58 lines
3.1 KiB
Markdown
58 lines
3.1 KiB
Markdown
#start of lec 35
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last problem of the course which we will finish today.
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$\frac{ \partial^2 u }{ \partial t^2 }=\frac{ \partial^2 u }{ \partial x^2 }+tx, 0\leq x\leq \pi$ $t>0$
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thats a driven wave equation. tx is the source term.
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$u(0,t)=u(\pi,t)=0 \quad t>0$
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$u(x,0)=\sin(x) \quad 0\leq x\leq \pi$
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$\frac{ \partial u }{ \partial t }(x,0)=5\sin(2x)-3\sin(5x)\quad 0\leq x\leq \pi$
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if any of the boundary conditions are non zero, then we have to split(?) into X and T.
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in this case there's a $\sin(x)$ term so its a nonhomogenous equation (?)
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when $tx$ wasn't there in last problem we had the solution:
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$u(t,x)=\sum_{n=1}^\infty \underbrace{ (a_{n}\cos(nt)+b_{n}\sin(nt)) }_{ u_{n}(t) }\sin(nt)$ notice $L=\pi$
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if we expand $tx$:
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$tx=\sum_{n=1}^\infty h_{n}\sin(nx)$ (this is called a formal expansion, the two arent exactly equal due to the discontinuouity in tx.)
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$h_{n}=\frac{2}{\pi}\int _{0} ^\pi tx\sin(nx)\, dx$
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continuous between 0 and pi but on the edge ponts, the foureir sin series will converge to the midpoint of the two edge points.
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$h_{n}=\frac{2t}{\pi}\int _{0}^\pi x\sin(nx)\, dx=- \frac{2t}{\pi n}\left( x\cos(nx)|_{0}^\pi-\int _{0}^\pi \cancel{ \cos(nx) }\, dx \right)$
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$h_{n(t)}=\frac{2t}{n}(-1)^{n+1}$
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$h(x,t)=tx=\sum_{n=1}^\infty h_{n}\sin(nx)=\sum_{n=1}^\infty \frac{2t}{n}(-1)^{n+1}\sin(nx)$
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$\frac{ \partial^2 u }{ \partial t^2 }=\sum_{n=1}^\infty u_{n}''(t)\sin(nx)$
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$\frac{ \partial^2 u }{ \partial x^2 }=\sum_{n=1}^\infty -u_{n}(t)n^2\sin(nx)$
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$\sum_{n=1}^\infty \underbrace{ \left( u_{n}''+n^2u_{n}+\frac{2t}{n}(-1)^n \right) }_{ =0 }\sin(nx)=0$
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$u''_{n}+n^2u_{n}=\frac{2}{n}(-1)^{n+1}t, \quad n=1,2,\dots$
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use #mouc (or laplace, but that'll take much longer.)
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characteristic eq:
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$r^2+n^2=0$
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$r_{1,2}=\pm in$
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$u_{n}^h(t)=a_{n}\cos(nt)+b_{n}\sin(nt)$
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$u_{n}^p(t)=At+B$
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$B=0$ because there's no constant term on the RHS
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$A=\frac{2(-1)^{n+1}}{n^3}$
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$u_{n}(t)=a_{n}\cos(nt)+b_{n}\sin(nt)+\frac{2(-1)^{n+1}}{n^3}t$
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$u(x,t)=\sum_{n=1}^\infty(a_{n}\cos(nt)+b_{n}\sin(nt)+\frac{2(-1)^{n+1}}{n^3}t)\sin(nx)$
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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.
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$u(x,0)=\sum_{n=1}^\infty a_{n}\sin(nx)=\sin(x)$
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$a_{1}=1, \quad a_{k}=0, \quad k=2,3,\dots$
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$\frac{ \partial u }{ \partial t }(x,0)=\sum_{n=1}^\infty\left( b_{n}n+\frac{2(-1)^{n+1}}{n^3} \right)\sin(nx)=5\sin(2x)-3\sin(5x)$
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coefficients =0 if $n\ne_{2},5$
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$\implies b_{n}=\frac{2}{n^4}(-1)^n, \quad n\ne_{2},5$
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$b_{2}=\frac{5}{2}+\frac{2(-1)^2}{2^4}$
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$b_{5}=-\frac{3}{5}+\frac{2(-1)^5}{5^4}$
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$u(x,t)=\cos(t)\sin(x)+\frac{5}{2}\sin(2t)\sin(2x)-\frac{3}{5}\sin(5t)\sin(5x)+2\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^3}\left( t-\frac{\sin(nt)}{n} \right)\sin(nx)$
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finished the solution. Man I got teary eyed from this lecture.
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#end of lec 35
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#start of lec 36
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What do you guys wanna do? Questions or summary of the course?
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Okay we do summary.
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# Summary of second half of Math 201
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(available on eclass)
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Laplace transforms:
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Definition of laplace,
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Properties (4 important ones)
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...
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</br>
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#end of lec 36
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#end of Math 201. Congratulations! |