zju博士资格考试考前复习（微分方程方向）pde 部分
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22-23《偏微分方程》课程笔记
Evens PDE
1. 一阶线性方程
定义
关于 \(u\) 的一阶线性方程
\[u_t + b\cdot Du = 0, (x,t)\in \mathbb{R}^n\times (0,\infty)
\]
通解：特征线法
固定 \((x,t)\)，令 \(z(s) := u(x+sb, t+s)\)，则
\[z'(s) = b\cdot Du(x+sb, t+s)+u_t(x+sb,t+s)=0
\]
因此 \(z(s) = C\)。即
\[u(x, t) = F(x - tb), F\in C^1(\mathbb R^n).
\]
齐次初值问题
\[\begin{aligned}
u_t + b\cdot Du = 0 ,& \text{in }\mathbb{R}^n\times (0,\infty) \\
u = g ,& \text{on }\mathbb{R}^n \times \{0\}
\end{aligned}
\]
解就是 \(u(x, t) = g(x - tb)\)。
非齐次初值问题
\[\begin{aligned}
u_t + b\cdot Du = &f, \text{in }\mathbb{R}^n\times (0,\infty) \\
u = &g, \text{on }\mathbb{R}^n\times \{0\} \\
\end{aligned}
\]
齐次化方法
固定 \((x,t)\)，令 \(z(s) = u(x+sb, t+s)\)，则
\[z'(s) = b\cdot Du(x+sb, t+s) + u_t(x+sb, t+s) = f(x+sb, t+s)
\]
因此
\[\begin{aligned}
& u(x,t) - g(x-tb) \\
= & z(0) - z(-t) \\
= & \int_{-t}^0 z'(s) ds \\
= & \int_{-t}^0 f(x+sb,t+s)ds \\
= & \int_0^t f(x+(s-t)b,s)ds \\
\end{aligned}
\]
即
\[u(x,t) = g(x-tb) + \int_0^t f(x+(s-t)b, s)ds.
\]
2. 一阶拟线性方程
定义
\[u_t + a(u)u_x = 0, u(0,x) = \varphi(x).
\]
解法
设 \(x = x(t)\) 是初值问题 \(x' = a(u(t, x(t)))\) 的解，则
\[U(t) = u(t, x(t))
\]
沿曲线 \(x = x(t)\) 为常数。
由初值条件得 \(u(0, x) = \varphi(x)\)，所以
\[x(t) = x(0) + a(u(0))t, U(t) = U(0).
\]
令 \(\alpha := x(0)\)，则 \(U(0) = u(0, x(0)) = \varphi(\alpha))\)。故 \(u\) 满足参数方程
\[x(t) = \alpha + a(\varphi(\alpha))t, u(t, x(t)) = \varphi(\alpha).
\]
特别地，当 \(\varphi\) 在 \(\mathbb R\) 上单增时，上述初值问题有唯一解。
3. 一维波动方程 - 初值问题
齐次初值问题
\[\begin{aligned}
&u_{tt} - c^2u_{xx} = 0,\\
t=0: &u = \varphi, u_t = \psi. \\
\end{aligned}
\]
D‘Alembert 公式
将方程重写为
\[\left(\frac{\partial}{\partial t} - c\frac{\partial}{\partial x}\right)\left(\frac{\partial}{\partial t} + c\frac{\partial}{\partial x}\right)u = 0.
\]
令 \(\xi = x-ct, \eta = x+ct\)，则
\[\frac{\partial^2 u}{\partial\xi\partial\eta} = 0.
\]
所以 \(u = F(\xi) + G(\eta) = F(x-ct) + G(x+ct)\)。
根据初值条件，有
\[\varphi(x) = F(x) + G(x), \psi(x) = -cF'(x) + cG'(x).
\]
联立解得
\[\begin{aligned}
F(x) &= \frac 12 \varphi(x) + \frac 1{2c}\int_0^x \psi(s)ds + C, \\
G(x) &= \frac 12 \varphi(x) - \frac 1{2c}\int_0^x \psi(s)ds - C. \\
\end{aligned}
\]
所以
\[u(t, x) = \frac 12[\varphi(x+ct) + \varphi(x-ct)] + \frac 1{2c}\int_{x-ct}^{x+ct}\psi(s)ds.
\]
非齐次初值问题
\[\begin{aligned}
&u_{tt} - c^2u_{xx} = f,\\
t=0: &u = 0, u_t = 0. \\
\end{aligned}
\]
齐次化
设 \(w(t,x;\tau)\) 是齐次问题
\[\begin{aligned}
&w_{tt} - c^2w_{xx} = 0,\\
t=0: &w = 0, w_t = f(\tau, x). \\
\end{aligned}
\]
则
\[u(t, x) = \int_0^t w(t, x; \tau)d\tau.
\]
注：此原理同样适用于边值问题和初边值问题。
叠加原理
\[\begin{aligned}
&u_{tt} - c^2u_{xx} = f,\\
t=0: &u = \varphi, u_t = \psi. \\
\end{aligned}
\]
将这个方程分解为两个子方程（只有右端项和只有初值），然后加权相加。
注：叠加原理适用于所有线性方程。
4. 一维波动方程 - 边值问题
Dirichlet 初边值问题
\[\begin{aligned}
&u_{tt} - c^2u_{xx} = f,\\
t=0: &u = \varphi, u_t = \psi, \\
x=0: &u = g, \\
x=L: &u = h. \\
\end{aligned}
\]
Step1：叠加原理分解为三个子问题
\[I_1:
\begin{aligned}
&u_{tt} - c^2u_{xx} = f,\\
t=0: &u = 0, u_t = 0, \\
x=0: &u = 0, \\
x=L: &u = 0. \\
\end{aligned}
\]
\[I_2:
\begin{aligned}
&u_{tt} - c^2u_{xx} = 0,\\
t=0: &u = \varphi, u_t = \psi, \\
x=0: &u = 0, \\
x=L: &u = 0. \\
\end{aligned}
\]
\[I_3:
\begin{aligned}
&u_{tt} - c^2u_{xx} = 0,\\
t=0: &u = 0, u_t = 0, \\
x=0: &u = g, \\
x=L: &u = h. \\
\end{aligned}
\]
Step2：统一转化为 \(I_2\) 的形式
\(I_1\) 可通过齐次化原理转化为 \(I_2\)；
对 \(I_3\)，令 \(\phi(t,x) = g(t) - \frac xL[h(t) - g(t)]\)，则 \(\phi\) 满足边值条件。再令 \(v = u - \phi\)，则 \(v\) 满足
\[I_4:
\begin{aligned}
&v_{tt} - c^2v_{xx} = -g'' - \frac xL[h'' - h''],\\
t=0: &v = -g(0) - \frac xL[h(0)-g(0)], v_t = -g'(0) - \frac xL[h'(0)-g'(0)] , \\
x=0: &v = 0, \\
x=L: &v = 0. \\
\end{aligned}
\]
进而通过叠加原理再次拆分即可。
Step3：分离变量法解 \(I_2\)
假设 \(u(t, x) = T(t)X(x)\)，则
\[T''(t)X(x) - c^2T(t)X''(x) = 0,
\]
即
\[\frac{T''(t)}{c^2T(t)} = \frac{X''(x)}{X(x)}.
\]
两端变量不同但相等，故只能为常数。设为 \(-\lambda\)，\(\lambda > 0\)。因此
\[T''(t) + c^2\lambda T(t) = 0, X''(x) + \lambda X(x) = 0.
\]
问题转化为边值问题
\[\begin{aligned}
X''(x) + \lambda X(x) = 0; \\
X(0) = 0, X(L) = 0.
\end{aligned}
\]
上述方程有非平凡解必须有 \(\sqrt \lambda L = k\pi, k\in \mathbb N\)。即
\[X_k(x) = C_k\sin \frac{k\pi}{L}x.
\]
故
\[T_k(t) = A_k\cos\frac{ck\pi}{L}t + B_k\sin\frac{ck\pi}{L}t.
\]
所以（系数可合并）
\[u_k(t,x) = T_k(t)X_k(x) = (A_k\cos\frac{ck\pi}{L}t + B_k\sin\frac{ck\pi}{L}t)\sin\frac{k\pi}{L}x.
\]
原边值问题的通解
\[u(t, x) = \sum_{k=1}^\infty u_k(t, x) = \sum_{k=1}^\infty(A_k\cos\frac{ck\pi}Lt + B_k\sin\frac{ck\pi}Lt)\sin\frac{k\pi}Lx.
\]
根据初值条件
\[\begin{aligned}
\varphi(x) &= \sum_{k=1}^\infty A_k\sin\frac{k\pi}Lx; \\
\psi(x) &= \sum_{k=1}^\infty \frac{ck\pi}LB_k\sin\frac{k\pi}Lx. \\
\end{aligned}
\]
因此 \(A_k, B_k\) 就是 Fourier 系数
\[\begin{aligned}
A_k &= \frac 2L\int_0^L \varphi(s)\sin\frac{k\pi s}Lds; \\
B_k &= \frac 2{ck\pi}\int_0^L \psi(s)\sin\frac{k\pi s}Lds. \\
\end{aligned}
\]
5. Fourier 变换
变换公式
\[\mathcal F[f](\xi) = \int_\mathbb R f(x)e^{-i\xi x}dx, \mathcal F^{-1}[g](x) = \frac 1{2\pi}\int_\mathbb Rg(\xi)e^{ix\xi}d\xi.
\]
性质
- \(\mathcal F^{-1}[\mathcal F[f]] = f\)
- \(\mathcal F[c_1f_1 + c_2f_2] = c_1\mathcal F[f_1] + c_2\mathcal F[f_2]\)
- \(\mathcal F[f_1 * f_2] = \mathcal F[f_1]\mathcal F[f_2]\)
- \(\mathcal F^{-1}[g_1 * g_2] = 2\pi\mathcal F^{-1}[g_1]\mathcal F^{-1}[g_2]\)
- \(\mathcal F[f_1f_2] = \frac 1{2\pi}\mathcal F[f_1]*\mathcal F[f_2]\)
- \(\mathcal F[f'] = i\xi\mathcal F[f]\)
- \(\mathcal F[-ixf] = (\mathcal F[f])'\)
6. 一维热方程 - 初值问题
齐次初值问题
\[\begin{aligned}
u_t &= c^2u_{xx}; \\
u|_{t=0} &= \varphi. \\
\end{aligned}
\]
Fourier 变换法
将 \(t\) 视为参数，关于 \(x\) 作 Fourier 变换。记 \(\tilde u(t, \xi) = \mathcal F[u], \tilde \varphi(\xi) := \mathcal F[\varphi]\)，则
\[\mathcal F[u_{xx}] = -c^2\xi^2\tilde u.
\]
因此
\[\tilde u_t = -c^2\xi^2\tilde u, \tilde u(0,\xi) = \tilde\varphi.
\]
直接解得
\[\tilde u(t, \xi) = \tilde\varphi(\xi)e^{-c^2\xi^2t}.
\]
两边关于 \(\xi\) 作 Fourier 逆变换，得
\[u = \mathcal F^{-1}[\tilde u] = \mathcal F^{-1}[\tilde\varphi e^{-c^2\xi^2t}].
\]
而
\[\mathcal F^{-1}[e^{-c^2\xi^2t}] = \frac 1{2\pi}\int_{\mathbb R}e^{-(c^2\xi^2t-i\xi x)}d\xi = \frac 1{2\pi}e^{-\frac{x^2}{4c^2t}}\int_{\mathbb R}e^{-c^2t(\xi-\frac{ix}{2c^2t})^2}d\xi = \frac 1{2\pi}e^{-\frac{x^2}{4c^2t}}\frac{\sqrt\pi}{c\sqrt t} = \frac 1{2c\sqrt\pi t}e^{-\frac{x^2}{4c^2t}}.
\]
所以
\[u(t, x) = \mathcal F^{-1}[\tilde\varphi]*\mathcal F^{-1}[e^{-c^2\xi^2t}] = \varphi * \left(\frac 1{2c\sqrt\pi t}e^{-\frac{x^2}{4c^2t}}\right) = \frac 1{2c\sqrt\pi t}\int_{\mathbb R}\varphi(\xi)e^{-\frac{(x-\xi)^2}{4c^2t}}d\xi.
\]
非齐次初值问题
\[\begin{aligned}
u_t &= c^2u_{xx} + f, \\
u|_{t=0} &= \varphi.
\end{aligned}
\]
由齐次化原理，当 \(\varphi=0\) 时方程的解为
\[u(t,x) = \int_0^t w(t,x;\tau)d\tau = \frac 1{2c\sqrt\pi t}\int_0^t\int_\mathbb R \frac{f(\tau,\xi)}{\sqrt{t-\tau}}e^{-\frac{(x-\xi)^2}{4c^2(t-\tau)}}d\xi d\tau.
\]
由叠加原理，原方程的解
\[u(t,x) = \frac 1{2c\sqrt\pi t}\int_{\mathbb R}\varphi(\xi)e^{-\frac{(x-\xi)^2}{4c^2t}}d\xi + \frac 1{2c\sqrt\pi t}\int_0^t\int_\mathbb R \frac{f(\tau,\xi)}{\sqrt{t-\tau}}e^{-\frac{(x-\xi)^2}{4c^2(t-\tau)}}d\xi d\tau.
\]
7. 一维热方程 - 初边值问题
齐次初边值问题
\[\begin{aligned}
u_t &= c^2u_{xx}, \\
t = 0: & u = \varphi(x), \\
x = 0: & u = 0, \\
x = L: & u_x + \sigma u = 0. \\
\end{aligned}
\]
分离变量法
假设 \(u(t, x) = T(t)X(x)\)，则
\[X(x)T'(t) = c^2X''(x)T(t),
\]
即
\[\frac{T'(t)}{c^2T(t)} = \frac{X''(x)}{X(x)} = -\lambda.
\]
所以
\[T'(t) + \lambda c^2T(t) = 0, X''(x) + \lambda c^2X(x) = 0.
\]
由边值条件得 \(X(0) = 0, X'(L) + \sigma X(L) = 0\)。
当 \(\lambda > 0\) 时，该 ODE 有非平凡解
\[X(x) = A\cos\sqrt\lambda x + B\sin\sqrt\lambda x.
\]
代入边值
\[X(0) = A = 0, X'(L) + \sigma X(L) = B(\sqrt\lambda\cos\sqrt\lambda L + \sigma\sin\sqrt\lambda L) = 0.
\]
\(\lambda\) 必须满足方程 \(\sqrt\lambda\cos\sqrt\lambda L + \sigma\sin\sqrt\lambda L = 0\)。即 \(\tan\sqrt\lambda L = -\frac{\sqrt\lambda}{\sigma}\)。该方程有无穷多个正解 \(\lambda_k(k\in \mathbb N)\)。
因此关于 \(X\) 的 ODE 有一列特解
\[X_k(x) = B_k\sin\sqrt{\lambda_k}x.
\]
所以
\[u(t,x) = \sum_{k=1}^\infty u_k(t,x) = \sum_{k=1}^\infty A_ke^{-c^2\lambda_k t}\sin\sqrt{\lambda_k}x.
\]
因为 \(u(0, x) = \varphi(x)\)，所以 \(\varphi(x) = \sum_{k=1}^\infty A_k\sin \sqrt{\lambda_k}x\)。
可证明特征函数系 \(\{\sin\sqrt{\lambda_k}x\}\) 在 \([0,L]\) 上正交。在 \(\varphi(x)\) 两边同时乘 \(\sin\sqrt{\lambda_k}x\) 并在 \([0,L]\) 上积分得
\[\int_0^L \sin\sqrt{\lambda_k}x\sum_{j=1}^\infty A_j\sin \sqrt{\lambda_j}xdx = A_k\int_0^L \sin^2\sqrt{\lambda_k}xdx = \int_0^L \varphi(x)\sin\sqrt{\lambda_k}xdx.
\]
而
\[\int_0^L\sin^2\sqrt{\lambda_k}xdx = \int_0^L \frac{1-\cos 2\sqrt{\lambda_k}x}{2}dx = \frac L2 - \frac{\sin 2\sqrt{\lambda_k}L}{4\sqrt{\lambda_k}} = \frac L2 - \frac{\sigma}{2(\sigma^2 + \lambda_k)}.
\]
所以
\[u(t, x) = \sum_{k=1}^\infty \frac{2(\sigma^2 + \lambda_k)}{L(\sigma^2 + \lambda_k)-\sigma}e^{-c^2\lambda_kt}\sin\sqrt{\lambda_k}t\int_0^L\varphi(s)\sin\sqrt{\lambda_k}sds.
\]