forked from Sasserisop/MATH201
43 lines
3.1 KiB
Markdown
43 lines
3.1 KiB
Markdown
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#start of lec 26
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We start with some thermodynamics
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# Heat equation
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Heat equation not only describes thermodynamics but it can also model the diffusion of gasses.
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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 (🤯).
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![[Drawing 2023-11-08 13.07.19.excalidraw.png]]
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>I'm sorry the image doesn't display properly :( I'm trying to get images to work on my notes. For now you can see the relevant .png files in the github repo under content/drawings/
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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.
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Fourier figured out that:
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$\text{Heat flux} = -k(x)a\frac{\partial u}{\partial x}(t,x) \Delta t$
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heat flux is in the positive x direction
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where du/dx is the opposite sign of the flux (because hot flows to cold.)
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where $u(t,x)$ is a function that describes the temperature in the tube.
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$-k(x+\Delta x)a\frac{\partial u}{\partial x}(t,x+\Delta x) \Delta t$
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$\Delta u=u(t+\Delta t,x)-u(t,x)$
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Amount of heat to change temperature over $\Delta t$ with $\Delta u$ is $\underbrace{ C(x) }_{ \text{specific heat capacity } }\underbrace{ \rho(x) }_{ \text{density} }\Delta u\underbrace{ a\Delta x }_{ \text{volume} }$
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$c(x)\rho(x)a\Delta x(u(t+\Delta t,x)-u(t,x))=a\Delta t(k(x+\Delta x) \frac{\partial u}{\partial x}(t_{1}x+\Delta x)-k(x) \frac{\partial u}{\partial x}(t,x))+Q(t,x)\Delta xa\Delta t$
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divide by $a\Delta x\Delta t$
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$c(x)\rho(x)\frac{(u(t+\Delta t,x)-u(t,x))}{\Delta t}=a\Delta t\left( k(x+\Delta x) \frac{\partial u}{\partial x}(t,x+\Delta x)-k(x) \frac{\partial u}{\partial x}(t,x) \right)+Q(t,x)$
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$\lim_{ \Delta x,\Delta t \to 0 }: c(x)\rho(x)\frac{ \partial u }{ \partial t }(t,x)=\frac{ \partial }{ \partial x }\left( k(x)\frac{ \partial u }{ \partial x }(t,x) \right)+Q(t,x)$
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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)
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the Maxwell equations that describes magnetic and electric fields are a system of partial differential equations.
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thermodynamics can be very important for electrical engineers, for instance the heat produced in a transformer, or a battery. It has applications.
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$\frac{ \partial u }{ \partial t }=D \frac{ \partial^2 u }{ \partial x^2 }, \quad 0\leq x\leq L, \quad t>0$
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boundary conditions:
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$u(t,0)=u(t,L)=0 , \quad t>0$ (simple case)
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$u(0,x)=f(x) , \quad 0\leq x\leq L$
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These three equations form an IBVP (initial boundary value problem)
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we will work on this equation next lecture:
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$u(t,x)=T(t)X(x)$
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$T'x=DTx''$
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divide by DTx:
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$\frac{T'}{DT}=\frac{x''}{x}$
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on the left is a function of time only, on the right is a function of x only.
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$\frac{T'}{DT}=\frac{x''}{x}=-\lambda$ where $\lambda$ is a constant
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$x''+\lambda x=0$
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$u(t,0)=\cancel{ T(t) } \quad x(0)=0$
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$x(0)=x(L)=0$
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this is called an eigen value problem.
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#end of lec 26
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