偏微分方程 (Partial Differential Equations, PDE) ，是指含有多元未知函数及其若干阶偏导数的微分方程。
F ( u , x 1 , x 2 , ⋯ , ∂ u ∂ x 1 , ∂ u ∂ x 2 , ⋯ , ∂ 2 u ∂ x 1 2 , ⋯ ) = 0 (1.1) F(u,x_1,x_2,\cdots,\cfrac{∂u}{∂x_1},\cfrac{∂u}{∂x_2},\cdots,\cfrac{∂^2u}{∂x_1^2},\cdots)=0\tag{1.1}
F ( u , x 1 , x 2 , ⋯ , ∂ x 1 ∂ u , ∂ x 2 ∂ u , ⋯ , ∂ x 1 2 ∂ 2 u , ⋯ ) = 0 ( 1 . 1 )
其中最高阶偏导数的阶称为偏微分方程的阶。
在此引入一些重要的微分算子
Hamilton 算子： ∇ = ( ∂ ∂ x , ∂ ∂ y , ∂ ∂ z ) ∇=(\cfrac{∂}{∂x},\cfrac{∂}{∂y},\cfrac{∂}{∂z}) ∇ = ( ∂ x ∂ , ∂ y ∂ , ∂ z ∂ ) Δ = ∂ 2 ∂ x 2 + ∂ 2 ∂ y 2 + ∂ 2 ∂ z 2 = ∇ ⋅ ∇ = ∇ 2 Δ=\cfrac{∂^2}{∂x^2}+\cfrac{∂^2}{∂y^2}+\cfrac{∂^2}{∂z^2}=∇\cdot∇=∇^2 Δ = ∂ x 2 ∂ 2 + ∂ y 2 ∂ 2 + ∂ z 2 ∂ 2 = ∇ ⋅ ∇ = ∇ 2
数学物理方程 (Equation of Mathematical Physics) ，通常指从物理学及其他各门自然科学、技术科学中产生的偏微分方程。数学物理方程有时还包括常微分方程和积分方程。本课程将着重讨论三类基本的二阶线性偏微分方程。
(1) 弦的横振动方程：未知函数 u ( x , t ) u(x,t) u ( x , t ) x x x
u t t − a 2 u x x = f ( x , t ) u_{tt}-a^2u_{xx}=f(x,t)
u t t − a 2 u x x = f ( x , t )
其中 a = T ρ , f = F ρ a=\sqrt{\cfrac{T}{\rho}},f=\cfrac{F}{\rho} a = ρ T , f = ρ F ρ \rho ρ T T T F F F
杆的纵振动方程：未知函数 u ( x , t ) u(x,t) u ( x , t ) x x x
u t t − a 2 u x x = f ( x , t ) u_{tt}-a^2u_{xx}=f(x,t)
u t t − a 2 u x x = f ( x , t )
其中 a = E ρ , f = F ρ a=\sqrt{\cfrac{E}{\rho}},f=\cfrac{F}{\rho} a = ρ E , f = ρ F ρ \rho ρ E E E F F F
更一般的，三维空间的波动方程 (wave equation)是
u t t − a 2 Δ u = f ( r , t ) (1.2) u_{tt}-a^2Δu=f(\mathbf r,t)\tag{1.2}
u t t − a 2 Δ u = f ( r , t ) ( 1 . 2 )
(2) 热传导方程 ：(heat equation)未知函数 u ( r , t ) u(\mathbf r,t) u ( r , t )
u t − a 2 Δ u = f ( r , t ) (1.3) u_t-a^2 Δu=f(\mathbf r,t) \tag{1.3}
u t − a 2 Δ u = f ( r , t ) ( 1 . 3 )
其中 κ ρ c , F ρ c \cfrac{\kappa}{\rho c},\cfrac{F}{\rho c} ρ c κ , ρ c F κ \kappa κ ρ \rho ρ c c c F ( x , y , z , t ) F(x,y,z,t) F ( x , y , z , t )
(3) 如果热传导方程热源不随时间变化，当温度达到稳定时 ( u t = 0 ) (u_t=0) ( u t = 0 ) Poisson方程
Δ u = − f / a 2 (1.4) Δu=-f/a^2\tag{1.4}
Δ u = − f / a 2 ( 1 . 4 )
如果没有热源 ( f = 0 ) (f=0) ( f = 0 ) Laplace方程 （也称调和方程）
Δ u = 0 (1.5) Δu=0\tag{1.5}
Δ u = 0 ( 1 . 5 )
静电场的电势 u ( r ) u(\mathbf r) u ( r )
Δ u = − ρ / ϵ 0 Δu=-\rho/\epsilon_0
Δ u = − ρ / ϵ 0
其中 ρ \rho ρ ϵ 0 \epsilon_0 ϵ 0
(4) 如果波动方程 u ( r , t ) u(\mathbf r,t) u ( r , t ) ω \omega ω u ( r , t ) = v ( r ) e − i ω t u(\mathbf r,t)=v(\mathbf r)e^{-i\omega t} u ( r , t ) = v ( r ) e − i ω t v ( r ) v(\mathbf r) v ( r ) Helmholtz 方程
Δ v + k 2 v = 0 (1.6) Δv+k^2v=0\tag{1.6}
Δ v + k 2 v = 0 ( 1 . 6 )
其中 k = ω / a k=\omega/a k = ω / a
通解和特解 ：如果多元函数 u u u 古典解 。和常微分方程类似，称 m m m m m m 通解 。通解中的任意函数一旦确定便成为特解。
示例 1：求解偏微分方程
u x y = 0 u_{xy}=0
u x y = 0
解：先关于 y y y u x = f ( x ) u_x=f(x) u x = f ( x ) x x x
u = ∫ f ( ξ ) d ξ + f 2 ( η ) = f 1 ( ξ ) + f 2 ( η ) u =\int f(ξ)\mathrm dξ+f_2(η)=f_1(ξ)+f_2(η)
u = ∫ f ( ξ ) d ξ + f 2 ( η ) = f 1 ( ξ ) + f 2 ( η )
其中 f 1 , f 2 f_1,f_2 f 1 , f 2
示例 2：二维Laplace方程的通解
u x x + u y y = 0 u_{xx}+u_{yy}=0
u x x + u y y = 0
引入变换
{ ξ = x + i y η = x − i y \begin{cases}
ξ=x+iy \\
η=x-iy
\end{cases}
{ ξ = x + i y η = x − i y
根据复合函数偏导法则，进一步求得
{ u x x = u ξ ξ + 2 u ξ η + u η η u y y = − ( u ξ ξ − 2 u ξ η + u η η ) \begin{cases}
u_{xx}=u_{ξξ}+2u_{ξη}+u_{ηη} \\
u_{yy}=-(u_{ξξ}-2u_{ξη}+u_{ηη})
\end{cases}
{ u x x = u ξ ξ + 2 u ξ η + u η η u y y = − ( u ξ ξ − 2 u ξ η + u η η )
原方程变为
u ξ η = 0 u_{ξη}=0
u ξ η = 0
于是可求得通解
u ( x , y ) = f ( x + i y ) + g ( x − i y ) u(x,y)=f(x+iy)+g(x-iy)
u ( x , y ) = f ( x + i y ) + g ( x − i y )
定解条件 ：通常把反应系统内部作用导出的偏微分方程称为泛定方程 。为了完全描写一个实际物理问题的确定解，我们需要在一定的制约条件，称为定解条件 。偏微分方程和相应的定解条件就构成一个定解问题 。常见的定解条件有以下几类：
初始条件 ：应该完全描写初始时刻（t = 0 t=0 t = 0 t t t m m m m m m
u ∣ t = 0 = ϕ ( r ) , ∂ u ∂ t ∣ t = 0 = ψ ( r ) u|_{t=0}=\phi(\mathbf r),\quad \cfrac{∂u}{∂t}|_{t=0}=\psi(\mathbf r)
u ∣ t = 0 = ϕ ( r ) , ∂ t ∂ u ∣ t = 0 = ψ ( r )
对于热传导方程，只用给出初始时刻物体温度的分布状态
u ∣ t = 0 = ϕ ( r ) u|_{t=0}=\phi(\mathbf r)
u ∣ t = 0 = ϕ ( r )
Laplace 方程和Possion 方程都是描述稳恒状态的，与时间无关，所以不需要初始条件。
边界条件 ：边界条件形式比较多样化，要由具体问题中描述的具体状态决定。总的原则是：边界条件应该完全描写边界上各点在任意时刻的状况。
如果弦的两端固定，那么边界条件就是
u ∣ x = 0 = 0 , u ∣ x = l = 0 u|_{x=0}=0,\quad u|_{x=l}=0
u ∣ x = 0 = 0 , u ∣ x = l = 0
如果一端 ( x = 0 ) (x=0) ( x = 0 ) ( x = l ) (x=l) ( x = l ) F ( t ) F(t) F ( t )
u ∣ x = 0 = 0 , ∂ u ∂ x ∣ x = l = 1 T F ( t ) u|_{x=0}=0,\quad \cfrac{∂u}{∂x}|_{x=l}=\cfrac{1}{T}F(t)
u ∣ x = 0 = 0 , ∂ x ∂ u ∣ x = l = T 1 F ( t )
如果一端固定，另一端外力为 0，即另一端自由，则
u ∣ x = 0 = 0 , ∂ u ∂ x ∣ x = l = 0 u|_{x=0}=0,\quad \cfrac{∂u}{∂x}|_{x=l}=0
u ∣ x = 0 = 0 , ∂ x ∂ u ∣ x = l = 0
如果外力是由弹簧提供的弹性力，即 F ( t ) = − k u ( l , t ) F(t)=-ku(l,t) F ( t ) = − k u ( l , t ) k k k
( ∂ u ∂ x + σ u ) ∣ x = l = 0 , σ = k T (\cfrac{∂u}{∂x}+\sigma u)|_{x=l}=0,\quad \sigma=\cfrac{k}{T}
( ∂ x ∂ u + σ u ) ∣ x = l = 0 , σ = T k
对于热传导问题，也有类似的边界条件（以 ∂ Ω ∂\Omega ∂ Ω Ω \Omega Ω
如果边界温度分布已知，则
u ∣ r ∈ ∂ Ω = f ( r , t ) u|_{\mathbf r\in∂\Omega}=f(\mathbf r,t)
u ∣ r ∈ ∂ Ω = f ( r , t )
如果物体边界和周围介质绝热，则
∂ u ∂ n ∣ r ∈ ∂ Ω = 0 \cfrac{∂u}{∂n}|_{\mathbf r\in∂\Omega}=0
∂ n ∂ u ∣ r ∈ ∂ Ω = 0
设周围介质的温度为 u 1 ( r , t ) u_1(\mathbf r,t) u 1 ( r , t )
( ∂ u ∂ n + σ u ) ∣ r ∈ ∂ Ω = σ u 1 (\cfrac{∂u}{∂n}+\sigma u)|_{\mathbf r\in∂\Omega}=\sigma u_1
( ∂ n ∂ u + σ u ) ∣ r ∈ ∂ Ω = σ u 1
其中 σ = h / k , h \sigma=h/k,h σ = h / k , h
常见的边界条件数学上分为三类（以 ∂ Ω ∂\Omega ∂ Ω Ω \Omega Ω f f f
第一类边界条件 (Dircichlet条件)：直接给出边界上各点未知函数 u u u
u ∣ ∂ Ω = f ( r , t ) u|_{∂\Omega}=f(\mathbf r,t)
u ∣ ∂ Ω = f ( r , t )
第二类边界条件 (Neu-mann条件)：给出边界外法线方向上方向导数的数值
∂ u ∂ n ∣ ∂ Ω = f ( r , t ) \cfrac{∂u}{∂n}|_{∂\Omega}=f(\mathbf r,t)
∂ n ∂ u ∣ ∂ Ω = f ( r , t )
第三类边界条件 (Robin条件)：给出边界上各点的函数值与外法线方向上方向导数的某一线性组合值
( ∂ u ∂ n + σ u ) ∣ ∂ Ω = f ( r , t ) (\cfrac{∂u}{∂n}+\sigma u)|_{∂\Omega}=f(\mathbf r,t)
( ∂ n ∂ u + σ u ) ∣ ∂ Ω = f ( r , t )
连接条件 ：从微分方程的推导知道，方程只在空间区域的内部成立，如果在区域内部出现结构上的跃变，那么在这些跃变点（线，面）上微分方程不再成立，因此需要补充上相应的条件，通常称为连接条件 。x 0 x_0 x 0 u 1 ( x 0 , t ) = u 2 ( x 0 , t ) , ∂ u 1 ∂ x ∣ x = x 0 − 0 = ∂ u 2 ∂ x ∣ x = x 0 + 0 u_1(x_0,t)=u_2(x_0,t),\cfrac{∂u_1}{∂x}|_{x=x_0-0}=\cfrac{∂u_2}{∂x}|_{x=x_0+0} u 1 ( x 0 , t ) = u 2 ( x 0 , t ) , ∂ x ∂ u 1 ∣ x = x 0 − 0 = ∂ x ∂ u 2 ∣ x = x 0 + 0
定解问题的适定性 ：定解问题来自于实际，它的解也应该回到实际中去。如果一个定解问题的解存在、唯一且稳定，则称该定解问题是适定的 (well-posed) 。稳定性指的是，如果定解条件的数值有细微的改变，解的数值也作细微的改变。
线性叠加原理 ：考虑 n n n t t t x 1 , x 2 , ⋯ , x n x_1,x_2,\cdots,x_n x 1 , x 2 , ⋯ , x n
∑ j = 1 n ∑ i = 1 n a i j u x i x j + ∑ i = 1 n b i u x i + c u = f \displaystyle\sum_{j=1}^{n}\sum_{i=1}^n a_{ij}u_{x_ix_j}+\sum_{i=1}^{n}b_iu_{x_i}+cu=f
j = 1 ∑ n i = 1 ∑ n a i j u x i x j + i = 1 ∑ n b i u x i + c u = f
其中 a i j , b i , c , f a_{ij},b_i,c,f a i j , b i , c , f x 1 , x 2 , ⋯ , x n x_1,x_2,\cdots,x_n x 1 , x 2 , ⋯ , x n f f f
引入微分算子
L = ∑ j = 1 n ∑ i = 1 n a i j ∂ ∂ x i ∂ x j + ∑ i = 1 n b i ∂ ∂ x i + c L=\sum_{j=1}^{n}\sum_{i=1}^n a_{ij}\cfrac{∂}{∂x_i∂x_j}
+\sum_{i=1}^{n}b_i\cfrac{∂}{∂x_i}+c
L = j = 1 ∑ n i = 1 ∑ n a i j ∂ x i ∂ x j ∂ + i = 1 ∑ n b i ∂ x i ∂ + c
则可简单的表示为
L [ u ] = f L[u]=f
L [ u ] = f
一般的，对任意常数 c 1 , c 2 c_1,c_2 c 1 , c 2 u 1 , u 2 u_1,u_2 u 1 , u 2
L [ c 1 u 1 + c 2 u 2 ] = c 1 L [ u 1 ] + c 2 L [ u 2 ] (2.1) L[c_1u_1+c_2u_2]=c_1L[u_1]+c_2L[u_2]\tag{2.1}
L [ c 1 u 1 + c 2 u 2 ] = c 1 L [ u 1 ] + c 2 L [ u 2 ] ( 2 . 1 )
称为线性微分算子。显然，L L L
L 0 [ u ] = ( α ∂ u ∂ n + β u ) ∣ ∂ Ω = ϕ L_0[u]=(\alpha\cfrac{∂u}{∂n}+\beta u)|_{∂\Omega}=\phi
L 0 [ u ] = ( α ∂ n ∂ u + β u ) ∣ ∂ Ω = ϕ
其中 α , β , ϕ \alpha,\beta,\phi α , β , ϕ L 0 L_0 L 0
叠加原理 u i u_i u i L [ u i ] = f i ( i = 1 , 2 , ⋯ , m ) L[u_i]=f_i\quad(i=1,2,\cdots,m) L [ u i ] = f i ( i = 1 , 2 , ⋯ , m ) u = ∑ i = 1 m c i u i u=\displaystyle\sum_{i=1}^mc_iu_i u = i = 1 ∑ m c i u i
L [ u ] = ∑ i = 1 m c i f i (2.2) L[u]=\displaystyle\sum_{i=1}^mc_if_i\tag{2.2}
L [ u ] = i = 1 ∑ m c i f i ( 2 . 2 )
(2) 级数叠加原理：若 u i u_i u i L [ u i ] = f i ( i = 1 , 2 , ⋯ ) L[u_i]=f_i\quad(i=1,2,\cdots) L [ u i ] = f i ( i = 1 , 2 , ⋯ ) u = ∑ i = 1 ∞ c i u i u=\displaystyle\sum_{i=1}^\infty c_iu_i u = i = 1 ∑ ∞ c i u i
L [ u ] = ∑ i = 1 ∞ c i f i (2.3) L[u]=\displaystyle\sum_{i=1}^\infty c_if_i\tag{2.3}
L [ u ] = i = 1 ∑ ∞ c i f i ( 2 . 3 )
(3) 积分叠加原理：若 u ( r ; r 0 ) u(\mathbf{r;r_0}) u ( r ; r 0 ) L [ u ] = f ( r ; r 0 ) L[u]=f(\mathbf{r;r_0}) L [ u ] = f ( r ; r 0 ) U ( r ) = ∫ V c ( r 0 ) u ( r ; r 0 ) d r 0 U(\mathbf r)=\displaystyle\int_V c(\mathbf r_0)u(\mathbf{r;r_0})\mathrm d\mathbf r_0 U ( r ) = ∫ V c ( r 0 ) u ( r ; r 0 ) d r 0
L [ U ] = ∫ V c ( r 0 ) f ( r ; r 0 ) d r 0 (2.4) L[U]=\int_Vc(\mathbf r_0)f(\mathbf{r;r_0})\mathrm d\mathbf r_0\tag{2.4}
L [ U ] = ∫ V c ( r 0 ) f ( r ; r 0 ) d r 0 ( 2 . 4 )
引例 ：求解右行单波方程
u t + a u x = 0 u_{t}+au_{x}=0
u t + a u x = 0
的通解，引入变换
{ ξ = x − a t η = x \begin{cases}
ξ=x-at \\
η=x
\end{cases}
{ ξ = x − a t η = x
根据复合函数偏导法则，进一步求得
{ u t = − a u ξ u x = u ξ + u η \begin{cases}
u_{t}=-au_{ξ}\\
u_{x}=u_{ξ}+u_{η}
\end{cases}
{ u t = − a u ξ u x = u ξ + u η
原方程变为
u η = 0 u_{η}=0
u η = 0
于是可求得通解
u ( x , t ) = g ( x − a t ) u(x,t)=g(x-at)
u ( x , t ) = g ( x − a t )
一阶线性偏微分方程
n n n 一阶线性偏微分方程 的一般形式为
∑ i = 1 n b i ∂ u ∂ x i + c u = f \sum_{i=1}^{n}b_i\cfrac{∂u}{∂x_i}+cu=f
i = 1 ∑ n b i ∂ x i ∂ u + c u = f
其中 b i , c , f b_i,c,f b i , c , f x 1 , x 2 , ⋯ , x n x_1,x_2,\cdots,x_n x 1 , x 2 , ⋯ , x n
先研究两自变量 x , y x,y x , y
a u x + b u y + c u + f = 0 (2.1) au_x+bu_y+cu+f=0\tag{2.1}
a u x + b u y + c u + f = 0 ( 2 . 1 )
其中 a , b , c , f a,b,c,f a , b , c , f x , y x,y x , y a ≡ 0 , b ≠ 0 a\equiv0,b\neq0 a ≡ 0 , b = 0
u y + c b u + f b = 0 u_y+\cfrac{c}{b}u+\cfrac{f}{b}=0
u y + b c u + b f = 0
利用一阶线性常微分方程的求解方法可求得通解
u = e − p ( x , y ) [ ∫ e p ( x , y ) f ( x , y ) b ( x , y ) d y + g ( x ) ] u=e^{-p(x,y)}[\int e^{p(x,y)}\cfrac{f(x,y)}{b(x,y)} \mathrm dy+g(x)]
u = e − p ( x , y ) [ ∫ e p ( x , y ) b ( x , y ) f ( x , y ) d y + g ( x ) ]
其中 p ( x , y ) = ∫ c ( x , y ) b ( x , y ) d y , g ( x ) p(x,y)=\displaystyle\int\cfrac{c(x,y)}{b(x,y)} \mathrm dy,g(x) p ( x , y ) = ∫ b ( x , y ) c ( x , y ) d y , g ( x ) a ( x , y ) b ( x , y ) ≠ 0 a(x,y)b(x,y)\neq0 a ( x , y ) b ( x , y ) = 0
{ ξ = ξ ( x , y ) η = η ( x , y ) \begin{cases}
ξ=ξ(x,y) \\
η=η(x,y)
\end{cases}
{ ξ = ξ ( x , y ) η = η ( x , y )
要求雅可比行列式
∣ J ( ξ , η ) ∣ = ∣ ∂ ( ξ , η ) ∂ ( x , y ) ∣ = ∣ ξ x ξ y η x η y ∣ ≠ 0 |J(ξ,η)|=|\cfrac{∂(ξ,η)}{∂(x,y)}|=\begin{vmatrix}ξ_x&ξ_y \\ η_x&η_y\end{vmatrix}\neq0
∣ J ( ξ , η ) ∣ = ∣ ∂ ( x , y ) ∂ ( ξ , η ) ∣ = ∣ ∣ ∣ ∣ ξ x η x ξ y η y ∣ ∣ ∣ ∣ = 0
以保证新变量 ξ , η ξ,η ξ , η
u x = ∂ u ∂ ξ ∂ ξ ∂ x + ∂ u ∂ η ∂ η ∂ x , u y = ∂ u ∂ ξ ∂ ξ ∂ y + ∂ u ∂ η ∂ η ∂ y u_x=\cfrac{∂u}{∂ξ}\cfrac{∂ξ}{∂x}+\cfrac{∂u}{∂η}\cfrac{∂η}{∂x},\quad
u_y=\cfrac{∂u}{∂ξ}\cfrac{∂ξ}{∂y}+\cfrac{∂u}{∂η}\cfrac{∂η}{∂y}
u x = ∂ ξ ∂ u ∂ x ∂ ξ + ∂ η ∂ u ∂ x ∂ η , u y = ∂ ξ ∂ u ∂ y ∂ ξ + ∂ η ∂ u ∂ y ∂ η
可得到新的方程
A u ξ + B u η + C u + F = 0 (3.2) Au_ξ+Bu_η+Cu+F=0\tag{3.2}
A u ξ + B u η + C u + F = 0 ( 3 . 2 )
其中
{ A = a ξ x + b ξ y B = a η x + b η y C = c , F = f (3.3) \begin{cases}
A=aξ_x+bξ_y \\
B=aη_x+bη_y \\
C=c,F=f
\end{cases} \tag{3.3}
⎩ ⎪ ⎨ ⎪ ⎧ A = a ξ x + b ξ y B = a η x + b η y C = c , F = f ( 3 . 3 )
新方程仍然是线性的。从上式看出，如果取 ξ = ξ ( x , y ) ξ=ξ(x,y) ξ = ξ ( x , y )
a ∂ z ∂ x + b ∂ z ∂ y = 0 (3.4) a\cfrac{∂z}{∂x}+b\cfrac{∂z}{∂y}=0\tag{3.4}
a ∂ x ∂ z + b ∂ y ∂ z = 0 ( 3 . 4 )
的解，则此时 A = 0 A=0 A = 0
B u η + C u + F = 0 Bu_η+Cu+F=0
B u η + C u + F = 0
从而可以像第一种类型一样积分求解。
( a , b ) ⋅ ( ∂ z ∂ x , ∂ z ∂ y ) = 0 (3.5) (a,b)\cdot(\cfrac{∂z}{∂x},\cfrac{∂z}{∂y})=0\tag{3.5}
( a , b ) ⋅ ( ∂ x ∂ z , ∂ y ∂ z ) = 0 ( 3 . 5 )
方程的解 z = ϕ ( x , y ) z=\phi(x,y) z = ϕ ( x , y ) x y z xyz x y z S : z = ϕ ( x , y ) S:z=\phi(x,y) S : z = ϕ ( x , y ) z = c z=c z = c L : ϕ ( x , y ) = c L:\phi(x,y)=c L : ϕ ( x , y ) = c L L L P ( x , y ) P(x,y) P ( x , y ) ( ∂ ϕ ∂ x , ∂ ϕ ∂ y ) (\cfrac{∂\phi}{∂x},\cfrac{∂\phi}{∂y}) ( ∂ x ∂ ϕ , ∂ y ∂ ϕ ) P P P ( a , b ) (a,b) ( a , b ) P P P ( d x , d y ) (\mathrm dx,\mathrm dy) ( d x , d y )
d x a ( x , y ) = d y b ( x , y ) (3.6) \cfrac{\mathrm dx}{a(x,y)}=\cfrac{\mathrm dy}{b(x,y)}\tag{3.6}
a ( x , y ) d x = b ( x , y ) d y ( 3 . 6 )
由于平面 z = c z=c z = c L L L S S S ϕ ( x , y ) = c \phi(x,y)=c ϕ ( x , y ) = c z = ϕ ( x , y ) z=\phi(x,y) z = ϕ ( x , y ) ϕ ( x , y ) = c \phi(x,y)=c ϕ ( x , y ) = c
d ϕ ( x , y ) = ∂ ϕ ∂ x d x + ∂ ϕ ∂ y d y = 0 \mathrm d\phi(x,y)=\cfrac{∂\phi}{∂x}\mathrm dx+\cfrac{∂\phi}{∂y}\mathrm dy=0
d ϕ ( x , y ) = ∂ x ∂ ϕ d x + ∂ y ∂ ϕ d y = 0
将常微分方程改写成 d y d x = b ( x , y ) a ( x , y ) \cfrac{\mathrm dy}{\mathrm dx}=\cfrac{b(x,y)}{a(x,y)} d x d y = a ( x , y ) b ( x , y )
a ∂ ϕ ∂ x + b ∂ ϕ ∂ y = 0 a\cfrac{∂\phi}{∂x}+b\cfrac{∂\phi}{∂y}=0
a ∂ x ∂ ϕ + b ∂ y ∂ ϕ = 0
可知 z = ϕ ( x , y ) z=\phi(x,y) z = ϕ ( x , y )
(4) 常微分方程 (3.6) 叫做偏微分方程 (3.1) 的特征方程 (characteristic equation)，特征方程的积分曲线叫做特征线 。如果求出了积分曲线簇 ϕ ( x , y ) = c \phi(x,y)=c ϕ ( x , y ) = c c c c ∣ J ( ξ , η ) ∣ ≠ 0 |J(ξ,η)|\neq0 ∣ J ( ξ , η ) ∣ = 0 ξ , η ξ,η ξ , η
一阶拟线性偏微分方程
n n n 一阶拟线性偏微分方程 的一般形式为
∑ i = 1 n b i ( x 1 , x 2 , ⋯ , x n , u ) ∂ u ∂ x i = c ( x 1 , x 2 , ⋯ , x n , u ) \sum_{i=1}^{n}b_i(x_1,x_2,\cdots,x_n,u)\cfrac{∂u}{∂x_i}=c(x_1,x_2,\cdots,x_n,u)
i = 1 ∑ n b i ( x 1 , x 2 , ⋯ , x n , u ) ∂ x i ∂ u = c ( x 1 , x 2 , ⋯ , x n , u )
它的求解归结到 n + 1 n+1 n + 1
∑ i = 1 n b i ( x 1 , x 2 , ⋯ , x n , u ) ∂ ϕ ∂ x i + c ( x 1 , x 2 , ⋯ , x n , u ) ∂ ϕ ∂ u = 0 \sum_{i=1}^{n}b_i(x_1,x_2,\cdots,x_n,u)\cfrac{∂\phi}{∂x_i}+c(x_1,x_2,\cdots,x_n,u)\cfrac{∂\phi}{∂u}=0
i = 1 ∑ n b i ( x 1 , x 2 , ⋯ , x n , u ) ∂ x i ∂ ϕ + c ( x 1 , x 2 , ⋯ , x n , u ) ∂ u ∂ ϕ = 0
以两自变量 x , y x,y x , y
a u x + b u y = c (3.7) au_x+bu_y=c\tag{3.7}
a u x + b u y = c ( 3 . 7 )
其中 a , b , c a,b,c a , b , c x , y , u x,y,u x , y , u u = u ( x , y ) u=u(x,y) u = u ( x , y ) a , b a,b a , b
a ∂ ϕ ∂ x + b ∂ ϕ ∂ y + c ∂ ϕ ∂ u = 0 (3.8) a\cfrac{∂\phi}{∂x}+b\cfrac{∂\phi}{∂y}+c\cfrac{∂\phi}{∂u}=0\tag{3.8}
a ∂ x ∂ ϕ + b ∂ y ∂ ϕ + c ∂ u ∂ ϕ = 0 ( 3 . 8 )
若 ϕ ( x , y , u ) \phi(x,y,u) ϕ ( x , y , u ) ϕ ( x , y , u ) = 0 \phi(x,y,u)=0 ϕ ( x , y , u ) = 0 ϕ ( x , y , u ) = 0 \phi(x,y,u)=0 ϕ ( x , y , u ) = 0 ϕ ( x , y , u ( x , y ) ) = 0 \phi(x,y,u(x,y))=0 ϕ ( x , y , u ( x , y ) ) = 0 x , y x,y x , y
∂ ϕ ∂ x + ∂ ϕ ∂ u ∂ u ∂ x = 0 , ∂ ϕ ∂ y + ∂ ϕ ∂ u ∂ u ∂ y = 0 \cfrac{∂\phi}{∂x}+\cfrac{∂\phi}{∂u}\cfrac{∂u}{∂x}=0,\quad \cfrac{∂\phi}{∂y}+\cfrac{∂\phi}{∂u}\cfrac{∂u}{∂y}=0
∂ x ∂ ϕ + ∂ u ∂ ϕ ∂ x ∂ u = 0 , ∂ y ∂ ϕ + ∂ u ∂ ϕ ∂ y ∂ u = 0
从中解出 ∂ ϕ ∂ x = − ∂ ϕ ∂ u ∂ u ∂ x , ∂ ϕ ∂ y = − ∂ ϕ ∂ u ∂ u ∂ y \cfrac{∂\phi}{∂x}=-\cfrac{∂\phi}{∂u}\cfrac{∂u}{∂x},\quad \cfrac{∂\phi}{∂y}=-\cfrac{∂\phi}{∂u}\cfrac{∂u}{∂y} ∂ x ∂ ϕ = − ∂ u ∂ ϕ ∂ x ∂ u , ∂ y ∂ ϕ = − ∂ u ∂ ϕ ∂ y ∂ u ϕ ( x , y , u ) = 0 \phi(x,y,u)=0 ϕ ( x , y , u ) = 0
d x a ( x , y , u ) = d y b ( x , y , u ) = d u c ( x , y , u ) (3.9) \cfrac{\mathrm dx}{a(x,y,u)}=\cfrac{\mathrm dy}{b(x,y,u)}=\cfrac{\mathrm du}{c(x,y,u)}\tag{3.9}
a ( x , y , u ) d x = b ( x , y , u ) d y = c ( x , y , u ) d u ( 3 . 9 )
的首次积分确定。上式称为方程 (3.7) 的完全特征方程组 ，它的积分曲线称为完全特征曲线 。
以下研究方程的分类，并把各类方程分别化为标准型，这样以后就只需讨论标准型的解法。
两个自变量方程的分类
先研究两自变量 x , y x,y x , y
a 11 u x x + a 12 u x y + a 22 u y y + b 1 u x + b 2 u y + c u + f = 0 (4.1) a_{11}u_{xx}+a_{12}u_{xy}+a_{22}u_{yy}+b_1u_x+b_2u_y+cu+f=0\tag{4.1}
a 1 1 u x x + a 1 2 u x y + a 2 2 u y y + b 1 u x + b 2 u y + c u + f = 0 ( 4 . 1 )
其中 a 11 , a 12 , a 22 , b 1 , b 2 , c , f a_{11},a_{12},a_{22},b_1,b_2,c,f a 1 1 , a 1 2 , a 2 2 , b 1 , b 2 , c , f x , y x,y x , y
作自变量非奇异变换（可逆）
{ ξ = ξ ( x , y ) η = η ( x , y ) \begin{cases}
ξ=ξ(x,y) \\
η=η(x,y)
\end{cases}
{ ξ = ξ ( x , y ) η = η ( x , y )
其中雅可比行列式
∣ J ∣ = ∣ ∂ ( ξ , η ) ∂ ( x , y ) ∣ = ∣ ξ x ξ y η x η y ∣ ≠ 0 |J|=|\cfrac{∂(ξ,η)}{∂(x,y)}|=\begin{vmatrix}ξ_x&ξ_y \\ η_x&η_y\end{vmatrix}\neq0
∣ J ∣ = ∣ ∂ ( x , y ) ∂ ( ξ , η ) ∣ = ∣ ∣ ∣ ∣ ξ x η x ξ y η y ∣ ∣ ∣ ∣ = 0
通过代换可得到新的方程
A 11 u ξ ξ + 2 A 12 u ξ η + A 22 u η η + B 1 u ξ + B 2 u η + C u + F = 0 (4.2) A_{11}u_{ξξ}+2A_{12}u_{ξη}+A_{22}u_{ηη}+B_1u_ξ+B_2u_η+Cu+F=0\tag{4.2}
A 1 1 u ξ ξ + 2 A 1 2 u ξ η + A 2 2 u η η + B 1 u ξ + B 2 u η + C u + F = 0 ( 4 . 2 )
其中
{ A 11 = a 11 ξ x 2 + 2 a 12 ξ x ξ y + a 22 ξ y 2 A 12 = a 11 ξ x η x + a 12 ( ξ x η y + ξ y η x ) + a 22 ξ y η y A 22 = a 11 η x 2 + 2 a 12 η x η y + a 22 η y 2 B 1 = a 11 ξ x x + 2 a 12 ξ x y + a 22 ξ y y + b 1 ξ x + b 2 ξ y B 2 = a 11 η x x + 2 a 12 η x y + a 22 η y y + b 1 η x + b 2 η y C = c , F = f (4.3) \begin{cases}
A_{11}=a_{11}ξ_x^2+2a_{12}ξ_xξ_y+a_{22}ξ_y^2 \\
A_{12}=a_{11}ξ_xη_x+a_{12}(ξ_xη_y+ξ_yη_x)+a_{22}ξ_yη_y \\
A_{22}=a_{11}η_x^2+2a_{12}η_xη_y+a_{22}η_y^2 \\
B_1=a_{11}ξ_{xx}+2a_{12}ξ_{xy}+a_{22}ξ_{yy}+b_1ξ_x+b_2ξ_y \\
B_2=a_{11}η_{xx}+2a_{12}η_{xy}+a_{22}η_{yy}+b_1η_x+b_2η_y \\
C=c,F=f
\end{cases} \tag{4.3}
⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ A 1 1 = a 1 1 ξ x 2 + 2 a 1 2 ξ x ξ y + a 2 2 ξ y 2 A 1 2 = a 1 1 ξ x η x + a 1 2 ( ξ x η y + ξ y η x ) + a 2 2 ξ y η y A 2 2 = a 1 1 η x 2 + 2 a 1 2 η x η y + a 2 2 η y 2 B 1 = a 1 1 ξ x x + 2 a 1 2 ξ x y + a 2 2 ξ y y + b 1 ξ x + b 2 ξ y B 2 = a 1 1 η x x + 2 a 1 2 η x y + a 2 2 η y y + b 1 η x + b 2 η y C = c , F = f ( 4 . 3 )
新方程仍然是线性的。A 11 , A 12 , A 22 A_{11},A_{12},A_{22} A 1 1 , A 1 2 , A 2 2
D = ( A 11 A 12 A 21 A 22 ) = ( ξ x ξ y η x η y ) ( a 11 a 12 a 21 a 22 ) ( ξ x ξ y η x η y ) T (4.4) \mathbf D=
\begin{pmatrix}A_{11}&A_{12} \\ A_{21}&A_{22}\end{pmatrix}=
\begin{pmatrix}ξ_x&ξ_y \\ η_x&η_y\end{pmatrix}
\begin{pmatrix}a_{11}&a_{12} \\ a_{21}&a_{22}\end{pmatrix}
\begin{pmatrix}ξ_x&ξ_y \\ η_x&η_y\end{pmatrix}^{T} \tag{4.4}
D = ( A 1 1 A 2 1 A 1 2 A 2 2 ) = ( ξ x η x ξ y η y ) ( a 1 1 a 2 1 a 1 2 a 2 2 ) ( ξ x η x ξ y η y ) T ( 4 . 4 )
其中 A 12 = A 21 , a 12 = a 21 A_{12}=A_{21},a_{12}=a_{21} A 1 2 = A 2 1 , a 1 2 = a 2 1
由 (4.4) 式看出，如果取一阶偏微分方程
a 11 z x 2 + 2 a 12 z x z y + a 22 z y 2 = 0 a_{11}z_x^2+2a_{12}z_xz_y+a_{22}z_y^2=0
a 1 1 z x 2 + 2 a 1 2 z x z y + a 2 2 z y 2 = 0
两个线性无关的特解作为新的自变量 ξ , η ξ,η ξ , η A 11 = A 22 = 0 A_{11}=A_{22}=0 A 1 1 = A 2 2 = 0
a 11 ( − z x z y ) 2 − 2 a 12 ( − z x z y ) + a 22 = 0 a_{11}(-\cfrac{z_x}{z_y})^2-2a_{12}(-\cfrac{z_x}{z_y})+a_{22}=0
a 1 1 ( − z y z x ) 2 − 2 a 1 2 ( − z y z x ) + a 2 2 = 0
如果把 z ( x , y ) = const z(x,y)=\text{const} z ( x , y ) = const y = y ( x ) y=y(x) y = y ( x ) d y / d x = − z x / z y dy/dx=-z_x/z_y d y / d x = − z x / z y
a 11 ( d y d x ) 2 − 2 a 12 ( d y d x ) + a 22 = 0 (4.5) a_{11}(\cfrac{dy}{dx})^2-2a_{12}(\cfrac{dy}{dx})+a_{22}=0\tag{4.5}
a 1 1 ( d x d y ) 2 − 2 a 1 2 ( d x d y ) + a 2 2 = 0 ( 4 . 5 )
常微分方程 (4.5) 叫做偏微分方程 (4.1) 的特征方程 ，特征方程的积分曲线叫做特征线 。
d y d x = a 12 ± a 12 2 − a 11 a 22 a 11 (4.6) \cfrac{dy}{dx}=\cfrac{a_{12} \pm \sqrt{a_{12}^2-a_{11}a_{22}}}{a_{11}}\tag{4.6}
d x d y = a 1 1 a 1 2 ± a 1 2 2 − a 1 1 a 2 2 ( 4 . 6 )
通常根据判别式 Δ = a 12 2 − a 11 a 22 Δ=a_{12}^2-a_{11}a_{22} Δ = a 1 2 2 − a 1 1 a 2 2 Δ > 0 Δ>0 Δ > 0 Δ = 0 Δ=0 Δ = 0 Δ > 0 Δ>0 Δ > 0
由 (4.4) 矩阵的性质得到
A 12 2 − A 11 A 22 = det ( D ) = ∣ J ∣ 2 ( a 12 2 − a 11 a 22 ) A_{12}^2-A_{11}A_{22}=\det(\mathbf D)=|J|^2(a_{12}^2-a_{11}a_{22})
A 1 2 2 − A 1 1 A 2 2 = det ( D ) = ∣ J ∣ 2 ( a 1 2 2 − a 1 1 a 2 2 )
这就是说，作自变量变换时，方程的类型不变。
下面就方程中 Δ = a 12 2 − a 11 a 22 Δ=a_{12}^2-a_{11}a_{22} Δ = a 1 2 2 − a 1 1 a 2 2
双曲型方程 ：方程 (4.6) 给出两族实的特征线 ξ ( x , y ) = c 1 , η ( x , y ) = c 2 ξ(x,y)=c_1,η(x,y)=c_2 ξ ( x , y ) = c 1 , η ( x , y ) = c 2 ξ = ξ ( x , y ) , η = η ( x , y ) ξ=ξ(x,y),η=η(x,y) ξ = ξ ( x , y ) , η = η ( x , y ) A 11 = A 22 = 0 A_{11}=A_{22}=0 A 1 1 = A 2 2 = 0
u ξ η = − 1 2 A 12 ( B 1 u ξ + B 2 u η + C u + F ) u_{ξη}=-\cfrac{1}{2A_{12}}(B_1u_ξ+B_2u_η+Cu+F)
u ξ η = − 2 A 1 2 1 ( B 1 u ξ + B 2 u η + C u + F )
或者，再做自变量变换 ξ = α + β , η = α − β ξ=α+β,η=α-β ξ = α + β , η = α − β
u α α − u β β = − 1 A 12 [ ( B 1 + B 2 ) u α + ( B 1 − B 2 ) u β + 2 C u + 2 F ] u_{αα}-u_{ββ}=-\cfrac{1}{A_{12}}[(B_1+B_2)u_α+(B_1-B_2)u_β+2Cu+2F]
u α α − u β β = − A 1 2 1 [ ( B 1 + B 2 ) u α + ( B 1 − B 2 ) u β + 2 C u + 2 F ]
上述两方程是双曲型方程的标准形式，弦振动方程即为标准型的双曲方程。
抛物型方程 ：特征方程只能给出一族特征线 ξ ( x , y ) = c ξ(x,y)=c ξ ( x , y ) = c ξ = ξ ( x , y ) ξ=ξ(x,y) ξ = ξ ( x , y ) η = η ( x , y ) η=η(x,y) η = η ( x , y ) ξ ξ ξ a 12 2 − a 11 a 22 = 0 a_{12}^2-a_{11}a_{22}=0 a 1 2 2 − a 1 1 a 2 2 = 0 A 11 = a 11 ξ x 2 + 2 a 12 ξ x ξ y + a 22 ξ y 2 = ( a 11 ξ x + a 22 ξ y ) 2 = 0 A 12 = a 11 ξ x η x + a 12 ( ξ x η y + ξ y η x ) + a 22 ξ y η y = ( a 11 ξ x + a 22 ξ y ) ( a 11 η x + a 22 η y ) = 0 A_{11}=a_{11}ξ_x^2+2a_{12}ξ_xξ_y+a_{22}ξ_y^2=(\sqrt{a_{11}}ξ_x+\sqrt{a_{22}}ξ_y)^2=0 \\
A_{12}=a_{11}ξ_xη_x+a_{12}(ξ_xη_y+ξ_yη_x)+a_{22}ξ_yη_y=(\sqrt{a_{11}}ξ_x+\sqrt{a_{22}}ξ_y)(\sqrt{a_{11}}η_x+\sqrt{a_{22}}η_y)=0 A 1 1 = a 1 1 ξ x 2 + 2 a 1 2 ξ x ξ y + a 2 2 ξ y 2 = ( a 1 1 ξ x + a 2 2 ξ y ) 2 = 0 A 1 2 = a 1 1 ξ x η x + a 1 2 ( ξ x η y + ξ y η x ) + a 2 2 ξ y η y = ( a 1 1 ξ x + a 2 2 ξ y ) ( a 1 1 η x + a 2 2 η y ) = 0
u η η = − 1 2 A 22 ( B 1 u ξ + B 2 u η + C u + F ) u_{ηη}=-\cfrac{1}{2A_{22}}(B_1u_ξ+B_2u_η+Cu+F)
u η η = − 2 A 2 2 1 ( B 1 u ξ + B 2 u η + C u + F )
这是抛物型方程的标准形式，扩散方程、热传导方程都是标准形式的抛物型方程。
椭圆形方程 ：特征方程给出两族复数的特征线 ξ ( x , y ) = c 1 , η ( x , y ) = c 2 ξ(x,y)=c_1,η(x,y)=c_2 ξ ( x , y ) = c 1 , η ( x , y ) = c 2 ξ = ξ ( x , y ) , η = η ( x , y ) ξ=ξ(x,y),η=η(x,y) ξ = ξ ( x , y ) , η = η ( x , y ) ξ , η ξ,η ξ , η ξ = α + i β , η = α − i β ξ=α+iβ,η=α-iβ ξ = α + i β , η = α − i β
u α α + u β β = − 1 A 12 [ ( B 1 + B 2 ) u α + i ( B 2 − B 1 ) u β + 2 C u + 2 F ] u_{αα}+u_{ββ}=-\cfrac{1}{A_{12}}[(B_1+B_2)u_α+i(B_2-B_1)u_β+2Cu+2F]
u α α + u β β = − A 1 2 1 [ ( B 1 + B 2 ) u α + i ( B 2 − B 1 ) u β + 2 C u + 2 F ]
这是椭圆型方程的标准形式，稳定温度分别、静电场方程都是椭圆型的双曲方程。
多自变量方程的分类
考虑多自变量 x 1 , x 2 , ⋯ , x n x_1,x_2,\cdots,x_n x 1 , x 2 , ⋯ , x n
∑ j = 1 n ∑ i = 1 n a i j u x i x j + ∑ i = 1 n b i u x i + c u + f = 0 (4.7) \displaystyle \sum_{j=1}^{n}\sum_{i=1}^n a_{ij}u_{x_ix_j}+\sum_{i=1}^{n}b_iu_{x_i}+cu+f=0 \tag{4.7}
j = 1 ∑ n i = 1 ∑ n a i j u x i x j + i = 1 ∑ n b i u x i + c u + f = 0 ( 4 . 7 )
做自变量的非奇异代换
ξ k = ξ k ( x 1 , x 2 , ⋯ , x n ) , k = 1 , 2 , ⋯ , n ξ_k=ξ_k(x_1,x_2,\cdots,x_n),\quad k=1,2,\cdots,n
ξ k = ξ k ( x 1 , x 2 , ⋯ , x n ) , k = 1 , 2 , ⋯ , n
代换的雅克比行列式
∣ J ∣ = ∣ ∂ ( ξ 1 , ξ 2 , ⋯ , ξ n ) ∂ ( x 1 , x 2 , ⋯ , x n ) ∣ ≠ 0 |J|=|\cfrac{∂(ξ_1,ξ_2,\cdots,ξ_n)}{∂(x_1,x_2,\cdots,x_n)}|\neq0
∣ J ∣ = ∣ ∂ ( x 1 , x 2 , ⋯ , x n ) ∂ ( ξ 1 , ξ 2 , ⋯ , ξ n ) ∣ = 0
通过代换可得到新的方程
∑ j = 1 n ∑ i = 1 n A i j u ξ i ξ j + ∑ i = 1 n B i u ξ i + C u + F = 0 (4.8) \displaystyle \sum_{j=1}^{n}\sum_{i=1}^n A_{ij}u_{ξ_iξ_j}+\sum_{i=1}^{n}B_iu_{ξ_i}+Cu+F=0 \tag{4.8}
j = 1 ∑ n i = 1 ∑ n A i j u ξ i ξ j + i = 1 ∑ n B i u ξ i + C u + F = 0 ( 4 . 8 )
注意到变换后的方程仍然是线性的，其中系数
{ A i j = ∑ l = 1 n ∑ k = 1 n a k l ∂ ξ i ∂ x k ∂ ξ j ∂ x l B i = ∑ k = 1 n ∂ ξ i ∂ x k + ∑ l = 1 n ∑ k = 1 n ∂ 2 ξ i ∂ x k ∂ x l \begin{cases}
A_{ij}=\displaystyle \sum_{l=1}^{n}\sum_{k=1}^na_{kl}\cfrac{∂ξ_i}{∂x_k}\cfrac{∂ξ_j}{∂x_l} \\
\displaystyle B_i=\sum_{k=1}^n\cfrac{∂ξ_i}{∂x_k}+\sum_{l=1}^{n}\sum_{k=1}^n\cfrac{∂^2ξ_i}{∂x_k∂x_l}
\end{cases}
⎩ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎧ A i j = l = 1 ∑ n k = 1 ∑ n a k l ∂ x k ∂ ξ i ∂ x l ∂ ξ j B i = k = 1 ∑ n ∂ x k ∂ ξ i + l = 1 ∑ n k = 1 ∑ n ∂ x k ∂ x l ∂ 2 ξ i
系数 A i j A_{ij} A i j
( A i j ) n × n = J A J T (A_{ij})_{n\times n}=JAJ^T
( A i j ) n × n = J A J T
其中 A = ( a i j ) n × n A=(a_{ij})_{n\times n} A = ( a i j ) n × n ∑ j = 1 n ∑ i = 1 n a i j λ i λ j \displaystyle \sum_{j=1}^{n}\sum_{i=1}^n a_{ij}λ_iλ_j j = 1 ∑ n i = 1 ∑ n a i j λ i λ j
( A i j ) n × n = B A B T = d i a g ( A 11 , A 22 , ⋯ , A n n ) (A_{ij})_{n\times n}=BAB^T=\mathrm{diag}(A_{11},A_{22},\cdots,A_{nn})
( A i j ) n × n = B A B T = d i a g ( A 1 1 , A 2 2 , ⋯ , A n n )
其中 A i i ∈ { − 1 , 0 , 1 } A_{ii}\in\{-1,0,1\} A i i ∈ { − 1 , 0 , 1 } A i i A_{ii} A i i − 1 , 0 , 1 -1,0,1 − 1 , 0 , 1 A i i A_{ii} A i i p p p q q q p = n , q = 0 p=n,q=0 p = n , q = 0
∑ i = 1 n ∂ 2 u ∂ ξ i 2 + ∑ i = 1 n B i ∂ u ∂ ξ i + C u + F = 0 \displaystyle\sum_{i=1}^n\cfrac{∂^2u}{∂ξ_i^2}+\sum_{i=1}^{n}B_i\cfrac{∂u}{∂ξ_i}+Cu+F=0
i = 1 ∑ n ∂ ξ i 2 ∂ 2 u + i = 1 ∑ n B i ∂ ξ i ∂ u + C u + F = 0
(2) 当 p = n − 1 , q = 0 p=n-1,q=0 p = n − 1 , q = 0
∑ i = 1 n − 1 ∂ 2 u ∂ ξ i 2 + ∑ i = 1 n B i ∂ u ∂ ξ i + C u + F = 0 \displaystyle\sum_{i=1}^{n-1}\cfrac{∂^2u}{∂ξ_i^2}+\sum_{i=1}^{n}B_i\cfrac{∂u}{∂ξ_i}+Cu+F=0
i = 1 ∑ n − 1 ∂ ξ i 2 ∂ 2 u + i = 1 ∑ n B i ∂ ξ i ∂ u + C u + F = 0
(3) 当 p = 1 , q = n − 1 p=1,q=n-1 p = 1 , q = n − 1
∂ 2 u ∂ ξ 1 2 − ∑ j = 2 n ∂ 2 u ∂ ξ j 2 + ∑ i = 1 n B i ∂ u ∂ ξ i + C u + F = 0 \displaystyle\cfrac{∂^2u}{∂ξ_1^2}-\sum_{j=2}^n\cfrac{∂^2u}{∂ξ_j^2}+\sum_{i=1}^{n}B_i\cfrac{∂u}{∂ξ_i}+Cu+F=0
∂ ξ 1 2 ∂ 2 u − j = 2 ∑ n ∂ ξ j 2 ∂ 2 u + i = 1 ∑ n B i ∂ ξ i ∂ u + C u + F = 0
(4) 当 p > 0 , q > 0 p>0,q>0 p > 0 , q > 0
∑ i = 1 p ∂ 2 u ∂ ξ i 2 − ∑ j = 1 p + q ∂ 2 u ∂ ξ i 2 + ∑ i = 1 n B i ∂ u ∂ ξ i + C u + F = 0 \displaystyle\sum_{i=1}^p\cfrac{∂^2u}{∂ξ_i^2}-\sum_{j=1}^{p+q}\cfrac{∂^2u}{∂ξ_i^2}+\sum_{i=1}^{n}B_i\cfrac{∂u}{∂ξ_i}+Cu+F=0
i = 1 ∑ p ∂ ξ i 2 ∂ 2 u − j = 1 ∑ p + q ∂ ξ i 2 ∂ 2 u + i = 1 ∑ n B i ∂ ξ i ∂ u + C u + F = 0
常系数二阶线性偏微分方程 ：方程 (2.12) 中的系数 a i j , b i , c , f a_{ij},b_i,c,f a i j , b i , c , f
u ( ξ , η ) = V ( ξ , η ) e λ ξ + μ η u(ξ,η)=V(ξ,η)e^{λξ+μη}
u ( ξ , η ) = V ( ξ , η ) e λ ξ + μ η
其中 λ , μ λ,μ λ , μ λ , μ λ,μ λ , μ
无限长弦振动初值问题 ：考虑两端为无限长的弦振动方程初值问题
{ u t t − a 2 u x x = 0 u ∣ t = 0 = ϕ ( x ) , ∂ u ∂ t ∣ t = 0 = ψ ( x ) (5.1) \begin{cases}
u_{tt}-a^2u_{xx}=0 \\
u|_{t=0}=\phi(x),\cfrac{∂u}{∂t}|_{t=0}=\psi(x)
\end{cases}\tag{5.1}
⎩ ⎨ ⎧ u t t − a 2 u x x = 0 u ∣ t = 0 = ϕ ( x ) , ∂ t ∂ u ∣ t = 0 = ψ ( x ) ( 5 . 1 )
其中 ϕ ( x ) , ψ ( x ) \phi(x),\psi(x) ϕ ( x ) , ψ ( x ) a > 0 a>0 a > 0
( ∂ ∂ t + a ∂ u ∂ x ) ( ∂ ∂ t − a ∂ ∂ x ) u = 0 (\cfrac{∂}{∂t}+a\cfrac{∂u}{∂x})(\cfrac{∂}{∂t}-a\cfrac{∂}{∂x})u=0
( ∂ t ∂ + a ∂ x ∂ u ) ( ∂ t ∂ − a ∂ x ∂ ) u = 0
等价于两个一阶线性偏微分方程组
{ ∂ u ∂ t + a ∂ u ∂ x = 0 ∂ u ∂ t − a ∂ u ∂ x = v \begin{cases}
\cfrac{∂u}{∂t}+a\cfrac{∂u}{∂x}=0 \\
\cfrac{∂u}{∂t}-a\cfrac{∂u}{∂x}=v
\end{cases}
⎩ ⎪ ⎨ ⎪ ⎧ ∂ t ∂ u + a ∂ x ∂ u = 0 ∂ t ∂ u − a ∂ x ∂ u = v
这两个方程有一组独立的特征线 x + a t = c 1 , x − a t = c 2 x+at=c_1,x-at=c_2 x + a t = c 1 , x − a t = c 2 ξ = x + a t , η = x − a t ξ=x+at,η=x-at ξ = x + a t , η = x − a t u t t , u x x u_{tt},u_{xx} u t t , u x x
u ξ η = 0 u_{ξη}=0
u ξ η = 0
上述方程容易求得通解
u = f 1 ( x + a t ) + f 2 ( x − a t ) (5.2) u=f_1(x+at)+f_2(x-at)\tag{5.2}
u = f 1 ( x + a t ) + f 2 ( x − a t ) ( 5 . 2 )
其中 f 1 , f 2 f_1,f_2 f 1 , f 2
通解具有明确的物理意义，以 f 2 ( x − a t ) f_2(x-at) f 2 ( x − a t ) a a a x x x X X X X = x − a t , T = t X=x-at,T=t X = x − a t , T = t f 2 ( x − a t ) = f ( X ) f_2(x-at)=f(X) f 2 ( x − a t ) = f ( X ) T T T a a a x x x f 1 ( x + a t ) f_1(x+at) f 1 ( x + a t ) a a a x x x a a a x x x
(2) 函数 f 1 , f 2 f_1,f_2 f 1 , f 2 { f 1 ( x ) + f 2 ( x ) = ϕ ( x ) a f 1 ′ ( x ) − a f 2 ′ ( x ) = ψ ( x ) \begin{cases}
f_1(x)+f_2(x)=\phi(x) \\af'_1(x)-af'_2(x)=\psi(x)
\end{cases} { f 1 ( x ) + f 2 ( x ) = ϕ ( x ) a f 1 ′ ( x ) − a f 2 ′ ( x ) = ψ ( x )
f 1 ( x ) = 1 2 ϕ ( x ) + 1 2 a ∫ x 0 x ψ ( ξ ) d ξ + 1 2 [ f 1 ( x 0 ) − f 2 ( x 0 ) ] f 2 ( x ) = 1 2 ϕ ( x ) − 1 2 a ∫ x 0 x ψ ( ξ ) d ξ − 1 2 [ f 1 ( x 0 ) − f 2 ( x 0 ) ] f_1(x)=\frac{1}{2}\phi(x)+\frac{1}{2a}\int_{x_0}^x\psi(ξ)dξ+\frac{1}{2}[f_1(x_0)-f_2(x_0)] \\
f_2(x)=\frac{1}{2}\phi(x)-\frac{1}{2a}\int_{x_0}^x\psi(ξ)dξ-\frac{1}{2}[f_1(x_0)-f_2(x_0)]
f 1 ( x ) = 2 1 ϕ ( x ) + 2 a 1 ∫ x 0 x ψ ( ξ ) d ξ + 2 1 [ f 1 ( x 0 ) − f 2 ( x 0 ) ] f 2 ( x ) = 2 1 ϕ ( x ) − 2 a 1 ∫ x 0 x ψ ( ξ ) d ξ − 2 1 [ f 1 ( x 0 ) − f 2 ( x 0 ) ]
将此代回 (5.2) 可得，初始问题的解
u ( x , t ) = 1 2 [ ϕ ( x + a t ) + ϕ ( x − a t ) ] + 1 2 a ∫ x − a t x + a t ψ ( ξ ) d ξ (5.3) u(x,t)=\frac{1}{2}[\phi(x+at)+\phi(x-at)]+\frac{1}{2a}\int_{x-at}^{x+at}\psi(ξ)dξ\tag{5.3}
u ( x , t ) = 2 1 [ ϕ ( x + a t ) + ϕ ( x − a t ) ] + 2 a 1 ∫ x − a t x + a t ψ ( ξ ) d ξ ( 5 . 3 )
这个公式称为达朗贝尔(d’Alembert)公式 。
(3) 物理意义示例：设初速度为零 ψ ( x ) = 0 \psi(x)=0 ψ ( x ) = 0 ϕ ( x ) \phi(x) ϕ ( x ) t = 0 t=0 t = 0 u ( x , t ) = 1 2 ϕ ( x + a t ) + 1 2 ϕ ( x − a t ) u(x,t)=\frac{1}{2}\phi(x+at)+\frac{1}{2}\phi(x-at) u ( x , t ) = 2 1 ϕ ( x + a t ) + 2 1 ϕ ( x − a t ) a a a
半无限弦振动定解问题 ——延拓法
{ u t t − a 2 u x x = 0 ( x > 0 ) u ∣ t = 0 = ϕ ( x ) , ∂ u ∂ t ∣ t = 0 = ψ ( x ) ( x ⩾ 0 ) u ∣ x = 0 = 0 (5.4) \begin{cases}
u_{tt}-a^2u_{xx}=0 \quad (x>0)\\
u|_{t=0}=\phi(x),\cfrac{∂u}{∂t}|_{t=0}=\psi(x) \quad(x\geqslant 0)\\
u|_{x=0}=0
\end{cases}\tag{5.4}
⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ u t t − a 2 u x x = 0 ( x > 0 ) u ∣ t = 0 = ϕ ( x ) , ∂ t ∂ u ∣ t = 0 = ψ ( x ) ( x ⩾ 0 ) u ∣ x = 0 = 0 ( 5 . 4 )
求解上述问题的基本思路是把半无限长当做无限长弦的 x ⩾ 0 x\geqslant0 x ⩾ 0 x = 0 x=0 x = 0 g ( x ) g(x) g ( x ) ( − ∞ , + ∞ ) (-\infty,+\infty) ( − ∞ , + ∞ ) g ( 0 ) = 0 g(0)=0 g ( 0 ) = 0 ϕ , ψ \phi,\psi ϕ , ψ
Φ ( x ) = { ϕ ( x ) ( x ⩾ 0 ) − ϕ ( x ) ( x < 0 ) ; Ψ ( x ) = { ψ ( x ) ( x ⩾ 0 ) − ψ ( x ) ( x < 0 ) \Phi(x)=\begin{cases}\phi(x)&(x\geqslant0)\\ -\phi(x)&(x<0)\end{cases};\quad
\Psi(x)=\begin{cases}\psi(x)&(x\geqslant0)\\ -\psi(x)&(x<0)\end{cases}
Φ ( x ) = { ϕ ( x ) − ϕ ( x ) ( x ⩾ 0 ) ( x < 0 ) ; Ψ ( x ) = { ψ ( x ) − ψ ( x ) ( x ⩾ 0 ) ( x < 0 )
将上式带入达朗贝尔公式，可求得
u ( x , t ) = { 1 2 [ ϕ ( x + a t ) + ϕ ( x − a t ) ] + 1 2 a ∫ x − a t x + a t ψ ( ξ ) d ξ ( t ⩽ x a ) 1 2 [ ϕ ( x + a t ) − ϕ ( x − a t ) ] + 1 2 a ∫ a t − x x + a t ψ ( ξ ) d ξ ( t > x a ) (5.5) u(x,t)=\begin{cases}\displaystyle
\frac{1}{2}[\phi(x+at)+\phi(x-at)]+\frac{1}{2a}\int_{x-at}^{x+at}\psi(ξ)dξ &(t⩽\cfrac{x}{a})\\
\displaystyle
\frac{1}{2}[\phi(x+at)-\phi(x-at)]+\frac{1}{2a}\int_{at-x}^{x+at}\psi(ξ)dξ &(t>\cfrac{x}{a})
\end{cases}\tag{5.5}
u ( x , t ) = ⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 2 1 [ ϕ ( x + a t ) + ϕ ( x − a t ) ] + 2 a 1 ∫ x − a t x + a t ψ ( ξ ) d ξ 2 1 [ ϕ ( x + a t ) − ϕ ( x − a t ) ] + 2 a 1 ∫ a t − x x + a t ψ ( ξ ) d ξ ( t ⩽ a x ) ( t > a x ) ( 5 . 5 )
物理意义示例：设初速度为零 ψ ( x ) = 0 \psi(x)=0 ψ ( x ) = 0 ϕ ( x ) \phi(x) ϕ ( x ) 半波损失 。
(2) 端点自由弦的自由振动：定解问题为
{ u t t − a 2 u x x = 0 u ∣ t = 0 = ϕ ( x ) , ∂ u ∂ t ∣ t = 0 = ψ ( x ) ∂ u ∂ x ∣ x = 0 = 0 ( 0 ⩽ x < ∞ ) (5.6) \begin{cases}
u_{tt}-a^2u_{xx}=0 \\
u|_{t=0}=\phi(x),\cfrac{∂u}{∂t}|_{t=0}=\psi(x) \\
\cfrac{∂u}{∂x}|_{x=0}=0
\end{cases}\quad(0\leqslant x<\infty)\tag{5.6}
⎩ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎧ u t t − a 2 u x x = 0 u ∣ t = 0 = ϕ ( x ) , ∂ t ∂ u ∣ t = 0 = ψ ( x ) ∂ x ∂ u ∣ x = 0 = 0 ( 0 ⩽ x < ∞ ) ( 5 . 6 )
同样对函数进行偶延拓
Φ ( x ) = { ϕ ( x ) ( x ⩾ 0 ) ϕ ( − x ) ( x < 0 ) ; Ψ ( x ) = { ψ ( x ) ( x ⩾ 0 ) ψ ( − x ) ( x < 0 ) \Phi(x)=\begin{cases}\phi(x)&(x\geqslant0)\\ \phi(-x)&(x<0)\end{cases};\quad
\Psi(x)=\begin{cases}\psi(x)&(x\geqslant0)\\ \psi(-x)&(x<0)\end{cases}
Φ ( x ) = { ϕ ( x ) ϕ ( − x ) ( x ⩾ 0 ) ( x < 0 ) ; Ψ ( x ) = { ψ ( x ) ψ ( − x ) ( x ⩾ 0 ) ( x < 0 )
将上式带入达朗贝尔公式，可求得
u ( x , t ) = { 1 2 [ ϕ ( x + a t ) + ϕ ( x − a t ) ] + 1 2 a ∫ x − a t x + a t ψ ( ξ ) d ξ ( t ⩽ x a ) 1 2 [ ϕ ( x + a t ) + ϕ ( a t − x ) ] + 1 2 a [ ∫ 0 x + a t ψ ( ξ ) d ξ + ∫ 0 a t − x ψ ( ξ ) d ξ ] ( t > x a ) (5.7) u(x,t)=\begin{cases}\displaystyle
\frac{1}{2}[\phi(x+at)+\phi(x-at)]+\frac{1}{2a}\int_{x-at}^{x+at}\psi(ξ)dξ &(t⩽\cfrac{x}{a})\\
\displaystyle
\frac{1}{2}[\phi(x+at)+\phi(at-x)]+\frac{1}{2a}[\int_{0}^{x+at}\psi(ξ)dξ+\int_{0}^{at-x}\psi(ξ)dξ] &(t>\cfrac{x}{a})
\end{cases}\tag{5.7}
u ( x , t ) = ⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 2 1 [ ϕ ( x + a t ) + ϕ ( x − a t ) ] + 2 a 1 ∫ x − a t x + a t ψ ( ξ ) d ξ 2 1 [ ϕ ( x + a t ) + ϕ ( a t − x ) ] + 2 a 1 [ ∫ 0 x + a t ψ ( ξ ) d ξ + ∫ 0 a t − x ψ ( ξ ) d ξ ] ( t ⩽ a x ) ( t > a x ) ( 5 . 7 )
自由端点的影响也是一种反射波，不同的是反射波的相位和入射波相同，没有半波损失。
三维波动方程的球面波 ：三维波动方程在求坐标系 ( r , θ , ϕ ) (r,θ,ϕ) ( r , θ , ϕ )
1 a 2 ∂ 2 u ∂ t 2 = 1 r 2 ∂ ∂ r ( r 2 ∂ u ∂ r ) + 1 r 2 sin θ ∂ ∂ θ ( sin θ ∂ u ∂ θ ) + 1 r 2 sin 2 θ ∂ 2 u ∂ ϕ 2 (5.8) \cfrac{1}{a^2}\cfrac{∂^2u}{∂t^2}=\cfrac{1}{r^2}\cfrac{∂}{∂r}\left(r^2\cfrac{∂u}{∂r}\right)
+\cfrac{1}{r^2\sinθ}\cfrac{∂}{∂θ}\left(\sinθ\cfrac{∂u}{∂θ}\right)
+\cfrac{1}{r^2\sin^2θ}\cfrac{∂^2u}{∂ϕ^2}\tag{5.8}
a 2 1 ∂ t 2 ∂ 2 u = r 2 1 ∂ r ∂ ( r 2 ∂ r ∂ u ) + r 2 sin θ 1 ∂ θ ∂ ( sin θ ∂ θ ∂ u ) + r 2 sin 2 θ 1 ∂ ϕ 2 ∂ 2 u ( 5 . 8 )
若函数 u u u u u u θ , ϕ θ,ϕ θ , ϕ u = u ( r , t ) u=u(r,t) u = u ( r , t )
∂ 2 u ∂ t 2 = a 2 ( ∂ 2 u ∂ r 2 + 2 r ∂ u ∂ r ) \cfrac{∂^2u}{∂t^2}=a^2(\cfrac{∂^2u}{∂r^2}+\cfrac{2}{r}\cfrac{∂u}{∂r})
∂ t 2 ∂ 2 u = a 2 ( ∂ r 2 ∂ 2 u + r 2 ∂ r ∂ u )
进一步重写为
∂ 2 ( r u ) ∂ t 2 = a 2 ∂ 2 ( r u ) ∂ r 2 \cfrac{∂^2(ru)}{∂t^2}=a^2\cfrac{∂^2(ru)}{∂r^2}
∂ t 2 ∂ 2 ( r u ) = a 2 ∂ r 2 ∂ 2 ( r u )
这时关于 r u ru r u
r u ( r , t ) = f 1 ( r + a t ) + f 2 ( r − a t ) ru(r,t)=f_1(r+at)+f_2(r-at)
r u ( r , t ) = f 1 ( r + a t ) + f 2 ( r − a t )
从而
u = f 1 ( r + a t ) + f 2 ( r − a t ) r (5.9) u=\cfrac{f_1(r+at)+f_2(r-at)}{r}\tag{5.9}
u = r f 1 ( r + a t ) + f 2 ( r − a t ) ( 5 . 9 )
其中 f 1 , f 2 f_1,f_2 f 1 , f 2 r r r
两端固定弦的自由振动 ：（第一类边界条件）设长度为 l l l
{ u t t − a 2 u x x = 0 x ∈ ( 0 , l ) u ∣ t = 0 = ϕ ( x ) , ∂ u ∂ t ∣ t = 0 = ψ ( x ) x ∈ [ 0 , l ] u ∣ x = 0 = 0 , u ∣ x = l = 0 (1.1) \begin{cases}
u_{tt}-a^2u_{xx}=0 & x\in(0,l)\\
u|_{t=0}=\phi(x),\cfrac{∂u}{∂t}|_{t=0}=\psi(x) & x\in[0,l]\\
u|_{x=0}=0,u|_{x=l}=0
\end{cases}\tag{1.1}
⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ u t t − a 2 u x x = 0 u ∣ t = 0 = ϕ ( x ) , ∂ t ∂ u ∣ t = 0 = ψ ( x ) u ∣ x = 0 = 0 , u ∣ x = l = 0 x ∈ ( 0 , l ) x ∈ [ 0 , l ] ( 1 . 1 )
(1) 分离变量 ：上节半无限弦的振动问题启发我们是否可以用驻波研究两端固定弦的振动问题，驻波的表达式为
u ( x , t ) = X ( x ) T ( t ) (1.2) u(x,t)=X(x)T(t)\tag{1.2}
u ( x , t ) = X ( x ) T ( t ) ( 1 . 2 )
带入弦振动方程可得
X T ′ ′ − a 2 X ′ ′ T = 0 XT''-a^2X''T=0
X T ′ ′ − a 2 X ′ ′ T = 0
可化为
T ′ ′ a 2 T = X ′ ′ X \cfrac{T''}{a^2T}=\cfrac{X''}{X}
a 2 T T ′ ′ = X X ′ ′
上式中左端只是 t t t x x x − λ -\lambda − λ
T ′ ′ ( t ) + λ a 2 T = 0 X ′ ′ ( x ) + λ X = 0 T''(t)+\lambda a^2T=0 \\
X''(x)+\lambda X=0
T ′ ′ ( t ) + λ a 2 T = 0 X ′ ′ ( x ) + λ X = 0
同样，表达式 (1.2) 带入边界条件可得
{ X ( 0 ) T ( t ) = 0 X ( l ) T ( t ) = 0 \begin{cases}
X(0)T(t)=0 \\
X(l)T(t)=0
\end{cases}
{ X ( 0 ) T ( t ) = 0 X ( l ) T ( t ) = 0
这时必须有 X ( 0 ) = X ( l ) = 0 X(0)=X(l)=0 X ( 0 ) = X ( l ) = 0 T ( t ) = 0 T(t)=0 T ( t ) = 0 u ( x , t ) u(x,t) u ( x , t ) (2) 本征值问题 ：先求解关于 X X X
{ X ′ ′ ( x ) + λ X ( x ) = 0 X ( 0 ) = 0 , X ( l ) = 0 (1.3) \begin{cases}
X''(x)+\lambda X(x)=0 \\
X(0)=0,X(l)=0
\end{cases}\tag{1.3}
{ X ′ ′ ( x ) + λ X ( x ) = 0 X ( 0 ) = 0 , X ( l ) = 0 ( 1 . 3 )
其中 λ \lambda λ 本征值问题 ，使问题有非零解的 λ \lambda λ 本征值 ，对应的函数称为本征函数 [^1]。分三种情形讨论
若 λ < 0 \lambda<0 λ < 0
X = c 1 e − λ x + c 2 e − − λ x X=c_1e^{\sqrt{-λ}x}+c_2e^{-\sqrt{-λ}x}
X = c 1 e − λ x + c 2 e − − λ x
积分常数 c 1 , c 2 c_1,c_2 c 1 , c 2 c 1 = c 2 = 0 c_1=c_2=0 c 1 = c 2 = 0 X ( x ) ≡ 0 X(x)\equiv0 X ( x ) ≡ 0 u ( x , t ) = X ( x ) T ( t ) ≡ 0 u(x,t)=X(x)T(t)\equiv0 u ( x , t ) = X ( x ) T ( t ) ≡ 0
若 λ = 0 \lambda=0 λ = 0
X = c 1 x + c 2 X=c_1x+c_2
X = c 1 x + c 2
积分常数 c 1 , c 2 c_1,c_2 c 1 , c 2 c 1 = c 2 = 0 c_1=c_2=0 c 1 = c 2 = 0 X ( x ) ≡ 0 X(x)\equiv0 X ( x ) ≡ 0
若 λ > 0 \lambda>0 λ > 0
X = c 1 cos λ x + c 2 sin λ x X=c_1\cos\sqrt{λ}x+c_2\sin\sqrt{λ}x
X = c 1 cos λ x + c 2 sin λ x
积分常数 c 1 , c 2 c_1,c_2 c 1 , c 2 c 1 = 0 , c 2 sin λ l = 0 c_1=0,c_2\sin\sqrt{λ}l=0 c 1 = 0 , c 2 sin λ l = 0 X X X sin λ l = 0 \sin\sqrt{λ}l=0 sin λ l = 0
λ n = ( n π l ) 2 , n = 1 , 2 , ⋯ (1.4) λ_n=(\cfrac{n\pi}{l})^2,\quad n=1,2,\cdots\tag{1.4}
λ n = ( l n π ) 2 , n = 1 , 2 , ⋯ ( 1 . 4 )
对应的本征函数
X n = C n sin n π x l (1.5) X_n=C_n\sin\cfrac{n\pi x}{l}\tag{1.5}
X n = C n sin l n π x ( 1 . 5 )
其中 C n C_n C n T ( t ) T(t) T ( t )
T ′ ′ ( t ) + λ n a 2 T = 0 T''(t)+\lambda_n a^2T=0
T ′ ′ ( t ) + λ n a 2 T = 0
其通解为
T n = A n cos n π a t l + B n sin n π a t l (1.6) T_n=A_n\cos\cfrac{n\pi at}{l}+B_n\sin\cfrac{n\pi at}{l}\tag{1.6}
T n = A n cos l n π a t + B n sin l n π a t ( 1 . 6 )
从而得到一组分离变量形式的特解
u n ( x , t ) = X n ( x ) T n ( t ) = ( a n cos n π a t l + b n sin n π a t l ) sin n π x l = A sin ( k n x ) sin ( ω n t + θ n ) (1.7) \begin{aligned}
u_n(x,t) & =X_n(x)T_n(t) \\
& =(a_n\cos\cfrac{n\pi at}{l}+b_n\sin\cfrac{n\pi at}{l})\sin\cfrac{n\pi x}{l} \\
&= A\sin(k_nx)\sin(ω_nt+θ_n)
\end{aligned}\tag{1.7}
u n ( x , t ) = X n ( x ) T n ( t ) = ( a n cos l n π a t + b n sin l n π a t ) sin l n π x = A sin ( k n x ) sin ( ω n t + θ n ) ( 1 . 7 )
其中 a n = C n A n , b n = C n B n a_n=C_nA_n,b_n=C_nB_n a n = C n A n , b n = C n B n n n n 本征振动 。
物理意义：本征振动中 A sin k n x A\sin k_nx A sin k n x A = a n 2 + b n 2 , k n = n π l A=\sqrt{a_n^2+b_n^2},k_n=\cfrac{n\pi}{l} A = a n 2 + b n 2 , k n = l n π sin ( ω n t + θ n ) \sin(ω_nt+θ_n) sin ( ω n t + θ n ) ω n = n π a l , θ n = arcsin a n A ω_n=\cfrac{n\pi a}{l},θ_n=\arcsin\cfrac{a_n}{A} ω n = l n π a , θ n = arcsin A a n ω n ω_n ω n k n k_n k n θ n θ_n θ n x = k l / n ( k = 0 , 1 , 2 , ⋯ , n ) x=kl/n\quad(k=0,1,2,\cdots,n) x = k l / n ( k = 0 , 1 , 2 , ⋯ , n ) n + 1 n+1 n + 1 sin ( n π x / l ) = sin k π = 0 \sin(n\pi x/l)=\sin k\pi=0 sin ( n π x / l ) = sin k π = 0 u n ( x , t ) = 0 u_n(x,t)=0 u n ( x , t ) = 0
(3) 特解的叠加 ：以上本征振动是满足弦振动方程和边界条件线性无关的特解，他们不一定满足初始条件。为此，根据叠加原理，对特解线性叠加
u ( x , t ) = ∑ n = 1 ∞ u n ( x , t ) = ∑ n = 1 ∞ ( a n cos n π a t l + b n sin n π a t l ) sin n π x l (1.8) \displaystyle u(x,t)=\sum_{n=1}^{\infty}u_n(x,t)=\sum_{n=1}^{\infty}
(a_n\cos\cfrac{n\pi at}{l}+b_n\sin\cfrac{n\pi at}{l})\sin\cfrac{n\pi x}{l}\tag{1.8}
u ( x , t ) = n = 1 ∑ ∞ u n ( x , t ) = n = 1 ∑ ∞ ( a n cos l n π a t + b n sin l n π a t ) sin l n π x ( 1 . 8 )
得到仍然满足弦振动方程和边界条件的一般解 。a n , b n a_n,b_n a n , b n u ( x , t ) u(x,t) u ( x , t )
{ u ∣ t = 0 = ∑ n = 1 ∞ a n cos n π x l = ϕ ( x ) ∂ u ∂ t ∣ t = 0 = ∑ n = 1 ∞ b n n π a l sin n π x l = ψ ( x ) ( 0 < x < l ) \begin{cases}\displaystyle
u|_{t=0}=\sum_{n=1}^{\infty}a_n\cos\cfrac{n\pi x}{l}=\phi(x) \\
\displaystyle
\cfrac{∂u}{∂t}|_{t=0}=\sum_{n=1}^{\infty}b_n\cfrac{n\pi a}{l}\sin\cfrac{n\pi x}{l}=\psi(x)
\end{cases}\quad(0<x<l)
⎩ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎧ u ∣ t = 0 = n = 1 ∑ ∞ a n cos l n π x = ϕ ( x ) ∂ t ∂ u ∣ t = 0 = n = 1 ∑ ∞ b n l n π a sin l n π x = ψ ( x ) ( 0 < x < l )
这两式正好是 ϕ ( x ) , ψ ( x ) \phi(x),\psi(x) ϕ ( x ) , ψ ( x ) [ 0 , l ] [0,l] [ 0 , l ]
{ a n = 2 l ∫ 0 l ϕ ( ξ ) sin n π ξ l d ξ b n = 2 n π a ∫ 0 l ψ ( ξ ) sin n π ξ l d ξ (1.9) \begin{cases}\displaystyle
a_n=\cfrac{2}{l}\int_0^l\phi(\xi)\sin\cfrac{n\pi\xi}{l}d\xi \\
\displaystyle
b_n=\cfrac{2}{n\pi a}\int_0^l\psi(\xi)\sin\cfrac{n\pi\xi}{l}d\xi
\end{cases}\tag{1.9}
⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ a n = l 2 ∫ 0 l ϕ ( ξ ) sin l n π ξ d ξ b n = n π a 2 ∫ 0 l ψ ( ξ ) sin l n π ξ d ξ ( 1 . 9 )
两端自由弦的振动 ：（第二类边界条件）设长度为 l l l
{ u t t − a 2 u x x = 0 x ∈ ( 0 , l ) u ∣ t = 0 = ϕ ( x ) , ∂ u ∂ t ∣ t = 0 = ψ ( x ) x ∈ [ 0 , l ] ∂ u ∂ x ∣ x = 0 = 0 , ∂ u ∂ x ∣ x = l = 0 (1.10) \begin{cases}
u_{tt}-a^2u_{xx}=0 & x\in(0,l)\\
u|_{t=0}=\phi(x),\cfrac{∂u}{∂t}|_{t=0}=\psi(x) & x\in[0,l]\\
\cfrac{∂u}{∂x}|_{x=0}=0,\cfrac{∂u}{∂x}|_{x=l}=0
\end{cases}\tag{1.10}
⎩ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎧ u t t − a 2 u x x = 0 u ∣ t = 0 = ϕ ( x ) , ∂ t ∂ u ∣ t = 0 = ψ ( x ) ∂ x ∂ u ∣ x = 0 = 0 , ∂ x ∂ u ∣ x = l = 0 x ∈ ( 0 , l ) x ∈ [ 0 , l ] ( 1 . 1 0 )
同样用分离变量法我们可以求得
u n ( x , t ) = { a 0 + b 0 t ( n = 0 ) ( a n cos n π a t l + b n sin n π a t l ) sin n π x l ( n = 1 , 2 , ⋯ ) (1.11) u_n(x,t)=\begin{cases}
a_0+b_0t \quad (n=0)\\
(a_n\cos\cfrac{n\pi at}{l}+b_n\sin\cfrac{n\pi at} {l})\sin\cfrac{n\pi x}{l} \quad(n=1,2,\cdots)
\end{cases}\tag{1.11}
u n ( x , t ) = ⎩ ⎨ ⎧ a 0 + b 0 t ( n = 0 ) ( a n cos l n π a t + b n sin l n π a t ) sin l n π x ( n = 1 , 2 , ⋯ ) ( 1 . 1 1 )
他们的线性叠加为
u ( x , t ) = a 0 + b 0 t + ∑ n = 1 ∞ ( a n cos n π a t l + b n sin n π a t l ) sin n π x l (1.12) \displaystyle u(x,t)=a_0+b_0t+\sum_{n=1}^{\infty}
(a_n\cos\cfrac{n\pi at}{l}+b_n\sin\cfrac{n\pi at}{l})\sin\cfrac{n\pi x}{l}\tag{1.12}
u ( x , t ) = a 0 + b 0 t + n = 1 ∑ ∞ ( a n cos l n π a t + b n sin l n π a t ) sin l n π x ( 1 . 1 2 )
由初始条件可确定系数为
{ a 0 = 1 l ∫ 0 l ϕ ( ξ ) b 0 = 1 l ∫ 0 l ψ ( ξ ) , { a n = 2 l ∫ 0 l ϕ ( ξ ) sin n π ξ l d ξ b n = 2 n π a ∫ 0 l ψ ( ξ ) sin n π ξ l d ξ (1.13) \begin{cases}\displaystyle
a_0=\cfrac{1}{l}\int_0^l\phi(\xi) \\
\displaystyle
b_0=\cfrac{1}{l}\int_0^l\psi(\xi)
\end{cases}
,\quad
\begin{cases}\displaystyle
a_n=\cfrac{2}{l}\int_0^l\phi(\xi)\sin\cfrac{n\pi\xi}{l}d\xi \\
\displaystyle
b_n=\cfrac{2}{n\pi a}\int_0^l\psi(\xi)\sin\cfrac{n\pi\xi}{l}d\xi
\end{cases}\tag{1.13}
⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ a 0 = l 1 ∫ 0 l ϕ ( ξ ) b 0 = l 1 ∫ 0 l ψ ( ξ ) , ⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ a n = l 2 ∫ 0 l ϕ ( ξ ) sin l n π ξ d ξ b n = n π a 2 ∫ 0 l ψ ( ξ ) sin l n π ξ d ξ ( 1 . 1 3 )
均匀杆的热传导问题 ：（第三类边界条件）设长度为 l l l x = 0 x=0 x = 0 x = l x=l x = l
{ u t − a 2 u x x = 0 x ∈ ( 0 , l ) u ∣ t = 0 = ϕ ( x ) x ∈ [ 0 , l ] u ∣ x = 0 = 0 , u x ( l , t ) + h u ( l , t ) = 0 (1.14) \begin{cases}
u_{t}-a^2u_{xx}=0 &x\in(0,l)\\
u|_{t=0}=\phi(x) & x\in[0,l]\\
u|_{x=0}=0,u_x(l,t)+hu(l,t)=0
\end{cases}\tag{1.14}
⎩ ⎪ ⎨ ⎪ ⎧ u t − a 2 u x x = 0 u ∣ t = 0 = ϕ ( x ) u ∣ x = 0 = 0 , u x ( l , t ) + h u ( l , t ) = 0 x ∈ ( 0 , l ) x ∈ [ 0 , l ] ( 1 . 1 4 )
其中常数 h > 0 h>0 h > 0
λ cos λ l + h sin λ l = 0 \sqrt{λ}\cos\sqrt{λ} l+h\sin\sqrt{λ} l=0
λ cos λ l + h sin λ l = 0
对应的函数族为
X n = B n sin λ n x , T n = A n e − λ n a 2 t X_n=B_n\sin\sqrt{λ_n}x,\quad T_n=A_ne^{-λ_na^2t}
X n = B n sin λ n x , T n = A n e − λ n a 2 t
进一步求得
u n ( x , t ) = C n e − λ n a 2 t sin λ n x ( n = 1 , 2 , ⋯ ) (1.15) u_n(x,t)=C_ne^{-λ_na^2t}\sin\sqrt{λ_n}x\quad (n=1,2,\cdots)\tag{1.15}
u n ( x , t ) = C n e − λ n a 2 t sin λ n x ( n = 1 , 2 , ⋯ ) ( 1 . 1 5 )
线性叠加为
u ( x , t ) = ∑ n = 1 ∞ C n e − λ n a 2 t sin λ n x (1.16) \displaystyle u(x,t)=\sum_{n=1}^{\infty}C_ne^{-λ_na^2t}\sin\sqrt{λ_n}x\tag{1.16}
u ( x , t ) = n = 1 ∑ ∞ C n e − λ n a 2 t sin λ n x ( 1 . 1 6 )
由初始条件可确定系数为
c n = 1 L n ∫ 0 l ϕ ( ξ ) sin λ n ξ d ξ (1.17) c_n=\cfrac{1}{L_n}\int_{0}^{l}\phi(\xi)\sin\sqrt{λ_n}\xi d\xi\tag{1.17}
c n = L n 1 ∫ 0 l ϕ ( ξ ) sin λ n ξ d ξ ( 1 . 1 7 )
其中
L n = ∫ 0 l sin λ n ξ d ξ \displaystyle L_n=\int_{0}^{l}\sinλ_n\xi d\xi
L n = ∫ 0 l sin λ n ξ d ξ
矩形区域拉普拉斯方程的边界值问题 ：
{ u x x + u y y = 0 0 < x < a , 0 < y < b u ∣ y = 0 = f ( x ) , u ∣ y = b = g ( x ) 0 ⩽ y ⩽ b ∂ u ∂ x ∣ x = 0 = 0 , ∂ u ∂ x ∣ x = a = 0 0 ⩽ x ⩽ a (1.18) \begin{cases}
u_{xx}+u_{yy}=0 & 0<x<a,0<y<b\\
u|_{y=0}=f(x),u|_{y=b}=g(x) & 0⩽y⩽b\\
\cfrac{∂u}{∂x}|_{x=0}=0,\cfrac{∂u}{∂x}|_{x=a}=0 &0⩽x⩽a
\end{cases}\tag{1.18}
⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ u x x + u y y = 0 u ∣ y = 0 = f ( x ) , u ∣ y = b = g ( x ) ∂ x ∂ u ∣ x = 0 = 0 , ∂ x ∂ u ∣ x = a = 0 0 < x < a , 0 < y < b 0 ⩽ y ⩽ b 0 ⩽ x ⩽ a ( 1 . 1 8 )
设解 u = X ( x ) Y ( y ) u=X(x)Y(y) u = X ( x ) Y ( y )
X ′ ′ X = − Y ′ ′ Y = − λ \cfrac{X''}{X}=-\cfrac{Y''}{Y}=-λ
X X ′ ′ = − Y Y ′ ′ = − λ
由此得到
{ X ′ ′ + λ X = 0 Y ′ ′ − λ y = 0 \begin{cases}
X''+λX=0 \\
Y''-λy=0
\end{cases}
{ X ′ ′ + λ X = 0 Y ′ ′ − λ y = 0
同样带入关于 x x x
{ X ′ ( 0 ) Y ( y ) = 0 X ′ ( a ) Y ( y ) = 0 \begin{cases}X'(0)Y(y)=0 \\X'(a)Y(y)=0\end{cases}
{ X ′ ( 0 ) Y ( y ) = 0 X ′ ( a ) Y ( y ) = 0
这时必须有 X ′ ( 0 ) = X ′ ( a ) = 0 X'(0)=X'(a)=0 X ′ ( 0 ) = X ′ ( a ) = 0
{ X ′ ′ ( x ) + λ X ( x ) = 0 X ′ ( 0 ) = 0 , X ′ ( a ) = 0 \begin{cases}
X''(x)+\lambda X(x)=0 \\
X'(0)=0,X'(a)=0
\end{cases}
{ X ′ ′ ( x ) + λ X ( x ) = 0 X ′ ( 0 ) = 0 , X ′ ( a ) = 0
可解得特征值和对应的特征函数，进一步可求得 u n ( x , y ) u_n(x,y) u n ( x , y ) y y y
u ( x , y ) = 1 2 ( π − y ) + 2 π ∑ n = 1 ∞ [ ( − 1 ) n − 1 ] n 2 sinh n ( π − y ) sinh n π cos n x \displaystyle u(x,y)=\frac{1}{2}(\pi-y)+
\frac{2}{\pi}\sum_{n=1}^{\infty}\cfrac{[(-1)^n-1]}{n^2}
\cfrac{\sinh n(\pi-y)}{\sinh n\pi}\cos nx
u ( x , y ) = 2 1 ( π − y ) + π 2 n = 1 ∑ ∞ n 2 [ ( − 1 ) n − 1 ] sinh n π sinh n ( π − y ) cos n x
圆形区域内拉普拉斯方程边界值问题
{ Δ = u x x + u y y = 0 ( x 2 + y 2 < a 2 ) u ∣ x 2 + y 2 = a 2 = f ( x , y ) (4.1) \begin{cases}
Δ=u_{xx}+u_{yy}=0 & (x^2+y^2<a^2) \\
u|_{x^2+y^2=a^2}=f(x,y)
\end{cases}\tag{4.1}
{ Δ = u x x + u y y = 0 u ∣ x 2 + y 2 = a 2 = f ( x , y ) ( x 2 + y 2 < a 2 ) ( 4 . 1 )
由于区域为圆域，边界条件不能直接分离变量，考虑在平面极坐标系中，上述边界值问题化为
{ u r r + 1 r u r + 1 r 2 u ϕ ϕ ( r < a , 0 ⩽ ϕ ⩽ 2 π ) u ∣ r = a = f ( ϕ ) ( 0 ⩽ ϕ ⩽ 2 π ) (4.2) \begin{cases}
u_{rr}+\cfrac{1}{r}u_r+\cfrac{1}{r^2}u_{ϕϕ} & (r<a,0⩽ϕ⩽2\pi) \\
u|_{r=a}=f(ϕ) & (0⩽ϕ⩽2\pi)
\end{cases}\tag{4.2}
⎩ ⎨ ⎧ u r r + r 1 u r + r 2 1 u ϕ ϕ u ∣ r = a = f ( ϕ ) ( r < a , 0 ⩽ ϕ ⩽ 2 π ) ( 0 ⩽ ϕ ⩽ 2 π ) ( 4 . 2 )
又因为在极坐标系中，( r , ϕ ) (r,ϕ) ( r , ϕ ) ( r , ϕ + 2 π ) (r,ϕ+2\pi) ( r , ϕ + 2 π ) 自然周期性条件
u ( r , ϕ ) = u ( r , ϕ + 2 π ) (4.3) u(r,ϕ)=u(r,ϕ+2\pi)\tag{4.3}
u ( r , ϕ ) = u ( r , ϕ + 2 π ) ( 4 . 3 )
令 u ( r , ϕ ) = R ( r ) Φ ( ϕ ) u(r,ϕ)=R(r)Φ(ϕ) u ( r , ϕ ) = R ( r ) Φ ( ϕ )
r 2 R ′ ′ + r R ′ R = − Φ ′ ′ Φ = λ \cfrac{r^2R''+rR'}{R}=-\cfrac{Φ''}{Φ}=λ
R r 2 R ′ ′ + r R ′ = − Φ Φ ′ ′ = λ
于是有
Φ ′ ′ + λ Φ = 0 (4.4) Φ''+λΦ=0 \tag{4.4}
Φ ′ ′ + λ Φ = 0 ( 4 . 4 )
r 2 R ′ ′ + r R ′ − λ R = 0 (4.5) r^2R''+rR'-λR=0 \tag{4.5}
r 2 R ′ ′ + r R ′ − λ R = 0 ( 4 . 5 )
将 u ( r , ϕ ) = R ( r ) Φ ( ϕ ) u(r,ϕ)=R(r)Φ(ϕ) u ( r , ϕ ) = R ( r ) Φ ( ϕ )
Φ ( ϕ ) = Φ ( ϕ + 2 π ) , Φ ′ ( ϕ ) = Φ ′ ( ϕ + 2 π ) (4.6) Φ(ϕ)=Φ(ϕ+2\pi),\quad Φ'(ϕ)=Φ'(ϕ+2\pi)\tag{4.6}
Φ ( ϕ ) = Φ ( ϕ + 2 π ) , Φ ′ ( ϕ ) = Φ ′ ( ϕ + 2 π ) ( 4 . 6 )
常微分方程 (4.4) 和周期条件 (4.6) 构成本征值问题，容易求得方程 (4.4) 的通解为
Φ = { A e − λ ϕ + B e − − λ ϕ ( λ < 0 ) A + B ϕ ( λ = 0 ) A cos λ ϕ + B sin λ ϕ ( λ > 0 ) Φ=\begin{cases}
Ae^{\sqrt{-λ}ϕ}+Be^{-\sqrt{-λ}ϕ} & (λ<0)\\
A+Bϕ & (λ=0)\\
A\cos\sqrt{λ}ϕ+B\sin\sqrt{λ}ϕ & (λ>0)
\end{cases}
Φ = ⎩ ⎪ ⎨ ⎪ ⎧ A e − λ ϕ + B e − − λ ϕ A + B ϕ A cos λ ϕ + B sin λ ϕ ( λ < 0 ) ( λ = 0 ) ( λ > 0 )
将上式带入周期条件 (4.6) 从而求得本征值（当 λ = 0 λ=0 λ = 0 Φ ≡ 0 Φ\equiv0 Φ ≡ 0
λ = m 2 ( m = 0 , 1 , 2 , ⋯ ) λ=m^2 \quad (m=0,1,2,\cdots)
λ = m 2 ( m = 0 , 1 , 2 , ⋯ )
对应的本征函数为
Φ = { A 0 ( m = 0 ) A m cos m ϕ + B m sin m ϕ ( m ≠ 0 ) Φ=\begin{cases}
A_0 & (m=0)\\
A_m\cos mϕ+B_m\sin mϕ & (m\neq0)
\end{cases}
Φ = { A 0 A m cos m ϕ + B m sin m ϕ ( m = 0 ) ( m = 0 )
然后将本征值带入 (4.5) 得到
r 2 R ′ ′ + r R ′ − m 2 R = 0 r^2R''+rR'-m^2R=0
r 2 R ′ ′ + r R ′ − m 2 R = 0
此方程为欧拉方程，其通解为
R = { C 0 + D 0 ln r ( m = 0 ) C m r m + D m 1 r m ( m ≠ 0 ) R=\begin{cases}
C_0+D_0\ln r & (m=0)\\
C_mr^m+D_m\cfrac{1}{r^m} & (m\neq0)
\end{cases}
R = ⎩ ⎨ ⎧ C 0 + D 0 ln r C m r m + D m r m 1 ( m = 0 ) ( m = 0 )
同时考虑在平面原点处 u ( x , y ) u(x,y) u ( x , y ) u ( 0 , ϕ ) < ∞ u(0,ϕ)<\infty u ( 0 , ϕ ) < ∞ D m = 0 , ( m = 0 , 1 , 2 ⋯ ) D_m=0,\quad(m=0,1,2\cdots) D m = 0 , ( m = 0 , 1 , 2 ⋯ )
u ( r , ϕ ) = a 0 2 + ∑ m = 1 ∞ r m ( a m cos m ϕ + b m sin m ϕ ) \displaystyle u(r,ϕ)=\cfrac{a_0}{2}+\sum_{m=1}^\infty r^m(a_m\cos mϕ+b_m\sin mϕ)
u ( r , ϕ ) = 2 a 0 + m = 1 ∑ ∞ r m ( a m cos m ϕ + b m sin m ϕ )
其中 a m , b m ( m = 0 , 1 , 2 , ⋯ ) a_m,b_m\quad (m=0,1,2,\cdots) a m , b m ( m = 0 , 1 , 2 , ⋯ )
{ a m = 1 a m π ∫ 0 2 π f ( t ) cos m t d t ( m = 0 , 1 , 2 , ⋯ ) b m = 1 a m π ∫ 0 2 π f ( t ) sin m t d t ( m = 1 , 2 , ⋯ ) \begin{cases}\displaystyle
a_m=\cfrac{1}{a^m\pi}\int_0^{2\pi}f(t)\cos mtdt & (m=0,1,2,\cdots) \\
\displaystyle
b_m=\cfrac{1}{a^m\pi}\int_0^{2\pi}f(t)\sin mtdt & (m=1,2,\cdots)
\end{cases}
⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ a m = a m π 1 ∫ 0 2 π f ( t ) cos m t d t b m = a m π 1 ∫ 0 2 π f ( t ) sin m t d t ( m = 0 , 1 , 2 , ⋯ ) ( m = 1 , 2 , ⋯ )
将系数带入上述一般解
u ( r , ϕ ) = 1 2 π ∫ 0 2 π f ( t ) [ 1 + 2 ∑ m = 1 ∞ ( r a ) m cos m ( ϕ − t ) ] d t \displaystyle u(r,ϕ)=\cfrac{1}{2\pi}\int_0^{2\pi}f(t)
[1+2\sum_{m=1}^\infty(\cfrac{r}{a})^m\cos m(ϕ-t)]dt
u ( r , ϕ ) = 2 π 1 ∫ 0 2 π f ( t ) [ 1 + 2 m = 1 ∑ ∞ ( a r ) m cos m ( ϕ − t ) ] d t
显然当 r < a r<a r < a
u ( r , ϕ ) = a 2 − r 2 2 π ∫ 0 2 π f ( t ) r 2 + a 2 − 2 a r cos ( ϕ − t ) d t (4.7) \displaystyle u(r,ϕ)=\cfrac{a^2-r^2}{2\pi}\int_0^{2\pi}
\cfrac{f(t)}{r^2+a^2-2ar\cos(ϕ-t)}dt\tag{4.7}
u ( r , ϕ ) = 2 π a 2 − r 2 ∫ 0 2 π r 2 + a 2 − 2 a r cos ( ϕ − t ) f ( t ) d t ( 4 . 7 )
这个结果称为圆域内泊松积分公式 。
上节只考虑了方程和边界条件都是齐次的情况，实际上还有方程和边界条件二者中至少有一个是非齐次的情况，称为非齐次问题。本节介绍齐次边界条件下非齐次方程的定解问题。
两端固定弦的强迫振动 ：定解问题为
{ u t t − a 2 u x x = f ( x , t ) x ∈ ( 0 , l ) u ∣ t = 0 = ϕ ( x ) , ∂ u ∂ t ∣ t = 0 = ψ ( x ) x ∈ [ 0 , l ] u ∣ x = 0 = 0 , u ∣ x = l = 0 (2.1) \begin{cases}
u_{tt}-a^2u_{xx}=f(x,t) & x\in(0,l)\\
u|_{t=0}=\phi(x),\cfrac{∂u}{∂t}|_{t=0}=\psi(x) & x\in[0,l]\\
u|_{x=0}=0,u|_{x=l}=0
\end{cases}\tag{2.1}
⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ u t t − a 2 u x x = f ( x , t ) u ∣ t = 0 = ϕ ( x ) , ∂ t ∂ u ∣ t = 0 = ψ ( x ) u ∣ x = 0 = 0 , u ∣ x = l = 0 x ∈ ( 0 , l ) x ∈ [ 0 , l ] ( 2 . 1 )
分离变量法给出的结果提示：不妨把所求的解本身展开为傅里叶级数，即
u ( x , t ) = ∑ n = 1 ∞ T n ( t ) X n ( x ) (2.2) \displaystyle u(x,t)=\sum_{n=1}^{\infty}T_n(t)X_n(x)\tag{2.2}
u ( x , t ) = n = 1 ∑ ∞ T n ( t ) X n ( x ) ( 2 . 2 )
把 { X n ( x ) } \{X_n(x)\} { X n ( x ) } T n ( t ) T_n(t) T n ( t ) u ( x , t ) u(x,t) u ( x , t ) f ( x , t ) f(x,t) f ( x , t ) { X n ( x ) } \{X_n(x)\} { X n ( x ) }
f ( x , t ) = ∑ n = 1 ∞ g n ( t ) X n ( x ) (2.3) \displaystyle f(x,t)=\sum_{n=1}^{\infty}g_n(t)X_n(x)\tag{2.3}
f ( x , t ) = n = 1 ∑ ∞ g n ( t ) X n ( x ) ( 2 . 3 )
原则上基本函数族的选取并无限制，但选择定解问题 (2.1) 对应的齐次定解问题的本征函数，会使后面求解待定系数变得简单可行。对应的齐次方程和齐次边界条件的本征值问题为
{ X ′ ′ ( x ) + λ X ( x ) = 0 X ( 0 ) = 0 , X ( l ) = 0 (1.3) \begin{cases}
X''(x)+\lambda X(x)=0 \\
X(0)=0,X(l)=0
\end{cases}\tag{1.3}
{ X ′ ′ ( x ) + λ X ( x ) = 0 X ( 0 ) = 0 , X ( l ) = 0 ( 1 . 3 )
根据上节求得本征函数，我们取 X n = sin n π x l X_n=\sin\cfrac{n\pi x}{l} X n = sin l n π x
g n ( t ) = 2 l ∫ 0 l f ( ξ , t ) sin n π ξ l d ξ g_n(t)=\cfrac{2}{l}\int_0^lf(ξ,t)\sin\cfrac{n\piξ}{l}dξ
g n ( t ) = l 2 ∫ 0 l f ( ξ , t ) sin l n π ξ d ξ
将u ( x , t ) , f ( x , t ) u(x,t),f(x,t) u ( x , t ) , f ( x , t )
∑ n = 1 ∞ T n ′ ′ ( t ) X n ( x ) − a 2 ∑ n = 1 ∞ T n ( t ) X n ′ ′ ( x ) = ∑ n = 1 ∞ g n ( t ) X n ( x ) \displaystyle
\sum_{n=1}^{\infty}T''_n(t)X_n(x)-a^2\sum_{n=1}^{\infty}T_n(t)X''_n(x)
=\sum_{n=1}^{\infty}g_n(t)X_n(x)
n = 1 ∑ ∞ T n ′ ′ ( t ) X n ( x ) − a 2 n = 1 ∑ ∞ T n ( t ) X n ′ ′ ( x ) = n = 1 ∑ ∞ g n ( t ) X n ( x )
利用 X n X_n X n
∑ n = 1 ∞ T n ′ ′ ( t ) X n ( x ) + a 2 ∑ n = 1 ∞ T n ( t ) λ n X n ( x ) = ∑ n = 1 ∞ g n ( t ) X n ( x ) \displaystyle
\sum_{n=1}^{\infty}T''_n(t)X_n(x)+a^2\sum_{n=1}^{\infty}T_n(t)\lambda_n X_n(x)
=\sum_{n=1}^{\infty}g_n(t)X_n(x)
n = 1 ∑ ∞ T n ′ ′ ( t ) X n ( x ) + a 2 n = 1 ∑ ∞ T n ( t ) λ n X n ( x ) = n = 1 ∑ ∞ g n ( t ) X n ( x )
根据傅里叶展开系数的唯一性，我们可以得到T n ( t ) T_n(t) T n ( t )
T n ′ ′ ( t ) + λ n a 2 T n ( t ) = g n ( t ) (2.4) T''_n(t)+\lambda_na^2T_n(t)=g_n(t)\tag{2.4}
T n ′ ′ ( t ) + λ n a 2 T n ( t ) = g n ( t ) ( 2 . 4 )
同样将 u ( x , t ) u(x,t) u ( x , t )
∑ n = 1 ∞ T n ( 0 ) X n ( x ) = ϕ ( x ) , ∑ n = 1 ∞ T n ′ ( 0 ) X n ( x ) = ψ ( x ) \displaystyle\sum_{n=1}^{\infty}T_n(0)X_n(x)=\phi(x),
\quad\sum_{n=1}^{\infty}T'_n(0)X_n(x)=\psi(x)
n = 1 ∑ ∞ T n ( 0 ) X n ( x ) = ϕ ( x ) , n = 1 ∑ ∞ T n ′ ( 0 ) X n ( x ) = ψ ( x )
根据傅里叶展开系数公式，可以导出
{ T n ( 0 ) = 2 l ∫ 0 l ϕ ( ξ ) sin n π ξ l d ξ T n ′ ( 0 ) = 2 l ∫ 0 l ψ ( ξ ) sin n π ξ l d ξ (2.5) \begin{cases}\displaystyle
T_n(0)=\cfrac{2}{l}\int_0^l\phi(ξ)\sin\cfrac{n\piξ}{l}dξ \\
\displaystyle
T'_n(0)=\cfrac{2}{l}\int_0^l\psi(ξ)\sin\cfrac{n\piξ}{l}dξ
\end{cases}\tag{2.5}
⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ T n ( 0 ) = l 2 ∫ 0 l ϕ ( ξ ) sin l n π ξ d ξ T n ′ ( 0 ) = l 2 ∫ 0 l ψ ( ξ ) sin l n π ξ d ξ ( 2 . 5 )
可用常数变易法求得 T n ( t ) T_n(t) T n ( t )
T n = l n π a ∫ 0 t g n ( τ ) sin 2 π a l ( t − τ ) d τ + T n ( 0 ) cos n π a t l + l T n ′ ( 0 ) n π a sin n π a t l T_n=\cfrac{l}{n\pi a}\int_0^tg_n(τ)\sin\cfrac{2\pi a}{l}(t-τ)dτ
+T_n(0)\cos\cfrac{n\pi at}{l}+\cfrac{lT'_n(0)}{n\pi a}\sin\cfrac{n\pi at}{l}
T n = n π a l ∫ 0 t g n ( τ ) sin l 2 π a ( t − τ ) d τ + T n ( 0 ) cos l n π a t + n π a l T n ′ ( 0 ) sin l n π a t
将 X n ( x ) , T n ( t ) X_n(x),T_n(t) X n ( x ) , T n ( t )
用同样的分离变量法也可解决热传导非齐次方程和齐次边界条件的定解问题、二维泊松方程的边界值问题。
非齐次边界条件处理的原则是利用叠加原理，把非齐次边界条件问题转化为另一未知函数的齐次边界条件。
{ u t t − a 2 u x x = f ( x , t ) x ∈ ( 0 , l ) u ∣ t = 0 = ϕ ( x ) , ∂ u ∂ t ∣ t = 0 = ψ ( x ) x ∈ [ 0 , l ] u ∣ x = 0 = p ( t ) , u ∣ x = l = q ( t ) (3.1) \begin{cases}
u_{tt}-a^2u_{xx}=f(x,t) & x\in(0,l)\\
u|_{t=0}=\phi(x),\cfrac{∂u}{∂t}|_{t=0}=\psi(x) & x\in[0,l]\\
u|_{x=0}=p(t),u|_{x=l}=q(t)
\end{cases}\tag{3.1}
⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ u t t − a 2 u x x = f ( x , t ) u ∣ t = 0 = ϕ ( x ) , ∂ t ∂ u ∣ t = 0 = ψ ( x ) u ∣ x = 0 = p ( t ) , u ∣ x = l = q ( t ) x ∈ ( 0 , l ) x ∈ [ 0 , l ] ( 3 . 1 )
为了使边界条件齐次化，设
u ( x , t ) = v ( x , t ) + w ( x , t ) (3.2) u(x,t)=v(x,t)+w(x,t)\tag{3.2}
u ( x , t ) = v ( x , t ) + w ( x , t ) ( 3 . 2 )
使得 v , w v,w v , w
{ v ∣ x = 0 = p ( t ) v ∣ x = l = q ( t ) , { w ∣ x = 0 = 0 w ∣ x = l = 0 (3.3) \begin{cases}v|_{x=0}=p(t) \\ v|_{x=l}=q(t) \end{cases},\quad
\begin{cases}w|_{x=0}=0 \\ w|_{x=l}=0 \end{cases}
\tag{3.3}
{ v ∣ x = 0 = p ( t ) v ∣ x = l = q ( t ) , { w ∣ x = 0 = 0 w ∣ x = l = 0 ( 3 . 3 )
这样 w ( x , t ) w(x,t) w ( x , t ) w ( x , t ) w(x,t) w ( x , t )
{ w t t − a 2 w x x = f 1 ( x , t ) x ∈ ( 0 , l ) w ∣ t = 0 = ϕ 1 ( x ) , ∂ u ∂ t ∣ t = 0 = ψ 1 ( x ) x ∈ [ 0 , l ] w ∣ x = 0 = 0 , w ∣ x = l = 0 (3.4) \begin{cases}
w_{tt}-a^2w_{xx}=f_1(x,t) & x\in(0,l)\\
w|_{t=0}=\phi_1(x),\cfrac{∂u}{∂t}|_{t=0}=\psi_1(x) & x\in[0,l]\\
w|_{x=0}=0,w|_{x=l}=0
\end{cases}\tag{3.4}
⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ w t t − a 2 w x x = f 1 ( x , t ) w ∣ t = 0 = ϕ 1 ( x ) , ∂ t ∂ u ∣ t = 0 = ψ 1 ( x ) w ∣ x = 0 = 0 , w ∣ x = l = 0 x ∈ ( 0 , l ) x ∈ [ 0 , l ] ( 3 . 4 )
其中
{ f 1 ( x , t ) = f ( x , t ) − ( v t t − a 2 v x x ) ϕ 1 ( x ) = ϕ ( x ) − v ( x , 0 ) ψ 1 ( x ) = ψ ( x ) − ∂ v ∂ t ∣ t = 0 \begin{cases}
f_1(x,t)=f(x,t)-(v_{tt}-a^2v_{xx}) \\
\phi_1(x)=\phi(x)-v(x,0) \\
\psi_1(x)=\psi(x)-\cfrac{∂v}{∂t}|_{t=0}
\end{cases}
⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ f 1 ( x , t ) = f ( x , t ) − ( v t t − a 2 v x x ) ϕ 1 ( x ) = ϕ ( x ) − v ( x , 0 ) ψ 1 ( x ) = ψ ( x ) − ∂ t ∂ v ∣ t = 0
关于函数 v ( x , t ) v(x,t) v ( x , t )
v ( x , t ) = A ( t ) x + B ( t ) v(x,t)=A(t)x+B(t)
v ( x , t ) = A ( t ) x + B ( t )
代入 (3.3) 中的边界条件，可求得 A ( t ) = 1 l [ q ( t ) − p ( t ) ] , B ( t ) = p ( t ) A(t)=\frac{1}{l}[q(t)-p(t)],\quad B(t)=p(t) A ( t ) = l 1 [ q ( t ) − p ( t ) ] , B ( t ) = p ( t ) w ( x , t ) w(x,t) w ( x , t ) u ( x , t ) u(x,t) u ( x , t ) v ( x , t ) v(x,t) v ( x , t ) w ( x , t ) w(x,t) w ( x , t ) u ( x , t ) u(x,t) u ( x , t )
一般的分类变量法 ：对于一般的二阶线性齐次偏微分方程定解问题
{ L t [ u ] + L x [ u ] = 0 x ∈ ( a , b ) ( α 1 ∂ u ∂ x + β 1 u ) ∣ x = a = 0 ( α 2 ∂ u ∂ x + β 2 u ) ∣ x = b = 0 关 于 t 的 定 解 条 件 (4.1) \begin{cases}
L_t[u]+L_x[u]=0 &x\in(a,b) \\
(α_1\cfrac{∂u}{∂x}+β_1u)|_{x=a}=0\\
(α_2\cfrac{∂u}{∂x}+β_2u)|_{x=b}=0 \\
关于 t 的定解条件
\end{cases}\tag{4.1}
⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ L t [ u ] + L x [ u ] = 0 ( α 1 ∂ x ∂ u + β 1 u ) ∣ x = a = 0 ( α 2 ∂ x ∂ u + β 2 u ) ∣ x = b = 0 关 于 t 的 定 解 条 件 x ∈ ( a , b ) ( 4 . 1 )
其中系数 ∣ α 1 ∣ + ∣ β 1 ∣ ≠ 0 , ∣ α 2 ∣ + ∣ β 2 ∣ ≠ 0 |α_1|+|β_1|\neq0,|α_2|+|β_2|\neq0 ∣ α 1 ∣ + ∣ β 1 ∣ = 0 , ∣ α 2 ∣ + ∣ β 2 ∣ = 0 L t , L x L_t,L_x L t , L x
L t = a 0 ( t ) ∂ 2 ∂ t 2 + a 1 ( t ) ∂ ∂ t + a 2 ( t ) L x = b 0 ( x ) ∂ 2 ∂ x 2 + b 1 ( x ) ∂ ∂ x + b 2 ( x ) L_t=a_0(t)\cfrac{∂^2}{∂t^2}+a_1(t)\cfrac{∂}{∂t}+a_2(t) \\
L_x=b_0(x)\cfrac{∂^2}{∂x^2}+b_1(x)\cfrac{∂}{∂x}+b_2(x)
L t = a 0 ( t ) ∂ t 2 ∂ 2 + a 1 ( t ) ∂ t ∂ + a 2 ( t ) L x = b 0 ( x ) ∂ x 2 ∂ 2 + b 1 ( x ) ∂ x ∂ + b 2 ( x )
(1) 分离变量：令 u ( x , t ) = X ( x ) T ( t ) u(x,t)=X(x)T(t) u ( x , t ) = X ( x ) T ( t )
{ L x [ X ] + λ X = 0 α 1 X ′ ( a ) + β 1 X ( a ) = 0 , α 2 X ′ ( b ) + β 2 X ( b ) = 0 (4.2) \begin{cases}
L_x[X]+λX=0 \\
α_1X'(a)+β_1X(a)=0,\ α_2X'(b)+β_2X(b)=0
\end{cases}\tag{4.2}
{ L x [ X ] + λ X = 0 α 1 X ′ ( a ) + β 1 X ( a ) = 0 , α 2 X ′ ( b ) + β 2 X ( b ) = 0 ( 4 . 2 )
和常微分方程
L t [ T ] − λ T = 0 L_t[T]-λT=0
L t [ T ] − λ T = 0
(2) 解本征值问题，获得分离变量形式的特解λ λ λ { λ n } \{λ_n\} { λ n } { X n ( x ) } \{X_n(x)\} { X n ( x ) } T ( t ) T(t) T ( t )
u n ( x , t ) = X n ( x ) T n ( t ) u_n(x,t)=X_n(x)T_n(t)
u n ( x , t ) = X n ( x ) T n ( t )
(3) 叠加定系数：令
u ( x , t ) = ∑ n = 1 ∞ C n X n ( x ) T n ( t ) u(x,t)=\sum_{n=1}^{\infty}C_nX_n(x)T_n(t)
u ( x , t ) = n = 1 ∑ ∞ C n X n ( x ) T n ( t )
带入关于 t t t C n C_n C n
以上三步中，分离变量是基础，求解本征值问题是核心。所谓本征值问题，就是在一定的边界条件下，求解含参数 λ λ λ
施图姆-刘维尔理论 ：考虑带有参数 λ λ λ
y ′ ′ + a ( x ) y ′ + b ( x ) y + λ c ( x ) y = 0 ( a < x < b ) y''+a(x)y'+b(x)y+λc(x)y=0\quad(a<x<b)
y ′ ′ + a ( x ) y ′ + b ( x ) y + λ c ( x ) y = 0 ( a < x < b )
如果方程两边同乘以 k ( x ) = exp [ ∫ a ( x ) d x ] \displaystyle k(x)=\exp[\int a(x)dx] k ( x ) = exp [ ∫ a ( x ) d x ]
d d x [ k ( x ) d y d x ] − q ( x ) y + λ ρ ( x ) y = 0 ( a < x < b ) (4.3) \cfrac{d}{dx}[k(x)\cfrac{dy}{dx}]-q(x)y+λ\rho(x)y=0\quad(a<x<b)\tag{4.3}
d x d [ k ( x ) d x d y ] − q ( x ) y + λ ρ ( x ) y = 0 ( a < x < b ) ( 4 . 3 )
其中 k ( x ) = exp [ ∫ a ( x ) d x ] , q ( x ) = − b ( x ) k ( x ) , ρ ( x ) = c ( x ) k ( x ) \displaystyle k(x)=\exp[\int a(x)dx],\quad q(x)=-b(x)k(x),\quad \rho(x)=c(x)k(x) k ( x ) = exp [ ∫ a ( x ) d x ] , q ( x ) = − b ( x ) k ( x ) , ρ ( x ) = c ( x ) k ( x ) ρ ( x ) \rho(x) ρ ( x ) 权重函数 ，这个方程称为施图姆-刘维尔方程 (Sturm-Liouville)，简称 S-L 方程 ，我们在分离变量法中遇到的常微分方程都是式 (4.3) 的特例。S-L 方程附加上适当的边界条件就就构成施图姆-刘维尔本征值问题 ， λ λ λ k ( x ) = 1 , q ( x ) = 0 , ρ ( x ) = 1 , a = 0 , b = l k(x)=1,q(x)=0,\rho(x)=1,a=0,b=l k ( x ) = 1 , q ( x ) = 0 , ρ ( x ) = 1 , a = 0 , b = l y ′ ′ + λ y = 0 , ( 0 < x < l ) y''+λy=0,\quad(0<x<l) y ′ ′ + λ y = 0 , ( 0 < x < l )
S-L 型方程，即 k ( x ) y ′ ′ + k ′ ( x ) y ′ − q ( x ) y + λ ρ ( x ) y = 0 k(x)y''+k'(x)y'-q(x)y+λ\rho(x)y=0 k ( x ) y ′ ′ + k ′ ( x ) y ′ − q ( x ) y + λ ρ ( x ) y = 0
y ′ ′ + k ′ ( x ) k ( x ) y ′ + − q ( x ) + λ ρ ( x ) k ( x ) y = 0 y''+\cfrac{k'(x)}{k(x)}y'+\cfrac{-q(x)+λ\rho(x)}{k(x)}y=0
y ′ ′ + k ( x ) k ′ ( x ) y ′ + k ( x ) − q ( x ) + λ ρ ( x ) y = 0
通常分为正则和奇异两种类型：
类型 1 如果区间 [ a , b ] [a,b] [ a , b ] k ( x ) , q ( x ) , ρ ( x ) k(x),q(x),\rho(x) k ( x ) , q ( x ) , ρ ( x ) [ a , b ] [a,b] [ a , b ] k ( x ) > 0 , ρ ( x ) > 0 k(x)>0,\rho(x)>0 k ( x ) > 0 , ρ ( x ) > 0 ( a , b ) (a,b) ( a , b ) k ( a ) , k ( b ) > 0 k(a),k(b)>0 k ( a ) , k ( b ) > 0 y ( x ) , y ′ ( x ) y(x),y'(x) y ( x ) , y ′ ( x ) a a a b b b
{ α 1 y ′ ( a ) + β 1 y ( a ) = 0 α 2 y ′ ( b ) + β 2 y ( b ) = 0 \begin{cases}
α_1y'(a)+β_1y(a)=0 \\
α_2y'(b)+β_2y(b)=0
\end{cases}
{ α 1 y ′ ( a ) + β 1 y ( a ) = 0 α 2 y ′ ( b ) + β 2 y ( b ) = 0
其中 ∣ α 1 ∣ + ∣ β 1 ∣ ≠ 0 , ∣ α 2 ∣ + ∣ β 2 ∣ ≠ 0 |α_1|+|β_1|\neq0,|α_2|+|β_2|\neq0 ∣ α 1 ∣ + ∣ β 1 ∣ = 0 , ∣ α 2 ∣ + ∣ β 2 ∣ = 0 k ( a ) = k ( b ) k(a)=k(b) k ( a ) = k ( b )
y ( a ) = y ( b ) y(a)=y(b)
y ( a ) = y ( b )
类型 2 或者区间 [ a , b ] [a,b] [ a , b ] k ( x ) k(x) k ( x ) ρ ( x ) \rho(x) ρ ( x ) [ a , b ] [a,b] [ a , b ] ( a , b ) (a,b) ( a , b ) ( 0 , 1 ) (0,1) ( 0 , 1 ) a , b a,b a , b k ( x ) k(x) k ( x ) k ( a ) = 0 k(a)=0 k ( a ) = 0 y ( a ) , y ′ ( a ) y(a),y'(a) y ( a ) , y ′ ( a ) k ( b ) = 0 k(b)=0 k ( b ) = 0 y ( b ) , y ′ ( b ) y(b),y'(b) y ( b ) , y ′ ( b )
共同性质 ：如果函数 k ( x ) , k ′ ( x ) , q ( x ) , ρ ( x ) k(x),k'(x),q(x),\rho(x) k ( x ) , k ′ ( x ) , q ( x ) , ρ ( x ) [ a , b ] [a,b] [ a , b ] λ m , λ n λ_m,λ_n λ m , λ n y m ( x ) , y n ( x ) y_m(x),y_n(x) y m ( x ) , y n ( x ) [ a , b ] [a,b] [ a , b ] ρ ( x ) \rho(x) ρ ( x ) 正交 的
∫ a b y m ( x ) y n ( x ) ρ ( x ) d x = 0 ( m ≠ n ) (4.4) \displaystyle \int_a^by_m(x)y_n(x)\rho(x)dx=0\quad(m\neq n)\tag{4.4}
∫ a b y m ( x ) y n ( x ) ρ ( x ) d x = 0 ( m = n ) ( 4 . 4 )
例如：对于本征值问题
X ′ ′ + λ X = 0 X ( 0 ) = X ( l ) = 0 X''+λX=0 \\
X(0)=X(l)=0
X ′ ′ + λ X = 0 X ( 0 ) = X ( l ) = 0
设X n , X m X_n,X_m X n , X m λ n ≠ λ m λ_n\neqλ_m λ n = λ m
X m ( X n ′ ′ + λ n X n ) = 0 X n ( X m ′ ′ + λ m X m ) = 0 X_m(X_n''+λ_n X_n)=0 \\
X_n(X_m''+λ_m X_m)=0
X m ( X n ′ ′ + λ n X n ) = 0 X n ( X m ′ ′ + λ m X m ) = 0
两式相减，并在 [ 0 , l ] [0,l] [ 0 , l ]
( λ n − λ m ) ∫ 0 l X n X m d x = ∫ 0 l [ X n X m ′ ′ − X m X n ′ ′ ] d x = [ X n X m ′ − X m X n ′ ] ∣ 0 l = 0 \begin{aligned}\displaystyle
&(λ_n-λ_m)\int_0^l X_nX_mdx \\
=&\int_0^l [X_nX_m''-X_mX_n'']dx \\
=&[X_nX_m'-X_mX_n']|_0^l \\
=&0
\end{aligned}
= = = ( λ n − λ m ) ∫ 0 l X n X m d x ∫ 0 l [ X n X m ′ ′ − X m X n ′ ′ ] d x [ X n X m ′ − X m X n ′ ] ∣ 0 l 0
因此得到本征函数的正交性
∫ 0 l X n ( x ) X m ( x ) d x = 0 ( n ≠ m ) \displaystyle\int_0^l X_n(x)X_m(x)dx=0\quad(n\neq m)
∫ 0 l X n ( x ) X m ( x ) d x = 0 ( n = m )
(2) 若 ρ ( x ) > 0 , x ∈ ( a , b ) \rho(x)>0,x\in(a,b) ρ ( x ) > 0 , x ∈ ( a , b )
λ 0 < λ 2 < λ 3 < ⋯ < λ n < ⋯ λ_0<λ_2<λ_3<\cdots<λ_n<\cdots
λ 0 < λ 2 < λ 3 < ⋯ < λ n < ⋯
其中当 n → ∞ n\to\infty n → ∞ y n ( x ) y_n(x) y n ( x ) ( a , b ) (a,b) ( a , b ) n n n { y n ( x ) } \{y_n(x)\} { y n ( x ) } f ( x ) f(x) f ( x ) [ a , b ] [a,b] [ a , b ] f ( x ) f(x) f ( x ) { y n ( x ) } \{y_n(x)\} { y n ( x ) }
f ( x ) = ∑ n = 0 ∞ C n y n ( x ) \displaystyle f(x)=\sum_{n=0}^{\infty}C_ny_n(x)
f ( x ) = n = 0 ∑ ∞ C n y n ( x )
如果函数 f ( x ) f(x) f ( x ) f ( x ) f(x) f ( x )
由于温度分布不均匀，热量从温度高的地方向温度低的地方转移，这种现象叫热传导。u ( x , y , z , t ) u(x,y,z,t) u ( x , y , z , t ) ∇ u ∇u ∇ u q \mathbf q q
q = − κ ∇ u \mathbf q=-κ∇u
q = − κ ∇ u
其中比例系数 κ κ κ
q x = − ∂ u ∂ x , q y = − ∂ u ∂ y , q z = − ∂ u ∂ z q_x=-\cfrac{∂u}{∂x},q_y=-\cfrac{∂u}{∂y},q_z=-\cfrac{∂u}{∂z}
q x = − ∂ x ∂ u , q y = − ∂ y ∂ u , q z = − ∂ z ∂ u
负号表示热量转移的方向和温度梯度相反。
S S S V V V d t \mathrm dt d t d S \mathrm dS d S
d Q = − κ ∇ u ⋅ d S d t \mathrm dQ=-κ∇u\cdot\mathrm d\mathbf S\mathrm dt
d Q = − κ ∇ u ⋅ d S d t
若区域内无热源，从t t t t + d t t+\mathrm dt t + d t
∭ V c ρ d V d u = ∯ S κ ∇ u ⋅ d S d t \iiint\limits_Vc\rho\mathrm dV\mathrm du=\oiint\limits_Sκ∇u\cdot\mathrm d\mathbf S\mathrm dt
V ∭ c ρ d V d u = S ∬ κ ∇ u ⋅ d S d t
其中c c c ρ \rho ρ
∭ V c ρ d V d u = ∭ V ∇ ⋅ ( κ ∇ u ) d V d t \iiint\limits_Vc\rho\mathrm dV\mathrm du
=\iiint\limits_V∇\cdot(κ∇u)\mathrm dV\mathrm dt
V ∭ c ρ d V d u = V ∭ ∇ ⋅ ( κ ∇ u ) d V d t
由于区域的随意性，取极限 V → M ( x , y , z ) V\to M(x,y,z) V → M ( x , y , z )
c ρ u t = ∇ ⋅ ( κ ∇ u ) c\rho u_t=∇\cdot(κ∇u)
c ρ u t = ∇ ⋅ ( κ ∇ u )
如果物体是均匀的，此时 c , ρ , κ c,\rho,κ c , ρ , κ
u t − a 2 Δ u = 0 u_t-a^2Δu=0
u t − a 2 Δ u = 0
其中系数 a 2 = κ ρ c a^2=\cfrac{κ}{\rho c} a 2 = ρ c κ
如果物体内有热源，其热源密度函数为 F ( x , y , z , t ) F(x,y,z,t) F ( x , y , z , t )
u t − a 2 Δ u = f u_t-a^2Δu=f
u t − a 2 Δ u = f
其中 f = F ρ c f=\cfrac{F}{\rho c} f = ρ c F
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