revised wave equation

content/Heat equation (lec 30-33).md

@@ -116,7 +116,7 @@ By the way, there are only 5 more lectures left.
 Imagine instead of holding the two ends of the rod with a temperature of zero, we insulated the two ends.
 How do we insulate the ends? Set the heat flux at the ends to $0$:
 $k\frac{ \partial u }{ \partial x }(t,0)=k\frac{ \partial u }{ \partial x }(t,L)=0$ <- $k$ is the thermal conductivity of the tube.
-This gives a new equation type: Neumann type. (Neumann type means information about the derivative is specified, rather than the actual value.)
+This gives a new equation type: Neumann type. (Neumann type means information about the derivative is specified, rather than the actual value. There's also such thing as a mixed boundary condition, where maybe one end of the tube is insulated and the maybe other end is held at $0$.)
 $$\frac{ \partial u }{ \partial t }=D\frac{ \partial^2 u }{ \partial x^2 }, \quad 0\leq x\leq L, \quad t>0$$
 $$\frac{ \partial u }{ \partial x }(t,0)=\frac{ \partial u }{ \partial x }(t,L)=0, \quad t>0$$
 $$u(0,x)=f(x) \quad 0\leq x\leq L$$
@@ -157,7 +157,8 @@ $X'(0)=0=-c_{1}\sqrt{ \lambda }\sin(0)+c_{2}\sqrt{ \lambda }\cos(\sqrt{ \lambda
 $X'(L)=0=-c_{1} \sqrt{ \lambda }\sin(\sqrt{ \lambda }L) \implies c_{1}\sin(\sqrt{ \lambda }L)=0$
 $$\implies \lambda_{n}=\left( \frac{n\pi}{L} \right)^2 \qquad n=1,2,\dots$$
 $$ \implies X_{n}(x)=c_{n}\cos\left( \frac{n\pi x}{L} \right)$$
->This is a different result than what we got last time, for comparison, the Dirichlet problem gave us the same expression for $\lambda_{n}$ however the $X_{n}(x)$ was $X_{n}(x)=c_{n}\sin\left( \frac{n\pi x}{L} \right)$
+We don't need the multiples of an eigen value, so we can set $c_{n}=1$ (or any arbitrary non zero constant). However, I will keep it just so you can see how it's redundant and gets removed in a later step anyways.
+>This result is different than what we got last time, for comparison, the Dirichlet problem gave us the same expression for $\lambda_{n}$ however the $X_{n}(x)$ was $X_{n}(x)=c_{n}\sin\left( \frac{n\pi x}{L} \right)$
 
 </br>
 plug $\lambda$ into: $\frac{T'}{DT}=\frac{X''}{X}=-\lambda$

content/Separation of variables & Eigen value problems (lec 26-28).md

@@ -112,7 +112,7 @@ but if $\sin(\sqrt{ \lambda }L)=0 \implies \sqrt{ \lambda }L=n\pi$
 where $n=1,2,3,\dots$ notice that $n\ne 0$ because that implies $\lambda=0$ but $\lambda>0$ in case 3 so that cant be.
 $\lambda_{n}=(\frac{n\pi}{L})^2$ <- These are called <u>eigen values</u>.
 The corresponding <u>eigen functions</u> are:
-$X_{n}(x)=c_{2}\sin(\sqrt{ \lambda_{n} }x)=c_{n}\sin\left( \frac{n\pi}{L}x \right)$
+$X_{n}(x)=c_{2}\sin(\sqrt{ \lambda_{n} }x)=c_{n}\sin\left( \frac{n\pi x}{L} \right)$
 ^They are the only non-trivial solutions.
 </br>
 lets go back to the problem and focus on $T$:

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

@@ -2,12 +2,12 @@
 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.
-$\frac{ \partial u^2 }{ \partial t^2 }=\alpha^2 \frac{ \partial^2 u }{ \partial x^2 } \quad 0\leq x\leq L,\quad t>0$
+$\frac{ \partial u^2 }{ \partial t^2 }=\alpha^2 \frac{ \partial^2 u }{ \partial x^2 }\qquad 0\leq x\leq L,\quad t>0$
 ^ Reminds me of the wave equation from Phys 130.
 Since the string is tied down on the ends we have the following initial conditions:
 $u(t,0)=u(t,L)=0 \qquad t>0$
-$u(0,x)=f(x)$  $0\leq x\leq L$
-$\frac{ \partial u }{ \partial t }(0,x)=g(x)$ $0\leq x\leq L$
+$u(0,x)=f(x) \qquad 0\leq x\leq L$ <- position of each segment of string at $t=0$
+$\frac{ \partial u }{ \partial t }(0,x)=g(x) \qquad 0\leq x\leq L$ <- velocity of each segment of string at at $t=0$
 ^IBVP of the system.
 #end of lec 33
 #start of lec 34
@@ -16,93 +16,113 @@ 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 !
+$\frac{T''}{\alpha^2T}=\frac{X''}{X}=-\lambda$  <- eigen value problem #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:
+We've solved this before (in lec 27).
+The only non-trivial solutions are:
 $\lambda_{n}=(\frac{n\pi}{L})^2$
 $X_{n}(x)=\sin\left( \frac{n\pi x}{L} \right)$ for $n=1,2,3,\dots$
 $\frac{T''}{\alpha^2T}=-\left( \frac{n\pi}{L} \right)^2$
 $T_{n}''+\left( \frac{\alpha n\pi}{L} \right)^2T_{n}=0$
+Unlike before where we could integrate both sides, we can't here! Last time it was a separable equation, now this one is a second order linear equation, and we must solve it as such.
 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?"
 $T_{n}(t)=b_{n}\cos\left( \frac{\alpha n\pi}{L}t \right)+a_{n}\sin\left( \frac{\alpha n\pi}{L}t \right)$
-$u_{n}(t,x)=\left( b_{n}\cos\left( \frac{\alpha n\pi}{L} \right)+a_{n}\sin\left( \frac{\alpha n\pi}{L}t \right) \right)\sin\left( \frac{n\pi x}{L} \right)$
+$u_{n}(t,x)=\left( b_{n}\cos\left( \frac{\alpha n\pi}{L}t \right)+a_{n}\sin\left( \frac{\alpha n\pi}{L}t \right) \right)\sin\left( \frac{n\pi x}{L} \right)$
 if you sum all these <u>modes</u>, you get the solution:
-$$u_{n}(t,x)=\sum_{n=1}^\infty\left( b_{n}\cos\left( \frac{\alpha n\pi}{L} \right)+a_{n}\sin\left( \frac{\alpha n\pi}{L}t \right) \right)\sin\left( \frac{n\pi x}{L} \right)$$
+$$u_{n}(t,x)=\sum_{n=1}^\infty\left( b_{n}\cos\left( \frac{\alpha n\pi}{L} t\right)+a_{n}\sin\left( \frac{\alpha n\pi}{L}t \right) \right)\sin\left( \frac{n\pi x}{L} \right)$$
 
-"mathematics and reality do align very well, if the speed of the string is different the solution differs aswell."
+"mathematics and reality do align very well, if the speed of the string is different [by varying $\alpha$] the solution differs as well."
+The first initial condition gives:
 $u(0,x)=f(x)=\sum_{n=1}^\infty b_{n}\sin\left( \frac{n\pi x}{L} \right)$
-^ 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.
-$\frac{ \partial u }{ \partial t }(0,x)=g(x)=\sum_{n=1}^\infty\underbrace{  a_{n} \frac{\alpha n\pi}{L} }_{  }\sin\left( \frac{\alpha n\pi}{L} \right)$
+$$\implies b_{n}=\frac{2}{L}\int _{0}^L f(x)\sin\left( \frac{n\pi x}{L} \right) \, dx$$
+Where does it converge? Well in real life your string, $f(x)$, is continuous and your $f'(x)$ is piecewise continuous, so it will converge everywhere. If $f(x)$ was discontinuous in real life that would mean your string has a break in it.
+The second initial condition gives:
+$\frac{ \partial u }{ \partial t }(0,x)=g(x)=\sum_{n=1}^\infty a_{n} \frac{\alpha n\pi}{L}\sin\left( \frac{\alpha n\pi}{L} \right)$
 $a_{n} \frac{\alpha n\pi}{L}$ are the Fourier $\sin$ coefficients of $g(x)$
-$a_{n}=\frac{2}{\alpha n\pi}\int _{0}^L g(x)\sin\left( \frac{n\pi x}{L} \right)\, dx$
+$\implies a_{n} \frac{\alpha n\pi}{L}=\frac{2}{L}\int _{0}^L g(x)\sin\left( \frac{n\pi x}{L} \right)\, dx$
+$$a_{n}=\frac{2}{\alpha n\pi}\int _{0}^L g(x)\sin\left( \frac{n\pi x}{L} \right)\, dx$$
 $\alpha^2$ 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.
+^ This is a good general formula for computing wave equations, as long as the initial conditions match up, you can just plug in your values and you're done.
+</br>
+Remember the heat equation, the amplitude is exponentially decreasing,
+here the amplitude is oscillatory and doesn't decrease in time. boi-oi-oi-oing.
+To make it more realistic 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 field, to some approximation.
+the lowest mode is called the fundamental mode, the following terms after are called harmonics.
+If two instruments play the same note (same fundamental frequency), they sound different! and that's because of their 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.
+"my claim, and this is not just 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 that's why its important to learn the wave equation. It's the prototype to waves.
 "waves are the fundamental object. [...]. So that's why it's important, these are the fundamental objects of nature here."
+</br>
+So, let's try an example using our formula above:
 $f(x)=\begin{cases}x, & 0\leq x\leq \frac{\pi}{2}\\ \ \pi-x, & \frac{\pi}{2}<x\leq \pi\end{cases}$
 $\alpha^2=1$
 $g(x)=\sin(x)$
+$L=\pi$
+</br>
+Plug in:
 $b_{n}=\frac{2}{\pi}\left( \int _{0} ^\frac{\pi}{2} x\sin(nx) \, dx +\int _{\frac{\pi}{2}} ^\pi (\pi-x)\sin(nx) \, dx\right)$
+Skip some integration:
 $b_{n}=\frac{4}{n^2\pi}\sin\left( \frac{n\pi}{2} \right)$
-$b_{2k}=0$
 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:
+$b_{2k}=0$ for $k=1,2,\dots$
+$b_{2k-1}=\frac{4}{(2k-1)^2\pi}(-1)^{k+1}$ for $k=1,2,\dots$
+Plug in g(x) to get $a_n$ terms
+however, there's a short cut here. From definition of $g(x)$ earlier:
+$g(x)=\sum_{n=1}^\infty a_{n} \frac{\alpha n\pi}{L}\sin\left( \frac{\alpha n\pi}{L} \right)$
 $g(x)=a_{1}\sin(x)+2a_{2}\sin(2x)+3a_{3}\sin(3x)+\dots$
-but g(x) is sin x.
+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
+plug this in to get the solution:
+$$u(t,x)=\sin (t)\sin(x)+\sum_{k=1}^\infty \frac{4(-1)^{k+1}}{\pi(2k-1)^2}\cos((2k-1)t)\sin((2k-1)x)$$
+Peep this gif:
+red is $f(x)$, purple is $u(t,x)$. Looks kinda stringy in behavior!
+![plot](drawings/wave_eq.gif)
+I was kinda confused from the plot, if the initial speed is $g(x)=\sin(x)$ then shouldn't that string peak pop up a bit before heading down? Because it has an initial speed of $g\left( \frac{\pi}{2} \right)=1$ in the upwards direction. Surprisingly the answer is no, the string experiences infinite acceleration in the downwards direction at that pinch point at $t=0$. So the initial speed at $x=\frac{\pi}{2}$ goes to zero as soon as the string is put into motion. How strange is that! Imagine being an ant on the string, as soon as any "pinch point" on the string reaches you, you would be knocked right off, either the floor suddenly drops from beneath your feet or the floor pushes up and crushes you.
 #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 (?)
-
+Last problem of the course which we will finish today.
+$$\frac{ \partial^2 u }{ \partial t^2 }=\frac{ \partial^2 u }{ \partial x^2 }+tx \qquad 0\leq x\leq \pi, \quad t>0$$
+That's a driven wave equation. $tx$ is the source term.
+$$u(0,t)=u(\pi,t)=0 \qquad t>0$$
+$$u(x,0)=\sin(x) \qquad 0\leq x\leq \pi$$
+$$\frac{ \partial u }{ \partial t }(x,0)=5\sin(2x)-3\sin(5x)\qquad 0\leq x\leq \pi$$
 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.)
+since $u(0,t)=u(\pi,t)=0$ we can expect a solution of the form:
+$u(x,t)=\sum_{n=1}^\infty u_{n}(t)\sin\left( \frac{n\pi x}{\pi} \right)$
+If any of the boundary conditions are non zero, then we have to split(?) into X and T. (needs verification)
+If we expand $tx$ as a Fourier series:
+$tx=\sum_{n=1}^\infty h_{n}\sin(nx)$ (this is called a formal expansion, the two actually aren't exactly equal due to the discontinuity in $tx$.)
 $h_{n}=\frac{2}{\pi}\int _{0} ^\pi tx\sin(nx)\, dx$
+here's a plot of the formal expansion of $tx$. The solid line represents $tx$ and the dotted line represents what $tx$ will look like in the future as $t$ increases. This value $tx$ is proportional to a pulling force placed on each point on the string. So we should expect the string to be stretched out! As you'll see in the last plot.
 ![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.
+Continuous between $0$ and $\pi$ but on the edge points, the Fourier sine series will converge to the midpoint of the two edge points.
+Integrate by parts:
 $h_{n}=\frac{2t}{\pi}\int _{0}^\pi x\sin(nx)\, dx=- \frac{2t}{\pi n}\left( x\cos(nx)|_{0}^\pi-\int _{0}^\pi \cancel{ \cos(nx) }\, dx \right)$
-$h_{n(t)}=\frac{2t}{n}(-1)^{n+1}$
+$h_{n}(t)=\frac{2t}{n}(-1)^{n+1}$
 $h(x,t)=tx=\sum_{n=1}^\infty h_{n}\sin(nx)=\sum_{n=1}^\infty \frac{2t}{n}(-1)^{n+1}\sin(nx)$
 $\frac{ \partial^2 u }{ \partial t^2 }=\sum_{n=1}^\infty u_{n}''(t)\sin(nx)$
 $\frac{ \partial^2 u }{ \partial x^2 }=\sum_{n=1}^\infty -u_{n}(t)n^2\sin(nx)$
+plug them into the equation
 $\sum_{n=1}^\infty \underbrace{ \left( u_{n}''+n^2u_{n}+\frac{2t}{n}(-1)^n \right) }_{ =0 }\sin(nx)=0$
 $u''_{n}+n^2u_{n}=\frac{2}{n}(-1)^{n+1}t, \quad n=1,2,\dots$
-use #mouc (or laplace, but that'll take much longer.)
+^Hey, we got a DE!
+use method of undetermined coefficients #mouc (or laplace, but that'll take much longer.)
 characteristic eq:
 $r^2+n^2=0$
 $r_{1,2}=\pm in$
+homogenous solution is simply:
 $u_{n}^h(t)=a_{n}\cos(nt)+b_{n}\sin(nt)$
+particular solution is guessed in the form:
 $u_{n}^p(t)=At+B$
 $B=0$ because there's no constant term on the RHS
 $A=\frac{2(-1)^{n+1}}{n^3}$
@@ -112,12 +132,14 @@ This is the last problem I'll be solving in my career. This is the last time he'
 $u(x,0)=\sum_{n=1}^\infty a_{n}\sin(nx)=\sin(x)$
 $a_{1}=1, \quad a_{k}=0, \quad k=2,3,\dots$
 $\frac{ \partial u }{ \partial t }(x,0)=\sum_{n=1}^\infty\left( b_{n}n+\frac{2(-1)^{n+1}}{n^3} \right)\sin(nx)=5\sin(2x)-3\sin(5x)$
-coefficients =0 if $n\ne_{2},5$
-$\implies b_{n}=\frac{2}{n^4}(-1)^n, \quad n\ne_{2},5$
+coefficients=0 if $n\ne2,5$
+$\implies b_{n}=\frac{2}{n^4}(-1)^n, \quad n\ne 2,5$
 $b_{2}=\frac{5}{2}+\frac{2(-1)^2}{2^4}$
 $b_{5}=-\frac{3}{5}+\frac{2(-1)^5}{5^4}$
-$u(x,t)=\cos(t)\sin(x)+\frac{5}{2}\sin(2t)\sin(2x)-\frac{3}{5}\sin(5t)\sin(5x)+2\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^3}\left( t-\frac{\sin(nt)}{n} \right)\sin(nx)$
-finished the solution. Man I got teary eyed from this lecture.
+$$u(x,t)=\cos(t)\sin(x)+\frac{5}{2}\sin(2t)\sin(2x)-\frac{3}{5}\sin(5t)\sin(5x)+2\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^3}\left( t-\frac{\sin(nt)}{n} \right)\sin(nx)$$
+Here's a plot showing the behavior of the string (graphed up to 40 harmonics):
+![plot](drawings/driven_wave_eq.gif)
+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?

content/Wave equations (lec 35).md

@@ -1,58 +0,0 @@
-#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)
-...
-</br>
-#end of lec 36
-#end of Math 201. Congratulations!

content/_index.md

@@ -26,7 +26,7 @@ I have written these notes for myself, I thought it would be cool to share them.
 [Separation of variables & Eigen value problems (lec 26-28)](separation-of-variables-eigen-value-problems-lec-26-28.html)
 [Fourier series (lec 28-29)](fourier-series-lec-28-29.html)
 [Heat equation (lec 30-33)](heat-equation-lec-30-33.html)
-[Wave equation (lec 33-36)](wave-equation-lec-33-36.html) (raw notes, not reviewed or revised yet.)
+[Wave equation (lec 33-36)](wave-equation-lec-33-36.html)
 
 </br>
 [How to solve any DE, a flow chart](Solve-any-DE.png) (Last updated Oct 1st, needs revision. But it gives a nice overview.)