## Existence and Blow-Up of a Parabolic Equation with Logarithmic Nonlinearity

Du Yuge,, Tian Shuying,

School of Science, Wuhan University of Technology, Wuhan 430070

 基金资助: 国家自然科学基金.  11601402中央高校基本科研业务费专项资金.  2020IA003

Received: 2020-07-8

 Fund supported: the NSFC.  11601402the Fundamental Research Funds for the Central Universities.  2020IA003

Abstract

In this paper, we consider the initial boundary value problem of a viscoelastic equation with logarithmic nonlinearity. Under some suitable conditions, we obtain the existence of global weak solutions. Otherwise, we get that the solution does not blow up in any finite time. This is different from the situation of the viscoelastic equation with a polynomial nonlinearity, in which case the solution blows up in finite time.

Keywords： Logarithmic nonlinearity ; Viscoelastic term ; Global existence ; Blow-up

Du Yuge, Tian Shuying. Existence and Blow-Up of a Parabolic Equation with Logarithmic Nonlinearity. Acta Mathematica Scientia[J], 2021, 41(6): 1816-1829 doi:

## 1 引言

$$$\left\{\begin{array}{ll} u_{t}-{\Delta}u+{ \int_{0}^{t}{g(t-s){\Delta}u(x, s){\rm d}s}} = u\log\vert u\vert, &\mbox{在{\Omega}{\times}(0, T) 中}, \\ u(x, t) = 0, &\mbox{在{\partial}{\Omega}{\times}(0, T)上}, \\ u(x, 0) = u_{0}(x), &\mbox{在\Omega中}, \\ \end{array}\right.$$$

$\begin{eqnarray} J(u)& = &\frac{1}{2}\int_0^t{g(t-s)\Vert{\nabla}u(\cdot, t)-\nabla u(\cdot, s)\Vert _{L^2(\Omega)}^2{\rm d}s} +\frac{1}{2}\Big(1-\int_0^t{g(s){\rm d}s}\Big)\Vert\nabla u(\cdot, t)\Vert_{L^2(\Omega)}^2 {}\\ &&-\frac{1}{2}\int_{\Omega}{u^2(x, t)\log\vert u(x, t)\vert {\rm d}x}+\frac{1}{4}\int_{\Omega}{u^2(x, t){\rm d}x}, \end{eqnarray}$

$\begin{eqnarray} I(u)& = &\int_0^t{g(t-s)\Vert{\nabla}u(\cdot, t)-\nabla u(\cdot, s)\Vert _{L^2(\Omega)}^2{\rm d}s} +\Big(1-\int_0^t{g(s){\rm d}s}\Big)\Vert\nabla u(\cdot, t)\Vert_{L^2(\Omega)}^2{}\\ &&-\int_{\Omega}{u^2(x, t)\log\vert u(x, t)\vert {\rm d}x}, \end{eqnarray}$

$$$J(u) = \frac{1}{2}I(u)+\frac{1}{4}\int_{\Omega}{u^2(x, t){\rm d}x}.$$$

(ⅰ) $\lim\limits_{\lambda\to 0}J(\lambda u) = 0, \lim\limits_{\lambda\to +\infty}J(\lambda u) = -\infty;$

(ⅱ) $J(\lambda u) $$0\leq \lambda \leq \lambda^* 上单调递增, 在 \lambda^*\leq \lambda <+\infty 上单调递减, 并且在 \lambda = \lambda^* 处取得最大值, 这里 (ⅲ) 对任意的 \lambda> 0 , 由于 \Vert u\Vert_{L^2(\Omega)}\neq 0 , 则 \lim\limits_{\lambda\to 0}J(\lambda u) = 0, \;\lim\limits_{\lambda\to +\infty}J(\lambda u) = -\infty , 且 因此, 可得 其中 故引理2.2得证. 下面的引理对于不变性的证明是非常重要的. 引理2.3 设 u = u(\cdot, t)\in H_0^1(\Omega), 则有下列性质 (ⅰ)若 0<\Vert\nabla u\Vert_{L^2(\Omega)}\leq r , 则 I(u)\geq 0; (ⅱ)若 I(u)<0 , 则 \Vert\nabla u\Vert_{L^2(\Omega)}>r; (ⅲ)若 I(u) = 0 , 则 \Vert u\Vert_{L^2(\Omega)} = 0 或者 \Vert\nabla u\Vert_{L^2(\Omega)}\geq r. 这里 r = \lambda_1^{\frac{1}{2}}(2\pi l)^{\frac{n}{4}}e^{\frac{n}{2 }} , \lambda_1 是下列方程的第一特征值 (ⅰ)由(2.1)式, 对任意的 a>0 , 有 \begin{eqnarray} I(u)&\geq &\int_0^t g(t-s)\Vert \nabla u(t)-\nabla u(s)\Vert _{L^2(\Omega)}^2{\rm d}s +\bigg[\Big(1-\int_0^t{g(s){\rm d}s}\Big)-\frac{a^2}{2\pi}\bigg]\Vert\nabla u\Vert_{L^2(\Omega)}^2 {}\\ &&+\Big(\frac{n(1+\log a)}{2}-\log\Vert u\Vert_{L^2(\Omega)}\Big)\Vert u \Vert_{L^2(\Omega)}^2. \end{eqnarray} 结合 g(s)\geq 0 , 在(2.3)式中, 取 a = \sqrt{2\pi l} , 可得 $$I(u)\geq \Big(\frac{n(1+\log\sqrt{2\pi l})}{2}-\log\Vert u\Vert_{L^2(\Omega)}\Big)\Vert u \Vert_{L^2(\Omega)}^2.$$ 0<\Vert\nabla u\Vert_{L^2(\Omega)}\leq r , 则 \Vert u \Vert_{L^2(\Omega)}\leq(2\pi l)^{\frac{n}{4}}e^{\frac{n}{2}} , 从而可得 I(u)\geq 0 . (ⅱ)根据(2.4)式和 I(u)\leq 0 , 可得 (ⅲ)若 I(u) = 0 , \Vert u\Vert_{L^2(\Omega)}\neq 0 , 则通过(2.4)式, 可得 故引理2.3得证. 以下解的不变性结论对于定理1.1和1.2的证明是必不可少的. 命题2.1 设 u = u(x, t) 是问题(1.1)的弱解, 其初值 u_0\in H_0^1(\Omega)$$ 0<J(u_0)<d,$则有

(ⅰ)若$I(u_0)>0,$则对任意的$0\leq t<T$, 有$u\in W;$

(ⅱ)若$I(u_0)<0,$则对任意的$0\leq t<T$, 有$u\in V$.

(ⅰ)根据引理2.1, 对任意的$t\in [0, T),$

$\begin{eqnarray} d>J(u_0)& = &J(u)+\int_0^t\Vert u_\tau \Vert_{L^2(\Omega)}^2{\rm d}\tau+\frac{1}{2}\int_0^t g(\tau)\Vert \nabla u(\tau) \Vert_{L^2(\Omega)}^2{\rm d}\tau {}\\ &&-\frac{1}{2}\int_0^t \int_0^{\tau} g'(\tau-s)\Vert \nabla u(\tau)-\nabla u(s) \Vert_{L^2(\Omega)}^2 {\rm d}s{\rm d}\tau. \end{eqnarray}$

(ⅰ)假设命题不成立, 则存在$t_1\in (0, T)$使得

$\begin{eqnarray} J(u(t_1))& = &J(u_0)-\int_0^{t_1}\Vert u_\tau \Vert_{L^2(\Omega)}^2{\rm d}\tau-\frac{1}{2}\int_0^{t_1} g(\tau)\Vert \nabla u(\tau) \Vert_{L^2(\Omega)}^2{\rm d}\tau{}\\ &&+\frac{1}{2}\int_0^{t_1} \int_0^{\tau} g'(\tau-s)\Vert \nabla u(\tau)-\nabla u(s) \Vert_{L^2(\Omega)}^2 {\rm d}s{\rm d}\tau<d. \end{eqnarray}$

$$$\big(u_{mt}, \omega_k\big)_2+\big({\nabla{u}}_m, {\nabla{\omega}}_k\big)_2- \Big(\int_{0}^{t}g(t-s){\nabla}u_m(x, s){\rm d}s, \nabla{\omega}_k\Big)_2 = \big(u_m {\log}\vert{u_m}\vert, \omega_k\big)_2, \\$$$

$$$u_m(x, 0) = \sum\limits_{j = 1}^{m}a^j_m\omega_j(x)\to u_0, \ \mbox{在 H_0^1(\Omega)中}.$$$

$\begin{eqnarray} l\Vert \nabla u_m(t)\Vert_{L^2(\Omega)}^2&\leq& \Big(1-\int_0^t g(s){\rm d}s\Big)\Vert \nabla u_m(t)\Vert_{L^2(\Omega)}^2{}\\ & = &2I(u_m(t))-2\int_0^tg(t-s)\Vert \nabla u_m(t)-\nabla u_m(s)\Vert_{L^2(\Omega)}^2 {\rm d}s{}\\ && +2\int_{\Omega}u_m^2(t)\log \vert u_m(t)\vert {\rm d}x-\Big(1-\int_0^t g(s){\rm d}s\Big)\Vert \nabla u_m(t)\Vert_{L^2(\Omega)}^2{}\\ &\leq& 2I(u_m(t))+2\int_{\Omega}u_m^2(t)\log \vert u_m(t)\vert {\rm d}x-l\Vert \nabla u_m(t)\Vert_{L^2(\Omega)}^2{}\\ &\leq &2I(u_m(t))+\big[2\log\Vert u_m(t)\Vert_{L^2(\Omega)}-n(1+\log \sqrt{\pi l})\big]\Vert u_m(t)\Vert_{L^2(\Omega)}^2{}\\ & = &4J(u_m(t))+\big[2\log\Vert u_m(t)\Vert_{L^2(\Omega)}-n(1+\log \sqrt{\pi l})-1\big]\Vert u_m(t)\Vert_{L^2(\Omega)}^2{}\\ &\leq& C_d. \end{eqnarray}$

$$$\int_0^t\Vert u_{m\tau}\Vert_{L^2(\Omega)}^2 {\rm d}\tau <d.$$$

$\begin{eqnarray} \int_{\Omega}(u_m\log\vert u_m\vert)^2{\rm d}x& = &\int_{\big\{x\in\Omega;u_m(x)\leq 1\big\}}(u_m\log\vert u_m\vert)^2{\rm d}x+ \int_{\big\{x\in\Omega;u_m(x)> 1\big\}}(u_m\log\vert u_m\vert)^2{\rm d}x{}\\ &\leq &e^{-2}\vert\Omega\vert+\big(\frac{n-2}{2}\big)^2\int_{\big\{x\in\Omega;u_m(x)> 1\big\}}u_m^{\frac{2n}{n-2}}{\rm d}x{}\\ &\leq &e^{-2}\vert\Omega\vert+\big(\frac{n-2}{2}\big)^2S^{2^*}\Vert \nabla u_m\Vert_{L^2(\Omega)}^{2^*}\leq C_d, \end{eqnarray}$

$\begin{eqnarray} I_1&\leq& 2l\int_0^{t}g(s){\rm d}s\Vert \nabla u(t)\Vert_{L^2(\Omega)}^2+\frac{1}{2l}\int_0^{t}g(t-s)\Vert \nabla u(t)-\nabla u(s)\Vert_{L^2(\Omega)}^2{}\\ &\leq& 2l(1-l)\Vert \nabla u(t)\Vert_{L^2(\Omega)}^2+I_2, \end{eqnarray}$

$\begin{eqnarray} I_2& = &\frac{1}{2l}\bigg[I(u)-\Big(1-\int_0^{t}g(s){\rm d}s\Big)\Vert \nabla u(t)\Vert_{L^2(\Omega)}^2 +\int_{\Omega}u^2(t)\log\vert u(t)\vert {\rm d}x\bigg]{}\\ &\leq &\frac{1}{2l}\int_{\Omega}u^2(t)\log\vert u(t)\vert {\rm d}x-\frac{1}{2}\Vert \nabla u(t)\Vert_{L^2(\Omega)}^2. \end{eqnarray}$

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