偏微分方程

中山大学数学学院2019学年秋季学期数学与应用数学专业《偏微分方程》课程网站

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第六周作业

本次作业题包括习题2.3第3、4题。

习题2.3

3

记$\mathcal{F} [u(x,y,z,t)](\xi,\eta,\zeta,t)=\int^\infty_{-\infty}\int^\infty_{-\infty}\int^\infty_{-\infty}u(x,y,z,t)e^{-i(x\xi+y\eta+z\zeta)}\mathrm{d}x\mathrm{d}y\mathrm{d}z$而$\mathcal{F}^{-1} [u(\xi,\eta,\zeta,t)](x,y,z,t)=\frac{1}{(2\pi)^3}\int^\infty_{-\infty}\int^\infty_{-\infty}\int^\infty_{-\infty}u(\xi,\eta,\zeta,t)e^{i(x\xi+y\eta+z\zeta)}\mathrm{d}\xi\mathrm{d}\eta\mathrm{d}\zeta$。

对方程$\frac{\partial u}{\partial t}=a^2(\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}+\frac{\partial^2 u}{\partial z^2})$两端应用Fourier变换,形式地得

$\begin{align}\frac{\partial}{\partial t}\mathcal{F} [u](\xi,\eta,\zeta,t)&=\mathcal{F} [\frac{\partial u}{\partial t}](\xi,\eta,\zeta,t)\\&=\mathcal{F}[a^2(\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}+\frac{\partial^2 u}{\partial z^2})](\xi,\eta,\zeta,t)\\&=a^2(\mathcal{F}[\frac{\partial^2 u}{\partial x^2}](\xi,\eta,\zeta,t)+\mathcal{F}[\frac{\partial^2 u}{\partial y^2}](\xi,\eta,\zeta,t)+\mathcal{F}[\frac{\partial^2 u}{\partial z^2}](\xi,\eta,\zeta,t))\\&=a^2(-\xi^2-\eta^2-\zeta^2)\mathcal{F}[u](\xi,\eta,\zeta,t)\end{align}$

可见应有$\mathcal{F} [u](\xi,\eta,\zeta,t)=C(\xi,\eta,\zeta)e^{-a^2(\xi^2+\eta^2+\zeta^2)t}$,特别地令$t=0$得$C(\xi,\eta,\zeta)=\mathcal{F} [\varphi](\xi,\eta,\zeta)$。形式地,

$\begin{align}u(x,y,z,t)&=\mathcal{F}^{-1} [\mathcal{F} [u](\xi,\eta,\zeta,t)](x,y,z,t)\\&=\mathcal{F}^{-1} [\mathcal{F} [\varphi](\xi,\eta,\zeta)e^{-a^2(\xi^2+\eta^2+\zeta^2)t}](x,y,z,t)\\&=(\mathcal{F}^{-1} [\mathcal{F} [\varphi]]\ast\mathcal{F}^{-1} [e^{-a^2(\xi^2+\eta^2+\zeta^2)t}])(x,y,z,t)\\&=(\varphi\ast\frac{1}{(2\pi)^3}\int^\infty_{-\infty}\int^\infty_{-\infty}\int^\infty_{-\infty}e^{-a^2(\xi^2+\eta^2+\zeta^2)t}e^{i(x\xi+y\eta+z\zeta)}\mathrm{d}\xi\mathrm{d}\eta\mathrm{d}\zeta)(x,y,z,t)\\&=(\varphi\ast\frac{1}{(2\pi)^3}\int^\infty_{-\infty}\int^\infty_{-\infty}\int^\infty_{-\infty}e^{-(a\sqrt{t}\xi-\frac{ix}{2a\sqrt{t}})^2-(a\sqrt{t}\eta-\frac{iy}{2a\sqrt{t}})^2-(a\sqrt{t}\zeta-\frac{iz}{2a\sqrt{t}})^2-\frac{x^2+y^2+z^2}{4a^2t}}\mathrm{d}\xi\mathrm{d}\eta\mathrm{d}\zeta)(x,y,z,t)\\&=(\varphi\ast\frac{e^{-\frac{x^2+y^2+z^2}{4a^2t}}}{(2\pi)^3}\int^\infty_{-\infty}e^{-(a\sqrt{t}\xi)^2}\mathrm{d}\xi\int^\infty_{-\infty}e^{-(a\sqrt{t}\eta)^2}\mathrm{d}\eta\int^\infty_{-\infty}e^{-(a\sqrt{t}\zeta)^2}\mathrm{d}\zeta)(x,y,z,t)\\&=(\varphi\ast\frac{e^{-\frac{x^2+y^2+z^2}{4a^2t}}}{(2\pi)^3}(\frac{\sqrt{\pi}}{a\sqrt{t}})^3)(x,y,z,t)\\&=\frac{1}{(2a\sqrt{\pi t})^3}\int^\infty_{-\infty}\int^\infty_{-\infty}\int^\infty_{-\infty}\varphi (\xi,\eta,\zeta)e^{-\frac{(x-\xi)^2+(y-\eta)^2+(z-\zeta)^2}{4a^2t}}\mathrm{d}\xi\mathrm{d}\eta\mathrm{d}\zeta\end{align}$

和一维情况类似,可以验证$\varphi\in C(\mathrm{R}^3)$有界时上式给出原问题的一个经典解。

4

假设$\varphi\in C((-\infty,+\infty))$和$f\in C((-\infty,+\infty)\times [0,+\infty))$有界$M$,再假设$\alpha\gt 0$且对任何$t>0$存在$x_t\in(-\infty,+\infty)$使对任何$x\in (-\infty,+\infty)$成立$\vert f(x_t+x,t)\vert\leq Mx^\alpha$。记$u_1(x,t)=\frac{1}{2a\sqrt{\pi t}}\int^\infty_{-\infty}\varphi (\xi)e^{-\frac{(x-\xi)^2}{4a^2t}}\mathrm{d}\xi$,$u_2(x,t)=\int^t_0 w(x,t;\tau)\mathrm{d}\tau$,$w(x,t;\tau)=\frac{1}{2a\sqrt{\pi (t-\tau)}}\int^\infty_{-\infty} f (\xi,\tau)e^{-\frac{(x-\xi)^2}{4a^2(t-\tau)}}\mathrm{d}\xi$,则$u_1\in C^1_t((0,+\infty)\times (-\infty,+\infty))\cap C^2_x((0,+\infty)\times (-\infty,+\infty))\cap C((-\infty,+\infty)\times [0,+\infty))$且满足问题:

$\begin{cases}\frac{\partial u_1}{\partial t}(x,t)=a^2\frac{\partial^2 u_1}{\partial x^2}(x,t)&,(x,t)\in (-\infty,+\infty)\times (0,+\infty)\\u_1(x,0)=\varphi (x)&,x\in(-\infty,+\infty)\end{cases}$

又对任何固定的$\tau\in [0,+\infty)$,$w(\cdot,\cdot;\tau)\in C^1_t((\tau,+\infty)\times (-\infty,+\infty))\cap C^2_x((\tau,+\infty)\times (-\infty,+\infty))\cap C((-\infty,+\infty)\times [\tau,+\infty))$且满足问题:

$\begin{cases}\frac{\partial w}{\partial t}(x,t;\tau)=a^2\frac{\partial^2 w}{\partial x^2}(x,t;\tau)&,(x,t)\in (-\infty,+\infty)\times (\tau,+\infty)\\w(x,\tau;\tau)=f (x,\tau)&,x\in(-\infty,+\infty)\end{cases}$

由$\vert w(x,t;\tau)\vert\leq\frac{1}{2a\sqrt{\pi (t-\tau)}}\int^\infty_{-\infty} M e^{-\frac{(x-\xi)^2}{4a^2(t-\tau)}}\mathrm{d}\xi\leq M$,同时

$\begin{align}\vert \frac{\partial w}{\partial x}(x,t;\tau)\vert &=\vert \frac{1}{2a\sqrt{\pi (t-\tau)}}\int^\infty_{-\infty} f(\xi,\tau) e^{-\frac{(x-\xi)^2}{4a^2(t-\tau)}}(\frac{\xi -x}{2a^2(t-\tau)})\mathrm{d}\xi\vert\\&=\frac{1}{a\sqrt{\pi (t-\tau)}}\int^\infty_{-\infty}\vert f(x+2a\sqrt{t-\tau}\eta,\tau)\vert e^{-\eta^2}\eta\mathrm{d}\eta\\&\leq \frac{M}{a\sqrt{\pi (t-\tau)}}\end{align}$

关于$\tau$在$[0,t]$可积,又

$\begin{align}\vert \frac{\partial^2 w}{\partial x^2}(x,t;\tau)\vert &=\vert \frac{1}{2a\sqrt{\pi (t-\tau)}}\int^\infty_{-\infty} f(\xi,\tau) e^{-\frac{(x-\xi)^2}{4a^2(t-\tau)}}(-\frac{1}{2a^2(t-\tau)}+\frac{\xi -x}{4a^4(t-\tau)^2})\mathrm{d}\xi\vert\\&=\frac{1}{a^2\sqrt{\pi }(t-\tau)}\int^\infty_{-\infty}\vert f(x+2a\sqrt{t-\tau}\eta,\tau) e^{-\eta^2}(\eta^2-\frac{1}{2})\vert\mathrm{d}\eta\\&\leq \frac{C}{(t-\tau)^{1-\frac{\alpha}{2}}}\end{align}$

关于$\tau$在$[0,t]$可积,因此$u_2\in C^1_t((0,+\infty)\times (-\infty,+\infty))\cap C^2_x((0,+\infty)\times (-\infty,+\infty))\cap C((-\infty,+\infty)\times [0,+\infty))$,而且相应导数可通过积分号下求导得到。于是对任何固定的$(x,t)\in (-\infty,+\infty)\times (0,+\infty)$,$u_2(x,0)=\int^0_0 w(x,t;\tau)\mathrm{d}\tau=0$,又

$\begin{align}\frac{\partial u_2}{\partial t}(x,t)&=\frac{\partial}{\partial t}(\int^t_0 w(x,t;\tau)\mathrm{d}\tau)\\&=w(x,t;t)+\int^t_0 \frac{\partial w}{\partial t}(x,t;\tau)\mathrm{d}\tau\\&=f(x,t)+\int^t_0 a^2\frac{\partial^2 w}{\partial x^2}(x,t;\tau)\\&=f(x,t)+a^2\frac{\partial^2}{\partial x^2}(\int^t_0 w(x,t;\tau)\mathrm{d}\tau)\\&=a^2\frac{\partial^2 u_2}{\partial x^2}(x,t)+f(x,t)\end{align}$

注意到$u=u_1+u_2\in C^1_t((0,+\infty)\times (-\infty,+\infty))\cap C^2_x((0,+\infty)\times (-\infty,+\infty))\cap C((-\infty,+\infty)\times [0,+\infty))$,对任何$(x,t)\in (-\infty,+\infty)\times (0,+\infty)$有$\frac{\partial u}{\partial t}(x,t)=\frac{\partial u_1}{\partial t}(x,t)+\frac{\partial u_2}{\partial t}(x,t)=a^2\frac{\partial^2 u_1}{\partial x^2}(x,t)+a^2\frac{\partial^2 u_2}{\partial x^2}(x,t)+f(x,t)=a^2\frac{\partial^2 u}{\partial x^2}(x,t)+f(x,t)$,又$u(x,0)=u_1(x,0)+u_2(x,0)=\varphi (x)+0=\varphi (x)$,即$u$是以下问题的经典解:

$\begin{cases}\frac{\partial u}{\partial t}(x,t)=a^2\frac{\partial^2 u}{\partial x^2}(x,t)+f(x,t)&,(x,t)\in (-\infty,+\infty)\times (0,+\infty)\\u(x,0)=\varphi (x)&,x\in(-\infty,+\infty)\end{cases}$

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