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$\omega \left( x, y \right)$ is an analytic function, we can find a function $\lambda \left( x, y \right)$ satisfying $$ \partial _{xx} \lambda -\partial _{yy} \lambda =\omega $$ (it's a wave equation with sources in physics).

Take Fourier transform for $x$ and $y$, we get $$ \tilde{\omega} \left( k_x, k_y \right) = -k_x^2 \tilde{\lambda} \left( k_x, k_y \right) + k_y^2 \tilde{\lambda} \left( k_x, k_y \right). $$

Now we have $$ \tilde{\omega} \left( k, k \right) = -k^2 \tilde{\lambda} \left( k, k \right) + k^2 \tilde{\lambda} \left( k, k \right) = 0. $$

But it's obvious that $\tilde{\omega} \left( k, k \right) = 0$ is not true in the most case. Where is the problem?

Daniele Tampieri
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xafeyog
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  • I apologize, I read your question too quickly. However, an answer could be that you usually do not apply the Fourier transform respect to all the variables when you deal with hyperbolic / parabolic PDEs since those kind of equations have a real characteristics i.e. they vanish naturally on some regions of the space (which is not the case for elliptic equations like Laplace's). Instead you use a partial fourier transform respect to the spatial variables ($x$ in our case) and reduce the PDE to an ODE for the partial Fourier transform of the sought for function. – Daniele Tampieri Nov 05 '19 at 11:37
  • Apart from my comment above, your question is interesting because it requires a non trivial understanding of the concept of characteristics of a PDE, so +1. – Daniele Tampieri Nov 05 '19 at 11:52
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    The point is that the solution $\tilde{\lambda}$ can be singular on the diagonal $k_x=k_y$ – lcv Nov 05 '19 at 11:55

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