一些二阶偏微分方程的解

热传导问题

无限长细杆

\[ \begin{gather} \begin{cases} \frac{\partial u}{\partial t}=a^2\frac{\partial^2u}{\partial x^2} & (x\in\mathbb R,t>0) \\ \left.u\right|_{t=0}=\phi(x) & (x\in\mathbb R) \end{cases} \\ u(x,t)=\phi(x)*K(x,t) \end{gather} \]

对流

\[ \begin{gather} \begin{cases} \frac{\partial u}{\partial t}=a^2\frac{\partial^2u}{\partial x^2}-k\frac{\partial u}{\partial x} & (x\in\mathbb R,t>0) \\ \left.u\right|_{t=0}=\phi(x) & (x\in\mathbb R) \end{cases} \\ u(x,t)=\phi(x)*K(x,t,k) \end{gather} \]

线性衰变

\[ \begin{gather} \begin{cases} \frac{\partial u}{\partial t}=a^2\frac{\partial^2u}{\partial x^2}-k\frac{\partial u}{\partial x}-\gamma u & (x\in\mathbb R,t>0) \\ \left.u\right|_{t=0}=\phi(x) & (x\in\mathbb R) \end{cases} \\ u(x,t)=\exp(-\gamma t)\phi(x)*K(x,t,k) \end{gather} \]

半导体载流子的输运方程

\[ \begin{gather} \begin{cases} \frac{\partial u}{\partial t}=D\frac{\partial^2u}{\partial x^2}-\mu E\frac{\partial u}{\partial x}-\frac{u}{\tau} & (x\in\mathbb R,t>0) \\ \left.u\right|_{t=0}=\delta(x) \end{cases} \\ u(x,t)=\frac{\exp(-t/\tau)}{\sqrt{4\pi Dt}}\exp\left[-\frac{(x-\mu Et)^2}{4Dt}\right] \end{gather} \]

非常数衰变

\[ \begin{gather} \begin{cases} \frac{\partial u}{\partial t}=a^2\frac{\partial^2u}{\partial x^2}-k\frac{\partial u}{\partial x}-tu & (x\in\mathbb R,t>0) \\ \left.u\right|_{t=0}=\phi(x) & (x\in\mathbb R) \end{cases} \\ u(x,t)=\exp\left(-\frac{t^2}{2}\right)\phi(x)*K(x,t,k) \end{gather} \]

有源

\[ \begin{gather} \begin{cases} \frac{\partial u}{\partial t}=a^2\frac{\partial^2u}{\partial x^2}+f(x,t) & (x\in\mathbb R,t>0) \\ \left.u\right|_{t=0}=\phi(x) & (x\in\mathbb R) \end{cases} \\ u(x,t)=\phi(x)*K(x,t)+\int_0^tf(x,\tau)*K(x,t-\tau)\mathrm{d}\tau \end{gather} \]

半无限长细杆

\[ \begin{gather} \begin{cases} \frac{\partial u}{\partial t}=a^2\frac{\partial^2u}{\partial x^2} & (x>0,t>0) \\ \left.u\right|_{t=0}=0 & (x\geq0) \\ \left.u\right|_{x=0}=f(t) & (t>0) \end{cases} \\ u(x,t)=\frac{x}{2a\sqrt{\pi}}\int_0^tf(\tau)\frac{1}{(t-\tau)^{3/2}}\exp\left[-\frac{x^2}{4a^2(t-\tau)}\right]\mathrm{d}\tau \end{gather} \]

波动方程

半无界弦强迫振动

\[ \begin{gather} \begin{cases} \frac{\partial^2u}{\partial t^2}=a^2\frac{\partial^2u}{\partial x^2}+f(t) & (x>0,t>0) \\ \left.u\right|_{x=0}=0 & (t>0) \\ \left.u\right|_{t=0}=0,\quad \left.\frac{\partial u}{\partial t}\right|_{t=0}=0 & (x\geq0) \end{cases} \\ u(x,t)= \begin{cases} \int_0^t\tau f(t-\tau)\mathrm{d}\tau & \left(t<\frac{x}{a}\right) \\ \int_0^t\tau f(t-\tau)\mathrm{d}\tau-\int_{x/a}^t\left(\tau-\frac{x}{a}\right)f(t-\tau)\mathrm{d}\tau & \left(t\geq\frac{x}{a}\right) \end{cases} \end{gather} \]

电报方程

\[ \begin{gather} \begin{cases} \frac{\partial^2u}{\partial t^2}-a^2\frac{\partial^2u}{\partial x^2}+2b\frac{\partial u}{\partial t}+c^2u=0 & (x\in\mathbb R,t>0) \\ \left.u\right|_{t=0}=\phi(x) & (x\in\mathbb R) \\ \left.\frac{\partial u}{\partial t}\right|_{t=0}=\psi(x) & (x\in\mathbb R) \end{cases} \\ u(x,t)=\frac{A}{\sqrt{c^2-b^2}}\exp(-bt)\sin\sqrt{c^2-b^2}t\quad(c>b,\phi(x)=0,\psi(x)=A) \end{gather} \]

\[ \begin{gather} \begin{cases} \frac{\partial^2u}{\partial t^2}-a^2\frac{\partial^2u}{\partial x^2}+B^2u=0 & (x\in\mathbb R,t>0) \\ \left.u\right|_{t=0}=\phi(x) & (x\in\mathbb R) \\ \left.\frac{\partial u}{\partial t}\right|_{t=0}=\psi(x)+b\phi(x) & (x\in\mathbb R) \end{cases} \\ u(x,t)=\frac{1}{2\pi}\int_{-\infty}^\infty\frac{\Psi(\omega)}{\sqrt{a^2\omega^2+B^2}}\sin t\sqrt{a^2\omega^2+B^2}\mathrm{e}^{i\omega x}\mathrm{d}\omega \\ u(x,t)=\frac{A}{B}\exp(-bt)\sin Bt\quad(c>b,\phi(x)=0,\psi(x)=A) \end{gather} \]