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 $y^{\prime }(x)=P(x)y(x)$ 分离变量,两边积分
  • $y(x)=C{e}^{\int P(x)dx}$
  • P(x)为常数 $y^{\prime }(x)=my(x)\to y(x)=C{e}^{mx}$
• 非齐次方程 $y^{\prime }(x)=P(x)y(x)+Q(x)$ 常数变异法
  • $y(x)=e^{\int P(x)dx}(\int Q(x)\cdot e^{-\int P(x)dx}dx+C)$

2.0.2二阶常系数齐次常微分方程

$a_2{y}^{\prime \prime }(x)+a_1{y}^{\prime }(x)+a_{0}y(x)=0$

• 特征方程: $a_2{r}^2+a_{1}r+a_0=0$, r特征根

• 2实根 $r_1\ne r_2$: 通解 $y(x)=A{e}^{r_{1}x}+B{e}^{r_{2}x}$

• 1实根 $r_1=r_2$: 通解 $y(x)=(Ax+B)e^{rx}$

• 复根 $r_1=\alpha +\beta i,r_2=\alpha -\beta i$: 通解 $y(x)=e^{\alpha x}(A\cos \beta x+B\sin \beta x)$
  

• ${\rho }^2{R}^{\prime \prime }(\rho )+\rho R^{\prime }(\rho )-\lambda R(\rho )=0$

  • $\rho =e^x$

2.1特征值问题

2.1.1常用特征值问题

• 常微分方程$X^{\prime \prime }(x)+\lambda X(x)=0$

  • $\lambda <0$, $X(x)=A{e}^{\sqrt{-\lambda }x}+B{e}^{-\sqrt{-\lambda }x}$
  • $\lambda =0$, $X(x)=Ax+B$
  • $\lambda >0$, $X(x)=A\cos \sqrt{\lambda }x+B\sin \sqrt{\lambda }x$

• 含待确定常数$\lambda$的特征值问题, $\lambda$特征值, $X(x)$特征函数

• { X ′ ′ ( x ) + λ X ( x ) = 0 X ( 0 ) = 0 , X ( l ) = 0 \left\{

  X″(x)+λX(x)=0X(0)=0,X(l)=0" role="presentation">X′′(x)+λX(x)=0X(0)=0,X(l)=0$$\begin{array}{c}{X}^{″}(x)+\lambda X(x)=0\\ X(0)=0,X(l)=0\end{array}$$
  \right. {X′′(x)+λX(x)=0X(0)=0,X(l)=0​
  1. $\lambda <0$, A=B=0, 平凡解
  2. $\lambda =0$, A=B=0, 平凡解
  3. $\lambda >0$, { λ n = ( n π l ) 2 X n ( x ) = B n sin ⁡ n π l x , n = 1 , 2 , 3... \left\{
     λn=(nπl)2Xn(x)=Bnsin⁡nπlx,n=1,2,3..." role="presentation" style="position: relative;">λn=(nπl)2Xn(x)=Bnsinnπlx,n=1,2,3...$$\begin{array}{c}{\lambda }_n=(\frac{n\pi }{l}{)}^2\\ X_n(x)=B_n\sin \frac{n\pi }{l}x,n=1,2,\mathrm{3...}\end{array}$$
     \right. {λn​=(lnπ​)2Xn​(x)=Bn​sinlnπ​x,n=1,2,3...​
     1. ${\sin \frac{n\pi }{l}x}_{n=1}^{\infty }$
     2. $C_n=\frac{2}{l}{\int }_0^{l}f(x)\sin \frac{n\pi }{l}xdx$

2.1.2关于特征值问题的理论

• S-L方程 $\frac{d}{dx}(k(x)\frac{dy}{dx})-q(x)y(x)+\lambda \rho (x)y(x)=0,x\in (a,b)$
  • 特征值:其次边界条件,周期性,自然边界条件
  • $k(x)=\rho (x)=1,q(x)=0$: $y^{\prime \prime }(x)+\lambda y(x)=0$
  • $k(x)=\rho (x)=x,q(x)=\frac{n^2}{x}$: $x^2{y}^{\prime \prime }(x)+x{y}^{\prime }(x)+(\lambda x^2-n^2)y(x)=0$
  • $k(x)=1-x^2,\rho (x)=1,q(x)=0$: $(1-x{)}^2{y}^{\prime \prime }(x)-2x{y}^{\prime }(x)+\lambda y(x)=0$
  1. 存在可数个实特征值->单调递增序列$0\le {\lambda }_1\le {\lambda }_2\le ...\le {\lambda }_n\le$
     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...
  2. 所有特征值非负 ${\lambda }_n\ge 0,n=1,2,\mathrm{3...}$
  3. 特征函数系 { y n ( x ) } n = 1 n = ∞ \{y_n(x)\}_{n=1}^{n=\infty} {yn​(x)}n=1n=∞​是$L_{\rho }^2[a,b]$上关于权函数$\rho (x)$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\{
     0n≠m||yn||22n=m" role="presentation" style="position: relative;">0||yn||22n≠mn=m$$\begin{array}{cc}0& n\ne m\\ ||y_n|{|}_2^2& n=m\end{array}$$
     \right. ∫ab​ρ(x)yn​(x)ym​(x)dx={0∣∣yn​∣∣22​​n=mn=m​
  4. 特征函数系 { y n ( x ) } n = 1 n = ∞ \{y_n(x)\}_{n=1}^{n=\infty} {yn​(x)}n=1n=∞​是$L_{\rho }^2[a,b]$上关于权函数$\rho (x)$的完备系

2.1.3Matlab

• 二阶常微分方程
  求$\frac{d^{2}y}{d{x}^2}+y=1-\frac{x^2}{\pi }$通解

  • y=dsolve('D2y+y=1-x^2/pi','x')
    1

  求 { d 2 y d x 2 + y = 1 − x 2 π y ( 0 ) = 0.2 , y ′ ( 0 ) = 0.5 \left\{
  d2ydx2+y=1−x2πy(0)=0.2,y′(0)=0.5" role="presentation" style="position: relative;">d2ydx2+y=1−x2πy(0)=0.2,y′(0)=0.5$$\begin{array}{l}\frac{d^{2}y}{d{x}^2}+y=1-\frac{x^2}{\pi }\\ y(0)=0.2,y^{\mathrm{\prime }}(0)=0.5\end{array}$$
  \right. {dx2d2y​+y=1−πx2​y(0)=0.2,y′(0)=0.5​特解,作图

  • 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])
    1
    2

• 常微分方程组 { d u d t = 3 u − 2 v d v d t + v = 2 u \left\{
  dudt=3u−2vdvdt+v=2u" role="presentation" style="position: relative;">dudt=3u−2vdvdt+v=2u$$\begin{array}{l}\frac{du}{dt}=3u-2v\\ \frac{dv}{dt}+v=2u\end{array}$$
  \right. {dtdu​=3u−2vdtdv​+v=2u​
  1.求通解

  • [u,v]=dsolve('Du=3*u-2*v','Dv+v=2*u')
    1

  2.求满足初始条件$u(0)=1,v(0)=0$的特解

  • [u,v]=dsolve('Du=3*u-2*v','Dv+v=2*u','u(0)=1,v(0)=0','t')
    1

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\{

\right. ⎩ ⎨ ⎧​∂t2∂2u​=a2∂x2∂2u​0<x<l,t>0u∣x=0​=u∣x=l​=0u∣t=0​=ϕ(x),∂t∂u​∣ ∣​t=0​=ψ(x)​

• $u_n(x,t)$在任意时刻为正弦曲线
  • $t=t_0$ $u_n(x,t_0)$为正弦曲线，振幅随时间变化
  • $x=x_0$ $u_n(x_0,t)$简谐振动
• 任意确定时刻$u_n(x,t)=A_n^{\prime }\sin \frac{n\pi }{l}x$
  • $n+1$个零点,$n$个极值点,$u_1,u_2,u_3$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

1. 定解问题$\frac{{\partial }^2\mathbf{u}}{\partial {\mathbf{x}}^2}+\frac{{\partial }^2\mathbf{u}}{\partial {\mathbf{y}}^2}=0$
2. 极坐标变换 { x = ρ cos ⁡ θ y = ρ sin ⁡ θ , 0 ≤ θ ≤ 2 π \left\{
   x=ρcos⁡θy=ρsin⁡θ," role="presentation" style="position: relative;">x=ρcosθy=ρsinθ,$$\begin{array}{l}x=\rho \cos \theta \\ y=\rho \sin \theta ,\end{array}$$
   \quad 0 \leq \theta \leq 2 \pi\right. {x=ρcosθy=ρsinθ,​0≤θ≤2π
3. $\frac{\partial \mathbf{u}}{\partial \mathbf{x}}=\frac{\partial \mathbf{u}}{\partial \rho }\frac{\partial \rho }{\partial \mathbf{x}}+\frac{\partial \mathbf{u}}{\partial \theta }\frac{\partial \theta }{\partial \mathbf{x}}=\mathbf{v}$代入得 $
   1ρ∂∂ρ(ρ∂u∂ρ)+1ρ2∂2u∂θ2=0frac∂2u∂ρ2+1ρ∂u∂ρ+1ρ2∂2u∂θ2=0" role="presentation" style="text-align: center; position: relative;">1ρ∂∂ρ(ρ∂u∂ρ)+1ρ2∂2u∂θ2=0frac∂2u∂ρ2+1ρ∂u∂ρ+1ρ2∂2u∂θ2=0$$\begin{array}{l}\frac{1}{\rho }\frac{\mathrm{\partial }}{\mathrm{\partial }\rho }\left(\rho \frac{\mathrm{\partial }\mathit{u}}{\mathrm{\partial }\rho }\right)+\frac{1}{{\rho }^2}\frac{{\mathrm{\partial }}^2\mathit{u}}{\mathrm{\partial }{\theta }^2}=0\\ frac{\mathrm{\partial }}^2\mathit{u}\mathrm{\partial }{\rho }^2+\frac{1}{\rho }\frac{\mathrm{\partial }\mathit{u}}{\mathrm{\partial }\rho }+\frac{1}{{\rho }^2}\frac{{\mathrm{\partial }}^2\mathit{u}}{\mathrm{\partial }{\theta }^2}=0\end{array}$$
   $

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\{

\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$u(x,t)=v(x,t)+w(x,t)$

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\{

\right. ⎩ ⎨ ⎧​∂t2∂2v​=a2∂x2∂2v​+f(x,t)v∣x=0​=v∣x=l​=0v∣t=0​=0,∂t∂v​∣ ∣​t=0​=0​

1. 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\{
   ∂2v∂t2=a2∂2v∂x2v|x=0=v|x=l=0" role="presentation" style="position: relative;">∂2v∂t2=a2∂2v∂x2v|x=0=v|x=l=0$$\begin{array}{l}\frac{{\mathrm{\partial }}^2\mathit{v}}{\mathrm{\partial }{\mathit{t}}^2}={\mathit{a}}^2\frac{{\mathrm{\partial }}^2\mathit{v}}{\mathrm{\partial }{\mathit{x}}^2}\\ {\mathit{v}|}_{\mathit{x}=0}={\mathit{v}|}_{\mathit{x}=l}=0\end{array}$$
   \right. {∂t2∂2v​=a2∂x2∂2v​v∣x=0​=v∣x=l​=0​
   1. 特征值问题 { X ′ ′ ( x ) + λ X ( x ) = 0 X ( 0 ) = X ( l ) = 0 \left\{
      X′′(x)+λX(x)=0X(0)=X(l)=0" role="presentation" style="position: relative;">X′′(x)+λX(x)=0X(0)=X(l)=0$$\begin{array}{l}{\mathit{X}}^{\mathrm{\prime }\mathrm{\prime }}(\mathit{x})+\lambda \mathit{X}(\mathit{x})=0\\ \mathit{X}(0)=\mathit{X}(\mathit{l})=0\end{array}$$
      \right. {X′′(x)+λX(x)=0X(0)=X(l)=0​
      1. 特征函数$X_n(x)=B_n\sin \frac{n\pi }{l}x,n=1,2,...$
      2. 假设非齐次方程解$v(x,t)={\sum }_{n=1}^{\infty }{v}_n(t)\sin \frac{n\pi }{l}x$
2. $f(x,t)$按特征函数序列 { sin ⁡ n π l x } n = 1 ∞ \{\sin\frac{n\pi}{l}x\}_{n=1}^{\infty} {sinlnπ​x}n=1∞​展开为级数形式
   1. $f(x,t)={\sum }_{n=1}^{\infty }{f}_n(t)\sin \frac{n\pi }{l}x$
   2. $f_n(t)=\frac{2}{l}{\int }_0^{l}f(x,t)\sin \frac{n\pi }{l}xdx$
3. 以上结果带入非齐次方程
   1. 
   2. 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\{
      vn′′(t)+(nπal)2vn(t)=fn(t)vn(0)=0,vn′(0)=0" role="presentation" style="position: relative;">v′′n(t)+(nπal)2vn(t)=fn(t)vn(0)=0,v′n(0)=0$$\begin{array}{l}{v}_n^{\mathrm{\prime }\mathrm{\prime }}(t)+{\left(\frac{n\pi a}{l}\right)}^2{v}_n(t)=f_n(t)\\ v_n(0)=0,v_n^{\mathrm{\prime }}(0)=0\end{array}$$
      \right. {vn′′​(t)+(lnπa​)2vn​(t)=fn​(t)vn​(0)=0,vn′​(0)=0​
   3. Laplace变换
      1. ${\mathbf{v}}_n(t)=\frac{l}{n\pi a}{\int }_0^t{f}_n(\tau )\sin \frac{n\pi a(t-\tau )}{l}d\tau$
      2. $\mathbf{v}(\mathbf{x},\mathbf{t})={\sum }_{n=1}^{\infty }\frac{l}{\mathbf{n}\pi a}{\int }_0^t{f}_n(\tau )\sin \frac{n\pi a(t-\tau )}{l}d\tau \sin \frac{n\pi }{l}\mathbf{x}$
         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))
1
2
3

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\{

\right. ⎩ ⎨ ⎧​∂t∂u​=a2∂x2∂2u​+sinωt0<x<l,t>0∂x∂u​∣ ∣​x=0​=∂x∂u​∣ ∣​x=l​=0u∣t=0​=0​

1. 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\{
   ∂u∂t=a2∂2u∂x2∂u∂x|x=0=∂u∂x|x=l=0" role="presentation" style="position: relative;">∂u∂t=a2∂2u∂x2∂u∂x∣∣x=0=∂u∂x∣∣x=l=0$$\begin{array}{l}\frac{\mathrm{\partial }\mathit{u}}{\mathrm{\partial }\mathit{t}}={\mathit{a}}^2\frac{{\mathrm{\partial }}^2\mathit{u}}{\mathrm{\partial }{\mathit{x}}^2}\\ {\frac{\mathrm{\partial }\mathit{u}}{\mathrm{\partial }\mathit{x}}|}_{x=0}={\frac{\mathrm{\partial }\mathit{u}}{\mathrm{\partial }\mathit{x}}|}_{x=l}=0\end{array}$$
   \right. {∂t∂u​=a2∂x2∂2u​∂x∂u​∣ ∣​x=0​=∂x∂u​∣ ∣​x=l​=0​
   1. 得特征值问题 { X ′ ′ ( x ) + λ X ( x ) = 0 X ′ ( 0 ) = X ′ ( l ) = 0 \left\{
      X′′(x)+λX(x)=0X′(0)=X′(l)=0" role="presentation" style="position: relative;">X′′(x)+λX(x)=0X′(0)=X′(l)=0$$\begin{array}{l}{\mathit{X}}^{\mathrm{\prime }\mathrm{\prime }}(\mathit{x})+\lambda \mathit{X}(\mathit{x})=0\\ {\mathit{X}}^{\mathrm{\prime }}(0)={\mathit{X}}^{\mathrm{\prime }}(\mathit{l})=0\end{array}$$
      \right. {X′′(x)+λX(x)=0X′(0)=X′(l)=0​
   2. 特征函数$X_n(x)=A_n\cos \frac{n\pi }{l}x,n=0,1,2,...$
   3. 设非齐次方程解$u(x,t)={\sum }_{n=1}^{\infty }{u}_n(t)\cos \frac{n\pi }{l}x$
2. $\sin wt$按特征函数序列 { cos ⁡ n π l x } n = 0 ∞ \{\cos\frac{n\pi}{l}x\}_{n=0}^\infty {coslnπ​x}n=0∞​展开为级数形式
   1. $\sin wt=f_0+{\sum }_{n=1}^{\infty }{f}_n(t)\cos \frac{n\pi }{l}x$
      1. $f_0(t)=\frac{1}{l}{\int }_0^l\sin wtdx=\sin wt$
      2. $f_n(t)=\frac{2}{l}{\int }_0^l\sin wt\cos \frac{n\pi }{l}xdx=0$
3. 代入非齐次方程得
   1. 
   2. 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\{
      un′(t)+(nπal)2un(t)=fn(t)un(0)=0" role="presentation" style="position: relative;">u′n(t)+(nπal)2un(t)=fn(t)un(0)=0$$\begin{array}{l}{u}_n^{\mathrm{\prime }}(t)+{\left(\frac{n\pi a}{l}\right)}^2{u}_n(t)=f_n(t)\\ u_n(0)=0\end{array}$$
      \right. {un′​(t)+(lnπa​)2un​(t)=fn​(t)un​(0)=0​
      1. $n=0$得 { u 0 ′ ( t ) = sin ⁡ w t u 0 ( 0 ) = 0 \left\{
         u0′(t)=sin⁡wtu0(0)=0" role="presentation" style="position: relative;">u′0(t)=sinwtu0(0)=0$$\begin{array}{l}{\mathit{u}}_0^{\mathrm{\prime }}(\mathit{t})=\sin \mathit{w}\mathit{t}\\ {\mathit{u}}_0(0)=0\end{array}$$
         \right. {u0′​(t)=sinwtu0​(0)=0​
         1. u 0 ( t ) = − 1 w cos ⁡ w t + C u 0 ( 0 ) = 0 } ⇒ u 0 ( t ) = 1 w ( 1 − cos ⁡ w t ) \left.
            u0(t)=−1wcos⁡wt+Cu0(0)=0" role="presentation" style="position: relative;">u0(t)=−1wcoswt+Cu0(0)=0$$\begin{array}{c}{u}_0(t)=-\frac{1}{w}\cos wt+C\\ u_0(0)=0\end{array}$$
            \right\} \Rightarrow u_{0}(t)=\frac{1}{w}(1-\cos w t) u0​(t)=−w1​coswt+Cu0​(0)=0​}⇒u0​(t)=w1​(1−coswt)
      2. $n\ne 0$得 { u n ′ ( t ) + ( n π a l ) 2 u n ( t ) = 0 u n ( 0 ) = 0 \left\{
         un′(t)+(nπal)2un(t)=0un(0)=0" role="presentation" style="position: relative;">u′n(t)+(nπal)2un(t)=0un(0)=0$$\begin{array}{l}{u}_n^{\mathrm{\prime }}(t)+{\left(\frac{n\pi a}{l}\right)}^2{u}_n(t)=0\\ u_n(0)=0\end{array}$$
         \right. {un′​(t)+(lnπa​)2un​(t)=0un​(0)=0​
         1. u n ( t ) = C e − a 2 n 2 π 2 l 2 t u n ( 0 ) = 0 } ⇒ u n ( t ) ≡ 0 \left.
            un(t)=Ce−a2n2π2l2tun(0)=0" role="presentation" style="position: relative;">un(t)=Ce−a2n2π2l2tun(0)=0$$\begin{array}{l}{u}_n(t)=C{e}^{-a^2\frac{n^2{\pi }^2}{l^2}t}\\ u_n(0)=0\end{array}$$
            \right\} \Rightarrow u_{n}(t) \equiv 0 un​(t)=Ce−a2l2n2π2​tun​(0)=0​}⇒un​(t)≡0
   3. $u(x,t)={\sum }_{n=0}^{\infty }{u}_n(t)\cos \frac{n\pi }{l}x$

2.5非齐次边界条件处理

2.4波方程$u(x,t)=v(x,t)+w(x,t)$

1. v在边界满足$v(o,t)=0,v(l,t)=0$则w在边界满足$w(0,t)=u_1(t),w(l,t)=u_2(t)$
2. w形式$w(x,t)=A(t)x+B(t)$
   1. 满足 { w ( x , t ) = A ( t ) x + B ( t ) w ( 0 , t ) = u 1 ( t ) , w ( l , t ) = u 2 ( t ) \left\{
      w(x,t)=A(t)x+B(t)w(0,t)=u1(t),w(l,t)=u2(t)" role="presentation" style="position: relative;">w(x,t)=A(t)x+B(t)w(0,t)=u1(t),w(l,t)=u2(t)$$\begin{array}{l}w(x,t)=A(t)x+B(t)\\ w(0,t)=u_1(t),w(l,t)=u_2(t)\end{array}$$
      \right. {w(x,t)=A(t)x+B(t)w(0,t)=u1​(t),w(l,t)=u2​(t)​
   2. 解得 { A ( t ) = u 2 ( t ) − u 1 ( t ) l B ( t ) = u 1 ( t ) \left\{
      A(t)=u2(t)−u1(t)lB(t)=u1(t)" role="presentation" style="position: relative;">A(t)=u2(t)−u1(t)lB(t)=u1(t)$$\begin{array}{l}A(t)=\frac{u_2(t)-u_1(t)}{l}\\ B(t)=u_1(t)\end{array}$$
      \right. {A(t)=lu2​(t)−u1​(t)​B(t)=u1​(t)​
   3. $\therefore w(x,t)=\frac{u_2(t)-u_1(t)}{l}x+u_1(t)$
3. v
   1. $v(x,t)$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\{
      ∂2v∂t2=a2∂2v∂x2+f1(x,t)v|x=0=0,v|x=l=0v|t=0=ϕ1(x),∂v∂t|t=0=ψ1(x)" role="presentation" style="position: relative;">∂2v∂t2=a2∂2v∂x2+f1(x,t)v|x=0=0,v|x=l=0v|t=0=ϕ1(x),∂v∂t∣∣t=0=ψ1(x)$$\begin{array}{l}\frac{{\mathrm{\partial }}^2\mathit{v}}{\mathrm{\partial }{\mathit{t}}^2}={\mathit{a}}^2\frac{{\mathrm{\partial }}^2\mathit{v}}{\mathrm{\partial }{\mathit{x}}^2}+{\mathit{f}}_1(\mathit{x},\mathit{t})\\ {\mathit{v}|}_{x=0}=0,{\mathit{v}|}_{x=l}=0\\ {\mathit{v}|}_{t=0}={\varphi }_1(\mathit{x}),{\frac{\mathrm{\partial }\mathit{v}}{\mathrm{\partial }\mathit{t}}|}_{t=0}={\psi }_1(\mathit{x})\end{array}$$
      \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
      f1(x,t)=f(x,t)−u2′′(t)−u1′′(t)lx−u1′′(t)ϕ1(x)=ϕ(x)−u1(0)−u2(0)−u1(0)lxψ1(x)=ψ(x)−u1′(0)−u2′(0)−u1′(0)lx" role="presentation" style="position: relative;">f1(x,t)=f(x,t)−u′′2(t)−u′′1(t)lx−u′′1(t)ϕ1(x)=ϕ(x)−u1(0)−u2(0)−u1(0)lxψ1(x)=ψ(x)−u′1(0)−u′2(0)−u′1(0)lx$$\begin{array}{l}{f}_1(\mathit{x},\mathit{t})=\mathit{f}(\mathit{x},\mathit{t})-\frac{{\mathit{u}}_2^{\mathrm{\prime }\mathrm{\prime }}(\mathit{t})-{\mathit{u}}_1^{\mathrm{\prime }\mathrm{\prime }}(\mathit{t})}{\mathit{l}}\mathit{x}-{\mathit{u}}_1^{\mathrm{\prime }\mathrm{\prime }}(\mathit{t})\\ {\varphi }_1(\mathit{x})=\varphi (\mathit{x})-{\mathit{u}}_1(0)-\frac{{\mathit{u}}_2(0)-{\mathit{u}}_1(0)}{\mathit{l}}\mathit{x}\\ {\psi }_1(\mathit{x})=\psi (\mathit{x})-{\mathit{u}}_1^{\mathrm{\prime }}(0)-\frac{{\mathit{u}}_2^{\mathrm{\prime }}(0)-{\mathit{u}}_1^{\mathrm{\prime }}(0)}{\mathit{l}}\mathit{x}\end{array}$$
      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​