Question

\[\frac{\partial u}{\partial t}=k\frac{\partial^2 u}{\partial x^2}\]\[u(x,0)=f(x),\]\[u(0,t)=0,\]\[u(L,t)=0.\]Let's have some fun.

Answer 1

Let's assume the solution will take the form\[u(x,t)=\phi(x)G(t).\]

Answer 2

\[u_t(x,t)=\phi(x)G'(t).\]

Answer 3

\[u_x(x,t)=\phi'(x)G(t)\]
\[u_{xx}(x,t)=\phi''(x)G(t)\]

Answer 4

\[\phi(x)G'(t)=k\phi''(x)G(t)\]

Answer 5

\[\phi \cdot G'=k \cdot \phi''G\]

Answer 6

thinking on how to solve this

Answer 7

i use to know

Answer 8

\[\phi(x)\frac{dG}{dt}=kG(t)\frac{d^2\phi}{dx^2}.\]Separation of variables?

Answer 9

\[\frac{1}{G}\frac{dG}{dt}=k\frac{1}{\phi}\frac{d^2\phi}{dx^2}\]

Answer 10

that's the only method I can see here. Haven't done it that much though.

Answer 11

i think we has use a Fourier transform

Answer 12

then we have to solve the transformed problem

Answer 13

then we find the inverse transform

Answer 14

I don't think you need the transform, it is practically identical to this one as far as I can tell...
http://tutorial.math.lamar.edu/Classes/DE/SeparationofVariables.aspx

Answer 15

ok you win lol

Answer 16

lol

Answer 17

i had to get my partial differential equation book out
i don't remember this stuff well :(

Answer 18

I'm going to have a serious talk with Dr. Jared. xd

Answer 19

Yeah, that's the only kind of PDE I have have even done. I wish I knew this stuff off the top of my head.

Answer 20

all that Fourier stuff was complicated to me

Answer 21

I've also always had trouble with series... perhaps because I dislike them so much. xd

Answer 22

ok i smell pizza later

Answer 23

They just hurt my eyes to look at for too long, but they're not so bad once you get acquainted with them.

Answer 24

\[u(x, t) = \phi(x)G(t)\]\[\frac{G'(t)}{G(t)}=k\frac{\phi''(x)}{\phi(x)}=-\lambda\]\[G'(t)+\lambda G(t)=0\]\[\phi''(x)+\frac{\lambda}{k} \phi(x)=0\]\[G(t)=ce^{-\frac{\sqrt{k}n\pi}{L}t}\]\[\phi(x)=c\sin(\frac{\sqrt{k}n\pi}{L}x)\]\[u(x, t)=e^{-\frac{\sqrt{k}n\pi}{L}t}\sum^{\infty}_{n= 1}B_n\sin(\frac{\sqrt{k}n\pi}{L}x)\]\[B_n = \frac{1}{2L}\int^{L}_{0}f(x)\sin(\frac{\sqrt{k}n\pi}{L}x)dx\]

Answer 25

Correction, \[B_n = \frac{2}{L}\int^{L}_{0}f(x)\sin(\frac{\sqrt{k}n\pi}{L}x)dx\]