# Linear Approximation and Asymptotic Expansion of Solutions for a Nonlinear Carrier Wave Equation in an Annular Membrane with Robin-Dirichlet Conditions

Authors: Le Thi Phuong Ngoc; Le Huu Ky Son; Tran Minh Thuyet; Nguyen Thanh Long
Journal: Mathematical Problems in Engineering
 (2016)

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## Abstract

This paper is devoted to the study of a nonlinear Carrier wave equation in an annular membrane associated with Robin-Dirichlet conditions. Existence and uniqueness of a weak solution are proved by using the linearization method for nonlinear terms combined with the Faedo-Galerkin method and the weak compact method. Furthermore, an asymptotic expansion of a weak solution of high order in a small parameter is established.


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## Body

## 1. Introduction

In this paper, we consider the following nonlinear Carrier wave equation in the annular membrane:(1)utt-μut02uxx+1xux=fx,t,u,ux,ut,ρ<x<1,0<t<T,associated with Robin-Dirichlet conditions(2)uρ,t=ux1,t+ζu1,t=0,and initial conditions(3)ux,0=u~0x,utx,0=u~1x,where μ, f, u~0, u~1 are given functions; ρ, ζ are given constants,with 0<ρ<1. In (1), nonlinear term μ(u(t)02) depends on integral u(t)02=∫ρ1‍xu2x,tdx.Equation (1) herein is the bidimensional nonlinear wave equation describing nonlinear vibrations of annular membrane Ω1={(x,y):ρ2<x2+y2<1}. In the vibration processing, the area of the annular membrane and the tension at various points change in time. The condition on boundary Γ1={(x,y):x2+y2=1}, that is, ux(1,t)+ζu(1,t)=0, describes elastic constraints where ζ constant has a mechanical signification. And with the boundary condition on Γρ={(x,y):x2+y2=ρ2} requiring u(ρ,t)=0, the annular membrane is fixed.In [1], Carrier established the equation which models vibrations of an elastic string when changes in tension are not small:(4)ρutt-1+EALT0∫0Lu2y,tdyuxx=0,where u(x,t) is x-derivative of the deformation, T0 is the tension in the rest position, E is the Young modulus, A is the cross section of a string, L is the length of a string, and ρ is the density of a material. Clearly, if properties of a material vary with x and t, then there is a hyperbolic equation of the type [2]:(5)utt-Bx,t,∫01u2y,tdyuxx=0.The Kirchhoff-Carrier equations of form (1) received much attention. We refer the reader to, for example, Cavalcanti et al. [3, 4], Ebihara et al. [5], Miranda and Jutuca [6], Lasiecka and Ong [7], Hosoya and Yamada [8], Larkin [2], Medeiros [9], Menzala [10], Park et al. [11, 12], Rabello et al. [13], and Santos et al. [14], for many interesting results and further references.The paper consists of four sections. Preliminaries are done in Section2, with the notations, definitions, list of appropriate spaces, and required lemmas. The main results are presented in Sections 3 and 4.First, by combining the linearization method for nonlinear terms, the Faedo-Galerkin method, and the weak compact method, we prove that problem (1)–(3) has a unique weak solution.Next, by using Taylor’s expansion of given functionsμ, μ1,f, and f1 up to high order N+1, we establish an asymptotic expansion of solution u=uε of order N+1 in small parameter ε for(6)utt-μut02+εμ1ut02uxx+1xux=fx,t,u,ux,ut+εf1x,t,u,ux,ut,ρ<x<1, 0<t<T, associated with (1) and (2) with μ∈CN+1(R+), μ1∈CN(R+), μ(z)≥μ∗>0, μ1(z)≥0, for all z∈R+, f∈CN+1([ρ,1]×R+×R3), f1∈CN([ρ,1]×R+×R3). Our results can be regarded as an extension and improvement of the corresponding results of [15, 16].


## 2. Preliminaries

First, putΩ=(ρ,1), QT=Ω×(0,T), T>0. We omit the definitions of the usual function spaces and denote them by notations Lp=Lp(Ω), Hm=HmΩ. Let (·,·) be a scalar product in L2. Notation · stands for the norm in L2 and we denote ·X the norm in Banach space X. We call X′ the dual space of X. We denote Lp(0,T;X), 1≤p≤∞ the Banach space of real functions u:(0,T)→X to be measurable, such that uLp(0,T;X)<+∞, with(7)
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.Withf∈Ck([ρ,1]×R+×R3), f=f(x,t,y1,y2,y3), we put D1f=∂f/∂x, D2f=∂f/∂t, Di+2f=∂f/∂yi with i=1,…,3, and Dαf=D1α1⋯D5α5f, α=(α1,…,α5)∈Z+5, α=α1+⋯+α5=k, D(0,…,0)f=f.OnH1, H2, we shall use the following norms:(8)vH1=v2+vx21/2,(9)vH2=v2+vx2+vxx21/2,respectively.We remark thatL2, H1, H2 are the Hilbert spaces with respect to the corresponding scalar products:(10)u,v=∫ρ1xuxvxdx,u,v+ux,vx,u,v+ux,vx+uxx,vxx.The norms inL2, H1, and H2 induced by the corresponding scalar products (10) are denoted by ·0, ·1, and ·2.Consider the following set:(11)V=v∈H1:vρ=0.It is obviously thatV is a closed subspace of H1 and on V two norms vH1 and vx are equivalent norms. On the other hand, V is continuously and densely embedded in L2. Identifying L2 with L2′ (the dual of L2), we have V↪L2↪V′. We note more that the notation 〈·,·〉 is also used for the pairing between V and V′.We then have the following lemmas.Lemma 1.
The following inequalities are fulfilled:(i)
ρv≤v0≤v for all v∈L2.(ii)
ρvH1≤v1≤vH1 for all v∈H1.Proof of Lemma1.
It is easy to verify the above inequalities via the following inequalities:(12)ρ∫ρ1v2xdx≤∫ρ1xv2xdx≤∫ρ1v2xdx,∀v∈L2,ρ∫ρ1vx2xdx≤∫ρ1xvx2xdx≤∫ρ1vx2xdx,∀v∈H1.Lemma 2.
EmbeddingV↪C0Ω¯ is compact and for all v∈V, we have (i)
vC0Ω¯≤1-ρvx,(ii)
v≤(1-ρ/2)vx,(iii)
v0≤(1-ρ/2ρ)vx0,(iv)
vx02+v21≥v02,(v)
v1≤3v1.Proof of Lemma2.
EmbeddingV↪H1 is continuous and embedding H1↪C0Ω¯ is compact, so embedding V↪C0Ω¯ is compact. In what follows, we prove (i)–(v).(i)
For allv∈V and x∈ρ,1, (13)vx=∫ρxvxydy≤∫ρ1vxydy≤1-ρvx.(ii)
For allv∈Vand x∈ρ,1, (14)v2x=∫ρxvxydy2≤x-ρ∫ρxvx2ydy≤x-ρvx2.Integrating overx from ρ to 1, we obtain (15)v2=∫ρ1v2xdx≤∫ρ1x-ρvx2dx=1-ρ22vx2.(iii)
For allv∈V, (16)v0≤v≤1-ρ2vx≤1-ρ2ρvx0.(iv)
Using integration by part, it leads to(17)v02=∫ρ1xv2xdx=12x2v2xρ1-∫ρ1x2vxvxxdx=12v21-∫ρ1x2vxvxxdx≤12v21+v0vx0≤12v21+12v02+vx02,for anyv∈V, so we get (iv).(v)
Byv02=(1/2)v21-∫ρ1‍x2vxvxxdx, we have (18)v21=2v02+2∫ρ1x2vxvxxdx≤2v02+2v0vx0≤2v02+v02+vx02≤3v12,implying (v).
Lemma2 is proved.Remark 3.
OnL2, two norms v↦v and v↦v0 are equivalent. So are two norms v↦vH1 and v↦v1 on H1, and five norms v↦vH1, v↦v1, v↦vx, v↦vx0, and v↦vx02+v21 on V.
Now, we define the following bilinear form:(19)au,v=ζu1v1+∫ρ1xuxxvxxdx,∀u,v∈V,where ζ≥0 is a constant.Lemma 4.
Symmetric bilinear forma·,· defined by (19) is continuous on V×V and coercive on V, that is, (i)
au,v≤C1u1v1,(ii)
av,v≥C0v12,for all u, v∈V, where C0=1/2min⁡{1,2ρ/1-ρ2} and C1=1+3ζ.Proof of Lemma4.
(i) By1-ρvx≥vC0Ω¯≥v1, and ρvx≤vx0 for all v∈V, we have (20)au,v≤ζu1v1+∫ρ1xuxxvxxdx≤3ζu1v1+ux0vx0≤3ζ+1u1v1.
(ii) By inequalityvx02≥(2ρ/1-ρ2)v02, we get (21)av,v=ζv21+∫ρ1xvx2xdx=ζv21+vx02≥vx02=12vx02+12vx02≥12vx02+122ρ1-ρ2v02≥12min⁡1,2ρ1-ρ2v12.
Lemma4 is proved.Lemma 5.
There exists Hilbert orthonormal basewj of space L2 consisting of eigenfunctions wj corresponding to eigenvalues λj such that (i)
0<λ1≤λ2≤⋯≤λj≤λj+1≤⋯, limj→+∞⁡λj=+∞,(ii)
awj,v=λjwj,v for all v∈V, j=1,2,….
Furthermore, sequence{wj/λj} is also the Hilbert orthonormal base of V with respect to scalar product a·,·.
On the other hand, we also havewj satisfying the following boundary value problem:(22)Awj≡-wjxx+1xwjx=-1x∂∂xxwjx=λjwj,inΩ,wjρ=wjx1+ζwj1=0,wj∈C∞ρ,1.Proof.
The proof of Lemma5 can be found in [17, p. 87, Theorem 7.7], with H=L2, and a(·,·) as defined by (19).We also note that operatorA:V→V′ in (22) is uniquely defined by Lax-Milgram’s lemma; that is,(23)au,v=Au,v∀u,v∈V.Lemma 6.
OnV∩H2, three norms v↦vH2, v↦v2=v02+vx02+vxx02, and v↦v2∗=vx02+Av02 are equivalent.Proof of Lemma6.
(i) It is easy to see that, onV∩H2, two norms v↦vH2, v↦v2=v02+vx02+vxx02 are equivalent, because(24)ρvH2≤v2≤vH2∀v∈H2.
(ii) For allx∈ρ,1, and v∈V∩H2, we have(25)xAux2=x1x2∂∂xxux2=xuxx2+2uxuxx+1xux2.
(a) Proofu2≤constu2∗.
It follows from (25) that (26)xuxx2≤xAux2+2uxuxx+1xux2.
Hence,(27)uxx02≤Au02+2ρux0uxx0+1ρ2ux02≤Au02+1ρ2ρux02+ρ2uxx02+1ρ2ux02=Au02+2ρ2ux02+12uxx02+1ρ2ux02.
This implies(28)uxx02≤2Au02+6ρ2ux02≤21+3ρ2Au02+ux02≤21+3ρ2u2∗2.
Byv0≤(1-ρ/2ρ)vx0, for all v∈V, we have (29)u22=u02+ux02+uxx02≤1-ρ22ρux02+ux02+uxx02≤1+1-ρ22ρu2∗2+21+3ρ2u2∗2=1-ρ22ρ+3+6ρ2u2∗2.
(b) Proofu2∗≤constu2.
It follows from (25) that (30)xAux2=x1x2∂∂xxux2=xuxx2+2uxuxx+1xux2.
Hence,(31)xAux2≤xuxx2+2uxuxx+1xux2.
Thus,(32)Au02≤uxx02+2ρux0uxx0+1ρ2ux02≤uxx02+1ρux02+uxx02+1ρ2ux02=1+1ρuxx02+1ρux02≤1+1ρ1ρuxx02+ux02≤1+1ρ1ρu22.
This implies(33)u2∗2=ux02+Au02≤u22+1+1ρ1ρu22=1+1ρ+1ρ2u22.
Lemma6 is proved.Remark 7.
The weak formulation of initial-boundary value problem (1)–(3) can be given in the following manner: find u∈W¯={u∈L∞(0,T;V∩H2):ut∈L∞(0,T;V), utt∈L∞(0,T;L2)}, such that u satisfies the following variational equation:(34)uttt,v+μut02aut,v=fx,t,u,ux,ut,v,for all v∈V, a.e., t∈(0,T), together with the initial conditions:(35)u0=u~0,ut0=u~1,where a(·,·) is the symmetric bilinear form on V defined by (19).


## 3. The Existence and Uniqueness Theorem

Now, we shall consider problem (1)–(3) with constant ζ≥0 and make the following assumptions:(H1)
u~0∈V∩H2, u~1∈V;(H2)
μ∈C1R+, with μ(z)≥μ∗>0,∀z∈R+;(H3)
f∈C0Ω¯×R+×R3 such that f(ρ,t,0,y2,0)=0,∀(t,y2)∈R+×R and Dif∈C0Ω¯×R+×R3,i=1,3,4,5.ConsideringT∗>0 fixed and letting T∈(0,T∗] and M>0, we put (36)K~Mμ=sup0≤z≤M2⁡μz+μ′z,KMf=fC0A∗M+D1fC0A∗M+∑3≤i≤5DifC0A∗M,fC0A∗M=sup⁡fx,t,y1,y2,y3:x,t,y1,y2,y3∈A∗M,where (37)A∗M=x,t,y1,y2,y3∈Ω¯×0,T∗×R3:yi≤1-ρρM,i=1,2,3.Also for eachM>0 and T∈(0,T∗], we set (38)WM,T=u∈L∞0,T;V∩H2:ut∈L∞0,T;V,utt∈L2QT,uL∞0,T;V∩H2≤M,utL∞0,T;V≤M,uttL2QT≤M,W1M,T=u∈WM,T:utt∈L∞0,T;L2.We choose first termu0≡0, suppose that(39)um-1∈W1M,T,and associate the following variational problem with problem (1)–(3): find um∈W1M,Tm≥1, so that(40)u¨mt,v+μmtaumt,v=Fmt,v,∀v∈V,um0=u~0,u˙m0=u~1,where(41)μmt=μum-1t02,Fmt=fx,t,um-1t,∇um-1t,u˙m-1t.Then, we have the following result.Theorem 8.
Let assumptionsH1–H3 hold. Then, there exist positive constants M, T such that the problem (40), (41) has solution um∈W1(M,T).Proof of Theorem8.
It consists of three steps.
Step 1 (the Faedo-Galerkin approximation (introduced by Lions [18])). Consider basis {wj} for V as in Lemma 5. Put(42)umkt=∑j=1kcmjktwj,where coefficients cmj(k) satisfy the system of linear differential equations:(43)u¨mkt,wj+μmtaumkt,wj=Fmt,wj,j=1,…,k,umk0=u0k,u˙mk0=u1k,with(44)u0k=∑j=1kαjkwj⟶u~0stronglyinV∩H2,u1k=∑j=1kβjkwj⟶u~1stronglyinV.
The system of (43) can be rewritten in form(45)c¨mjkt+λjμmtcmjkt=Fmjt,1≤j≤k,cmjk0=αjk,c˙mjk0=βjk,in which(46)Fmjt=Fmt,wj,1≤j≤k.
Note that, by (39), it is not difficult to prove that system (45), (46) has a unique solution cmjkt,1≤j≤k on interval [0,T], so let us omit the details.
Step 2 (a priori estimates). We put(47)Smkt=Xmkt+Ymkt+∫0tu¨mks02ds,where(48)Xmkt=u˙mkt02+μmtaumkt,umkt,Ymkt=au˙mkt,u˙mkt+μmtAumkt02.
Then, it follows from (43), (47), and (48) that(49)Smkt=Smk0+∫0tμm′saumks,umks+Aumks02ds+2∫0tFms,u˙mksds+2∫0taFms,u˙mksds+∫0tu¨mks02ds≡Smk0+∑j=14Ij.
We shall estimate termsIj on the right-hand side of (49) as follows.
First Term I 1 . By the following inequalities,(50)μm′t=2μ′um-1t02um-1t,um-1′t≤2μ′um-1t02um-1t0um-1′t0≤2M2K~Mμ,Smkt≥μmtaumkt,umkt+Aumkt02≥μ∗aumkt,umkt+Aumkt02,we have(51)I1=∫0tμm′saumks,umks+Aumks02ds≤2μ∗M2K~Mμ∫0tSmksds.
Second Term I 2 . By the Cauchy-Schwartz inequality, it gives(52)I2=2∫0tFms,u˙mksds≤2∫0tFms0u˙mks0ds≤21-ρ22KMf∫0tu˙mks0ds≤1-ρ22TKM2f+∫0tSmksds.
Third Term I 3 . Similarly, we have(53)I3=2∫0taFms,u˙mksds≤2∫0taFms,Fmsau˙mks,u˙mksds≤2∫0taFms,FmsSmksds≤∫0taFms,Fmsds+∫0tSmksds.
Note that(54)av,v≤C1v12=C1v02+vx02≤C11-ρ22ρvx02+vx02=C11+ρ22ρvx02∀v∈V,so(55)aFms,Fms≤C11+ρ22ρFmxs02.
We also have(56)Fmxt=D1fum-1+D3fum-1∇um-1t+D4fum-1Δum-1t+D5fum-1∇u˙m-1t,where Dif[um-1]=Difx,t,um-1(t),∇um-1(t),u˙m-1(t),i = 1,…,5.
It implies from (39), (56) that(57)Fmxt0≤KMf1-ρ22+∇um-1t0+Δum-1t0+∇u˙m-1t0≤1-ρ22+3MKMf.
Combining (53), (55), and (57), we obtain(58)I3≤C11+ρ22ρ∫0tFmxs02ds+∫0tSmksds≤C11+ρ22ρ1-ρ22+3M2TKM2f+∫0tSmksds.
Fourth Term I 4 . Equation (43)1 can be rewritten as follows:(59)u¨mkt,wj+μmtAumkt,wj=Fmt,wj,j=1,…,k.
Hence, it follows after replacingwj with u¨m(k)(t), that(60)u¨mkt02=-μmtAumkt,u¨mkt+Fmt,u¨mkt≤μmtAumkt0+Fmt0u¨mks0≤μmtAumkt0+Fmt02≤2μm2tAumkt02+2Fmt02≤2μmtSmkt+2Fmt02≤2K~MμSmkt+21-ρ22KM2f.
Integrate intot to get(61)I4=∫0tu¨mks02ds≤1-ρ2TKM2f+2K~Mμ∫0tSmksds.
It follows from (49), (51), (52), (58), and (61) that(62)Smkt≤Smk0+TD1M+D2M∫0tSmksds,where(63)D1M=321-ρ2+C11+ρ22ρ1-ρ22+3M2KM2f,D2M=21+1+1μ∗M2K~Mμ.
By means of the convergences in (44), we can deduce the existence of constant M>0 independent of k and m such that(64)Smk0=u1k02+au1k,u1k+μu~002au0k,u0k+Au0k02≤12M2,for all m, k∈N.
Therefore, from (63) and (64), we can choose T∈(0,T∗], such that(65)12M2+TD1Mexp⁡TD2M≤M2,(66)kT=1+1μ∗C0Texp⁡T1+1μ∗M2K~Mμ4M4K~M2μ+KM2f1+1-ρ2ρ21/2<1.
Finally, it follows from (62), (64), and (65) that(67)Smkt≤M2exp⁡-TD2M+D2M∫0tSmksds.
By using Gronwall’s Lemma, (67) yields(68)Smkt≤M2exp⁡-TD2Mexp⁡tD2M≤M2,for all t∈[0,T], for all m and k. Therefore, we have(69)umk∈WM,T,∀m,k.
Step 3 (limiting process). From (69), there exists a subsequence of {um(k)}, still so denoted, such that(70)umk⟶uminL∞0,T;V∩H2weakly∗,u˙mk⟶um′inL∞0,T;Vweakly∗,u¨mk⟶um′′inL2QTweakly,um∈WM,T.
Passing to limit in (43), we have um satisfying (40), (41) in L2(0,T). On the other hand, it follows from (40)1 and (70)4 that um′′=-μmtAum+Fm∈L∞(0,T;L2), and hence um∈W1(M,T) and the proof of Theorem 8 is complete.We will use the result obtained in Theorem8 and the compact imbedding theorems to prove the existence and uniqueness of a weak solution of problem (1)–(3). Hence, we get the main result in this section.Theorem 9.
LetH1–H3 hold. Then, there exist positive constants M, T satisfying (64)–(66) such that problem (1)–(3) has unique weak solution u∈W1(M,T). Furthermore, the linear recurrent sequence {um} defined by (40), (41) converges to solution u strongly in space W1(T)={v∈L∞(0,T;V):v′∈L∞(0,T;L2)}, with estimate(71)um-uW1T≤M1-kTkTm,∀m∈N.Proof of Theorem9.
(a) The Existence. First, we note that W1(T) is a Banach space with respect to norm vW1(T)=vL∞(0,T;V)+v′L∞(0,T;L2) (see Lions [18]).
We shall prove that{um} is a Cauchy sequence in W1(T). Let wm=um+1-um. Then, wm satisfies the variational problem:(72)wm′′t,v+μm+1tawmt,v+μm+1t-μmtAumt,v=Fm+1t-Fmt,v,∀v∈V,wm0=wm′0=0.
Takingv=wm′ in (72)1, after integrating into t, we get(73)Zmt=∫0tμm+1′sawms,wmsds-2∫0tμm+1s-μmsAums,wm′sds+2∫0tFm+1s-Fms,wm′sds≡J1+J2+J3,where(74)Zmt=wm′t02+μm+1tawmt,wmt≥wm′t02+μ∗awmt,wmt≥wm′t02+μ∗C0wmt12.All integrals on the right-hand side of (73) will be estimated as below.
First Integral J 1 . By (50) and (74), we have(75)J1≤∫0tμm+1′sawms,wmsds≤1μ∗2M2K~Mμ∫0tZmsds.
Second Integral J 2 . By (H2), it is clear to see that(76)μm+1t-μmt=μumt02-μum-1t02≤K~Mμumt02-um-1t02≤2MK~Mμwm-1t0≤2MK~Mμwm-1W1T.
Hence,(77)J2≤2∫0tμm+1s-μmsAums,wm′sds≤4MK~Mμwm-1W1T∫0tAums0wm′s0ds≤4M2K~Mμwm-1W1T∫0twm′s0ds≤4TM4K~M2μwm-1W1T2+∫0tZmsds.
Second Integral J 3 . By (H3), it yields(78)Fm+1t-Fmt0≤KMfwm-1t0+∇wm-1t0+wm-1′t0≤KMf1-ρ2ρ∇wm-1t0+∇wm-1t0+wm-1′t0≤KMf1+1-ρ2ρ∇wm-1t0+wm-1′t0≤KMf1+1-ρ2ρwm-1W1T.
Hence,(79)J3≤2∫0tFm+1s-Fms,wm′sds≤2∫0tFm+1s-Fms0wm′s0ds≤2KMf1+1-ρ2ρwm-1W1T∫0twm′s0ds≤TKM2f1+1-ρ2ρ2wm-1W1T2+∫0tZmsds.
Combining (73), (75), (77), and (79), we obtain(80)Zmt≤T4M4K~M2μ+KM2f1+1-ρ2ρ2wm-1W1T2+21+1μ∗M2K~Mμ∫0tZmsds.
Using Gronwall’s lemma, we deduce from (80) that(81)wmW1T≤kTwm-1W1T∀m∈N,which implies that(82)um-um+pW1T≤u0-u1W1T1-kT-1kTm≤M1-kTkTm∀m,p∈N.
It follows that{um} is a Cauchy sequence in W1(T). Then, there exists u∈W1(T) such that(83)um⟶ustronglyinW1T.
Note thatum∈W1(M,T), and then there exists subsequence {umj} of {um} such that(84)umj⟶uinL∞0,T;V∩H2weakly∗,umj′⟶u′inL∞0,T;Vweakly∗,umj′′⟶u′′inL2QTweakly,u∈WM,T.
We also note that(85)Fmt-fx,t,u,ux,utL∞0,T;L2≤KMf1+1-ρ2ρum-1-uW1T.
Hence, from (83) and (85), we obtain(86)Fmt⟶fx,t,u,ux,utstronglyinL∞0,T;L2.
On the other hand, we have(87)μmt-μut02≤2MK~Mμwm-1t0≤2MK~Mμum-1-uW1T.
Hence, it follows from (83) and (87) that(88)μmt⟶μut02stronglyinL∞0,T.
Finally, passing to limit in (40), (41) as m=mj→∞, it implies from (83), (84)1,3, (86), and (88) that there exists u∈W(M,T) satisfying(89)uttt,v+μut02aut,v=fx,t,u,ux,ut,v,for all v∈V and the initial conditions(90)u0=u~0,u′0=u~1.
Furthermore, from assumptions(H2), (H3) we obtain from (84)4, (86), (88), and (89), that(91)u′′=-μut02Aut+fx,t,u,ux,ut∈L∞0,T;L2,and thus we have u∈W1(M,T). The existence of a weak solution of problem (1)–(3) is proved.
(b) The Uniqueness. Let u1,u2∈W1(M,T) be two weak solutions of problem (1)–(3). Then, u=u1-u2 satisfies the variational problem:(92)u′′t,v+μ1taut,v+μ1t-μ2tAu2t,v=F1t-F2t,v,∀v∈V,u0=u′0=0,where Fi(x,t)=fx,t,ui,∇ui,ui′, μit=μui(t)02,i=1,2.
We takew=u′ in (92)1 and integrate in t to get(93)Zt=∫0tμ1′saus,usds+2∫0tF1s-F2s,u′sds-2∫0tμ1s-μ2sAu1s,u′sds,with Z(t)=u′t02+μ1taut,ut.
PuttingKM∗=2KMf1+(1-ρ)/2ρ1+1/μ∗+1/μ∗+2(1-ρ)/2μ∗ρM2K~Mμ, it follows from (93) that (94)Zt≤KM∗∫0tZsds,∀t∈0,T.
Using Gronwall’s lemma, it follows thatZ(t)≡0, that is, u1≡u2.
Therefore, Theorem9 is proved.


## 4. Asymptotic Expansion of the Solution with respect to a Small Parameter

In this section, let(H1)–(H4) hold. We make more the following assumptions:(H2′)
μ1∈C1R+, with μ1(z)≥0,∀z∈R+.(H3′)
f1∈C0Ω¯×R+×R3 such that f1ρ,t,0,y2,0=0,∀t,y2∈R+×R and Dif1∈C0Ω¯×R+×R3,i=1,3,4,5.Considering the following perturbed problem, whereε is a small parameter and ε≤1: Pεutt+μεuAu=Fεux,t,ρ<x<1,0<t<T,uρ,t=ux1,t+ζu1,t=0,ux,0=u~0x,utx,0=u~1x,with (95)Au=-1x∂∂xxux=-uxx+1xux,μεu=μεut02=μut02+εμ1ut02,Fεux,t=fux,t+εf1ux,t,fux,t=fx,t,u,ut,ux,f1ux,t=f1x,t,u,ut,ux.First, we note that if functionsμ,μ1,f,f1 satisfy (H2), (H2′), (H3), (H3′), then a priori estimates of the Galerkin approximation sequence {um(k)} for problem (1)–(3) corresponding to μ=με, f=Fε[u], ε≤1, leads to um(k)∈W1(M,T), where constants M, T, independent of ε, are chosen as in (63)–(66), in which μ, K~Mμ, KMf are replaced with μ+μ1, K~M(μ)+K~M(μ1), KM(f)+KM(f1), respectively. Hence, limit uε in suitable function spaces of sequence {um(k)} as k→+∞, after m→+∞, is a unique weak solution of problem Pε satisfying uε∈W1(M,T).We can prove in a manner similar to the proof of Theorem9 that limit u0 in suitable function spaces of family {uε} as ε→0 is a unique weak solution of problem P0 (corresponding to ε=0) satisfying u0∈W1(M,T).Next, we shall study the asymptotic expansion of solutionuε with respect to a small parameter ε. For multi-index α=(α1,…,αN)∈Z+N, and x=(x1,…,xN)∈RN, we put (96)α=α1+⋯+αN,α!=α1!⋯αN!,α,β∈Z+N,α≤β⟺αi≤βi∀i=1,…,N,xα=x1α1⋯xNαN.We need the following lemma.Lemma 10.
Letm,N∈N and x=(x1,…,xN)∈RN, ε∈R. Then,(97)∑i=1Nxiεim=∑k=mmNPkmN,xεk,where coefficients Pk(m)[N,x], m≤k≤mN, depending on x=(x1,…,xN), are defined by the following formulas:(98)Pk1N,x=xk,1≤k≤N,PkmN,x=∑α∈AkmNm!α!xα,m≤k≤mN,m≥2,AkmN=α∈Z+N:α=m,∑i=1Niαi=k.Proof.
The proof of Lemma10 is easy; hence, we omit the details.Now, we assume that(H2(N))
μ∈CN+1(R+),μ1∈CN(R+), with μ(z)≥μ∗>0,μ1(z)≥0,∀z∈R+;(H3(N))
f∈CN+1(Ω¯×R+×R3),f1∈CN(Ω¯×R+×R3) such that f(ρ,t,0,y2,0)=f1(ρ,t,0,y2,0)=0,∀(t,y2)∈R+×R.Letu0 be a unique weak solution of problem P0; that is, P0u0′′+μu0Au0=fu0≡F0,ρ<x<1,0<t<T,u0ρ,t=u0x1,t+ζu01,t=0,u0x,0=u~0x,u0′x,0=u~1x,u0∈W1M,T.Let us consider the sequence of weak solutionsuk, 1≤k≤N, defined by the following problems: P~kuk′′+μu0Auk=Fk,ρ<x<1,0<t<T,ukρ,t=ukx1,t+ζuk1,t=0,ukx,0=uk′x,0=0,uk∈W1M,T,where Fk, 1≤k≤N, are defined by the following formulas:(99a)Fk=f1u0+Φ¯1N,f,u0,u→-μ1u0+Φ^1N,μ,u0,u→Au0,k=1,(99b)Fk=Φ¯kN,f,u0,u→+Φ¯k-1N-1,f1,u0,u→-μ1u0+Φ^1N,μ,u0,u→Auk-1-∑i+j=k,1≤i≤N+1,1≤j≤N-1Φ^j+1N,μ,u0,u→+Φ^jN-1,μ1,u0,u→Aui-1,2≤k≤N,with Φ¯k[N,f,u0,u→], Φ^k[N,μ,u0,u→], 1≤k≤N, are defined by the following formulas:(100)Φ¯kN,f,u0,u→=∑1≤γ≤k1γ!Dγfu0Ψkγ,N,u→,1≤k≤N,Ψkγ,N,u→=∑k1,k2,k3∈A~γ,N,k1+k2+k3=kPk1γ1N,u→Pk2γ2N,u→′Pk3γ3N,∇u→,1≤k≤Nγ,u→=u1,…,uN,u→′=u1′,…,uN′,∇u→=∇u1,…,∇uN,A~γ,N=k1,k2,k3∈Z+3:γi≤ki≤Nγi,∀i=1,2,3,γ=γ1,γ2,γ3∈Z+3,1≤γ≤N,(101)Φ^kN,μ,u0,u→=∑m=1k1m!μmu0Pkm2N,σ→,1≤k≤N,σ→=σ1,…,σ2N,σk=2u0t,u1t,k=1,2u0t,ukt+∑j≤kujt,uk-jt,2≤k≤N,∑j≤kujt,uk-jt,N+1≤k≤2N.Then, we have the following theorem.Theorem 11.
Let(H1), (H2(N)), and (H3(N)) hold. Then, there exist constants M>0 and T>0 such that, for every ε∈[-1,1], problem Pε has unique weak solution uε∈W1(M,T) satisfying the asymptotic estimation up to order N+1 as follows:(102)uε-∑k=0NukεkW1T≤CTεN+1,where functions uk,0≤k≤N are the weak solutions of problems P0, P~k, 1≤k≤N, respectively, and CT is a constant depending only on N, T, ρ, ζ, f, f1, μ, μ1, uk,0≤k≤N.In order to prove Theorem11, we need the following lemmas.Lemma 12.
LetΦ¯k[N,f,u0,u→],1≤k≤N, be the functions defined by the formulas (100). Put h=∑k=0Nukεk, then we have(103)fh=fu0+∑k=1NΦ¯kN,f,u0,u→εk+εN+1R¯Nf,u0,u→,ε,with R¯N[f,u0,u→,ε]L∞(0,T;L2)≤C, where C is a constant depending only on N, T, f, uk, 0≤k≤N.Proof of Lemma12.
In the case ofN=1, the proof of (103) is easy; hence, we omit the details, and we only prove with N≥2. Put h=u0+∑k=1Nukεk≡u0+h1. By using Taylor’s expansion of function f[h]=f[u0+h1]=f(x,t,u0+h1,u0′+h1′,∇u0+∇h1) around point [u0]≡(x,t,u0,u0′,∇u0) up to order N+1, we obtain(104)fh=fx,t,u0+h1,u0′+h1′,∇u0+∇h1=fu0+∑1≤γ≤N1γ!Dγfu0h1γ1h1′γ2∇h1γ3+RNf,u0,h1,where(105)RNf,u0,h1=∑γ=N+1N+1γ!∫011-θNDγfx,t,u0+θh1,u0′+θh1′,∇u0+θ∇h1h1γ1h1′γ2∇h1γ3dθ≡εN+1RN1f,u0,h1,ε,γ=(γ1,γ2,γ3)∈Z+3, |γ|=γ1+γ2+γ3, γ!=γ1!γ2!γ3!, Dγf=D3γ1D4γ2D5γ3f, Dγf[u0]=Dγfx,t,u0,u0′,∇u0.
By formula (97), we get(106)h1γ1=∑k=1Nukεkγ1=∑k=γ1Nγ1Pkγ1N,u→εk,where u→=(u1,…,uN).
Similarly, with(h1′)γ2, ∇h1γ3, we also have(107)h1′γ2=∑k=1Nuk′εkγ2=∑k=γ2Nγ2Pkγ2N,u→′εk,∇h1γ3=∑k=1N∇ukεkγ3=∑k=γ3Nγ3Pkγ3N,∇u→εk,where u→′=(u1′,…,uN′), ∇u→=(∇u1,…,∇uN).
Hence, we deduce from (106)-(107) that(108)h1γ1h1′γ2∇h1γ3=∑k=γNΨkγ,N,u→εk+∑k=N+1NγΨkγ,N,u→εk,where Ψk[γ,N,u→],1≤k≤Nγ, are defined by (100).
We deduce from (104), (108) that(109)fh=fu0+∑1≤γ≤N1γ!Dγfu0h1γ1h1′γ2∇h1γ3+εN+1RN1f,u0,h1,ε=fu0+∑1≤γ≤N1γ!Dγfu0∑k=γNΨkγ,N,u→εk+∑1≤γ≤N1γ!Dγfu0∑k=N+1NγΨkγ,N,u→εk+εN+1RN1f,u0,h1,ε=fu0+∑k=1N∑1≤γ≤k1γ!Dγfu0Ψkγ,N,u→εk+εN+1R¯Nf,u0,u→,ε=fu0+∑k=1NΦ¯kN,f,u0,u→εk+εN+1R¯Nf,u0,u→,ε,where Φ¯k[N,f,u0,u→],1≤k≤N, are defined by (100) and(110)εN+1R¯Nf,u0,u→,ε=∑1≤γ≤N1γ!Dγfu0∑k=N+1NγΨkγ,N,u→εk+εN+1RN1f,u0,h1,ε.
By the boundedness of functionsuk, uk′, ∇uk,1≤k≤N in the function space L∞(0,T;H1), we obtain from (100), (105), (110) that R¯N[f,u0,u→,ε]L∞(0,T;L2)≤C, where C is a constant depending only on N, T, f, uk,0≤k≤N. Thus, Lemma 12 is proved.Lemma 13.
LetΦ^k[N,μ,u0,u→],1≤k≤N, be the functions defined by formulas (101). Put h=∑k=0Nukεk, and then we have(111)μh=μu0+∑k=1NΦ^kN,μ,u0,u→εk+εN+1R^Nμ,u0,u→,ε,with R^N[μ,u0,u→,ε]L∞(0,T;L2)≤C, where C is a constant depending only on N, T, μ, uk,0≤k≤N.Proof of Lemma13.
In the case ofN=1, the proof of (111) is easy; hence we omit the details, and we only prove with N≥2.
Putξ=ht02-u0t02. By using Taylor’s expansion of function μ[h]=μht02=μu0t02+ξ around point u0t02 up to order N+1, we obtain(112)μh=μht02=μu0t02+ξ=μu0t02+∑m=1N1m!μmu0ξm+RNμ,u0,ξ,where(113)RNμ,u0,ξ=1N!∫011-θNμN+1u0t02+θξξN+1dθ=εN+1RN1μ,u0,ξ,ε,μmu0=dmμdzmu0=dmμdzmu0t02.
On the other hand, we also get(114)ξ=u0t+h1t02-u0t02=2u0t,h1t+h1t02=∑1≤k≤2Nσkεk,where σk,1≤k≤2N, are defined by (101).
Using formula (97) again, it follows from (114) that(115)ξm=∑1≤k≤2Nσkεkm=∑k=m2mNPkm2N,σ→εk=∑k=mNPkm2N,σ→εk+∑k=N+12mNPkm2N,σ→εk,where σ→=(σ1,…,σ2N).
We deduce from (112), (115) that(116)μh=μu0+∑m=1N1m!μmu0ξm+εN+1RN1μ,u0,ξ,ε=μu0+∑m=1N1m!μmu0∑k=mNPkm2N,σ→εk+∑m=1N1m!μmu0∑k=N+12mNPkm2N,σ→εk+εN+1RN1μ,u0,ξ,ε=μu0+∑k=1N∑m=1k1m!μmu0Pkm2N,σ→εk+εN+1R^Nμ,u0,u→,ε=μu0+∑k=1NΦ^kN,μ,u0,u→εk+εN+1R^Nμ,u0,u→,ε,where Φ^k[N,μ,u0,u→], 1≤k≤N, are defined by (101) and(117)εN+1R^Nμ,u0,u→,ε=∑m=1N1m!μmu0∑k=N+12mNPkm2N,σ→εk+εN+1RN1μ,u0,ξ,ε.
By the boundedness of functionsuk, uk′, ∇uk,1≤k≤N in function space L∞(0,T;H1), we obtain from (101), (113), (117) that R^N[μ,u0,u→,ε]L∞(0,T;L2)≤C, where C is a constant depending only on N, T, μ, uk,0≤k≤N. Thus, Lemma 13 is proved.Remark 14.
Lemmas12 and 13 are a generalization of the formula contained in [19, p. 262, formula (4.38)] and it is useful to obtain Lemma 15 below. These lemmas are the key to establish the asymptotic expansion of weak solution uε of order N+1 in small parameter ε.
Letu=uε∈W1(M,T) be the unique weak solution of problem (Pε). Then, v=uε-∑k=0Nukεk≡uε-h satisfies the problem:(118)v′′+μεv+hAv=Fεv+h-Fεh-μεv+h-μεhAh+Eεx,t,ρ<x<1,0<t<T,vρ,t=vx1,t+ζv1,t=0,vx,0=v′x,0=0,where(119)Eεx,t=fh-fu0+εf1h-μh-μu0+εμ1hAh-∑k=1NFkεk.Lemma 15.
Let(H1), (H2(N)), and (H3(N)) hold. Then, there exists constant C∗ such that(120)EεL∞0,T;L2≤C∗εN+1,where C∗ is a constant depending only on N, T, f, f1, μ, μ1, uk,0≤k≤N.Proof of Lemma15.
We only prove withN≥2.
By using formula (103) for function f1[h], we obtain(121)f1h=f1u0+∑k=1N-1Φ¯kN-1,f1,u0,u→εk+εNR¯N-1f1,u0,u→,ε,where R¯N-1[f1,u0,u→,ε]L∞(0,T;L2)≤C, with constant C depending only on N, T, f1, g1, uk, 0≤k≤N.
By (121), we rewrite εf1[h] as follows(122)εf1h=εf1u0+∑k=2NΦ¯k-1N-1,f1,u0,u→εk+εεNR¯N-1f1,u0,u→,ε.
Hence, we deduce from (103) and (122) that(123)fh-fu0+εf1h=f1u0+Φ¯1N,f,u0,u→ε+∑k=2NΦ¯kN,f,u0,u→+Φ¯k-1N-1,f1,u0,u→εk+εN+1R¯Nf,f1,u0,u→,ε,where R¯N[f,f1,u0,u→,ε]=R¯N[f,u0,u→,ε]+(ε/ε)R¯N-1[f1,u0,u→,ε] is bounded in function space L∞(0,T;L2) by a constant depending only on N, T, f, f1, uk, 0≤k≤N.
On the other hand, we putη1=μ1[u0]+Φ^1[N,μ,u0,u→], ηk=Φ^k[N,μ,u0,u→]+Φ^k-1[N-1,μ1,u0,u→], 2≤k≤N, and we deduce from (111) that (124)-μh-μu0+εμ1hAh=-Ahμ1u0+Φ^1N,μ,u0,u→ε+∑k=2NΦ^kN,μ,u0,u→+Φ^k-1N-1,μ1,u0,u→εk-AhεN+1R^Nμ,u0,u→,ε+εεNR^N-1μ1,u0,u→,ε=-Ahμ1u0+Φ^1N,μ,u0,u→ε+∑k=2NΦ^kN,μ,u0,u→+Φ^k-1N-1,μ1,u0,u→εk+εN+1R~Nμ,μ1,u0,u→,ε≡-∑i=0NAuiεiη1ε+∑j=2Nηjεj+εN+1R~Nμ,μ1,u0,u→,ε=-∑i=0NAuiεiη1ε-∑i=0NAuiεi+1∑j=2Nηjεj-1+εN+1R~Nμ,μ1,u0,u→,ε=-∑i=1NAui-1εiη1-η1AuNεN+1-∑i=1N+1Aui-1εi∑j=1N-1ηj+1εj+εN+1R~Nμ,μ1,u0,u→,ε=-∑k=1NAuk-1εkη1-∑i=1N+1∑j=1N-1ηj+1Aui-1εi+j-η1AuNεN+1+εN+1R~Nμ,μ1,u0,u→,ε=-η1Au0ε-∑k=2NAuk-1εkη1-∑k=22N∑i+j=k,1≤i≤N+1,1≤j≤N-1ηj+1Aui-1εk-η1AuNεN+1+εN+1R~Nμ,μ1,u0,u→,ε=-η1Au0ε-∑k=2Nη1Auk-1εk-∑k=2N∑i+j=k,1≤i≤N+1,1≤j≤N-1ηj+1Aui-1εk-∑k=N+12N∑i+j=k,1≤i≤N+1,1≤j≤N-1ηj+1Aui-1εk-η1AuNεN+1+εN+1R~Nμ,μ1,u0,u→,ε=-η1Au0ε-∑k=2Nη1Auk-1+∑i+j=k,1≤i≤N+1,1≤j≤N-1ηj+1Aui-1εk+εN+1R~N1μ,μ1,u0,u→,ε,where(125)εN+1R~Nμ,μ1,u0,u→,ε=-AhεN+1R^Nμ,u0,u→,ε+εεNR^N-1μ1,u0,u→,ε,εN+1R~N1μ,μ1,u0,u→,ε=-∑k=N+12N∑i+j=k,1≤i≤N+1,1≤j≤N-1ηj+1Aui-1εk-η1AuNεN+1+εN+1R~Nμ,μ1,u0,u→,ε.
Combining (99a)-(99b), (119), (123), and (125) leads to(126)Eεx,t=fh-fu0+εf1h-μh-μu0+εμ1hAh-∑k=1NFkεk=f1u0+Φ¯1N,f,u0,u→ε+∑k=2NΦ¯kN,f,u0,u→+Φ¯k-1N-1,f1,u0,u→εk+εN+1R¯Nf,f1,u0,u→,ε-η1Au0ε-∑k=2Nη1Auk-1+∑i+j=k,1≤i≤N+1,1≤j≤N-1ηj+1Aui-1εk+εN+1R~N1μ,μ1,u0,u→,ε-∑k=1NFkεk=f1u0+Φ¯1N,f,u0,u→-η1Au0-F1ε+∑k=2NΦ¯kN,f,u0,u→+Φ¯k-1N-1,f1,u0,u→-η1Auk-1-∑i+j=k,1≤i≤N+1,1≤j≤N-1ηj+1Aui-1-Fkεk+εN+1R¯Nf,f1,u0,u→,ε+R~N1μ,μ1,u0,u→,ε=εN+1R¯Nf,f1,u0,u→,ε+R~N1μ,μ1,u0,u→,ε.By the boundedness of functions uk, uk′, ∇uk, 1≤k≤N in function space L∞(0,T;H1), we obtain from (125), (123), and (126) that(127)EεL∞0,T;L2≤C∗εN+1,where C∗ is a constant depending only on N, T, ρ, ζ, f, f1, μ, μ1, uk, 0≤k≤N.
The proof of Lemma15 is complete.Proof of Theorem11.
Consider sequence{vm} defined by(128)v0≡0,vm′′+μεvm-1+hAvm=Fεvm-1+h-Fεh-μεvm-1+h-μεhAh+Eεx,t,ρ<x<1,0<t<T,vmρ,t=vmx1,t+ζvm1,t=0,vmx,0=vm′x,0=0,m≥1.
By multiplying two sides of (128)1 with vm′ and after integrating in t, we have(129)Zmt=∫0tμm′savms,vmsds+2∫0tEεs,vm′sds+2∫0tFεvm-1+h-Fεh,vm′sds-2∫0tμεvm-1+h-μεhAh,vm′sds,where (130)μ¯mt=μεvm-1+ht=μvm-1t+ht02+εμ1vm-1t+ht02,Zmt=vm′t02+μ¯mtavmt,vmt.
Note that(131)μ¯m′t=2μ′vm-1t+ht02+εμ1′vm-1t+ht02vm-1t+ht,vm-1′t+h′t≤2K~M2μ+K~M2μ1vm-1t+ht0vm-1′t+h′t0≤2K~M2μ+K~M2μ1M22≡σ2M,where M2=(N+2)M, and (132)Zmt≥vm′t02+μ∗avmt,vmt≥vm′t02+μ∗C0vmt12.
Using Lemma15, (129) gives (133)Zmt≤TC∗2ε2N+2+∫0tvm′s02ds+1μ∗σmM∫0tZmsds+2∫0tfvm-1+h-fh0vm′s0ds+2ε∫0tf1vm-1+h-f1h0vm′s0ds+2∫0tμεvm-1+h-μεhAh0vm′s0ds≤TC∗2ε2N+2+1+1μ∗σ2M∫0tZmsds+2∫0tfvm-1+h-fh0vm′s0ds+2∫0tf1vm-1+h-f1hvm′s0ds+2N+1M∫0tμvm-1+h-μhvm′s0ds+2εN+1M∫0tμ1vm-1+h-μ1hvm′s0ds=TC∗2ε2N+2+1+1μ∗σ2M∫0tZmsds+∑i=14J¯i.
We estimate the integrals on the right-hand side of (133) as follows.
Estimating J ¯ 1 . We note that(134)fvm-1+h-fh0≤KM2f1+1-ρ2ρ∇vm-1t0+vm-1′t0≤KM2f1+1-ρ2ρvm-1W1T,with M2=(N+2)M.
It follows from (134) that(135)J¯1=2∫0tfvm-1+h-fh0vm′s0ds≤TKM22f1+1-ρ2ρ2vm-1W1T2+∫0tvm′s02ds.
Estimating J ¯ 2 . Similarly,(136)J¯2=2∫0tf1vm-1+h-f1h0vm′s0ds≤TKM22f11+1-ρ2ρ2vm-1W1T2+∫0tvm′s02ds.
Estimating J ¯ 3 . We have (137)μvm-1+h-μh≤K~M2μvm-1t+ht02-ht02≤2N+3MK~M2μvm-1t0≤2N+3MK~M2μ1-ρ2ρvm-1W1T.
Hence,(138)J¯3=2N+1M∫0tμvm-1+h-μhvm′s0ds≤2N+12N+3M2K~M2μ1-ρ2ρvm-1W1T∫0tvm′s0ds≤N+122N+32M4TK~M22μ1-ρ2ρvm-1W1T2+∫0tvm′s02ds.
Estimating J ¯ 4 . Similarly,(139)J¯4=2N+1M∫0tμ1vm-1+h-μ1hvm′s0ds≤N+122N+32M4TK~M2μ11-ρ2ρvm-1W1T2+∫0tvm′s02ds.
Combining (133), (135), (136), (138), (139), it leads to(140)Zmt≤TρMvm-1W1T2+TC∗2ε2N+2+5+1μ∗σ2M∫0tZmsds,where ρM=KM22f+KM22f11+(1-ρ)/2ρ2+K~M22μ+K~M22μ1(N+1)2(2N+3)2M4((1-ρ)2/ρ).
By using Gronwall’s lemma, we deduce from (140) that(141)vmW1T≤σTvm-1W1T+δTε,∀m≥1,where σT=ηTρM, δT(ε)=ηTC∗εN+1, ηT=1+1/μ∗C0Texp⁡[T5+(1/μ∗)σ2(M)].
We can assume that(142)σT<1,withthesuitableconstantT>0.
We require the following lemma, and its proof is immediate, so we omit the details.
Lemma  16.Let sequence{ζm}satisfy(143)ζm≤σζm-1+δ∀m≥1,ζ0=0,where0≤σ<1, δ≥0are the given constants. Then,(144)ζm≤δ1-σ∀m≥1.
Applying Lemma16 with ζm=vmW1(T), σ=σT<1, δ=δT(ε)=C∗ηTεN+1, it follows from (144) that(145)vmW1T≤δTε1-σT=CTεN+1,where CT=ηTC∗/(1-ηTρM).
On the other hand, linear recurrent sequence{vm} defined by (128) converges strongly in space W1(T) to solution v of problem (118). Hence, letting m→+∞ in (145), we get (146)vW1T≤CTεN+1.
This implies (102). The proof of Theorem 11 is complete.



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*Source: 8031638-2016-10-24.xml*