## The Discrete Possion Equation and the Heat Equation with the Exponential Nonlinear Term

Li Yafeng1, Xin Qiao,1, Mu Chunlai2

 基金资助: 国家自然科学基金.  11461075

 Fund supported: the NSFC.  11461075

Abstract

This paper mainly study the relations between the solution of the discrete Poisson equation and the solution of the discrete heat equation with exponential nonlinear term by monotone iterative method and comparison principle. When the solutions of the discrete Poisson equation exist, we discuss the asymptotic stability of the solutions to the discrete heat equation with exponential nonlinear term.

Keywords： Discrete Poisson equation ; Discrete heat equation ; Graph ; Comparison principle ; Blow-up

Li Yafeng, Xin Qiao, Mu Chunlai. The Discrete Possion Equation and the Heat Equation with the Exponential Nonlinear Term. Acta Mathematica Scientia[J], 2019, 39(4): 773-784 doi:

## 1 引言

(ⅰ)任意$x\in V$,有$\omega(x, x) = 0$;

(ⅱ)任意$x, y\in V$,有$\omega(x, y) = \omega(y, x)\geq 0$;

(ⅲ)若$(x, y)\notin E$,则$\omega(x, y) = 0$.

$$$\Delta_{\omega}u(x) = \sum\limits_{y\in V}\left[u(y)-u(x)\right]\cdot\omega(x, y).$$$

## 2 指数型非线性项的离散泊松方程

关于解的存在唯一性的证明可以参考文献[1],事实上,方程解的表示式也在文献中给出了,为了表示方便,在下文中仅简单的记作$f(x) = (- \Delta_{\omega})^{-1} g(x)$.

$$$-\Delta_{\omega}f(x_0) = -\sum\limits_{y\in V}[f(y)-f(x_0)]\omega(x_0, y) = g(x_0) \geq 0,$$$

$-\sum\limits_{y\in V}(f(y)-f(x_0))\omega(x_0, y) \leq 0$,进而,若$y \sim x_0$,则有$f(y) = f(x_0)$并且$g(x_0) = 0$,然后由图$G$的连通性和$f(x) = 0, \forall x \in \partial S$,可得$f(x) \equiv 0 $$g(x) \equiv 0 ,这与 g(x) \not \equiv 0 矛盾,所以 f(x) > 0, x \in S .证毕. 定理2.1 设具有齐次Dirichlet边界条件的算子 -\Delta_{\omega} 最小的非平凡特征值为 \lambda_0 . \lambda_0 < e ,那么方程 (1.1) 解的集合 \Phi 是空集. 首先,由引理2.2可知,算子 -\Delta_{\omega} 的最小特征值 \lambda_0 所对应的特征向量 \phi_0 >0, x\in S$$ \sum\limits_{x\in S}\phi_0 = 1$.接下来,对方程$(1.1)$两边同乘$\phi_0$并且在$V$进行积分,得到

$$$\sum\limits_{x\in V}(-\Delta_{\omega}u)\phi_0 = \sum\limits_{x\in V}e^u\phi_0.$$$

$$$-\sum\limits_{x\in V}\Delta_{\omega}u\phi_0 = \frac{1}{2}\sum\limits_{x\in V}\nabla_{\omega}u\cdot\nabla_{\omega}\phi_0 = -\sum\limits_{x\in V}u\Delta_{\omega}\phi_0 ,$$$

$$$\sum\limits_{x\in S}(-\Delta_{\omega}u)\phi_0 +\sum\limits_{x\in \partial S}(-\Delta_{\omega}u)\phi_0 = \sum\limits_{x\in S}e^u\phi_0 +\sum\limits_{x\in \partial S}e^u\phi_0.$$$

$$$-\sum\limits_{x\in S}u \Delta_{\omega}\phi_0 = \sum\limits_{x\in S}e^u\phi_0,$$$

$(2.2)$式代入$(2.8)$式可得

$$$\lambda_0\sum\limits_{x\in S}\phi_0 u = \sum\limits_{x\in S}e^u\phi_0 .$$$

$J = \sum\limits_{x\in S}\phi_0u$,通过詹生不等式可得

$$$-\lambda_0J+e^J\leq0,$$$

$\lambda_0 <e$,上述方程无解,可得结论.证毕.

假设$u(x)$在某一些内部顶点的值不大于零,可令$u(x) $$x_0 \in S 顶点处取得最小值且 u(x_0) \leq 0 ,这样在 x_0 处有 u(x)$$ x_0 \in S$顶点处取得最小值,故等式的左边满足

$$$u_{n+1}(x) = \left\{ \begin{array}{lll} (-\Delta_{\omega})^{-1}e^{u_n(x)}, \quad &x\in S, \\ 0, &x\in \partial S \end{array} \right.$$$

首先,证明$u_0 < u_1$.由函数列构造的定义可知$u_1 = (-\Delta_{\omega})^{-1}e^{u_0}, x\in S$,进而可得

$$$\left\{ \begin{array}{lll} -\Delta_{\omega} u_1 = e^{u_0} = 1, \quad&x \in S, \\ u_1(x) = 0, &x \in \partial S, \end{array} \right.$$$

反证法.若$\varphi \geq \phi$,且$\varphi \not\equiv \phi$.$\psi = \varphi - \phi$,则对于任意的$x\in S$,有$\psi(x) \geq 0 $$\psi(x)\not\equiv 0 ,不妨设 \psi 在内部顶点 x_0 处取得最小值,即 \psi(x_0) = 0 ,在该顶点 x_0 处有 因此,可得 再由 \psi(x) \geq 0 可得,若 \omega(y, x_0) \neq 0 ,则必有 \psi(y) = 0 ,最后,由图 G 的连通性可得 \psi(x) \equiv 0,$$ \forall x \in V$,即$\varphi \equiv \phi$矛盾.证毕.

取定$u_0(x) = 0, x \in V$,由引理2.4可知,迭代公式

$$$\left\{ \begin{array}{lll} -\Delta_{\omega}(\phi-u_{n+1}) = e^{\phi}-e^{u_n} > 0, \quad &x\in S, \\ \phi-u_{n+1} = 0, \quad &x\in \partial S, \end{array} \right.$$$

(ⅰ)令$0<\alpha<1$, $\nu_0 = (1-\alpha)\varphi+\alpha\psi$,迭代公式

(ⅱ)令$\beta > 0$, $w_0(x) = (1+\beta)\psi-\beta\varphi$,迭代公式

(ⅰ)先证明$\nu_1 \ll \nu_0$.$\varphi, \psi\in \Phi$,可知$-\Delta_{\omega}\varphi = e^ {\varphi} $$-\Delta_{\omega}\psi = e^ {\psi} ,又因为 \Delta_{\omega} 是一个线性算子可得 - \Delta_{\omega}\nu_0 = -\Delta_{\omega}\left[(1-\alpha)\varphi+\alpha\psi\right] = (1-\alpha)e^{\varphi} + \alpha e^{\psi} ,进一步,再由迭代公式 可得到 再由指数函数 e^x 是一个严格凸函数可得 $$e^{\nu_0(x)} = e^{(1-\alpha)\varphi+\alpha\psi} < (1-\alpha)e^{\varphi} + \alpha e^{\psi},$$ 再次利用引理2.3,可以得出 \nu_1\ll\nu_0 .现假设 \nu_{n-1} \gg \nu_{n} ,下证 \nu_{n} \gg\nu_{n+1} ,其证明方法类似于引理2.4的证明过程,这里不再重复. 接下来,证明(ⅱ).类似于(ⅰ)的证明,先证明 w_1 \gg w_0 . w_0(x) = (1+\beta)\psi-\beta\phi ,可知 -\Delta_{\omega}w_0(x) = (1+\beta)e^{\psi}- \beta e^ {\varphi} ,这样由函数列 \{w_{n}\} 的迭代公式可得 事实上,由于 w_0 = (1+\beta)\psi-\beta\varphi 和指数函数 e^x 是严格的凸函数,可得 进而可知 e^{w_0(x)} - [(1+\beta)e^{\psi} - \beta e^{\varphi}] \geq 0 ,再由引理2.3可知, w_1 \gg w_0 .类似于(ⅰ)的证明过程,利用数学归纳法容易证明 \{ w_n \} 是一个单调递增的函数列.证毕. 接下来,讨论如何确定方程 (1.1) 的极小解或者极小解的性质.为此,先引入增算子的概念.设 其上的范数 \|f\|_{C_0(V)} = \max\limits_{x\in V}|f(x)| ,则 C_0(V) 是一个Banach空间.令 容易验证 P 正则锥,进而可定义 C_0(V) 上的半序, f \leq g, (f, g\in C_0(V)) ,如果 g-f \in P .定义非线性算子 F 如下 显然 F: C_0(V) \rightarrow C_0(V) 的连续算子,此外, F 也是一个增算子.事实上,对于任意的 f, g \in C_0(V)$$ f \leq g$.$h(x) = F[g]- F[f]$,所以$h(x)$满足

反证法.假设存在三元数组$u_1\ll u_2\ll u_3 \in \Phi$.第一步:根据引理2.6可知,对于任意的$0<\alpha<1$,令$\nu_0 = (1-\alpha)u_2+\alpha u_3$, $x\in V$,迭代公式

$$$\left\{ \begin{array}{lll} -\Delta_{\omega}\nu_{n+1} = e^{\nu_n(x)}, &x\in S, \\ \nu_{n+1}(x) = 0, &x\in \partial S, \end{array} \right.$$$

$$$\left\{ \begin{array}{lll} -\Delta_{\omega}u_2(x) = e^{u_2(x)}, &x\in S, \\ u_2(x) = 0, &x\in \partial S, \end{array} \right.$$$

$$$\left\{ \begin{array}{lll} -\Delta_{\omega}[\nu_{n+1}- u_2(x)] = e^{\nu_n(x)}- e^{u_2(x)}, &x\in S, \\ \nu_{n+1}(x)-u_2(x) = 0, &x\in \partial S, \end{array} \right.$$$

对于任意$0\leq t_0<T$,设$m = \min\limits_{S\times[0, t_0]}\{\overline{v}, \underline{v}\} $$M = \max\limits_{S\times[0, t_0]}\{\overline{v}, \underline{v}\} ,这里 m, M 是正的常数.设 V(x, t) = \underline{v} - \overline{v} ,对于任意 x \in S , V(x, 0) <0 ,通过对问题 (1.2) 上下解的定义可知 $$V_t \leq\Delta_{\omega}V + (e^{\underline{v}}-e^{\overline{v}}).$$ V^{+}(x, t) = \max\{v(x, t), 0\}\geq 0 , (3.2) 式两边同乘 V^{+} 并且在 S 进行积分,可得 $$\frac{1}{2}\left(\int_{x\in S}(V^{+}(x, t))^2\right)_t \\ \leq \int_{x\in S}\Delta_{\omega}V(x, t)V^{+}(x, t) + \int_{x\in S}(e^{\underline{v}}-e^{\overline{v}}) V^{+}(x, t).$$ 对上式的不等式 (3.3) 右边第一部分,得到 $$\int_{x\in S}\Delta_{\omega}V(x, t)V^{+}(x, t)\leq 0.$$ 事实上,令 J(t) = \{x\in V: V(x, t)>0\} ,如果 J(t) 为空集合,此定理自然得证.现在,假设 J(t) 为非空集合,因为对于任意 x \in \partial S$$ 0\leq t \leq t_0$, $\underline{v}(x, t) \leq 0$, $\overline{v}(x, t) \geq 0$,所以对于任意$x \in \partial S $$0\leq t \leq t_0 , V(x, t) = \underline{v}(x, t) - \overline{v}(x, t) \leq 0 .现在,得到 J(t) \subset S .因此,如果 x \in J(t)$$ y\in V \setminus J(t)$,得到$V(x, t) > 0 $$V(y, t)-V(x, t) < 0 ,因此,可得 $$\sum\limits_{x\in J(t)}\sum\limits_{y\in V \setminus J(t)}V(x, t)[V(y, t)-V(x, t)]\omega(x, y)<0.$$ 进一步,得到 \begin{eqnarray} & &\sum\limits_{x\in S}\sum\limits_{y\in V}V^+(x, t)[V(y, t)-V(x, t)]\omega(x, y)\\ & = &\sum\limits_{x\in J(t)}\sum\limits_{y\in J(t)}V(x, t)[V(y, t)-V(x, t)]\omega(x, y)\\ &&+\sum\limits_{x\in J(t)}\sum\limits_{y\in V \setminus J(t)}V(x, t)[V(y, t)-V(x, t)]\omega(x, y)\\ & = &-\frac{1}{2}\sum\limits_{x\in J(t)}\sum\limits_{y\in J(t)}[V(y, t)-V(x, t)]^2\omega(x, y)\\ &&+\sum\limits_{x\in J(t)}\sum\limits_{y\in V \setminus J(t)}V(x, t)[V(y, t)-V(x, t)]\omega(x, y)< 0. \end{eqnarray} 另一方面,对于任意 x\in S ,通过中值定理,得到 这里 \xi(x, t) = \theta(x)\underline{v}(x, t)+(1-\theta(x))\overline{v}(x, t)$$ 0\leq \theta(x)\leq 1$,因此$m \leq \xi(x, t) \leq M$.进而,得到

$$$\int_{x\in S}(e^{\underline{v}}-e^{\overline{v}}) V^{+}(x, t) \leq e^{M}\int_{x\in S}(V^{+}(x, t))^2.$$$

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