revised Separation of variables & Eigen value problems

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Sasserisop 2023-12-03 17:47:59 -07:00
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@ -23,7 +23,7 @@ $u'(t-a)=\delta(t-a)$
What is $\mathcal{L}\{\delta(t-a)\}$?
$\mathcal{L}\{\delta(t-a)\}=\int _{0} ^\infty \delta(t-a)e^{-st} \, dt$ for $a>0$
Using the definition earlier:
Using the definition of Dirak $\int _{-\infty} ^{\infty} \delta(t-a)f(t)\, dt=f(a)$:
$$\mathcal{L}\{\delta(t-a)\}=\int _{-\infty} ^{\infty} e^{-st} \delta(t-a) \, dt=e^{-as}$$
## Examples of DE's with Dirak
#ex #second_order_nonhomogenous #dirak_delta #IVP

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@ -49,10 +49,10 @@ If $f$ and $f'$ are piecewise continuous on $[-L,L]$, then the fourier series co
$\frac{1}{2}(f(x^-)+f(x^+))$ for all $x \in (-L,L)$
Basically meaning, the fourier series converges.
At $x=\pm L$ the fourier series converges to $\frac{1}{2}(f(-L^+)+f(L^-))$
![[Partial differential equations (lec 26-27) 2023-11-22 13.15.26.excalidraw]]
![draw](drawings/Drawing-2023-11-22-13.15.26.excalidraw.png)
### Theorem:
If f(x) is continuous on $(-\infty,\infty)$ and $2L$ periodic and if $f'$ is continuous, then the taylor series converges to $f(x)$ everywhere
![[Partial differential equations (lec 26-27) 2023-11-22 13.14.05.excalidraw]]
![draw](drawings/Drawing-2023-11-22-13.14.05.excalidraw.png)
#ex lets compute the fourier transform of:
$f(x)=\begin{cases}1, & -\pi\leq x\leq 0 \\x, & 0<x\leq \pi\end{cases}$
$L$ here is $\pi$ clearly.
@ -124,7 +124,7 @@ then $\frac{a_{0}}{2}+\sum_{n=1}^\infty a_{n}\cos\left( \frac{n\pi x}{L} \right)
#ex Fourier sine series for $f(x)=x^2$ from $0\leq x\leq \pi$
well that means we want the odd extension:
![[Lec 30 2023-11-24 13.15.17.excalidraw]]
![draw](drawings/Drawing-2023-11-24-13.15.17.excalidraw.png)
the $\cos()$ ($a_{n}$) terms are zero.
the b terms are:
$b_{n}=\frac{2}{\pi}\int _{0}^\pi x^2\sin(nx) \, dx$
@ -134,7 +134,7 @@ $b_{n}=-\frac{2}{n\pi}\left[ \pi^2(-1)^n-\frac{2}{n^2}\cos(nx)|_{0}^\pi \right]=
for $n=1,2,3,\dots$ note no $n=0$ so no divison by zero problems here.
#ex fourier cosine series of $f(x)=\sin(x)$ for $0\leq x\leq \pi$
![[Lec 30 2023-11-24 13.23.08.excalidraw]]
![draw](drawings/Drawing-2023-11-24-13.23.08.excalidraw.png)
$a_{n}=\frac{2}{\pi}\int _{0}^\pi \sin(x)\cos(nx)\, dx$ for $n=0,1,2,\dots$
use trig identity: (by the way the identites will be provided in the exam.)

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@ -35,7 +35,7 @@ $k=1,l=-3$
so $x=u+1 \quad y=v-3$
$(-3u+v)du+(u+v)dv=0$ //Beautiful! It's homogenous now
$\frac{ dv }{ du }=\frac{{3u-v}}{u+v}$
divide top and bottom by u so we turn the homogenous equation into the form #de_h_type1 and solve it using the tools we developed from lecture 2.
divide top and bottom by $u$ so we turn the homogenous equation into the form #de_h_type1 and solve it using the tools we developed from lecture 2.
$\frac{ dv }{ du }=\frac{{3-\frac{v}{u}}}{1+\frac{v}{u}}$
$\frac{v}{u}=w \quad v=uw \quad \frac{ dv }{ du }=w+u\frac{ dw }{ du }$

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@ -8,7 +8,7 @@ $u(t,0)=u(t,L)=0, \quad t>0$
$u(0,x)=f(x), \quad 0\leq x\leq L$
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]]
![draw](drawings/Drawing-2023-11-24-13.42.29.excalidraw.png)
So we have a non-uniformly heated rod with both ends insulated. What happens to the temperature inside the rod over time? <i>"\[...\]. Very interesting problem."</i> -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: $u(t,x)=T(t)X(x)$
@ -229,7 +229,7 @@ ok i spent 5 minutes talking nonsense but
you complaining that your exams are hard theyre not hard. I'm talking about 40 years ago, no computers no cellphones, and it wasnt so bad because we had to use our brain more.
now we consider a guitar string:
![[Partial differential equations (lec 30-32) 2023-12-01 13.49.58.excalidraw]]
![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.
$\frac{ \partial u^2 }{ \partial t^2 }=\alpha^2 \frac{ \partial^2 u }{ \partial x^2 } \quad 0\leq x\leq L, t>0$
^ Reminds me of the wave equation from phys 130.

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@ -189,7 +189,6 @@ if $p(x)$ and $q(x)$ are <u>analytic</u> functions in a vicinity of $x_{0}$ then
we expect that the solution $y$ can be represented by a power series. This is true according to the following theorem:
Theorem: If $x_{0}$ is an ordinary point then the differential equation above has two linearly independent solution of the form $\sum_{n=0} ^\infty a_{n}(x-x_{0})^n, \qquad\sum_{n=0}^\infty b_{n}(x-x_{0})^n$.
The radius of convergence for them is at least as large as the distance between $x_{0}$ and the closest singular point (which can be real or complex).
![draw](drawings/Drawing-2023-10-30-13.12.57.excalidraw.png)
## Examples for calculating $\rho$
#ex
@ -306,7 +305,7 @@ $\left( 30a_{6}+a_{3}-\cancelto{ 0 }{ \frac{a_{1}}{6} } \right)t^4=0 \implies a_
</br>
substitute back $x-\pi=t$
$$y(x)=1-\frac{1}{6}(x-\pi)^3+\frac{1}{120}(x-\pi)^5+\frac{1}{180}(x-\pi)^6+\dots$$
there's no general formula here for the constants (or maybe he said no formula for y(x)?), but we can write the solution in the following form^.
there's no general formula here for the constants (or maybe he said no formula for $y(x)$?), but we can write the solution in the following form^.
#ex #powseries
Here's a non-homogenous example: (RHS$\ne 0$)

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@ -5,140 +5,185 @@ We start with some thermodynamics
Heat equation not only describes thermodynamics but it can also model the diffusion of gasses. It is a partial differential equation.
Strikingly, it can also model option prices in the stock market. However, using it as a strategy to make money is not so simple, because if it worked then everyone would try to use it to make money, which would cause the overall strategy to be less effective as the option prices start to get priced to accommodate for the prediction (🤯).
![draw](drawings/Drawing-2023-11-08-13.07.19.excalidraw.png)
We assume that the tube is perfectly insulating along its surface, this helps reduce the problem into a one dimensional problem. Heat can only travel inside and along the x axis.
We assume that the tube is perfectly insulating along its cylindrical surface, this helps reduce the problem into a one dimensional problem. Heat can only travel inside and along the x axis.
We can express the heat along the tube as $u(t,x)$
>I like to visualize $u(t,x)$ as a wavetable that smooths out as the time variable progresses. Something like this: [(image source)](https://markmoshermusic.com/2015/07/29/intro-to-loading-custom-waldorf-blofeld-wavetables/)
![wavetable.jpg](drawings/wavetable.jpg)
## Derivations (I'd skip this)
Fourier figured out that:
$\text{Heat flux} = -k(x)a\frac{\partial u}{\partial x}(t,x) \Delta t$
heat flux is in the positive $x$ direction
where du/dx is the opposite sign of the flux (because hot flows to cold.)
where $u(t,x)$ is a function that describes the temperature in the tube.
$-k(x+\Delta x)a\frac{\partial u}{\partial x}(t,x+\Delta x) \Delta t$
$k(x)$ is the thermal conductivity of the tube at point $x$
$\frac{ \partial u }{ \partial x }$ is always opposite in sign of flux (because hot flows to cold.)
$a$ is the area of the cross-section.
$=-k(x+\Delta x)a\frac{\partial u}{\partial x}(t,x+\Delta x) \Delta t$
$\Delta u=u(t+\Delta t,x)-u(t,x)$
Amount of heat to change temperature over $\Delta t$ with $\Delta u$ is $\underbrace{ C(x) }_{ \text{specific heat capacity } }\underbrace{ \rho(x) }_{ \text{density} }\Delta u\underbrace{ a\Delta x }_{ \text{volume} }$
$c(x)\rho(x)a\Delta x(u(t+\Delta t,x)-u(t,x))=a\Delta t(k(x+\Delta x) \frac{\partial u}{\partial x}(t_{1}x+\Delta x)-k(x) \frac{\partial u}{\partial x}(t,x))+Q(t,x)\Delta xa\Delta t$
$C(x)\rho(x)a\Delta x(u(t+\Delta t,x)-u(t,x))=a\Delta t(k(x+\Delta x) \frac{\partial u}{\partial x}(t,x+\Delta x)-k(x) \frac{\partial u}{\partial x}(t,x))+Q(t,x)\Delta xa\Delta t$
>I'm gonna be honest, I'm lost in this derivation already. What is $Q(t,x)$?
divide by $a\Delta x\Delta t$
$c(x)\rho(x)\frac{(u(t+\Delta t,x)-u(t,x))}{\Delta t}=a\Delta t\left( k(x+\Delta x) \frac{\partial u}{\partial x}(t,x+\Delta x)-k(x) \frac{\partial u}{\partial x}(t,x) \right)+Q(t,x)$
$\lim_{ \Delta x,\Delta t \to 0 }: c(x)\rho(x)\frac{ \partial u }{ \partial t }(t,x)=\frac{ \partial }{ \partial x }\left( k(x)\frac{ \partial u }{ \partial x }(t,x) \right)+Q(t,x)$
$C(x)\rho(x)\frac{(u(t+\Delta t,x)-u(t,x))}{\Delta t}=\frac{ k(x+\Delta x) \frac{\partial u}{\partial x}(t,x+\Delta x)-k(x) \frac{\partial u}{\partial x}(t,x)}{\nabla x}+Q(t,x)$
$\lim_{ \Delta x,\Delta t \to 0 }: C(x)\rho(x)\frac{ \partial u }{ \partial t }(t,x)=\frac{ \partial }{ \partial x }\left( k(x)\frac{ \partial u }{ \partial x }(t,x) \right)+Q(t,x)$
Any differential equation, to his knowledge, is to balance some volume and to take the limit to produce a pointwise / instantaneous balance (in the form of a differential equation)
the Maxwell equations that describes magnetic and electric fields are a system of partial differential equations.
thermodynamics can be very important for electrical engineers, for instance the heat produced in a transformer, or a battery. It has applications.
## Separation of variables & Eigen value problems
#ex #SoV #evp
(This is more of a case study than an example.)
Rewrite the equation we derived by grouping the constant terms into one constant $D$
$\frac{ \partial u }{ \partial t }=D \frac{ \partial^2 u }{ \partial x^2 }, \quad 0\leq x\leq L, \quad t>0$
boundary conditions:
$u(t,0)=u(t,L)=0 , \quad t>0$ (simple case)
$u(0,x)=f(x) , \quad 0\leq x\leq L$
These three equations form an IBVP (initial boundary value problem)
we will work on this equation next lecture:
Separation of variables starts with the assumption that the solution can be written as:
$u(t,x)=T(t)X(x)$
$T'x=DTx''$
divide by DTx:
$\frac{T'}{DT}=\frac{x''}{x}$
on the left is a function of time only, on the right is a function of x only.
$\frac{T'}{DT}=\frac{x''}{x}=-\lambda$ where $\lambda$ is a constant
$x''+\lambda x=0$
$u(t,0)=\cancel{ T(t) } \quad x(0)=0$
$x(0)=x(L)=0$
this is called an eigen value problem.
plug into equation:
$\frac{ \partial}{ \partial t }(T(t)X(x))=D \frac{ \partial^2 }{ \partial x^2 }(T(t)X(x))$
$T'X=DTX''$
divide by $DTX$:
$\frac{T'}{DT}=\frac{X''}{X}$
recall $T$ and $X$ are single variable functions ($u(t,x)=T(t)X(x)$)
that means the left is a function of $t$ only, on the right is a function of $x$ only.
$\frac{T'}{DT}=\frac{X''}{X}=-\lambda$ where $\lambda$ is a constant
we first consider the problem:
$\frac{X''}{X}=-\lambda$
$X''+\lambda X=0$ (called an eigen value problem on a second derivative operator)
$u(t,0)=0=T(t)X(0)$
$\implies X(0)=0$
$T(t)$ cannot be the 0 term because if it was then you get $u(t,x)=T(t)X(x)=0$ for all $t$ and $x$ which can't be.
$u(t,L)=0=T(t)X(L)$
$\implies X(L)=0$
</br>
$\begin{cases}X''+\lambda X=0 \\X(0)=0 \\X(L)=0\end{cases}$ <-This is called an eigen value problem.
#end of lec 26 #start of lec 27
Recall from last lec:
$\frac{ \partial u }{ \partial t }=D \frac{ \partial^2 u }{ \partial x^2 }, \quad 0\leq x\leq L, \quad t>0$
boundary conditions:
$u(t,0)=u(t,L)=0 , \quad t>0$ (simple case)
$u(0,x)=f(x) , \quad 0\leq x\leq L$
This is an IBVP
$\frac{T'}{DT}=\frac{x''}{x}=-\lambda$
$x''+\lambda x=0$ (called an eigen value problem on a second derivative operator)
$x(0)=x(L)=0$
its an eigen value problem because lambda is not fixed, it can be any number.
recall in linear algebra $Ax+\lambda x=0$ where A is a matrix
eigen is a germanic word meaning intrinsic, important
(i)$\lambda<0$
$r^2+\lambda=0 \implies r_{1,2}=\pm \sqrt{ -\lambda }$
$X(x)=c_{1}e^{\sqrt{ -\lambda }x}+c_{2}^{-\sqrt{ -\lambda }x}$
It's an eigen value problem because unlike before when we did second order linear equations with constant coefficients, lambda is not fixed, it can be any number.
recall in linear algebra $Ax+\lambda x=0$ where $A$ is a matrix
eigen is a Germanic word meaning intrinsic, important
We find the characteristic equations by splitting into three cases, just like we've done since lecture 7
$X''+\lambda X=0$
case 1) $\lambda<0$
$r^2+\lambda=0 \implies r_{1,2}=\pm \sqrt{ -\lambda }$ (real solutions only since $\sqrt{ -(-) }=\sqrt{ + }$)
$X(x)=c_{1}e^{\sqrt{ -\lambda }x}+c_{2}e^{-\sqrt{ -\lambda }x}$
but we have a boundary condition:
$X(0)=0=c_{1}+c_{2}$
$c_{1}+c_{2}=0$
$X(L)=c_{1}e^{\sqrt{ -\lambda }L}+c_{2}^{-\sqrt{ -\lambda }L}=0$
this has a solution, as the determinant $\ne 0$
the solution is $c_{1}=0,\ c_{2}=0$
$X(L)=c_{1}e^{\sqrt{ -\lambda }L}+c_{2}e^{-\sqrt{ -\lambda }L}=0$
this has a unique solution, as the determinant is non-zero: $\det\left(\begin{matrix}1 & 1 \\e^{\sqrt{ -\lambda }L} & e^{-\sqrt{ -\lambda }L}\end{matrix}\right)=e^{-\sqrt{ -\lambda }L}-e^{\sqrt{ -\lambda }L}\ne 0$ (as long as $L\ne 0$, which is true since we are assuming the tube has non-zero length.)
the only solution is $c_{1}=0,\ c_{2}=0$
which gives $X(x)=0$ for all $x\in[0,L]$ very boring solution!
other option:
(ii) $\lambda=0$
$x''=0$
integrate both sides twice:
$X(x)=c_{1}x+c_{2}$
boundary conditions:
$X(0)=0\implies c_{2}=0$
$X(L)=0=c_{1}L \implies c_{1}=0$
again a boring solution! $X(x)=0$ this is called a trivial solution.
We are looking for non-trival solutions.
try the last other possible case:
(iii) $\lambda>0$
$r_{1,2}=\pm \sqrt{ \lambda }i$
</br>
case 2) $\lambda=0$
$X''=0$
$r^2=0$
$r=0$ (repeated root)
$X(x)=c_{1}e^{0}+c_{2}0e^0=c_{1}$
$X(0)=0\implies c_{1}=0$
$X(L)=0 \implies c_{1}=0$
there are no restrictions on $c_{2}$
but our solution is $X(x)=0$ again for all $x$. This is called a trivial solution.
>alternatively, we could have found this in a different approach:
>$X''=0$
>integrate both sides twice:
>$X(x)=c_{1}x+c_{2}$
>boundary conditions:
>$X(0)=0\implies c_{2}=0$
>$X(L)=0=c_{1}L \implies c_{1}=0$
>$\implies X(x)=0$
>This method worked because the LHS is just one term of $X^n$ so we treated it as a separable equation.
We are looking for non-trival solutions. Try the last other possible case:
case 3) $\lambda>0$
$r^2+\lambda=0$
$r_{1,2}=\pm \sqrt{ -\lambda }=\pm i\sqrt{ \lambda }$
$X(x)=c_{1}\cos(\sqrt{ \lambda }x)+c_{2}\sin(\sqrt{ \lambda }x)$
boundary conditions:
$X(0)=0\implies c_{1}=0$
$X(L)=0 \implies c_{2}\sin(\sqrt{ \lambda }L)=0$
if you take $c_{2}=0$ you lose, another boring solution.
if you take $c_{2}=0$ you lose, another boring solution ($X(x)=0$).
but if $\sin(\sqrt{ \lambda }L)=0 \implies \sqrt{ \lambda }L=n\pi$
where $n=1,2,3,\dots$ notice that $n\ne 0$ becuase that implies lambda=0 but lambda is >0 so that cant be.
$\lambda_{n}=\frac{n\pi}{L}$
$X_{n}(x)=c_{2}\sin(\sqrt{ \lambda }x)=\sin\left( \frac{n\pi}{L}x \right)$ this is an eigen function. It's the only non-trivial solution.
lets go back to the problem:
$\frac{T'}{DT}=\frac{x''}{x}=-\lambda$
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:
$X_{n}(x)=c_{2}\sin(\sqrt{ \lambda_{n} }x)=c_{n}\sin\left( \frac{n\pi}{L}x \right)$
^They are the only non-trivial solutions.
</br>
lets go back to the problem and focus on $T$:
$\frac{T'}{DT}=\frac{X''}{X}=-\lambda$
$\frac{T'}{T}=-\left( \frac{n\pi}{L} \right)^2D$
this is a separable equation.
$T_{n}(t)=c_{n}e^{-(n\pi/2)^2Dt}$
you might have thought you could forget the material before the midterm to make more space in your brain, your brains are a lot more emptier than mine, mine is filled with garbage [...] so don't forget anything you learned before the midterm!
you'll see that as you continue in education, you'll see a lot of completely new things and you'll have to find shortcuts and tricks to make the content fit what you already know.
okay enough with life stuff, back to mathematics:
$\implies u_{n}(t,x)=c_{n}e^{-(n\pi/2)^2Dt}\sin\left( \frac{n\pi x}{L} \right)$ $n=1,2,3,\dots$
applying superposition (sum of any solutions is also a solution)
$u(t,x)=\sum_{n=1}^\infty c_{n}e^{-(n\pi/L)^2Dt}\sin\left( \frac{n\pi x}{L} \right)$ this is the most general form of the solution.
$u(0,x)=f(x)=\sum_{n=1}^\infty c_{n}\sin\left( \frac{n\pi x}{L} \right),\ 0\leq x\leq L$ we have never seen anything like this before, an infinite number of sin terms added together, usually its polynonmials we sum.
this is when fourier (?) steps in. he proved that you can represent various functions as a sum of sines.
fun stories about pioncre and cauchy euler: cauchy euler was an engineer, and pioncre had his theorem released around 2008 and it was really long, like 400 pages, he posted it online and asked if anyone wanted to prove it, after a while 4-5 or so mathematicians checked his proof and said, yep okay the proof looks correct. His theorem has a lot to do with the material world.
f doesn't even have to be analytic in order for it to be expressed as a sum of sines.
We can treat the function T as a variable:
$\frac{dT}{dt} \frac{1}{T}=-\left( \frac{n\pi}{L} \right)^2D$
$\int{dT} \frac{1}{T}=\int-\left( \frac{n\pi}{L} \right)^2Ddt$
$\ln(T)=-\left( \frac{n\pi}{L} \right)Dt+c_{n}$
$T_{n}(t)=c_{n}e^{-(\frac{n\pi}{L})^2Dt}$
>Yes this looks illegal, but it works, you could also integrate more rigorously if you did a u-sub: $u=T(t) \quad \frac{du}{dt}=T'(t)$)
</br>
><i>"You might have thought you could forget the material before the midterm to make more space in your brain, your brains are a lot more emptier than mine, (class laughs) mine is filled with garbage [...] so don't forget anything you learned before the midterm!"</i>
><i>"You'll see that as you continue in education, you'll see a lot of completely new things and you'll have to find shortcuts and tricks to make the content fit what you already know."</i>
take $f(x)=\sin(x)$ and $L=\pi$ can we pick the coefficients $c_{n}$ to make the expressian above true?
of course $c_{1}=1$ and all the other coefficients equals 0
done with class! more history time:
why is this important? in 1979 a team of engineers and mathematicians from a company philips they discovered, or practically implemented: that an audio signal has billions of datapoints over time if you represent it as a fourier seriers, and truncate some of the coeffecients we can represent many signals really well. which condenses down the data to just a handful of coefficients
this is how EQ's are made too to filter out noise, just set the c_n of the frequencies you dont want to 0
so philips used math, math that is simillar to what we are discussing in the lecture, to make a digital record, the first digital cd. Not only that but fourier series are used for image and video compression as well, although they often use a sum of wavelets instead of a sum of trigonometric functions.
#end of lec 27
<i>reading week</i>
Okay enough with life stuff, back to mathematics:
$\implies u_{n}(t,x)=c_{n}e^{-(n\pi/L)^2Dt}\sin\left( \frac{n\pi x}{L} \right)$ $n=1,2,3,\dots$
applying superposition (sum of any solutions is also a solution):
$$u(t,x)=\sum_{n=1}^\infty c_{n}e^{-(n\pi/L)^2Dt}\sin\left( \frac{n\pi x}{L} \right)$$ ^This is the most general form of the solution.
$u(0,x)=f(x)=\sum_{n=1}^\infty c_{n}\sin\left( \frac{n\pi x}{L} \right),\ 0\leq x\leq L$ we have never seen anything like this before, an infinite number of sin terms added together, usually its polynomials that are summed.
This is when Fourier (?) steps in. He proved that you can represent various functions as a sum of sines.
fun stories about Pioncre and Cauchy Euler: Cauchy Euler was an engineer, and Pioncre had his theorem released around 2008 and it was really long, like 400 pages, he posted it online and asked if anyone wanted to prove it, after a while 4-5 or so mathematicians checked his proof and said, yep okay the proof looks correct. His theorem has a lot to do with the material world.
</br>
$f$ doesn't even have to be analytic in order for it to be expressed as a sum of sines.
Take $f(x)=\sin(x)$ and $L=\pi$ can we pick the coefficients $c_{n}$ to make the expression above true?
of course! $c_{1}=1$ and all the other coefficients equals 0.
</br>
Done with class! More history time:
Why is this important? in 1979 a team of engineers and mathematicians from a company Philips they discovered, or practically implemented that an audio signal has billions of datapoints over time if you represent it as a Fourier series, and truncate some of the coefficients we can represent many signals really well. Which condenses down the data to just a handful of coefficients
This is how filters are made too to filter out noise, just set the $c_{n}$ of the frequencies you don't want to $0$.
So Philips used math, math that is similar to what we are discussing in the lecture, to make a digital record, the first digital cd. Not only that but Fourier series are used for image and video compression as well, although they often use a sum of wavelets instead of a sum of trigonometric functions.
#end of lec 27 (Nov 10)
<i>*reading week*</i>
#start of lec 28 (Nov 20)
## Example Eigen value problem
#ex #evp
$y''-2y'+\lambda y=0 \qquad y(0)=0, \quad y'(\pi)+y(\pi)=0$
eigen value problem, find $\lambda$ such that the initial conditions are true.
find the characteristic polynomial.
Find $\lambda$ such that the initial conditions are true.
find the characteristic polynomial:
$r^2-2r+\lambda=0$
$r_{1,2}=\frac{2\pm \sqrt{ 4-4\lambda }}{2}=1\pm \sqrt{ 1-\lambda }$
</br>
case 1: $1-\lambda>0$
$y(x)=c_{1}e^{(1+\sqrt{ 1-\lambda })x}+c_{2}e^{ (1-\sqrt{ 1-\lambda })x }$
$y(0)=0=c_{1}+c_{2}$
$c_{1}(1+\sqrt{ 1-\lambda })e^{(1+\sqrt{ 1-\lambda })\pi}+c_{2}(1-\sqrt{ 1-\lambda })e^{ (1-\sqrt{ 1-\lambda })\pi }+c_{1}e^{(1+\sqrt{ 1-\lambda })\pi}+c_{2}e^{ (1-\sqrt{ 1-\lambda })\pi }=0$
$y'(\pi)+y(\pi)=0$
$\implies c_{1}(1+\sqrt{ 1-\lambda })e^{(1+\sqrt{ 1-\lambda })\pi}+c_{2}(1-\sqrt{ 1-\lambda })e^{ (1-\sqrt{ 1-\lambda })\pi }+c_{1}e^{(1+\sqrt{ 1-\lambda })\pi}+c_{2}e^{ (1-\sqrt{ 1-\lambda })\pi }=0$
substitute $c_{1}=-c_{2}$
$-c_{2}(1+\sqrt{ 1-\lambda })e^{(1+\sqrt{ 1-\lambda })\pi}+c_{2}(1-\sqrt{ 1-\lambda })e^{ (1-\sqrt{ 1-\lambda })\pi }-c_{2}e^{(1+\sqrt{ 1-\lambda })\pi}+c_{2}e^{ (1-\sqrt{ 1-\lambda })\pi }=0$
$c_{2}=0=c_{1}$ boring solution :( (also called a trivial solution)
$\implies c_{2}=0$
$\implies y(x)=0$ boring solution :( (also called a trivial solution)
</br>
case 2: $1-\lambda=0$
$y(x)=c_{1}e^x+c_{2}xe^x$
$r=r_{1,2}=1\pm \sqrt{ 0 }$ (repeated root)
$y(x)=c_{1}e^{rx}+c_{2}xe^{rx}$
$y(0)=0=c_{1}$
$y'(\pi)+y(\pi)=c_{2}(\pi e^\pi+e^\pi+\pi e^\pi)=0 \implies c_{2}=0$
(same trivial solution.)
</br>
case 3: $1-\lambda<0$
$r_{1,2}=1\pm i\sqrt{\lambda-1}$
</br>
$y(x)=e^x(c_{1}\cos(\sqrt{ \lambda-1 }x)+c_{2}\sin(\sqrt{ \lambda-1 }x))$
$y(0)=0=c_{1}$
$y(x)=c_{2}e^x\sin \sqrt{ \lambda-1 }x$
$y'(\pi)+y(\pi)=0=c_{2}(e^\pi \sin \sqrt{ \lambda-1 }\pi+e^\pi \cos( \sqrt{ x-1 }\pi)\sqrt{ \lambda-1 })+c_{2}e^\pi \sin \sqrt{ \lambda-1 }\pi$
$$2\sin(\sqrt{ \lambda-1 }\pi)-\sqrt{ \lambda-1 }\cos(\sqrt{ \lambda-1 }\pi)=0$$
$y(x)=c_{2}e^x\sin (\sqrt{ \lambda-1 }x)$
$y'(\pi)+y(\pi)=0=c_{2}\underbrace{ (e^\pi \sin \sqrt{ \lambda-1 }\pi+e^\pi \cos( \sqrt{ x-1 }\pi)\sqrt{ \lambda-1 }+e^\pi \sin \sqrt{ \lambda-1 }\pi) }_{ =0 }$
$$2\sin(\sqrt{ \lambda-1 }\pi)+\sqrt{ \lambda-1 }\cos(\sqrt{ \lambda-1 }\pi)=\frac{0}{e^\pi}=0$$
This is a transcendental equation, non algebraic, cannot be solved explicitly.
Only option we have is to approximate the eigen values. there are an infinite number of them.
We have to use software in order to obtain a finite number of approximations. I like software, software is good, as long as your being mindful of how you're using it.
"engineers over design things." first they use software and add a 50%, sometimes 300% margin on it. But if your using that big of a margin I can just tell you how far you need to build your house away from the river. -Prof
Btw the eigen function is $y(x)=ce^x\sin(\sqrt{\lambda-1}x)$
Only option we have is to approximate the eigen values. There are an infinite number of them.
We have to use software in order to obtain a finite number of approximations. I like software, software is good, as long as your being mindful of how you're using it. (I'm not putting quotes as I can't recall for certain what he said.)
"Engineers over design things." first they use software and add a 50%, sometimes 300% margin on it. But if your using that big of a margin I can just tell you how far you need to build your house away from the river. -Prof
</br>
Here's a plot, every crossing of red line with the $x$ axis is a solution for $\lambda$. Blue is a solution curve corresponding with $\lambda\approx8.305027$. Green shows the DE is being satisfied. And the equation on the bottom left shows the second initial condition is being met. The first condition is also met as you can see $Y(0)=0$ on the plot.
![evp.png](drawings/evp.png)
We are done.

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@ -70,7 +70,7 @@ $y'=z$
$\begin{cases}y'-z=0 \\az'+bz+cy=f\end{cases}$
last example of the chapter:
#ex
#ex #SoLE #IVP
$x'+y=0, \qquad x(0)=0$
$x+y'=1-u(t-2) \qquad y(0)=0$
This is a review problem in chapter 7 of the textbook.

View File

@ -23,7 +23,7 @@ I have written these notes for myself, I thought it would be cool to share them.
[Dirak δ-function (lec 21)](dirak-δ-function-lec-21.html)
[Systems of linear equations (lec 21-22)](systems-of-linear-equations-lec-21-22.html)
[Power series (lec 22-25)](power-series-lec-22-25.html)
[Separation of variables & Eigen value problems (lec 26-28)](separation-of-variables-eigen-value-problems-lec-26-28.html) (raw notes, not reviewed or revised yet.)
[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) (raw notes, not reviewed or revised yet.)
[Partial differential equations (lec 30-33)](partial-differential-equations-lec-30-33.html) (raw notes, not reviewed or revised yet.)
</br>

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excalidraw-plugin: parsed
tags: [excalidraw]
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---
excalidraw-plugin: parsed
tags: [excalidraw]
---
==⚠ Switch to EXCALIDRAW VIEW in the MORE OPTIONS menu of this document. ⚠==
# Text Elements
%%
# Drawing
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```
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---
excalidraw-plugin: parsed
tags: [excalidraw]
---
==⚠ Switch to EXCALIDRAW VIEW in the MORE OPTIONS menu of this document. ⚠==
# Text Elements
%%
# Drawing
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A===
```
%%

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---
excalidraw-plugin: parsed
tags: [excalidraw]
---
==⚠ Switch to EXCALIDRAW VIEW in the MORE OPTIONS menu of this document. ⚠==
# Text Elements
%%
# Drawing
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```
%%

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---
excalidraw-plugin: parsed
tags: [excalidraw]
---
==⚠ Switch to EXCALIDRAW VIEW in the MORE OPTIONS menu of this document. ⚠==
# Text Elements
%%
# Drawing
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---
excalidraw-plugin: parsed
tags: [excalidraw]
---
==⚠ Switch to EXCALIDRAW VIEW in the MORE OPTIONS menu of this document. ⚠==
# Text Elements
%%
# Drawing
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