第七周作业

本次作业题包括习题2.4第3题。

习题2.4

3

令$v(x,t)=e^{-\lambda t}u(x,t)$，则

$\begin{cases}v_t-a^2v_{xx}+\lambda v=e^{-\lambda t}f(x,t)\\v(0,t)=e^{-\lambda t}\mu_{1}(t),(\frac{\partial v}{\partial x}+hv)(l,t)=e^{-\lambda t}\mu_{2}(t)\\v(x,0)=\varphi (x)\end{cases}$

因$v$在有界闭集$R_T$连续，它有至少一个最大值点$(x^\ast,t^\ast)$。假如$v(x^\ast,t^\ast)\gt 0$，则所有可能情况以下：

• $t^\ast=0$：这时$v(x^\ast,t^\ast)=\varphi (x^\ast)\leq\displaystyle\max_{x\in [0,l]}\varphi (x)$
• $x^\ast=0$：这时$v(x^\ast,t^\ast)=e^{-\lambda t^\ast}\mu_1 (t^\ast)\leq\displaystyle\max_{t\in [0,T]}e^{-\lambda t}\mu_1 (t)$
• $x^\ast=l$：这时$\frac{\partial v}{\partial x}(l,t^\ast)\geq 0$，于是$v(x^\ast,t^\ast)=\frac{1}{h}(e^{-\lambda t^\ast}\mu_2 (t^\ast)-\frac{\partial v}{\partial x}(l,t^\ast))\leq\frac{1}{h}e^{-\lambda t^\ast}\mu_2 (t^\ast)\leq\displaystyle\max_{t\in [0,T]}\frac{1}{h}e^{-\lambda t}\mu_2 (t)$
• 其它情况：这时$\frac{\partial^2 v}{\partial x^2}(x^\ast,t^\ast)\leq 0$且$\frac{\partial v}{\partial t}(x^\ast,t^\ast)\geq 0$，于是$v(x^\ast,t^\ast)=\frac{1}{\lambda}(-v_t(x^\ast,t^\ast)+a^2v_{xx}(x^\ast,t^\ast)+e^{-\lambda t^\ast}f(x^\ast,t^\ast))\leq\frac{1}{\lambda}e^{-\lambda t^\ast}f(x^\ast,t^\ast)\leq\frac{1}{\lambda}\displaystyle\max_{(x,t)\in R_T}e^{-\lambda t}f(x,t)$

综上所述，$\displaystyle\max_{(x,t)\in R_T}v(x,t)\leq\max\{0,\max_{x\in [0,l]}\varphi (x),\max_{t\in [0,T]}e^{-\lambda t}\mu_1 (t),\max_{t\in [0,T]}\frac{1}{h}e^{-\lambda t}\mu_2 (t),\frac{1}{\lambda}\max_{(x,t)\in R_T}e^{-\lambda t}f(x,t)\}$，从而$\displaystyle\max_{(x,t)\in R_T}u(x,t)=\max_{(x,t)\in R_T}e^{\lambda t}v(x,t)\leq e^{\lambda T}\max_{(x,t)\in R_T}v(x,t)\leq e^{\lambda T}\max\{0,\max_{x\in [0,l]}\varphi (x),\max_{t\in [0,T]}e^{-\lambda t}\mu_1 (t),\max_{t\in [0,T]}\frac{1}{h}e^{-\lambda t}\mu_2 (t),\frac{1}{\lambda}\max_{(x,t)\in R_T}e^{-\lambda t}f(x,t)\}$