# On a Nonlinear Wave Equation of Kirchhoff-Carrier Type: Linear Approximation and Asymptotic Expansion of Solution in a Small Parameter

Authors: Nguyen Huu Nhan; Le Thi Phuong Ngoc; Nguyen Thanh Long
Journal: Mathematical Problems in Engineering
 (2018)

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## Abstract

We consider the Robin-Dirichlet problem for a nonlinear wave equation of Kirchhoff-Carrier type. Using the Faedo-Galerkin method and the linearization method for nonlinear terms, the existence and uniqueness of a weak solution are proved. An asymptotic expansion of high order in a small parameter of a weak solution is also discussed.


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## Body

## 1. Introduction

In this paper, we consider the following Robin-Dirichlet problem for a nonlinear wave equation of Kirchhoff-Carrier type:(1)utt-∂∂xμx,t,∫01gx,y,t,uy,t,uxy,tdyux=fx,t,0<x<1,0<t<T,(2)ux0,t-h0u0,t=u1,t=0,(3)ux,0=u~0x,utx,0=u~1x,where μ, f, g, u~0, u~1 are given functions and h0≥0 is a given constant.Equation (1) can be considered as a general equation containing relatively some classical equations; for example, when g(x,y,t,u,ux)=ux2,f=0, (1) has a relation to the Kirchhoff wave equation:(4)ρhutt=P0+Eh2L∫0L∂u∂yy,t2dyuxx(see [1]). This equation is a generalization of the well-known D’Alembert’s wave equation for free vibrations of elastic strings. Kirchhoff’s model takes into account the changes in length of the string produced by transverse vibrations. The parameters in (4) have the following meanings: u is the lateral deflection, L is the length of the string, h is the area of the cross section, E is the Young modulus of the material, ρ is the mass density, and P0 is the initial tension.In another case, withg(x,y,t,u,ux)=u2,f=0, (1) contains the form of Carrier equation. In [2], Carrier established the equation modeling the vibration of an elastic string when the changes in tension are not small:(5)ρhutt-1+EALT0∫0Lu2dxuxx=0,where u(x,t) is the 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, ρ is the density of a material. Therefore, it is clear that (1) considered here contains (4) and (5) as special cases.Moreover, with various boundary conditions, the particular forms of (1) have been extensively studied by many authors; for example, we refer to [3–15] and the references given therein. In these works, many interesting results about existence, regularity, asymptotic behavior, asymptotic expansion, and decay of solutions were obtained.Cavalcanti et al., in [4–7], investigated a series of four papers in which the results of existence, global existence, exponential or uniform decay rates, and asymptotic behavior for Kirchhoff-Carrier models are considered.In [10], the unique existence and asymptotic expansion of solutions of (1) with μ=1 associated with the boundary conditions(6)ux0,t-h0u0,t=ux1,t+h1u1,t=0and the initial conditions are also studied.In [15], de Lima Santos studied the asymptotic behavior of solutions of (1) with f(x,t)≡0, μ=μ(t), associated with the Dirichlet boundary condition at x=0 and a boundary condition of memory type at x=1; that is, u(1,t)+∫0tg(t-s)μ(s)ux(1,s)ds=0,t>0.In [3], Beilin investigated the existence and uniqueness of a generalized solution for the following wave equation with an integral nonlocal condition(7)utt-Δu+cx,tu=fx,t,x,t∈Ω×0,T,∂u∂η+∫0t∫Ωkx,ξ,τuξ,τdξdτ=0,x,t∈∂Ω×0,T,ux,0=u~0x,utx,0=u~1x,x∈Ω,where Ω is a bounded domain in RN with a smooth boundary, η is the unit outward normal on ∂Ω, and f, u~0, u~1, k(x,ξ,τ) are given functions. Nonlocal conditions come up when values of the function on the boundary are connected to values inside the domain. There are various types of nonlocal boundary conditions of integral form for hyperbolic, parabolic, or elliptic equations; the ones were introduced in [3].The well-posedness and optimal decay rate estimates of the energy associated with the Kirchhoff-Carrier problem with memory(8)u′′-M∇ut22Δu+∫0tgt-sΔusds=0,inΩ×R+,u=0,onΓ×R+,ux,0=u0x,utx,0=u1x,inΩ,where Ω is a bounded domain in RN, with a smooth boundary ∂Ω≔Γ, are proved in [8].In [11], the following nonlinear wave equation with initial conditions and boundary conditions of two-point type has been investigated:(9)utt-∂∂xμx,tux+fu,ut=Fx,t,0<x<1,0<t<T.In [12], by combining the linearization method for the nonlinear term, the Faedo-Galerkin method, and the weak compact method, the existence of a unique weak solution of an initial and boundary value problem for nonlinear wave equation utt-∂/∂xμ(x,t,u,ux2)ux=F(x,t,u,ux,ut) with the nonhomogeneous boundary conditions is proved.Very recently, in [13, 14], with the same method used in [12], the authors proved the results of existence and uniqueness for the wave equations with nonlinear sources containing the nonlocal terms. In [13], the linearization method together with Taylor’s expansion is used for both of the source term and the nonlinear integral in it. These techniques have not been used before.In the same spirit of [10–14], we establish the local existence and uniqueness for prob. (1)–(3) by using the Faedo-Galerkin method and the weak compact method. These results are presented in Section 3. In Section 4, the perturbed solution uε(x,t) is approximated by the polynomial of N+1 degree in a small parameter ε for the following perturbed equation:(10)utt-∂∂xμεux,tux=fx,t,0<x<1,0<t<T,associated with (2), (3), where(11)μεux,t=μx,t,∫01gux,y,tdy+εμ1x,t,∫01g1ux,y,tdy,gux,y,t=gx,y,t,uy,t,uxy,t,g1ux,y,t=g1x,y,t,uy,t,uxy,t.


## 2. Preliminaries

PutΩ=(0,1) and denote the usual function spaces used in this paper by the notations Lp=Lp(Ω), Hm=HmΩ. Let 〈·,·〉 be either the scalar product in L2 or the dual pairing of a continuous linear functional and an element of a function space. The notation · stands for the norm in L2, ·X is the norm in the Banach space X, and X′ is the dual space of X.We denoteLp(0,T;X),1≤p≤∞ for the Banach space of real functions u:(0,T)→X measurable, such that (12)uLp0,T;X=∫0TutXpdt1/p<∞,for1≤p<∞,uL∞0,T;X=esssup0<t<TutXforp=∞.Letu(t), u′(t)=ut(t)=u˙(t), u′′(t)=utt(t)=u¨(t), uxt=∇u(t), uxx(t)=Δu(t), denote u(x,t), (∂u/∂t)(x,t), ∂2u/∂t2(x,t), ∂u/∂x(x,t), ∂2u/∂x2(x,t), respectively.With∈Ck([0,1]2×R+×R2), g=g(x,y,t,z1,z2), we put D1g=∂g/∂x, D2g=∂g/∂y, D3g=∂g/∂t, Di+3g=∂g/∂zi, with i=1,2 and Dβg=D1β1⋯D5β5g; β=(β1,…,β5)∈Z+5,β=β1+⋯+β5=k, D(0,…,0)g=g.Similarly, withμ=μ(x,t,z), we also put D1μ=∂μ/∂x,D2μ=∂μ/∂t,D3μ=∂μ/∂z.We shall use the following norm onH1: (13)vH1=v2+vx21/2.We put(14)V=v∈H1:v1=0,(15)au,v=∫01uxxvxxdx+h0u0v0,∀u,v∈V.V is a closed subspace of H1 and on V three norms vH1, vx, and va=a(v,v) are equivalent norms.We have the following lemmas, the proofs of which are straightforward and hence we omit the details.Lemma 1.
The imbeddingH1↪C0(Ω¯) is compact and(16)vC0Ω¯≤2vH1∀v∈H1,where vC0(Ω¯)=supx∈[0,1]⁡v(x) (see [16]).Lemma 2.
Leth0≥0. The imbedding V↪C0(Ω¯) is compact and(17)vC0Ω¯≤vx≤va,12vH1≤vx≤va≤1+h0vH1,for all v∈V.Lemma 3.
Leth0≥0. Then the symmetric bilinear form a(·,·) defined by (15) is continuous on V×V and coercive on V.Lemma 4.
Leth0≥0. Then there exists the Hilbert orthonormal base {w~j} of L2 consisting of the eigenfunctions w~j corresponding to the eigenvalues λj such that (18)limj→+∞λj=+∞,0<λ1≤λ2≤⋯≤λj≤⋯,aw~j,v=λjw~j,v∀v∈V,j=1,2,….
Furthermore, the sequencew~j/λj is also a Hilbert orthonormal base of V with respect to the scalar product a(·,·).
On the other hand, we also havew~j satisfying the following boundary value problem:(19)-Δw~j=λjw~j,in0,1,w~jx0-h0w~j0=w~j1=0,w~j∈V∩C∞Ω¯.The proof of Lemma4 can be found in ([17], p.87, Theorem (7.7)), with H=L2 and V,a(·,·) as defined by (14), (15).Remark 5.
The weak formulation of the 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:(20)uttt,w+Aut;ut,w=ft,w,for all w∈V, a.e., t∈(0,T), together with the initial conditions(21)u0=u~0,ut0=u~1,where, for each w∈W~, {A[w](t;·,·)}0≤t≤T is the family of symmetric bilinear forms on V×V defined by(22)Awt;u,v=μwtux,vx+h0μw0,tu0v0,for all u,v∈V, 0≤t≤T, with h0≥0 being given constant, and(23)μwx,t=μx,t,∫01gwx,y,tdy,gwx,y,t=gx,y,t,wy,t,wxy,t.


## 3. The Existence and Uniqueness

LetT∗>0. We make the following assumptions:(H1)
(u~0,u~1)∈V∩H2×V satisfying the condition u~0x(0)-h0u~0(0)=0.(H2)
g∈C2([0,1]2×[0,T∗]×R2).(H3)
μ∈C2([0,1]×[0,T∗]×R) and there exists a constant μ0>0 such that μ(x,t,z)≥μ0, for all x,t,z∈0,1×[0,T∗]×R.(H4)
f,f′∈L2(QT∗).For eachM>0 given, we set the constants K¯M(g), K¯0(M,g), K~M(μ), K~0(M,μ,g), as follows:(24)K¯Mg=∑α≤2K¯0M,Dαg,K~Mμ=∑β≤2K~0M,Dβμ,g,where(25)K¯0M,g=supx,y,t,z1,z2∈A1M⁡gx,y,t,z1,z2,K~0M,μ,g=supx,t,z∈A2g,M⁡μx,t,z,A1M=x,t,y,z1,z2:0≤x,y≤1,0≤t≤T∗,max1≤i≤2⁡zi≤M,A2g,M=x,t,v:0≤x≤1,0≤t≤T∗,v≤K¯0M,g.For everyT∈(0,T∗] and M>0, we put (26)VT=v∈L∞0,T;V∩H2:vt∈L∞0,T;V,vtt∈L2QT, in which QT=Ω×(0,T).ThenVT is a Banach space with respect to the norm (27)vVT=max⁡vL∞0,T;V∩H2,vtL∞0,T;V,vttL2QT. (See Lions [18]) We also put(28)WM,T=v∈VT:vVT≤M,W1M,T=v∈WM,T:vtt∈L∞0,T;L2.Now, we establish the recurrent sequence{um}. The first term is chosen as u0≡u~0, and supposing that(29)um-1∈W1M,T,we associate problem (1) with the following problem.Findum∈W1(M,T)(m≥1) satisfying the linear variational problem(30)um′′t,v+Amt;umt,v=ft,v,∀v∈V,um0=u~0,um′0=u~1,where(31)Amt;u,v=Aum-1t;u,v=μmtux,vx+h0μm0,tu0v0,∀u,v∈V,μmx,t=μx,t,∫01gum-1x,y,tdy,gum-1x,y,t=gx,y,t,um-1y,t,∇um-1y,t.Theorem 6.
Suppose thatH1–H4 hold. Then, there exist positive constants M,T>0 such that the recurrent sequence {um} is defined by (29)–(31).Proof.
The proof consists of several steps.
Step  1(the Faedo-Galerkin approximation (introduced by Lions [18])). Consider the basis {wj} for V as in Lemma 4. Approximate solution of (29)–(31) problem which will be found in form(32)umkt=∑j=1kcmjktwj,where the coefficients cmj(k)(t) satisfy the system of linear differential equations(33)u¨mkt,wj+Amt;umkt,wj=ft,wj,1≤j≤k,umk0=u~0k,u˙mk0=u~1k,where(34)u~0k=∑j=1kαjkwj⟶u~0strongly  inV∩H2,u~1k=∑j=1kβjkwj⟶u~1strongly  inV.
The system of (33) can be rewritten in form(35)c¨mjkt+∑i=1kAijmtcmikt=fjt,cmjk0=αjk,c˙mjk0=βjk,1≤j≤k,where(36)Aijmt=Amt;wi,wj,fjt=ft,wj,1≤i,j≤k.
By (29), it is not difficult to prove that system (35), (36) has a unique solution cmj(k)(t),1≤j≤k on interval [0,T], so let us omit the details (see [19]).
Step  2(a priori estimates). First, we need the following lemma.
Lemma  7.
Putting
μ
∗
=
K
~
M
(
μ
)
1
+
(
1
+
2
M
)
K
¯
M
(
g
), one has(37)iAmt;u,v≤K~μuava∀u,v∈V,0≤t≤T∗,iiAmt;v,v≥μ0va2∀v∈V,0≤t≤T∗,iii∂Am∂tt;u,v=μm′tux,vx+h0μm′0,tu0v0,∀u,v∈V,iv∂Am∂tt;v,v≤μ∗va2∀v∈V,0≤t≤T∗,vddtAmt;umkt,umkt=2Amt;umkt,u˙mkt+∂Am∂tt;umkt,umkt.
The proof of Lemma7 is easy; hence we omit the details.
Next, we put(38)Smkt=Xmkt+Ymkt+∫0tu¨mks2ds,where(39)Xmkt=u˙mkt2+Amt;umkt,umkt,Ymkt=u˙mkta2+μmtΔumkt2.
Then, it follows from (33), (37)(iii),(v), (38), (39) that(40)Smkt=Smk0+2μmx0u~0kx,Δu~0k+2f0,Δu~0k+∫0tds∫01μm′x,sΔumkx,s2dx+∫0t∂Am∂ts;umks,umksds+2∫0t∂∂sμmxsumxks,Δumksds-2μmxtumxkt,Δumkt+2∫0tfs,u˙mksds+2∫0tf′s,Δumksds-2ft,Δumkt+∫0tu¨mks2ds≡Smk0+2μmx0u~0kx,Δu~0k+2f0,Δu~0k+∑j=18Ij.
We shall estimate the termsIj on the right-hand side of (40) as follows.
First Term
I
1. We note that(41)μm′x,t=D2μum-1+D3μum-1∫01∂gum-1∂tx,y,tdy,where we use the notations (42)Diμum-1=Diμx,t,∫01gx,y,t,um-1y,t,∇um-1y,tdy,i=1,2,3,∂gum-1∂tx,y,t=D3gum-1+D4gum-1um-1′y,t+D5gum-1∇um-1′y,t,Digum-1x,y,t=Digx,y,t,um-1y,t,∇um-1y,t,i=1,…,5.
So, by (24), (25), and (41), we obtain (43)μm′x,t≤μ∗.
Hence,(44)I1=∫0tds∫01μm′x,sΔumkx,s2dx≤μ∗μ0∫0tSmksds.
Second Term
I
2. By Lemma  7 (ii) and (iv), we have(45)I2=∫0t∂Am∂ts;umks,umksds≤μ∗∫0tumkta2ds≤μ∗μ0∫0tSmksds.
Third Term
I
3. The Cauchy-Schwartz inequality leads to(46)I3=2∫0t∂∂sμmxsumxks,Δumksds≤2μ0∫0tJmksSmksds,where Jm(k)(s)=∂/∂sμmx(s)umx(k)(s).
We shall estimate the termJm(k)(s) as follows.
BySm(k)(t)≥u˙mx(k)(t)2+umx(k)(t)2, we have(47)Jmks=∂∂sμmxsumxks≤μmxsC0Ω¯u˙mxks+μmx′sumxksC0Ω¯≤μmxsC0Ω¯+1μ0μmx′sSmks.
On the other hand, byμmx(x,t)=D1μ[um-1]+D3μ[um-1]∫01D1g[um-1](x,y,t)dy, it implies that(48)μmxtC0Ω¯≤K~Mμ1+K¯Mg.
Similarly, from the following equality(49)μmx′x,t=∂∂t∂μm∂xx,t=D2D1μum-1+D3D1μum-1∫01∂gum-1∂tx,y,tdy+D2D3μum-1+D32μum-1∫01∂gum-1∂tx,y,tdy∫01D1gum-1x,y,tdy+D3μum-1∫01∂D1gum-1∂tx,y,tdy;∂gum-1∂tx,y,t=D3gum-1x,y,t+D4gum-1x,y,tum-1′y,t+D5gum-1x,y,t∇um-1′y,t;∂D1gum-1∂tx,y,t=D3D1gum-1x,y,t+D4D1gum-1x,y,tum-1′y,t+D5D1gum-1x,y,t∇um-1′y,t, we obtain that(50)μmx′t≤K~Mμ1+K¯Mg∫011+um-1′t+∇um-1′tdy+K~MμK¯Mg1+K¯Mg∫011+um-1′t+∇um-1′tdy+K~MμK¯Mg∫011+um-1′t+∇um-1′tdy≤K~Mμ1+21+MK¯Mg+21+2MK¯M2g.
By (48) and (50), it follows from (47) that(51)Jmks≤ζ1MSmks,where(52)ζ1M=K~Mμ1+K¯Mg+1μ01+21+MK¯Mg+21+2MK¯M2g.
Therefore, from (46) and (51), we obtain(53)I3≤2μ0ζ1M∫0tSmksds.
Fourth Term
I
4. Applying the Cauchy-Schwartz inequality again, we have(54)I4=-2μmxtumxkt,Δumkt≤1βμmxtumxkt2+βμ0Smkt,for all β>0. On the other hand, it follows from (51) that(55)μmxtumxkt=μmx0∇u~0k+∫0t∂∂sμmxsumxksds≤μmx0C0Ω¯∇u~0k+∫0tJmksds≤μmx0C0Ω¯∇u~0k+ζ1M∫0tSmksds.
Hence, we obtain from (54) and (55) that(56)I4≤βμ0Smkt+2βμmx0C0Ω¯2∇u~0k2+2βTζ12M∫0tSmksds.
Fifth Term
I
5
(57)
I
5
=
2
∫
0
t
f
s
,
u
˙
m
k
s
d
s
≤
2
∫
0
t
f
s
u
˙
m
k
s
d
s
≤
T
f
L
∞
0
,
T
;
L
2
2
+
∫
0
t
S
m
k
s
d
s
.
Sixth Term
I
6. Similarly, we obtain(58)I6=2∫0tf′s,Δumksds≤f′L2QT∗2+1μ0∫0tSmksds.
Seventh Term
I
7. We have(59)I7=-2ft,Δumkt≤1βft2+βΔumkt2≤2βf02+T∫0tf′s2ds+βμ0Smkt=2βf02+Tf′L2QT∗2+βμ0Smkt,∀β>0.
Eighth Term
I
8. We note that (33)1 can be rewritten as follows:(60)u¨mkt,wj-∂∂xμmtumxkt,wj=ft,wj,1≤j≤k.
Hence, it follows after replacingwj with u¨m(k)(t) and integrating that(61)I8=∫0tu¨mks2ds≤2∫0t∂∂xμmsumxks2ds+2∫0tfs2ds≤2∫0t∂∂xμmsumxks2ds+2TfL∞0,T∗;L22.
We estimate the term∂/∂xμm(s)umx(k)(s)2.
By (48), we obtain(62)∂∂xμmsumxks2≤μmxsumxks+μmsΔumks2≤2K~MμK~Mμ1+K¯Mg2umxks2+μmsΔumks2≤2K~Mμ1+K~Mμ1+K¯Mg2umxks2+μmsΔumks2≤2K~Mμ1+K~Mμ1+K¯Mg21+μ0μ0Smks.
Therefore, by Lemma  7 (ii), (61) and (62), we obtain(63)I8≤2TfL∞0,T∗;L22+ζ2M∫0tSmksds,where (64)ζ2M=4K~Mμ1+μ0μ01+K~Mμ1+K¯Mg2.
Choosingβ>0, with 2β/μ0≤1/2, it follows from (40), (44), (45), (53), (56)–(59), and (63) that(65)Smkt≤C~0k+2T2βf′L2QT∗2+3fL∞0,T∗;L22+C~1M,T∫0tSmksds,where(66)C~0k=C~0kμ,f,g,u~0k,u~1k=2Smk0+4μmx0u~0kx,Δu~0k+4f0,Δu~0k+4βμmx0C0Ω¯2∇u~0k2+4βf02+2f′L2QT∗2,C~1M,T=21+1+2μ∗μ0+2βTζ12M+2μ0ζ1M+ζ2M.
By means of the convergences in (34), we can deduce the existence of a constant M>0 independent of k and m such that(67)C~0kμ,f,g,u~0k,u~1k≤12M2.
So, from (66)2, we can choose T∈(0,T∗], such that(68)12M2+2T2βf′L2QT∗2+3fL∞0,T∗;L22exp⁡TC~1M,T≤M2,(69)kT=2TMK~MμK¯Mg1+22+K¯Mg1+1μ0exp⁡Tμ0+μ∗2μ0<1.
Finally, it follows from (65), (67), and (68) that(70)Smkt≤M2exp⁡-TC~1M,T+C~1M,T∫0tSmksds.
By using Gronwall’s Lemma, we deduce from (70) that(71)Smkt≤M2exp⁡-TC~1M,Texp⁡tC~1M,T≤M2,for all t∈[0,T], for all m and k. Therefore, we have(72)umk∈WM,T,∀m,k.
Step  3 (limiting process). From (72), we deduce the existence of a subsequence of {um(k)} still so denoted, such that(73)umk⟶uminL∞0,T;V∩H2weak∗,u˙mk⟶um′inL∞0,T;Vweak∗,u¨mk⟶um′′inL2QTweak,um∈WM,T.
Passing to limit in (33), we have um satisfying (30), (31) in L2(0,T). On the other hand, it follows from (30)1 and (73)4 that um′′=∂/∂xμm(t)umx+f∈L∞(0,T;L2); hence um∈W1(M,T) and the proof of Theorem 6 is complete.We note thatW1(T)={v∈L∞(0,T;V):v′∈L∞(0,T;L2)} is a Banach space with respect to the norm (see Lions [18]). (74)vW1T=vL∞0,T;V+v′L∞0,T;L2.We use the result given in Theorem6 and the compact imbedding theorems to prove the existence and uniqueness of a weak solution of prob. (1)–(3). Hence, we get the main result in this section as follows.Theorem 8.
Let(H1)–(H4) hold. Then one has the following.
(i) Prob. (1)–(3) has a unique weak solution u∈W1(M,T), where the constants M>0 and T>0 are chosen as in Theorem 6.
(ii) Furthermore, the recurrent sequence {um} defined by (29)-(30) converges to the solution u of prob. (1)–(3) strongly in W1(T).
And one has the estimate(75)um-uW1T≤CTkTm,∀m∈N,where the constant kT∈[0,1) is defined as in (69) and CT is a constant depending only on T, h0, f, g, μ, u~0, u~1, and kT.Proof.
(a) Existence of the Solution. We shall prove that {um} is a Cauchy sequence in W1(T). Let wm=um+1-um. Then wm satisfies the variational problem(76)wm′′t,w+Am+1t;wmt,w=-Am+1t;umt,w+Amt;umt,w,∀w∈V,wm0=wm′0=0.
Note that(77)ddtAm+1t;wmt,wmt=2Am+1t;wmt,wm′t+∂Am+1∂tt;wmt,wmt,Am+1t;umt,wm′t-Amt;umt,wm′t=-∂∂xμm+1t-μmtumxt,wm′t.
Takingw=wm′(t) in (76)1, after integrating in t, we get(78)Zmt=∫0t∂Am+1∂ts;wms,wmsds+2∫0t∂∂xμm+1s-μmsumxs,wm′sds≡J1+J2,where(79)Zmt=wm′t2+Am+1t;wmt,wmt≥wm′t2+μ0wmta2,and the integrals on the right-hand side of (78) are estimated as follows.
First Integral
J
1. By (37)(iv) and (79), we have(80)J1≤∫0t∂Am+1∂ts;wms,wmsds≤μ∗μ0∫0tZmsds.
Second Integral
J
2. By the inequalities(81)ΔumsH2≤umsH2≤M,umxsC0Ω¯≤2umxsH1≤2umsH2≤2M,DiμumsC0Ω¯≤K~Mμ,i=1,3,D1gumsC0Ω¯≤K¯Mg,μm+1s-μmsC0Ω¯≤2K~MμK¯Mg∇wm-1s≤2K~MμK¯Mgwm-1W1T,Diμums-Diμum-1sC0Ω¯≤2K~MμK¯Mgwm-1W1T,i=1,3,D1gums-D1gum-1sC0Ω¯≤2K¯Mgwm-1W1Tand from the equation(82)∂∂xμm+1s-μms∇ums=μm+1s-μmsΔums+D1μums-D1μum-1sumxs+D3μums-D3μum-1s∫01D1gumx,y,sdyumxs+D3μum-1s∫01D1gum-D1gum-1dyumxs,we obtain that(83)∂∂xμm+1s-μms∇ums≤2MK~MμK¯Mg1+22+K¯Mgwm-1W1T.
This implies that(84)J2≤2∫0t∂∂xμm+1s-μms∇ums,wm′sds≤4TM2K~M2μK¯M2g1+22+K¯Mg2wm-1W1T2+∫0tZmsds.
Combining (78), (80), and (84), we obtain(85)Zmt≤4TM2K~M2μK¯M2g1+22+K¯Mg2wm-1W1T2+μ0+μ∗μ0∫0tZmsds.
Using Gronwall’s Lemma, we deduce from (85) that(86)wmW1T≤kTwm-1W1T,∀m∈N,where kT∈(0,1) is defined as in (69), which implies that(87)um-um+pW1T≤u0-u1W1T1-kT-1kTm,∀m,p∈N.
It follows that{um} is a Cauchy sequence in W1(T). Then there exists u∈W1(T) such that(88)um⟶ustrongly  inW1T.
Note thatum∈W1(M,T); then there exists a subsequence {umj} of {um} such that(89)umj⟶uinL∞0,T;V∩H2weak∗,umj′⟶u′inL∞0,T;Vweak∗,umj′′⟶u′′inL2QTweak,u∈WM,T.
We also note that(90)μmx,t-μux,t≤2K~MμK¯Mgum-1-uW1T,a.e.x,t∈QT.
Hence, from (88) and (90), we obtain(91)μm⟶μustrongly  inL∞QT.
On the other hand, for allv∈V, we have (92)Amt;um,v-Aut;u,v≤1+h0K~Mμ2K¯MgMum-1-uW1T+um-uW1Tvx.
Hence(93)∫0TAmt;um,v-Aut;u,vϕtdt⟶0∀v∈V,∀ϕ∈L10,T.
Finally, passing to limit in (30)-(31) as m=mj→∞, it is implied from (88), (89)1,3, and (93) that there exists u∈W(M,T) satisfying the equation(94)u′′t,w+Aut;ut,w=ft,wfor all w∈V and the initial conditions(95)u0=u~0,u′0=u~1.
On the other hand, from the assumptions(H2)–(H4) we obtain from (89)4, (93), and (94) that(96)u′′=∂∂xμutux+f∈L∞0,T;L2,and thus we have u∈W1(M,T). The existence result follows.
(b) Uniqueness of the Solution. Let u1, u2∈W1(M,T) be two weak solutions of prob. (1)–(3). Then u=u1-u2 satisfies the variational problem(97)u′′t,w+Au1t;ut,w=-Au1t;u2t,w+Au2t;u2t,w,∀w∈V,u0=u′0=0,where(98)Auit;u,w=μuitux,wx+h0μui0,tu0w0,u,w∈V,μuix,t=μx,t,∫01gx,y,t,uiy,t,∇uiy,tdy,i=1,2.
We takew=u′ in (97)1 and integrate in t to get(99)Zt=∫0t∂Au1∂ts;us,usds+2∫0t∂∂xμu1s-μu2s∇u2s,u′sds,where Z(t)=u′(t)2+A[u1](t;u(t),u(t)).
PutZ~M=μ∗/μ0+2/μ0MK~M(μ)K¯M(g)1+2(2+K¯M(g)), and then it follows from (99) that (100)Zt≤Z~M∫0tZsds.
By Gronwall’s Lemma, we deduceZ(t)=0; that is, u1≡u2. Theorem 8 is proved completely.


## 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′)
g1∈C2([0,1]2×R+×R2).(H3′)
μ1∈C2([0,1]×R+×R) and μ1(x,t,z)≥0,for  all(x,t,z)∈[0,1]×R+×R.We consider the following perturbed problem, whereε is a small parameter such that ε≤1:Pεutt-∂∂xμεux,tux=fx,t,0<x<1,0<t<T,ux0,t-h0u0,t=u1,t=0,ux,0=u~0x,utx,0=u~1x, where (101)μεux,t=μx,t,∫01gux,y,tdy+εμ1x,t,∫01g1ux,y,tdy,gux,y,t=gx,y,t,uy,t,uxy,t,g1ux,y,t=g1x,y,t,uy,t,uxy,t.By Theorem8, problem Pε has a unique weak solution uε depending on ε, satisfying uε∈W1(M,T), in which M,T are independent of ε; these constants are chosen as in (67), (68), and (69), with K¯M(g)+K¯M(g1), K~M(μ)+K~M(μ1) stand for K¯M(g), K~M(μ), respectively.Moreover, we can prove that the limitu0 in suitable function spaces of the family {uε} as ε→0 is a unique weak solution of the problem P0 (corresponding to ε=0) also satisfying u0∈W1(M,T).We shall study the asymptotic expansion of the solution of the problemPε with respect to a small parameter ε.We use the following notations. For a multi-indexα=(α1,…,αp)∈Z+N and x=(x1,…,xN)∈RN, we put (102)α=α1+⋯+αN,α!=α1!⋯αN!,α,β∈Z+N,α≤β⟺αi≤βi,∀i=1,…,N,xα=x1α1⋯xNαN.First, we need the following lemma.Lemma 9.
Letm, N∈N and  x=x1,…,xN∈RN, ε∈R. Then(103)∑i=1Nxiεim=∑k=mmNPkmN,xεk,where the coefficients Pk(m)[N,x],m≤k≤mN, depend on x=(x1,…,xN) defined by the formulas(104)PkmN,x=xk,1≤k≤N,m=1,∑α∈AkmNm!α!xα,m≤k≤mN,m≥2,AkmN=α∈Z+N:α=m,∑i=1Niαi=k.Proof of Lemma9.
The proof of Lemma9 is easy; hence we omit the details.Now, we assume that(H2N)
g∈CN+2([0,1]2×R+×R2), g1∈CN+1([0,1]2×R+×R2);(H3N)
μ∈CN+2([0,1]×R+×R), μ1∈CN+1([0,1]×R+×R),μ≥μ0>0 and μ1≥0,for  all(x,t,z)∈[0,1]×R+×R.We use notationsμ[u](x,t)=μ(x,t,∫01g[u](x,y,t)dy), g[u](x,y,t)=g(x,y,t,u(y,t),ux(y,t)).Letu0 be a unique weak solution of the problem P0 corresponding to ε=0; that is, P0u0′′-∂∂xμu0u0x=f,0<x<1,0<t<T,u0x0,t-h0u00,t=u01,t=0,u0x,0=u~0x,u0′x,0=u~1x,u0∈W1M,T.Let us consider the sequence of the weak solutionsuk,1≤k≤N, defined by the following problems: P~kuk′′-∂∂xμu0ukx=F~kuk,0<x<1,0<t<T,ukx0,t-h0uk0,t=uk1,t=0,ukx,0=uk′x,0=0,uk∈W1M,T, where F~kuk, 1≤k≤N, are defined by the formulas(105)F~k=F~kuk=∑i=1k∂∂xρiN,μ,g+ρi-1N-1,μ1,g1∇uk-i,1≤k≤N,with ρkN,μ,g, which are defined by the formulas(106)ρkN,μ,g=μu0,k=0,∑j=1k1j!Djμu0Φkj,N,g,u0,u→,1≤k≤N,where u→=(u1,…,uN) and(107)Φkj,N,g,u0,u→=PkjN,Ψ→N,g,u0,u→,∇u→=Ψ¯kN,g,u0,u→,∇u→,j=1,∑α∈AkjNj!α!Ψ→αN,g,u0,u→,∇u→,j≤k≤jN,j≥2,with Ψ→[N,g,u0,u→,∇u→]=Ψ¯1[N,g,u0,u→,∇u→],…,Ψ¯N[N,g,u0,u→,∇u→], are defined by(108)Ψ¯kN,g,u0,u→,∇u→=∑1≤β≤k1β!∫01Dβgu0Ψkβ,N,u→,∇u→dy,1≤k≤N,Ψkβ,N,u→,∇u→=∑i,j∈A~β,N,i+j=kPiβ1N,u→Pjβ2N,∇u→,A~β,N=i,j∈Z+2:β1≤i≤Nβ1,β2≤j≤Nβ2.Then, we have the following theorem.Theorem 10.
Let(H1), (H4), (H2N), and (H3N) hold. Then there exist constants M>0 and T>0 such that, for every 0≤ε≤1, the problem Pε has a unique weak solution uε∈W1(M,T) satisfying the asymptotic estimation up to order N+1 as follows:(109)uε-∑k=0NukεkW1T≤CTεN+1,where the functions uk, 0≤k≤N, are the weak solutions of the problems P0, P~k, 1≤k≤N, respectively, and CT is a constant depending only on N, T, μ, μ1, f, g, g1, uk, 0≤k≤N.In order to prove Theorem10, we need the following lemmas.Lemma 11.
LetρkN,μ,g, 1≤k≤N, be the functions defined by the formulas (106)–(108). Putting h=∑k=0Nukεk, then one has(110)μh=∑k=0NρkN,μ,gεk⁡+εN+1R^Nμ,g,u0,u→,ε,with R^N[μ,g,u0,u→,ε]L∞(0,T;L2)≤C, where C is a constant depending only on N, T, μ, g, uk, 0≤k≤N.Proof of Lemma11.
In the case ofN=1, the proof of (110) is easy; hence we omit the details, which we only prove with N≥2. Putting h=u0+∑k=1Nukεk≡u0+h1, we rewrite as follows:(111)μhx,t=μx,t,∫01gx,y,t,u0+h1,∇u0+∇hdy=μx,t,∫01gu0x,y,tdy⁡+ξ,where ξ=∫01g[u0+h1](x,y,t)-g[u0](x,y,t)dy and g¯u0(x,t)=∫01g(x,y,t,u0(y,t),∇u0(y,t))dy.
By using Taylor’s expansion of the functionμ(x,t,∫01gu0(x,y,t)dy+ξ) around the point [u0]≡x,t,∫01g[u0](x,y,t)dy up to order N+1, we obtain(112)μhx,t=μx,t,∫01gu0x,y,tdy+ξ=μu0+∑j=1N1j!D3jμu0ξj+RNμ,u0,h1,ξ,where(113)RNμ,u0,h1,ξ=1N!ξN+1∫011-θND3N+1μx,t,∫01gu0x,y,tdy+θξdθ=εN+1R^N1μ,u0,h1,ξ,ε.
Similarly, withg[h], by using Taylor’s expansion of the function g[h](x,y,t)=gx,y,t,u0+h1,∇u0+∇h1 around the point [u0]≡(x,y,t,u0,∇u0) up to order N+1, we obtain(114)ghx,y,t=gu0+∑1≤β≤N1β!Dβgu0h1β1∇h1β2+RNg,u0,h1,ε,where(115)RNg,u0,h1,ε=∑β=N+1N+1β!∫011-θNDβgx,y,t,u0+θh1,∇u0+θ∇h1h1β1∇h1β2dθ=εN+1RN1ε,g,u0,h1,β=(β1,β2)∈Z+2, |β|=β1+β2, β!=β1!β2!, Dβg=D4β1D5β2g, Dβg[u0]=Dβgx,y,t,u0,∇u0.
Note, by formula (103), we get(116)h1β1=∑k=1Nukεkβ1=∑i=β1Nβ1Piβ1N,u→εi,where u→=(u1,…,uN).
Similarly, with(∇h1)γ2, we also have(117)∇h1β2=∑k=1N∇ukεkβ2=∑j=β2Nβ2Pjβ2N,∇u→εj,where ∇u→=(∇u1,…,∇uN).
Hence, we deduce from (116), (117) that(118)h1β1∇h1β2=∑k=βNΨkβ,N,u→,∇u→εk+∑k=N+1NβΨkβ,N,u→,∇u→εk,where(119)Ψkβ,N,u→,∇u→=∑i,j∈A~β,N,i+j=kPiβ1N,u→Pjβ2N,∇u→,A~β,N=i,j∈Z+2:β1≤i≤Nβ1,β2≤j≤Nβ2.
Hence, it follows from (114), (118) that(120)gh=gu0+∑1≤β≤N1β!Dβgu0∑k=βNΨkβ,N,u→,∇u→εk+∑1≤β≤N1β!Dβgu0∑k=N+1NβΨkβ,N,u→,∇u→εk+εN+1RN1ε,g,u0,h1=gu0+∑k=1N∑1≤β≤k1β!Dβgu0Ψkβ,N,u→,∇u→εk+εN+1RN2ε,g,u0,h1,u→,where(121)εN+1RN2ε,g,u0,h1,u→=∑1≤β≤N1β!Dβgu0∑k=N+1NβΨkβ,N,u→,∇u→εk+εN+1RN1ε,g,u0,h1.
Therefore(122)ξ=∫01ghx,y,t-gu0x,y,tdy=∑k=1N∑1≤β≤k1β!∫01Dβgu0Ψkβ,N,u→,∇u→dyεk+εN+1∫01RN2ε,g,u0,h1,u→dy=∑k=1NΨ¯kN,g,u0,u→,∇u→εk+εN+1RN3ε,g,u0,h1,u→,with(123)Ψ¯kN,g,u0,u→,∇u→=∑1≤β≤k1β!∫01Dβgu0Ψkβ,N,u→,∇u→dy,1≤k≤N,RN3ε,g,u0,h1,u→=∫01RN2ε,g,u0,h1,u→dy.
On the other hand, we also have(124)ξj=∑k=1NΨ¯kN,g,u0,u→,∇u→εk+εN+1RN3ε,g,u0,h1,u→j=∑k=1NΨ¯kN,g,u0,u→,∇u→εkj+εN+1RN4ε,j,g,u0,h1,u→,∇u→=∑k=jjNPkjN,Ψ→N,g,u0,u→,∇u→εk+εN+1RN4ε,j,g,u0,h1,u→,∇u→≡∑k=jjNΦkj,N,g,u0,u→εk+εN+1RN4ε,j,g,u0,h1,u→,∇u→,where(125)Φkj,N,g,u0,u→=PkjN,Ψ→N,g,u0,u→,∇u→=Ψ¯kN,g,u0,u→,∇u→,j=1,∑α∈AkjNj!α!Ψ→αN,g,u0,u→,∇u→,j≤k≤jN,j≥2,AkjN=α∈Z+N:α=j,∑i=1Niαi=k,with Ψ→[N,g,u0,u→,∇u→]=Ψ¯1[N,g,u0,u→,∇u→],…,Ψ¯N[N,g,u0,u→,∇u→], which are defined by (108).
Decompose the sum∑k=jjN into the sum of two sums ∑k=jN and ∑k=N+1jN‍; therefore, we deduce from (124) that(126)ξj=∑k=jNΦkj,N,g,u0,u→εk+εN+1RN5ε,j,g,u0,u→,where(127)εN+1RN5ε,j,g,u0,u→=∑k=N+1jNΦkj,N,g,u0,u→εk+εN+1RN4ε,j,g,u0,h1,u→,∇u→.
Hence, it follows from (112) and (124) that(128)μh=μu0+∑1≤j≤N1j!Djμu0∑k=jNΦkj,N,g,u0,u→εk+εN+1∑1≤j≤N1j!Djμu0RN5ε,j,g,u0,h1,u→+R^N1μ,u0,h1,ξ,ε=μu0+∑k=1N∑1≤j≤k1j!Djμu0Φkj,N,g,u0,u→εk+εN+1R^Nμ,g,u0,u→,ε=∑k=0NρkN,μ,gεk+εN+1R^Nμ,g,u0,u→,ε,where ρkN,μ,g, 1≤k≤N, are defined by (106)–(108) and(129)R^Nμ,g,u0,u→,ε=∑1≤j≤N1j!Djμu0RN5ε,j,g,u0,h1,u→+R^N1μ,u0,h1,ξ.
By the boundedness of the functionsuk, ∇uk, 1≤k≤N, in the function space L∞(0,T;L2), we obtain from (113), (115), (121), (123)2, (124), and (127) that R^N[μ,g,u0,u→,ε]L∞(0,T;L2)≤C, where C is a constant depending only on N, T, μ, g, uk,1≤k≤N. Thus, Lemma 11 is proved.Remark 12.
Lemma11 is a generalization of the formula contained in ([9], p.262, formula (4.38)) and it is useful to obtain the following Lemma 13. These lemmas are the key to establish the asymptotic expansion of the weak solution uε of order N+1 in a small parameter ε as follows.
Letu=uε∈W1(M,T) be the unique weak solution of problem Pε. Then v=uε-∑k=0Nukεk≡uε-h satisfies the problem(130)v′′-∂∂xμεv+hvx=∂∂xμεv+h-μεhhx+Eεx,t,0<x<1,0<t<T,vx0,t-h0v0,t=v1,t=0,vx,0=v′x,0=0,where(131)Eεx,t=∂∂xμh-μu0+εμ1hhx-∑k=1NF~kεk.Then, we have the following lemma.Lemma 13.
Let(H1), (H4), (H2N), and (H3N) hold. Then there exists a constant C∗ such that(132)EεL∞0,T;L2≤C∗εN+1,where C∗ is a constant depending only on N, T, μ, μ1, g, g1, uk,1≤k≤N.Proof of Lemma13.
In the case ofN=1, the proof of Lemma 13 is easy; hence we omit the details, which we only prove with N≥2.
By using formula (110) for the function μ1[h] we obtain(133)μ1h=∑k=0N-1ρkN-1,μ1,g1εk+εNR^N-1μ1,g1,u0,u→,ε,where R^N-1[μ1,g1,u0,u→,ε]L∞(0,T;L2)≤C, with C being a constant depending only on N, T, μ1, g1, uk, 0≤k≤N.
By (133), we rewrite εμ1[h] as follows:(134)εμ1h=∑k=1Nρk-1N-1,μ1,g1εk+εN+1R^N-1μ1,g1,u0,u→,ε.
Hence, we deduce from (110) and (134) that(135)μh-μu0+εμ1hhx=∑k=1N∇u0ρkN,μ,g+ρk-1N-1,μ1,g1εk+∑k=22N∑i,j=1,i+j=kNρiN,μ,g+ρi-1N-1,μ1,g1∇ujεk+εN+1R~N1μ,g,μ1,g1,u0,u→,ε,where(136)R~N1μ,g,μ1,g1,u0,u→,ε=R^Nμ,g,u0,u→,ε+R^N-1μ1,g1,u0,u→,εhx.
We decompose the sum∑i=22N into the sum of two sums ∑i=2N‍ and ∑i=N+12N; therefore, we deduce from (135) that(137)μh-μu0+εμ1hhx=∑k=1N∇u0ρkN,μ,g+ρk-1N-1,μ1,g1εk+∑k=2N∑i,j=1,i+j=kNρiN,μ,g+ρi-1N-1,μ1,g1∇ujεk+∑k=N+12N∑i,j=1,i+j=kNρiN,μ,g+ρi-1N-1,μ1,g1∇ujεk+εN+1R~N1μ,g,μ1,g1,u0,u→,ε=∑k=1N∑i=1kρiN,μ,g+ρi-1N-1,μ1,g1∇uk-iεk+εN+1R~N2μ,g,μ1,g1,u0,u→,ε,where(138)εN+1R~N2μ,g,μ1,g1,u0,u→,ε=∑k=N+12N∑i,j=1,i+j=kNρiN,μ,g+ρi-1N-1,μ1,g1∇ujεk+εN+1R~N1μ,g,μ1,g1,u0,u→,ε.
Combining (105), (131), and (137) leads to(139)Eεx,t=εN+1∂∂xR~N2μ,g,μ1,g1,u0,u→,ε.
By the boundedness of the functionsuk, ∇uk, 1≤k≤N, in the function space L∞(0,T;H1), we obtain from (110), (136), (138), and (139) that(140)EεL∞0,T;L2≤C∗εN+1,where C∗ is a constant depending only on N, T, μ, μ1, g, g1, uk, 1≤k≤N.
Lemma13 is proved.Proof of Theorem10.
Consider the sequence{vm} defined by(141)v0≡0,vm′′-∂∂xμεvm-1+hvmx=∂∂xμεvm-1+h-μεhhx+Eεx,t,0<x<1,0<t<T,vmx0,t-h0vm0,t=vm1,t=0,vmx,0=vm′x,0=0,m≥1.
By multiplying two sides of (141) with vm′ and after integration in t, we have(142)Zmt=2∫0tEεs,vm′sds+∫0t∂Am,ε∂ts;vms,vmsds+2∫0t∂∂xμεvm-1+h-μεhhx,vm′sds,where(143)Zmt=vm′t2+Am,εt;vmt,vmt,Am,εt;u,v=μm,εtux,vx+h0μm,ε0,tu0v0,u,v∈V,μm,εx,t=μx,t,∫01gvm-1+hx,y,tdy+εμ1x,t,∫01g1vm-1+hx,y,tdy,gvm-1+hx,y,t=gx,y,t,vm-1+h,∇vm-1+∇h,g1vm-1+hx,y,t=g1x,y,t,vm-1+h,∇vm-1+∇h.
By using Lemma13, we deduce from (142) that(144)vm′t2+μ0vmta2≤TC∗2ε2N+2+∫0tvm′s2ds+∫0t∂Am,ε∂ts;vms,vmsds+2∫0t∂∂xμεvm-1+h-μεhhxvm′sds=TC∗2ε2N+2+∫0tvm′s2ds+J^1+J^2.
We estimate the integrals on the right-hand side of (144) as follows.
Estimating
J
^
1. We note that (145)μm,ε′x,t=D2μvm-1+hx,t+D3μvm-1+hx,t∫01D3gvm-1+h+D4gvm-1+hvm-1′+h′y,t+D5gvm-1+h∇vm-1′+∇h′y,tdy+εD2μ1vm-1+hx,t+εD3μ1vm-1+hx,t∫01D3g1vm-1+h+D4g1vm-1+hvm-1′+h′y,t+D5g1vm-1+h∇vm-1′+∇h′y,tdy, and we have(146)μm,ε′x,t≤ζ¯1,with ζ¯1=K~M∗(μ)1+K¯M∗(g)1+2M∗+K~M∗(μ1)1+K¯M∗(g1)1+2M∗, M∗=(N+2)M.
It follows from (146) that(147)J^1≤∫0t∂Am,ε∂ts;vms,vmsds≤ζ¯1∫0tvmsa2ds.
Estimating
J
^
2. First, we need an estimation ∂/∂xμ[vm-1+h]-μ[h]hx.
We also note that(148)μvm-1+h-μhC0Ω¯≤2K~M∗μK¯M∗gvm-1W1T,Diμvm-1+h-Diμh≤2K~M∗μK¯M∗gvm-1W1T,i=1,3,D1gvm-1+h-D1gh≤2K¯M∗gvm-1W1T.
From the equation(149)∂∂xμvm-1+h-μhhx=μvm-1+h-μhhxx+D1μvm-1+h-D1μhhx+D3μvm-1+h∫01D1gvm-1+h-D1ghdyhx+D3μvm-1+h-D3μh∫01D1ghdyhx,it follows that(150)∂∂xμvm-1+h-μhhx≤dμ,g,M∗vm-1W1T,where d(μ,g,M∗)=2M∗K~M∗(μ)K¯M∗(g)1+22+K¯M∗(g).
Next, byμε=μ+εμ1, it follows that(151)∂∂xμεvm-1+h-μεhhx≤ζ¯2vm-1W1T,where ζ¯2=d(μ,g,M∗)+d(μ1,g1,M∗).
By (151), we obtain(152)J^2=2∫0t∂∂xμεvm-1+h-μεhhxvm′sds≤Tζ¯22vm-1W1T2+∫0tvm′s2ds.
Combining (144), (147), and (152), this leads to(153)Ymt≤TC∗2ε2N+2+Tζ¯22vm-1W1T2+2+ζ¯1μ0∫0tYmsds,where Ym(t)=vm′(t)2+μ0vm(t)a2.
By using Gronwall’s Lemma, we deduce from (153) that(154)vmW1T≤σTvm-1W1T+δTε,∀m≥1,where σT=ηTζ¯2,δT(ε)=C∗ηTεN+1, ηT=1+1/μ0Texp⁡1+ζ¯1/2μ0T.
We can assume that(155)σT<1,with  the  suitable  constantT>0.
We require the following lemma whose proof is immediate.
Lemma 14.Let the sequence{γm}satisfy(156)γm≤σγm-1+δ∀m≥1,γ0=0,where0≤σ<1, δ≥0are the given constants. Then(157)γm≤δ1-σ∀m≥1.
Applying Lemma 14 withγm=vmW1(T), σ=σT<1, δ=δT(ε), it follows from (154) that(158)vmW1T≤δTε1-σT=CTεN+1,where CT is a constant depending only on T.
On the other hand, the linear recurrent sequence{vm} defined by (141) converges strongly in the space W1(T) to the solution v of problem (130). Hence, letting m→+∞ in (158), we get(159)vW1T≤CTεN+1.
This implies (109).
The proof of Theorem10 is complete.



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*Source: 3626543-2018-01-22.xml*