## 具有非线性记忆项的半线性双波动方程解的全局非存在性

1 广州华商学院 广州 511300

2 广东农工商职业技术学院 广州 510507

## Nonexistence of Global Solutions for a Semilinear Double-Wave Equation with a Nonlinear Memory Term

Ouyang Baiping,1, Xiao Shengzhong,2

1 Guangzhou Huashang College, Guangzhou 511300

2 Guangdong AIB Polytechnic College, Guangzhou 510507

 基金资助: 国家自然科学基金.  11371175广东省普通高校创新团队.  2020WCXTD008广东财经大学华商学院校内项目.  2020HSDS01广州华商学院科研团队.  2021HSKT01

 Fund supported: the NSFC.  11371175the Innovation Team Project in Colleges and Universities of Guangdong Province.  2020WCXTD008the Science Foundation of Huashang College Guangdong University of Finance & Economics.  2020HSDS01the Science Research Team Project in Guangzhou Huashang College.  2021HSKT01

Abstract

In this paper, we investigate the blow-up of solutions to a semilinear double-wave equation with a nonlinear memory term. By establishing some auxiliary functions and using iteration methods associated with a nonlinear integral inequality, the estimate of upper bound for the lifespan is obtained.

Keywords： Semilinear double-wave equation ; Blow-up ; Nonlinear memory term ; Lifespan

Ouyang Baiping, Xiao Shengzhong. Nonexistence of Global Solutions for a Semilinear Double-Wave Equation with a Nonlinear Memory Term. Acta Mathematica Scientia[J], 2021, 41(5): 1372-1381 doi:

## 1 引言

$\begin{eqnarray} \left\{\begin{array}{ll} u_{tt}-\Delta u = {\cal N}_{\gamma, p}(u), & (x, t)\in {{\Bbb R}} ^n \times (0, T), \\ (u, u_t)(0, x) = \varepsilon (u_0, u_1)(x), & x\in {{\Bbb R}} ^n, \end{array}\right. \end{eqnarray}$

$\begin{eqnarray} \begin{array}{ll} { } {\cal N}_{\gamma, p}(u) = C_{\gamma}\int_0^t (t-s)^{-\gamma} |u(s, x)|^p {\rm d}s, C_{\gamma} = \frac{1}{\Gamma(1-\gamma)}, \\ p>1, \gamma\in (0, 1), \varepsilon>0, \ \Gamma\ \mbox{为欧拉积分.} \end{array} \end{eqnarray}$

$\begin{eqnarray} \left\{\begin{array}{ll} (\partial^2_t-\Delta)^2u = {\cal N}_{\gamma, p}(u), & (x, t)\in {{\Bbb R}} ^n \times (0, T), \\ (u, u_t, u_{tt}, u_{ttt})(0, x) = \varepsilon (u_0, u_1, u_2, u_3)(x), & x\in {{\Bbb R}} ^n, \end{array}\right. \end{eqnarray}$

## 2 主要结果

$\begin{array}{l} \int_{{{\Bbb R}} ^n} u_{ttt}(t, x)\phi(t, x) {\rm d}x-\int_{{{\Bbb R}} ^n}u_{ttt}(0, x)\phi(0, x){\rm d}x-\int^t_0\int_{{{\Bbb R}} ^n}u_{ttt}(s, x)\phi_t(s, x){\rm d}x{\rm d}s\\+2\int^t_0\int_{{{\Bbb R}} ^n}\nabla u_{tt}(s, x)\cdot\nabla\phi(s, x) {\rm d}x{\rm d}s-\int^t_0\int_{{{\Bbb R}} ^n}\nabla \Delta u(s, x)\cdot \nabla \phi(s, x){\rm d}x{\rm d}s\\ = C_{\gamma}\int_0^t \int_{{{\Bbb R}} ^n}\phi(s, x)\int_0^s(s-\tau)^{-\gamma} |u(\tau, x)|^p {\rm d}\tau {\rm d}x{\rm d}s, \end{array}$

(2.1)式中运用分部积分, 有

$\begin{array}{l}\int_{{{\Bbb R}} ^n} u_{ttt}(t, x)\phi(t, x) {\rm d}x-\varepsilon\int_{{{\Bbb R}} ^n}u_3(x)\phi(0, x){\rm d}x-\int_{{{\Bbb R}} ^n} u_{tt}(t, x)\phi_t(t, x) {\rm d}x\\+\varepsilon\int_{{{\Bbb R}} ^n}u_2(x)\phi_t(0, x){\rm d}x+\int_{{{\Bbb R}} ^n} u_t(t, x)\phi_{tt}(t, x){\rm d}x-\varepsilon\int_{{{\Bbb R}} ^n}u_1(x)\phi_{tt}(0, x){\rm d}x\\-\int_{{{\Bbb R}} ^n} u(t, x)\phi_{ttt}(t, x) {\rm d}x+\varepsilon\int_{{{\Bbb R}} ^n}u_0(x)\phi_{ttt}(0, x){\rm d}x\\+\int^t_0\int_{{{\Bbb R}} ^n}u(s, x)\phi_{tttt}(s, x){\rm d}x{\rm d}s-2\int_{{{\Bbb R}} ^n} u_t(t, x)\Delta\phi(t, x){\rm d}x+2\varepsilon\int_{{{\Bbb R}} ^n} u_1(x)\Delta\phi(0, x){\rm d}x \\+2\int_{{{\Bbb R}} ^n} u(t, x)\Delta\phi_t(t, x){\rm d}x-2\varepsilon\int_{{{\Bbb R}} ^n} u_0(x)\Delta\phi_t(0, x){\rm d}x-2\int^t_0\int_{{{\Bbb R}} ^n} u(s, x)\Delta\phi_{tt}(s, x) {\rm d}x{\rm d}s\\-\int^t_0\int_{{{\Bbb R}} ^n}\nabla \Delta u(s, x)\cdot \nabla \phi(s, x){\rm d}x{\rm d}s\\ = C_{\gamma}\int_0^t \int_{{{\Bbb R}} ^n}\phi(s, x)\int_0^s(s-\tau)^{-\gamma} |u(\tau, x)|^p {\rm d}\tau {\rm d}x{\rm d}s. \end{array}$

$t\rightarrow T$, 易知$u$满足(1.3)式弱解的定义.

$(u_0, u_1, u_2, u_3)\in H^3({{\Bbb R}} ^n)\times H^2({{\Bbb R}} ^n)\times H^1({{\Bbb R}} ^n)\times L^2({{\Bbb R}} ^n)$是非负的紧支集函数, 其支集包含于$B_R(R>0)$, 使得$u_0 $$u_1$$ u_2 $$u_3 不恒为 0 . 其中, B_R 是一个原点为球心半径为 R 的球. 特别地, u_3(x)+u_2(x)>u_1(x)+u_0(x)$$ 2u_1(x)>u_0(x)+u_2(x)$. 如果$u$是问题(0.3)式的能量解, 其生命跨度$T(\varepsilon)$满足supp $u(t, \cdot)\subset B_{t+R}, t\in (0, T)$, 则存在一个正常数$\varepsilon_0 = \varepsilon_0(u_0, u_1, u_2, u_3, n, p, \gamma, R)$, 使得当$\varepsilon\in (0, \varepsilon_0] $$u 在有限时间内爆破. 进一步, 可得解的生命跨度上界估计, 即 其中 \widetilde{\widetilde{C}} 是独立于 \varepsilon 的正常数并且 $$\Upsilon(p, n, \gamma) = (n+5-2\gamma)p+2-(n-3)p^2.$$ ## 3 主要结果的证明 $$F(t) = \int_{{{\Bbb R}} ^n} u(t, x){\rm d}x.$$ \phi 满足 \phi\equiv1, \{(s, x)\in [0, t]\times {{\Bbb R}} ^n:|x|\leq R+s\} . 于是, 由(2.2)式, 得 $$\int_{{{\Bbb R}} ^n} u_{ttt}(t, x){\rm d}x-\varepsilon\int_{{{\Bbb R}} ^n}u_3(x){\rm d}x = C_{\gamma}\int_0^t \int_{{{\Bbb R}} ^n}\int_0^s(s-\tau)^{-\gamma} |u(\tau, x)|^p {\rm d}\tau {\rm d}x{\rm d}s.$$ 从而有 $$F'''(t) = F'''(0)+ C_{\gamma}\int_0^t \int_{{{\Bbb R}} ^n}\int_0^s(s-\tau)^{-\gamma} |u(\tau, x)|^p {\rm d}\tau {\rm d}x{\rm d}s.$$ 对(3.3)式关于时间 t 积分三次, 得到 \begin{eqnarray} F(t)& = &F(0)+F'(0)t+\frac{1}{2}F''(0)t^2+\frac{1}{6}F'''(0)t^3\\+C_{\gamma}\int_0^t \int_0^s\int_0^\tau\int_0^\sigma\int_0^\eta\int_{{{\Bbb R}} ^n}(\eta-\xi)^{-\gamma} |u(\xi, x)|^p {\rm d}x{\rm d}\xi{\rm d}\eta{\rm d}\sigma{\rm d}\tau{\rm d}s. \end{eqnarray} 由supp u(t, \cdot)\subset B_R+t, t\in (0, T) 和Hölder不等式, 可得 \begin{eqnarray} \int_{{{\Bbb R}} ^n} |u(\xi, x)|^p{\rm d}x\geq C(R+\xi)^{-n(p-1)}(F(\xi))^p. \end{eqnarray} 根据(3.4)式及初始值的非负假设, 易知 F(\xi)\geq0, \xi \in [0, t]. 联立(3.4)–(3.5)式, 有 \begin{eqnarray} F(t)\geq CC_{\gamma}(R+t)^{-n(p-1)-\gamma}\int_0^t \int_0^s\int_0^\tau\int_0^\sigma\int_0^\eta(F(\xi))^p {\rm d}\xi{\rm d}\eta{\rm d}\sigma{\rm d}\tau{\rm d}s. \end{eqnarray} 接下来, 通过对 F(t) 的下界序列进行迭代完成定理的证明. 实际上, (3.6) 式已给出了下界序列的迭代框架. 为了推导 F(t) 的第一下界估计, 引入如下函数 其中 \Phi 是正的光滑函数且有下面的性质 \Psi = \Psi(t, x) = {\rm e}^{-t}\Phi(x). 易知, \Psi 是方程 (\partial^2_t-\Delta)^2\Psi = 0 的解. 给定辅助函数 \begin{eqnarray} F_0(t) = \int_{{{\Bbb R}} ^n}u(t, x)\Psi(t, x){\rm d}x. \end{eqnarray} (3.3)式对 t 求导数, 得 \begin{eqnarray} F''''(t) = C_{\gamma}\int_0^t\int_{{{\Bbb R}} ^n}(t-s)^{-\gamma} |u(s, x)|^p {\rm d}x{\rm d}s. \end{eqnarray} 运用Hölder不等式于(3.7)式, 有 \begin{eqnarray} \int_{{{\Bbb R}} ^n} |u(s, x)|^p {\rm d}x\geq |F_0(s)|^p\bigg(\int_{B_{R+s}} |\Psi(s, x)|^{\frac{p}{p-1}} {\rm d}x\bigg)^{-(p-1)}. \end{eqnarray} 将测试函数 \Psi 应用于(2.1)式, 可推出 \begin{array}{l} \int_{{{\Bbb R}} ^n} u_{ttt}(t, x)\Psi(t, x) {\rm d}x-\int_{{{\Bbb R}} ^n}u_{ttt}(0, x)\Psi(0, x){\rm d}x-\int^t_0\int_{{{\Bbb R}} ^n}u_{ttt}(s, x)\Psi_t(s, x){\rm d}x{\rm d}s\\+2\int^t_0\int_{{{\Bbb R}} ^n}\nabla u_{tt}(s, x)\cdot\nabla\Psi(s, x) {\rm d}x{\rm d}s-\int^t_0\int_{{{\Bbb R}} ^n}\nabla \Delta u(s, x)\cdot \nabla \Psi(s, x){\rm d}x{\rm d}s\\ = C_{\gamma}\int_0^t \int_{{{\Bbb R}} ^n}\Psi(s, x)\int_0^s(s-\tau)^{-\gamma} |u(\tau, x)|^p {\rm d}\tau {\rm d}x{\rm d}s. \end{array} 由分部积分并注意到 \Psi 的性质, (3.10)式化为 \begin{array}{l} \int_{{{\Bbb R}} ^n} u_{ttt}(t, x)\Psi(t, x) {\rm d}x+\int_{{{\Bbb R}} ^n} u_{tt}(t, x)\Psi(t, x) {\rm d}x\\-\int_{{{\Bbb R}} ^n} u_t(t, x)\Psi(t, x) {\rm d}x -\int_{{{\Bbb R}} ^n} u(t, x)\Psi(t, x) {\rm d}x \\ = \varepsilon I[u_0, u_1, u_2, u_3]+ C_{\gamma}\int_0^t \int_{{{\Bbb R}} ^n}\Psi(s, x)\int_0^s(s-\tau)^{-\gamma} |u(\tau, x)|^p {\rm d}\tau {\rm d}x{\rm d}s, \end{array} 其中 联立(3.7)–(3.11)式, 可得 \begin{array}{l} F'''_0(t)+4F''_0(t)+4F'_0(t)\\ =\varepsilon I[u_0, u_1, u_2, u_3]+ C_{\gamma}\int_0^t \int_{{{\Bbb R}} ^n}\Psi(s, x)\int_0^s(s-\tau)^{-\gamma} |u(\tau, x)|^p {\rm d}\tau {\rm d}x{\rm d}s\\ \geq \varepsilon I[u_0, u_1, u_2, u_3]. \end{array} G(t) = F''_0(t)+2F'_0(t) , 重写(3.12)式为 \begin{eqnarray} G'(t)+2G(t)\geq \varepsilon I[u_0, u_1, u_2, u_3]. \end{eqnarray} 积分(3.13)式, 得到 \begin{eqnarray} G(t)\geq \left(G(0)-\frac{1}{2}\varepsilon I[u_0, u_1, u_2, u_3]\right){\rm e}^{-2t}+\frac{1}{2}\varepsilon I[u_0, u_1, u_2, u_3]. \end{eqnarray} 从而有 \begin{eqnarray} F''_0(t)+2F'_0(t)\geq \left(G(0)-\frac{1}{2}\varepsilon I[u_0, u_1, u_2, u_3]\right){\rm e}^{-2t}+\frac{1}{2}\varepsilon I[u_0, u_1, u_2, u_3]. \end{eqnarray} 对(3.15)式求积分, 可得 \begin{array}{l} F_0(t)\geq\varepsilon \int_{{{\Bbb R}} ^n}u_0(x)\Phi(x){\rm d}x+\frac{\varepsilon}{4}(1-{\rm e}^{-2t}) \int_{{{\Bbb R}} ^n}(3u_1(x)-2u_0(x)-u_3(x))\Phi(x){\rm d}x\\+\frac{\varepsilon}{4}t \int_{{{\Bbb R}} ^n}(u_3(x)+u_2(x)-u_1(x)-u_0(x))\Phi(x){\rm d}x\\+\frac{\varepsilon}{4} t{\rm e}^{-2t}\int_{{{\Bbb R}} ^n}(u_3(x)+u_0(x)-u_1(x)-u_2(x))\Phi(x){\rm d}x\\ \geq\varepsilon \int_{{{\Bbb R}} ^n}u_0(x)\Phi(x){\rm d}x+2\eta \int_{{{\Bbb R}} ^n}(2u_1(x)-u_0(x)-u_2(x))\Phi(x){\rm d}x\\+\frac{\varepsilon}{4}t \int_{{{\Bbb R}} ^n}(u_3(x)+u_2(x)-u_1(x)-u_0(x))\Phi(x){\rm d}x, \end{array} 其中 \eta = \min\left\{\frac{\varepsilon}{4}(1-{\rm e}^{-2t}), \frac{\varepsilon}{4} t{\rm e}^{-2t}\right\}>0. 由(3.16) 式, 存在一个 t_0 使得 $$F_0(t)\geq \widetilde{C}\varepsilon t ,$$ 其中 t\geq t_0, \widetilde{C} 为正常数. 又由假设有 另外, 由 \Psi 的渐近性可得 $$\int_{B_{R+s}} |\Psi(s, x)|^{\frac{p}{p-1}} {\rm d}x\leq \widetilde{K}(R+s)^{(n-1)(1-\frac{p'}{2})},$$ 其中 \widetilde{K}>0. 联立(3.9)式和(3.17)–(3.18)式, 有 \begin{eqnarray} \int_{{{\Bbb R}} ^n} |u(s, x)|^p {\rm d}x\geq C_0\varepsilon^p(R+s)^{(n-1)-\frac{(n-1)p}{2}}s^p, \end{eqnarray} 其中 C_0 = \widetilde{C}^p\widetilde{K}^{-(p-1)}, s\geq t_0. 由(3.8)–(3.19)式, 可得 \begin{eqnarray} F''''(t)\geq C_1\varepsilon^p(R+t)^{(n-1)-\frac{(n-1)p}{2}-\gamma}(t-t_0)^{p+1}, \end{eqnarray} 其中 t\geq t_0, C_1 = \frac{C_0C_{\gamma}}{p+1}. 对(3.20)式求积分四次, 得到 \begin{array}{l} F(t)\geq F(t_0)+F'(t_0)t+F''(t_0)\frac{t^2}{2}+F'''(t_0)\frac{t^3}{6}\\ +C_1\varepsilon^p\int^t_{t_0}\int^s_{t_0}\int^\tau_{t_0}\int^\sigma_{t_0}(R+\eta)^{(n-1)-\frac{(n-1)p}{2}-\gamma}(\eta-t_0)^{p+1}{\rm d}\eta{\rm d}\sigma{\rm d}\tau{\rm d}s\\ \geq C_1\varepsilon^p\int^t_{t_0}\int^s_{t_0}\int^\tau_{t_0}\int^\sigma_{t_0}(R+\eta)^{(n-1)-\frac{(n-1)p}{2}-\gamma}(\eta-t_0)^{p+1}{\rm d}\eta{\rm d}\sigma{\rm d}\tau{\rm d}s\\ \geq C_2\varepsilon^p(R+t)^{-\frac{(n-1)p}{2}-\gamma}(t-t_0)^{n+p+4}, \end{array} 其中 C_2 = \frac{C_1}{(n+p+1)(n+p+2)(n+p+3)(n+p+4)}, t\geq t_0. 重记(3.21)式为 \begin{eqnarray} F(t)\geq K_0(R+t)^{-\alpha_0}(t-t_0)^{\beta_0}, \quad t\geq t_0, \end{eqnarray} 其中 K_0 = C_2\varepsilon^p, \alpha_0 = \frac{(n-1)p}{2}+\gamma, \beta_0 = n+p+4 . 接下来, 将运用迭代框架(3.6) 式推出 F(t) 的下界序列, 即 \begin{eqnarray} F(t)\geq K_j(R+t)^{-\alpha_j}(t-t_0)^{\beta_j}, \end{eqnarray} 其中 \{K_j\}_{j\in N} , \{\alpha_j\}_{j\in N} , \{\beta_j\}_{j\in N} 是后面会定义的非负实数列. 因而后面的任务变成 t\rightarrow \infty$$ F(t)$的爆破研究.

$\begin{array}{l} F(t)\geq CC_{\gamma} K^p_j\int^t_{t_0}\int^s_{t_0}\int^{\tau}_{t_0}\int^{\sigma}_{t_0}\int^\eta_{t_0}(R+\xi)^{-n(p-1)-\gamma-\alpha_jp}(\xi-t_0)^{p\beta_j}d \xi {\rm d}\eta{\rm d}\sigma{\rm d}\tau{\rm d}s\\ \geq\frac{ CC_{\gamma} K^p_j(R+t)^{-n(p-1)-\gamma-\alpha_jp}(t-t_0)^{p\beta_j+5}}{(p\beta_j+1)(p\beta_j+2)(p\beta_j+3)(p\beta_j+4)(p\beta_j+5)}. \end{array}$

$$$\begin{array}{ll} { } K_{j+1} = \frac{ C C_{\gamma} K^p_j}{(p\beta_j+1)(p\beta_j+2)(p\beta_j+3)(p\beta_j+4)(p\beta_j+5)}, \\ \alpha_{j+1} = n(p-1)+\gamma+\alpha_jp, \\ \beta_{j+1} = p\beta_j+5, \end{array}$$$

$\begin{eqnarray} F(t)\geq K_{j+1}(R+t)^{\alpha_{j+1}}(t-t_0)^{\beta_{j+1}}. \end{eqnarray}$

$\begin{array}{l} \alpha_j=[n(p-1)+\gamma](1+p+p^2+\cdots+p^{j-1})+\alpha_0p^j\\ =(n(p-1)+\gamma)\frac{p^j-1}{p-1}+\alpha_0p^j\\ =\left(\alpha_0+n+\frac{\gamma}{p-1}\right)p^j-\left(n+\frac{\gamma}{p-1}\right), \\ \beta_{j}=\beta_0p^j+5p^{j-1}+5p^{j-2}+\cdots+5p+5\\ =\beta_0p^j+\frac{5(p^j-p)}{p-1}+5\\ =\left(\frac{5}{p-1}+\beta_0\right)p^j-\frac{5}{p-1}. \end{array}$

$\begin{eqnarray} K_j\geq CC_{\gamma}\left(\beta_0+\frac{5}{p-1}\right)^{-5}p^{-5j}K^p_{j-1} = Dp^{-5j}K^p_{j-1}. \end{eqnarray}$

$$$\log K_j\geq p^j\left(\log K_0-\frac{5p\log p}{(p-1)^2}+\frac{\log D}{p-1}\right)+\frac{5j\log p}{p-1}+\frac{5p\log p}{(p-1)^2}-\frac{\log D}{p-1}, \ \ \forall j\in N.$$$

$j_0 = j_0(n, p, \gamma)\in N$是最小非负整数, 使得

$\begin{eqnarray} \log K_j\geq p^j\log \left(K_0p^{-\frac{5p}{(p-1)^2}}D^{\frac{1}{p-1}}\right) = p^j\log(E_0\varepsilon^p), \end{eqnarray}$

$\begin{array}{l} F(t)\geq \exp\left(p^j\log(E_0\varepsilon^p)\right)(R+t)^{-(\alpha_0+n+\frac{\gamma}{p-1})p^j+(n+\frac{\gamma}{p-1})}(t-t_0)^{(\beta_0+\frac{5}{p-1})p^j-\frac{5}{p-1}}\\ =\exp\left(p^j\left(\log(E_0\varepsilon^p)-\left(\alpha_0+n+\frac{\gamma}{p-1}\right)\log(R+t)+\left(\beta_0+\frac{5}{p-1}\right)\log (t-t_0)\right)\right)\\\times(t+R)^{n+\frac{\gamma}{p-1}}(t-t_0)^{-\frac{5}{p-1}}, \end{array}$

$t\geq R+2t_0$时, 有$\log(R+t)\leq\log(2(t-t_0)).$因此(3.31)式可化为

$$$F(t)\geq \exp\left(p^j\left(\log(E_0\varepsilon^p2^{-(\alpha_0+n+\frac{\gamma}{p-1})}(t-t_0)^{\beta_0+\frac{5}{p-1}-(\alpha_0+n+\frac{\gamma}{p-1})})\right)\right) (R+t)^n(t-t_0)^{-\frac{5}{p-1}}.$$$

(3.32)式中, $t-t_0$的指数为

$\begin{eqnarray} \beta_0+\frac{5}{p-1}-\left(\alpha_0+n+\frac{\gamma}{p-1}\right) = \frac{(n+5-2\gamma)p+2-(n-3)p^2}{2(p-1)} = \frac{\Upsilon(n, p, \gamma)}{2(p-1)}. \end{eqnarray}$

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