# L12
###### tags: `微分方程特論`
###### date: `27 May 2020`

## §8.5 Separation of Variables
:film_frames: https://reurl.cc/7XNZQb 00:19
### Classes of PDEs
- Elliptic: Laplace eqn./Poisson eqn.
- Parabolic: heat eqn.
- Hyperbolic: e.g. wave eqn.

### Heat (Conduction) Equation
:film_frames: https://reurl.cc/7XNZQb 04:36

Consider a rod of length $L$.
Let $u(x,t)$ be the temparature on the rod at position $x$ $(0<x<L)$ and time $t$.

Then $u(x,t)$ satisfies the heat equation $$u_{t}=ku_{xx}\quad (0<x<L,\ t>0)\tag{1}$$ or $$\dfrac{\partial u}{\partial t}=k\dfrac{\partial^2 u}{\partial x^2}\quad (0<x<L,\ t>0).\tag{1$'$}$$

Given (homogeneous) boundary/endpoint conditions $$u(0,t)=0,\quad u(L,t)=0\quad (t>0)\tag{2}$$ and a initial condition $$u(x,0)=f(x)\quad (0\leq x\leq L),\tag{3}$$
:film_frames: https://reurl.cc/7XNZQb 11:09
Recall the **principle of superposition**: If $u_1,u_2.\ldots,u_n$ are solutions of Eqn. (1), then $c_1u_1+c_2u_2+\ldots+c_nu_n$ is a solution, too. For homogeneous boundary conditions (2), if $u_1,u_2.\ldots,u_n$ satisfy (2), then so does $c_1u_1+c_2u_2+\ldots+c_nu_n$.

### Example
:film_frames: https://reurl.cc/ZOgR0V 00:00
(Example 1, p.589)

Supposed that $u_1=e^{-t}\sin x$, $u_2=e^{-4t}\sin 2x$, $u_3=e^{-4t}\sin 3x$ are all solutions of the PDE $$u_t=u_{xx}.\tag{4}$$ Let $u=c_1u_1+c_2u_2+c_3u_3$, then prove that $u(x,t)$ satisfies Eqn. (4) and conditions $$\left\{\begin{array}{l}u(0,t)=0,\quad u(\pi, t)=0;\\u(x,0)=80\sin^3 x.\end{array}\right.$$
#### Solution
$u(x,0)=c_1u_1(x,0)+c_2u_2(x,0)+c_3u_3(x,0)=80\sin^3 x$
$\implies c_1\sin x+ c_2\sin 2x+c_3\sin 3x=80\sin^3 x\\=80\left(\dfrac{3\sin x-\sin 3x}{4}\right)=60\sin x-20\sin 3x$
$\therefore$ $c_1=60,\ c_2=0,\ c_3=-20$.
Thus, a solution of the given boundary value problem is $\boxed{u(x,t)=60e^{-t}\sin x-20e^{-9t}\sin 3x}$. $\blacksquare$

:film_frames: https://reurl.cc/ZOgR0V 08:50

### Separation of Variables
:film_frames: https://reurl.cc/ZOgR0V 11:30

Now consider the heat equation again (Eqn. 1) with $k=1$.
#### Solution
1. Suppose the solution is a product of a function of position $x$ (only) and a function of time $t$ (only): $$u(x,t)=X(x)T(t).\tag{5}$$ Then the partial derivatives become ordinary ones $$u_t=X\frac{dT}{dt}\quad \text{and}\quad u_{xx}=T\frac{d^2X}{dx^2}.$$
Also Eqn. 1 becomes $$X\frac{dT}{dt}=T\frac{d^2X}{dx^2}.$$ After divided by $XT$, because both sides of the equality depends on only one variable, they must equal to the same constant, say, $-\lambda$: $$\frac{1}{T}\frac{dT}{dt}=\dfrac{1}{X}\frac{d^2X}{dx^2}=-\lambda\tag{6}$$

This yields two ODEs: (Primes denote ordinary derivative)
- spatial part $X''+\lambda X=0$
- temporal part $T'+\lambda T=0$

2. Apply B.C.s:
$u(0,t)=0\implies X(0)T(t)=0\implies X(0)=0$
$u(L,t)=0\implies X(L)T(t)=0\implies X(L)=0$

3. Solve the two ODEs:
- spatial part (S-L prob.)
$X''+\lambda X=0;\ X(0)=0,\ X(L)=0$
$\implies \lambda_n=\left(\frac{n\pi}{L}\right)^2$, $\phi_n(x)=\sqrt{\frac{2}{L}}\sin\frac{n\pi x}{L}$, $n=1,2,3,\ldots$
- temporal part
$T'+\left(\dfrac{n\pi}{L}\right)^2T=0$
$\displaystyle \implies \int \frac{dT}{T}=-\int \left(\dfrac{n\pi}{L}\right)^2 dt$
$\implies \ln T=-\left(\dfrac{n\pi}{L}\right)^2+C_n$
$\implies T(t)=C_ne^{-(n\pi/L)^2t}$

Therefore, $$\displaystyle u_n(x,t)=C_n\sqrt{\frac{2}{\pi}}\sin\left(\frac{n\pi x}{L}\right)e^{-(n\pi/L)^2t}$$ is a solution of the B.V.P., and its linear combination $$\begin{align}\sum^\infty_{n=1}\kappa_nu_n(x,t)&=\sum^\infty_{n=1}\underbrace{\kappa_nC_n\sqrt{\frac{2}{\pi}}}_{\equiv\ c_n}\sin\left(\frac{n\pi x}{L}\right)e^{-(n\pi/L)^2t}\\&=\boxed{\sum^\infty_{n=1}c_n\sin\left(\frac{n\pi x}{L}\right)e^{-(n\pi/L)^2t}=u(x,t)}\end{align}$$ is a solution, too.

4. Apply I.C. and evaluate $c_n$: $$\require{cancel} \sum^\infty_{n=1}c_n\sin\left(\frac{n\pi x}{L}\right)\cancelto{1}{e^{-(n\pi/L)^2\cdot 0}}=f(x),$$ where $c_n$ happens to be the coefficients of Fourier sine series and can be evaluated by $$\boxed{c_n=\frac{2}{L}\int^L_0 f(x)\sin \dfrac{n\pi x}{L} dx,\quad n=1,2,3\ldots.}\ \blacksquare$$

#### General case
:film_frames: https://reurl.cc/rxVdob 00:00

If we generalize the result for any ${\color{blue}k}$, the solutions differ by a ${\color{blue}k}$ factor in the exponential: $$\require{color}\displaystyle \boxed{u(x,t)=\sum^{\infty}_{n=1}c_n\sin\left(\dfrac{n\pi x}{L}\right)e^{-(n\pi/L)^2{\color{blue}k}t}}.$$

#### Special case
(Example 2, p.593)

Consider $$\left\{\begin{array}{ll}u_y=ku_{xx},\\u(0,t)=0,\quad u(\pi, t)=0,\\u(x,0)=f(x),\end{array}\right.$$
where $u_0$ is constant.
:film_frames: https://reurl.cc/rxVdob 03:50 $$\displaystyle \begin{align}c_n&=\frac{2}{L}\int^L_0 u_0\sin \tfrac{n\pi x}{L} dx\\&=\cdots=\frac{2u_0}{n\pi}\left[1-(-1)^n\right]=\left\{\begin{array}{l}0,& n\text{ even},\\\frac{4u_0}{n\pi},&n\text{ odd}.\end{array}\right.\end{align}$$

$\displaystyle \boxed{u(x,t)=\frac{4u_0}{\pi}\sum^\infty_{j=1}\dfrac{1}{2j-1}\sin\left[\frac{(2j-1)\pi x}{L}\right]e^{-[(2j-1)\pi/L]^2{\color{blue}k}t}}.\ \blacksquare$

### Nonhomogeneous B.C.s
:film_frames: https://reurl.cc/rxVdob 08:33

Now we consider $$\left\{\begin{array}{ll}u_t=ku_{xx},&(0<x<L,\ t>0)\\u(0,t)=T_1,\quad u(L,t)=T_2&(t>0)\\u(x,0)=f(x),&(0<x<L)\end{array}\right.$$ where $T_1$, $T_2$ are constants.
#### Solution
1. Use the ansatz $$u(x,t)=v(x)+w(x,t),\tag{7}$$ where $w$ satifies $w_t=kw_{xx}$ and $w(0,t)=w(L,t)=0$. Then, Eqn. (7) yields $$u_t=0+w_t$$ and $$u_{xx}=v''+w_{xx}.$$

Substituting these into Eqn. (1) and using Eqn. (7) to cancel some terms, we have $$(0+w_t)=k(v''+w_{xx})\implies \cancel{w_t}=kv''+k\cancel{w_{xx}}\implies v''=0.$$
2. Apply B.C.s:
- $u(0,t)=T_1\implies v(0)+\underbrace{w(0,t)}_{=\,0}=T_1\implies v(0)=T_1$
- $u(L,t)=T_2\implies v(L)+\underbrace{w(L,t)}_{=\,0}=T_2\implies v(L)=T_2$

3. Solve $v$: $$v''=0\implies v(x)=c_1 x+c_2,\\ v(0)=T_1,\ v(L)=T_2\implies c_2=T_2,\ c_1=\dfrac{T_2-T_1}{L}.$$ Thus $$v(x)=\left(\dfrac{T_2-T_1}{L}\right)x+T_1.$$

4. Apply I.C. and evaluate $c_n$: $u(x,0)=f(x)=v(x)+w(x,0)$
$\implies f(x)-v(x)=\sum c_n\sin\frac{n\pi x}{L}$

### Example (Nonhomg.)
:film_frames: https://reurl.cc/rxVdob 19:17

$\left\{\begin{array}{ll}u_t=u_{xx},&(0<x< 30,\ t>0)\\u(0,t)=20,\quad u(30,t)=50&(t>0)\\u(x,0)=60-2x&(0< x< 30)\end{array}\right.$
#### Solution
Let $u(x,t)=v(x)+w(x,t)$ so that $w_t=w_{xx}$ and $w(0,t)=0$, $w(30,t)=0$
$0+w_t=v''+w_{xx}\implies v''=0$
$u(0,t)=20\implies v(0)+w(0,t)=20\implies v(0)=20$
$u(30,t)=50\implies v(30)+w(30,t)=50\implies v(30)=50$

$\left\{\begin{array}{ll}v''=0\\v(0)=20,\ v(30)=50\end{array}\right.\implies v(x)=x+20$

$\boxed{\displaystyle u(x,t)=(x+20)+\sum_{n=1}^\infty c_n\sin\dfrac{n\pi x}{30}e^{-(n\pi/30)^2t}}.$

$u(x,0)=60-2x=20+\sum_{n=1}^\infty c_n\sin\dfrac{n\pi x}{30}$
$\displaystyle\implies 40-3x=\sum_{n=1}^\infty c_n\sin\dfrac{n\pi x}{30}$
$\displaystyle\implies \boxed{c_n=\dfrac{2}{30}\int_0^{30}(40-3x)\sin\dfrac{n\pi x}{30}dx=\cdots}.$ $\blacksquare$

## §8.6 Wave equation
:film_frames: https://reurl.cc/rxVdob 31:18
### One-dimensional vibrating string
A string of length $L$
$u(x,t)$ is the (transverse) position at longitudinal position $x$ $(0< x<L)$ and time $t$.
$\left\{\begin{array}{ll}u_{tt}=a^2 u_{xx},&(0< x< L,\ t>0)\\u(0,t)=0,\quad u(L,t)=0&(t>0)\\u(x,0)=f(x),\quad u_t(x,0)=g(x)&(0< x< L)\end{array}\right.\tag{8}$
#### Solution
1. Ansatz $u(x,t)=X(x)T(t)$ yields $$\dfrac{1}{X}\dfrac{d^2X}{dx^2}=\dfrac{1}{a^2T}\dfrac{d^2T}{dt^2}=-\lambda,$$ where $\lambda$ is a constant.
And we obtain two ODEs:
- $X''(x)+\lambda X(x)=0,$
- $T''(t)+a^2\lambda T(t)=0.$

2. Apply B.C.s:
- $u(0,t)=0\implies X(0)T(0)=0\implies X(0)=0$
- $u(L,t)=0\implies X(L)T(0)=0\implies X(L)=0$

3. Solve the two ODEs:
- $\left\{\begin{array}{ll}X''+\lambda X=0\\X(0)=0,\ X(L)=0\end{array}\right.\implies X_n(x)=\sin\dfrac{n\pi x}{L},\quad n=1,2,3,\ldots$
- $T''+a^2\underbrace{\left(\dfrac{n\pi}{L}\right)^2}_{=\lambda}T=0\implies T_n(t)=a_n\cos\left(\dfrac{an\pi}{L}t\right)+b_n\sin\left(\dfrac{an\pi}{L}t\right)$
Thus the gen. sol. of Eqn. (8) is $$\begin{align}u(x,t)&=\sum_{n=1}^\infty c_nX_n(x)T_n(t)\\&=\boxed{\sum_{n=1}^\infty\left[A_n\sin\left(\dfrac{n\pi}{L}x\right)\cos\left(\dfrac{an\pi}{L}t\right)+B_n\sin\left(\dfrac{n\pi}{L}x\right)\sin\left(\dfrac{an\pi}{L}t\right)\right]},\tag{9}\end{align}$$ where $A_n=c_na_n,\ B_n=c_nb_n,\ n=1,2,3,\ldots$.

4. Apply I.C.s:
:film_frames: https://reurl.cc/7Xq2bb 00:28
- $\displaystyle u(x,0)=f(x)=\sum_{n=1}^\infty A_n\sin\left(\dfrac{n\pi}{L}x\right)$, with $\displaystyle \boxed{A_n=\dfrac{2}{L}\int_0^L f(x)\sin\left(\dfrac{n\pi x}{L}\right) dx}$.
- $\displaystyle u_t(x,0)=g(x)=\sum_{n=1}^\infty B_n\dfrac{an\pi}{L}\sin\left(\dfrac{n\pi}{L}x\right)$, with $\displaystyle \boxed{B_n=\dfrac{2}{an\pi}\int_0^L g(x)\sin\left(\dfrac{n\pi x}{L}\right) dx}$.

### Example
:film_frames: https://reurl.cc/7Xq2bb 06:00
(Example 1, p.603)

$\left\{\begin{array}{ll}u_{tt}=4 u_{xx},\qquad(0<x<\pi,\ t>0);\\u(0,t)=0,\quad u(\pi,t)=0,\\u(x,0)=\frac{1}{10}\sin^3x,\quad u_t(x,0)=0,\end{array}\right.$
#### Solution
:film_frames: https://reurl.cc/7Xq2bb 09:20
$a=2,\quad L=\pi$
1. Use Eqn. (9) to write $\displaystyle u(x,t)=\sum_{n=1}^\infty\left[A_n\sin(nx)\cos(2nt)+ B_n\sin(nx)\sin(2nt)\right]$

2. Apply I.C.s:
- $\displaystyle u(x,0)=\frac{1}{10}\sin^3x=\dfrac{3}{40}\sin x-\dfrac{1}{40}\sin 3x=\sum_{n=1}^\infty A_n\sin(nx)$
$\implies A_1=\dfrac{3}{40},\ A_3=-\dfrac{1}{40}$, and $A_n=0$ for any other $n$'s.
- $\displaystyle u_t(x,0)=0=\sum_{n=1}^\infty 2nB_n\sin(nx)$
$\implies B_n=0$, for all $n=1,2,3,\ldots$

3. $\boxed{u(x,t)=\dfrac{3}{40}\sin(x)\cos(2t)-\dfrac{1}{40}\sin(3x)\cos(6t)}.\ \blacksquare$

4. [Desmos simulation](https://www.desmos.com/calculator/eazx2gnphj)

### Example (Nonhomg.)
:film_frames: https://reurl.cc/7Xq2bb 19:38
Consider $$\left\{\begin{array}{ll}u_{tt}=a^2 u_{xx},\qquad(0<x<L,\ t>0);\\u(0,t)=T_1,\quad u(L,t)=T_2,\\u(x,0)=f(x),\quad u_t(x,0)=g(x),\end{array}\right.$$ where $T_1$, $T_2$ are constants.
#### Solution
1. $u(x,t)=v(x)+w(x,t)$
where $w_{tt}=a^2w_{xx}$, $w(0,t)=0$, $w(L,t)=0$.

The equation becomes $w_{tt}=a^2(v''+w_{xx})\implies v''=0$.

2. Apply B.C.s:
- $u(0,t)=T_1\implies v(0)+w(0,t)=T_1\implies v(0)=T_1$
- $u(L,t)=T_2\implies v(L)+w(L,t)=T_2\implies v(L)=T_2$

3. Solve $v(x)$:
$\left\{\begin{array}{ll}v''=0\\v(0)=T_1,\ v(L)=T_2\end{array}\right.\implies v(x)=\left(\dfrac{T_2-T_1}{L}\right)x+T_1$

4. A gen. sol is $$\boxed{ u(x,t)=\small\left[\left(\dfrac{T_2-T_1}{L}\right)x+T_1\right]+\sum_{n=1}^\infty\sin\frac{n\pi x}{L}\left[A_n\cos\left(\frac{an\pi t}{L}\right)+B_n\sin\left(\frac{an\pi t}{L}\right)\right]}.\ \blacksquare$$