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 (🤯).
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/)
Amount of heat to change temperature over $\Delta t$ with $\Delta u$ is $\underbrace{ C(x) }_{ \text{specific heat capacity } }\underbrace{ \rho(x) }_{ \text{density} }\Delta u\underbrace{ a\Delta x }_{ \text{volume} }$
$\lim_{ \Delta x,\Delta t \to 0 }: C(x)\rho(x)\frac{ \partial u }{ \partial t }(t,x)=\frac{ \partial }{ \partial x }\left( k(x)\frac{ \partial u }{ \partial x }(t,x) \right)+Q(t,x)$
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.
Rewrite the equation we derived by grouping the constant terms into one constant $D$ and assume for this problem that the internally generated heat is $0$ (i.e., $Q(x,t)=0$)
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 $\sqrt{ -(-) }=\sqrt{ + }$)
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.)
>Yes this looks illegal, but it works, you could also integrate more rigorously if you did a u-sub: $u=T(t) \quad \frac{du}{dt}=T'(t)$)
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><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>
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.
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$f$ 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.
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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.
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
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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.
