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$u_{tt} = c^2u_{xx}+sin(\alpha t)$

$u(0,t)=0=u(\pi,t)$

$u(x,0) = 0 = u_t(x,0)$

where $0<x< \pi $ and $t>0$

I know how to solve this problem using Fourier series, but I also encountered another solution method where let $u(x,t) = v(x)+w(x,t)$. I want to ask how should I choose $v(x)$ in this case? Thanks for any help.

user59036
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  • are you sure it is not $u(x,t) = v(t) + w(x,t)$? in this case, choosing $v$ such as $v'' = \sin\alpha t$. – mookid Mar 05 '14 at 02:26
  • @mookid but the boundary is $0<x< \pi$, if $v$ is dependent on $t$, there will be problem with the boundary conditions – user59036 Mar 06 '14 at 02:51
  • ... and that just means that your ansatz $u(x,t) = v(x) + w(x,t)$ does not work in this case. That specific ansatz is useful for time independent sources, which is not the case of your equation. – Willie Wong Mar 07 '14 at 10:07
  • @WillieWong Can you explain more why the specific ansatz $u(x,t) = v(x) +w(x,t)$ is only useful for time independent sources? Also, the PDE in the question above is a wave equation, what if it's a heat equation like $u_t = \alpha ^2 u_{xx}$? Will the splitting method $u(x,t) = v(t) +w(x,t) $works in this case? – user59036 Mar 09 '14 at 18:30
  • @user59036: you plug your ansatz in, and what do you find? $v_{xx} = 0$ (since it cannot depend on $t$) and $w_{tt} = c^2 w_{xx} + \sin(at)$. So the equation doesn't simplify at all. For the heat equation the splitting still won't work. The $v(x)$ ansatz doesn't work for the same reason (plug it in, and it doesn't simplify) and the $v(t)$ ansatz also doesn't work as it is incompatible with the boundary conditions unless $v\equiv 0$. – Willie Wong Mar 10 '14 at 08:38

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