133 lines
7.6 KiB
Markdown
133 lines
7.6 KiB
Markdown
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Let's consider a guitar string:
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![draw](drawings/Drawing-2023-12-01-13.49.58.excalidraw.png)
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assuming the thickness of the string is much smaller than the length of the string, which is true.
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$\frac{ \partial u^2 }{ \partial t^2 }=\alpha^2 \frac{ \partial^2 u }{ \partial x^2 } \quad 0\leq x\leq L,\quad t>0$
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^ Reminds me of the wave equation from Phys 130.
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Since the string is tied down on the ends we have the following initial conditions:
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$u(t,0)=u(t,L)=0 \qquad t>0$
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$u(0,x)=f(x)$ $0\leq x\leq L$
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$\frac{ \partial u }{ \partial t }(0,x)=g(x)$ $0\leq x\leq L$
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^IBVP of the system.
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#end of lec 33
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#start of lec 34
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The wave equation follows many phenomena in electrical engineering.
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separation of variables:
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$u(t,x)=X(x)T(t)$
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plug in to equation:
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$T''X=\alpha^2TX''$
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$\frac{T''}{\alpha^2T}=\frac{X''}{X}=-\lambda$
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^ #evp !
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consider the $X$ side:
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$X''+\lambda X=0, \quad X(0)=X(L)=0$
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We've solved this before.
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the only non-trivial solutions for that Eigen value problem is:
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$\lambda_{n}=(\frac{n\pi}{L})^2$
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$X_{n}(x)=\sin\left( \frac{n\pi x}{L} \right)$ for $n=1,2,3,\dots$
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$\frac{T''}{\alpha^2T}=-\left( \frac{n\pi}{L} \right)^2$
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$T_{n}''+\left( \frac{\alpha n\pi}{L} \right)^2T_{n}=0$
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characteristic equation:
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$r^2+\left( \frac{\alpha n\pi}{L} \right)^2=0$
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$r_{1,2}=\pm i \frac{\alpha n\pi}{L}$
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"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?"
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$T_{n}(t)=b_{n}\cos\left( \frac{\alpha n\pi}{L}t \right)+a_{n}\sin\left( \frac{\alpha n\pi}{L}t \right)$
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$u_{n}(t,x)=\left( b_{n}\cos\left( \frac{\alpha n\pi}{L} \right)+a_{n}\sin\left( \frac{\alpha n\pi}{L}t \right) \right)\sin\left( \frac{n\pi x}{L} \right)$
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if you sum all these <u>modes</u>, you get the solution:
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$$u_{n}(t,x)=\sum_{n=1}^\infty\left( b_{n}\cos\left( \frac{\alpha n\pi}{L} \right)+a_{n}\sin\left( \frac{\alpha n\pi}{L}t \right) \right)\sin\left( \frac{n\pi x}{L} \right)$$
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"mathematics and reality do align very well, if the speed of the string is different the solution differs aswell."
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$u(0,x)=f(x)=\sum_{n=1}^\infty b_{n}\sin\left( \frac{n\pi x}{L} \right)$
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^ that's starting to look familiar.
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$\implies b_{n}=\frac{2}{L}\int _{0}^L f(x)\sin\left( \frac{n\pi x}{L} \right) \, dx$
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where does it converge? well $f(x)$ and $f'(x)$ are both continuous, so it converges everywhere.
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$\frac{ \partial u }{ \partial t }(0,x)=g(x)=\sum_{n=1}^\infty\underbrace{ a_{n} \frac{\alpha n\pi}{L} }_{ }\sin\left( \frac{\alpha n\pi}{L} \right)$
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$a_{n} \frac{\alpha n\pi}{L}$ are the Fourier $\sin$ coefficients of $g(x)$
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$a_{n}=\frac{2}{\alpha n\pi}\int _{0}^L g(x)\sin\left( \frac{n\pi x}{L} \right)\, dx$
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$\alpha^2$ is the Hooke modulus of the string btw.
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remember the heat equation, the amplitude is exponentionally decreasing,
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here the amplitude is oscillatory and doesnt increae in time. boi-oi-oi-oing
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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.
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btw this equation models the electromagnetic feild, to some approximation.
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the lowest mode is called the fundemental mode, the following terms after are called harmonics.
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If two instruments play the same note (same fundemental frequency), they still sound different! and that's because of the difference in harmonics.
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The modes are standing waves in the string.
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"my claim is that any object, including social objects ,can be described by waves. Everything is a wave."
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you can model elementary particle behaviours with solitons (non linear waves.)
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in life in the real world, all waves have finite speed.
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So thats why its important to learn the wave equation. its the prototype to waves.
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"waves are the fundamental object. [...]. So that's why it's important, these are the fundamental objects of nature here."
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$f(x)=\begin{cases}x, & 0\leq x\leq \frac{\pi}{2}\\ \ \pi-x, & \frac{\pi}{2}<x\leq \pi\end{cases}$
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$\alpha^2=1$
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$g(x)=\sin(x)$
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$b_{n}=\frac{2}{\pi}\left( \int _{0} ^\frac{\pi}{2} x\sin(nx) \, dx +\int _{\frac{\pi}{2}} ^\pi (\pi-x)\sin(nx) \, dx\right)$
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$b_{n}=\frac{4}{n^2\pi}\sin\left( \frac{n\pi}{2} \right)$
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$b_{2k}=0$
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half of the b coefficients are 0.
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$b_{2k-1}=\frac{4}{(2k-1)^2\pi}(-1)^{k+1}$
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plug in g(x) to get a_n terms
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however theres a short cut here. from definition of g(x) earlier:
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$g(x)=a_{1}\sin(x)+2a_{2}\sin(2x)+3a_{3}\sin(3x)+\dots$
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but g(x) is sin x.
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so $a_{1}=1$ and every other term is 0
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plug this in to get solution:
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$$u(t,x)=\left( \frac{4}{\pi}\cos t+\sin t \right)\sin(x)+\sum_{k=1}^\infty \frac{4(-1)^{k+1}}{\pi(2k+1)^2}\cos((2k+1)t)\sin((2k+1)x)$$
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typo in his notes, not 2k-1 its 2k+1
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#end of lec 34
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#start of lec 35
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last problem of the course which we will finish today.
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$\frac{ \partial^2 u }{ \partial t^2 }=\frac{ \partial^2 u }{ \partial x^2 }+tx, 0\leq x\leq \pi$ $t>0$
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thats a driven wave equation. tx is the source term.
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$u(0,t)=u(\pi,t)=0 \quad t>0$
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$u(x,0)=\sin(x) \quad 0\leq x\leq \pi$
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$\frac{ \partial u }{ \partial t }(x,0)=5\sin(2x)-3\sin(5x)\quad 0\leq x\leq \pi$
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if any of the boundary conditions are non zero, then we have to split(?) into X and T.
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in this case there's a $\sin(x)$ term so its a nonhomogenous equation (?)
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when $tx$ wasn't there in last problem we had the solution:
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$u(t,x)=\sum_{n=1}^\infty \underbrace{ (a_{n}\cos(nt)+b_{n}\sin(nt)) }_{ u_{n}(t) }\sin(nt)$ notice $L=\pi$
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if we expand $tx$:
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$tx=\sum_{n=1}^\infty h_{n}\sin(nx)$ (this is called a formal expansion, the two arent exactly equal due to the discontinuouity in tx.)
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$h_{n}=\frac{2}{\pi}\int _{0} ^\pi tx\sin(nx)\, dx$
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![draw](drawings/2023-12-06-13.14.28.excalidraw.png)
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continuous between 0 and pi but on the edge ponts, the foureir sin series will converge to the midpoint of the two edge points.
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$h_{n}=\frac{2t}{\pi}\int _{0}^\pi x\sin(nx)\, dx=- \frac{2t}{\pi n}\left( x\cos(nx)|_{0}^\pi-\int _{0}^\pi \cancel{ \cos(nx) }\, dx \right)$
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$h_{n(t)}=\frac{2t}{n}(-1)^{n+1}$
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$h(x,t)=tx=\sum_{n=1}^\infty h_{n}\sin(nx)=\sum_{n=1}^\infty \frac{2t}{n}(-1)^{n+1}\sin(nx)$
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$\frac{ \partial^2 u }{ \partial t^2 }=\sum_{n=1}^\infty u_{n}''(t)\sin(nx)$
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$\frac{ \partial^2 u }{ \partial x^2 }=\sum_{n=1}^\infty -u_{n}(t)n^2\sin(nx)$
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$\sum_{n=1}^\infty \underbrace{ \left( u_{n}''+n^2u_{n}+\frac{2t}{n}(-1)^n \right) }_{ =0 }\sin(nx)=0$
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$u''_{n}+n^2u_{n}=\frac{2}{n}(-1)^{n+1}t, \quad n=1,2,\dots$
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use #mouc (or laplace, but that'll take much longer.)
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characteristic eq:
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$r^2+n^2=0$
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$r_{1,2}=\pm in$
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$u_{n}^h(t)=a_{n}\cos(nt)+b_{n}\sin(nt)$
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$u_{n}^p(t)=At+B$
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$B=0$ because there's no constant term on the RHS
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$A=\frac{2(-1)^{n+1}}{n^3}$
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$u_{n}(t)=a_{n}\cos(nt)+b_{n}\sin(nt)+\frac{2(-1)^{n+1}}{n^3}t$
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$u(x,t)=\sum_{n=1}^\infty(a_{n}\cos(nt)+b_{n}\sin(nt)+\frac{2(-1)^{n+1}}{n^3}t)\sin(nx)$
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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.
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$u(x,0)=\sum_{n=1}^\infty a_{n}\sin(nx)=\sin(x)$
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$a_{1}=1, \quad a_{k}=0, \quad k=2,3,\dots$
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$\frac{ \partial u }{ \partial t }(x,0)=\sum_{n=1}^\infty\left( b_{n}n+\frac{2(-1)^{n+1}}{n^3} \right)\sin(nx)=5\sin(2x)-3\sin(5x)$
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coefficients =0 if $n\ne_{2},5$
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$\implies b_{n}=\frac{2}{n^4}(-1)^n, \quad n\ne_{2},5$
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$b_{2}=\frac{5}{2}+\frac{2(-1)^2}{2^4}$
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$b_{5}=-\frac{3}{5}+\frac{2(-1)^5}{5^4}$
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$u(x,t)=\cos(t)\sin(x)+\frac{5}{2}\sin(2t)\sin(2x)-\frac{3}{5}\sin(5t)\sin(5x)+2\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^3}\left( t-\frac{\sin(nt)}{n} \right)\sin(nx)$
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finished the solution. Man I got teary eyed from this lecture.
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#end of lec 35
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#start of lec 36
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What do you guys wanna do? Questions or summary of the course?
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Okay we do summary.
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# Summary of second half of Math 201
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(available on eclass)
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Laplace transforms:
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Definition of laplace,
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Properties (4 important ones)
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... (I decided to listen rather than note down.)
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</br>
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#end of lec 36
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#end of Math 201. Congratulations!
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