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International Journal of Differential Equations
Volume 2011 (2011), Article ID 679528, 31 pages
Research Article

A High Order Iterative Scheme for a Nonlinear Kirchhoff Wave Equation in the Unit Membrane

1Nhatrang Educational College, 01 Nguyen Chanh Street, Nhatrang City, Vietnam
2Department of Mathematics and Computer Science, University of Natural Science, Vietnam National University Ho Chi Minh City, 227 Nguyen Van Cu Street, District 5, Ho Chi Minh City, Vietnam

Received 5 May 2011; Accepted 16 October 2011

Academic Editor: Bashir Ahmad

Copyright © 2011 Le Thi Phuong Ngoc and Nguyen Thanh Long. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.


A high-order iterative scheme is established in order to get a convergent sequence at a rate of order 𝑁 (𝑁1) to a local unique weak solution of a nonlinear Kirchhoff wave equation in the unit membrane. This extends a recent result in (EJDE, 2005, No. 138) where a recurrent sequence converges at a rate of order 2.

1. Introduction

In this paper we consider the initial and boundary value problem𝑢𝑡𝑡𝑢𝐵𝑟20𝑢𝑟𝑟+1𝑟𝑢𝑟||||=𝑓(𝑟,𝑡,𝑢),0<𝑟<1,0<𝑡<𝑇,lim𝑟0+𝑟𝑢𝑟||||𝑢(𝑟,𝑡)<,𝑟(1,𝑡)+𝑢(1,𝑡)=0,𝑢(𝑟,0)=̃𝑢0(𝑟),𝑢𝑡(𝑟,0)=̃𝑢1(𝑟),(1.1) where 𝐵,𝑓,̃𝑢0,̃𝑢1 are given functions satisfying conditions specified later, 𝑢𝑟20=10𝑟|𝑢𝑟(𝑟,𝑡)|2𝑑𝑟, and >0 is a given constant.

Equation (1.1)1 herein is the bidimensional nonlinear wave equation describing nonlinear vibrations of the unit membrane Ω1={(𝑥,𝑦)𝑥2+𝑦2<1}. In the vibration process, the area of the unit membrane and the tension at various points change in time. The condition on the boundary 𝜕Ω1 describes elastic constraints, where the constant 1 has a mechanical signification. The boundary condition |lim𝑟0+𝑟𝑢𝑟(𝑟,𝑡)|< is satisfied automatically if 𝑢 is a classical solution of the problem (1.1), for example, with 𝑢C1([0,1]×(0,𝑇))𝐶2((0,1)×(0,𝑇)). This condition is also used in connection with Sobolev spaces with weight 𝑟 (see [13]).

Equation (1.1)1 is related to the Kirchhoff equation𝜌𝑢𝑡𝑡=𝑃0+𝐸2𝐿𝐿0||||𝜕𝑢||||𝜕𝑦(𝑦,𝑡)2𝑢𝑑𝑦𝑥𝑥(1.2) presented by Kirchhoff in 1876 (see [4]). This equation is an extension of the classical D’Alembert wave equation which considers the effects of the changes in the length of the string during the vibrations. The parameters in (1.2) have the following meanings: 𝑢 is the lateral deflection, 𝐿 is the length of the string, 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.

The Kirchhoff wave equation of the form (1.1)1 received much attention. Many interesting results about the existence, stability, regularity in time variable, asymptotic behavior, and asymptotic expansion of solutions can be found, for example, in [2, 3, 514] and references therein.

In [2], in a special case, sufficient conditions were established for a quadratic convergence to the solution of (1.1) with 𝑓(𝑟,𝑡,𝑢)=𝑓(𝑟,𝑢) and 𝐵(𝑢𝑟20)=𝑏0+𝑢𝑟20,𝑏0>0. Based on the ideas about recurrence relations for a third-order method for solving the nonlinear operator equation 𝐹(𝑢)=0 in [15], we extend the above result by the construction of a high-order iterative scheme for (1.1)1, where 𝑓 and 𝐵 are more generalized.

In this paper, we associate with (1.1)1 a recurrent sequence {𝑢𝑚} defined by𝜕2𝑢𝑚𝜕𝑡2𝑢𝐵𝑚𝑟20𝜕2𝑢𝑚𝜕𝑟2+1𝑟𝜕𝑢𝑚=𝜕𝑟𝑁1𝑖=01𝜕𝑖!𝑖𝑓𝜕𝑢𝑖𝑟,𝑡,𝑢𝑚1𝑢𝑚𝑢𝑚1𝑖,(1.3)0<𝑟<1,0<𝑡<𝑇, with 𝑢𝑚 satisfying (1.1)2-3. The first term 𝑢0 is chosen as 𝑢00. If𝐵𝐶1(+) and 𝑓𝐶𝑁([0,1]×+×), we prove that the sequence {𝑢𝑚} converges at a rate of order 𝑁 to a unique weak solution of the problem (1.1). This result is a relative generalization of [2, 3, 8, 9, 14, 16].

2. Preliminary Results, Notations, Function Spaces

Put Ω=(0,1). We omit the definitions of the usual function spaces 𝐶𝑚(Ω),𝐿𝑝(Ω),𝐻𝑚(Ω), and 𝑊𝑚,𝑝(Ω). For any function 𝑣𝐶0(Ω) we define 𝑣0 as 𝑣0=(10𝑟𝑣2(𝑟)𝑑𝑟)1/2 and define the space 𝑉0 as completion of the space 𝐶0(Ω) with respect to the norm 0. Similarly, for any function 𝑣𝐶1(Ω) we define 𝑣1 as 𝑣1=(𝑣20+𝑣𝑟20)1/2 and define the space 𝑉1 as completion of the space 𝐶1(Ω) with respect to the norm 1. Note that the norms 0 and 1 can be defined, respectively, from the inner products 𝑢,𝑣=10𝑟𝑢(𝑟)𝑣(𝑟)𝑑𝑟,𝑢,𝑣+𝑢𝑟,𝑣𝑟.(2.1) Identifying 𝑉0 with its dual 𝑉0 we obtain the dense and continuous embedding 𝑉1𝑉0𝑉0𝑉1. The inner product notation will be reutilized to denote the duality pairing between 𝑉1 and 𝑉1.

We then have the following lemmas, the proofs of which can be found in [1].

Lemma 2.1. There exist two constants 𝐾1>0 and 𝐾2>0 such that, for all 𝑣𝐶1(Ω), we have(i)𝑣𝑟20+𝑣2(1)𝑣20, (ii)|𝑣(1)|𝐾1𝑣1, (iii)𝑟|𝑣(𝑟)|𝐾2𝑣1,forall𝑟Ω.

Lemma 2.2. The embedding 𝑉1𝑉0 is compact.

Remark 2.3. In Lemma 2.1, the two constants 𝐾1 and 𝐾2 can be given explicitly as 𝐾1=1+2 and 𝐾2=1+5. We also note that lim𝑟0+𝑟𝑣(𝑟)=0 for all 𝑣𝑉1 (see [17, page 128/Lemma 5.40]). On the other hand, by 𝐻1(𝜀,1)𝐶0([𝜀,1]),0<𝜀<1 and 𝜀𝑣𝐻1(𝜀,1)𝑣1 for all 𝑣𝑉1, it follows that 𝑣|[𝜀,1]𝐶0([𝜀,1]). From both relations we deduce that 𝑟𝑣𝐶0(Ω) for all 𝑣𝑉1.
Now, let the bilinear form 𝑎(,) be defined by𝑎(𝑢,𝑣)=𝑢(1)𝑣(1)+10𝑟𝑢𝑟(𝑟)𝑣𝑟(𝑟)𝑑𝑟,𝑢,𝑣𝑉1,(2.2) where is a positive constant. Then, there exists a unique bounded linear operator 𝐴𝑉1𝑉1 such that 𝑎(𝑢,𝑣)=𝐴𝑢,𝑣 for all 𝑢,𝑣𝑉1. We then have the following lemma.

Lemma 2.4. The symmetric bilinear form 𝑎(,) defined by (2.2) is continuous on 𝑉1×𝑉1 and coercive on 𝑉1, that is,(i)|𝑎(𝑢,𝑣)|𝐶1𝑢1𝑣1, (ii)𝑎(𝑣,𝑣)𝐶0𝑣21, for all 𝑢,𝑣𝑉1, where 𝐶0=(1/2)min{1,} and 𝐶1=1+(1+2).

The proof of Lemma 2.4 is straightforward and we omit it.

Lemma 2.5. There exists an orthonormal Hilbert basis {𝑤𝑗} of the space 𝑉0 consisting of eigenfunctions 𝑤𝑗 corresponding to eigenvalues 𝜆𝑗 such that(i)0<𝜆1𝜆2𝜆𝑗+as𝑗, (ii)𝑎(𝑤𝑗,𝑣)=𝜆𝑗𝑤𝑗,𝑣forall𝑣𝑉1and𝑗. Note that it follows from (ii) that {𝑤𝑗/𝜆𝑗} is automatically an orthonormal set in 𝑉1 with respect to 𝑎(,) as inner product. The eigensolutions 𝑤𝑗 are indeed eigensolutions for the boundary value problem 𝐴𝑤𝑗1𝑟𝑑𝑟𝑑𝑟𝑑𝑤𝑗𝑑𝑟=𝜆𝑗𝑤𝑗||||,inΩ,lim𝑟0+𝑟𝑑𝑤𝑗||||𝑑r(𝑟)<+,𝑑𝑤𝑗𝑑𝑟(1)+𝑤𝑗(1)=0.(2.3)

The proof of Lemma 2.5 can be found in ([18, page 87, Theorem 7.7]) with 𝑉=𝑉1,𝐻=𝑉0 and 𝑎(,) as defined by (2.2).

For any function 𝑣𝐶2(Ω) we define 𝑣2 as 𝑣2=𝑣20+𝑣𝑟20+𝐴𝑣201/2(2.4) and define the space 𝑉2 as completion of 𝐶2(Ω) with respect to the norm 2. Note that 𝑉2 is also a Hilbert space with respect to the scalar product 𝑢,𝑣+𝑢𝑟,𝑣𝑟+𝐴𝑢,𝐴𝑣(2.5) and that 𝑉2 can be defined also as 𝑉2={𝑣𝑉1𝐴𝑣𝑉0}.

We then have the following two lemmas the proof of which can be found in [1].

Lemma 2.6. The embedding 𝑉2𝑉1 is compact.

Lemma 2.7. For all 𝑣𝑉2 we have 𝑣(i)𝑟𝐿(Ω)12𝐴𝑣0,𝑣(ii)𝑟𝑟032𝐴𝑣0,(iii)𝑣2𝐿(Ω)2𝑣0+12𝐴𝑣0𝑣0.(2.6)

For a Banach space 𝑋, we denote by 𝑋 its norm, by 𝑋 its dual space, and by 𝐿𝑝(0,𝑇;𝑋),1𝑝 the Banach space of all real measurable functions 𝑢(0,𝑇)𝑋 such that 𝑢𝐿𝑝(0,𝑇;𝑋)=𝑇0𝑢(𝑡)𝑝𝑋𝑑𝑡1/𝑝<,for1𝑝<,𝑢𝐿(0,𝑇;𝑋)=esssup0<𝑡<𝑇𝑢(𝑡)𝑋for𝑝=.(2.7)

Let 𝑢(𝑡),𝑢(𝑡)=𝑢𝑡(𝑡)=̇𝑢(𝑡),𝑢(𝑡)=𝑢𝑡𝑡(𝑡)=̈𝑢(𝑡),𝑢𝑟(𝑡)=𝑢(𝑡),𝑢𝑟𝑟(𝑡)(2.8) denote 𝑢(𝑟,𝑡),𝜕𝑢𝜕𝜕𝑡(𝑟,𝑡),2𝑢𝜕𝑡2(𝑟,𝑡),𝜕𝑢𝜕𝜕𝑟(𝑟,𝑡),2𝑢𝜕𝑟2(𝑟,𝑡),(2.9) respectively.

With 𝑓𝐶𝑘(Ω×+×),𝑓=𝑓(𝑟,𝑡,𝑢), we put 𝐷1𝑓=𝜕𝑓/𝜕𝑟,𝐷2𝑓=𝜕𝑓/𝜕𝑡,𝐷3𝑓=𝜕𝑓/𝜕𝑢, and 𝐷𝛾𝑓=𝐷𝛾11𝐷𝛾22𝐷𝛾33𝑓,𝛾=(𝛾1,𝛾2,𝛾3)3+,|𝛾|=𝛾1+𝛾2+𝛾3=𝑘.

3. The Hight Order Iterative Schemes

Fix 𝑇>0, we make the following assumptions: (𝐻1)̃𝑢0𝑉2and̃𝑢1𝑉1; (𝐻2)𝐵𝐶1(+)andthereexistconstants𝑏>0,𝛼>1,𝑑0,𝑑1>0suchthat(i)𝑏𝐵(𝜂)𝑑0(1+𝜂𝛼),forall𝜂0, (ii)|𝐵(𝜂)|𝑑1(1+𝜂𝛼1),forall𝜂0; (𝐻3)𝑓𝐶𝑁(Ω×[0,𝑇]×)andsatisesthefollowingconditionforall𝑀>0, 𝐾(1)𝑖(𝑀,𝑓)=sup(𝑟,𝑡,𝑢)𝐴𝑀||||𝑟𝑖𝜕𝑖𝑓𝜕𝑢𝑖||||𝐾(𝑟,𝑡,𝑢)<+,𝑖=0,1,,𝑁1,(2)𝑖(𝑀,𝑓)=sup(𝑟,𝑡,𝑢)𝐴𝑀||||𝑟𝑖𝜕𝑖𝑓𝜕𝑟𝜕𝑢𝑖1||||(𝑟,𝑡,𝑢)<+,𝑖=1,,𝑁1,(3.1)

where 𝐴𝑀={(𝑟,𝑡,𝑢)[0,1]×[0,𝑇]×|𝑢|𝑀2+1/2}. We put 𝐾𝑖𝐾(𝑀,𝑓)=(1)0𝐾(𝑀,𝑓),𝑖=0,max(1)𝑖𝐾(𝑀,𝑓),(2)𝑖(𝑀,𝑓),𝑖=1,,𝑁1.(3.2)

With 𝐵 and 𝑓 satisfying assumptions (𝐻2) and (𝐻3), respectively, for each 𝑀>0 given, we introduce the following constants:𝐾𝑀(𝐵)=sup0𝜂𝑀2𝐵||𝐵(𝜂)+||,(𝜂)𝐾0(𝑀,𝑓)=sup(𝑟,𝑡,𝑢)𝐴𝑀||||,𝑓(𝑟,𝑡,𝑢)𝐾𝑁(𝑀,𝑓)=||𝛾||𝑁𝐾0(𝑀,𝐷𝛾𝑓).(3.3)

For each 𝑇(0,𝑇] and 𝑀>0 we get 𝑊(𝑀,𝑇)=𝑣𝐿0,𝑇;𝑉2𝑣𝐿0,𝑇;𝑉1,𝑣𝐿20,𝑇;𝑉0,with𝑣𝐿(0,𝑇;𝑉2),𝑣𝐿(0,𝑇;𝑉1),𝑣𝐿2(0,𝑇;𝑉0),𝑊𝑀1(𝑀,𝑇)=𝑣𝑊(𝑀,𝑇)𝑣𝐿0,𝑇;𝑉0.(3.4)

We will choose as first initial term 𝑢00, suppose that𝑢𝑚1𝑊1(𝑀,𝑇),(3.5) and associate with the problem (1.1) the following variational problem.

Find 𝑢𝑚𝑊1(𝑀,𝑇)(𝑚1) so that𝑢𝑚(𝑡),𝑣+𝑏𝑚𝑢(𝑡)𝑎𝑚(𝑡),𝑣=𝐹𝑚(𝑡),𝑣,𝑣𝑉1,𝑢𝑚(0)=̃𝑢0,𝑢𝑚(0)=̃𝑢1,(3.6) where𝑏𝑚(𝑡)=𝐵𝑢𝑚(𝑡)20,𝐹𝑚(𝑟,𝑡)=𝑁1𝑖=01𝐷𝑖!𝑖3𝑓𝑟,𝑡,𝑢𝑚1𝑢𝑚𝑢𝑚1𝑖.(3.7)

Then, we have the following theorem.

Theorem 3.1. Let assumptions (𝐻1)-(𝐻3) hold. Then there exist a constant 𝑀>0 depending on 𝑇,̃𝑢0,̃𝑢1,𝐵, and a constant 𝑇>0 depending on 𝑇,̃𝑢0,̃𝑢1,𝐵,,𝑓 such that, for 𝑢00, there exists a recurrent sequence {𝑢𝑚}𝑊1(𝑀,𝑇) defined by (3.6), (3.7).

Proof. The proof consists of several steps.
Step 1. The Faedo-Galerkin approximation (introduced by Lions [19]). Consider as in Lemma 2.5 the basis {𝑤𝑗} for 𝑉1 and put 𝑢𝑚(𝑘)(𝑡)=𝑘𝑗=1𝑐(𝑘)𝑚𝑗(𝑡)𝑤𝑗,(3.8) where the coefficients 𝑐(𝑘)𝑚𝑗 satisfy the system of the following nonlinear differential equations: ̈𝑢𝑚(𝑘)(𝑡),𝑤𝑗+𝑏𝑚(𝑘)𝑢(𝑡)𝑎𝑚(𝑘)(𝑡),𝑤𝑗=𝐹𝑚(𝑘)(𝑡),𝑤𝑗𝑢,1𝑗𝑘,𝑚(𝑘)(0)=̃𝑢0𝑘,̇𝑢𝑚(𝑘)(0)=̃𝑢1𝑘,(3.9) where ̃𝑢0𝑘=𝑘𝑗=1𝛼𝑗(𝑘)𝑤𝑗̃𝑢0stronglyin𝑉2,̃𝑢1𝑘=𝑘𝑗=1𝛽𝑗(𝑘)𝑤𝑗̃𝑢1stronglyin𝑉1.𝑏(3.10)𝑚(𝑘)(𝑡)=𝐵𝑢𝑚(𝑘)(𝑡)20,𝐹𝑚(𝑘)(𝑟,𝑡)=𝑁1𝑖=01𝐷𝑖!𝑖3𝑓𝑟,𝑡,𝑢𝑚1𝑢𝑚(𝑘)𝑢𝑚1𝑖=𝑁𝑗=0Ψ𝑗𝑟,𝑡,𝑢𝑚1𝑢𝑚(𝑘)𝑗,Ψ𝑗𝑟,𝑡,𝑢𝑚1=𝑁1𝑖=𝑗1𝑗!(𝑖𝑗)!(1)𝑖𝑗𝐷𝑖3𝑓𝑟,𝑡,𝑢𝑚1𝑢𝑖𝑗𝑚1,0𝑗𝑁1.(3.11)
Let us suppose that 𝑢𝑚1 satisfies (3.5). Then we have the following lemma.

Lemma 3.2. Let assumptions (𝐻1)-(𝐻3) hold. For fixed 𝑀>0 and 𝑇>0, then, the system (3.8)–(3.11) has a unique solution 𝑢𝑚(𝑘)(𝑡) on an interval [0,𝑇𝑚(𝑘)][0,𝑇].
Proof of Lemma 3.2. The system of (3.8)–(3.11) is rewritten in the form ̈𝑐(𝑘)𝑚𝑗(𝑡)=𝜆𝑗𝑏𝑚(𝑘)(𝑡)𝑐(𝑘)𝑚𝑗𝐹(𝑡)+𝑚(𝑘)(𝑡),𝑤𝑗𝑐,1𝑗𝑘,(𝑘)𝑚𝑗(0)=𝛼𝑗(𝑘),̇𝑐(𝑘)𝑚𝑗(0)=𝛽𝑗(𝑘),(3.12) and it is equivalent to the system of integral equations 𝑐(𝑘)𝑚𝑗(𝑡)=𝛼𝑗(𝑘)+𝛽𝑗(𝑘)𝑡𝜆𝑗𝑡0𝑑𝜏𝜏0𝑏𝑚(𝑘)(𝑠)𝑐(𝑘)𝑚𝑗(𝑠)𝑑𝑠+𝑡0𝑑𝜏𝜏0𝐹𝑚(𝑘)(𝑠),𝑤𝑗𝑑𝑠,(3.13) for 1𝑗𝑘. Omitting the index 𝑚, it is written as follows: [𝑐],𝑐=(3.14) where [𝑐]=(1[𝑐],,𝑘[𝑐]),𝑐=(𝑐1,,𝑐𝑘), 𝑗[𝑐](𝑡)=𝑞𝑗(𝑡)𝜆𝑗𝑡0𝑑𝜏𝜏0̃𝑏(𝑠)𝑐𝑗(+𝑠)𝑑𝑠𝑁1𝑖=1𝑡0𝑑𝜏𝜏0Ψ𝑖,𝑠,𝑢𝑚1𝑢𝑖(𝑠),𝑤𝑗𝑞𝑑𝑠,1𝑗𝑘,𝑗(𝑡)=𝛼𝑗(𝑘)+𝛽𝑗(𝑘)𝑡+𝑡0𝑑𝜏𝜏0Ψ0,𝑠,𝑢𝑚1,𝑤𝑗̃̃𝑏[𝑐]𝑑𝑠,1𝑗𝑘,𝑏(𝑡)=(𝑡)=𝐵𝑢(𝑡)20,𝑢(𝑡)=𝑘𝑗=1𝑐𝑗(𝑡)𝑤𝑗.(3.15)
For every 𝑇𝑚(𝑘)(0,𝑇] and 𝜌>0 that will be chosen later, we put 𝑋=𝐶0([0,𝑇𝑚(𝑘)];𝑘),𝑆={𝑐𝑌𝑐𝑋𝜌}, where 𝑐𝑋=sup0𝑡𝑇𝑚(𝑘)|𝑐(𝑡)|1,|𝑐(𝑡)|1=𝑘𝑗=1|𝑐𝑗(𝑡)|, for each 𝑐=(𝑐1,,𝑐𝑘)𝑋. Clearly 𝑆 is a closed nonempty subset in 𝑋, and we have the operator 𝑋𝑋. In what follows, we will choose 𝜌>0 and 𝑇𝑚(𝑘)>0 such that(i)𝑆ismappedintoitselfby, (ii)𝑆𝑆iscontractive.
Proof (i). First we note that, for all 𝑐=(𝑐1,,𝑐𝑘)𝑆, 𝑢(𝑡)=𝑘𝑗=1𝑐𝑗(𝑡)𝑤𝑗,𝑢(𝑡)0=𝑘𝑗=1𝑐2𝑗||||(𝑡)𝑐(𝑡)1,𝑢(𝑡)20𝑎(𝑢(𝑡),𝑢(𝑡))=𝑘𝑖,𝑗=1𝑐𝑖(𝑡)𝑐𝑗𝑤(𝑡)𝑎𝑖,𝑤𝑗=𝑘𝑗=1𝜆𝑗𝑐2𝑗(𝑡)𝜆𝑘𝑢(𝑡)20,𝑢(𝑡)0||||𝑐(𝑡)1𝑐𝑋𝜌,𝑢(𝑡)0𝜆𝑘||||𝑐(𝑡)1𝜆𝑘𝜌,𝑢(𝑡)11𝐶0𝑎(𝑢(𝑡),𝑢(𝑡))1𝐶0𝜆𝑘𝑢(𝑡)0𝜆𝑘𝐶0||||𝑐(𝑡)1𝜆𝑘𝐶0𝜌,(3.16) so ̃𝑏(𝑡)=𝐵𝑢(𝑡)20𝑑01+𝑢(𝑡)02𝛼𝑑01+𝜆𝑘2𝛼𝜌2𝛼.(3.17) On the other hand, by ||𝑢𝑚1||𝑀12+2||Ψ𝜃,0𝑟,𝑡,𝑢𝑚1||=|||||𝑁1𝑖=01𝑖!(1)𝑖𝐷𝑖3𝑓𝑟,𝑡,𝑢𝑚1𝑢𝑖𝑚1|||||𝑁1𝑖=0|||1𝐷𝑖!𝑖3𝑓𝑟,𝑡,𝑢𝑚1𝑢𝑖𝑚1|||𝐾𝑁(𝑀,𝑓)𝑁1𝑖=01𝑀𝑖!12+2𝑖=𝐾𝑁(𝑀,𝑓)𝑁1𝑖=0𝜃𝑖,𝑖!(3.18) we have ||Ψ0𝑡,𝑢𝑚1,𝑤𝑗||Ψ0𝑡,𝑢𝑚10𝑤𝑗0=Ψ0𝑡,𝑢𝑚1012𝐾𝑁(𝑀,𝑓)𝑁1𝑖=0𝜃𝑖.𝑖!(3.19) By Lemma 2.1, (iii), and the assumption (𝐻3), we deduce from (3.16) that ||Ψ𝑖𝑠,𝑢𝑚1𝑢𝑖(𝑠),𝑤𝑗||=|||||𝑁1𝑙=𝑖1𝑖!(𝑙𝑖)!(1)𝑙𝑖𝐷𝑙3𝑓𝑟,𝑡,𝑢𝑚1𝑢𝑙𝑖𝑚1𝑢𝑖(𝑠),𝑤𝑗|||||=|||||𝑁1𝑙=𝑖1𝑖!(𝑙𝑖)!(1)𝑙𝑖𝑟𝑙𝐷𝑙3𝑓𝑟,𝑠,𝑢𝑚1𝑟𝑙𝑖𝑢𝑙𝑖𝑚1𝑟𝑖𝑢𝑖(𝑠),𝑤𝑗|||||𝑁1𝑙=𝑖1𝐾𝑖!(𝑙𝑖)!𝑙(𝑀,𝑓)𝜃𝑙𝑖𝐾𝑖2𝑢(𝑠)𝑖1𝑟𝑙𝑖,||𝑤𝑗||𝑁1𝑙=𝑖1𝐾𝑖!(𝑙𝑖)!𝑙(𝑀,𝑓)𝜃𝑙𝑖𝐾𝑖2𝜆𝑘𝐶0𝜌𝑖12+𝑙𝑖𝑁1𝑙=𝑖11𝑖!(𝑙𝑖)!𝐾2+𝑙𝑖𝑙(𝑀,𝑓)𝜃𝑙𝑖𝐾𝑖2𝜆𝑘𝐶0𝜌𝑖,1𝑖𝑁1.(3.20)
It follows that||𝑗[𝑐](||||𝑞𝑡)𝑗(||𝑡)+𝜆𝑘𝑑01+𝜆𝑘2𝛼𝜌2𝛼𝑡0𝑑𝜏𝜏0||𝑐𝑗(||+1𝑠)𝑑𝑠2𝑡2𝑁1𝑖=1𝑁1𝑙=𝑖11𝑖!(𝑙𝑖)!𝐾2+𝑙𝑖𝑙(𝑀,𝑓)𝜃𝑙𝑖𝐾𝑖2𝜆𝑘𝐶0𝜌𝑖.(3.21) Thus ||[𝑐](||𝑡)1||||𝑞(𝑡)1+𝜆𝑘𝑑01+𝜆𝑘2𝛼𝜌2𝛼𝑡0𝑑𝜏𝜏0||||𝑐(𝑠)1+1𝑑𝑠2𝑘𝑡2𝑁1𝑖=1𝑁1𝑙=𝑖11𝑖!(𝑙𝑖)!𝐾2+𝑙𝑖𝑙(𝑀,𝑓)𝜃𝑙𝑖𝐾𝑖2𝜆𝑘𝐶0𝜌𝑖𝑞𝑇+12𝐷𝜌𝑇𝑚(𝑘)2,(3.22) where 𝑞𝑇=sup0𝑡𝑇|𝑞(𝑡)|1 and 𝐷𝜌=𝐷𝜌(𝑓,𝜌,𝑘,𝑀,𝑚,𝑁)=𝜆𝑘𝑑01+𝜆𝑘2𝛼𝜌2𝛼𝜌+𝑘𝑁1𝑖=1𝑁1𝑙=𝑖11𝑖!(𝑙𝑖)!𝐾2+𝑙𝑖𝑙(𝑀,𝑓)𝜃𝑙𝑖𝐾𝑖2𝜆𝑘𝐶0𝜌𝑖.(3.23)
Hence, we obtain[𝑐]𝑋𝑞𝑇+12𝐷𝜌𝑇𝑚(𝑘)2,(3.24)
choosing 𝜌>𝑞𝑇 and𝑇𝑚(𝑘)(0,𝑇], such that12𝐷𝜌𝑇𝑚(𝑘)2𝜌𝑞𝑇,12𝐷𝜌𝑇𝑚(𝑘)2<1,(3.25) where 𝐷𝜌=𝐷𝜌(𝜌,𝑘,𝑀,𝑇,𝑚,𝑁,𝑓)𝑑0𝜆𝑘1+𝜆𝑘2𝛼𝜌2𝛼+2𝑑1𝜆2𝑘𝜌21+𝜆𝑘𝛼1𝜌2𝛼2+𝑘𝑁1𝑖=1𝑖𝜆𝑘𝐶0𝐾2𝑖𝜌𝑖1𝑁1𝑙=𝑖11𝑖!(𝑙𝑖)!𝐾2+𝑙𝑖𝑙(𝑀,𝑓)𝜃𝑙𝑖.(3.26) Then [𝑐]𝑋𝑞𝑇+12𝐷𝜌𝑇𝑚(𝑘)2𝜌,𝑐𝑆,(3.27) which means that maps 𝑆 into itself.

Proof (ii). We now prove that, for all 𝑐,𝑑𝑆, for all 𝑡[0,𝑇𝑚(𝑘)], ||[𝑐][𝑑]||(𝑡)(𝑡)112𝐷𝜌𝑡2𝑐𝑑𝑋,𝑛,(3.28) where 𝐷𝜌 is defined as (3.26).
Proof of (3.28) is as follows.
For all 𝑗=1,2,,𝑘, for all 𝑡[0,𝑇𝑚(𝑘)], we have||𝑗[𝑐](𝑡)𝑗[𝑑](||𝑡)𝜆𝑗𝑡0𝑑𝜏𝜏0||̃𝑏[𝑐](𝑐𝑠)𝑗(𝑠)𝑑𝑗(||𝑠)𝑑𝑠+𝜆𝑗𝑡0𝑑𝜏𝜏0||̃𝑏[𝑐]̃𝑏[𝑑]𝑑(𝑠)(𝑠)𝑗||+(𝑠)𝑑𝑠𝑁1𝑖=1𝑡0𝑑𝜏𝜏0||Ψ𝑖𝑠,𝑢𝑚1𝑢𝑖(𝑠)𝑣𝑖(𝑠),𝑤𝑗||𝑑𝑠,(3.29) where ̃𝑏[𝑐](𝑡)=𝐵𝑢(𝑡)20,̃𝑏[𝑑](𝑡)=𝐵𝑣(𝑡)20,𝑢(𝑡)=𝑘𝑗=1𝑐𝑗(𝑡)𝑤𝑗,𝑣(𝑡)=𝑘𝑗=1𝑑𝑗(𝑡)𝑤𝑗,(3.30) so ||[𝑐]([𝑑](||𝑡)𝑡)1𝜆𝑘𝑡0𝑑𝜏𝜏0̃𝑏[𝑐](||||𝑠)𝑐(𝑠)𝑑(𝑠)1𝑑𝑠+𝜆𝑘𝑡0𝑑𝜏𝜏0||̃𝑏[𝑐]̃𝑏[𝑑]||||||(𝑠)(𝑠)𝑑(𝑠)1+𝑑𝑠𝑘𝑗=1𝑁1𝑖=1𝑡0𝑑𝜏𝜏0||Ψ𝑖𝑠,𝑢𝑚1𝑢𝑖(𝑠)𝑣𝑖(𝑠),𝑤𝑗||𝑑𝑠𝜆𝑘𝑡0𝑑𝜏𝜏0̃𝑏[𝑐]||||(𝑠)𝑐(𝑠)𝑑(𝑠)1𝑑𝑠+𝜆𝑘𝜌𝑡0𝑑𝜏𝜏0||̃𝑏[𝑐]̃𝑏[𝑑]||+(𝑠)(𝑠)𝑑𝑠𝑘𝑗=1𝑁1𝑖=1𝑡0𝑑𝜏𝜏0||Ψ𝑖𝑠,𝑢𝑚1𝑢𝑖(𝑠)𝑣𝑖(𝑠),𝑤𝑗||𝑑𝑠𝐽1+𝐽2+𝐽3,(3.31) in which 𝐽1=𝜆𝑘𝑡0𝑑𝜏𝜏0̃𝑏[𝑐](||||𝑠)𝑐(𝑠)𝑑(𝑠)1𝑑𝑠𝜆𝑘𝑑01+𝜆𝑘2𝛼𝜌2𝛼𝑡0𝑑𝜏𝜏0||||𝑐(𝑠)𝑑(𝑠)1𝑑𝑠𝜁1𝑡0𝑑𝜏𝜏0||||𝑐(𝑠)𝑑(𝑠)1𝑑𝑠.(3.32) In order to consider 𝐽2, we also note that ̃𝑏[𝑐]̃𝑏[𝑑](𝑠)(𝑠)=𝐵(𝜉)𝑢(𝑠)20𝑣(𝑠)20,(3.33) where 𝜉=𝜃𝑢(𝑠)20+(1𝜃)𝑣(𝑠)20,0𝜉𝜆𝑘𝜌2,0<𝜃<1,(3.34) and 𝐵(𝜉) satisfy the following inequality: ||𝐵||(𝜉)𝑑11+𝜉𝛼1𝑑11+𝜆𝑘𝛼1𝜌2𝛼2.(3.35)
It implies that||̃𝑏[𝑐]̃𝑏[𝑑]||=||𝐵(𝑠)(𝑠)(𝜉)𝑢(𝑠)20𝑣(𝑠)20||𝑑11+𝜆𝑘𝛼1𝜌2𝛼22𝜆𝑘𝜌||||𝑐(𝑠)𝑑(𝑠)1=2𝜆𝑘𝜌𝑑11+𝜆𝑘𝛼1𝜌2𝛼2||||𝑐(𝑠)𝑑(𝑠)1,(3.36) and then 𝐽2=𝜆𝑘𝜌𝑡0𝑑𝜏𝜏0||̃𝑏[𝑐](̃𝑏[𝑑](||𝑠)𝑠)𝑑𝑠2𝜆2𝑘𝜌2𝑑11+𝜆𝑘𝛼1𝜌2𝛼2𝑡0𝑑𝜏𝜏0||||𝑐(𝑠)𝑑(𝑠)1𝑑𝑠𝜁2𝑡0𝑑𝜏𝜏0||||𝑐(𝑠)𝑑(𝑠)1𝑑𝑠.(3.37)
It remains to estimate 𝐽3. By𝑟𝑢(𝑠)𝑖𝑟𝑣(𝑠)𝑖=𝑖1𝑗=0𝑟𝑢(𝑠)𝑗𝑟𝑣(𝑠)𝑖𝑗1𝑟(𝑢(𝑠)𝑣(𝑠)),(3.38) we obtain |||𝑟𝑢(𝑠)𝑖𝑟𝑣(𝑠)𝑖|||𝐾𝑖2𝑖1𝑗=0𝑢(𝑠)𝑗1𝑣(𝑠)1𝑖𝑗1𝑢(𝑠)𝑣(𝑠)1𝐾𝑖2𝑖1𝑗=0𝑢(𝑠)𝑗1𝑣(𝑠)1𝑖𝑗1𝑢(𝑠)𝑣(𝑠)1𝐾𝑖2𝑖1𝑗=0𝜆𝑘𝐶0𝜌𝑗𝜆𝑘𝐶0𝜌𝑖𝑗1𝜆𝑘𝐶0||||𝑐(𝑠)𝑑(𝑠)1=𝐾𝑖2𝑖1𝑗=0𝜆𝑘𝐶0𝑖𝜌𝑖1||||𝑐(𝑠)𝑑(𝑠)1=𝑖𝜆𝑘𝐶0𝐾2𝑖𝜌𝑖1||||𝑐(𝑠)𝑑(𝑠)1.(3.39) On the other hand, Ψ𝑖𝑟,𝑡,𝑢𝑚1=𝑁1𝑙=𝑖1𝑖!(𝑙𝑖)!(1)𝑙𝑖𝐷𝑙3𝑓𝑟,𝑡,𝑢𝑚1𝑢𝑙𝑖𝑚1,𝑟𝑖||Ψ𝑖𝑟,𝑡,𝑢𝑚1||=|||||𝑁1𝑙=𝑖1(𝑖!(𝑙𝑖)!1)𝑙𝑖𝑟𝑙𝐷𝑙3𝑓𝑟,𝑡,𝑢𝑚1𝑟𝑙𝑖𝑢𝑙𝑖𝑚1||||||||||𝑁1𝑙=𝑖1𝑖!(𝑙𝑖)!𝑟𝑙||𝐷𝑙3𝑓𝑟,𝑡,𝑢𝑚1||𝑟𝑙𝑖||𝑢𝑙𝑖𝑚1|||||||𝑁1𝑙=𝑖1𝐾𝑖!(𝑙𝑖)!𝑙(𝑀,𝑓)𝜃𝑙𝑖𝑟𝑙𝑖.(3.40)
Hence, we deduce from (3.39), (3.40) that𝐽3=𝑘𝑗=1𝑁1𝑖=1𝑡0𝑑𝜏𝜏0||||𝑟𝑖Ψ𝑖𝑠,𝑢𝑚1𝑟𝑢(𝑠)𝑖𝑟𝑣(𝑠)𝑖,𝑤𝑗||||𝑑𝑠𝑘𝑗=1𝑁1𝑖=1𝑁1𝑙=𝑖1𝐾𝑖!(𝑙𝑖)!𝑙(𝑀,𝑓)𝜃𝑙𝑖𝑖𝜆𝑘𝐶0𝐾2𝑖𝜌𝑖1×𝑡0𝑑𝜏𝜏0||||𝑟𝑙𝑖,||𝑤𝑗||||||||||𝑐(𝑠)𝑑(𝑠)1=𝑑𝑠𝑘𝑗=1𝑁1𝑖=1𝑁1𝑙=𝑖1𝐾𝑖!(𝑙𝑖)!𝑙(𝑀,𝑓)𝜃𝑙𝑖𝑖𝜆𝑘𝐶0𝐾2𝑖𝜌𝑖11×2+𝑙𝑖𝑡0𝑑𝜏𝜏0||||𝑐(𝑠)𝑑(𝑠)1𝑑𝑠=𝑘𝑁1𝑖=1𝑖𝜆𝑘𝐶0𝐾2𝑖𝜌𝑖1𝑁1𝑙=𝑖1𝐾𝑖!(𝑙𝑖)!𝑙(𝑀,𝑓)𝜃𝑙𝑖1×2+𝑙𝑖𝑡0𝑑𝜏𝜏0||||𝑐(𝑠)𝑑(𝑠)1𝑑𝑠𝜁3𝑡0𝑑𝜏𝜏0||||𝑐(𝑠)𝑑(𝑠)1𝑑𝑠.(3.41)
We deduce that||[𝑐]([𝑑](||𝑡)𝑡)1𝜁1+𝜁2+𝜁3𝑡0𝑑𝜏𝜏0||||𝑐(𝑠)𝑑(𝑠)11𝑑𝑠2𝐷𝜌𝑡2𝑐𝑑𝑋.(3.42) We note that 𝜁1+𝜁2+𝜁3=𝐷𝜌𝐷(𝜌,𝑘,𝑀,𝑇,𝑚,𝑁,𝑓)=𝜌.(3.43)
It follows from (3.28) that[𝑐][𝑑]𝑋12𝐷𝜌𝑇𝑚(𝑘)2𝑐𝑑𝑋,𝑐,𝑑𝑆.(3.44)
By (3.25), it follows that 𝑆𝑆 is contractive. We deduce that has a unique fixed point in 𝑆; that is, the system (3.8)–(3.11) has a unique solution 𝑢𝑚(𝑘)(𝑡) on an interval [0,𝑇𝑚(𝑘)]. The proof of Lemma 3.2 is complete.

The following estimates allow one to take constant 𝑇𝑚(𝑘)=𝑇 for all 𝑚 and 𝑘.
Step 2. A priori estimates. Put 𝑆𝑚(𝑘)(𝑡)=𝑋𝑚(𝑘)(𝑡)+𝑌𝑚(𝑘)(𝑡)+𝑡0̈𝑢𝑚(𝑘)(𝑠)20𝑑𝑠,(3.45) where 𝑋𝑚(𝑘)(𝑡)=̇𝑢𝑚(𝑘)(𝑡)20+𝑏𝑚(𝑘)𝑢(𝑡)𝑎𝑚(𝑘)(𝑡),𝑢𝑚(𝑘),𝑌(𝑡)𝑚(𝑘)(𝑡)=𝑎̇𝑢𝑚(𝑘)(𝑡),̇𝑢𝑚(𝑘)(𝑡)+𝑏𝑚(𝑘)(𝑡)𝐴𝑢𝑚(𝑘)(𝑡)20,(3.46) with 𝐴 is defined by (2.2). Then it follows that 𝑆𝑚(𝑘)(𝑡)=𝑆𝑚(𝑘)(0)+𝑡0̇𝑏𝑚(𝑘)(𝑎𝑢𝑠)𝑚(𝑘)(𝑠),𝑢𝑚(𝑘)(+𝑠)𝐴𝑢𝑚(𝑘)(𝑠)20𝑑𝑠+2𝑡0𝐹𝑚(𝑘)(𝑠),̇𝑢𝑚(𝑘)(𝑠)𝑑𝑠+2𝑡0𝑎𝐹𝑚(𝑘)(𝑠),̇𝑢𝑚(𝑘)+(𝑠)𝑑𝑠𝑡0𝐹𝑚(𝑘)(𝑠),̈𝑢𝑚(𝑘)(𝑠)𝑑𝑠𝑡0𝑏𝑚(𝑘)(𝑠)𝐴𝑢𝑚(𝑘)(𝑠),̈𝑢𝑚(𝑘)(𝑠)𝑑𝑠𝑆𝑚(𝑘)(0)+5𝑗=1𝐼𝑗.(3.47)
We will now require the following lemma.

Lemma 3.3. We have (i)0<𝑏𝑏𝑚(𝑘)(𝑡)𝑑01+𝑢𝑚(𝑘)(𝑡)02𝛼,||̇𝑏(ii)𝑚(𝑘)||(𝑡)2𝑑1𝑏𝑆𝑚(𝑘)(𝑡)+𝑏1𝛼𝑆𝑚(𝑘)(𝑡)𝛼,𝐹(iii)𝑚(𝑘)(𝑡)0𝑁1𝑗=0̃𝑎𝑗(0)𝑆𝑚(𝑘)(𝑡)𝑗,(𝜕iv)𝐹𝜕𝑟𝑚(𝑘)(𝑡)0𝑁1𝑗=0̃𝑎𝑗(1)𝑆𝑚(𝑘)(𝑡)𝑗,(3.48) where ̃𝑎𝑗(0),̃𝑎𝑗(1),𝑗=0,1,,𝑁1 are defined as follows: ̃𝑎𝑗(0)=̃𝑎0(0)=12𝐾𝑁(𝑀,𝑓)𝑁1𝑖=0𝜃𝑖𝑖!,𝑗=0,̃𝑎𝑗(0)=𝐾𝑗2𝑏𝐶0𝑗𝑁1𝑖=𝑗𝜃𝑖𝑗1𝑗!(𝑖𝑗)!𝐾2+𝑖𝑗𝑖(𝑀,𝑓),1𝑗𝑁1,̃𝑎𝑗(1)=12+2𝑀𝐾𝑁(𝑀,𝑓)𝑁1𝑖=0𝜃𝑖𝐾𝑖!,𝑗=0,𝑗2𝑏𝐶0𝑗𝑁1𝑖=𝑗𝜃𝑖𝑗𝑗!(𝑖𝑗)!𝑗𝐾21𝐾𝑖+1(𝑀,𝑓)𝐾𝑖𝑗+3+2𝑀𝑖+1(𝑀,𝑓),1𝑗𝑁1,𝜃=𝑀12+2.(3.49)
Proof of Lemma 3.3. Proof (i), (ii). Note that𝑆𝑚(𝑘)(𝑡)𝑋𝑚(𝑘)(𝑡)𝑏𝑎𝑢𝑚(𝑘)(𝑡),𝑢𝑚(𝑘)(𝑡)𝑏𝑢𝑚(𝑘)(𝑡)20,𝑆𝑚(𝑘)(𝑡)𝑌𝑚(𝑘)(𝑡)𝑎̇𝑢𝑚(𝑘)(𝑡),̇𝑢𝑚(𝑘)(𝑡)̇𝑢𝑚(𝑘)(𝑡)20.(3.50) We deduce that 𝑏𝑚(𝑘)(𝑡)=𝐵𝑢𝑚(𝑘)(𝑡)20𝑑01+𝑢𝑚(𝑘)(𝑡)02𝛼,||̇𝑏𝑚(𝑘)|||||𝐵(𝑡)=2𝑢𝑚(𝑘)(𝑡)20||||||𝑢𝑚(𝑘)(𝑡),̇𝑢𝑚(𝑘)|||(𝑡)2𝑑11+𝑢𝑚(𝑘)(𝑡)02𝛼2𝑢𝑚(𝑘)(𝑡)0̇𝑢𝑚(𝑘)(𝑡)02𝑑11+𝑏1𝛼𝑆𝑚(𝑘)(𝑡)𝛼11𝑏𝑆𝑚(𝑘)=(𝑡)2𝑑1𝑏𝑆𝑚(𝑘)(𝑡)+𝑏1𝛼𝑆𝑚(𝑘)(𝑡)𝛼.(3.51)
Proof (iii). We have 𝐹𝑚(𝑘)(𝑡)0Ψ0𝑟,𝑡,𝑢𝑚10+𝑁1𝑗=1Ψ𝑗𝑟,𝑡,𝑢𝑚1𝑢𝑚(𝑘)𝑗0.(3.52) By (3.18)3, we have Ψ0𝑟,𝑡,𝑢𝑚1012𝐾𝑁(𝑀,𝑓)𝑁1𝑖=0𝜃𝑖𝑖!̃𝑎0(0).(3.53) On the other hand, it follows from (3.49) and 𝑢𝑚1𝑊1(𝑀,𝑇) that Ψ𝑗𝑟,𝑡,𝑢𝑚1𝑢𝑚(𝑘)𝑗0𝑁1𝑖=𝑗1𝐷𝑗!(𝑖𝑗)!𝑖3𝑓𝑟,𝑡,𝑢𝑚1𝑢𝑖𝑗𝑚1𝑢𝑚(𝑘)𝑗0=𝑁1𝑖=𝑗1𝑗!(𝑖𝑗)!𝑟𝑖𝑗𝑟𝑖𝐷𝑖3𝑓𝑟,𝑡,𝑢𝑚1𝑢𝑖𝑗𝑚1𝑟𝑗𝑢𝑚(𝑘)𝑗0𝑁1𝑖=𝑗1𝑗!(𝑖𝑗)!𝑟𝑖𝑗0𝐾𝑖(𝑀,𝑓)𝜃𝑖𝑗𝐾𝑗2𝑢𝑚(𝑘)(𝑡)𝑗1=𝑁1𝑖=𝑗11𝑗!(𝑖𝑗)!𝐾2+𝑖𝑗𝑖(𝑀,𝑓)𝜃𝑖𝑗𝐾𝑗2𝑢𝑚(𝑘)(𝑡)𝑗1𝐾𝑗2𝑏𝐶0𝑗𝑁1𝑖=𝑗𝜃𝑖𝑗1𝑗!(𝑖𝑗)!𝐾2+𝑖𝑗𝑖(𝑀,𝑓)𝑆𝑚(𝑘)(𝑡)𝑗̃𝑎𝑗(0)𝑆𝑚(𝑘)(𝑡)𝑗.(3.54)
It follows from (3.52)–(3.54) that 𝐹𝑚(𝑘)(𝑡)0𝑁1𝑗=0̃𝑎𝑗(0)𝑆𝑚(𝑘)(𝑡)𝑗,(3.55) where ̃𝑎𝑗(0),0𝑗𝑁1 are defined by (3.49)1.

Proof (iv). We have 𝜕𝐹𝜕𝑟𝑚(𝑘)𝜕(𝑟,𝑡)=Ψ𝜕𝑟0𝑟,𝑡,𝑢𝑚1+𝑁1𝑗=1𝑗Ψ𝑗𝑟,𝑡,𝑢𝑚1𝑢𝑚(𝑘)𝑗1𝑢𝑚(𝑘)+𝑁1𝑗=1𝜕Ψ𝜕𝑟𝑗𝑟,𝑡,𝑢𝑚1𝑢𝑚(𝑘)𝑗.(3.56)
Hence𝜕𝐹𝜕𝑟𝑚(𝑘)(𝑡)0𝜕Ψ𝜕𝑟0𝑟,𝑡,𝑢𝑚10+𝑁1𝑗=1𝑗Ψ𝑗𝑟,𝑡,𝑢𝑚1𝑢𝑚(𝑘)𝑗1𝑢𝑚(𝑘)0+𝑁1𝑗=1𝜕Ψ𝜕𝑟𝑗𝑟,𝑡,𝑢𝑚1𝑢𝑚(𝑘)𝑗0𝐿1+𝐿2+𝐿3.(3.57) We shall estimate step by step the terms on the right-hand side of (3.57) as follows.(iv.1) Estimating 𝐿1=(𝜕/𝜕𝑟)Ψ0(𝑟,𝑡,𝑢𝑚1)0. We have𝜕Ψ𝜕𝑟0𝑟,𝑡,𝑢𝑚1=𝑁1𝑖=11𝑖!(1)𝑖𝐷𝑖3𝑓𝑟,𝑡,𝑢𝑚1𝑖𝑢𝑖1𝑚1𝑢𝑚1+𝑁1𝑖=01𝑖!(1)𝑖𝐷1𝐷𝑖3𝑓𝑟,𝑡,𝑢𝑚1𝑢𝑖𝑚1+𝑁1𝑖=01𝑖!(1)𝑖𝐷3𝑖+1𝑓𝑟,𝑡,𝑢𝑚1𝑢𝑖𝑚1𝑢𝑚1=𝑎(1)+𝑎(2)+𝑎(3).(3.58) We will estimate step by step the terms 𝑎(1),𝑎(2),𝑎(3) as follows.(iv.1.1) Estimating 𝑎(1)0. We have||||=|||||𝑎(1)𝑁1𝑖=11𝑖!(1)𝑖𝐷𝑖3𝑓𝑟,𝑡,𝑢𝑚1𝑖𝑢𝑖1𝑚1𝑢𝑚1|||||𝐾𝑁(𝑀,𝑓)𝑁1𝑖=11𝑖!𝑖𝜃𝑖1||𝑢𝑚1||=𝐾𝑁(𝑀,𝑓)𝑁2𝑖=0𝜃𝑖||𝑖!𝑢𝑚1||𝐾𝑁(𝑀,𝑓)𝑁1𝑖=0𝜃𝑖||𝑖!𝑢𝑚1||.(3.59) Hence 𝑎(1)0𝐾𝑁(𝑀,𝑓)𝑁1𝑖=0𝜃𝑖𝑖!𝑢𝑚10𝑀𝐾𝑁(𝑀,𝑓)𝑁1𝑖=0𝜃𝑖.𝑖!(3.60)(iv.1.2) Estimating 𝑎(2)0. It follows from||||=|||||𝑎(2)𝑁1𝑖=01𝑖!(1)𝑖𝐷1𝐷𝑖3𝑓𝑟,𝑡,𝑢𝑚1𝑢𝑖𝑚1|||||𝐾𝑁(𝑀,𝑓)𝑁1𝑖=0𝜃𝑖𝑖!(3.61) that 𝑎(2)012𝐾𝑁(𝑀,𝑓)𝑁1𝑖=0𝜃𝑖.𝑖!(3.62)(iv.1.3) Estimating 𝑎(3)0. Similarly, with ||||=|||||𝑎(3)𝑁1𝑖=01𝑖!(1)𝑖𝐷3𝑖+1𝑓𝑟,𝑡,𝑢𝑚1𝑢𝑖𝑚1𝑢𝑚1|||||𝐾𝑁(𝑀,𝑓)𝑁1𝑖=0𝜃𝑖||𝑖!𝑢𝑚1||,(3.63) we obtain𝑎(3)0𝐾𝑁(𝑀,𝑓)𝑁1𝑖=0𝜃𝑖𝑖!𝑢𝑚10𝑀𝐾𝑁(𝑀,𝑓)𝑁1𝑖=0𝜃𝑖.𝑖!(3.64)
It follows from (3.58), (3.60), (3.62), (3.64) that 𝐿1=𝜕Ψ𝜕𝑟0𝑟,𝑡,𝑢𝑚10𝑎(1)0+𝑎(2)0+𝑎(3)012𝐾𝑁(𝑀,𝑓)𝑁1𝑖=0𝜃𝑖𝑖!+2𝑀𝐾𝑁(𝑀,𝑓)𝑁1𝑖=0𝜃𝑖1𝑖!2+2𝑀𝐾𝑁(𝑀,𝑓)𝑁1𝑖=0𝜃𝑖.𝑖!(3.65)
(iv.2) Estimating 𝐿2=𝑁1𝑗=1𝑗Ψ𝑗(𝑟,𝑡,𝑢𝑚1)(𝑢𝑚(𝑘))𝑗1𝑢𝑚(𝑘)0. By the assumption (𝐻3), we deduce that𝐿2=𝑁1𝑗=1𝑗Ψ𝑗𝑟,𝑡,𝑢𝑚1𝑢𝑚(𝑘)𝑗1𝑢𝑚(𝑘)0𝑁1𝑗=1𝑗𝑁1𝑖=𝑗1𝐷𝑗!(𝑖𝑗)!𝑖3𝑓𝑟,𝑡,𝑢𝑚1𝑢𝑖𝑗𝑚1𝑢𝑚(𝑘)𝑗1𝑢𝑚(𝑘)0=𝑁1𝑗=1𝑗𝑁1𝑖=𝑗1𝑗!(𝑖𝑗)!𝑟𝑖𝐷𝑖3𝑓𝑟,𝑡,𝑢𝑚1𝑟𝑖𝑗𝑢𝑖𝑗𝑚1𝑟𝑢𝑚(𝑘)𝑗1𝑟𝑢𝑚(𝑘)0=𝑁1𝑗=1𝑗𝑁1𝑖=𝑗1𝐾𝑗!(𝑖