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Results on existence for generalized nD Navier-Stokes equations

  • Khaled Zennir EMAIL logo
Published/Copyright: December 31, 2019

Abstract

In this paper we consider a class of nD Navier-Stokes equations of Kirchhoff type and prove the global existence of solutions by using a new approach introduced in [Jday R., Zennir Kh., Georgiev S.G., Existence and smoothness for new class of n-dimentional Navier-Stokes equations, Rocky Mountain J. Math., 2019, 49(5), 1595–1615].

MSC 2010: 35Q30; 76D05; 46E35; 35B65; 35K55

1 Introduction

In this article we investigate a Navier-Stokes equations of Kirchhoff type

uit+j=1nujuixj(m+aM(Rn|A1/2ui|2dx))Aui+pxi=fi(t,x),i=1nuixi=0,xRn,t>0, (1.1)

for i ∈ {1, …, n}, subject to the initial condition

u(0,x)=u0(x),xRn, (1.2)

where a > 0, n ≥ 2 and A is a self-adjoint non-positive operator. The function M satisfies

(Hyp0)m0>0:MC1(R+) and M(λ)>m0,λ0,γ,δ>0:M(λ)γRn|u0(x)|2δdx,λ>0.

Here u0(x) = (u10(x), …, un0(x)) is a given 𝓒 divergence-free vector field on ℝn, fi(t, x), i = 1, …, n, are the components of a given externally applied force, m is a positive constant, w = (w1, …, wn) and p are independent unknown. In the case when m = 0, the equations (1.1), (1.2) are well known as the Euler equations. The first equation of (1.1) is inspired from the Newton law for a fluid element subject to the external force f = (f1, …, fn) and to the forces arising from pressure and friction. The second equation of (1.1) says that the fluid is incompressible. (See [1, 2])

By the last equation of the system (1.1), we get

uji=1nuixi=0foranyj{1,,n}.

Then

j=1nujuixj=j=1nxjujuiuij=1nujxj=j=1nxjujui.

In the case a = 0, system (1.1) has been treated in [3]. Therefore the system (1.1), for the case a = 1 can be rewritten for i ∈ {1, …, n} and A = –Δ, as

uit+j=1nxjuiuj(m+M(j=1n|uixj|2))j=1n2xj2ui+pxi=fi(t,x),i=1nuixi=0,(t,x)(0,)×Rn. (1.3)

There is a large literature regarding this type of problem. It is stated an open problem for (1.1), (1.2): if n = 3 and u0(x) be any smooth, divergence-free vector field satisfying the condition

xαu0(x)Cα,K1+|x|KonRn, (1.4)

for any α and K, and f to be identically equal to zero, then there exist smooth functions p, ui, i = 1, …, n, on ℝn × [0, ∞) that satisfy (1.1), (1.2) and p, u ∈ 𝓒(ℝn × [0, ∞)), and the energy is bounded for all t ≥ 0. (See [4, 5, 6, 7].)

In this paper we extend the previous works to find for any n ≥ 3 a new class of smooth initial data u0 satisfying (1.4) and a new class of functions fi, including fi = 0, i = 1, …, n, such that the problem (1.1), (1.2) has a solution p, u ∈ 𝓒(ℝn × [0, ∞)).

This kind of systems appears in the models of nonlinear Kirchhoff-type. It is a generalization of a model introduced by Kirchhoff [8] in the case n = 1 this type of problem describes a small amplitude vibration of an elastic string. The original equation is:

ρhutt+τut=P0+Eh2L0L|ux(x,t)|2dsuxx+f, (1.5)

where 0 ≤ xL, t > 0 and

u is the lateral deflection x is the space coordenate variable whiletdenotes the time variable E is the Young modulus ρ is the mass density L is the lenght h is the cross section area P0 is the initial axial tension τ is the resistence modulus f is the external force  (1.6)

(For more see [9]). Here we assume that the initial data u0 and the force term f are as follows

(Hyp1)ui0C0(Rn),suppui0B,Rn|u0(x)|κdxnQκμ(B),κ2,xαui0(x)Cα,K1+|x|KonRn,i{1,,n},

for any multi-index α and for any positive constant K, where Cα,K is a positive constant depending on α and K, Q is a positive constant such that

|u0(x)|Q,|u0xi(x)|Q,|u0xixj(x)|Q,xB,i,j{1,,n},

μ(B) is the meassure of B,

(Hyp2)fiC[0,),C0(Rn),suppxfiB,tmxαfi(t,x)Cα,m,K1+|x|+tKon[0,)×Rn,i{1,,n},

for any α, m, K, where B is a compact subset of ℝn, respectively.

Several numerical methods for solving of (1.1) are used. In [10] Lagrangian and semi-Lagrangian velocity and displacement methods are introduced for the numerical solution of (1.1). In the scalar case, methods of characteristics for time discretization of convention diffusion problems are extensively used (see [11] and references therein). These methods are based on time discretization of the material time derivative combined with finite differences or finite elements for space discretization. When the characteristic methods are formulated in a fixed reference domain they are called pure Lagrangian methods. The classical methods of characteristics are semi-Lagrangian and first-order in time. There exists an extensive literature for these methods (see [12, 13] and references therein). The error estimates of the norm O(hk) + O(Δt) + O(hk+1/δt) in l(L2(Ω))- norm are obtained under the assumption that the normal velocity vanishes on the boundary Ω, where h is the space step and δt is the time step (See [14, 15, 16]).

In order to increase the order of time and space approximations, higher order schemes for the discretization of the material derivative and higher order finite element spaces are used(see [17, 18, 19] and references therein). Second order characteristic method for solving of (1.1) is used in [20, 21].

In this paper we propose a new method for investigation of the Cauchy problem (1.1), (1.2) which is different than the well-known methods. In Section 2 we give some auxiliary results. In Section 3 we proof the main result introduced in Theorem (3.1)

2 Preliminaries

We will start with the following useful Lemma.

Lemma 2.1

Let x0=(x10,,xn0)B and fi satisfies (Hyp2), i ∈ {1, …, n}. If p and ui ∈ 𝓒1([a, b], C02 (ℝn)), suppx uiB, suppxpB, i ∈ {1, …, n}, satisfy the system

x0xx0sui(t,σ)gi(σ)dσds+j=1natx0xx¯j0s¯juiτ,σ~sjujτ,σ~sjdσ¯jdsdτ(m+M(j=1n|uixj|2))j=1natx¯j0x¯jx¯j0s¯juiτ,σ~xjdσ¯jds¯jdτ+atx0xx¯i0s¯ipτ,σ~sidσ¯idsdτ=atx0xx0sfi(τ,σ)dσdsdτ,i{1,,n},j=1natx0xx¯j0s¯jujτ,σ~sjdσ¯jdsdτ=0,t[a,b],xRn, (2.1)

then p and ui, i ∈ {1, …, n}, satisfy the problem

uit+j=1nxjuiuj(m+M(j=1n|uixj|2))j=1n2xj2ui+pxi=fi(t,x),i=1nuixi=0,(t,x)[a,b]×Rn,u(a,x)=g(x),g(x)=(g1(x),,gn(x)),xRn (2.2)

where gi C02 (ℝn), supp giB, i ∈ {1, …, n}.

Here

x0x=x10x1xn0xn,dσ=dσndσ1,x¯j0s¯j=x10s1xj10sj1xj+10sj+1xn0sn,x¯j0x¯j=x10x1xj10xj1xj+10xj+1xn0xn,σ~sj=σ1,,σj1,sj,σj+1,,σn,dσ¯j=dσndσj+1dσj1dσ1,σ~xj=σ1,,σj1,xj,σj+1,,σn,ds¯j=dsndsj+1dsj1ds1,σ~si=σ1,,σi1,si,σi+1,,σn,i,j{1,,n}.

Proof

We differentiate once in t and twice in x1, …, xn, all equations of the system (2.1) and we see that the function u satisfies the system (1.3). Now we put t = a in the first n equations of the system (2.1) and we get

x0xx0sui(a,σ)gi(σ)dσds=0,i{1,,n}.

We differentiate the last system twice in x1, …, xn, and we obtain

ui(a,x)=gi(x),xRn,i{1,,n},

i.e., the function ui, i ∈ {1, …, n}, satisfies the initial condition (2.2). This completes the proof.□

The proof of the existence result is based on a fixed point theorem for sum of two operators one of which is expansive.

Definition 2.2

Let (X, d) be a metric space and Y be a subset of X. The mapping T : YX is said to be expansive if there exists a constant h > 1 such that

d(Tx,Ty)hd(x,y)

for any x, yM.

Next result we will use to prove our fixed point theorem.

Theorem 2.3

[22] Let X be a nonempty closed convex subset of a Banach space Y. Suppose that T and S map X into Y such that

  1. S is continuous and S(X) resides in a compact subset of Y.

  2. T : XY is expansive and onto.

Then there exists a point x*X such that

Sx+Tx=x.

Theorem 2.4

Let X be a nonempty closed convex subset of a Banach space E and Y is a nonempty compact subset of E such that XY, YX. Suppose that T and S map X into E such that

  1. S is continuous and S(X) resides in Y.

  2. T : XE is linear, continuous and expansive, and T : XY is onto.

Then there exists an x*X such that

Tx+Sx=x.

Proof

Since Y is compact and S(X) resides in Y, we have that the first condition of Theorem 2.3 holds. Because T : XE is expansive, we have that the second condition of Theorem 2.3 holds. Note that T–1 : YE exists, it is linear and contractive with a constant l ∈ (0, 1). Let zS(X) be arbitrarily chosen and fixed. Set

A={yz:yY}.

Take y0Y arbitrarily. Define the sequence {yn}n∈ℕ as follows.

yn+1=T1ynz,nN{0}.

Then

y2y1=T1y1T1y0ly1y0,y3y2=T1y2T1y1ly2y1l2y1y0.

Using the principle of the mathematical induction, we get

yn+1ynlny1y0,nN.

Now, for m > n, m, n ∈ ℕ, we find

ymynymym1++yn+1ynlm1++lny1y0lnj=0ljy1y0=ln1ly1y0.

Therefore {yn}n∈ℕ is a Cauchy sequence of elements of YE. Since E is a Banach space, it follows that the sequence {yn}n∈ℕ is convergent to an element y*E. Because {yn}n∈ℕY and YE is compact, we have that y*Y. Thus

y=T1yz

or

z=Tz+z,z=T1yX.

Because zS(X) was arbitrarily chosen, we conclude that S(X) ⊂ (IT)(X), i.e., the third condition of Theorem 2.2 holds. Hence and Theorem 2.3, it follows that there exists an x*X such that

Tx+Sx=x.

This completes the proof.□

Below we will suppose that x0=x10,,xn0B is arbitrarily chosen and fixed. Also, we will use the following notations.

x¯j,m0s¯j,m=x10s1xj10sj1xj+10sj+1xm10sm1xm+10sm+1xn0sn,x¯j,m,r0s¯j,m,r=x10s1xj10sj1xj+10sj+1xm10sm1xm+10sm+1xr10sr1xr+10sr+1xn0sn,σ~sj,sm=σ1,,σj1,sj,σj+1,,σm1,sm,σm+1,,σn,σ~si,sm,sr=σ1,,σi1,si,σi+1,,σm1,sm,σm+1,,σr1,sr,σr+1,,σn,i,j,m,r{1,,n}.

With μ(A) we will denote the measure of a set A ⊂ ℝk, k ≥ 1.

Let

B=max{1,μ(B),μBRx1××Rxk11×Rxk1+1××Rxn,μBRx1××Rxk11×Rxk1+1××Rxk21×Rxk2+1××Rxn,μ(B(Rx1××Rxk11×Rxk1+1××Rxk21×Rxk2+1××Rxk31×Rxk3+1××Rxn)),k1,k2,k3{1,,n}},Rn=Rx1××Rxn,

as we set

μ(B(Rx1××Rxk11×Rxk1+1××Rxk21×Rxk2+1××Rxk31×Rxk3+1××Rxn)):=1,k1,k2,k3{1,2,3}forn=3.

3 Main result – proof

Our main result is as follows.

Theorem 3.1

Suppose that wi0 satisfies (Hyp1) and fi satisfies (Hyp2), i ∈ {1, …, n}. Then the Cauchy problem (1.1), (1.2) has a solution p, ui ∈ 𝓒([0, ∞), C0 (ℝn)), i ∈ {1, …, n}, such that

Rn|u(t,x)|2dxCforallt0,

where |u(t,x)|=u1(t,x)2++un(t,x)2.

Firstly, we will prove that the Cauchy problem for i ∈ {1, …, n}

uit+j=1nujuixj(m+M(Rn|xui|2dx))Δui+pxi=fi(t,x),i=1nuixi(t,x)=0in[0,1]×Rn, (3.1)
w(0,x)=w0(x)inRn, (3.2)

has a 𝓒1([0, 1], C02 (ℝn))-solution (u1, p1) such that

suppxui1,suppxp1B,i{1,,n}.

Let

N1=maxt[0,1],xB|f(t,x)|.

We choose the constants γ, δ, l > 0 as follows

lnQ+3n(1+(m+γQ2δ1))(B)2Q+n2Q2(B)2+(B)2N1Q. (3.3)

Let

E1={v=(v1,,vn+1):viC1([0,1],C02(Rn)),suppxviB,i{1,,n+1}}

be endowed with the norm

||v||=maxq{1,,n+1}{maxt[0,1],xB|vq(t,x)|,maxt[0,1],xB|vqt(t,x)|,maxt[0,1],xB|vqxi(t,x)|,maxt[0,1],xB|vqxixj(t,x)|,i,j{1,,n}}.

With 1 we denote the set of all equi-continuous families in E1, i.e., if 𝓕 ⊂ 1 is a family of elements of E1, then for every ϵ > 0 there exists a δ = δ(ϵ) > 0 such that

|vq(t1,x1)vq(t2,x2)|<ϵ,|vqt(t1,x1)vqt(t2,x2)|<ϵ,|vqxi(t1,x1)vqxi(t2,x2)|<ϵ,|vqxixj(t1,x1)vqxixj(t2,x2)|<ϵ

for any i, j ∈ {1, …, n}, for any q ∈ {1, …, n + 1}, and for any v ∈ 𝓕, whenever |t1t2| < δ, |x1x2| < δ. Let also,

K¯1=K~1¯,K1={vK¯1:||v||Q},Q1={vK¯1:||v||(1+l)Q}.

Note that K1 is a compact subset of Q1. For (u, p) ∈ Q1 we define the operators.

Li1(u,p)(t,x)=lui(t,x)+lx0xx0sui(t,σ)ui0(σ)dσds+lj=1n0tx0xx¯j0s¯juiτ,σ~sjujτ,σ~sjdσ¯jdsdτl(m+M(j=1n|uixj|2))j=1n0tx¯j0x¯jx¯j0s¯juiτ,σ~xjdσ¯jds¯jdτ+l0tx0xx¯i0s¯ipτ,σ~sidσ¯idsdτl0tx0xx0sfi(τ,σ)dσdsdτ,Ln+11(u,p)(t,x)=lp(t,x)+lj=1n0tx0xx¯j0s¯jujτ,σ~sjdσ¯jdsdτ,Ni1(u,p)(t,x)=(1+l)ui(t,x),i{1,,n},Nn+11(u,p)(t,x)=(1+l)p(t,x),L1(u,p)(t,x)=L11(u,p)(t,x),,Ln+11(u,p)(t,x),N¯1(u,p)(t,x)=N11(u,p)(t,x),,Nn+11(u,p)(t,x),(t,x)[0,1]×Rn.

Proposition 3.2

L1 : K1K1 is continuous and L1(K1) resides in a compact subset of Q1.

Proof

  1. For (u, p) ∈ K1 we have that Li1 (u, p) ∈ 𝓒1([0, 1], 𝓒2(ℝn)), i ∈ {1, …, n + 1} and for x ∈ ℝnB, using the definition of L1, we have that L1(u, p)(t, x) = 0 for all t ∈ [0, 1]. Hence, Li1 (u, p) ∈ 𝓒1([0, 1], C02 (ℝn)) and suppx Li1 (u, p) ⊂ B, i ∈ {1, …, n + 1}.

  2. Let (u, p) ∈ K1. Then

    tLi1(u,p)(t,x)=ltui(t,x)+lx0xx0stui(t,σ)dσds+lj=1nx0xx¯j0s¯juit,σ~sjujt,σ~sjdσ¯jdsl(m+M(j=1n|uixj|2))j=1nx¯j0x¯jx¯j0s¯juit,σ~xjdσ¯jds¯j+lx0xx¯i0s¯ipt,σ~sidσ¯idslx0xx0sfi(t,σ)dσds,i{1,,n},(t,x)[0,1]×Rn.

    Hence, using (3.3) and (Hyp0) for some canstantes γ, δ > 0,

    tLi1(u,p)(t,x)ltui(t,x)+l|x0xx0stui(t,σ)dσds|+lj=1n|x0xx¯j0s¯juit,σ~sjujt,σ~sjdσ¯jds|+l(m+γ(j=1n|uixj|2)δ)j=1n|x¯j0x¯jx¯j0s¯juit,σ~xjdσ¯jds¯j|+l|x0xx¯i0s¯ipt,σ~sidσ¯ids|+l|x0xx0sf(t,σ)dσds|lQ+l(B)2Q+nl(B)2Q2+ln(m+γQ2δ1)(B)2Q+l(B)2Q+l(B)2N1lQ+n(B)2Q2+(n(m+γQ2δ1)+2)(B)2Q+(B)2N1Q,i{1,,n},(t,x)[0,1]×Rn.
  3. Let (u, p) ∈ K1. Then

    tLn+11(u,p)(t,x)=ltp(t,x)+lj=1nx0xx¯j0s¯jujt,σ~sjdσ¯jds,(t,x)[0,1]×Rn,

    and using (3.3), we get

    tLn+11(u,p)(t,x)ltp(t,x)+lj=1n|x0xx¯j0s¯jujt,σ~sjdσ¯jds|lQ+l(B)2Q=lQ+(B)2QQ,(t,x)[0,1]×Rn.
  4. Let (u, p) ∈ K1 and ki, k, i ∈ {1, …, n}. Then

    xkLi1(u,p)(t,x)=lxkui(t,x)+lx0xx¯k0s¯kut,σ~skui0σ~skdσ¯kds+lj=1,jkn0tx0xx¯k,j0s¯k,juiτ,σ~sk,sjujτ,σ~sk,sjdσ¯k,jdsdτ+l0tx¯k0x¯kx¯k0s¯kuiτ,σ~xkukτ,σ~xkdσ¯kds¯kdτl(m+M(j=1n|uixj|2))j=1,jkn0tx¯j0x¯jx¯k,j0s¯k,juiτ,σ~sk,xjdσ¯j,kds¯jdτl(m+M(j=1n|uixj|2))0tx¯k0x¯kx¯k0s¯kxkuiτ,σ~xkdσ¯kds¯kdτ+l0tx0xx¯k,i0s¯k,ipτ,σ~sk,sidσ¯i,kdsdτl0tx0xx¯k0s¯kfiτ,σ~skdσ¯kdsdτ,(t,x)[0,1]×Rn.

    Again using (3.3) and (Hyp0) for some canstantes γ, δ > 0, we obtain

    xkLi1(u,p)(t,x)lkui(t,x)+l|x0xx¯k0s¯kuit,σ~skui0σ~skdσ¯kds|+lj=1,jkn|0tx0xx¯k,j0s¯k,juiτ,σ~sk,sjujτ,σ~sk,sjdσ¯k,jdsdτ|+l|0tx¯k0x¯kx¯k0s¯muiτ,σ~xkukτ,σ~xkdσ¯kds¯kdτ|+l(m+M(j=1n|uixj|2))j=1,jkn|0tx¯j0x¯jx¯k,j0s¯k,juiτ,σ~sk,xjdσ¯j,kds¯jdτ|+l(m+M(j=1n|uixj|2))|0tx¯k0x¯kx¯k0s¯kxkuiτ,σ~xkdσ¯kds¯kdτ|+l|0tx0xx¯k,i0s¯k,ipτ,σ~sk,sidσ¯i,kdsdτ|+l|0tx0xx¯k0s¯kfτ,σ~skdσ¯kdsdτ|lQ+2l(B)2Q+(n1)l(B)2Q2+l(B)2Q2+l(m+γQ2δ1)(n1)(B)2Q+(m+γQ2δ1)l(B)2Q+l(B)2Q+l(B)2N1lQ+(3+n(m+γQ2δ1))(B)2Q+n(B)2Q2+(B)2N1Q,(t,x)[0,1]×Rn.
  5. Let (u, p) ∈ K1. Then

    xiLi1(u,p)(t,x)=lxiui(t,x)lx0xx¯i0s¯iuit,σ~siui0σ~sidσ¯ids+lj=1,jin0tx0xx¯i,j0s¯i,juiτ,σ~si,sjujτ,σ~si,sjdσ¯i,jdsdτ+l0tx¯i0x¯ix¯i0s¯iuiτ,σ~xi2dσ¯ids¯idτl(m+M(j=1n|uixj|2))j=1,jin0tx¯j0x¯jx¯i,j0s¯i,juiτ,σ~si,xjdσ¯i,jds¯jdτl(m+M(j=1n|uixj|2))0tx¯i0x¯ix¯i0s¯ixiuiτ,σ~xidσ¯ids¯idτ+l0tx¯i0x¯ix¯i0s¯ipτ,σ~xidσ¯ids¯idτl0tx0xx¯i0s¯ifiτ,σ~sidσ¯idsdτ,(t,x)[0,1]×Rn.

    From here, by (Hyp0) for some canstantes γ, δ > 0

    xiLi1(u,p)(t,x)lxiui(t,x)+l|x0xx¯i0s¯iuit,σ~siui0σ~sidσ¯ids|+lj=1,jin|0tx0xx¯i,j0s¯i,juiτ,σ~si,sjujτ,σ~si,sjdσ¯i,jdsdτ|+l|0tx¯i0x¯ix¯i0s¯iuiτ,σ~xi2dσ¯ids¯idτ|+l(m+M(j=1n|uixj|2))j=1,jin|0tx¯j0x¯jx¯i,j0s¯i,juiτ,σ~si,xjdσ¯i,jds¯jdτ|+l(m+M(j=1n|uixj|2))|0tx¯i0x¯ix¯i0s¯ixiuiτ,σ~xidσ¯ids¯idτ|+l|0tx¯i0x¯ix¯i0s¯ipτ,σ~xidσ¯ids¯idτ|+l|0tx0xx¯i0s¯ifiτ,σ~sidσ¯idsdτ|lQ+2l(B)2Q+(n1)l(B)2Q2+l(B)2Q2+(m+γQ2δ1)l(n1)(B)2Q+l(m+γQ2δ1)(B)2Q+l(B)2Q+l(B)2N1lQ+(3+n(m+γQ2δ1))(B)2Q+n(B)2Q2+(B)2N1Q,(t,x)[0,1]×Rn.
  6. Let (u, p) ∈ K1. Then

    xkLn+11(u,p)(t,x)=lxkp(t,x)+lj=1,jkn0tx0xx¯k,j0s¯k,jujτ,σ~sk,sjdσ¯j,kdsdτ+l0tx¯k0x¯kx¯k0s¯kukτ,σ~xkdσ¯kds¯kdτ,(t,x)[0,1]×Rn,

    and

    xkLn+11(u,p)(t,x)lxkp(t,x)+lj=1,jkn|0tx0xx¯k,j0s¯k,jujτ,σ~sk,sjdσ¯j,kdsdτ|+l|0tx¯k0x¯kx¯k0s¯kukτ,σ~xkdσ¯kds¯kdτ|lQ+l(n1)(B)2Q+l(B)2QlQ+n(B)2QQ,(t,x)[0,1]×Rn.
  7. Let (u, p) ∈ K1 and ki, k, i ∈ {1, …, n}. Then

    2xk2Li1(u,p)(t,x)=l2xk2ui(t,x)+lx¯k0x¯kx¯k0s¯kuit,σ~xkui0σ~xkdσ¯kds¯k+lj=1,jkn0tx¯k0x¯kx¯k,j0s¯k,juiτ,σ~xk,sjujτ,σ~xk,sjdσ¯j,kds¯kdτ+l0tx¯k0x¯kx¯k0s¯kxkuiτ,σ~kujτ,σ~xkdσ¯kds¯kdτl(m+M(j=1n|uixj|2)j=1,jkn0tx¯k,j0x¯k,jx¯k,j0s¯k,juiτ,σ~xk,xjdσ¯j,kds¯j,kdτl(m+M(j=1n|uixj|2))0tx¯k0x¯kx¯k0s¯k2xk2uiτ,σ~xkdσ¯kds¯kdτ+l0tx¯k0x¯kx¯k,i0s¯k,ipτ,σ~xk,sidσ¯k,ids¯kdτl0tx¯k0x¯kx¯k0s¯kfiτ,σ~xkdσ¯kds¯kdτ,(t,x)[0,1]×Rn.

    Now, using (3.3) and (Hyp0) for some canstantes γ, δ > 0,

    2xk2Li1(u,p)(t,x)l2xk2ui(t,x)+l|x¯k0x¯kx¯k0s¯kuit,σ~xkui0σ~xkdσ¯kds¯k|+lj=1,jkn|0tx¯k0x¯kx¯k,j0s¯k,juiτ,σ~xk,sjujτ,σ~xk,sjdσ¯j,kds¯kdτ|+l|0tx¯k0x¯kx¯k0s¯kxkuiτ,σ~xkujτ,σ~xkdσ¯kds¯kdτ|+l(m+M(j=1n|uixj|2))j=1,jkn|0tx¯k,j0x¯k,jx¯k,j0s¯k,juiτ,σ~xk,xjdσ¯j,kds¯j,kdτ|+l(m+M(j=1n|uixj|2))|0tx¯k0x¯kx¯k0s¯k2xk2uiτ,σ~xkdσ¯kds¯kdτ|+l|0tx¯k0x¯kx¯k,i0s¯k,ipτ,σ~xk,sidσ¯k,ids¯kdτ|+l|0tx¯k0x¯kx¯k0s¯kfiτ,σ~xkdσ¯kds¯kdτ|lQ+2l(B)2Q+(n1)l(B)2Q2+2l(B)2Q2+ln(m+γQ2δ1)(B)2Q+(m+γQ2δ1)l(B)2Q+l(B)2Q+l(B)2N1lQ+(3+(m+γQ2δ1)(n+1))(B)2Q+(n+1)l(B)2Q2+(B)2N1Q,(t,x)[0,1]×Rn.
  8. Let (u, p) ∈ K1. Then

    2xi2Li1(u,p)(t,x)=l2xi2ui(t,x)+lx¯i0x¯ix¯i0s¯iuit,σ~xiui0σ~xidσ¯ids¯i+j=1,jinl0tx¯i0x¯ix¯i,j0s¯i,juiτ,σ~xi,sjujτ,σ~xi,sjdσ¯i,jds¯idτ+2l0tx¯i0x¯ix¯i0s¯iuiτ,σ~xixiuiτ,σ~xidσ¯ids¯idτl(m+M(j=1n|uixj|2))j=1,jin0tx¯i,j0x¯i,jx¯i,j0s¯i,juiτ,σ~xi,xjdσ¯i,jds¯i,jdτl(m+M(j=1n|uixj|2))0tx¯i0x¯ix¯i0s¯i2xi2uiτ,σ~xidσ¯ids¯idτ+l0tx¯i0x¯ix¯i0s¯ixipτ,σ~xidσ¯ids¯idτl0tx¯i0x¯ix¯i0s¯ifiτ,σ~xidσ¯ids¯idτ,(t,x)[0,1]×Rn.

    Using (3.3), we arrive to the following estimate

    2xi2Li1(u,p)(t,x)l2xi2pi(t,x)+l|x¯i0x¯ix¯i0s¯iuit,σ~xiui0σ~xidσ¯ids¯i|+j=1,jinl|0tx¯i0x¯ix¯i,j0s¯i,juiτ,σ~xi,sjujτ,σ~xi,sjdσ¯i,jds¯idτ|+2l|0tx¯i0x¯ix¯i0s¯iuiτ,σ~xixiuiτ,σ~xidσ¯ids¯idτ|+l(m+M(j=1n|uixj|2))j=1,jin|0tx¯i,j0x¯i,jx¯i,j0s¯i,juiτ,σ~xi,xjdσ¯i,jds¯i,jdτ|+l(m+M(j=1n|uixj|2))|0tx¯i0x¯ix¯i0s¯i2xi2uiτ,σ~xidσ¯ids¯idτ|+l|0tx¯i0x¯ix¯i0s¯ixipτ,σ~xidσ¯ids¯idτ|+l|0tx¯i0x¯ix¯i0s¯ifiτ,σ~xidσ¯ids¯idτ|lQ+2l(B)2Q+(n1)l(B)2Q2+2l(B)2Q2+l(m+γQ2δ1)(n1)(B)2Q+l(m+γQ2δ1)(B)2Q+l(B)2Q+l(B)2N1lQ+(3+n(m+γQ2δ1))(B)2Q+(n+1)(B)2Q2+(B)2N1Q,(t,x)[0,1]×Rn.
  9. Let (u, p) ∈ K1 and rk, ki, ri, r, k, i ∈ {1, …, n}. Then

    2xrxkLi1(u,p)(t,x)=l2xkxrui(t,x)+lx0xx¯r,k0s¯r,kuit,σ~sr,skui0σ~sr,skdσ¯r,kds+lj=1,jk,jrn0tx0xx¯r,k,j0s¯r,k,juiτ,σ~sr,sk,sjujτ,σ~sr,sk,sjdσ¯r,k,jdsdτ+l0tx¯r0x¯rx¯r,k0s¯r,kuiτ,σ~xr,skurτ,σ~xr,skdσ¯r,kds¯rdτ+l0tx¯k0x¯kx¯r,k0s¯r,kuiτ,σ~sr,xkukτ,σ~sr,xkdσ¯r,kds¯kdτl(m+M(j=1n|uixj|2))j=1,jr,jkn0tx¯j0x¯jx¯r,k,js¯r,k,juiτ,σ~sr,sk,xjdσ¯j,k,rds¯jdτl(m+M(j=1n|uixj|2))0tx¯r0x¯rx¯r,k0s¯r,kxruiτ,σ~sk,xrdσ¯r,kds¯rdτl(m+M(j=1n|uixj|2))0tx¯k0x¯kx¯r,k0s¯r,k2xrxkuiτ,σ~sr,xkdσ¯k,rds¯kdτ+l0tx0xx¯r,k,is¯r,k,ipτ,σ~sr,sk,sidσ¯r,k,idsdτl0tx0xx¯r,k0x¯r,kfiτ,σ~sr,skdσ¯k,rdsdτ,(t,x)[0,1]×Rn,

    and

    2xrxkLi1(u,p)(t,x)l2xkxrui(t,x)+l|x0xx¯r,k0s¯r,kuit,σ~sr,skui0σ~sr,skdσ¯r,kds|+lj=1,jk,jrn|0tx0xx¯r,k,j0s¯r,k,juiτ,σ~sr,sk,sjujτ,σ~sr,sk,sjdσ¯r,k,jdsdτ|+l|0tx¯r0x¯rx¯r,k0s¯r,kuiτ,σ~xr,skurτ,σ~xr,skdσ¯r,kds¯rdτ|+l|0tx¯k0x¯kx¯r,k0s¯r,kuiτ,σ~sr,xkukτ,σ~sr,xkdσ¯r,kds¯kdτ|+l(m+M(j=1n|uixj|2))j=1,jr,jkn|0tx¯j0x¯jx¯r,k,js¯r,k,juiτ,σ~sr,sk,xjdσ¯j,k,rds¯jdτ|+l(m+M(j=1n|uixj|2))|0tx¯r0x¯rx¯r,k0s¯r,kxruiτ,σ~sk,xrdσ¯r,kds¯rdτ|+l(m+M(j=1n|uixj|2))|0tx¯k0x¯kx¯r,k0s¯r,k2xrxkuiτ,σ~sr,xkdσ¯k,rds¯kdτ|+l|0tx0xx¯r,k,is¯r,k,ipτ,σ~sr,sk,sidσ¯r,k,idsdτ|+l|0tx0xx¯r,k0x¯r,kfiτ,σ~sr,skdσ¯k,rdsdτ|lQ+2l(B)2Q+(n2)l(B)2Q2+l(B)2Q2+l(B)2Q2+l(m+γQ2δ1)(n2)(B)2Q+l(m+γQ2δ1)(B)2Q+l(m+γQ2δ1)(B)2Q+l(B)2Q+l(B)2N1lQ+(3+n(m+γQ2δ1))(B)2Q+n(B)2Q2+(B)2N1Q,(t,x)[0,1]×Rn.
  10. Let (u, p) ∈ K1 and ki, k, i ∈ {1, …, n}. Then

    2Li1xixk(u,p)(t,x)=l2xixkui(t,x)+lx0xx¯i,k0s¯i,kuit,σ~si,skui0σ~si,skdσ¯k,ids+lj=1,jk,jin0tx0xx¯i,k,j0s¯i,k,juiτ,σ~si,sk,sjujτ,σ~si,sk,sjdσ¯j,k,idsdτ+l0tx¯i0x¯ix¯k,i0s¯k,iuiτ,σ~sk,xi2dσ¯i,kds¯idτ+l0tx¯k0x¯kx¯i,k0s¯i,kuiτ,σ~si,xkujτ,σ~si,xkdσ¯k,ids¯kdτl(m+M(j=1n|uixj|2))j=1,jk,jin0tx¯j0x¯jx¯i,k,j0s¯i,k,juiτ,σ~si,sk,xjdσ¯j,k,ids¯jdτl(m+M(j=1n|uixj|2))0tx¯i0x¯ix¯k,i0s¯k,iuixiτ,σ~sk,xidσ¯i,kds¯idτl(m+M(j=1n|uixj|2))0tx¯k0x¯kx¯i,k0s¯i,kxkuiτ,σ~si,xkdσ¯k,ids¯kdτ+l0tx¯i0x¯ix¯k,i0s¯k,ipτ,σ~sk,xidσ¯i,kds¯idτl0tx0xx¯i,mk0s¯i,kfiτ,σ~si,skdσ¯i,kdsdτ,(t,x)[0,1]×Rn.

    From here and from (3.3) and (Hyp0) for some canstantes γ, δ > 0, we obtain

    2Li1xixk(u,p)(t,x)l2xixkui(t,x)+l|x0xx¯i,k0s¯i,kuit,σ~si,skui0σ~si,skdσ¯k,ids|+lj=1,jk,jin|0tx0xx¯i,k,j0s¯i,k,juiτ,σ~si,sk,sjujτ,σ~si,sk,sjdσ¯j,k,idsdτ|+l|0tx¯i0x¯ix¯k,i0s¯k,iuiτ,σ~sk,xi2dσ¯i,kds¯idτ|+l|0tx¯k0x¯kx¯i,k0s¯i,kuiτ,σ~si,xkujτ,σ~si,xkdσ¯k,ids¯kdτ|+l(m+M(j=1n|uixj|2))j=1,jk,jin|0tx¯j0x¯jx¯i,k,j0s¯i,k,juiτ,σ~si,sk,xjdσ¯j,k,ids¯jdτ|+l(m+M(j=1n|uixj|2))|0tx¯i0x¯ix¯k,i0s¯k,iuixiτ,σ~sk,xidσ¯i,kds¯idτ|+l(m+M(j=1n|uixj|2))|0tx¯k0x¯kx¯i,k0s¯i,kxkuiτ,σ~si,xkdσ¯k,ids¯kdτ|+l|0tx¯i0x¯ix¯k,i0s¯k,ipτ,σ~sk,xidσ¯i,kds¯idτ|+l|0tx0xx¯i,k0s¯i,kfiτ,σ~si,skdσ¯i,kdsdτ|lQ+2l(B)2Q+(n2)l(B)2Q2+l(B)2Q2+l(B)2Q2+l(m+γQ2δ1)(B)2Q+l(m+γQ2δ1)(n2)(B)2Q+ml(B)2Q+l(B)2Q+l(B)2N1lQ+(2+n(m+γQ2δ1))(B)2Q+n(B)2Q2+(B)2N1Q,(t,x)[0,1]×Rn.
  11. Let (w, p) ∈ K1 and ri, k = i, k, i ∈ {1, …, n}. Then

    2xixrLi1(u,p)(t,x)=l2xixrui(t,x)+lx0xx¯r,i0s¯r,iuit,σ~sr,siui0σ~sr,sidσ¯i,rds+lj=1,ji,jrn0tx0xx¯r,i,j0s¯r,i,juiτ,σ~sr,si,sjujτ,σ~sr,si,sjdσ¯j,i,rdsdτ+l0tx¯r0x¯rx¯i,r0s¯i,ruiτ,σ~si,xrujτ,σ~si,xrdσ¯r,ids¯rdτ+l0tx¯i0x¯ix¯r,i0s¯r,iuiτ,σ~sr,xi2dσ¯i,rds¯idτl(m+M(j=1n|uixj|2))j=1,ji,jrn0tx¯j0x¯jx¯r,i,j0s¯r,i,juiτ,σ~sr,si,xjdσ¯r,i,jds¯jdτl(m+M(j=1n|uixj|2))0tx¯r0x¯rx¯i,r0s¯i,rxruiτ,σ~si,xrdσ¯r,ids¯rdτl(m+M(j=1n|uixj|2))0tx¯i0x¯ix¯i,r0s¯i,rxiuiτ,σ~xi,srdσ¯r,ids¯idτ+l0tx¯i0x¯ix¯i,r0s¯i,rpτ,σ~xi,srdσ¯i,rds¯idτl0tx0xx¯i,r0s¯i,rfiτ,σ~si,srdσ¯i,rdsdτ,(t,x)[0,1]×Rn,

    and

    2xixrLi1(u,p)(t,x)l2xixrui(t,x)+l|x0xx¯r,i0s¯r,iuit,σ~sr,siui0σ~sr,sidσ¯i,rds|+lj=1,ji,jrn|0tx0xx¯r,i,j0s¯r,i,juiτ,σ~sr,si,sjujτ,σ~sr,si,sjdσ¯j,i,rdsdτ|+l|0tx¯r0x¯rx¯i,r0s¯i,ruiτ,σ~si,xrujτ,σ~si,xrdσ¯r,ids¯rdτ|+l|0tx¯i0x¯ix¯r,i0s¯r,iuiτ,σ~sr,xi2dσ¯i,rds¯idτ|+l(m+M(j=1n|uixj|2))j=1,ji,jrn|0tx¯j0x¯jx¯r,i,j0s¯r,i,juiτ,σ~sr,si,xjdσ¯r,i,jds¯jdτ|+l(m+M(j=1n|uixj|2))|0tx¯r0x¯rx¯i,r0s¯i,rxruiτ,σ~si,xrdσ¯r,ids¯rdτ|+l(m+M(j=1n|uixj|2))|0tx¯i0x¯ix¯i,r0s¯i,rxiuiτ,σ~xi,srdσ¯r,ids¯idτ|+l|0tx¯i0x¯ix¯i,r0s¯i,rpτ,σ~xi,srdσ¯i,rds¯idτ|+l|0tx0xx¯i,r0s¯i,rfiτ,σ~si,srdσ¯i,rdsdτ|lQ+2l(B)2Q+(n2)l(B)2Q2+l(B)2Q2+l(B)2Q2+l(m+γQ2δ1)(n2)(B)2Q+l(m+γQ2δ1)(B)2Q+(m+γQ2δ1)l(B)2Q+l(B)2Q+l(B)2N1lQ+(3+(m+γQ2δ1)n)(B)2Q2+n(B)2Q2+(B)2N1Q,(t,x)[0,1]×Rn.
  12. Let (u, p) ∈ K1 and k ∈ {1, …, n}. Then

    2xk2Ln+11(u,p)(t,x)=l2xk2p(t,x)+lj=1,jkn0tx¯k0x¯kx¯k,j0s¯k,jujτ,σ~xk,sjdσ¯j,kds¯kdτ+l0tx¯k0x¯kx¯k0s¯kxkukτ,σ~xkdσ¯kds¯kdτ,(t,x)[0,1]×Rn.

    Using the last equality and (3.3), we go to

    2xk2Ln+11(u,p)(t,x)l2xk2p(t,x)+lj=1,jkn|0tx¯k0x¯kx¯k,j0s¯k,jujτ,σ~xk,sjdσ¯j,kds¯kdτ|+l|0tx¯k0x¯kx¯k0s¯kxkukτ,σ~xkdσ¯kds¯kdτ|lQ+l(n1)(B)2Q+l(B)2Q=lQ+n(B)2QQ,(t,x)[0,1]×Rn.
  13. Let (u, p) ∈ K1 and kr, k, r ∈ {1, …, n}. Then

    2xkxrLn+11(u,p)(t,x)=l2xkxrp(t,x)+lj=1,jk,jrn0tx0xx¯k,j,r0s¯k,j,rujτ,σ~sk,sj,srdσ¯j,k,rdsdτ+l0tx¯r0x¯rx¯k,r0s¯k,rujτ,σ~sk,xrdσ¯k,rds¯rdτ+l0tx¯k0x¯kx¯k,r0s¯k,rukτ,σ~xk,srdσ¯k,rds¯kdτ.(t,x)[0,1]×Rn.

    From here,

    2xkxrLn+11(u,p)(t,x)l2xkxrp(t,x)+lj=1,jk,jrn|0tx0xx¯k,j,r0s¯k,j,rujτ,σ~sk,sj,srdσ¯j,k,rdsdτ|+l|0tx¯r0x¯rx¯k,r0s¯k,rujτ,σ~sk,xrdσ¯k,rds¯rdτ|+l|0tx¯k0x¯kx¯k,r0s¯k,rukτ,σ~xk,srdσ¯k,rds¯kdτ|lQ+l(n2)(B)2Q+l(B)2Q+l(B)2Q=lQ+n(B)2QQ,(t,x)[0,1]×Rn.

    From (1)-(13) it follows that L1 : K1K1 and it is continuous. Since K1 is a compact subset of Q1, we have that L1(K1) resides in a compact subset of Q1. This completes the proof.□

Proposition 3.3

N1 : K1Q1 is an expansive operator and onto.

Proof

For U = (u, p), Ũ = (ũ, ) ∈ K1 we have

||N¯1(U)N¯1(U~)||=(1+l)||UU~||,

i.e., N1 : K1Q1 is an expansive operator with h = 1 + l. Let (ũ, ) ∈ Q1. Then u~1+l,p~1+lK1 and

N¯1u~1+l,p~1+l=(u~,p~),

i.e., the operator N1 : K1Q1 is onto. This completes the proof.□

By Proposition 3.2, Proposition 3.3 and Theorem 2.4, it follows that the operator L1 + N1 has a fixed point (u1, p1) ∈ K1. For it we have

x0xx0sui1(t,σ)ui0(σ)dσds+j=1n0tx0xx¯j0s¯jui1τ,σ~sjuj1τ,σ~sjdσ¯jdsdτ(m+M(j=1n|uixj|2))j=1n0tx¯j0x¯jx¯j0s¯jui1τ,σ~xjdσ¯jds¯jdτ+0tx0xx¯i0s¯ip1τ,σ~sidσ¯idsdτ=0tx0xx0sfi(τ,σ)dσdsdτ,i{1,,n},
j=1n0tx0xx¯j0s¯juj1τ,σ~sjdσ¯jdsdτ=0,t[0,1],xRn.

Hence and Lemma 2.1, it follows that (u1, p1) is a 𝓒1([0, 1], C02 (ℝn)) solution of the problem (3.1), (3.2) for which suppxp1, suppx ui1 B, i ∈ {1, …, n}.

Now we consider the Cauchy problem

uit+j=1nujuixj(m+M(Rn|xui|2dx))Δui+pxi=fi(t,x),i{1,,n},i=1nuixi(t,x)=0in[1,2]×Rn, (3.4)
u(1,x)=u1(1,x)inRn. (3.5)

Let

M2=maxt[1,2],xB|f(t,x)|.

We choose the constant l1 > 0 such that

l1nQ+3n(1+(m+γQ2δ1))(B)2Q+n2Q2(B)2+(B)2N2Q.

Let

E2={v=(v1,,vn+1):viC1([1,2],C02(Rn)),suppxviB,i{1,,n+1}}

be endowed with the norm

||v||=maxq{1,,n+1}{maxt[1,2],xB|vq(t,x)|,maxt[1,2],xB|vqt(t,x)|,maxt[1,2],xB|vqxi(t,x)|,maxt[1,2],xB|vqxixj(t,x)|,i,j{1,,n}}.

With 2 we denote the set of all equi-continuous families in E2. Let also,

K¯2=K~2¯,K2={vK¯2:||v||Q},Q2={vK¯2:||v||(1+l1)Q}.

Note that K2 is a compact subset of Q2. For (u, p) ∈ Q2 we define the operators.

Li2(u,p)(t,x)=l1ui(t,x)+l1x0xx0sui(t,σ)ui1(1,σ)dσds+l1j=1n1tx0xx¯j0s¯juiτ,σ~sjujτ,σ~sjdσ¯jdsdτl1(m+M(j=1n|uixj|2))j=1n1tx¯j0x¯jx¯j0s¯juiτ,σ~xjdσ¯jds¯jdτ+l11tx0xx¯i0s¯ipτ,σ~sidσ¯idsdτl11tx0xx0sfi(τ,σ)dσdsdτ,Ln+12(u,p)(t,x)=l1p(t,x)+l1j=1n1t1yx0xx¯j0s¯jujτ,σ~sjdσ¯jdsdτdy+l1(t1)x0xx0spt(1,σ)pt1(1,σ)dσds+l1x0xx0sp(1,σ)p1(1,σ)dσds,Mi2(u,p)(t,x)=(1+l1)ui(t,s),i{1,,n},Mn+12(u,p)(t,x)=(1+l1)p(t,x),L2(u,p)(t,x)=L12(u,p)(t,x),,Ln+12(u,p)(t,x),M¯2(u,p)(t,x)=M12(u,p)(t,x),,Mn+12(u,p)(t,x),(t,x)[1,2]×Rn.

As in above, we obtain that the operator L2 + M2 has a fixed point (u2, p2) ∈ K2. For it we have

0=x0xx0sui2(t,σ)ui1(1,σ)dσds+j=1n1tx0xx¯j0s¯jui2τ,σ~sjuj2τ,σ~sjdσ¯jdsdτ(m+M(j=1n|uixj|2))j=1n1tx¯j0x¯jx¯j0s¯jui2τ,σ~xjdσ¯jds¯jdτ+1tx0xx¯i0s¯ip2τ,σ~sidσ¯idsdτ1tx0xx0sfi(τ,σ)dσdsdτ,i{1,,n},=j=1n1t1yx0xx¯j0s¯juj2τ,σ~sjdσ¯jdsdτdy+(t1)x0xx0spt2(1,σ)pt1(1,σ)dσds+x0xx0sp2(1,σ)p1(1,σ)dσds,(t,x)[1,2]×Rn. (3.6)

We differentiate the first n equations of the system (3.6) once in t, twice in x1, and so on, twice in xn, and we arrive to

tui2+j=1nxjui2uj2(m+M(j=1n|uixj|2))j=1n2xj2ui2+xip2=fi(t,x), (3.7)

i ∈ {1, …, n}, (t, x) ∈ [1, 2] × ℝn.

Now we differentiate the last equation of (3.6) twice in t, twice in x1, and so on, twice in xn, and we get

j=1nxjuj2(t,x)=0,(t,x)[1,2]×Rn.

For i ∈ {1, …, n} we put in (3.6) t = 1 and we find

x0xx0sui2(1,σ)ui1(1,σ)dσds=0,

which we differentiate twice in x1, and so on, twice in xn, and we find

ui2(1,x)=ui1(1,x),xRn, (3.8)

whereupon

xjui2(1,x)=xjui1(1,x),xRn, (3.9)

and

2xmxrui2(1,x)=2xmxrui1(1,x),i,m,r{1,,n},xRn. (3.10)

In the (n + 1)-th equation of the system (3.6) we put t = 1 and we find

x0xx0sp2(1,σ)p1(1,σ)dσds=0,

which we differentiate twice in x1, and so on, twice in xn, and we find

p2(1,x)=p1(1,x),xRn.

Hence,

p2xi(1,x)=p1xi(1,x),xRn,i{1,,n}, (3.11)

and

2p2xmxr(1,x)=2p1xmxr(1,x),xRn,r,m{1,,n}. (3.12)

Now we differentiate in t the n + 1-th equation of the system (3.6) and we get

j=1n1tx0xx¯j0s¯juj2τ,σ~sjdσ¯jdsdτ+x0xx0spt2(1,σ)pt1(1,σ)dσds=0,

in which we put t = 1 and we get

x0xx0spt2(1,σ)pt1(1,σ)dσds=0.

The last equation we differentiate twice in x1, and so on, twice in xn, and we get

pt2(1,x)=pt1(1,x),xRn. (3.13)

By (3.8)-(3.13), using (3.7), we obtain

tui2(1,x)j=1nxjui2(1,x)uj2(1,x)+(m+M(j=1n|ui(1,x)xj|2))j=1n2xj2ui2(1,x)p2xi(1,x)+fi(1,x)j=1nxjui1(1,x)uj1(1,x)+(m+M(j=1n|ui(1,x)xj|2))j=1n2xj2ui1(1,x)p2xi(1,x)+fi(1,x)tui1(1,x),xRn,i{1,,n}.

Consequently the function

u¯,p¯=(u1,p1)fort[0,1],xRn,u2,p2fort[1,2],xRn,

is a solution to the problem

uit+j=1nujuixj(m+M(Rn|xui|2dx))Δui+pxi=fi(t,x),i{1,,n},i=1nuixi(t,x)in[0,2]×Rn,u(0,x)=u0(x)inRn,

such that ui, p ∈ 𝓒1([0, 2], C02 (ℝn)), suppx ui, suppx pB, i ∈ {1, …, n}. Note that

|u¯i|,|p¯|Qon[0,2]×Rn,i{1,,n}.

Now we consider the Cauchy problem

uit+j=1nujuixj(m+M(Rn|xui|2dx))Δui+pxi=fi(t,x),i{1,,n},i=1nuixi(t,x)=0in[2,3]×Rn,u(2,x)=u2(2,x)inRn.

Let

M3=maxt[2,3],xB|f(t,x)|.

We choose the constant l2 > 0 such that

l2nQ+3n(1+(m+γQ2δ1))(B)2Q+n2Q2(B)2+(B)2M3Q.

Let

E3={v=(v1,,vn+1):viC1([2,3],C02(Rn)),suppxviB,i{1,,n+1}}

be endowed with the norm

||v||=maxq{1,,n+1}{maxt[2,3],xB|vq(t,x)|,maxt[2,3],xB|vqt(t,x)|,maxt[2,3],xB|vqxi(t,x)|,maxt[2,3],xB|vqxixj(t,x)|,i,j{1,,n}}.

With 3 we denote the set of all equi-continuous families in E3. Let also,

K¯3=K~3¯,K3={vK¯3:||v||Q},Q3={vK¯3:||v||(1+l2)Q}.

Note that K3 is a compact subset of Q3. For (u, p) ∈ Q3 we define the operators.

Li3(u,p)(t,x)=l2ui(t,x)+l2x0xx0sui(t,σ)ui2(2,σ)dσds+l2j=1n2tx0xx¯j0s¯juiτ,σ~sjujτ,σ~sjdσ¯jdsdτl2(m+M(j=1n|uixj|2))j=1n2tx¯j0x¯jx¯j0s¯juiτ,σ~xjdσ¯jds¯jdτ+l22tx0xx¯i0s¯ipτ,σ~sidσ¯idsdτl22tx0xx0sfi(τ,σ)dσdsdτ,Ln+13(u,p)(t,x)=l2p(t,x)+l2j=1n2t2yx0xx¯j0s¯jujτ,σ~sjdσ¯jdsdτdy+l2(t2)x0xx0spt(2,σ)pt2(2,σ)dσds+l2x0xx0sp(2,σ)p2(2,σ)dσds,Mi3(u,p)(t,x)=(1+l2)ui(t,s),i{1,,n},Mn+13(u,p)(t,x)=(1+l2)p(t,x),L3(u,p)(t,x)=L13(u,p)(t,x),,Ln+13(u,p)(t,x),M¯3(u,p)(t,x)=M13(u,p)(t,x),,Mn+13(u,p)(t,x),(t,x)[2,3]×Rn.

Consequently the function

u~,p~=(u1,p1)fort[0,1],xRn,u2,p2fort[1,2],xRn,u3,p3fort[2,3],xRn,

is a solution to the problem (1.1), (1.2) such that ũi, ∈ 𝓒1([0, ∞), C02 (ℝn)), suppx ũi, suppxB, i ∈ {1, …, n}. Since fi ∈ 𝓒2([0, ∞), C0 (ℝn)), using (1.1), we conclude that , ũi ∈ 𝓒([0, ∞) × ℝn), i ∈ {1, …, n}. Note that

|u~i|,|p~|Qon[0,)×Rn,i{1,,n}.

Therefore

Rn|u~(t,x)|2dx=B|u~(t,x)|2dxnQ2μ(B)<.

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Received: 2018-04-17
Accepted: 2019-10-28
Published Online: 2019-12-31

© 2019 Khaled Zennir, published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 Public License.

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