- #1
Jncik
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Homework Statement
solve uxx = utt
for 0<x<π when u(0,t) = u(π,t) = 0 and u(x,0) = sin(2x) and ut(π/4,0) = 0
The Attempt at a Solution
i used the method of separation of variables and have found out that
[URL]http://latex.codecogs.com/gif.latex?u%28x,t%29%20=%20%28c_{1}e^{x\sqrt{\lambda}}%20+%20c_{2}e^{-x\sqrt{\lambda}}%29%28C_{1}e^{t\sqrt{\lambda}}%20+%20C_{2}e^{-t\sqrt{\lambda}}%29[/URL]the problem is that i don't know how to use the given equations to find the final result
for example
[URL]http://latex.codecogs.com/gif.latex?u%280,t%29%20=%200%20=%3E%20%28c_{1}%20+%20c_{2}%29%28C_{1}e^{t\sqrt{\lambda}}%20+%20C_{2}e^{-t\sqrt{\lambda}}%29%20=%200%20=%3E%20?[/URL]
my professor says that we assume that there is a t for which
[URL]http://latex.codecogs.com/gif.latex?%28C_{1}e^{t\sqrt{\lambda}}%20+%20C_{2}e^{-t\sqrt{\lambda}}%29%20\neq%200[/URL]
and he finds that c1 = -c2
for
u(π,t) he says
[URL]http://latex.codecogs.com/gif.latex?u%28%CF%80,t%29%20=%20%28c_{1}e^{\pi%20\sqrt{\lambda}}%20-%20c_{1}e^{-\pi\sqrt{\lambda}}%29%28C_{1}e^{t\sqrt{\lambda}}%20+%20C_{2}e^{-t\sqrt{\lambda}}%29%20=%200%20=%3E%20%28c_{1}e^{\pi%20\sqrt{\lambda}}%20-%20c_{1}e^{-\pi\sqrt{\lambda}}%29%20=%200%20=%3E%20\lambda%20=%20-n^{2},%20n%20=%200,1,2,3...[/URL]
so we have
[URL]http://latex.codecogs.com/gif.latex?%28e^{\pi%20\sqrt{\lambda}}%20-%20e^{-\pi\sqrt{\lambda}}%29%20=%200%20=%3E%20e^{in\pi}%20-%20e^{-in\pi]}%20=%200%20=%3E%202sin%28n\pi%29%20=%200[/URL]
and from this he says the following which i can't understand
from u(x,0) = sin(2x) we get
λ = -4 and c2 = 1/2
how did we get this?
Thanks in advance
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