MATH201/content/Partial differential equations (lec 30-32).md

Let's revisit the heat equation: The first time the heat equation was introduced, we figured out that it's solution was of the form u(t,x)=\sum_{n=1}^\infty c_{n}e^{-(n\pi/L)^2Dt}\sin\left( \frac{n\pi x}{L} \right) Now that we learned about fourier series and eigen value problems, we can finally solve it! (for a given specific case.) IBVP of heat eq: \frac{ \partial u }{ \partial t }=D\frac{ \partial^{2} u }{ \partial x^{2} } for 0\leq x\leq L for t>0 u(t,0)=u(t,L)=0, \quad t>0 u(0,x)=f(x), \quad 0\leq x\leq L lets choose L=\pi f(x)=\begin{cases}-x & 0\leq x\leq \frac{\pi}{2} \\1-x & \frac{\pi}{2}<x\leq \pi\end{cases} !Lec 30 2023-11-24 13.42.29.excalidraw So we have a non-uniformly heated rod with both ends insulated. What happens to the temperature inside the rod over time? "[...]. Very interesting problem." -Prof (I agree.) If we made this a series, where would it converge? Well it's continuous from 0 to pi and its windowed form when repeated will be convergent everywhere, this is good news for us. Separation of variables: u(t,x)=T(t)X(x) theres a theorem that this will give a unique solution. T'X=DTX'' \frac{T'}{DT}=\frac{X''}{X} LHS is a function of t only , RHS is function of x only. "I don't know what is time, I know space, I can take a step and see the step I take, but can you see time? Can you see the future? Some can but I can't." -prof. Very philosophical. X''+\lambda X=0 where X(0)=X(L=\pi)=0 u(t,0)=T(t)X(0)=0 case 1) \lambda<0, r_{1,2}=\pm \sqrt{ -\lambda } X(x)=c_{1}e^{ \sqrt{ -\lambda }x }+c_{2}e^{ -\sqrt{ -\lambda }x } X(0)=c_{1}+c_{2}=0 X(\pi)=c_{1}e^{ \sqrt{ -\lambda }\pi }+c_{2}e^{ -\sqrt{-\lambda }\pi }=0 c_{1}=c_{2}=0 we will continue the problem in the next lecture. #end of lec 30 #start of lec 31 "Can we skip the assignment?" Petar replies: You have a choice, not just in math but in life, confusions says there's three things you cant take back opportunity to do something, time, and your word so do your last assignment and don't pass up the opportunity!

recall last lec: \frac{ \partial u }{ \partial t }=\frac{ \partial u^2 }{ \partial x^2 }, 0\leq x\leq \pi, t>0 u(t,0)=u(t,\pi)=0, t>0 <- Dirichlet (french mathematician) IVP. u(0,x)=f(x), \quad 0\leq x<\pi f(x)=\begin{cases}-x, & 0\leq x\leq \frac{\pi}{2} \\x-\pi, & \frac{\pi}{2}{\leq x\leq \pi}\end{cases} graph of tube when t=0. here.

u(t,x)=T(t)X(x) =X+\lambda X=0, \quad X(0)=X(\pi)=0 case 1) \lambda<0 \implies X(x)=0 case 2) \lambda=0 \implies X(x)==0 X(0)=0=c_{2} X(\pi)=0=c_{1}\pi case 3) \lambda>0 X(x)=c_{1}\cos(\sqrt{ \lambda }x)+c_{2}\sin(\sqrt{ \lambda }x) X(0)=c_{1}=0 X(\pi)=0=c_{2}\sin(\sqrt{ \lambda }\pi)\implies \sqrt{ \lambda }\pi=n\pi\quad n=1,2,\dots \lambda_{n}=n^2 a countable infinitely many lambdas (taggable with an index) real numbers are uncountable "what is a point? a point on a screen on a peice of paper?" (We have lost our minds) X_{n}(x)=c_{2}\sin(nx) but obviously, we don't need the multiples of an eigen value so we set c_{2}=1 (or any arbitrary non zero constant) X_{n}(x)=\sin(nx) u(t,x)=T(t)X(x) \frac{T'}{T}=-n^2 integrate both sides: T_{n}(t)=b_{n}e^{-n^2t} u_{n}(t,x)=b_{n}e^{-n^2t}\sin(nx), n=1,2,\dots u(t,x)=\sum_{n=1}^\infty b_{n}e^{-n^2t}\sin(nx) u(0,x)=f(x)=\sum_{n=1}^\infty b_{n}\sin(nx) b_{n}=\frac{2}{\pi}\int _{0}^\pi f(x)\, dx b_{n}=\frac{2}{\pi}[-\int _{0} ^\frac{\pi}{2} x\sin(nx) \, dx+\int _{\frac{\pi}{2}} ^\pi x\sin(nx) \, dx-\pi \int _{\frac{\pi}{2}} ^\pi \sin(nx)\, dx] \int x\sin(nx) \, dx=-\frac{1}{n}\left( x\cos(nx)-\int \cos(nx) \, dx \right) =-\frac{1}{n}\left( x\cos(nx)-\frac{1}{n}\sin(nx) \right)=\frac{1}{n^2}\sin(nx)-\frac{1}{n}x\cos(nx) bla bla bla, we get that b_{n} is: b_{n}=\frac{2}{\pi}\left[ \frac{\pi}{2n}\cos \frac{n\pi}{2}-\frac{1}{n^2}\sin \frac{n\pi}{2}-\frac{1}{n^2}\sin \frac{n\pi}{2}-\frac{\pi}{n}(-1)^n +\frac{\pi}{2n}\cos \frac{n\pi}{2}+\frac{\pi}{n}(-1)^n-\frac{\pi}{n}\cos \frac{n\pi}{2} \right] b_{n}=-\frac{4}{n^2\pi}\sin \frac{n\pi}{2}, \quad n=1,2,\dots n=2k\implies b_{2k}=0, \quad k=1,2,\dots n=2k-1\implies b_{2k-1}=-\frac{4}{(2k+1)^2\pi}(-1)^{k+1}, \quad k=1,2,\dots b_{2k-1}=\frac{4}{(2k+1)^2\pi}(-1)^{k}

u(t,x)=\sum_{n=1}^\infty \frac{(-1)^k}{(2k-1)^2}e^{-(2k-1)^2t}\sin((2k-1)x)

notice that the limit of the sum as t\to \infty is 0 graph time! But oh no! He forgot the plot from the last lecture, i went to grab it for him. While I was gone, I heard he shared stories about his family. Unfortunately I missed it. if you imagine the plot was showing conc. of co2 in a room, it follows a similar curve as time goes on as the gas diffuses. (5 more lectures, left)

k\frac{ \partial u }{ \partial x }=0 where k is the thermal conductivity, \frac{ \partial u }{ \partial x }=0|_{0<x<L} this gives a new equation type: Neumann type. \frac{ \partial u }{ \partial t }=D\frac{ \partial^2 u }{ \partial x^2 }, \quad 0\leq x\leq L, \quad t>0 \frac{ \partial u }{ \partial x }(t,0)=\frac{ \partial u }{ \partial x }(t,L)=0, \quad t>0 u(0,x)=f(x) \quad 0\leq x\leq \pi

#end of lec 31 #start of lec 32 back to the heat problem we do the Neumann problem, start with separation of variables. assume we can seperate the variable into two different functions: u(t,x)=T(t)X(x) plug into: \frac{ \partial u }{ \partial t }=D\frac{ \partial^2 u }{ \partial x^2 } T'X=DTX'' \frac{T'}{DT}=\frac{X''}{X}=-\lambda X''+\lambda X=-\lambda second order DE. u(t,x)=T(t)X(x) \implies T'(0)\cancel{ T(t) }=0 divide out T then the boundary conditions are X''+\lambda X=0 \quad X'(0)=0 \quad X'(L)=0 reminisient of dirchett problem conditions. the two edges are insulated here.
case 1) \lambda<0 \implies X(x)=c_{1}e^{\sqrt{ -\lambda }x}+c_{2}e^{-\sqrt{ -\lambda }x} X'(0)=0+c_{1}\sqrt{ -\lambda }-c_{2}\sqrt{ -\lambda }=0 X'(L)=0 c_{1}\sqrt{ -\lambda }e^{\sqrt{ -\lambda }L}-c_{2}\sqrt{ -\lambda }e^{-\sqrt{ -\lambda }L} \implies c_{1}=c_{2}
case 2) \lambda=0 X(x)=c_{1}x+c_{2} X'(0)=c_{1}=0 X'(L)=c_{1}=0 \implies X_{0}(x)=1 the rest are constant multiples, we could pick any non zero constant.
case 3) \lambda>0 X(x)=c_{1}\cos(\sqrt{ \lambda }x)+c_{2}\sin(\sqrt{ \lambda }x) X'(0)=-c_{1}\sqrt{ \lambda }\sin(0)+c_{2}\sqrt{ \lambda }\cos(\sqrt{ \lambda }0)=0 \implies c_{2}=0 X'(L)=-c_{1} \sqrt{ \lambda }\sin(\sqrt{ \lambda }L)=0 \implies c_{1}\sin(\sqrt{ \lambda }L)=0

\implies \lambda_{n}=\left( \frac{n\pi}{L} \right)^2 \qquad n=1,2,\dots

X_{n}(x)=\cancel{ c_{1} }\cos\left( \frac{n\pi x}{L} \right)

Any constant multiple of an eigen function is an eigen function so we get rid of the c_{1}. This is a different result than what we got last time, good to note.
\frac{T_{n}'}{DT_{n}}=-(\frac{n\pi}{L})^2 for n=0,1,2,\dots T_{n}(t)=c_{n}e^{-D(n\pi/L)^2t} u(t,x)=\frac{a_{0}}{2}+\sum_{n=1}^\infty a_{n}e^{-D(n\pi/L)^2t}\cos\left( \frac{n\pi x}{L} \right) a_{0}=2c_{0} a_{n}=c_{n} u(0,x)=\frac{a_{0}}{2}+\sum_{n=1}^\infty a_{n }\cos\left( \frac{n\pi x}{L} \right)=f(x) this is a fourier cos series, last time we had a fourier sin series. a_{0}=\frac{2}{L}\int _{0}^Lf(x) \, dx a_{n}=\frac{2}{L}\int _{0}^L f(x)\cos\left( \frac{n\pi x}{L} \right)\, dx plug these into the solution kaboom we have a solution to the problem in the form of a fourier series. u(t,x)=\frac{a_{0}}{2}+\sum_{n=1}^\infty a_{n}e^{-D(n\pi/L)^2t}\cos\left( \frac{n\pi x}{L} \right) and this is called a formal solution. do you remember the two rules for convergance for fourier series? theorems are not rules they are not axioms. a rule is given, you dont prove it, theorems are what you prove.

D=1, L=\pi, f(x)=\begin{cases}1, & 0\leq x\leq \frac{\pi}{2} \\0, & \frac{\pi}{2}<x\leq \pi\end{cases} this gives a similar solution: u(t,x)=\frac{a_{0}}{2}+\sum_{n=1}^\infty a_{n}e^{-n^2t}\cos\left({n x} \right) u(0,x)=\frac{a_{0}}{2}+\sum_{n=1}^ \infty a_{n}\cos\left( \frac{n\pi x}{L} \right)=f(x) a_{0}=\frac{2}{\pi}\int _{0} ^{\pi/2} \, dx=1 a_{n}=\frac{2}{\pi}\int _{0} ^{\pi/2} \cos(nx) \, dx=\frac{2}{n\pi}\sin(nx)|_{0} ^\frac{\pi}{2}=\frac{2}{n\pi}\sin\left( \frac{n\pi}{2} \right) \quad n=1,2,\dots u(t,x)=\frac{1}{2}+\frac{2}{\pi}\sum_{k=1}^\infty \frac{(-1)^{k+1}}{2k-1}e^{-(2k-1)^2t}\cos((2k-1)x) we are done, and we didn't need to take out the zero terms but if you want to be diligent, then there you go. "Can we just memorize the formula and plug in the values" His reply was something along the lines of: No! Please don't, you'll put in some arbitrary values with the wrong boundary conditions and arive with some crap result. You would need to redo the calculations and get completely new eigen values and eigen functions. I know how much you guys love formulas. But you need to understand what's happening, I don't know who thought it would be a good idea to make people memorize formulas, certainly not my idea. I want to be confident in the future engineers and the bridges that are built. ... Reminds me of what he said when he was talking about George Green, and how nowadays everything is McDonalds style, even our education. Maybe this is what he is referring to. #end of lec 32