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For a wave equation

$\begin{equation} \begin{cases} u_{tt}- \Delta u =0 \\ u|_{t=0} = f(x) \\ u_{tt}|{t=0} = g(x) \\ \end{cases} \end{equation}$

If we take Fourier transform in time $t$, then we have the Helmholtz equation

$\begin{equation} \begin{cases} -\omega^2v(\omega,x) - \Delta v(\omega,x) =0 \\ \int v(\omega,x)d\omega = f(x) \\ \int i\omega v(\omega,x)d\omega = g(x) \\ \end{cases} \end{equation}$

Now if we have a Helmholtz equation whoes solution $v$ does not satiesfy the conditions $\int v(\omega,x)d\omega = f(x)$ or $\int i\omega v(\omega,x)d\omega = g(x)$, will it still be equivalent to some wave equation?

I read some textbooks. All of them only state that Helmholz equation is the time-harmonic form of wave equation, but do not say anything about the initial conditions.

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  • $\begingroup$ Please note that in general $\omega$ can take any values. Any linear combination of particular solutions with different frequencies is also a solution of the original equation (for some unspecified initial conditions). You can satisfy any kind of initial condition that way (as your formulas show) $\endgroup$ – Yuriy S Feb 19 '18 at 9:21
  • $\begingroup$ @YuriyS Sorry, I did not catch your point. $\omega$ can be any value, but it is also the variable of integration. Can you please give me a more detailed example? $\endgroup$ – N N Feb 19 '18 at 17:44

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