equivalence of wave equations

Question

I wonder if the following 2 PDEs are equivalent:

1. $$\frac{\partial^2}{\partial t^2}\psi(\vec{r},t)-c(\vec{r})^2\nabla^2\psi(\vec{r},t)=s(\vec{r})\delta'(t)$$ subjects to zero initial conditions $$\psi(\vec{r},0)=0, \quad \left.\frac{\partial}{\partial t}\psi(\vec{r},t)\right|_{t=0}=0,$$ where $\delta'(t)$ denotes the derivative of Dirac delta function.

2. $$\frac{\partial^2}{\partial t^2}\psi(\vec{r},t)-c(\vec{r})^2\nabla^2\psi(\vec{r},t)=0$$ subjects to initial conditions $$\psi(\vec{r},0)=s(\vec{r}), \quad \left.\frac{\partial}{\partial t}\psi(\vec{r},t)\right|_{t=0}=0$$

I don't how to prove or disprove it. Any help will be appreciated. Thanks!

Answer 1

I would say the initial conditions are technically the same for the two PDEs, but the PDEs are indeed inequivalent if $s$ does not vanish identically: while $\psi(\overrightarrow{r},t)\equiv 0$ is a solution of the second PDE, it is not a solution of the first PDE. To get the solution of the first PDE, you may wish to use Fourier transform in space and Laplace transform in time.

EDIT (01/12/2012) I misread the initial conditions, so the above is wrong, and I apologize. You can prove that the PDEs are in general inequivalent in a similar way though: if $s(\overrightarrow{r})\equiv 1$, then $\psi(\overrightarrow{r},t)\equiv 1$ is a solution of the second PDE, but not of the first PDE.