Jncik said:
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]
This is wrong. I presume that, separating variables so that u(x,t)= X(x)T(t), you found that X''/X= \lambda= T''/T so that you got the two ordinary differential equations X''= \lambda X and T''= \lambda T. What you have here is the solution
if \lambda> 0 and you have no reason to believe that.
But you can also say that u(0,t)= X(0)T(t)= 0 means either X(0)= 0 or T(t)= 0. Since the latter would make the solution 0 for all t, it cannot satisfy the initial conditions so we must have X(0)= 0. Similarly, u(\pi, t)= X(\pi)T(t)= 0 gives X(\pi)= 0.
So we are solving X''= \lambda X with boundary conditions X(0)= X(\pi)= 0.
Consider three cases: \lambda= 0, \lambda> 0, \lambda< 0
If \lambda= 0, the equation is just X''= 0 and, integrating twice, X(x)= C_1x+ C_2. Then X(0)= C_2= 0 so X(\pi)= C_1\pi= 0 which implies that X is identically 0. But then we cannot satisfy the intial condition that u(x, 0)= sin(2x) so that is impossible,
If \lambda> 0, we can let \lambda= \alpha^2 for \alpha non-zero. The differential equation is X''= \alpha^2X and the general solution is X(x)= C_1e^{\alpha x}+ C_2e^{-\alpha x} (this is what you had). Now we must have X(0)= C_1+ C_2= 0 and X(\pi)= C_1e^{\alpha\pi}+ C_2e^{-\alpha\pi}. From the first equation, C_2= -C_1 and, putting that into the second, C_1e^{\alpha\pi}- C_1e^{-\alpha\pi}= C_1(e^{\alpha\pi}- e^{-\alpha\pi})= 0.
Now, since \alpha \ne 0, e^{\alpha\pi} and e^{-\alpha\p} are not equal (one is larger than 1 and the other less than 1) and their difference cannot be 0, so we must have C_1= 0 and then C_2= 0. That is, X(x) is identically 0 so it cannot give a u that satisifies the initial condition.
Finally, if \lambda< 0, we can let \lambda= -\alpha^2 for \alpha non-zer. The differential equation is X''= -\alpha^2X and the general solution to that is X(x)= C_1cos(\alpha x)+ C_2 sin(\alpha x). Now we must have X(0)= C_1= 0 and X(\pi)= C_2 sin(\alpha\pi)= 0. This would give C_2= 0 and so again a "trivial" solution
unless sin(\alpha\pi)= 0 which
does happen if \alpha\pi is a multiple of \pi- that is, if \alpha= n and \lambda= -n^2 with n and integer.
That is, in order to have a non-trivial solution, we must have \lambda= -n^2 for n some integer, in that case, X(x)= Csin(nx) for unknown constant C and unknown integer n.
Now, go back to your equation for T: T''= \lambda T= -n
2T. What is the general solution to that?
( Your condition "u_t(\pi/4, 0)= 0" looks very peculiar- usually it is something like u(x, 0)= 0 for all x, but I think u_t(\pi/4, 0)= 0 will give the same thing. u_t(\pi/4, 0)= X(\pi/4)T&#039;(0)= 0 and X(\pi/4)\ne 0 so you must have T&#039;(\pi/4)= 0.<br />
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the problem is that i don't know how to use the given equations to find the final result<br />
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for example<br />
<br />
[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]<br />
<br />
my professor says that we assume that there is a t for which <br />
<br />
[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]<br />
<br />
and he finds that c<sub>1</sub> = -c<sub>2</sub><br />
<br />
for<br />
<br />
u(π,t) he says<br />
<br />
[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]<br />
<br />
so we have<br />
<br />
[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]<br />
<br />
and from this he says the following which i can't understand<br />
<br />
from u(x,0) = sin(2x) we get<br />
<br />
λ = -4 and c<sub>2</sub> = 1/2<br />
<br />
how did we get this?<br />
<br />
Thanks in advance
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