文章目录
2分离变量法
2.0常微分方程
2.0.1齐次&非齐次方程
- 齐次方程 y ′ ( x ) = P ( x ) y ( x ) y'(x)=P(x)y(x) y′(x)=P(x)y(x) 分离变量,两边积分
- y ( x ) = C e ∫ P ( x ) d x y(x)=Ce^{\int P(x)dx} y(x)=Ce∫P(x)dx
- P(x)为常数 y ′ ( x ) = m y ( x ) → y ( x ) = C e m x y'(x)=my(x)\rightarrow y(x)=Ce^{mx} y′(x)=my(x)→y(x)=Cemx
- 非齐次方程 y ′ ( x ) = P ( x ) y ( x ) + Q ( x ) y'(x)=P(x)y(x)+Q(x) y′(x)=P(x)y(x)+Q(x) 常数变异法
- y ( x ) = e ∫ P ( x ) d x ( ∫ Q ( x ) ⋅ e − ∫ P ( x ) d x d x + C ) y(x)=e^{\int P(x)dx}(\int Q(x)\cdot e^{-\int P(x)dx}dx+C) y(x)=e∫P(x)dx(∫Q(x)⋅e−∫P(x)dxdx+C)
2.0.2二阶常系数齐次常微分方程
a 2 y ′ ′ ( x ) + a 1 y ′ ( x ) + a 0 y ( x ) = 0 a_2y''(x)+a_1y'(x)+a_0y(x)=0 a2y′′(x)+a1y′(x)+a0y(x)=0
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特征方程: a 2 r 2 + a 1 r + a 0 = 0 a_2r^2+a_1r+a_0=0 a2r2+a1r+a0=0, r特征根
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2实根 r 1 ≠ r 2 r_1\neq r_2 r1=r2: 通解 y ( x ) = A e r 1 x + B e r 2 x y(x)=Ae^{r_1x}+Be^{r_2x} y(x)=Aer1x+Ber2x
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1实根 r 1 = r 2 r_1=r_2 r1=r2: 通解 y ( x ) = ( A x + B ) e r x y(x)=(Ax+B)e^{rx} y(x)=(Ax+B)erx
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复根 r 1 = α + β i , r 2 = α − β i r_1=\alpha+\beta i, r_2=\alpha-\beta i r1=α+βi,r2=α−βi: 通解 y ( x ) = e α x ( A cos β x + B sin β x ) y(x)=e^{\alpha x}(A\cos\beta x+B\sin\beta x) y(x)=eαx(Acosβx+Bsinβx)
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ρ 2 R ′ ′ ( ρ ) + ρ R ′ ( ρ ) − λ R ( ρ ) = 0 \rho^2R''(\rho)+\rho R'(\rho)-\lambda R(\rho)=0 ρ2R′′(ρ)+ρR′(ρ)−λR(ρ)=0
- ρ = e x \rho=e^x ρ=ex
2.1特征值问题
2.1.1常用特征值问题
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常微分方程 X ′ ′ ( x ) + λ X ( x ) = 0 X''(x)+\lambda X(x)=0 X′′(x)+λX(x)=0
- λ < 0 \lambda<0 λ<0, X ( x ) = A e − λ x + B e − − λ x X(x)=Ae^{\sqrt{-\lambda}x}+Be^{-\sqrt{-\lambda}x} X(x)=Ae−λx+Be−−λx
- λ = 0 \lambda=0 λ=0, X ( x ) = A x + B X(x)=Ax+B X(x)=Ax+B
- λ > 0 \lambda>0 λ>0, X ( x ) = A cos λ x + B sin λ x X(x)=A\cos\sqrt{\lambda}x+B\sin\sqrt{\lambda}x X(x)=Acosλx+Bsinλx
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含待确定常数 λ \lambda λ的特征值问题, λ \lambda λ特征值, X ( x ) X(x) X(x)特征函数
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{ X ′ ′ ( x ) + λ X ( x ) = 0 X ( 0 ) = 0 , X ( l ) = 0 \left\{\begin{matrix}X''(x)+\lambda X(x)=0 \\X(0)=0, X(l)=0\end{matrix}\right. { X′′(x)+λX(x)=0X(0)=0,X(l)=0
- λ < 0 \lambda<0 λ<0, A=B=0, 平凡解
- λ = 0 \lambda=0 λ=0, A=B=0, 平凡解
- λ > 0 \lambda>0 λ>0, { λ n = ( n π l ) 2 X n ( x ) = B n sin n π l x , n = 1 , 2 , 3... \left\{\begin{matrix}\lambda _n=(\frac{n\pi}{l} )^2 \\ X_n(x)=B_n\sin\frac{n\pi}{l}x,n=1,2,3... \end{matrix}\right. {
λn=(lnπ)2Xn(x)=Bnsinlnπx,n=1,2,3...
- sin n π l x n = 1 ∞ {\sin\frac{n\pi}{l}x}_{n=1}^\infty sinlnπxn=1∞
- C n = 2 l ∫ 0 l f ( x ) sin n π l x d x C_n=\frac{2}{l}\int_0^l f(x)\sin\frac{n\pi}{l}xdx Cn=l2∫0lf(x)sinlnπxdx
2.1.2关于特征值问题的理论
- S-L方程 d d x ( k ( x ) d y d x ) − q ( x ) y ( x ) + λ ρ ( x ) y ( x ) = 0 , x ∈ ( a , b ) \frac{d}{dx}(k(x)\frac{dy}{dx})-q(x)y(x)+\lambda\rho(x)y(x)=0,x\in(a,b) dxd(k(x)dxdy)−q(x)y(x)+λρ(x)y(x)=0,x∈(a,b)
- 特征值:其次边界条件,周期性,自然边界条件
- k ( x ) = ρ ( x ) = 1 , q ( x ) = 0 k(x)=\rho(x)=1,q(x)=0 k(x)=ρ(x)=1,q(x)=0: y ′ ′ ( x ) + λ y ( x ) = 0 y''(x)+\lambda y(x)=0 y′′(x)+λy(x)=0
- k ( x ) = ρ ( x ) = x , q ( x ) = n 2 x k(x)=\rho(x)=x,q(x)=\frac{n^2}{x} k(x)=ρ(x)=x,q(x)=xn2: x 2 y ′ ′ ( x ) + x y ′ ( x ) + ( λ x 2 − n 2 ) y ( x ) = 0 x^2y''(x)+xy'(x)+(\lambda x^2-n^2)y(x)=0 x2y′′(x)+xy′(x)+(λx2−n2)y(x)=0
- k ( x ) = 1 − x 2 , ρ ( x ) = 1 , q ( x ) = 0 k(x)=1-x^2,\rho(x)=1,q(x)=0 k(x)=1−x2,ρ(x)=1,q(x)=0: ( 1 − x ) 2 y ′ ′ ( x ) − 2 x y ′ ( x ) + λ y ( x ) = 0 (1-x)^2y''(x)-2xy'(x)+\lambda y(x)=0 (1−x)2y′′(x)−2xy′(x)+λy(x)=0
- 存在可数个实特征值->单调递增序列 0 ≤ λ 1 ≤ λ 2 ≤ . . . ≤ λ n ≤ {0\leq\lambda_1\leq\lambda_2\leq...\leq\lambda_n\leq} 0≤λ1≤λ2≤...≤λn≤
可数个特征函数 { y x ( x ) } , n = 1 , 2 , 3... \{y_x(x)\}, n=1,2,3... { yx(x)},n=1,2,3... - 所有特征值非负 λ n ≥ 0 , n = 1 , 2 , 3... \lambda_n\geq 0, n=1,2,3... λn≥0,n=1,2,3...
- 特征函数系 { y n ( x ) } n = 1 n = ∞ \{y_n(x)\}_{n=1}^{n=\infty} { yn(x)}n=1n=∞是 L ρ 2 [ a , b ] L_\rho^2[a,b] Lρ2[a,b]上关于权函数 ρ ( x ) \rho(x) ρ(x)的正交系 ∫ a b ρ ( x ) y n ( x ) y m ( x ) d x = { 0 n ≠ m ∣ ∣ y n ∣ ∣ 2 2 n = m \int_a^b\rho(x)y_n(x)y_m(x)dx=\left\{\begin{matrix}0 &n\neq m \\||y_n||_2^2 &n=m\end{matrix}\right. ∫abρ(x)yn(x)ym(x)dx={ 0∣∣yn∣∣22n=mn=m
- 特征函数系 { y n ( x ) } n = 1 n = ∞ \{y_n(x)\}_{n=1}^{n=\infty} { yn(x)}n=1n=∞是 L ρ 2 [ a , b ] L_\rho^2[a,b] Lρ2[a,b]上关于权函数 ρ ( x ) \rho(x) ρ(x)的完备系
2.1.3Matlab
- 二阶常微分方程
求 d 2 y d x 2 + y = 1 − x 2 π \frac{d^{2} y}{d x^{2}}+y=1-\frac{x^{2}}{\pi} dx2d2y+y=1−πx2通解-
y=dsolve('D2y+y=1-x^2/pi','x')
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y=dsolve('D2y+y=1-x^2/pi','y(0)=0.2,Dy(0)=0.5','x') ezplot(y),aixs([-3 3 -0.5 2])
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- 常微分方程组 { d u d t = 3 u − 2 v d v d t + v = 2 u \left\{\begin{array}{l}\frac{d u}{d t}=3 u-2 v \\\frac{d v}{d t}+v=2 u\end{array}\right. {
dtdu=3u−2vdtdv+v=2u
1.求通解-
[u,v]=dsolve('Du=3*u-2*v','Dv+v=2*u')
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[u,v]=dsolve('Du=3*u-2*v','Dv+v=2*u','u(0)=1,v(0)=0','t')
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2.2一维波方程,一维热方程解法
2.2.1一维波方程
{ ∂ 2 u ∂ t 2 = a 2 ∂ 2 u ∂ x 2 0 < x < l , t > 0 u ∣ x = 0 = u ∣ x = l = 0 u ∣ t = 0 = ϕ ( x ) , ∂ u ∂ t ∣ t = 0 = ψ ( x ) \left\{\begin{array}{l} \frac{\partial^{2} \boldsymbol{u}}{\partial \boldsymbol{t}^{2}}=\boldsymbol{a}^{2} \frac{\partial^{2} \boldsymbol{u}}{\partial \boldsymbol{x}^{2}} \quad 0<\boldsymbol{x}<\boldsymbol{l}, \boldsymbol{t}>0 \\ \left.\boldsymbol{u}\right|_{\boldsymbol{x}=0}=\left.\boldsymbol{u}\right|_{\boldsymbol{x}=l}=0 \\ \left.\boldsymbol{u}\right|_{t=0}=\phi(\boldsymbol{x}),\left.\frac{\partial \boldsymbol{u}}{\partial \boldsymbol{t}}\right|_{t=0}=\psi(\boldsymbol{x}) \end{array}\right. ⎩
⎨
⎧∂t2∂2u=a2∂x2∂2u0<x<l,t>0u∣x=0=u∣x=l=0u∣t=0=ϕ(x),∂t∂u∣
∣t=0=ψ(x)
- u n ( x , t ) u_n(x,t) un(x,t)在任意时刻为正弦曲线
- t = t 0 t=t_0 t=t0 u n ( x , t 0 ) u_n(x,t_0) un(x,t0)为正弦曲线,振幅随时间变化
- x = x 0 x=x_0 x=x0 u n ( x 0 , t ) u_n(x_0,t) un(x0,t)简谐振动
- 任意确定时刻 u n ( x , t ) = A n ′ sin n π l x u_{n}(x, t)=A_{n}^{\prime} \sin \frac{n \pi}{l} x un(x,t)=An′sinlnπx
- n + 1 n+1 n+1个零点, n n n个极值点, u 1 , u 2 , u 3 u_1,u_2,u_3 u1,u2,u3是一系列驻波
2.2.2一维热方程解法
2.3Laplace方程定解问题解法
2.3.1直角坐标系下Laplace
2.3.2二维圆域Laplace定解问题解法
- 定解问题 ∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 = 0 \frac{\partial^{2} \boldsymbol{u}}{\partial \boldsymbol{x}^{2}}+\frac{\partial^{2} \boldsymbol{u}}{\partial \boldsymbol{y}^{2}}=0 ∂x2∂2u+∂y2∂2u=0
- 极坐标变换 { x = ρ cos θ y = ρ sin θ , 0 ≤ θ ≤ 2 π \left\{\begin{array}{l}x=\rho \cos \theta \\y=\rho \sin \theta,\end{array} \quad 0 \leq \theta \leq 2 \pi\right. { x=ρcosθy=ρsinθ,0≤θ≤2π
- ∂ u ∂ x = ∂ u ∂ ρ ∂ ρ ∂ x + ∂ u ∂ θ ∂ θ ∂ x = v \frac{\partial \boldsymbol{u}}{\partial \boldsymbol{x}}=\frac{\partial \boldsymbol{u}}{\partial \rho} \frac{\partial \rho}{\partial \boldsymbol{x}}+\frac{\partial \boldsymbol{u}}{\partial \theta} \frac{\partial \theta}{\partial \boldsymbol{x}}=\boldsymbol{v} ∂x∂u=∂ρ∂u∂x∂ρ+∂θ∂u∂x∂θ=v代入得 $ \begin{array}{l}\frac{1}{\rho} \frac{\partial}{\partial \rho}\left(\rho \frac{\partial \boldsymbol{u}}{\partial \rho}\right)+\frac{1}{\rho^{2}} \frac{\partial^{2} \boldsymbol{u}}{\partial \theta^{2}}=0 \\frac{\partial^{2} \boldsymbol{u}}{\partial \rho^{2}}+\frac{1}{\rho} \frac{\partial \boldsymbol{u}}{\partial \rho}+\frac{1}{\rho^{2}} \frac{\partial^{2} \boldsymbol{u}}{\partial \theta^{2}}=0\end{array} $
2.4非齐次方程
波方程 { ∂ 2 u ∂ t 2 = a 2 ∂ 2 u ∂ x 2 + f ( x , t ) , 0 < x < l , t > 0 u ∣ x = 0 = u ∣ x = l = 0 u ∣ t = 0 = ϕ ( x ) , ∂ u ∂ t ∣ t = 0 = Ψ ( x ) \left\{\begin{array}{l}\frac{\partial^{2} \boldsymbol{u}}{\partial \boldsymbol{t}^{2}}=\boldsymbol{a}^{2} \frac{\partial^{2} \boldsymbol{u}}{\partial \boldsymbol{x}^{2}}+\boldsymbol{f}(\boldsymbol{x}, \boldsymbol{t}), \quad 0<\boldsymbol{x}<\boldsymbol{l}, \boldsymbol{t}>0 \\\left.\boldsymbol{u}\right|_{\boldsymbol{x}=0}=\left.\boldsymbol{u}\right|_{\boldsymbol{x}=l}=0 \\\left.\boldsymbol{u}\right|_{t=0}=\phi(\boldsymbol{x}),\left.\frac{\partial \boldsymbol{u}}{\partial \boldsymbol{t}}\right|_{t=0}=\varPsi(\boldsymbol{x})\end{array}\right. ⎩
⎨
⎧∂t2∂2u=a2∂x2∂2u+f(x,t),0<x<l,t>0u∣x=0=u∣x=l=0u∣t=0=ϕ(x),∂t∂u∣
∣t=0=Ψ(x)
拆解 u ( x , t ) = v ( x , t ) + w ( x , t ) u(x,t)=v(x,t)+w(x,t) u(x,t)=v(x,t)+w(x,t)
2.4.1特征函数法
波方程
{ ∂ 2 v ∂ t 2 = a 2 ∂ 2 v ∂ x 2 + f ( x , t ) v ∣ x = 0 = v ∣ x = l = 0 v ∣ t = 0 = 0 , ∂ v ∂ t ∣ t = 0 = 0 \left\{\begin{array}{l}\frac{\partial^{2} v}{\partial t^{2}}=a^{2} \frac{\partial^{2} v}{\partial x^{2}}+f(x, t) \\\left.v\right|_{x=0}=\left.v\right|_{x=l}=0 \\\left.v\right|_{t=0}=0,\left.\frac{\partial v}{\partial t}\right|_{t=0}=0\end{array}\right. ⎩ ⎨ ⎧∂t2∂2v=a2∂x2∂2v+f(x,t)v∣x=0=v∣x=l=0v∣t=0=0,∂t∂v∣ ∣t=0=0
- 齐次方程+齐次边界条件 { ∂ 2 v ∂ t 2 = a 2 ∂ 2 v ∂ x 2 v ∣ x = 0 = v ∣ x = l = 0 \left\{\begin{array}{l}\frac{\partial^{2} \boldsymbol{v}}{\partial \boldsymbol{t}^{2}}=\boldsymbol{a}^{2} \frac{\partial^{2} \boldsymbol{v}}{\partial \boldsymbol{x}^{2}} \\\left.\boldsymbol{v}\right|_{\boldsymbol{x}=0}=\left.\boldsymbol{v}\right|_{\boldsymbol{x}=l}=0\end{array}\right. {
∂t2∂2v=a2∂x2∂2vv∣x=0=v∣x=l=0
- 特征值问题 { X ′ ′ ( x ) + λ X ( x ) = 0 X ( 0 ) = X ( l ) = 0 \left\{\begin{array}{l}\boldsymbol{X}^{\prime \prime}(\boldsymbol{x})+\lambda \boldsymbol{X}(\boldsymbol{x})=0 \\\boldsymbol{X}(0)=\boldsymbol{X}(\boldsymbol{l})=0\end{array}\right. {
X′′(x)+λX(x)=0X(0)=X(l)=0
- 特征函数 X n ( x ) = B n sin n π l x , n = 1 , 2 , . . . X_n(x)=B_n\sin\frac{n\pi}{l}x,n=1,2,... Xn(x)=Bnsinlnπx,n=1,2,...
- 假设非齐次方程解 v ( x , t ) = ∑ n = 1 ∞ v n ( t ) sin n π l x v(x,t)=\sum_{n=1}^{\infty}v_n(t)\sin \frac{n\pi}{l}x v(x,t)=∑n=1∞vn(t)sinlnπx
- 特征值问题 { X ′ ′ ( x ) + λ X ( x ) = 0 X ( 0 ) = X ( l ) = 0 \left\{\begin{array}{l}\boldsymbol{X}^{\prime \prime}(\boldsymbol{x})+\lambda \boldsymbol{X}(\boldsymbol{x})=0 \\\boldsymbol{X}(0)=\boldsymbol{X}(\boldsymbol{l})=0\end{array}\right. {
X′′(x)+λX(x)=0X(0)=X(l)=0
- f ( x , t ) f(x,t) f(x,t)按特征函数序列 { sin n π l x } n = 1 ∞ \{\sin\frac{n\pi}{l}x\}_{n=1}^{\infty} {
sinlnπx}n=1∞展开为级数形式
- f ( x , t ) = ∑ n = 1 ∞ f n ( t ) sin n π l x f(x,t)=\sum_{n=1}^{\infty}f_n(t)\sin\frac{n\pi}{l}x f(x,t)=∑n=1∞fn(t)sinlnπx
- f n ( t ) = 2 l ∫ 0 l f ( x , t ) sin n π l x d x f_n(t)=\frac{2}{l}\int_0^lf(x,t)\sin\frac{n\pi}{l}xdx fn(t)=l2∫0lf(x,t)sinlnπxdx
- 以上结果带入非齐次方程
- 得常微分方程问题 { v n ′ ′ ( t ) + ( n π a l ) 2 v n ( t ) = f n ( t ) v n ( 0 ) = 0 , v n ′ ( 0 ) = 0 \left\{\begin{array}{l}v_{n}^{\prime \prime}(t)+\left(\frac{n \pi a}{l}\right)^{2} v_{n}(t)=f_{n}(t) \\v_{n}(0)=0, \quad v_{n}^{\prime}(0)=0\end{array}\right. { vn′′(t)+(lnπa)2vn(t)=fn(t)vn(0)=0,vn′(0)=0
- Laplace变换
- v n ( t ) = l n π a ∫ 0 t f n ( τ ) sin n π a ( t − τ ) l d τ \boldsymbol{v}_{n}(t)=\frac{l}{n \pi a} \int_{0}^{t} f_{n}(\tau) \sin \frac{n \pi a(t-\tau)}{l} d \tau vn(t)=nπal∫0tfn(τ)sinlnπa(t−τ)dτ
- v ( x , t ) = ∑ n = 1 ∞ l n π a ∫ 0 t f n ( τ ) sin n π a ( t − τ ) l d τ sin n π l x \boldsymbol{v}(\boldsymbol{x}, \boldsymbol{t})=\sum_{n=1}^{\infty} \frac{l}{\boldsymbol{n} \pi a} \int_{0}^{t} f_{n}(\tau) \sin \frac{n \pi a(t-\tau)}{l} d \tau \sin \frac{n \pi}{l} \boldsymbol{x} v(x,t)=∑n=1∞nπal∫0tfn(τ)sinlnπa(t−τ)dτsinlnπx
Matlab解二阶常微分方程
syms V n a L
S=dsolve(`D2V+(n*pi*a/L)^2*V=5`,`V(0)=0,DV(0)=0`,`t`)
pretty(simple(S))
热方程
{ ∂ u ∂ t = a 2 ∂ 2 u ∂ x 2 + sin ω t 0 < x < l , t > 0 ∂ u ∂ x ∣ x = 0 = ∂ u ∂ x ∣ x = l = 0 u ∣ t = 0 = 0 \left\{\begin{array}{l} \frac{\partial \boldsymbol{u}}{\partial \boldsymbol{t}}=\boldsymbol{a}^{2} \frac{\partial^{2} \boldsymbol{u}}{\partial \boldsymbol{x}^{2}}+\sin \omega \boldsymbol{t} \quad 0<\boldsymbol{x}<\boldsymbol{l}, \boldsymbol{t}>0 \\ \left.\frac{\partial \boldsymbol{u}}{\partial \boldsymbol{x}}\right|_{x=0}=\left.\frac{\partial \boldsymbol{u}}{\partial \boldsymbol{x}}\right|_{x=l}=0 \\ \left.\boldsymbol{u}\right|_{t=0}=0 \end{array}\right. ⎩ ⎨ ⎧∂t∂u=a2∂x2∂2u+sinωt0<x<l,t>0∂x∂u∣ ∣x=0=∂x∂u∣ ∣x=l=0u∣t=0=0
- 齐次方程+齐次边界条件 { ∂ u ∂ t = a 2 ∂ 2 u ∂ x 2 ∂ u ∂ x ∣ x = 0 = ∂ u ∂ x ∣ x = l = 0 \left\{\begin{array}{l}\frac{\partial \boldsymbol{u}}{\partial \boldsymbol{t}}=\boldsymbol{a}^{2} \frac{\partial^{2} \boldsymbol{u}}{\partial \boldsymbol{x}^{2}} \\\left.\frac{\partial \boldsymbol{u}}{\partial \boldsymbol{x}}\right|_{x=0}=\left.\frac{\partial \boldsymbol{u}}{\partial \boldsymbol{x}}\right|_{x=l}=0\end{array}\right. {
∂t∂u=a2∂x2∂2u∂x∂u∣
∣x=0=∂x∂u∣
∣x=l=0
- 得特征值问题 { X ′ ′ ( x ) + λ X ( x ) = 0 X ′ ( 0 ) = X ′ ( l ) = 0 \left\{\begin{array}{l}\boldsymbol{X}^{\prime \prime}(\boldsymbol{x})+\lambda \boldsymbol{X}(\boldsymbol{x})=0 \\\boldsymbol{X}^{\prime}(0)=\boldsymbol{X}^{\prime}(\boldsymbol{l})=0\end{array}\right. { X′′(x)+λX(x)=0X′(0)=X′(l)=0
- 特征函数 X n ( x ) = A n cos n π l x , n = 0 , 1 , 2 , . . . X_n(x)=A_n\cos\frac{n\pi}{l}x,n=0,1,2,... Xn(x)=Ancoslnπx,n=0,1,2,...
- 设非齐次方程解 u ( x , t ) = ∑ n = 1 ∞ u n ( t ) cos n π l x u(x,t)=\sum_{n=1}^\infty u_n(t)\cos\frac{n\pi}{l}x u(x,t)=∑n=1∞un(t)coslnπx
- sin w t \sin wt sinwt按特征函数序列 { cos n π l x } n = 0 ∞ \{\cos\frac{n\pi}{l}x\}_{n=0}^\infty {
coslnπx}n=0∞展开为级数形式
- sin w t = f 0 + ∑ n = 1 ∞ f n ( t ) cos n π l x \sin wt=f_0+\sum_{n=1}^\infty f_n(t)\cos\frac{n\pi}{l}x sinwt=f0+∑n=1∞fn(t)coslnπx
- f 0 ( t ) = 1 l ∫ 0 l sin w t d x = sin w t f_0(t)=\frac{1}{l}\int_0^l\sin wtdx=\sin wt f0(t)=l1∫0lsinwtdx=sinwt
- f n ( t ) = 2 l ∫ 0 l sin w t cos n π l x d x = 0 f_n(t)=\frac{2}{l}\int_0^l\sin wt\cos\frac{n\pi}{l}xdx=0 fn(t)=l2∫0lsinwtcoslnπxdx=0
- sin w t = f 0 + ∑ n = 1 ∞ f n ( t ) cos n π l x \sin wt=f_0+\sum_{n=1}^\infty f_n(t)\cos\frac{n\pi}{l}x sinwt=f0+∑n=1∞fn(t)coslnπx
- 代入非齐次方程得
- 得常微分方程问题 { u n ′ ( t ) + ( n π a l ) 2 u n ( t ) = f n ( t ) u n ( 0 ) = 0 \left\{\begin{array}{l}u_{n}^{\prime}(t)+\left(\frac{n \pi a}{l}\right)^{2} u_{n}(t)=f_{n}(t) \\u_{n}(0)=0\end{array}\right. {
un′(t)+(lnπa)2un(t)=fn(t)un(0)=0
- n = 0 n=0 n=0得 { u 0 ′ ( t ) = sin w t u 0 ( 0 ) = 0 \left\{\begin{array}{l}\boldsymbol{u}_{0}^{\prime}(\boldsymbol{t})=\sin \boldsymbol{w} \boldsymbol{t} \\\boldsymbol{u}_{0}(0)=0\end{array}\right. {
u0′(t)=sinwtu0(0)=0
- u 0 ( t ) = − 1 w cos w t + C u 0 ( 0 ) = 0 } ⇒ u 0 ( t ) = 1 w ( 1 − cos w t ) \left.\begin{array}{c}u_{0}(t)=-\frac{1}{w} \cos w t+C \\u_{0}(0)=0\end{array}\right\} \Rightarrow u_{0}(t)=\frac{1}{w}(1-\cos w t) u0(t)=−w1coswt+Cu0(0)=0}⇒u0(t)=w1(1−coswt)
- n ≠ 0 n\neq 0 n=0得 { u n ′ ( t ) + ( n π a l ) 2 u n ( t ) = 0 u n ( 0 ) = 0 \left\{\begin{array}{l}u_{n}^{\prime}(t)+\left(\frac{n \pi a}{l}\right)^{2} u_{n}(t)=0 \\u_{n}(0)=0\end{array}\right. {
un′(t)+(lnπa)2un(t)=0un(0)=0
- u n ( t ) = C e − a 2 n 2 π 2 l 2 t u n ( 0 ) = 0 } ⇒ u n ( t ) ≡ 0 \left.\begin{array}{l}u_{n}(t)=C e^{-a^{2} \frac{n^{2} \pi^{2}}{l^{2}} t} \\u_{n}(0)=0\end{array}\right\} \Rightarrow u_{n}(t) \equiv 0 un(t)=Ce−a2l2n2π2tun(0)=0}⇒un(t)≡0
- n = 0 n=0 n=0得 { u 0 ′ ( t ) = sin w t u 0 ( 0 ) = 0 \left\{\begin{array}{l}\boldsymbol{u}_{0}^{\prime}(\boldsymbol{t})=\sin \boldsymbol{w} \boldsymbol{t} \\\boldsymbol{u}_{0}(0)=0\end{array}\right. {
u0′(t)=sinwtu0(0)=0
- u ( x , t ) = ∑ n = 0 ∞ u n ( t ) cos n π l x u(x,t)=\sum_{n=0}^\infty u_n(t)\cos\frac{n\pi}{l}x u(x,t)=∑n=0∞un(t)coslnπx
- 得常微分方程问题 { u n ′ ( t ) + ( n π a l ) 2 u n ( t ) = f n ( t ) u n ( 0 ) = 0 \left\{\begin{array}{l}u_{n}^{\prime}(t)+\left(\frac{n \pi a}{l}\right)^{2} u_{n}(t)=f_{n}(t) \\u_{n}(0)=0\end{array}\right. {
un′(t)+(lnπa)2un(t)=fn(t)un(0)=0
2.5非齐次边界条件处理
2.4波方程 u ( x , t ) = v ( x , t ) + w ( x , t ) u(x,t)=v(x,t)+w(x,t) u(x,t)=v(x,t)+w(x,t)
- v在边界满足 v ( o , t ) = 0 , v ( l , t ) = 0 v(o,t)=0,v(l,t)=0 v(o,t)=0,v(l,t)=0则w在边界满足 w ( 0 , t ) = u 1 ( t ) , w ( l , t ) = u 2 ( t ) w(0,t)=u_1(t),w(l,t)=u_2(t) w(0,t)=u1(t),w(l,t)=u2(t)
- w形式 w ( x , t ) = A ( t ) x + B ( t ) w(x,t)=A(t)x+B(t) w(x,t)=A(t)x+B(t)
- 满足 { w ( x , t ) = A ( t ) x + B ( t ) w ( 0 , t ) = u 1 ( t ) , w ( l , t ) = u 2 ( t ) \left\{\begin{array}{l}w(x, t)=A(t) x+B(t) \\w(0, t)=u_{1}(t), w(l, t)=u_{2}(t)\end{array}\right. { w(x,t)=A(t)x+B(t)w(0,t)=u1(t),w(l,t)=u2(t)
- 解得 { A ( t ) = u 2 ( t ) − u 1 ( t ) l B ( t ) = u 1 ( t ) \left\{\begin{array}{l}A(t)=\frac{u_2(t)-u_1(t)}{l} \\B(t)=u_1(t)\end{array}\right. { A(t)=lu2(t)−u1(t)B(t)=u1(t)
- ∴ w ( x , t ) = u 2 ( t ) − u 1 ( t ) l x + u 1 ( t ) \therefore w(x,t)=\frac{u_2(t)-u_1(t)}{l}x+u_1(t) ∴w(x,t)=lu2(t)−u1(t)x+u1(t)
- v
- v ( x , t ) v(x,t) v(x,t)满足 { ∂ 2 v ∂ t 2 = a 2 ∂ 2 v ∂ x 2 + f 1 ( x , t ) v ∣ x = 0 = 0 , v ∣ x = l = 0 v ∣ t = 0 = ϕ 1 ( x ) , ∂ v ∂ t ∣ t = 0 = ψ 1 ( x ) \left\{\begin{array}{l}\frac{\partial^{2} \boldsymbol{v}}{\partial \boldsymbol{t}^{2}}=\boldsymbol{a}^{2} \frac{\partial^{2} \boldsymbol{v}}{\partial \boldsymbol{x}^{2}}+\boldsymbol{f}_{1}(\boldsymbol{x}, \boldsymbol{t}) \\\left.\boldsymbol{v}\right|_{x=0}=0,\left.\quad \boldsymbol{v}\right|_{x=l}=0 \\\left.\boldsymbol{v}\right|_{t=0}=\phi_{1}(\boldsymbol{x}),\left. \frac{\partial \boldsymbol{v}}{\partial \boldsymbol{t}}\right|_{t=0}=\psi_{1}(\boldsymbol{x})\end{array}\right. ⎩ ⎨ ⎧∂t2∂2v=a2∂x2∂2v+f1(x,t)v∣x=0=0,v∣x=l=0v∣t=0=ϕ1(x),∂t∂v∣ ∣t=0=ψ1(x)其中 f 1 ( x , t ) = f ( x , t ) − u 2 ′ ′ ( t ) − u 1 ′ ′ ( t ) l x − u 1 ′ ′ ( t ) ϕ 1 ( x ) = ϕ ( x ) − u 1 ( 0 ) − u 2 ( 0 ) − u 1 ( 0 ) l x ψ 1 ( x ) = ψ ( x ) − u 1 ′ ( 0 ) − u 2 ′ ( 0 ) − u 1 ′ ( 0 ) l x \begin{array}{l} f_{1}(\boldsymbol{x}, \boldsymbol{t})=\boldsymbol{f}(\boldsymbol{x}, \boldsymbol{t})-\frac{\boldsymbol{u}_{2}^{\prime \prime}(\boldsymbol{t})-\boldsymbol{u}_{1}^{\prime \prime}(\boldsymbol{t})}{\boldsymbol{l}} \boldsymbol{x}-\boldsymbol{u}_{1}^{\prime \prime}(\boldsymbol{t}) \\ \phi_{1}(\boldsymbol{x})=\phi(\boldsymbol{x})-\boldsymbol{u}_{1}(0)-\frac{\boldsymbol{u}_{2}(0)-\boldsymbol{u}_{1}(0)}{\boldsymbol{l}} \boldsymbol{x} \\ \psi_{1}(\boldsymbol{x})=\psi(\boldsymbol{x})-\boldsymbol{u}_{1}^{\prime}(0)-\frac{\boldsymbol{u}_{2}^{\prime}(0)-\boldsymbol{u}_{1}^{\prime}(0)}{\boldsymbol{l}} \boldsymbol{x} \end{array} f1(x,t)=f(x,t)−lu2′′(t)−u1′′(t)x−u1′′(t)ϕ1(x)=ϕ(x)−u1(0)−lu2(0)−u1(0)xψ1(x)=ψ(x)−u1′(0)−lu2′(0)−u1′(0)x