数学物理方程之通解

波动方程 u t t = a 2 u x x u_{tt}=a^2u_{xx} utt​=a2uxx​

齐次波动方程分离变量得到

{ T ′ ′ a 2 T = X ′ ′ X = − λ X ′ ′ + λ X = 0 T ′ ′ + λ a 2 T = 0 \left\{\begin{array}{l}\frac{T^{\prime \prime}}{a^{2} T}=\frac{X^{\prime \prime}}{X}=-\lambda \\ X^{\prime \prime}+\lambda X=0 \\ T^{\prime \prime}+\lambda a^{2} T=0\end{array}\right. ⎩⎨⎧​a2TT′′​=XX′′​=−λX′′+λX=0T′′+λa2T=0​

在不同边界条件下有不同的特征值和特征函数

{ λ n = n 2 π 2 l 2 X n ( x ) = sin ⁡ λ n x , u ∣ x = 0 = 0 = u ∣ x = l \left\{\left.\begin{array}{l}\lambda_{n}=\frac{n^{2} \pi^{2}}{l^{2}} \\ X_{n}(x)=\sin \sqrt{\lambda_{n}} x\end{array} ,\quad \left.u\right|_{x=0}\right.=0=\left.u\right|_{x=l}\right. {λn​=l2n2π2​Xn​(x)=sinλn​ ​x​,u∣x=0​=0=u∣x=l​

{ λ n = [ ( 2 n + 1 ) π 2 l ] 2 X n ( x ) = sin ⁡ λ n x , u ∣ x = 0 = 0 = u x ∣ x = l \left\{\left.\begin{array}{l}\lambda_{n}=\left[\frac{(2n+1)\pi}{{2l}}\right]^{2} \\ X_{n}(x)=\sin \sqrt{\lambda_{n}} x\end{array}, \quad \left.u\right|_{x=0}\right.=0=\left.u_x\right|_{x=l}\right. {λn​=[2l(2n+1)π​]2Xn​(x)=sinλn​ ​x​,u∣x=0​=0=ux​∣x=l​

{ λ n = n = [ ( 2 n + 1 ) π 2 l ] 2 X n ( x ) = cos ⁡ λ n x , u x ∣ x = 0 = 0 = u ∣ x = l \left\{\begin{array}{l}\lambda_{n}={n}=\left[\frac{(2n+1)\pi}{{2l}}\right]^{2} \\ X_{n}(x)=\cos \sqrt{\lambda_{n}} x\end{array}, \quad \left.u_x\right|_{x=0}\right.=0=\left.u\right|_{x=l} {λn​=n=[2l(2n+1)π​]2Xn​(x)=cosλn​ ​x​,ux​∣x=0​=0=u∣x=l​

{ λ n = n 2 π 2 l 2 X n ( x ) = cos ⁡ λ n x , u x ∣ x = 0 = 0 = u x ∣ x = l \left\{\begin{array}{l}\lambda_{n}=\frac{n^{2} \pi^{2}}{l^{2}} \\ X_{n}(x)=\cos \sqrt{\lambda_{n}} x\end{array}, \quad \left.u_x\right|_{x=0}\right.=0=\left.u_x\right|_{x=l} {λn​=l2n2π2​Xn​(x)=cosλn​ ​x​,ux​∣x=0​=0=ux​∣x=l​

∫ 0 l X n ( x ) X m ( x ) d x = l 2 \int_0^l X_n(x)X_m(x)dx=\frac{l}{2} ∫0l​Xn​(x)Xm​(x)dx=2l​

【也可令 X n ( x ) = 2 l sin ⁡ λ n x X_{n}(x)=\sqrt{\frac{2}{l}}\sin{\sqrt{\lambda_{n}}x} Xn​(x)=l2​ ​sinλn​ ​x或 X n ( x ) = 2 l cos ⁡ λ n x X_{n}(x)=\sqrt{\frac{2}{l}}\cos{\sqrt{\lambda_{n}}x} Xn​(x)=l2​ ​cosλn​ ​x以满足归一化条件 ∫ 0 l X n ( x ) X m ( x ) d x = 1 \int_0^l X_n(x)X_m(x)dx=1 ∫0l​Xn​(x)Xm​(x)dx=1，】

则波动方程的解为 u ( x , t ) = ∑ n = 1 + ∞ X n ( x ) T n ( t ) = ∑ n = 1 ∞ ( C n cos ⁡ λ n a t + D n sin ⁡ λ n a t ) X n ( x ) u(x, t)=\sum_{n=1}^{+\infty} X_{n}(x) T_{n}(t)=\sum_{n=1}^{\infty}\left(C_{n} \cos \sqrt{\lambda_{n}} a t+D_{n} \sin \sqrt{\lambda}_{n} a t\right) X_{n}(x) u(x,t)=∑n=1+∞​Xn​(x)Tn​(t)=∑n=1∞​(Cn​cosλn​ ​at+Dn​sinλ ​n​at)Xn​(x)

其中 { C n = 2 l ∫ 0 l φ ( x ) X n ( x ) d x D n = 2 l ⋅ 1 λ n a ∫ 0 l ψ ( x ) X n ( x ) d x \left\{\begin{array}{l}C_{n}=\frac{2}{l} \int_{0}^{l} \varphi(x) X_{n}(x) d x \\ D_{n}=\frac{2}{l}\cdot\frac{1}{\sqrt{\lambda_n}a} \int_{0}^{l} \psi(x) X_{n}(x) d x\end{array}\right. {Cn​=l2​∫0l​φ(x)Xn​(x)dxDn​=l2​⋅λn​ ​a1​∫0l​ψ(x)Xn​(x)dx​， φ ( x ) 、 ψ ( x ) \varphi(x)、\psi(x) φ(x)、ψ(x)分别为初始条件中的初位移、初速度。

扩散方程 u t = a 2 u x x u_t=a^2u_{xx} ut​=a2uxx​

齐次扩散方程分离变量得到

{ T ′ a 2 T = X ′ ′ X = − λ X ′ ′ + λ X = 0 T ′ + λ a 2 T = 0 \left\{\begin{array}{l}\frac{T^{ \prime}}{a^{2} T}=\frac{X^{\prime \prime}}{X}=-\lambda \\ X^{\prime \prime}+\lambda X=0 \\ T^{ \prime}+\lambda a^{2} T=0\end{array}\right. ⎩⎨⎧​a2TT′​=XX′′​=−λX′′+λX=0T′+λa2T=0​

X n ( x ) X_n(x) Xn​(x)同上， T n = C n e − a 2 λ n t T_{n}=C_{n} e^{-a^{2} \lambda_{n} t} Tn​=Cn​e−a2λn​t

则热传导方程的解为 u ( x , t ) = ∑ n = 1 ∞ C n e − λ n a 2 t X n u(x, t)=\sum_{n=1}^{\infty} C_{n} e^{-\lambda_{n} a^{2} t} X_{n} u(x,t)=∑n=1∞​Cn​e−λn​a2tXn​

其中 C n = 2 l ∫ 0 l φ ( x ) X n ( x ) d x C_{n}=\frac{2}{l} \int_{0}^{l} \varphi(x) X_{n}(x) d x Cn​=l2​∫0l​φ(x)Xn​(x)dx， φ ( x ) \varphi(x) φ(x)为初始条件中的初"位移"。

矩形区域Laplace 方程

Laplace 方程/调和方程： u x x + u y y = 0 ( 0 < x < a , 0 < y < b ) u_{xx}+u_{yy}=0\quad (0