)(k=1,...,m),(3.3)d k(0)=βk=(ψ,w k)L2(Ω)(k=1,...,m).(3.4) Substituting(3.1)into(3.2),we obtainΩmk=1(d k w k w l+d k∇w k∇w l+cd k w k w l)dx+∂Ωw l(x)tΩKx,ξ,τ,mk=1d k(τ)w k(ξ)dξdτds=Ωfw l dx(k,l=1,...,m).Denote,for short,γkl(t)=(∇w k,∇w l)L2(Ω)+(cw k,w l)L2(Ω),(3.5)f l(t)=(f,w l)L2(Ω),(3.6)g l(t)=∂Ωw l(x)tΩKx,ξ,τ,mk=1d k(τ)w k(ξ)dξdτds,(3.7)we haved l(t)+mk=1d k(t)γkl(t)+g l(t)=f l(t)(l=1,...,m).(3.8)Letb l(t)=d l(t),(l=1,...,m)z(t)=b1(t),b2(t),...,b m(t),d1(t),d2(t),...,d m(t)TandΦ(t,z)=f1(t)−ml=1d k(t)γk1(t)−g1(t)f2(t)−ml=1d k(t)γk2(t)−g2(t)...f m(t)−ml=1d k(t)γkm(t)−g m(t)b1(t)b2(t)...b m(t)7we haved z(t)dt=Φ(t,z)(3.9) z(0)=(β1,β2,...,βm,α1,α2,...,αm)T(3.10)ThenT|Φ(t,z)|2dt=ml=1Tf l(t)−mk=1d k(t)γkl(t)−g l(t)2dt+mk=1Tb2k(t)dt≤3ml=1Tg l(t)2dt+3ml=1Tmk=1d k(t)γkl(t)2dt+3ml=1Tf l(t)2dt+ml=1Tb k(t)2dt.From(3.7)and(A2)we have3ml=1Tg l(t)2dt=3ml=1T∂Ωw l(x)tΩKx,ξ,τ,mk=1d k(τ)w k(ξ)dξdτds2dt≤3ml=1T∂Ωw2l(x)ds∂ΩtΩKx,ξ,τ,mk=1d k(τ)w k(ξ)dξdτ2dsdt≤3|Ω|ml=1Tδ1|∇w l(x)|2+C(δ1)w2l(x)dx·δ2tΩ|∇Kx,ξ,τ,mk=1d k(τ)w k(ξ)|2dξdτ+C(δ2)tΩK2x,ξ,τ,mk=1d k(τ)w k(ξ)dξdτdt≤C2mk=1Td k(t)2dt,where C2is a constant;From(3.5)-(3.6)and(A1)we have3ml=1Tmk=1d k(t)γkl(t)2dt+3ml=1Tf l(t)2dt≤C3+C4mk=1Td k(t)2dt,8where C 3,C 4are constants;let C 5=max {1,C 2,C 4}we haveT0|Φ(t,z )|2dt ≤C 3+C 5T|z (t )|2dt.(3.11)From (3.1)we know that z (t )∈L 2([0,T ];R 2m ),we haveΦ∈C [0,T ]×L 2([0,T ];R 2m );L 2([0,T ];R 2m ) (3.12)Letz i = b i 1(t ),b i2(t ),...,b i m (t ),d i 1(t ),d i 2(t ),...,d i m (t ) T,(i =1,2)then for ∀t ∈[0,T ],Φ(t,z 1)−Φ(t,z 2) 2L ([0,T ];R 2m )= T0|Φ(t,z 1)−Φ(t,z 2)|2dt =m l =1T 0b 1l (t )−b 2l (t ) 2dt +3ml =1 Tm k =1d 1k (t )−d 2k (t )γkl (t ) 2dt+3m l =1T 0∂Ωw l (x )t 0ΩKx,ξ,τ,m k =1d 1k(τ)w k (ξ)−Kx,ξ,τ,mk =1d 2k (τ)w k (ξ)2dt≤C 5 z 1−z 2 2L ([0,T ];R 2m )(3.13)Applying lemma(1.3)and lemma(1.4),we known that (3.9)-(3.10)isuniquely solvable.Thus,for every m there exists a unique u m (x,t ),that satisfies (3.2),in other words,the sequence {u m }is defined.We shall investigate it’s conver-gence.Multiply (3.2)by d l (t ),sum by l from 1to m and,finally,integrate by t from 0to τ,τ0Ω(u m tt u mt +∇u m ∇u m t +cu m u m t )dxdt + τ0∂Ωu m tt 0 ΩK (x,ξ,η,u m )dξdηdsdt = τ0Ωfu m t dxdt.9Integration by parts on the left will give usτ0 Ωu m tt u mt dxdt =12 Ω u m t (x,τ) 2dx −12Ω u m t (x,0) 2dx,τ0 Ω∇u m ∇u m tdxdt =12 Ω ∇u m(x,τ) 2dx −12Ω ∇u m (x,0) 2dx, τ0Ωcu m u m t dxdt =12 Ωc (x,τ) u m (x,τ) 2dx −12Ωc (x,τ) u m (x,0) 2dx−12 τ0 Ωc t (u m )2dxdt, ∂Ωτu m t (x,t )t 0ΩK (x,ξ,η,u m )dξdηdsdt=− ∂Ωτ0u m (x,t ) ΩK (x,ξ,t,u m )dξdtds+ ∂Ωu m (x,τ) τ0ΩK (x,ξ,η,u m )dξdηds.Finally,we obtain 12 Ωu m t (x,τ) 2+|∇u m (x,τ)|2+c (x,τ) u m (x,τ) 2 dx=12 Ωu m t (x,0) 2+|∇u m (x,0)|2+c (x,0) u m (x,0) 2 dx+12 τΩc t (u m )2dxdt + τ0 Ωfu m t dxdt + τ0 ∂Ωu m (x,t ) ΩK (x,ξ,t,u m )dξdsdt − ∂Ωu m (x,τ) τ0ΩK (x,ξ,η,u m )dξdηds.(3.14)Consider the right-most integral in (3.13).Applying the Cauchy inequality,we haveτ0 ∂Ωu m(x,t )ΩK (x,ξ,t,u m )dξdsdt ≤12 τ0 ∂Ω u m (x,t ) 2dsdt +12 τ0 ∂Ω ΩK (x,ξ,t,u m)dξ 2dsdt ;10next,using the inequality (2.3),12 τ0 ∂Ω u m (x,t ) 2dsdt +12 τ0 ∂Ω ΩK (x,ξ,t,u m )dξ 2dsdt ;≤12 τ0 Ω |∇u m (x,t )|2+C ( )u 2 dxdt +12 τ0 Ω Ω∇K (x,ξ,t,u m )dξ 2+C ( ) ΩK (x,ξ,t,u m )dξ 2 dxdt ≤12 τ0 Ω |∇u m (x,t )|2+C ( )u 2 dxdt +12 τ0 Ω |Ω| Ω|∇K (x,ξ,t,u m )|2dξdxdt +12 τ0 ΩC ( )|Ω| ΩK (x,ξ,t,u m ) 2dξdxdt.Thusτ0∂Ωu m (x,t ) ΩK (x,ξ,t,u m )dξdsdt≤12 τ0 Ω |∇u m (x,t )|2+C ( )u 2dxdt+12 τ0 Ω |Ω|K 21 Ω u m (ξ,t ) 2dξdxdt +12 τ0 ΩC ( )|Ω|K 20 Ω u m (ξ,t ) 2dξdxdt ≤12 τ0 Ω |∇u m |2+ C ( )+ |Ω|K 21+C ( )|Ω|K 20 (u m )2 dxdt ≤ 2 τ0 Ω|∇u m (x,t )|2dxdt +C ( )+ |Ω|K 21+C ( )|Ω|K 202 τ0 Ω u m (x,t ) 2dxdt.(3.15)Applying the modified Cauchy inequality to the second integral in (3.13),we have∂Ωu m (x,τ)τ0 ΩK (x,ξ,η,u m )dξdηds ≤ 2 ∂Ωu m (x,τ)ds +12 ∂Ω τ0 ΩK (x,ξ,η,u m )dξdη 2ds 11Using again the inequality(2.3),we have ∂Ωu m (x,τ)τ0 ΩK (x,ξ,η,u m )dξdηds ≤ 2 Ω µ|∇u m (x,τ)|2+C (µ) u m (x,τ) 2 dx +12 Ω τµ|Ω| τ0 Ω|∇K (x,ξ,η,u m )|2dξdη+τC (µ)|Ω| τ0 Ω K (x,ξ,η,u m ) 2dξdη dx≤ µ2 Ω|∇u m (x,τ)|2dx + C (µ)2 Ωu m (x,τ) 2dx +τ|Ω|2(µK 21+C (µ)K 20)2 τ0 Ωu m (x,t ) 2dxdt.(3.16)From (A1)we know that 0<c 1≤c (x,t )≤c 2,|∂c ∂t|≤c 3;∀(x,t )∈Q.Choose and µso that µ<1, C (µ)≤c 1.Inequalities (3.14)and (3.15)give us: Ω u m t (x,τ) 2+(1− µ)|∇u m (x,τ)|2+ c 1− C (µ) u m (x,τ) 2 dx≤Ω u m t (x,0) 2+|∇u m (x,0)|2+c 2 u m (x,0) 2 dx + τ0 Ωf 2dxdt + τ0 Ω(u m t )2dxdt +τ0 Ω|∇u m (x,t )|2dxdt + c 3+C ( )+ |Ω|K 21+C ( )|Ω|K 20+τ|Ω|2(µK 21+C (µ)K 20) τ0 Ω(u m )2dxdt.(3.17)Thus,denotingc 4=max {1,c 2},M 1=1c 4·min {1,1− µ,c 1− C (µ)},M 2=1c 4·max {c 3,C ( )+ |Ω|K 21+C ( )|Ω|K 20,τ|Ω|2(µK 21+C (µ)K 20)},we obtain the inequality M 1 Ω u m t (x,τ) 2+|∇u m (x,τ)|2+ u m (x,τ) 2 dx≤ Ω u m t (x,0) 2+|∇u m (x,0)|2+ u m (x,0) 2 dx+M 2 τ0 Ω (u m t )2+|∇u m (x,t )|2+(u m )2 dxdt + τ0 Ωf 2dxdt.12Applying the Gronwall inequality and integrating on t from 0to τ,we finally obtainu m W 1,12(Q τ)≤C (T )( f L 2(Q )+ φ W 12(Ω)+ ψ L 2(Q )).(3.18)Since f L 2(Q ), φ W 12(Ω), ψ L 2(Q )are bounded,the sequence {u m }∞m =1is bounded in W 1,12(Q ),therefore,there exists a weakly converging subse-quence (we will denote it {u m }∞m =1again for simplicity),now we shall show that its limit u ∈W 1,12(Q )is a desired solution.To prove this,we fix an integer N ,select m ≥N ,multiply (3.2)by thefunction h l (t )∈W 12[0,T ],h l(T )=0,sum by l from 1to N ,and integrate on t from 0to T ;denoting ηN (x,t )=N l =1w l (x )h l (t ),we obtainT 0Ω(−u m t ηN t +∇u m ∇ηN +cu m ηN )dxdt+ T 0 ∂ΩηNt 0 ΩK (x,ξ,τ,u m )dξdτdsdt = Ωu m t (x,0)ηN (x,0)dx +T 0 ΩfηN dxdt.It is no a problem to go to the limit at m →∞in the latter expression but the one addendT 0 ∂ΩηN t 0 ΩK (x,ξ,τ,u m )dξdτdsdt (3.19)Consider the integralT 0 ∂ΩηN t 0 ΩK (x,ξ,τ,u m )−K (x,ξ,τ,u ) dξdτdsdt.Applying (A2),we have for the “internal”integralt 0 Ω K (x,ξ,τ,u m )−K (x,ξ,τ,u ) dξdτdsdt≤ T 0 Ω|K (x,ξ,τ,u m )−K (x,ξ,τ,u )|dξdτdsdt≤K T 0 Ω|u m −u |dxdt,from W 1,12(Q ) →L 1(Q )we known thatT 0Ω|u m −u |dxdt →013therefore,we can go to the limit in(3.18).Hence,the limit function usatisfies(3.2)for everyηN(x,t)=Nl=1w l(x)h l(t).Denote S N a set of the functions of the formηN(x,t)=Nl=1w l(x)h l(t),h l(t)∈W12[0,T],h l(T)=0.While ∞N=1S N is dense in W1,12(Q),then(3.2)holds true for every function from W1,12(Q).This proves that u(x,t)is a generalized solution of(1.1)-(1.2)-(1.3).References[1]Beilin,S.,On a mixed nonlocal problem for a wave equation,ElectronicJournal of Differential Equations.2006Vol.2006.N.103.Pp.1-10.[2]Beilin,S.,Existence of solutions for one-dimensional wave equationswith nonlocal conditions,Electronic Journal of Differential Equations.2001.Vol.2001.N.76.Pp.1-8.[3]Cannon J.R.The solution of heat equation subject to the specificationof energy.Quart.Appl.Math.1963.Vol.21.N.2.Pp.155-160.[4]Deimling,K.,Ordinary differential equations in Banach spaces,Springer-Verlag(1977).[5]D.-Q.Dai,Y.Huang,Nonlocal boundary problems for a third-orderone-dimensional nonlinear pseudoparabolic equation.Nonlinear Analy-sis66(2007)179-191.[6]D.-Q.Dai,Y.Huang,On a nonlocal problem for nonlinear pseu-doparabolic equations.Nonlinear Analysis64(2006)499-512.[7]Gordeziani D.G.Avalishvili G.A.,Solutions to nonlocal problems ofone-dimensional oscillation of medium.Matem,Modelirovanie.2000.Vol.12.N.1.Pp.94-103.[8]Lars H¨o rmander,Lectures on nonlinear hyperbolic differential equa-tions,Springer-Verlag(1997).[9]Ladizhenskaja O. A.,Boundary value problems of mathematicalphysics,M.:Nauka.1973.[10]Pulkina L.S.,On a certain nonlocal problem for degenerate hyperbolicequation,Matem.Zametki.1992.Vol.51,N.3..Pp.91-96.[11]Pulkina L.S.,Mixed problem with integral conditions for a hyperbolicequation Matem,Zametki.2003.Vol.74,N.3.Pp.435-445.14 [12]Pulkina L.S.,Nonlocal problem with integral conditions for a hyper-bolic equation,Differenz.Uravnenija.2004.Vol.40.N.7.Pp.887-892.15。