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This is a famous problem, but I am stuck at the very last step. Given the wave equation $ y_{xx} = \frac{1}{v^2} y_{tt}$, I need to use Fourier transform to solve for $y(x,t)$ given the initial condition $y(x,0) = f(x)$ for some known $f(x)$ with FT $F(k)$.

After taking the FT of both the wave equation and its initial condition, I get $y(k,t) = A(k) e^{i k v t} + B(k) e^{- i k v t}$ where $A(k) + B(k) = F(k)$.

I now need to invert $y(k,t)$. If I look at $A(k) e^{i k v t}$ first, this would give:

$FT^{-1} [A(k) e^{i k v t}] = FT^{-1} [\, \,FT[A(x)] \, FT[y(x,t)]\, \,] $ which is $A(x) * y(x,t)$, $*$ denoting convolution.

But since I only have a relation for $A(k)$ in terms of $B(k)$ and $F(k)$, how can I solve for $A(x)$ in this case? Just a bit confused.

drg
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  • The somewhat shady solution is to say that $\exp(+ikvt)$ cannot occur, i.e. chose $A(k) = 0$ and $B(k) = F(k)$. This is justified by energy conservation. – AlexR Nov 20 '13 at 22:58

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