Subcritical and critical generalized Zakharov-Kuznetsov equation posed on bounded rectangles

TL;DR

This paper investigates the initial-boundary value problem for the generalized Zakharov-Kuznetsov equation on bounded rectangles, focusing on the effects of critical and subcritical nonlinear powers.

Contribution

It analyzes the well-posedness and behavior of solutions for different nonlinear powers in a bounded domain, extending previous work to new boundary conditions.

Findings

Identified conditions for existence of solutions in subcritical and critical cases

Established boundary behavior and solution regularity

Extended analysis to bounded rectangular domains

Abstract

Initial-boundary value problem for the generalized Zakharov-Kuznetsov equation posed on a bounded rectangle is considered. Critical and subcritical powers in nonlinearity are studied.

Equations

$$u_t+u_x+u^{1+\delta }{u}_x+u_{xxx}+u_{xyy}=0,$$

$$u_t+u_x+u^2{u}_x+u_{xxx}=0.$$

$$\Omega =\{(x,y)\in {\mathbb{R}}^2:x\in (0,L),y\in (-B,B)\},Q_T=\Omega \times (0,T).$$

$$A$$

$$u(x,-B,t)=u(x,B,t)=0,x\in (0,L),t>0;$$

$$u(0,y,t)=u(L,y,t)=u_x(L,y,t)=0,y\in (-B,B),t>0;$$

$$u(x,y,0)=u_0(x,y),(x,y)\in \Omega ,$$

$$u\in L^{\infty }(0,T;L^2(\Omega ))\cap L^2(0,T;H_0^1(\Omega )).$$

$$A^{\epsilon }{u}_{\epsilon }$$

$$u_{\epsilon }(x,-B,t)=u_{\epsilon }(x,B,t)={\partial }_y^2{u}_{\epsilon }(x,-B,t)={\partial }_y^2{u}_{\epsilon }(x,B,t)=0,x\in (0,L),t>0;$$

$$u_{\epsilon }(0,y,t)=u_{\epsilon }(L,y,t)={\partial }_x^2{u}_{\epsilon }(0,y,t)={\partial }_x{u}_{\epsilon }(L,y,t)=0,y\in (-B,B),t>0;$$

$$u_{\epsilon }(x,y,0)=u_0(x,y),(x,y)\in \Omega .$$

$$\|u\|^{2}(t)+\int_{0}^{t}\int_{-B}^{B}u^{2}_{x}(0,y,\tau)\,dyd\tau+2\epsilon\int_{0}^{t}\big{(}\|u_{xx}\|^{2}(\tau)+\|u_{yy}\|^{2}(\tau)\big{)}d\tau=\|u_{0}\|^{2},\,\,t\in(0,T).$$

$$\displaystyle\dfrac{d}{dt}\|\sqrt{x}u\|^{2}(t)+\frac{1}{2}\|\nabla u\|^{2}(t)+2\|u_{x}\|^{2}(t)+2\varepsilon\Big{(}\|\sqrt{x}u_{xx}\|^{2}(t)+\|\sqrt{x}u_{yy}\|^{2}(t)\Big{)}$$

$$\le \Vert u{\Vert }^2(t)+2\epsilon {\int }_{-B}^B{u}_x^2(0,y,t)dy+\frac{C(\xi ,\delta )C_{\Omega }^{\frac{2}{1-\delta }}}{3+\delta }\Vert u{\Vert }^{\frac{4}{1-\delta }}(t).$$

$$\displaystyle\|\sqrt{x}u\|^{2}(t)+\frac{1}{2}\int_{0}^{t}\|\nabla u\|^{2}(\tau)\,d\tau+2\int_{0}^{t}\|u_{x}\|^{2}(\tau)\,d\tau+2\varepsilon\int_{0}^{t}\Big{(}\|\sqrt{x}u_{xx}\|^{2}(\tau)+\|\sqrt{x}u_{yy}\|^{2}(\tau)\Big{)}\,d\tau$$

$$\le {\int }_0^t\Vert u_0{\Vert }^{2}d\tau +{\int }_0^t{\int }_{-B}^B{u}_x^2(0,y,\tau )dyd\tau +\frac{C(\xi ,\delta )C_{\Omega }^{\frac{2}{1-\delta }}}{3+\delta }\cdot {\int }_0^t\Vert u_0{\Vert }^{\frac{4}{1-\delta }}d\tau$$

$$\le (T+1)\Vert u_0{\Vert }^2+\frac{C(\xi ,\delta )C_{\Omega }^{\frac{2}{1-\delta }}}{3+\delta }\cdot T\Vert u_0{\Vert }^{\frac{4}{1-\delta }}.$$

$$\begin{array}[]{c}u_{\varepsilon}\;\;\;\;\text{is bounded in }\;\;\;\;L^{\infty}\big{(}0,T;L^{2}(\Omega)\big{)},\\ u_{\varepsilon x}(0,y,t)\;\;\;\;\text{is bounded in }\;\;\;\;L^{2}\big{(}0,T;L^{2}(-B,B)\big{)},\\ \nabla u_{\varepsilon}\;\;\;\;\text{is bounded in }\;\;\;\;L^{2}\big{(}0,T;L^{2}(\Omega)\big{)},\end{array}$$

$$\Vert u^{1+\delta }{u}_x{\Vert }_{L^m(0,T;L^m(\Omega ))}^m$$

$$u^{1+\delta }{u}_x\text{is bounded in}{L}^{\frac{4}{3+\delta }}(0,T;H^{-1}(\Omega )).$$

$$\frac{\partial u_{ϵ}}{\partial t}\text{is bounded (independently of ε) in}{L}^{\frac{4}{3+\delta }}(0,T;H^{-3}(\Omega ))$$

$$u_{x},u_{xx}\in C(0,L;V)\ \text{ with }\ V=H^{-2}\big{(}(0,T)\times(-B,B)\big{)},$$

$$u_{x}\big{|}_{x=0,1},\;\;u_{xx}\big{|}_{x=0,1}$$

$$u_{xxx}=-u_x-u_{xyy}-u^{1+\delta }{u}_x-u_t,$$

$$\displaystyle u_{t}\in L^{2}(0,L;H^{-1}\big{(}0,T;L^{2}(-B,B)\big{)},$$

$$\displaystyle u_{xyy}\in L^{2}(0,L;L^{2}\big{(}0,T;H^{-2}(-B,B)\big{)}.$$

$$u^{1+\delta}u_{x}\in L^{\frac{4}{3+\delta}}\big{(}0,L;L^{\frac{4}{3+\delta}}((0,T)\times(-B,B))\big{)}\hookrightarrow L^{\frac{4}{3+\delta}}\big{(}0,L;V\big{)}.$$

$$u_{xxx}\in L^{\frac{4}{3+\delta}}\big{(}0,L;V\big{)}$$

$$\left.\begin{array}[]{c}u_{\varepsilon xxx}+\varepsilon u_{\varepsilon xxxx}=g_{\varepsilon},\\ u_{\varepsilon}(0)=u_{\varepsilon}(L)=u_{\varepsilon x}(L)=u_{\varepsilon xx}(0)=0,\end{array}\right.$$

$$g_{\epsilon }:=-u_{\epsilon t}-\epsilon u_{\epsilon x}-u_{\epsilon xyy}-u_{\epsilon }^{1+\delta }{u}_{\epsilon \epsilon }-\epsilon u_{\epsilon yyyy},$$

$$U=H^{-1}(0,T;L^2(-B,B))+L^2(0,T;H^{-4}(-B,B))+L^{\frac{4}{3+\delta }}(0,T;L^{\frac{4}{3+\delta }}(-B,B)),$$

$$p=\frac{4}{3+\delta }.$$

$$\frac{d}{dt}\left((1+x),u^2\right)(t)+\Vert \nabla u{\Vert }^2(t)+2\Vert u_x{\Vert }^2(t)+(1-2\epsilon ){\int }_{-B}^B{u}_x^2(0,y,t)dy$$

$$\le \Vert u{\Vert }^2(t)+2\Vert u{\Vert }_{L^4(\Omega )}^4\le \Vert u{\Vert }^2(t)+2\Vert \nabla u{\Vert }^2(t)\Vert u{\Vert }^2(t).$$

$$u\in L^{\infty }\left(0,T;L^2(\Omega )\right)\cap L^2\left(0,T;H_0^1(\Omega )\right)$$

Full text

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1. Introduction

We are concerned with initial-boundary value problems (IBVPs) posed on bounded rectangles located at the right half-plane $\{(x,y)\in\mathbb{R}^{2}:\ x>0\}$ for the generalized Zakharov-Kuznetsov [9] equation

[TABLE]

with $\delta\in[0,1].$ When $\delta=0,$ (1.1) turns the classical Zakharov-Kuznetsov (ZK) equation [16], while $\delta=1$ corresponds to so-called modified Zakharov-Kuznetsov (mZK) equation [10] which is a two-dimensional analog of the well-known modified Korteweg-de Vries (mKdV) equation [1]

[TABLE]

Notes that both ZK and mZK possess real plasma physics applications [16].

As far as ZK is concerned, the results on both IVP and IBVPs can be found in [4, 5, 6, 9, 11, 12, 14, 15]. For IVP to mZK, see [10]; at the same time we do not know solid results concerning IBVP to mZK. The main difference between initial and initial-boundary value problems is that IVP provides (almost immediately) good estimates in $(L^{\infty}_{t};H^{1}_{xy})$ by the conservation laws, while IBVP does not possesses this advantage.

Our work is a natural continuation of [2] where (1.1) with $\delta=0$ has been considered. There one can find out a more detailed background, descriptions of main features and the deployed reference list.

In the present note we put forward an analysis of (1.1) for $\delta\in(0,1].$ When $\delta=1,$ the power is critical (see [9, 10]) and a challenge concerning the well-posedness of IBVPs appears. For one-dimensional dispersive models the critical nonlinearity has been treated in [13].

Once $\delta\in(0,1)$ the existence of a weak solution in $\left((L^{\infty}_{T};L^{2})\cap(L^{2}_{T};H^{1}_{0})\right)$ with $u_{0}\in L^{2}_{xy}$ is proved in our work via parabolic regularization. If $\delta=1,$ we apply the fixed point arguments to prove the local existence and uniqueness of solutions with more regular initial data. We also show the exponential decay of $L^{2}$ norm of solutions as $t\to\infty$ if $u\in(L^{\infty}_{\mathbb{R}^{+}};H^{1}_{0}),$ under domain’s size restrictions. These are the main results of the paper.

2. Problem and notations

Let $L,B,T$ be finite positive numbers. Define $\Omega$ and $Q_{T}$ to be spatial and time-spatial domains

[TABLE]

In $Q_{T}$ we consider the following IBVP:

[TABLE]

where $u_{0}:\Omega\to\mathbb{R}$ is a given function.

Hereafter subscripts $u_{x},\ u_{xy},$ etc. denote the partial derivatives, as well as $\partial_{x}$ or $\partial_{xy}^{2}$ when it is convenient. Operators $\nabla$ and $\Delta$ are the gradient and Laplacian acting over $\Omega.$ By $(\cdot,\cdot)$ and $\|\cdot\|$ we denote the inner product and the norm in $L^{2}(\Omega),$ and $\|\cdot\|_{H^{k}}$ stands for the norm in $L^{2}$-based Sobolev spaces. Abbreviations like $(L^{s}_{t};L^{l}_{xy})$ are also used for anisotropic spaces.

3. Existence in sub-critical case

In this section we state the existence result in sub-critical case, i.e., for $\delta\in(0,1).$ We provide a short motivation for this study at the final of the section.

3.1. Sub-critical nonlinearity

Theorem 3.1**.**

Let $\delta\in(0,1)$ and $u_{0}\in L^{2}(\Omega)$ be a given function. Then for all finite positive $B,\ L,\ T$ there exists a weak solution to (2.1)-(2.4) such that

[TABLE]

To prove this theorem we consider for all real $\varepsilon>0$ the following parabolic regularization of (2.1)-(2.4):

[TABLE]

For all $\varepsilon>0,$ (3.1)-(3.4) admits a unique regular solution in $Q_{T}$ [8]. In what follows we omit the subscript $\varepsilon$ whenever it is unambiguous.

Multiplying $A^{\varepsilon}u_{\varepsilon}$ by $u_{\varepsilon}$ and integrating over $Q_{T},$ we have

[TABLE]

Multiplying $A^{\varepsilon}u_{\varepsilon}$ by $xu_{\varepsilon}$, integrating over $\Omega$ with the use of the Nirenberg, Hölder and Young inequalities yields

[TABLE]

Integrating with respect to $t>0$ in (3.1) and taking $\varepsilon<1/2$ gives

[TABLE]

Remark 3.1**.**

Note that (3.1) does not hold for critical case, i.e., while $\delta\to 1.$

Estimates (3.5) and (3.1) thus become

[TABLE]

where limitations do not depend on $\varepsilon$ but depend only on $T$, $\delta$, $\Omega$ and $\|u_{0}\|$.

Thanks to (3.8) we have boundness of $u_{\varepsilon}^{1+\delta}u_{\varepsilon x}$ for all $\delta\in(0,1)$. In fact, given $\delta\in(0,1)$ take $m=\frac{4}{3+\delta}$ and $\kappa(\delta)=\frac{1+\delta}{3+\delta}.$ Then Hölder’s and Nirenberg’s inequality yield

[TABLE]

Therefore, due to (3.9) and (3.8) we conclude that $u^{1+\delta}u_{x}$ is bounded in $L^{m}(0,T;L^{m}(\Omega)).$ Since $L^{\frac{4}{1-\delta}}$ is the dual space of $L^{\frac{4}{3+\delta}}$ and $H^{1}\subset L^{\frac{4}{1-\delta}}$ in dimension 2, we have as well

[TABLE]

Thanks to (3.8) and (3.10) jointly with the equation, we get

[TABLE]

which assures the family $u_{\varepsilon}$ to be relatively compact in $L^{2}(0,T;L^{2}(\Omega))$. This is sufficiently to obtain the existence of $\lim u_{\varepsilon}$ as $\varepsilon\rightarrow 0$, using the compactness argument in the nonlinear term.

The initial condition $u(x,y,0)=u_{0}(x,y)$ is fulfilled; indeed, due to (3.11) $u_{\varepsilon}$ converges to $u$ in C\big{(}[0,T];H^{-3}_{w}(\Omega)\big{)}, where $H^{-3}_{w}$ is $H^{-3}$ equipped with the weak topology.

By the same way, the Dirichlet condition $u=0$ onto $\partial\Omega$ is satisfied since $u_{\varepsilon}$ converges to $u$ weakly in $L^{2}(0,T;H^{1}_{0}(\Omega)).$ It remains to show that $u_{x}(L,y,t)=0,$ which is done by the following two lemmas (cf. [14, 15]).

Lemma 3.1**.**

If $u\in L^{\infty}(0,T;L^{2}(\Omega))\cap L^{2}(0,T;H^{1}_{0}(\Omega))$ solves (2.1), then

[TABLE]

and, in particular,

[TABLE]

*are well defined in $V$. Moreover, these traces depend continuously of $u$ in an appropriate sense. *

To prove this lemma, write (2.1) in the form

[TABLE]

and observe that

[TABLE]

Accordingly with (3.10) and definition of $V$ in (3.12), it holds

[TABLE]

Thus we have

[TABLE]

and (3.12) and (3.13) follow. Moreover, if a sequence of functions $u_{m}$ satisfies (LABEL:eq_mZK) and $u_{m}\rightarrow u$ in $L^{\infty}(0,T;L^{2}(\Omega))\cap L^{2}(0,T;H^{1}_{0}(\Omega))$ strongly, then u_{mx}\big{|}_{x=0,1},\;\;u_{mxx}\big{|}_{x=0,1} converge to u_{x}\big{|}_{x=0,1},\;\;u_{xx}\big{|}_{x=0,1} in $V.$ If a convergence of $u_{m}$ being weak (star-weak for $L^{\infty}$,) then a convergence take place in $C(0,L;V_{w})$ and $Y_{w}$. This is based on compactness arguments justified by (3.11), used to prove that $u_{m}^{1+\delta}u_{mx}\rightarrow u^{1+\delta}u_{x}$.

Lemma 3.2**.**

Let $U$ be a reflexive Banach space and $p\geq 1$. Suppose that two function sequences $u_{\varepsilon},\ g_{\varepsilon}\in L^{p}(0,L;U)$ satisfy

[TABLE]

with $g_{\varepsilon}$ being bounded in $L^{p}(0,L;U)$ as $\varepsilon\to 0.$ Then $u_{\varepsilon xx}$ (consequently $u_{\varepsilon x},$ and $u_{\varepsilon}$) is bounded in $L^{\infty}(0,L;U)$ as $\varepsilon\rightarrow 0.$ Moreover, for a subsequence $u_{\varepsilon}\to u$ converging (strongly or weakly) in $L^{q}(0,L;U),$ $1\leq q<\infty,$ it holds that $u_{\varepsilon x}(L)$ converges to $u_{x}(L)$ in $U$ (at least weakly), and therefore $u_{x}(L)=0.$

See [15] for the proof.

To prove Theorem 3.1, apply the above lemmas with

[TABLE]

[TABLE]

and

[TABLE]

The proof is completed.

3.2. Motivation and explanation of the main difficulty

Note that inclusions (3.8) can be obtained also for $\delta=1$ with $\|u_{0}\|<1/2.$ Using embedding machinery and interpolation theory for anisotropic spaces, one could pass to the limit as $\varepsilon\to 0$ in nonlinear term, as well. Indeed, let $\delta=1.$ Multiplying $A^{\varepsilon}u_{\varepsilon}=0$ by $2(1+x)u_{\varepsilon}$ and integrating over $\Omega,$ we have

[TABLE]

Bearing in mind that $\|u\|(t)\leq\|u_{0}\|(t)<1/2$ and integrating in $t>0,$ Gronwall’s lemma gives

[TABLE]

with both estimates independent of $\varepsilon<1/4.$

Now we observe that

[TABLE]

and by estimate above this implies $u^{3}\in L^{\frac{4}{3}}(Q_{T}).$ Since $L^{\frac{4}{3}}(\Omega)\hookrightarrow H^{-1}(\Omega),$ we conclude that

[TABLE]

whence

[TABLE]

and passage to the limit as $\varepsilon\to 0$ in nonlinear term can be justified as above.

It is difficult, however, to obtain explicit estimates like (3.9) with $m>1$ for $\delta=1.$ In fact, let $r,s\geq 1.$ We are going to determine conditions upon $r$ and $s$ such that $u^{2}u_{x}$ lies in $L^{r}\left((0,T;L^{s}(\Omega)\right).$ Consider $p,q>1$ with $1/p+1/q=1.$ Then

[TABLE]

By Nirenberg’s inequality with $\alpha=\frac{sp-1}{sp}$ one has

[TABLE]

Supposing $sq\leq 2,$ estimate (3.18) reads

[TABLE]

In order to gain $r(2\alpha+1)=2,$ it should be $\alpha=1/r-1/2.$ Therefore, $\frac{1}{sp}=\frac{3}{2}-\frac{1}{r},$ which implies

[TABLE]

Since $sq\leq 2,$ it follows that $\frac{2rs}{2(r+s)-3rs}\leq 2$ which means $sr\leq\frac{r+s}{2}.$ Observe that for $r,s>1$ this condition does not hold. The only possibility thus reads $r=s=1,$ i.e., $u^{2}u_{x}\in L^{1}\left((0,T;L^{1}(\Omega)\right).$

The space $(L^{1}_{t};L^{1}_{xy})$ is known to be difficult to deal with. For example, it is not clear even whether the condition $u_{x}(L,y,t)=0$ being satisfied. We leave it here only to illustrate a challenge appearing in the critical case.

4. Local result for critical case

Consider the following Cauchy problem in abstract form:

[TABLE]

where $f\in L^{1}(0,T;L^{2}(\Omega))$ and $A:L^{2}(\Omega)\to L^{2}(\Omega)$ defined as $A\equiv\partial_{x}+\Delta\partial_{x}$ with the domain

[TABLE]

endowed with its natural Hilbert norm $\|u\|_{D(A)}(t)=\left(\|u\|^{2}_{L^{2}(\Omega)}(t)+\|\Delta u_{x}+u_{x}\|^{2}_{L^{2}(\Omega)}(t)\right)^{1/2}$ for all $t\in(0,T)$.

Proposition 4.1**.**

Assume $u_{0}\in D(A)$ and $f\in L^{1}_{loc}({\mathbb{R}}^{+};L^{2}(\Omega))$ with $f_{t}\in L^{1}_{loc}({\mathbb{R}}^{+};L^{2}(\Omega))$. Then problem (4.1) possesses the unique solution $u(t)$ such that

[TABLE]

Moreover, if $u_{0}\in L^{2}(\Omega)$ and $f\in L^{1}_{loc}({\mathbb{R}}^{+};L^{2}(\Omega)),$ then (4.1) possesses a unique (mild) solution $u\in C([0,T];L^{2}(\Omega))$ given by

[TABLE]

Corollary 4.1**.**

Under the hypothesys of Proposition 4.1, the solution $u$ in (4.2) satisfies

[TABLE]

For the proof, see [15].

Furthermore, one can get (see [7], for instance) the estimate for strong solution (4.2):

[TABLE]

and

[TABLE]

Since $D(A){\hookrightarrow}H^{1}_{0}(\Omega)\cap H^{2}(\Omega)$ compactly (see [15] for instance), we have the estimate

[TABLE]

where $C$ depends only on $\Omega$. Next, we define

[TABLE]

with the norm

[TABLE]

Remark 4.1**.**

If $f\in Y_{T},$ then $f\in C([0,T];L^{2}(\Omega))$, with the constant $C_{T}$ from $\|f\|_{C_{t}L^{2}_{xy}}\leq C_{T}\|f\|_{Y_{T}}$ which is proportional to $T$ and its positive powers [3].

Consider X_{T}^{0}=L^{\infty}\big{(}0,T;H^{1}_{0}(\Omega)\cap H^{2}(\Omega)\big{)} and define the Banach space

[TABLE]

with the norm

[TABLE]

Theorem 4.1**.**

Let $u_{0}\in D(A)$. Then there exists $T>0$ such that IBVP (2.1)-(2.4) possesses a unique solution in $X_{T}$.

The proof of the Theorem consists in three lemmas below.

Lemma 4.1**.**

The function $Y_{T}\longrightarrow X_{T};\ \ f\mapsto\int_{0}^{t}S(t-s)f(s)ds$ is well defined and continuous.

For the proof, note that this function maps $f$ to the solution of homogeneous linear problem with zero initial datum. Estimates (4.5) and (4.7) then give

[TABLE]

where $C$ is as above. Thus, it rests to estimate the term $\|\nabla u_{t}\|_{L^{2}_{T}L^{2}_{xy}}$ in (4.10).

Differentiate the equation in (4.1) with respect to $t,$ multiply it by $(1+x)u_{t}$ and integrate the outcome over $\Omega.$ The result reads

[TABLE]

Hölder’s inequality and (4.5) imply

[TABLE]

Using the equation from (4.1) and taking in mind that $u_{0}\equiv 0$, we get

[TABLE]

Inserting (4.15) into (4.14) provides

[TABLE]

where $K_{T}=\max\{1,C_{T}\}$. Therefore, estimates (4.12) and (4.16) read

[TABLE]

Lemma 4.2**.**

The function

[TABLE]

is well defined and continuous.

The proof follows the same steps as Lemma 4.1, taking into account that now $f\equiv 0$. The resulting estimate is

[TABLE]

where $M$ is given by

[TABLE]

and $C$ (which depends only on $\Omega$) is defined by continuous immersion $D(A)\hookrightarrow H^{1}_{0}(\Omega)\cap H^{2}(\Omega).$

Lemma 4.3**.**

Given $R>0$, consider the closed ball $B_{R}=\{u\in X_{T};\|u\|_{X_{T}}\leq R\}.$ Then the operator

[TABLE]

is the contraction.

Fix $R>0$ and $u,v\in B_{R}.$ We have

[TABLE]

so that (4.17) implies

[TABLE]

We study the right-hand norm in detail:

[TABLE]

First, we write

[TABLE]

For the integral $I_{1}$ one has

[TABLE]

Nirenberg’s inequality gives

[TABLE]

where $D$ is the Poincare’s constant from $\|w\|\leq D\|\nabla w\|.$ Since $u$ and $v$ lie in $B_{R},$ we conclude

[TABLE]

The integral $I_{2}$ can be treated in the similar way as $I_{1}$. It rests to estimate the integral $J$.

[TABLE]

For $J_{1}$ we have

[TABLE]

Niremberg’s inequality implies

[TABLE]

The integrals $J_{3}$ and $J_{4}$ are analogous to $J_{1}$. To get bound for $J_{5}$ we observe that

[TABLE]

The integral $J_{2}$ follows like $J_{5}$. Thus,

[TABLE]

Finally, choosing $T>0$ such that $KK^{*}T^{\frac{1}{2}}R^{2}<1,$ we conclude that $\Phi$ is a contraction map.

Lemma 4.3 is proved.

Let $u\in B_{R}$. If $R=2M\|u_{0}\|_{D(A)},$ then estimates (4.18) and (4.30) with $v\equiv 0$ assure

[TABLE]

Setting $T>0$ such that $KK^{*}T^{\frac{1}{2}}R^{3}<\frac{R}{2},$ one get

[TABLE]

Choose $T>0$ such that $KK^{*}T^{\frac{1}{2}}R^{2}<1$ and $KK^{*}T^{\frac{1}{2}}R^{3}<\frac{R}{2}.$ Then $\Phi$ is the contraction from the ball $B_{R}$ into itself. Therefore, the Banach fixed point theorem assures the existence of a unique element $u\in B_{R}$ such that $\Phi(u)=u.$

This completes the proof of Theorem 4.1.

5. Decay

Theorem 5.1**.**

Let $B,L>0$ satisfy

[TABLE]

If there exists solution

[TABLE]

to (2.1)-(2.4), then

[TABLE]

To prove this result we will use

Lemma 5.1**.**

(V. A. Steklov)* Let $L,B>0$ and $\omega\in H_{0}^{1}(\Omega)$. Then*

[TABLE]

*and *

[TABLE]

See [2] for the proof. We start the proof of (5.1), multiplying (2.1) by $u$ and integrating over $Q_{t},$ which easily gives

[TABLE]

Multiplying (2.1) by $(1+x)u$ and integrating over $\Omega,$ we have

[TABLE]

For the integral $I_{1}=\frac{1}{2}\int_{\Omega}u^{4}=\frac{1}{2}\|u\|^{4}_{L^{4}(\Omega)}(t),$ Nirenberg’s inequality implies

[TABLE]

Take

[TABLE]

For all $\varepsilon>0$ we have

[TABLE]

Lemma 5.1 jointly with (5) and (5.6) provides

[TABLE]

Define

[TABLE]

The result for (5) reads

[TABLE]

If $0\leq\varepsilon-2\|u_{0}\|^{2},$ then

[TABLE]

and consequently

[TABLE]

The proof is completed.