MATH201/content/Wave equation (lec 33-36).md

Let's consider a guitar string:
![draw](drawings/Drawing-2023-12-01-13.49.58.excalidraw.png)
assuming the thickness of the string is much smaller than the length of the string, which is true.
$\frac{ \partial u^2 }{ \partial t^2 }=\alpha^2 \frac{ \partial^2 u }{ \partial x^2 } \quad 0\leq x\leq L,\quad t>0$
^ Reminds me of the wave equation from Phys 130.
Since the string is tied down on the ends we have the following initial conditions:
$u(t,0)=u(t,L)=0 \qquad t>0$
$u(0,x)=f(x)$  $0\leq x\leq L$
$\frac{ \partial u }{ \partial t }(0,x)=g(x)$ $0\leq x\leq L$
^IBVP of the system.
#end of lec 33
#start of lec 34
The wave equation follows many phenomena in electrical engineering.
separation of variables:
$u(t,x)=X(x)T(t)$
plug in to equation:
$T''X=\alpha^2TX''$
$\frac{T''}{\alpha^2T}=\frac{X''}{X}=-\lambda$
^ #evp !
consider the $X$ side:
$X''+\lambda X=0, \quad X(0)=X(L)=0$
We've solved this before.
the only non-trivial solutions for that Eigen value problem is:
$\lambda_{n}=(\frac{n\pi}{L})^2$
$X_{n}(x)=\sin\left( \frac{n\pi x}{L} \right)$ for $n=1,2,3,\dots$
$\frac{T''}{\alpha^2T}=-\left( \frac{n\pi}{L} \right)^2$
$T_{n}''+\left( \frac{\alpha n\pi}{L} \right)^2T_{n}=0$
characteristic equation:
$r^2+\left( \frac{\alpha n\pi}{L} \right)^2=0$
$r_{1,2}=\pm i \frac{\alpha n\pi}{L}$
"Don't memorize the steps. If you try to memorize you will mess up the final for sure. Ask yourself, why am I doing this here?"
$T_{n}(t)=b_{n}\cos\left( \frac{\alpha n\pi}{L}t \right)+a_{n}\sin\left( \frac{\alpha n\pi}{L}t \right)$
$u_{n}(t,x)=\left( b_{n}\cos\left( \frac{\alpha n\pi}{L} \right)+a_{n}\sin\left( \frac{\alpha n\pi}{L}t \right) \right)\sin\left( \frac{n\pi x}{L} \right)$
if you sum all these <u>modes</u>, you get the solution:
$$u_{n}(t,x)=\sum_{n=1}^\infty\left( b_{n}\cos\left( \frac{\alpha n\pi}{L} \right)+a_{n}\sin\left( \frac{\alpha n\pi}{L}t \right) \right)\sin\left( \frac{n\pi x}{L} \right)$$

"mathematics and reality do align very well, if the speed of the string is different the solution differs aswell."
$u(0,x)=f(x)=\sum_{n=1}^\infty b_{n}\sin\left( \frac{n\pi x}{L} \right)$
^ that's starting to look familiar.
$\implies b_{n}=\frac{2}{L}\int _{0}^L f(x)\sin\left( \frac{n\pi x}{L} \right) \, dx$
where does it converge? well $f(x)$ and $f'(x)$ are both continuous, so it converges everywhere.
$\frac{ \partial u }{ \partial t }(0,x)=g(x)=\sum_{n=1}^\infty\underbrace{  a_{n} \frac{\alpha n\pi}{L} }_{  }\sin\left( \frac{\alpha n\pi}{L} \right)$
$a_{n} \frac{\alpha n\pi}{L}$ are the Fourier $\sin$ coefficients of $g(x)$
$a_{n}=\frac{2}{\alpha n\pi}\int _{0}^L g(x)\sin\left( \frac{n\pi x}{L} \right)\, dx$
$\alpha^2$ is the Hooke modulus of the string btw.
remember the heat equation, the amplitude is exponentionally decreasing,
here the amplitude is oscillatory and doesnt increae in time. boi-oi-oi-oing
to make it more releasitic we have to add a term for resistance, and we end up with a b in the characteristic equation for T.
btw this equation models the electromagnetic feild, to some approximation.
the lowest mode is called the fundemental mode, the following terms after are called harmonics.
If two instruments play the same note (same fundemental frequency), they still sound different! and that's because of the difference in harmonics.
The modes are standing waves in the string.
"my claim is that any object, including social objects ,can be described by waves. Everything is a wave."
you can model elementary particle behaviours with solitons (non linear waves.)
in life in the real world, all waves have finite speed.
So thats why its important to learn the wave equation. its the prototype to waves.
"waves are the fundamental object. [...]. So that's why it's important, these are the fundamental objects of nature here."
$f(x)=\begin{cases}x, & 0\leq x\leq \frac{\pi}{2}\\ \ \pi-x, & \frac{\pi}{2}<x\leq \pi\end{cases}$
$\alpha^2=1$
$g(x)=\sin(x)$
$b_{n}=\frac{2}{\pi}\left( \int _{0} ^\frac{\pi}{2} x\sin(nx) \, dx +\int _{\frac{\pi}{2}} ^\pi (\pi-x)\sin(nx) \, dx\right)$
$b_{n}=\frac{4}{n^2\pi}\sin\left( \frac{n\pi}{2} \right)$
$b_{2k}=0$
half of the b coefficients are 0.
$b_{2k-1}=\frac{4}{(2k-1)^2\pi}(-1)^{k+1}$
plug in g(x) to get a_n terms
however theres a short cut here. from definition of g(x) earlier:
$g(x)=a_{1}\sin(x)+2a_{2}\sin(2x)+3a_{3}\sin(3x)+\dots$
but g(x) is sin x.
so $a_{1}=1$ and every other term is 0
plug this in to get solution:
$$u(t,x)=\left( \frac{4}{\pi}\cos t+\sin t \right)\sin(x)+\sum_{k=1}^\infty \frac{4(-1)^{k+1}}{\pi(2k+1)^2}\cos((2k+1)t)\sin((2k+1)x)$$
typo in his notes, not 2k-1 its 2k+1
#end of lec 34
#start of lec 35
last problem of the course which we will finish today.
$\frac{ \partial^2 u }{ \partial t^2 }=\frac{ \partial^2 u }{ \partial x^2 }+tx, 0\leq x\leq \pi$ $t>0$
thats a driven wave equation. tx is the source term.
$u(0,t)=u(\pi,t)=0 \quad t>0$
$u(x,0)=\sin(x) \quad 0\leq x\leq \pi$
$\frac{ \partial u }{ \partial t }(x,0)=5\sin(2x)-3\sin(5x)\quad 0\leq x\leq \pi$
if any of the boundary conditions are non zero, then we have to split(?) into X and T.
in this case there's a $\sin(x)$ term so its a nonhomogenous equation (?)

when $tx$ wasn't there in last problem we had the solution:
$u(t,x)=\sum_{n=1}^\infty \underbrace{ (a_{n}\cos(nt)+b_{n}\sin(nt)) }_{ u_{n}(t) }\sin(nt)$ notice $L=\pi$
if we expand $tx$:
$tx=\sum_{n=1}^\infty h_{n}\sin(nx)$ (this is called a formal expansion, the two arent exactly equal due to the discontinuouity in tx.)
$h_{n}=\frac{2}{\pi}\int _{0} ^\pi tx\sin(nx)\, dx$
![draw](drawings/2023-12-06-13.14.28.excalidraw.png)

continuous between 0 and pi but on the edge ponts, the foureir sin series will converge to the midpoint of the two edge points.
$h_{n}=\frac{2t}{\pi}\int _{0}^\pi x\sin(nx)\, dx=- \frac{2t}{\pi n}\left( x\cos(nx)|_{0}^\pi-\int _{0}^\pi \cancel{ \cos(nx) }\, dx \right)$
$h_{n(t)}=\frac{2t}{n}(-1)^{n+1}$
$h(x,t)=tx=\sum_{n=1}^\infty h_{n}\sin(nx)=\sum_{n=1}^\infty \frac{2t}{n}(-1)^{n+1}\sin(nx)$
$\frac{ \partial^2 u }{ \partial t^2 }=\sum_{n=1}^\infty u_{n}''(t)\sin(nx)$
$\frac{ \partial^2 u }{ \partial x^2 }=\sum_{n=1}^\infty -u_{n}(t)n^2\sin(nx)$
$\sum_{n=1}^\infty \underbrace{ \left( u_{n}''+n^2u_{n}+\frac{2t}{n}(-1)^n \right) }_{ =0 }\sin(nx)=0$
$u''_{n}+n^2u_{n}=\frac{2}{n}(-1)^{n+1}t, \quad n=1,2,\dots$
use #mouc (or laplace, but that'll take much longer.)
characteristic eq:
$r^2+n^2=0$
$r_{1,2}=\pm in$
$u_{n}^h(t)=a_{n}\cos(nt)+b_{n}\sin(nt)$
$u_{n}^p(t)=At+B$
$B=0$ because there's no constant term on the RHS
$A=\frac{2(-1)^{n+1}}{n^3}$
$u_{n}(t)=a_{n}\cos(nt)+b_{n}\sin(nt)+\frac{2(-1)^{n+1}}{n^3}t$
$u(x,t)=\sum_{n=1}^\infty(a_{n}\cos(nt)+b_{n}\sin(nt)+\frac{2(-1)^{n+1}}{n^3}t)\sin(nx)$
This is the last problem I'll be solving in my career. This is the last time he's teaching math 201 :( or any course for that matter.
$u(x,0)=\sum_{n=1}^\infty a_{n}\sin(nx)=\sin(x)$
$a_{1}=1, \quad a_{k}=0, \quad k=2,3,\dots$
$\frac{ \partial u }{ \partial t }(x,0)=\sum_{n=1}^\infty\left( b_{n}n+\frac{2(-1)^{n+1}}{n^3} \right)\sin(nx)=5\sin(2x)-3\sin(5x)$
coefficients =0 if $n\ne_{2},5$
$\implies b_{n}=\frac{2}{n^4}(-1)^n, \quad n\ne_{2},5$
$b_{2}=\frac{5}{2}+\frac{2(-1)^2}{2^4}$
$b_{5}=-\frac{3}{5}+\frac{2(-1)^5}{5^4}$
$u(x,t)=\cos(t)\sin(x)+\frac{5}{2}\sin(2t)\sin(2x)-\frac{3}{5}\sin(5t)\sin(5x)+2\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^3}\left( t-\frac{\sin(nt)}{n} \right)\sin(nx)$
finished the solution. Man I got teary eyed from this lecture.
#end of  lec 35
#start of lec 36
What do you guys wanna do? Questions or summary of the course?
Okay we do summary.
# Summary of second half of Math 201
(available on eclass)
Laplace transforms:
Definition of laplace,
Properties (4 important ones)
... (I decided to listen rather than note down.)
</br>
#end of lec 36
#end of Math 201. Congratulations!