# A Semilinear Wave Equation with a Boundary Condition of Many-Point Type: Global Existence and Stability of Weak Solutions

Authors: Giai Giang Vo
Journal: Abstract and Applied Analysis
 (2015)

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## Abstract

This paper is devoted to the study of a wave equation with a boundary condition of many-point type. The existence of weak solutions is proved by using the Galerkin method. Also, the uniqueness and the stability of solutions are established.


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## Body

## 1. Introduction

Recently, initial-boundary value problems of wave equations have appeared more and more in mechanics, they have been deeply studied by many authors, and we can refer to the works [1–13]. In this paper, we consider the following nonlinear wave equation with a boundary condition of many-point type:(1)utt-ρuxx+fx,u=0,0<x<1,0<t<T,u1,t=0,ρux0,t=∫0tkt-shuξ0,s,uξ1,s,…,uξN,sds+guξ0,t+σt,ux,0=u0x,utx,0=u1x,where f,g,h,k,σ,u0,u1 are given real functions and ρ,ξ1,ξ2,…,ξN are positive constants such that 0=ξ0<ξ1<⋯<ξN<1.Dang and Alain [4] studied the global existence of the following problem:(2)utt-uxx+utαsign⁡ut=0,u1,t=0,ux0,t=gt,ux,0=u0x,utx,0=u1x,where 0<α<1 is a constant and g,u0,u1 are given functions.In [11], Santos considered the following problem:(3)utt-ρtuxx=0,u0,t=0,u1,t+∫0tgt-sρsux1,sds=0,ux,0=u0x,utx,0=u1x,in which g,ρ,u0,u1 are given functions. He studied the asymptotic behavior of the solutions of problem (3) with respect to the time variable. In this case, problem (3) is a mathematical model for a linear one-dimensional motion of an elastic bar connected with a viscoelastic element at an end of the bar.Applying the Mikusinski operational calculus, D. Takači and A. Takači [12] gave the formula for finding the exact solutions of the linear wave equation given by(4)utt-uxx+Ku+λut=fx,t,ux0,t=vt,-ux1,t=δu1,t+εut1,t,ux,0=u0x,utx,0=u1x,where K,λ,δ,ε,h,ω are given nonnegative constants and f,g,u0,u1 are given functions. Also, the unknown function u(x,t) and the unknown boundary value v(t) satisfy the following integral equation:(5)vt=gt+hu0,t-ωh∫0tsin⁡ωt-su0,sds.It is worth noting that the function v(t) is deduced from a Cauchy problem for an ordinary differential equation at the boundary condition t=0. Indeed, if v(t) satisfies the following Cauchy problem(6)v′′t+2pv′t+p0vt=qutt0,t+q0ut0,t+q1u0,t,v0=v0,v′0=v1,where p,p0,q,q0,q1,v0,v1 are constants such that p0>p2, we can easily show that(7)vt=gt+hu0,t+∫0tkt-su0,sds,where(8)gt=exp⁡-ptv1-pv0+pqu00-qu10sin⁡ωtω+exp⁡-ptv0-qu00cos⁡ωt-q0ωu00,kt=exp⁡-ptqp2-ω2-pq0+q1sin⁡ωtω+q0-2pqcos⁡ωt,with ω=p0-p2,h=q+q0/ω. In the special case of p=q0=q1=0, we obtain (5).Besides, Nguyen and Giang Vo [8] obtained the asymptotic behavior of the weak solution of the following initial-boundary value problem as ε→0+:(9)utt-uxx+Ku+λut=fx,t,u1,t=0,ux0,t=gt+hu0,t+εut0,t-∫0tkt-su0,sds,ux,0=u0x,utx,0=u1x,in which K,λ,h∈R,ε>0 are given constants and f,g,k,u0,u1 are given functions. Problem (9) is said to be a mathematical model describing a shock problem involving a linear viscoelastic bar.The organization of this paper is as follows. First of all, we establish the global existence and uniqueness of weak solutions of problem (1). The proof is based on the Galerkin method associated with a priori estimates and the weak compact method. Finally, here we prove that this solution is stable in the sense of continuous dependence on the given data (f,g,h,k,σ). This paper is a relative generalization of the works [4, 6, 11, 12].


## 2. Preliminaries

For convenience, we denote by·,· and ·, respectively, the scalar product and the norm in L2(0,1). Also, we define a closed subspace of the Sobolev space H1(0,1) as follows:(10)H=u∈H10,1:u1=0,with the following scalar product and norm:(11)u,vH=ux,vx,uH=ux.Then it is easy to show the following.Lemma 1.
The embeddingH↪C0([0,1]) is compact and(12)uC00,1≤uH≤uH10,1≤2uH,∀u∈H.On the other hand, we also have the following result.Lemma 2.
Letε>0. Then(13)vC00,12≤εvx2+1+1εv2,∀u∈H10,1.The proof of the lemma is straightforward; we omit the details.Remark 3.
Letε=1, we get(14)vC00,1≤vx2+2v21/2≤2vH10,1,∀u∈H10,1.Next, ifW is a real Banach space with norm ·W, Lp(0,T;W) consists of equivalence classes of strongly measurable functions u:(0,T)→W such that uLp(0,T;W)<∞, with(15)uLp0,T;W=∫0TutWpdt1/p,if1≤p<∞,esssup0<t<T⁡utW,ifp=∞.It is not difficult to prove that Lp(0,T;W) is a Banach space.LetW0, W1, and W2 be Banach spaces satisfying W0⊂W1⊂W2. We further assume that W0 and W2 are reflexive, and the imbedding W0↪W1 is compact. Set(16)W0,T=u:u∈Lp0,T;W0,ut∈Lq0,T;W2,where 1≤p,q≤∞. Then W(0,T) is a Banach space with the norm(17)uW0,T=uLp0,T;W0+utLq0,T;W2.We also have the following lemma.Lemma 4 (see [14]).
Let1<p,q<∞. The embedding W(0,T)↪Lp(0,T;W1) is compact.


## 3. Global Existence and Uniqueness of Weak Solutions

To investigate the existence of a unique weak solution of problem (1), the following assumptions are needed:(A1)
u0∈H and u1∈L2(0,1).(A2)
σ∈H1(0,T),k∈W1,1(0,T).(A3)
f,D2f∈C0([0,1]×R) satisfy the following conditions:
(i)
There exist positive functionsσ1,σ2∈L1(0,1) such that (18)∫0ufx,sds≥-σ1xu2-σ2x,a.e.x∈0,1,∀u∈R.(ii)
There exist positive functionsσ^∈L2(0,1) and f^∈C0(R) such that (19)D2fx,u≤σ^xf^u,a.e.x∈0,1,∀u∈R.(A4)
g∈C2(R); there exist constants a,b>0 such that (20)∫0ugsds≥-au2-b,∀u∈R.(A5)
h∈C0(RN+1); there exist constants c,d>0 such that (21)hu0,u1,…,uN≤c∑i=0Nui+d,∀ui∈R,i=0,N¯.(A6)
For eachM>0, there exists a constant cM>0 such that (22)hu0,…,uN-hv0,…,vN≤cM∑i=0Nui-vi∀ui,vi≤M,i=0,N¯.With these assumptions, we have the following theorem.Theorem 5.
Let assumptions(A1)–(A6) hold. Then problem (1) has a unique weak solution u such that(23)u∈L∞0,T;H,ut∈L∞0,T;L20,1,uξi,·∈H10,T,i=0,N¯.Remark 6.
In the special case ofN=0 and g′(t)>-1, for all t∈R, we have obtained the same results in the paper [6].Proof of Theorem5. The main tool of this proof is the Galerkin method. The procedure includes four steps as follows:(i)
Galerkin approximation.(ii)
A priori estimates.(iii)
Limiting process.(iv)
Uniqueness of the weak solutions.Step 1 (Galerkin approximation).
We use a special orthonormal base ofH: (24)φkx=21+μk2cos⁡μkx,μk=2k-1π2,k=1,2,….Now we are looking for the approximate solution of problem (1) in the form(25)umx,t=∑k=1mωmktφkx,where the coefficient functions ωmk(t) satisfy the following system of nonlinear differential equations(26)uttm·,t,φk+ρum·,t,φkH+vmtφk0+f·,um·,t,φk=0,k=1,m¯,with(27)vmt=∫0tkt-shumξ0,s,umξ1,s,…,umξN,sds+gumξ0,t+σt,(28)um·,0≡u0m=∑k=1mamkφk⟶u0stronglyinH10,1,(29)utm·,0≡u1m=∑k=1mbmkφk⟶u1stronglyinL20,1.By substituting ρk=ρμk, we can rewrite the system of (26)–(29) as follows:(30)ωmk′′t+ρk2ωmkt=-1φk2vmtφk0+f·,um·,t,φk,vmt=∫0tkt-shumξ0,s,umξ1,s,…,umξN,sds+gumξ0,t+σt,ωmk0=amk,ωmk′0=bmk,k=1,m¯.Therefore, we obtain(31)ωmkt=amkcos⁡ρkt+bmksin⁡ρktρk-2φk0∫0tds∫0ssin⁡ρkt-sρkks-τhumξ0,τ,…,umξN,τdτ-2φk0∫0tsin⁡ρkt-sρkσs+gumξ0,sds-2φk20∫0tsin⁡ρkt-sρkf·,ums,φkds,k=1,m¯.Applying the Schauder fixed-point theorem, it is not difficult that system (31) has a solution (ωm1,ωm2,…,ωmm) on an interval [0,Tm]. This implies that in system (26)–(29) there exists a solution to um(t) on [0,Tm]. Moreover, we can extend the approximate solution um to the whole interval [0,T] (see [15]).Step 2 (a priori estimates).
In (26), we replace φk(x) by utm(x,t). Then integrating from 0 to t, we have after some calculations(32)Emt=Em0+2σ0u0m0+2∫0u0m0gsds+2∫01dx∫0u0mxfx,sds-2σtum0,t+2∫0tσ′sum0,sds-2∫0um0,tgsds-2∫01∫0umx,tfx,sdsdx-2∫0tutm0,rdr∫0rkr-shumξ0,s,…,umξN,sds≡Em0+2σ0u0m0+2∫0u0m0gsds+2∫01dx∫0u0mxfx,sds+∑k=15Ikt,where(33)Emt=utm·,t2+ρum·,tH2.We will estimate, respectively, the following terms on the right-hand side of (32).Estimating
I
1
(
t
). Using (33) and Lemma 1, we infer that(34)um0,t≤um·,tC00,1≤um·,tH≤1ρEmt.Hence,(35)I1t=-2σtum0,t≤εum0,t2+1εσ2t≤εEmt+1εσL∞0,T2,∀ε>0.EstimatingI2(t). Using Lemma 1 and inequality (34), we arrive at(36)I2t=2∫0tσ′sum0,sds≤1ρ∫0tσ′sds+ρ∫0tσ′sum0,s2ds≤∫0tσ′sEmsds+1ρσ′L10,T.EstimatingI3(t). From assumption (A4), we have(37)I3t=-2∫0um0,tgsds≤2aum0,t2+2b.On the other hand, we see that(38)um·,t2=u0m+∫0tutm·,sds2≤2u0m2+2t∫0tEmsds.It follows from (37), (38), and Lemma 2 that(39)I3t≤2εaρEmt+4at1+1ε∫0tEmsds+4a1+1εu0m2+2b.EstimatingI4(t). Owing to assumption (A3)-(i), (38), and Lemma 2, it is not difficult to show that(40)I4t=-2∫01dx∫0umx,tfx,sds≤2∫01σ1xumx,t2dx+2σ2L10,1≤ε2ρσ1L10,1Emt+4t1+1εσ1L10,1∫0tEmsds+41+1εu0m2σ1L10,1+2σ2L10,1.EstimatingI5(t). Applying integration by parts, it follows that(41)I5t=-2∫0tutm0,rdr∫0rkr-shumξ0,s,…,umξN,sds=-2um0,t∫0tkt-shumξ0,s,…,umξN,sds+2k0∫0tum0,shumξ0,s,…,umξN,sds+2∫0tum0,rdr∫0rk′r-shumξ0,s,…,umξN,sds≡J1t+J2t+J3t.We can estimate the integrals in the right-hand side of (41) as follows:(42)J1t=-2um0,t∫0tkt-shumξ0,s,…,umξN,sds≤2cum0,t∑i=0N∫0tkt-sumξi,sds+2dum0,t∫0tkt-sds≤2N+1cρEmt∫0tkt-sEmsds+2dρEmt∫0tksds≤2εEmt+N+12c2ερkL20,T2∫0tEmsds+d2ερkL10,T2,(43)J2t=2k0∫0tum0,shumξ0,s,…,umξN,sds≤2N+1cρk0∫0tEmsds+2dρk0∫0tEmsds≤1+2N+1cρk0∫0tEmsds+d2ρTk20,(44)J3t=2∫0tum0,rdr∫0rk′r-shumξ0,s,…,umξN,sds≤2N+1cρ∫0tEmrdr∫0rk′r-sEmsds+2dρ∫0tEmrdr∫0rk′r-sds≤2N+1cρ∫0tEmrdr+∫0t∫0rk′r-sEmsds2dr+∫0tEmrdr+d2tρk′L10,T2.Going in for the Cauchy-Schwartz inequality, we arrive at(45)∫0rk′r-sEmsds2≤∫0rk′r-sds∫0rk′r-sEmsds≤k′L10,T∫0rk′r-sEmsds.Consequently,(46)∫0t∫0rk′r-sEmsds2dr≤k′L10,T∫0tdr∫0rk′r-sEmsds=k′L10,T∫0tEmsds∫stk′r-sdr≤k′L10,T2∫0tEmsds.It follows from (44) and (46) that(47)J3t≤1+2N+1cρ1+k′L10,T2∫0tEmrdr+d2Tρk′L10,T2.We deduce from the estimates of J1(t),J2(t),J3(t) that(48)I5t≤2εEmt+DT1∫0tEmsds+DT2,with (49)DT1=N+12c2ερkL20,T2+2N+1cρk′L10,T2+k0+1+2,DT2=d2ερεTk′L10,T2+kL10,T2+εTk20.On the other hand, by (28), (29), and assumptions (A1)–(A3), (50)Em0+2σ0u0m0+2∫0u0m0gsds+2∫01dx∫0u0mxfx,sds≤C,where C is a positive constant depending only on u0,u1,f,g,σ.Combining (32), (35), (36), (39), (40), (48), and (50), we obtain after some rearrangements(51)Emt≤εDT3Emt+∫0tDT4sEmsds+DT5,where(52)DT3=2ρσ1L10,1+2aρ+3,DT4s=σ′s+4T1+1ερ1L10,1+a+DT1,DT5=1εσL∞0,T2+1ρσ′L10,T+2σ2L10,1+41+1εσ1L10,1+au0m2+DT2+C+2b.Choosing εDT3=1/2, by Gronwall’s inequality, we have(53)Emt≤2DT5exp⁡2∫0tDT4sds≤DT,where DT is a positive constant depending on T.Next, we will require the following lemma.Lemma 7.
There exists a positive constantCT depending only on T such that (54)∫0tutmξi,s2ds≤CT,∀t∈0,T,i=0,N¯.Proof of Lemma7. We put(55)kmit=∑k=1mcos⁡μkξisin⁡ρktρk,(56)gmit=∑k=1mφkξiamkcos⁡ρkt+bmksin⁡ρktρk-2∑k=1mcos⁡μkξiφk0∫0tsin⁡ρkt-sρkf·,um·,s,φkds.In view of (25) and (31), um(ξi,t) can be rewritten as follows:(57)umξi,t=gmit-2∫0tkmit-svmsds.In connection with gmi(t), we have the following lemma.Lemma 8.
There exists a positive constantC^T depending only on T such that(58)∫0tgmi′s2ds≤C^T,∀t∈0,T,i=0,N¯.Proof of Lemma8. We define(59)gmi′t=-pmit+qmit-rmit-smit,with(60)pmit=∑k=1mφk0ρkamkcos⁡μkξisin⁡ρkt,(61)qmit=∑k=1mφk0bmkcos⁡μkξicos⁡ρkt,(62)rmit=2∑k=1mcos⁡μkξiφk0sin⁡ρktρkf·,u0m,φk,(63)smit=2∑k=1mcos⁡μkξiφk0∫0tsin⁡ρkt-sρkD2f·,um·,sutm·,s,φkds.On the other hand, use the inequality (64)a+b+c+d2≤4a2+b2+c2+d2,∀a,b,c,d∈R.Hence,(65)∫0tgmi′s2ds≤4∑j∈p,q,r,s∫0tjmis2ds≡K1t+K2t+K3t+K4t.We will estimate each term on the right-hand side of this inequality.Estimating
K
1
(
t
)
=
4
∫
0
t
‍
p
m
i
(
s
)
2
d
s. Thanks to (60) and (65), (66)K1t=ρ∫0t∑k=1mφk0μkamksin⁡μkρs+ξi+sin⁡μkρs-ξi2ds≤2ρ∫0t∑k=1mφk0μkamksin⁡μkρs+ξi2ds+2ρ∫0t∑k=1mφk0μkamksin⁡μkρs-ξi2ds=2ρ∫ξiρt+ξi∑k=1mφk0μkamksin⁡μks2ds+2ρ∫-ξiρt-ξi∑k=1mφk0μkamksin⁡μks2ds.Now, we will need the following lemma.Lemma 9.
Leta,b∈R,a<b and ck∈R,k=1,m¯. Then(67)∫ab∑k=1mcksin⁡μks2ds≤2max⁡a,b+2∫01∑k=1mcksin⁡μks2ds.The proof of this lemma is simple; we omit the details.Applying Lemma9, we deduce from (28), (66), and assumption (A1) that(68)K1t≤8ρρT+ξi+2∫01∑k=1mφk0μkamksinμks2ds≤8ρρT+ξi+2u0mH2≤CT,where CT always indicates a constant depending on T.Estimating
K
2
(
t
)
=
4
∫
0
t
‍
q
m
i
(
s
)
2
d
s. Similarly, we also obtain(69)K2t≤8T+ξi+2ρu1m2≤CT.EstimatingK3(t)=4∫0t‍rmi(s)2ds. We see that(70)rmis=2ρ∑k=1m1μk∫01sin⁡μkρscos⁡μkξicos⁡μkxfx,u0mxdx=12ρ∫01∑k=1m1μksin⁡μkρs+ξi+x+sin⁡μkρs-ξi-xfx,u0mxdx+12ρ∫01∑k=1m1μksin⁡μkρs+ξi-x+sin⁡μkρs-ξi+xfx,u0mxdx.To estimate K3(t), we need the following lemma.Lemma 10.
Letm∈N. One always has(71)Smz=∑k=0msin⁡μkzμk<1+4π,∀z∈R.Proof of Lemma10. First, we assume that 0<z≤1. Set M=z-1 (which is an integer part of z-1). We consider two cases of m.Case 1 (m>M+1).
Then(72)Smz≤∑k=1Msin⁡μjzμj+∑k=M+1msin⁡μkzμk≡Sm1z+Sm2z.Since sin⁡φ≤φ, for all φ∈R, we get(73)Sm1z=∑k=1Msin⁡μkzμk≤∑k=1Mμkzμk=Mz≤1.Moreover, the function φ↦(sin⁡φ)/φ is decreasing on (0,π/2]; hence, sin⁡(π/2)φ≥φ; for all φ∈[0,1], it follows that(74)∑i=kmsin⁡μiz=sin⁡m+k-1/2πzsin⁡m-k+1/2πzsin⁡πz/2≤1sin⁡πz/2≤1z,k=1,m¯.On the other hand, it is easy to verify the following equality by the induction:(75)∑k=M+1msin⁡μkzμk=1μM+1∑k=M+1msin⁡μkz+∑k=M+1m-11μk+1-1μk∑i=k+1msin⁡μiz.Using (74) and (75), we arrive at(76)Sm2z=∑k=M+1msin⁡μkzμk≤1μM+1∑k=M+1msin⁡μkz+∑k=M+1m-11μk-1μk+1∑i=k+1msin⁡μiz≤1μM+11z+∑k=M+1m-11μk-1μk+11z=2μM+1-1μm1z<4π.Consequently, it follows from (72), (73), and (76) that(77)Smz≤Sm1z+Sm2z<1+4π.Case 2 (1≤m≤M+1).
We have(78)Smz≤∑k=1msin⁡μkzμk≤∑k=1M+1sin⁡μkzμk≤∑k=1M+1μkzμk=M+1z<1+4π.Combining (77) and (78), we conclude that(79)Smz<1+4π,∀m∈N,z∈0,1.Since Sm(0)=0 and the function Sm is even, periodic with the period 2, thus inequality (79) holds for all m∈N,z∈R. The proof of Lemma 10 is complete.By (70) and Lemma 10, it leads to(80)rmis≤12ρ∫0141+4πfx,u0mxdx=2ρ1+4πf·,u0mL10,1.Hence, it follows from (28), (80), and assumptions (A1), (A3) that(81)K3t=4∫0trmis2ds≤16Tρ1+4π2f·,u0mL10,12≤CT.EstimatingK4(t)=4∫0t‍smi(s)2ds. Proving in the same way as (81), we get(82)K4t=4∫0tsmis2ds≤8T2ρ1+4π2∫0tD2f·,um·,sutm·,sL10,12ds.On the other hand, using assumption (A3)-(ii), then(83)D2fx,umx,s≤σ^xsupz≤D^T⁡f^z=C^Tσ^x,with D^T=DT/ρ.Applying the Cauchy-Schwartz inequality, we clearly get(84)K4t≤8T2ρ1+4π2C^T2σ^2∫0tutm·,s2ds≤CT.Combining (65), (68), (69), (81), and (84), we obtain Lemma 8.Remark 11.
Lemma3 in [4] is a special case of Lemma 8 with g=h=0 and ρ=1.We now return to the proof of Lemma7.Note that it follows from (27) that(85)vm′t=σ′t+g′umξ0,tutmξ0,t+k0humξ0,t,umξ1,t,…,umξN,t+∫0tk′t-shumξ0,s,umξ1,t,…,umξN,sds.On account of (53) and assumptions (A2), (A4), and (A5), we get(86)vm′t≤σ′t+supz≤D^T⁡g′zutmξ0,t+k0+k′L10,TN+1cD^T+d≤σ′t+CTutmξ0,t+1.Therefore, it is easy to see that(87)vm′t2≤2σ′t2+4CT2utmξ0,t2+1.Applying Lemma 10 and the imbedding H1(0,1)↪C0([0,1]), we deduce from (27) and (57) that(88)utmξi,t=gmi′t-2kmitvm0-2∫0tkmit-svm′sds≤gmi′t+2ρ1+4πvm0+2ρ1+4π∫0tvm′sds≤gmi′t+CT1+∫0tvm′sds.Thus,(89)utmξi,t2≤2gmi′t2+4CT21+t∫0tvm′s2ds.Using the Cauchy-Schwartz inequality and Lemma 8, then we obtain from (87) and (89) that(90)∫0tutmξi,s2ds≤4CT2t+2∫0tgmi′s2ds+4CT2∫0tτdτ∫0τvm′s2ds≤4CT2t+163CT4t3+4CT2tσ′L20,T2+2∫0tgmi′s2ds+16CT4∫0tτdτ∫0τutmξ0,s2ds≤CT+CT∫0tdτ∫0τutmξ0,s2ds.Hence,(91)∑i=0N∫0tutmξi,s2ds≤N+1CT1+∫0t∑i=0N∫0τutmξi,s2dsdτ.By the Gronwall inequality, we get Lemma 7. This completes the proof of Lemma 7.Step 3 (limiting process).
Due to (53) and (54), applying the Banach-Alaoglu theorem, we can extract a subsequence of sequence um, still labeled by the same notations, such that(92)um⟶uweakly∗inL∞0,T;H10,1,utm⟶utweakly∗inL∞0,T;L20,1,umξi,·⟶uξi,·weaklyinH10,T.Thanks to Lemma 4 and the compactness of the imbedding H1(0,T)↪C0([0,T]), they lead us to(93)um⟶ustronglyinL20,1×0,T,a.e.  in0,1×0,T,umξi,·⟶uξi,·stronglyinC00,T.By (93)2 and assumptions (A2), (A4), and (A6), we have(94)vmt⟶∫0tkt-shuξ0,s,…,uξN,sds+guξ0,t+σt=vt,strongly in C0([0,T]).Also, we apply the inequality(95)fx,umx,t-fx,ux,t≤dTumx,t-ux,t,where dT=D2fC0([0,1]×[-D^T,D^T]). By (93)1 and (95), we get(96)f·,um⟶f·,ustronglyinL20,1×0,T.Passing to the limit in (26) by (92)1,2, (94), and (96), we have u satisfying the equation(97)ddtut·,t,φ+ρu·,t,φH+vtφ0+f·,u·,t,φ=0,for all φ∈H, a.e. t∈[0,T].On the other hand, it is easy to show a similar way as in [4, p. 588](98)ux,0=u0x,utx,0=u1x.The existence of global solutions is proved.Step 4 (uniqueness of the weak solutions).
Letu1 and u2 be two weak solutions of problem (1). Then u=u1-u2 is a weak solution of the following problem:(99)utt-ρuxx+f·,u1-f·,u2=0,ρux0,t=vt,u1,t=0,ux,0=utx,0=0,in which(100)vt=∑i=12-1i-1guiξ0,t+∫0tkt-s∑i=12-1i-1huiξ0,s,…,uiξN,sds.To prove u1=u2, then the following lemma is needed.Lemma 12.
Letu be the weak solution of the following problem:(101)utt-ρuxx+F=0,0<x<1,0<t<T,u1,t=0,ρux0,t=vt,ux,0=u0x,utx,0=u1x,u∈L∞0,T;H,ut∈L∞0,T;L20,1,u0,·∈H10,T,v∈L20,T,F∈L10,T;L20,1.Then we have(102)ut·,t2+ρu·,tH2+2∫0tvsut0,sds+2∫0tF·,s,ut·,sds≥u12+ρu0H2,a.e.t∈0,T.Equality holds in case of u0=u1=0.
The proof of Lemma12 is the same as Lemma  2.1 in [16].Applying Lemma12 with u0=u1=0,F=f(·,u1)-f(·,u2), we get(103)Et=-2∫0tf·,u1-f·,u2,u′sds-2∫0tut0,s∑i=12-1i-1guiξ0,sds-2ρu0,t∫0tkt-s∑i=12-1i-1huiξ0,s,…,uiξN,sds+2ρk0∫0tu0,s∑i=12-1i-1huiξ0,s,…,uiξN,sds+2ρ∫0tu0,rdr∫0rk′r-s∑i=12-1i-1huiξ0,s,…,uiξN,sds≡J1t+J2t+J3t+J4t+J5t,with E(t)=ut(·,t)2+ρu(·,t)H2.Now we can estimate the integrals in the right-hand side of (103) as follows.First termJ1(t): using assumption (A3), it follows from Lemma 1 and (103) that(104)J1t=-2∫0tf·,u1·,s-f·,u2·,s,ut·,sds≤dM1ρ∫0tut·,s2+ρu·,s2ds≤dM1ρ∫0tEsds,where(105)dM1=D2fC00,1×-M,M,M=∑i=12uiL∞0,T;H.Second term J2(t): set dM2=gC2([-M,M]). Integrating by parts, then we arrive at(106)J2t=-2∫0tut0,s∑i=12-1i-1guiξ0,sds=-2∫0tut0,sds∫01ddθgu2ξ0,s+θuξ0,sdθ=-u20,t∫01g′u2ξ0,t+θuξ0,tdθ+∫0tu20,sds∫01g′′u2ξ0,s+θuξ0,su2′ξ0,s+θu′ξ0,sdθ≤dM2u20,t+∫0tu1′ξ0,s+u2′ξ0,su20,sds.On the other hand, we easily show that(107)u·,t2≤t∫0tut·,s2ds≤T∫0tEsds.Applying Lemma 2, we deduce from (107) that(108)u20,t≤u·,tC00,12≤εu·,tH2+1+1εu·,t2≤ερEt+T1+1ε∫0tEsds,∀ε>0.Thus, it follows from (106) and (108) that(109)J2t≤εdM2ρEt+∫0tdsEsds,in which(110)ds=ερdM2u1′ξ0,s+u2′ξ0,s+1+1εTdM2u1′ξ0,·L10,T+u2′ξ0,·L10,T+1.Third term J3(t): using the Cauchy-Schwartz inequality, we have from (103), (A2), and (A6) that(111)J3t=-2ρu0,t∫0tkt-s∑i=12-1i-1huiξ0,s,…,uiξN,sds≤2Et∫0tkt-s∑i=0NcMiEsds≤14Et+4N+12cM2TkL∞0,T2∫0tEsds.In addition, we can similarly prove as in (111) for the fourth and fifth terms as follows:(112)J4t=2ρk0∫0tu0,s∑i=12-1i-1huiξ0,s,…,uiξN,sds≤2ρk0∫0t1ρEsN+1cM1ρEsds=2N+1cMk0∫0tEsds,J5t=2ρ∫0tu0,rdr∫0rk′r-s∑i=12-1i-1huiξ0,s,…,uiξN,sds≤2∫0tEr∫0rk′r-sN+1cMEsdsdr≤N+1cMk′L10,T2+1∫0tEsds.Choosing 4εdM2≤ρ, the combination of (103), (104), (109), and (111)-(112) shows that(113)Et≤∫0td^sEsds,where(114)d^s=2ds+8N+12cM2TkL∞0,T2+2N+1cMk′L10,T2+2k0+1+2dM1ρ.By the Gronwall inequality, we see that E(t)≡0; that is, u1≡u2.This completes the proof of Theorem5.


## 4. Stability of the Weak Solutions

In this section, let(u0,u1)∈H1(0,1)×L2(0,1) be fixed functions. Also, we assume that σ1,σ2∈L1(0,1),σ^∈L2(0,1),f^∈C0(R), and a,b, d,c,cM are fixed functions, constants satisfying assumptions (A3)–(A6) (independent of f, g, and h). Applying Theorem 5, then problem (1) has a unique weak solution u depending on σ,k,f,g,h. We denote(115)u=uσ,k,f,g,h,where σ,k,f,g,h satisfy assumptions (A2)–(A6).Then the stability of the solutions of problem (1) is given as follows.Theorem 13.
Let (A1)–(A6) hold. Then the solutions of problem (1) are stable with respect to the data (σ,k,f,g,h) in the following sense.
If(σj,kj,fj,gj,hj) and (σ,k,f,g,h) satisfy the assumptions (A2)–(A6) such that(116)σj,kj⟶σ,kstronglyinW1,10,T2,fj,gj,hj⟶f,g,hstrongly in C0([0,1]×[-M,M])×C1([-M,M])×C0([-M,M]N+1), as j→∞, for all M>0.
Then(117)uj,utj⟶u,utstronglyinL∞0,T;H10,1L∞0,T;L20,1,asj⟶∞,where uj=u(σj,kj,fj,gj,hj),u=u(σ,k,f,g,h).Proof of Theorem13.
Firstly, we assume that(118)σjW1,10,T+σW1,10,T≤σ∗,kjW1,10,T+kW1,10,T≤k∗,where σ∗,k∗ are fixed positive constants.
On the other hand, by the proof of Theorem5, the a priori estimates of the sequence um satisfy(119)utm·,t2+ρum·,tH2+∑i=0N∫0tutmξi,s2ds≤CT,∀t∈0,T,where CT is a constant depending only on u0,u1,σ1,σ2,σ^,f,f^,σ∗,k∗,a,b,d,c,ρ,T.
Due to (119) and (92), we conclude that(120)CT≥liminfm→∞⁡utm·,t2+ρum·,tH2+∑i=0N∫0tutmξi,s2ds≥liminfm→∞⁡utm·,t2+ρliminfm→∞⁡um·,tH2+∑i=0Nliminfm→∞⁡∫0tutmξi,s2ds≥ut·,t2+ρu·,tH2+∑i=1N∫0tutξi,s2ds.In addition, we can prove in a similar way above that the solution uj of problem (1) corresponding to the data (σj,kj,fj,gj,hj) also satisfies(121)utj·,t2+ρuj·,tH2+∑i=0N∫0tutjξi,s2ds≤C^T,∀t∈0,T,with C^T being a constant depending only on u0,u1,σ1,σ2,σ^,f,f^,σ∗,k∗,a,b,d,c,ρ,T.
We set(122)fj^=fj-f,gj^=gj-g,hj^=hj-h,kj^=kj-k,σj^=σj-σ,vj=uj-u.Then vj satisfies the following problem:(123)vttj-ρvxxj+Fj=0,0<x<1,0<t<T,vj1,t=0,vtj0,t=wjt,vjx,0=vtjx,0=0,where(124)Fjx,t=fj^x,uj+fx,uj-fx,u,wjt=σj^t+gj^ujξ0,t+gujξ0,t-guξ0,t+∫0tkjt-shj^ujξ0,s,…,ujξN,sds+∫0tkj^t-shujξ0,s,…,ujξN,sds+∫0tkt-shujξ0,s,…,ujξN,sds-∫0tkt-shuξ0,s,…,uξN,sds.Applying Lemma 12 with u0=u1=0, we see that(125)Ejt=-2∫0tf·,uj-f·,u,vtj·,sds-2∫0tfj^·,uj,vtj·,sds-2∫0tσj^svtj0,sds-2∫0tgj^ujξ0,svtj0,sds-2∫0tgujξ0,t-guξ0,tvtj0,sds-2∫0tvtj0,rdr∫0rkjr-shj^ujξ0,s,…,ujξN,sds-2∫0tvtj0,rdr∫0rkj^r-shujξ0,s,…,ujξN,sds-2∫0tvtj0,rdr∫0rkr-shujξ0,s,…,ujξN,sds-∫0tvtj0,rdr∫0rkr-shuξ0,s,…,uξN,sds≡K1t+K2t+⋯+K8t,where(126)Ejt=vtj·,t2+ρvj·,tH2.Let M=(CT+C^T)/ρ. Now we can estimate eight integrals in the right-hand side of (125) as follows.
Estimating
K
1
(
t
). From assumption (A3), it yields(127)K1t=-2∫0tf·,uj·,s-f·,u·,s,vtj·,sds≤dMρ∫0tEjsds,with dM=D2fC0([0,1]×[-M,M]).
Estimating
K
2
(
t
)
. It is easy to show that(128)K2t=-2∫0tfj^·,uj,vtj·,sds≤Tfj^C00,1×-M,M2+∫0tEjsds.EstimatingK3(t). Since ρvj2(0,t)≤Ej(t), by Cauchy-Schwartz inequality, then(129)K3t=-2∫0tσj^svtj0,sds≤εEjt+1ρσj^′L10,T+1ερσj^t2+∫0tσj^′sEjsds.Moreover, using the imbedding W1,1(0,T)↪C0([0,T]), there exists a positive constant kT such that(130)vC00,T≤kTvW1,10,T,∀v∈W1,10,T.Thus,(131)K3t≤εEjt+1ρσj^W1,10,T+kT2ερσj^W1,10,T2+∫0tσj^′sEjsds.EstimatingK4(t). By help of assumption (A4) and (121), we get(132)K4t=-2∫0tgj^ujξ0,svtj0,sds≤εEt+1ρ∫0tEjsds+1ερgj^C0-M,M2+gj^′C0-M,M2∫0tutjξ0,s2ds≤εEt+1ρ∫0tEjsds+C^T+1ερgj^C1-M,M2,∀ε>0.EstimatingK5(t). Proving in a similar way to (109), we also obtain(133)K5t=-2∫0tgujξ0,s-guξ0,svtj0,sds≤ερEjtgC2-M,M+∫0tdjsEsds,where(134)djs=gC2-M,Mερutjξ0,s+utξ0,s+1+1εTutjξ0,·L10,T+utξ0,·L10,T+1.On the other hand, we have(135)vL10,T≤TvL20,T,∀v∈L20,T.Therefore,(136)djs≤ερgC2-M,Mutjξ0,s+utξ0,s+T3/21+1ε1+CT+C^TgC2-M,M=d^js.It implies(137)K5t≤ερEjtgC2-M,M+∫0td^jsEsds.EstimatingK6(t). By reusing the inequalities (121) and (135) and assumption (A2), then(138)K6t=-2∫0tvtj0,rdr∫0rkjr-shj^ujξ0,s,…,ujξN,sds≤εEjt+2ρ∫0tEjsds+1+TkT2+1ερkjW1,10,T2hj^C0-M,MN+12≤εEjt+2ρ∫0tEjsds+k∗21+TkT2+1ερhj^C0-M,MN+12.EstimatingK7(t). Similarly, from assumption (A5), we also obtain(139)K7t=-2∫0tvtj0,rdr∫0rkj^r-shujξ0,s,…,ujξN,sds≤εEjt+2ρ∫0tEjsds+1+TkT2+1ερkj^W1,10,T2hC0-M,MN+12≡εEjt+2ρ∫0tEjsds+1+TkT2+1ερd^Mkj^W1,10,T2.EstimatingK8(t). Due to assumptions (A2) and (A6), we get(140)K8t=-2∫0tvtj0,rdr∫0rkr-shujξ0,s,…,ujξN,sds+2∫0tvtj0,rdr∫0rkr-shuξ0,s,…,uξN,sds≤εEjt+MT1∫0tEjsds,where(141)MT1=1ρN+12cM2k′L10,T2+1ερkL20,T2+2N+1cMk0+1.Thus, combining (125)–(128), (131), (132), and (137)–(140), (142)Ejt≤εMT2Ejt+MT3fj^C00,1×-M,M2+gj^C1-M,M2+hj^C0-M,MN+12+kj^W1,10,T2+σj^W1,10,T2+σj^W1,10,T+∫0tερgC2-M,Mutjξ0,s+utξ0,s+σj^′s+MT4Ejsds,in which(143)MT2=1ρgC2-M,M+5,MT3=k∗2+d^M1+TkT2+1ερ+1+kT2+εερ+T+C^T,MT4=T3/21+1ε1+CT+C^TgC2-M,M+dMρ+MT1+5ρ+1.Choosing εMT2=1/2 and applying the Gronwall inequality, we deduce from (120), (121), and (142) that(144)Ejt≤2MT3exp⁡2TMT4+2TερCT+C^TgC2-M,M+2σ∗fj^C00,1×-M,M2+gj^C1-M,M2+hj^C0-M,MN+12+kj^W1,10,T2+σj^W1,10,T2+σj^W1,10,T.This shows that(145)vtjL∞0,T;L20,12+vjL∞0,T;H10,12≤MTfj^C00,1×-M,M2+gj^C1-M,M2+hj^C0-M,MN+12+kj^W1,10,T2+σj^W1,10,T2+σj^W1,10,T,with MT being a constant depending only on u0,u1,σ1,σ2,σ^,f,f^,σ∗,k∗,a,b,d,c,ρ,T.
This completes the proof of Theorem13.Remark 14.
If we use the inequalityρvj2(ξi,t)≤Ej(t),i=0,N¯, then, with regard to (126) and (145), conclusion (117) in Theorem 13 can be extended as follows:(146)uj,utj,ujξ0,·,ujξ1,·,…,ujξN,·⟶u,ut,uξ0,·,uξ1,·,…,uξN,·,strongly in L∞(0,T;H1(0,1))×L∞(0,T;L2(0,1))×[C0([0,T])]N+1,  as  j→∞.



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*Source: 531872-2015-12-27.xml*