---
# System prepended metadata

title: L13
tags: [微分方程特論]

---

# L13
###### tags: `微分方程特論`
###### date: `3 June 2020`
## Recitation
![](https://i.imgur.com/QuNq4pp.png)


## Review of §8.5
### Example
:film_frames: https://reurl.cc/lVOVdj 00:00
(Insulated Endpoint Conditions, p.594)

$\left\{\begin{array}{l}u_t=ku_{xx},\qquad(0<x<L, t>0);\\u_x(0,t)=u_x(L,t)=0\\u(x,0)=f(x)\end{array}\right.\tag{1}$
![](https://i.imgur.com/hNb4mGE.png)

#### Solution
1. Let $u(x,t)=X(x)T(t)$.
    Then the equation becomes $XT'=kX''T$
    $\implies\dfrac{X''}{X}=\dfrac{T'}{kT}=-\lambda$
    $\implies X''+\lambda X=0\quad$ and $\quad T'+k\lambda  T=0$.
2. Apply B.C.s:
    - $u_x(0,t)=0\implies X'(0)=0$
    - $u_x(L,t)=0\implies X'(L)=0$

3. Solve B.V.P.s:
    - $X''+\lambda X=0;\ X'(0)=0,\quad X'(L)=0\implies$
        - e-val.: $\lambda_n=\left(\dfrac{n\pi}{L}\right)^2,\quad n=0,1,2,3,\ldots$,
        - e-func.: $X_n(x)=\cos\left(\dfrac{n\pi x}{L}\right),\quad n=0,1,2,3,\ldots$.
    - If $\lambda\neq 0$, then $T_n'+k\left(\dfrac{n\pi}{L}\right)T_n=0$
        $\implies T_n(t)=C_ne^{-k(n\pi /L)^2t},\quad n=0,1,2,3,\ldots$.
    - If $\lambda=0$, then $T_0'+0T=0\implies T_0(t)\equiv \dfrac{C_0}{2}$ (constant).
    Thus, the gen. sol is $$\begin{align}u(x,t)&=\dfrac{C_0\kappa_0}{2}+\sum^\infty_{n=1}\kappa_nX_n(x)T_n(t)\\&=\boxed{ \dfrac{c_0}{2}+\sum^\infty_{n=1} c_n\cos\left(\dfrac{n\pi x}{L}\right)e^{-k(n\pi /L)^2t}},\end{align}$$ where $c_n\equiv\kappa_nC_n$ for $n=0,1,2,\ldots$.

4. Apply I.C.:
    $\displaystyle u(x,0)=f(x)=\dfrac{c_0}{2}+\sum^\infty_{n=0}c_n \cos\left(\dfrac{n\pi x}{L}\right)$ (Fourier cosine series) with coefficients being $$ \boxed{c_0=\dfrac{2}{L}\int^{L}_0 \cos\left(\dfrac{n\pi x}{L}\right)\text{d}x},$$ and $$\boxed{c_n=\dfrac{2}{L}\int^{L}_0 f(x)\cos\left(\dfrac{n\pi x}{L}\right)\text{d}x}.\ \blacksquare$$

---
### Example (extra)
:film_frames: https://reurl.cc/lVOVdj 15:33

$u_t=ku_{xx}, (0<x<50, t>0);$
$u_x(0,t)=u_x(50,t)=0,\ u(x,0)=x.$

---

## §8.7  Laplace Equation
:film_frames: https://reurl.cc/lVOVdj 16:56

e.g. $\nabla^2 u\equiv u_{xx}+u_{yy}=0\tag{2}$
- thin plate, $u$: temp. -- heat eqn. $\nabla^2 u=u_{t}$
- thin plate, **steady state**. $u_t=0$ so that $\nabla^2 u=0$


### Example (Dirichlet's Problem)
:film_frames: https://reurl.cc/lVOVdj 19:35
(p.612)

Consider a rectangular plate. The temperature on the plate $u(x,t)$ satisfies $$\left\{\begin{array}{ll}u_{xx}+u_{yy}=0,&(0<x<a,\quad 0<y<b);\\u(x,b)=u(a,y)=u(0,y)=0&(\text{3 edges of homog. cond.}),\\u(x,0)=f(x)&(\text{1 edge of non-homog. cond.}).\end{array}\right.\tag{3}$$

#### Solution
:film_frames: https://reurl.cc/lVOVdj 20:20

1. Let $u(x,y)=X(x)Y(y)$
    The PDE in Eqn. (3) becomes $X''Y+XY''=0\implies\dfrac{X''}{X}=-\dfrac{Y''}{Y}=-\lambda\implies$
    - $X''+\lambda X=0$
    - $Y''-\lambda Y=0$

2. Apply B.C.s on $y$:
    - $u(0,y)=0\implies X(0)=0$
    - $u(a,y)=0\implies X(a)=0$

3. Solve B.V.P.s:
    - $X''+\lambda X=0;\ X(0)=0,\ X(a)=0\implies$
        - e-val.: $\lambda_n=\left(\dfrac{n\pi}{a}\right)^2,\quad n=1,2,3\ldots$
        - e-func.: $X_n(x)=\sin\left(\dfrac{n\pi x}{a}\right),\quad n=1,2,3\ldots$
    - $Y_n''-\left(\dfrac{n\pi}{a}\right)^2 Y_n=0\implies Y_n(y)=C_ne^{n\pi y/a}+D_ne^{-n\pi y/a},\quad n=1,2,3\ldots$
    Thus a gen. sol of the PDE in Eqn. (3) is $$\begin{align}u(x,y)&=\sum^\infty_{n=1}\kappa_nX_n(x)Y_n(y)\\&=\boxed{\sum^\infty_{n=1}\sin\left(\dfrac{n\pi x}{a}\right)\left[A_ne^{n\pi y/a}+B_ne^{-n\pi y/a}\right]},\tag{4}\end{align}$$ where $A\equiv n=\kappa_nC_n,\ B_n\equiv \kappa_nD_n$ for $n=1,2,3,\ldots.$
    - :film_frames: https://reurl.cc/lVOVdj 33:23
        There's another solution set: *hyperbolic functions*
         $\sinh(n\pi y/a)$ and $\cosh(n\pi y/a)$
        ![](https://i.imgur.com/W3BpZoY.png =300x)


4. Apply B.C.s on $x$: 
    :film_frames: https://reurl.cc/lVOVdj 36:05
    - $\displaystyle u(x,b)=0\implies \sum^\infty_{n=1}\sin\left(\dfrac{n\pi x}{a}\right)\left[A_ne^{n\pi b/a}+B_ne^{-n\pi b/a}\right]=0$. Since $\{\sin (n\pi x/a)\}^{\infty}_{n=1}$ are orthogonal,
        $A_ne^{n\pi b/a}+B_ne^{-n\pi b/a}=0\implies \boxed{B_n=-\dfrac{e^{n\pi b/a}}{e^{-n\pi b/a}}A_n}.\tag{5}$
        ![](https://i.imgur.com/lbnQ62u.png =200x)
    - $\displaystyle u(x,0)=f(x)\implies \sum^\infty_{n=1}(A_n+B_n)\sin\dfrac{n\pi x}{a}=f(x)$
$A_n+B_n$ are the coefficients of Fourier sine series of $u$: $$\boxed{A_n+B_n=\dfrac{2}{a}\int^a_0f(x)\sin(n\pi x/a)\text{d}x}\tag{6}$$

5. Solve $A_n$ and $B_n$ with Eqs. (5) and (6). $\blacksquare$

---
### Example (an semi-infinite strip)
:film_frames: https://reurl.cc/lVOVdj 43:21

$\left\{\begin{array}{l} 
u_{xx}+u_{yy}=0,&(0<x<\infty,\quad 0<y<b);\\ 
u(x,b)=u(x,0)=0,&(0<x<\infty),\\ 
u(x,t)\text{ is bounded as }x\to+\infty,\\ u(0,y)=g(y).\end{array}\right.\tag{7}$

#### Solution
:film_frames: https://reurl.cc/kdWd3n 00:25

1. Let $u(x,y)=X(x)Y(y)$
    The PDE becomes $X''Y+XY''=0\implies-\dfrac{X''}{X}=\dfrac{Y''}{Y}=-\lambda\implies$
    - $X''-\lambda X=0$, and 
    - $Y''+\lambda Y=0$.

2. Apply B.C.s on $y$:
    - $u(x,0)=0\implies Y(0)=0$
    - $u(x,b)=0\implies Y(b)=0$

3. Solve B.V.P.s:
    - $Y''+\lambda Y=0;\ Y(0)=0,\ Y(b)=0\implies$
        - e-val.: $\lambda_n=\left(\dfrac{n\pi}{b}\right)^2,\quad n=1,2,3\ldots$
        - e-func.: $Y_n(y)=\sin\left(\dfrac{n\pi y}{b}\right),\quad n=1,2,3\ldots$

    - $X_n''-\left(\dfrac{n\pi}{b}\right)^2 X_n=0\implies X_n(x)=C_ne^{n\pi x/b}+D_ne^{-n\pi x/b}$

4. Apply B.C.s on $x$:
    :film_frames: https://reurl.cc/kdWd3n 06:02
    
    - Because $u(x,t)$ is bounded as $x\to+\infty$, $C_n=0$ for all $n$.
        Thus a gen. sol. of PDE is $$u(x,y)=\sum^\infty_{n=1}\kappa_nX_n(x)Y_n(y)=\boxed{\sum^\infty_{n=1} B_ne^{-n\pi x/b}\sin\left(\dfrac{n\pi y}{b}\right)},$$ where $B_n\equiv \kappa_nD_n$ for $n=1,2,3,\ldots.$
    - $\displaystyle u(0,y)=\sum^\infty_{n=1}B_n\sin\left(\dfrac{n\pi y}{b}\right)=g(y)$, with coefficients being $$B_n=\boxed{\dfrac{2}{b}\int^b_0 g(y)\sin\left(\dfrac{n\pi y}{b}\right)\text{d}y}.\ \blacksquare$$

---
### Laplace eqn. in polar coord.
:film_frames: https://reurl.cc/kdWd3n 11:41

Laplacian $\nabla^2u=u_{rr}+\dfrac{1}{r}u_r +\dfrac{1}{r^2}u_{\theta\theta}\tag{8}$

Wave eqn. $\alpha\left(u_{rr}+\dfrac{1}{r}u_r +\dfrac{1}{r^2}u_{\theta\theta}\right)=u_{tt}\tag*{}$

Laplace eqn. $u_{rr}+\dfrac{1}{r}u_r +\dfrac{1}{r^2}u_{\theta\theta}=0\tag{9}$
### Example
:film_frames: https://reurl.cc/kdWd3n 14:09

$\left\{\begin{array}{l}u_{rr}+\dfrac{1}{r}u_r +\dfrac{1}{r^2}u_{\theta\theta}=0,\qquad (0\leq r<p);\\u(p,\theta)=f(\theta),\text{ where}\\f(\theta) \text{ is periodic with period }2\pi.\end{array}\right.\tag{10}$
#### Solution
1. Let $u(r,\theta)=R(r)\Theta(\theta)$
    The PDE becomes $R''\Theta+\dfrac{1}{r}R'\Theta+\dfrac{1}{r^2}R\Theta''=0$
    $\implies \dfrac{r^2R''+rR'}{R}=\dfrac{-\Theta''}{\Theta}\equiv\lambda$

2. Solve the ODE of $\Theta$: $$\Theta''+\lambda\Theta=0\tag{11}$$ $\because$ $f(\theta)$ is periodic with period $2\pi$, $\therefore$ $\Theta(\theta)$ is also periodic.

| situation             | form of solution                                                                                   |
| --------------------- | ------------------------------------------------------------ |
| $\lambda=-\alpha^2<0$ | $\Theta(\theta)=c_1e^{\alpha\theta}+c_2e^{-\alpha\theta}$ is ==never periodic==.   |
| $\lambda=0$           | $\Theta(\theta)=c_1+c_2\theta$  is ==periodic only if $c_2=0$==.          |
| $\lambda=\alpha^2>0$  | $\Theta(\theta)=c_1\cos(\alpha\theta)+c_2\sin(\alpha\theta)$  is ==periodic only if $\alpha$ is an integer==. [Desmos simulation](https://www.desmos.com/calculator/lfstqcmw2v).|

Therefore,

| index $n$            | eigenvalue of <br>Eqn. (11)     | eigenfunction of <br>Eqn. (11)                                              |
| -------------- | --------------- | ---------------------------------------------------------- |
| $n=0$          | $\lambda_0=0$   | $\Theta_0(\theta)=1$                                       |
| $n=1,2,\ldots$ | $\lambda_n=n^2$ | $\Theta_n(\theta)=A_n\cos(n\theta)+B_n\sin(n\theta)$ |

3. Solve the ODE of $R$:
    :film_frames: https://reurl.cc/kdWd3n 28:50
    - For $\lambda_0=0$, then the ODE of $R$ is  $$r^2R_0''+rR_0'=0.\tag{12}$$
        >    This is a **Cauchy–Euler diff. eqn.**.
        >    :film_frames: https://reurl.cc/kdWd3n 29:41
        >    ![](https://i.imgur.com/OPQQOYi.png)
        >    Let $r=e^t,\ t=\ln r$ ...
        >    $m(m-1)+m=0\implies m=0,0$
        >    $R_0(t)=a_0e^{0t}+b_0te^{0t}$
        >    $R_0(r)=a_0+b_0\ln r$

        $\because$ $u$ is cont. at $r=0,\quad$     $\therefore\,b_0=0$ and $R_0(r)=a_0$
        Thus, a sol. is $\boxed{u_0(r,\theta)=R_0(t)\Theta_0(\theta)\equiv a_0/2}$.
    
    - :film_frames: https://reurl.cc/V6oXDn 00:00
For $\lambda_n>0\ (n=1,2,\ldots)$, then the ODE of $R$ is  $$r^2R_n''+rR_n'-n^2R_n=0.\tag{13}$$ 
        >    **Cauchy–Euler diff. eqn.**
        >    Let $r=e^t,\ t=\ln r$ ...
        >    $m(m-1)+m-n^2=0\implies m=\pm n$
        >    $R_n(t)=a_{n}e^{nt}+b_ne^{-nt}$
        >    $R_n(r)=a_nr^n+b_nr^{-n}$
    
        $\because$ $u$ is cont. at $r=0,\quad$ $\therefore\, b_n=0$ and $R_n(r)=a_nr^n$ for  $n=1,2,\ldots$
         Thus, a sol. is $\boxed{u_n(r,\theta)=R_n(r)\Theta_n(\theta)=a_nr^n[A_n\cos(n\theta)+B_n\sin(n\theta)]}.$
    
    The gen. sol. is $$\boxed{u(r,\theta)=a_0+\sum^\infty_{n=1}r^n[C_{1n}\cos(n\theta)+C_{2n}\sin(n\theta)]}.$$

4. Apply B.C.
    - $\displaystyle u(p,\theta)=f(\theta)=\dfrac{1}{2}a_0+\sum^\infty_{n=1}p^n[C_{1n}\cos(n\theta)+C_{2n}\sin(n\theta)]$
        - $\displaystyle \boxed{a_0=\frac{1}{\pi}\int^\pi_{-\pi}  f(\theta)\ \text{d}\theta},$
        - $\displaystyle \boxed{C_{1n}=\frac{1}{p^n\pi}\int^\pi_{-\pi}  f(\theta)\cos (n\theta)\ \text{d}\theta},\quad n=1,2,3\ldots,$
        - $\displaystyle \boxed{C_{2n}=\frac{1}{p^n\pi}\int^\pi_{-\pi} f(\theta)\sin (n\theta)\ \text{d}\theta},\quad n=1,2,3\ldots.$
$\blacksquare$

---
## Non-homog. S-L BVP
:film_frames: https://reurl.cc/V6oXDn 11:12
:film_frames: https://reurl.cc/V6oXDn 13:02
> **S-L BVP**
> $\dfrac{\text{d}}{\text{d}x}\left[p(x)\dfrac{\text{d}y}{\text{d}x}\right]- q(x)y+\lambda r(x)y=0;$
> $\alpha_1y(a)+\alpha_2y'(a)=0,\quad\beta_1y(b)+\beta_2 y'(b)=0.$

### Example
:film_frames: https://reurl.cc/V6oXDn 15:16

Consider $$\left\{\begin{array}{l}-(py')'+qy=\mu ry+f; 
\\\alpha_1y(0)+\alpha_2y'(0)=0,\quad\beta_1y(1)+\beta_2 y'(1)=0,\end{array}\right.\tag{14}$$ where $y,p,q,r,\mu,f$ are all defined on $x\in(0,1)$. Other conditions satisfy those in homog. S-L problem.
#### Solution (proposed)
Let $L[y]\equiv -(py')'+qy$
First, suppose the solutions of ***homog.*** S-L B.V.P. associated with Eqn. (14) are
- e-val. $\lambda_n,\quad n=1,2,3,\ldots,$ and
- (normalized) e-func. $\phi_n,\quad n=1,2,3,\ldots.$

Let $\phi$ be any solution of Eqn. (14); i.e. $L[\phi]=\mu r\phi+f$.

Then, it appears that $\phi$ can be expanded with $\{\phi_n\}$: $$\phi(x)=\sum^\infty_{n=1}c_n\phi(x),$$ where $$c_n=\int^1_0\phi(x)\phi_n(x)r(x)\text{d}x.$$

However, there is a problem --
since $\phi$ is unknown, we are not able to evaluate $c_n$...


