Article directory
- 2 Separation of variables
2 Separation of variables
2.0 Ordinary Differential Equations
2.0.1 Homogeneous & Inhomogeneous Equations
- Homogeneous equation y ′ ( x ) = P ( x ) y ( x ) y'(x)=P(x)y(x)Y′(x)=P ( x ) y ( x ) separate variables and integrate on both sides
- y ( x ) = C e ∫ P ( x ) d x y(x)=Ce^{\int P(x)dx} y ( x )=This _∫P(x)dx
- P(x) is constant y ′ ( x ) = my ( x ) → y ( x ) = C emx y'(x)=my(x)\rightarrow y(x)=Ce^{mx}Y′(x)=my(x)→y ( x )=This _mx
- Inhomogeneous equation y ′ ( x ) = P ( x ) y ( x ) + Q ( x ) y'(x)=P(x)y(x)+Q(x)Y′(x)=P(x)y(x)+Q ( x ) constant variation method
- 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 )=and∫P(x)dx(∫Q(x)⋅and−∫P(x)dxdx+C)
2.0.2 Second-order homogeneous ordinary differential equations with constant coefficients
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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Special Operation: a 2 r 2 + a 1 r + a 0 = 0 a_2r^2+a_1r+a_0=0a2r2+a1r+a0=0 , r eigenroot
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2 real roots r 1 ≠ r 2 r_1\neq r_2r1=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 )=But your1x+Ber2x
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1 real root r 1 = r 2 r_1=r_2r1=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 ir1=a+β i ,r2=a−β i : 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 )=andαx(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) =0r2 R" (p)+ρR′ (p)−λ R ( ρ )=0
- ρ = ex \ rho = e ^ xr=andx
2.1 Eigenvalue problem
2.1.1 Common eigenvalue problems
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Ordinary Differential Equation X ′ ′ ( x ) + λ X ( x ) = 0 X''(x)+\lambda X(x)=0X′′(x)+λX(x)=0
- λ < 0 \lambda<0 l<0, X ( x ) = A e − λ x + B e − − λ x X(x)=Ae^{\sqrt{-\lambda}x}+Be^{-\sqrt{-\lambda}x} X(x)=But you− lx+Be−− lx
- λ = 0 \lambda=0l=0, X ( x ) = A x + B X(x)=Ax+B X(x)=Ax+B
- λ > 0 \lambda>0l>0 ,X ( x ) = A cos λ x + B sin λ x X(x)=A\cos\sqrt{\lambda}x+B\sin\sqrt{\lambda}xX(x)=Acoslx+Bsinlx
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contains the constant λ \lambda to be determinedThe eigenvalue problem of λ, λ \lambdaλ eigenvalue,X ( x ) X(x)X ( x ) characteristic function
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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 l<0 , A=B=0, trivial solution
- λ = 0 \lambda=0l=0 , A=B=0, trivial solution
- λ > 0 \lambda>0l>0 ,{ λ n = ( n π l ) 2 X n ( x ) = B n sin n π lx , 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.{
ln=(lnπ)2Xn(x)=Bnsinlnπx,n=1 ,2 ,3 ...
- sin n π lxn = 1 ∞ {\sin\frac{n\pi}{l}x}_{n=1}^\inftysinlnπxn=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)sinlnπx d x
2.1.2 Theory on the Eigenvalue Problem
- SLequationddx ( k ( x ) dydx ) − 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)dxd y)−q(x)y(x)+λρ(x)y(x)=0 ,x∈(a,b)
- Eigenvalues: Minor Boundary Condition, Periodicity, Natural Boundary Condition
- k ( x ) = ρ ( x ) = 1 , q ( x ) = 0 k(x)=\rho(x)=1,q(x)=0k(x)=p ( 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)=p ( 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 x2 y′′(x)+xy′(x)+(λx2−n2 )and(x)=0
- k ( x ) = 1 − x 2 , ρ ( x ) = 1 , q ( x ) = 0 k(x)=1-x^2,\rho(x)=1,q(x)=0k(x)=1−x2 ,p ( 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)2 y′′(x)−2xy′(x)+λy(x)=0
- There are countable real eigenvalues -> monotonically increasing sequence 0 ≤ λ 1 ≤ λ 2 ≤ . . . ≤ λ n ≤ {0\leq\lambda_1\leq\lambda_2\leq...\leq\lambda_n\leq}0≤l1≤l2≤...≤ln≤
countable eigenfunctions{ yx ( x ) } , n = 1 , 2 , 3... \{y_x(x)\}, n=1,2,3...{ andx(x)},n=1 ,2 ,3... - All eigenvalues are non-negative λ n ≥ 0 , n = 1 , 2 , 3... \lambda_n\geq 0, n=1,2,3...ln≥0 ,n=1 ,2 ,3...
- Characteristic function system { yn ( x ) } n = 1 n = ∞ \{y_n(x)\}_{n=1}^{n=\infty}{ andn(x)}n=1n=∞是L ρ 2 [ a , b ] L_\rho^2[a,b]Lr2[a,b ] on the weight functionρ ( 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. ∫abp ( x ) yn( x ) ym(x)dx={ 0∣∣ yn∣ ∣22n=mn=m
- Characteristic function system { yn ( x ) } n = 1 n = ∞ \{y_n(x)\}_{n=1}^{n=\infty}{ andn(x)}n=1n=∞是L ρ 2 [ a , b ] L_\rho^2[a,b]Lr2[a,b ] on the weight functionρ ( x ) \rho(x)The complete system of ρ ( x )
2.1.3Matlab
- Second Order Ordinary Differential Equation
Find d 2 ydx 2 + y = 1 − x 2 π \frac{d^{2} y}{dx^{2}}+y=1-\frac{x^{2}}{ \pi}dx2d2 y+Y=1−Pix2General solution-
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. {
dtof _=3u−2 vdtd v+in=2 and
1. Seek a general solution-
[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 One-dimensional wave equation, one-dimensional heat equation solution
2.2.1 One-dimensional wave equation
{ ∂ 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∂2 and=a2∂ x2∂2 and0<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) inn(x,t ) is sinusoidal at any time
- t = t 0 t=t_0 t=t0 u n ( x , t 0 ) u_n(x,t_0) inn(x,t0) is a sinusoid whose amplitude varies with time
- x = x 0 x=x_0 x=x0 u n ( x 0 , t ) u_n(x_0,t) inn(x0,t ) Simple harmonic vibration
- Arbitrarily determined time un ( x , t ) = A n ′ sin n π lx u_{n}(x, t)=A_{n}^{\prime} \sin \frac{n \pi}{l} xinn(x,t)=An′sinlnπx
- n+1 n+1n+1 zero,nnn个极值点,u 1 , u 2 , u 3 u_1,u_2,u_3in1,in2,in3is a series of standing waves
2.2.2 One-dimensional heat equation solution
2.3 The solution method of Laplace equation definite solution problem
2.3.1 Laplace in Cartesian Coordinate System
2.3.2 The solution of Laplace definite solution problem in two-dimensional circular domain
- 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∂2 and+∂ y2∂2 and=0
- Rationalization { x = ρ cos θ y = ρ sin θ , 0 ≤ θ ≤ 2 π \left\{\begin{array}{l}x=\rho \cos \theta \\y=\rho\; without \theta,\end{array}\quad 0 \leq \theta \leq 2 \pi\right.{ x=rcosiY=rsini ,0≤i≤2 p.m
- ∂ 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 Inhomogeneous equations
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∂2 and=a2∂ x2∂2 and+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 Eigenfunction method
wave equation
{ ∂ 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∂2 v=a2∂ x2∂2 v+f(x,t)v ∣x=0=v ∣x=l=0v ∣t=0=0 ,∂ t∂ v∣ ∣t=0=0
- Homogeneous equation + homogeneous boundary condition { ∂ 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∂2 v=a2∂ x2∂2 vv ∣x=0=v ∣x=l=0
- Eigenvalue problem { 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
- Characteristic function X n ( x ) = B n sin n π lx , n = 1 , 2 , . . . X_n(x)=B_n\sin\frac{n\pi}{l}x,n=1,2 ,...Xn(x)=Bnsinlnπx,n=1 ,2 ,...
- Suppose the inhomogeneous equation solution v ( x , t ) = ∑ n = 1 ∞ vn ( t ) sin n π lxv(x,t)=\sum_{n=1}^{\infty}v_n(t)\ sin \frac{n\pi}{l}xv(x,t)=∑n=1∞inn(t)sinlnπx
- Eigenvalue problem { 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 ) according to the sequence ofeigenfunctions { sin n π lx } n = 1 ∞ \{\sin\frac{n\pi}{l}x\}_{n=1}^{\infty}{
sinlnπx}n=1∞Expand into series form
- 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)sinlnπx d x
- The above results are brought into the inhomogeneous equation
- Get the ordinary differential equation problem { vn ′ ′ ( t ) + ( n π al ) 2 vn ( t ) = fn ( t ) vn ( 0 ) = 0 , vn ′ ( 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.{ inn′ ′(t)+(lnπa)2inn(t)=fn(t)inn( 0 )=0 ,inn′( 0 )=0
- Laplace Transform
- vn ( t ) = ln π a ∫ 0 tfn ( τ ) sin n π a ( t − τ ) ld 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 \tauinn(t)=nπal∫0tfn( t )sinlnpa ( t − τ )dτ
- v ( x , t ) = ∑ n = 1 ∞ ln π a ∫ 0 tfn ( τ ) sin n π a ( t − τ ) ld τ sin n π lx \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∞n πal∫0tfn( t )sinlnpa ( t − τ )dτsinlnπx
Matlab Solving Second Order Ordinary Differential Equations
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))
heat equation
{ ∂ 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∂2 and+sint _0<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∂2 and∂ x∂ u∣
∣x=0=∂ x∂ u∣
∣x=l=0
- Eigenvalue problem { 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
- Characteristic function X n ( x ) = A n cos n π lx , 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 ,...
- Let inhomogeneous equation solution u ( x , t ) = ∑ n = 1 ∞ un ( t ) cos n π lxu(x,t)=\sum_{n=1}^\infty u_n(t)\cos\ frac{n\pi}{l}xu(x,t)=∑n=1∞inn(t)coslnπx
- sin w t \sin wt sinwt according to the sequence ofeigenfunctions { cos n π lx } n = 0 ∞ \{\cos\frac{n\pi}{l}x\}_{n=0}^\infty{
coslnπx}n=0∞Expand into series form
- sin wt = f 0 + ∑ n = 1 ∞ fn ( t ) cos n π lx \sin wt=f_0+\sum_{n=1}^\infty f_n(t)\cos\frac{n\pi}{ l}xsinwt=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∫0lsinwtcoslnπx d x=0
- sin wt = f 0 + ∑ n = 1 ∞ fn ( t ) cos n π lx \sin wt=f_0+\sum_{n=1}^\infty f_n(t)\cos\frac{n\pi}{ l}xsinwt=f0+∑n=1∞fn(t)coslnπx
- Substitute into the inhomogeneous equation to get
- 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. {
inn′(t)+(lnπa)2inn(t)=fn(t)inn( 0 )=0
- n=0 n=0n=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. {
in0′(t)=sinwtin0( 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) in0(t)=−in1coswt+Cin0( 0 )=0}⇒in0(t)=in1( 1−coswt)
- n ≠ 0 n\neq 0n=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. {
inn′(t)+(lnπa)2inn(t)=0inn( 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 inn(t)=This _−a2l2n2 p.m2tinn( 0 )=0}⇒inn(t)≡0
- n=0 n=0n=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. {
in0′(t)=sinwtin0( 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∞inn(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. {
inn′(t)+(lnπa)2inn(t)=fn(t)inn( 0 )=0
2.5 Treatment of inhomogeneous boundary conditions
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 at the boundary satisfies v ( o , t ) = 0 , v ( l , t ) = 0 v(o,t)=0,v(l,t)=0v ( o ,t)=0 ,v(l,t)=0 then w satisfiesw ( 0 , t ) = u 1 ( t ) , w ( l , t ) = u 2 ( t ) w(0,t)=u_1(t),w(l,t)= u_2(t)in ( 0 ,t)=in1(t),w(l,t)=in2(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)in ( 0 ,t)=in1(t),w(l,t)=in2(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)=lin2(t)−u1(t)B(t)=in1(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)=lin2(t)−u1(t)x+in1(t)
- in
- 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∂2 v=a2∂ x2∂2 v+f1(x,t)v ∣x=0=0 ,v ∣x=l=0v ∣t=0=ϕ1(x),∂ t∂ v∣ ∣t=0=p1(x)in 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)−lin2′ ′(t)−u1′ ′(t)x−in1′ ′(t)ϕ1(x)=ϕ ( x )−in1( 0 )−lin2( 0 ) − u1( 0 )xp1(x)=ψ ( x )−in1′( 0 )−lin2′( 0 ) − u1′( 0 )x