revised lecture 30 and 31
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@ -86,7 +86,7 @@ $\ln\mid x-y\mid=C+y+2x$ < this is where he moved the C to the left
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$\mid x-y\mid=e^Ce^ye^{2x}$
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$x-y=Ae^ye^{2x}$
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$A(x-y)=e^{y+2x}$
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> I know that above step looks illegal, but the prof did this (indirectly, he moved C to the LHS in a prior step without regarding it's sign). I wonder what happens if A was 0 though? Do we get divide by zero errors? Thinking about it more, we are changing $x-y=0$ to $e^{y+2x}=0$ when $A=0$ The first one has a solution (y=x) the second loses that solution because of ln(0) issues (gives a function that's undefined for all x). when checking y(x)=x in the DE, it is a valid solution. So perhaps it is an illegal step! Because we lost a valid solution. I'll have to check with the prof.
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> I know that above step looks illegal, but the prof did this (indirectly, he moved C to the LHS in a prior step without regarding it's sign). I wonder what happens if A was 0 though? Do we get divide by zero errors? Thinking about it more, we are changing $x-y=0$ to $e^{y+2x}=0$ when $A=0$ The first one has a solution ($y=x$) the second loses that solution because of $\ln(0)$ issues (gives a function that's undefined for all $x$). when checking $y(x)=x$ in the DE, it is a valid solution. So perhaps it is an illegal step! Because we lost a valid solution. I'll have to check with the prof.
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> Interestingly, if we act like $e^{y+2x}=0$ is defined, we get $\frac{dy}{dx}=-2$
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>
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>Proof:
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@ -98,7 +98,7 @@ $A(x-y)=e^{y+2x}$
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>$1=\frac{2(x-y+1)}{2x-2y-1}$
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>$2x-2y-1=2(x-y+1)$
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>$-1=2$
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>So what does this all mean? I think it means that even if we imagine that $\frac{dy}{dx}$ exists, the equation is not satisfied and $y=x$ is definitely not a solution even when we try to cheat a little.
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>So what does this all mean? I think it means that even if we imagine that $\frac{dy}{dx}$ exists, the equation is not satisfied and $e^{y+2x}=0$ is definitely not a solution even when we try to cheat a little.
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We can rearrange to our liking, but we have found the general solution to the DE:
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#### $$x-y=Ae^{2x+y}$$
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@ -1,91 +1,121 @@
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Let's revisit the heat equation:
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The first time the heat equation was introduced, we figured out that it's solution was of the form $u(t,x)=\sum_{n=1}^\infty c_{n}e^{-(n\pi/L)^2Dt}\sin\left( \frac{n\pi x}{L} \right)$
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Now that we learned about fourier series and eigen value problems, we can finally solve it! (for a given specific case.)
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Now that we learned about Fourier series, we can compute the values of $c_{n}$ and finally solve it!
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#ex #PDE
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Solve the following PDE (find $u(t,x)$)
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IBVP of heat eq:
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$\frac{ \partial u }{ \partial t }=D\frac{ \partial^{2} u }{ \partial x^{2} }$ for $0\leq x\leq L$ for $t>0$
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$u(t,0)=u(t,L)=0, \quad t>0$
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$u(0,x)=f(x), \quad 0\leq x\leq L$
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lets choose $L=\pi$
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$f(x)=\begin{cases}-x & 0\leq x\leq \frac{\pi}{2} \\1-x & \frac{\pi}{2}<x\leq \pi\end{cases}$
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$\begin{cases}\frac{ \partial u }{ \partial t }=D\frac{ \partial^{2} u }{ \partial x^{2} } \text{ for } 0\leq x\leq L, \quad t>0\\u(t,0)=u(t,L)=0, \quad t>0\\u(0,x)=f(x), \quad 0\leq x\leq L\quad\end{cases}$
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where:
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$D=1$
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$L=\pi$
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$f(x)=\begin{cases}-x & 0\leq x\leq \frac{\pi}{2} \\x-\pi & \frac{\pi}{2}<x\leq \pi\end{cases}$
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>Psst, There's a shortcut, we have the same boundary conditions, the exact same IBVP that we solved for a general case. We can just plug in using the formula we derived earlier.
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>$u(t,x)=\sum_{n=1}^\infty c_{n}e^{-(n\pi/L)^2Dt}\sin\left( \frac{n\pi x}{L} \right)$
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>$u(t,x)=\sum_{n=1}^\infty c_{n}e^{-n^2t}\sin(nx)$
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>All you have to do now is compute the Fourier transform of $f(x)$ to find the $c_{n}$ coefficients.
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>Besides, the computations below for finding the eigen values and eigen function follow the exact same steps we had in lectures 26 and 27. So if you saw that and understood it, feel free to skip ahead to where we compute $c_{n}$
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The initial temperature of the tube (at $t=0$) looks like this:
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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.)
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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.
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Separation of variables: $u(t,x)=T(t)X(x)$
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theres a theorem that this will give a unique solution.
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So we have a non-uniformly heated rod with both ends held at zero temperature. What happens to the temperature inside the rod over time? <i>"[...]. Very interesting problem."</i> -Prof (I agree.)
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If we expressed $f(x)$ as a Fourier series, where would it converge? Well it's continuous from $0$ to $\pi$ and its windowed form when repeated will be continuous on $(-\infty,\infty)$ furthermore, the derivative of this function is piecewise continuous on $[-\pi,\pi]$ therefore, it will be convergent everywhere (second convergence theorem). This is good news for us.
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We do separation of variables as we did before: $u(t,x)=T(t)X(x)$
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There's a theorem that this will give a unique solution. (Not proved in class.)
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Plug into $\frac{ \partial u }{ \partial t }=D\frac{ \partial^{2} u }{ \partial x^{2} }$ where in this case, $D=1$
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$T'X=DTX''$
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$\frac{T'}{DT}=\frac{X''}{X}$
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LHS is a function of t only , RHS is function of x only.
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<i>"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."</i> -prof. Very philosophical.
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$X''+\lambda X=0$ where $X(0)=X(L=\pi)=0$
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$u(t,0)=T(t)X(0)=0$
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case 1) $\lambda<0$, $r_{1,2}=\pm \sqrt{ -\lambda }$
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$X(x)=c_{1}e^{ \sqrt{ -\lambda }x }+c_{2}e^{ -\sqrt{ -\lambda }x }$
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$X(0)=c_{1}+c_{2}=0$
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$X(\pi)=c_{1}e^{ \sqrt{ -\lambda }\pi }+c_{2}e^{ -\sqrt{-\lambda }\pi }=0$
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$c_{1}=c_{2}=0$
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we will continue the problem in the next lecture.
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$\frac{T'}{DT}=\frac{X''}{X}=-\lambda$
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LHS is a function of $t$ only , RHS is function of $x$ only. The only time they can ever equate to each other if both of them equal some constant value(s) $\lambda$. The reason it's $-\lambda$ is because it doesn't matter, it's a constant and making it negative helps make rearranging the equation easier.
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<i>"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."</i> -prof.
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We will continue the problem in the next lecture.
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#end of lec 30
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#start of lec 31
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"Can we skip the assignment?"
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Petar replies:
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You have a choice, not just in math but in life,
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confusions says there's three things you cant take back
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Confucius says there's three things you cant take back:
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opportunity to do something, time, and your word
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so do your last assignment and don't pass up the opportunity!
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recall last lec:
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$\frac{ \partial u }{ \partial t }=\frac{ \partial u^2 }{ \partial x^2 }, 0\leq x\leq \pi, t>0$
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$u(t,0)=u(t,\pi)=0, t>0$ <- Dirichlet (french mathematician) IVP.
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$u(0,x)=f(x), \quad 0\leq x<\pi$
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$f(x)=\begin{cases}-x, & 0\leq x\leq \frac{\pi}{2} \\x-\pi, & \frac{\pi}{2}{\leq x\leq \pi}\end{cases}$
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graph of tube when t=0. here.
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Recall last lec:
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$\frac{ \partial u }{ \partial t }=\frac{ \partial u^2 }{ \partial x^2 } \qquad 0\leq x\leq \pi \qquad t>0$
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$u(t,0)=u(t,\pi)=0 \quad t>0$ <- Called a Dirichlet boundary condition
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$u(0,x)=f(x) \quad 0\leq x<\pi$
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$f(x)=\begin{cases}-x & 0\leq x\leq \frac{\pi}{2} \\x-\pi & \frac{\pi}{2}{\leq x\leq \pi}\end{cases}$
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</br>
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$u(t,x)=T(t)X(x)$
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$=X+\lambda X=0, \quad X(0)=X(\pi)=0$
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case 1) $\lambda<0 \implies$ $X(x)=0$
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case 2) $\lambda=0 \implies X(x)==0$
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$X(0)=0=c_{2}$
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$X(\pi)=0=c_{1}\pi$
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case 3) $\lambda>0$
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plug into equation:
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$T'X=DTX''$
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$\frac{T'}{DT}=\frac{X''}{X}=-\lambda$
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$X''+\lambda X=0, \quad X(0)=X(\pi)=0$
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characteristic equation: $r^2+\lambda=0 \implies r_{1,2}=\pm\sqrt{ -\lambda }$
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</br>
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Case 1) $\lambda<0 \implies$ $r_{1,2}=\pm \sqrt{ -\lambda }$
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$X(x)=c_{1}e^{\sqrt{ -\lambda }x}+c_{2}e^{-\sqrt{ -\lambda }x}$
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$X(0)=0=c_{1}+c_{2} \implies c_{1}=-c_{2}$
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$X(\pi)=0=c_{1}e^{\sqrt{ -\lambda }\pi}+c_{2}e^{-\sqrt{ -\lambda }\pi}=-c_{2}e^{\sqrt{ -\lambda }\pi}+c_{2}e^{-\sqrt{ -\lambda }\pi}$
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$0=c_{2}(e^{\sqrt{ -\lambda }\pi}-e^{-\sqrt{ -\lambda }\pi})$
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either $c_2=0$ which gives a trivial solution, or
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$e^{\sqrt{ -\lambda }\pi}-e^{-\sqrt{ -\lambda }\pi}=0\implies \sqrt{ -\lambda }=-\sqrt{ -\lambda } \implies\lambda=0$
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but then $X(x)=c_{1}e^{\sqrt{ -0 }x}+c_{2}e^{-\sqrt{ -0 }x}=c_{1}+c_{2}=0 \implies X(x)=0$ Trivial solution again.
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</br>
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Case 2) $\lambda=0 \quad r_{1}=r_{2}=0$
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$X(x)=c_{1}e^{0x}+c_{2}xe^{0x}=c_{1}+c_{2}x$
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$X(0)=0=c_{1}$
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$X(\pi)=0=c_{2}\pi$
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$\implies X(x)=0$ Trivial solution.
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</br>
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Case 3) $\lambda>0 \implies r_{1,2}=0\pm i\sqrt{ \lambda }$
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$X(x)=c_{1}\cos(\sqrt{ \lambda }x)+c_{2}\sin(\sqrt{ \lambda }x)$
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$X(0)=c_{1}=0$
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$X(\pi)=0=c_{2}\sin(\sqrt{ \lambda }\pi)\implies \sqrt{ \lambda }\pi=n\pi\quad n=1,2,\dots$
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$\lambda_{n}=n^2$ a countable infinitely many lambdas (taggable with an index)
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real numbers are uncountable
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"what is a point? a point on a screen on a peice of paper?" (We have lost our minds)
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</br>
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$\lambda_{n}=n^2$ <- a countable infinitely many lambdas (it's countably infinite because each lambda is taggable with an index, real numbers are uncountable.)
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"What is a point? A point on a screen, on a piece of paper?" Loosy quote, and I can't recall what the intention behind this quote was unfortunately.
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$X_{n}(x)=c_{2}\sin(nx)$
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but obviously, we don't need the multiples of an eigen value so we set $c_{2}=1$ (or any arbitrary non zero constant)
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but obviously, we don't need the multiples of an eigen value, so we set $c_{2}=1$ (or any arbitrary non zero constant)
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$X_{n}(x)=\sin(nx)$
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$u(t,x)=T(t)X(x)$
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Using $\frac{T'}{T}=\frac{X''}{X}=-\lambda$ we plug in the value we got for $\lambda$:
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$\frac{T'}{T}=-n^2$
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integrate both sides:
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$T_{n}(t)=b_{n}e^{-n^2t}$
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$u_{n}(t,x)=b_{n}e^{-n^2t}\sin(nx), n=1,2,\dots$
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$u(t,x)=\sum_{n=1}^\infty b_{n}e^{-n^2t}\sin(nx)$
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$u(0,x)=f(x)=\sum_{n=1}^\infty b_{n}\sin(nx)$
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$b_{n}=\frac{2}{\pi}\int _{0}^\pi f(x)\, dx$
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$b_{n}=\frac{2}{\pi}[-\int _{0} ^\frac{\pi}{2} x\sin(nx) \, dx+\int _{\frac{\pi}{2}} ^\pi x\sin(nx) \, dx-\pi \int _{\frac{\pi}{2}} ^\pi \sin(nx)\, dx]$
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$\int x\sin(nx) \, dx=-\frac{1}{n}\left( x\cos(nx)-\int \cos(nx) \, dx \right)$
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$=-\frac{1}{n}\left( x\cos(nx)-\frac{1}{n}\sin(nx) \right)=\frac{1}{n^2}\sin(nx)-\frac{1}{n}x\cos(nx)$
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bla bla bla, we get that $b_{n}$ is:
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$b_{n}=\frac{2}{\pi}\left[ \frac{\pi}{2n}\cos \frac{n\pi}{2}-\frac{1}{n^2}\sin \frac{n\pi}{2}-\frac{1}{n^2}\sin \frac{n\pi}{2}-\frac{\pi}{n}(-1)^n +\frac{\pi}{2n}\cos \frac{n\pi}{2}+\frac{\pi}{n}(-1)^n-\frac{\pi}{n}\cos \frac{n\pi}{2} \right]$
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$b_{n}=-\frac{4}{n^2\pi}\sin \frac{n\pi}{2}, \quad n=1,2,\dots$
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$n=2k\implies b_{2k}=0, \quad k=1,2,\dots$
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$n=2k-1\implies b_{2k-1}=-\frac{4}{(2k+1)^2\pi}(-1)^{k+1}, \quad k=1,2,\dots$
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$b_{2k-1}=\frac{4}{(2k+1)^2\pi}(-1)^{k}$
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Integrate both sides:
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$\int \frac{1}{T}\,dT=\int -n^2 \, dt$
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$T_{n}(t)=c_{n}e^{-n^2t}$
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> ^Side note, he uses $b_{n}$ in this lecture, but I want it to be consistent with previous lectures where he uses $c_{n}$
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$$u(t,x)=\sum_{n=1}^\infty \frac{(-1)^k}{(2k-1)^2}e^{-(2k-1)^2t}\sin((2k-1)x)$$
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notice that the limit of the sum as $t\to \infty$ is $0$
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graph time! But oh no! He forgot the plot from the last lecture, i went to grab it for him.
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$u(x,t)=X(x)T(t)$
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$u_{n}(t,x)=c_{n}e^{-n^2t}\sin(nx) \qquad n=1,2,\dots$
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$u(t,x)=\sum_{n=1}^\infty c_{n}e^{-n^2t}\sin(nx)$
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> THIS IS WHERE WE START TO COMPUTE THE FOURIER COEFFICIENTS :D
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bring in the initial condition babyyy:
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$u(0,x)=f(x)=\sum_{n=1}^\infty c_{n}\sin(nx)$
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$c_{n}=\frac{2}{\pi}\int _{0}^\pi f(x)\, dx$
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recall the definition of $f \quad$ $f(x)=\begin{cases}-x & 0\leq x\leq \frac{\pi}{2} \\x-\pi & \frac{\pi}{2}{\leq x\leq \pi}\end{cases}$
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$c_{n}=\frac{2}{\pi}[-\int _{0} ^\frac{\pi}{2} x\sin(nx) \, dx+\int _{\frac{\pi}{2}} ^\pi x\sin(nx) \, dx-\pi \int _{\frac{\pi}{2}} ^\pi \sin(nx)\, dx]$
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Integrate by parts:
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$\int \underbrace{x}_{ u }\underbrace{ \sin(nx) }_{ v' } \, dx=x\frac{-\cos(nx)}{n}-\int 1\cdot\frac{-\cos(nx)}{n} \, dx$
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$=-\frac{1}{n} x\cos(nx)+\frac{\sin(nx)}{n^2} =\frac{\sin(nx)}{n^2}-\frac{x\cos(nx)}{n}$
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$c_{n}=\frac{2}{\pi}\left[ -\frac{\sin(nx)}{n^2}+\frac{x\cos(nx)}{n}|_{0}^{\frac{\pi}{2}}+\frac{\sin(nx)}{n^2}-\frac{x\cos(nx)}{n}|_{\frac{\pi}{2}}^\pi+\frac{\pi \cos(nx)}{n}|_{\frac{\pi}{2}}^\pi \right]$
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$c_{n}=\frac{2}{\pi}\left[ -\frac{1}{n^2}\sin \frac{n\pi}{2}+\cancel{ \frac{\pi}{2n}\cos \frac{n\pi}{2} } \quad +\quad0-\frac{\pi}{n}(-1)^n-\frac{1}{n^2}\sin \frac{n\pi}{2} +\cancel{ \frac{\pi}{2n}\cos \frac{n\pi}{2} }\quad+\quad\frac{\pi}{n}(-1)^n-\cancel{ \frac{\pi}{n}\cos \frac{n\pi}{2} } \right]$
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$c_{n}=-\frac{4}{n^2\pi}\sin \frac{n\pi}{2}, \quad n=1,2,\dots$
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$n=2k\implies c_{2k}=0, \quad k=1,2,\dots$
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$n=2k-1\implies c_{2k-1}=-\frac{4}{(2k-1)^2\pi}(-1)^{k+1}, \quad k=1,2,\dots$
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$c_{2k-1}=\frac{4}{(2k-1)^2\pi}(-1)^{k}$
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plug it in:
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$u(t,x)=\sum_{n=1}^\infty c_{n}e^{-n^2t}\sin(nx)$
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$$u(t,x)=\sum_{k=1}^\infty \frac{4(-1)^k}{\pi(2k-1)^2}e^{-(2k-1)^2t}\sin((2k-1)x)$$
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Notice that the limit of the sum as $t\to \infty$ is $0$. The bar cools down eventually.
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Graph time! But oh no! He forgot the plot from the last lecture, I went to grab it for him.
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<i>I have a sick picture of the whiteboard showing the plot, but I will have to ask for permission to put it here.</i>
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While I was gone, I heard he shared stories about his family. Unfortunately I missed it.
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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.
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(5 more lectures, left)
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$k\frac{ \partial u }{ \partial x }=0$
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where $k$ is the thermal conductivity,
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$\frac{ \partial u }{ \partial x }=0|_{0<x<L}$
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this gives a new equation type: Neumann type.
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If you imagine the plot was showing the concentration of CO$_{2}$ in a room, it follows a similar curve as time goes on and the gas diffuses.
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By the way, there are only 5 more lectures left.
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Imagine instead of holding the two ends of the rod with a temperature of zero, we insulated the two ends.
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How do we insulate an end? Set the heat flux at the ends to $0$:
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$k\frac{ \partial u }{ \partial x }(t,0)=k\frac{ \partial u }{ \partial x }(t,L)=0$ <-$k$ is the thermal conductivity of the tube.
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This gives a new equation type: Neumann type.
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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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$\frac{ \partial u }{ \partial x }(t,0)=\frac{ \partial u }{ \partial x }(t,L)=0, \quad t>0$
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$u(0,x)=f(x) \quad 0\leq x\leq \pi$
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@ -93,8 +123,8 @@ $u(0,x)=f(x) \quad 0\leq x\leq \pi$
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#end of lec 31
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#start of lec 32
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back to the heat problem
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we do the Neumann problem, start with separation of variables.
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assume we can seperate the variable into two different functions:
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Let's solve the Neumann problem, start with separation of variables.
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assume we can separate the variable into two different functions:
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$u(t,x)=T(t)X(x)$
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plug into: $\frac{ \partial u }{ \partial t }=D\frac{ \partial^2 u }{ \partial x^2 }$
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$T'X=DTX''$
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@ -63,6 +63,7 @@ then: $y(x)=\sum_{n=0}^\infty \frac{y^{(n)}(x_{0})}{n!}(x-x_{0})^n$ (Taylor seri
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>"I mean engineering is all about approximations, do you know how your calculator computes [...]? Taylor series!"
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>"Applied mathematics is all about approximating and then measuring how good your approximation is, it's what engineering is all about." -Prof (loosy quotes, can't keep up with how enthusiastic he is!)
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</br>
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Definition: If $y(x)$ can be represented with a power series on $\mid x-x_{0}\mid$ then $y(x)$ is an analytic function on $(x_{0}-\rho,x_{0}+\rho)$
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btw analytic functions are very important in complex calculus.
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@ -18,7 +18,7 @@ $\frac{ \partial u }{ \partial x }$ is always opposite in sign of flux (because
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$a$ is the area of the cross-section.
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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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Amount of heat to change temperature over $\Delta t$ with $\Delta u$ is $\overbrace{ C(x) }^{ \text{specific heat capacity } }\hspace{ -0.8cm}\underbrace{ \rho(x) }_{ \text{density} }\Delta u\underbrace{ a\Delta x }_{ \text{volume} }$
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$C(x)\rho(x)a\Delta x(u(t+\Delta t,x)-u(t,x))=a\Delta t(k(x+\Delta x) \frac{\partial u}{\partial x}(t,x+\Delta x)-k(x) \frac{\partial u}{\partial x}(t,x))+Q(t,x)\Delta xa\Delta t$
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$Q(t,x)$ is a source term, a term to model internally produced heat.
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@ -36,7 +36,9 @@ thermodynamics can be very important for electrical engineers, for instance the
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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., $Q(x,t)=0$)
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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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$u(t,0)=u(t,L)=0 , \quad t>0$ <- (a simple case, this is a Dirichlet boundary condition)
|
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> Btw, Dirichlet boundary condition means the value at the two endpoints are given, A Neumann boundary condition provides information about the endpoints by giving you the derivative of the value.
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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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Separation of variables starts with the assumption that the solution can be written as:
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@ -65,7 +67,7 @@ $\begin{cases}X''+\lambda X=0 \\X(0)=0 \\X(L)=0\end{cases}$ <-This is called an
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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.
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recall in linear algebra $Ax+\lambda x=0$ where $A$ is a matrix
|
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eigen is a Germanic word meaning intrinsic, important
|
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We find the characteristic equations by splitting into three cases, just like we've done since lecture 7
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We find the characteristic equations by splitting into three cases, just like we've done back in lecture 7
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$X''+\lambda X=0$
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case 1) $\lambda<0$
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$r^2+\lambda=0 \implies r_{1,2}=\pm \sqrt{ -\lambda }$ (real solutions only since $\sqrt{ -(-) }=\sqrt{ + }$)
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@ -127,11 +129,12 @@ $T_{n}(t)=c_{n}e^{-(\frac{n\pi}{L})^2Dt}$
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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>
|
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><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>
|
||||
|
||||
$u(t,x)=X(t)T(t)$
|
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Okay enough with life stuff, back to mathematics:
|
||||
$\implies u_{n}(t,x)=c_{n}e^{-(n\pi/L)^2Dt}\sin\left( \frac{n\pi x}{L} \right)$ $n=1,2,3,\dots$
|
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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.
|
||||
$u(0,x)=f(x)=\sum_{n=1}^\infty c_{n}\sin\left( \frac{n\pi x}{L} \right),\ 0\leq x\leq L$ we have never seen anything like this before, an infinite number of $\sin$ terms added together, usually its polynomials that are summed.
|
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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>
|
||||
|
|
@ -143,9 +146,9 @@ 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)
|
||||
#end of lec 27 (Nov 10 2023)
|
||||
<i>*reading week*</i>
|
||||
#start of lec 28 (Nov 20)
|
||||
#start of lec 28 (Nov 20 2023)
|
||||
## Example Eigen value problem
|
||||
#ex #evp
|
||||
$y''-2y'+\lambda y=0 \qquad y(0)=0, \quad y'(\pi)+y(\pi)=0$
|
||||
|
|
|
|||
|
|
@ -9,7 +9,7 @@
|
|||
{"id":"1044e4c2c0610a1e","type":"text","text":"if $y'=G\\left( \\frac{y}{x} \\right)$ \n\nsubstitute $u=\\frac{y}{x}$\n$\\frac{dy}{dx}=u+x{\\frac{du}{dx}}$","x":150,"y":121,"width":315,"height":165},
|
||||
{"id":"c280d80abc2ea256","type":"file","file":"Math 201/Lectures/Homogenous equations (lec 2).md","x":-91,"y":380,"width":441,"height":149},
|
||||
{"id":"ffbe5ed5493f9419","type":"text","text":"find $\\mu(x)$","x":-561,"y":156,"width":250,"height":60},
|
||||
{"id":"78362e72fb0d54af","type":"file","file":"Math 201/Lectures/Linear equations (lec 2-3).md","x":-777,"y":323,"width":466,"height":170},
|
||||
{"id":"78362e72fb0d54af","type":"file","file":"Math 201/Lectures/Linear equations (lec 2-3).md","x":-777,"y":354,"width":466,"height":139},
|
||||
{"id":"e71a5b824ac543f8","type":"text","text":"combine terms using product rule","x":-590,"y":-30,"width":309,"height":103},
|
||||
{"id":"70316d4131dc52c3","type":"file","file":"Math 201/Lectures/Bernoulli equations (lec 3).md","x":-884,"y":973,"width":465,"height":136},
|
||||
{"id":"e70d1e3eea85e227","type":"text","text":"substitute","x":-669,"y":601,"width":250,"height":60},
|
||||
|
|
@ -47,7 +47,7 @@
|
|||
{"id":"0daa567ae142c34f","type":"text","text":"split into the homogenous solution and particular solution","x":2313,"y":927,"width":352,"height":131},
|
||||
{"id":"e5369edab4d8a607","type":"text","text":"find $y_{h}(t)$\n(the easy part)","x":1911,"y":895,"width":229,"height":98},
|
||||
{"id":"208cafba1dc71798","type":"text","text":"Use intuition and guess and check. (Practice improves intuition)","x":2068,"y":247,"width":332,"height":132},
|
||||
{"id":"eded7f0b26c3ebc5","type":"text","text":"use generalized algorithm:\ncase 1) $ay''+by'+cy=P_{m}(t)e^{rt}$\nthen: $y_{p}(t)=t^s(b_{m}t^m+b_{m-1}t^{m-1}+\\dots+b_{0})e^{rt}$\nwhere:\ns=0, if r is not a root,\ns=1 if r is a single root,\ns=2 if r is a double root\n\ncase 2)\n$ay''+by'+cy=P_{m}(t)e^{\\alpha t}\\cos(\\beta t)+P_{m}(t)e^{\\alpha t}\\sin(\\beta t)$\nthen: $y_{p}(t)=t^s[(A_{k}t^k+A_{K-1}t^{k-1}+\\dots+A_{0})e^{\\alpha t}\\cos(\\beta t)+(B_{k}t^k+B_{k-1}t^{k-1}+\\dots+B_{0})e^{\\alpha t}\\sin(\\beta t)]$\nwhere:\ns=0 if $\\alpha+i\\beta$ is not a root,\ns=1 if $\\alpha+i\\beta$ is a root.","x":2489,"y":-156,"width":1054,"height":575},
|
||||
{"id":"eded7f0b26c3ebc5","type":"text","text":"use generalized algorithm:\ncase i) $ay''+by'+cy=P_m(t)e^{rt}$\nwhere $P_{m}(t)=a_{m}t^m+a_{m-1}t^{m-1}+ \\dots +a_{0}$ i.e., $P$ is a polynomial degree $m$.\nThen we guess the particular solution is of the form: $y_{p}(t)=t^s(b_{m}t^m+b_{m-1}t^{m-1}+\\dots+b_{0})e^{rt}$\nwhere:\ns=0, if r is not a root,\ns=1 if r is a single root,\ns=2 if r is a double root.\n\ncase ii) $ay''+by'+cy=P_{m}(t)e^{\\alpha t}\\cos(\\beta t)+Q_{n}(t)e^{\\alpha t}\\sin(\\beta t)$\nThen we guess the particular solution is of the form: $y_{p}(t)=t^s[(A_{k}t^k+A_{k-1}t^{k-1}+\\dots+A_{0})e^{\\alpha t}\\cos(\\beta t)+(B_{k}t^k+B_{k-1}t^{k-1}+\\dots+B_{0})e^{\\alpha t}\\sin(\\beta t)]$\nwhere:\n$k=\\max(m,n)$\n$s=0$ if $\\alpha+i\\beta$ is not a root,\n$s=1$ if $\\alpha+i\\beta$ is a root. ","x":2489,"y":-156,"width":1054,"height":596},
|
||||
{"id":"d327b2b2e59fc4cc","type":"text","text":"general solution will be $y(t)=y_{h}(t)+y_{p}(t)$","x":2108,"y":-261,"width":336,"height":105},
|
||||
{"id":"b9455e68aa57c635","type":"text","text":"$ay''+by'+cy=f(t)$","x":2239,"y":1339,"width":283,"height":59},
|
||||
{"id":"5ef326cd80c96f9a","type":"text","text":"if $f(t)$ is a sum of two functions, split it up and find $y_{p_{1}}$, $y_{p_{2}}, \\dots$ separately, then $y_{p}$ will be the sum of the all the $y_{p_{n}}$'s you find.","x":2276,"y":658,"width":426,"height":188},
|
||||
|
|
|
|||
|
|
@ -36,14 +36,23 @@ $u(x,t)=\sum_{n=1}^\infty(a_{n}\cos(nt)+b_{n}\sin(nt)+\frac{2(-1)^{n+1}}{n^3}t)\
|
|||
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.
|
||||
$u(x,0)=\sum_{n=1}^\infty a_{n}\sin(nx)=\sin(x)$
|
||||
$a_{1}=1, \quad a_{k}=0, \quad k=2,3,\dots$
|
||||
*audio recording started*
|
||||
$\frac{ \partial u }{ \partial t }(x,0)=\sum_{n=1}^\infty\left( b_{n}n+\frac{2(-1)^{n+1}}{n^3} \right)\sin(nx)=5\sin(2x)-3\sin(5x)$
|
||||
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
|
||||
finished the solution. Man I got teary eyed from this lecture.
|
||||
#end of lec 35
|
||||
|
||||
man I got teary eyed from this lecture.
|
||||
#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!
|
||||
Loading…
Reference in New Issue