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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!

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what do you mean by equivalent? The initial conditions for $\psi(\vec r,0)$ are different provided $s\not \equiv 0,$ so there exists no function that solves both of them – Lacek Jan 9 '13 at 8:22
Possibly better suited for Math.SE. – Qmechanic Jan 9 '13 at 14:26
@Lacek Equivalence means they have the same solution. – chaohuang Jan 9 '13 at 16:11
up vote 1 down vote accepted

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.

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how is it possible that $\psi(vec r, t)$ can be both equal to $s(\vec r)$ and on the other hand to zero? Except for $s=0$ that is... – Lacek Jan 9 '13 at 17:08
@Lacek: I am not sure such reasoning is appropriate here. Indeed, there is a distribution ( ) in the right-hand side of the first PDE, so it is reasonable to expect that the solution will be a distribution, not a function. A distribution does not have to be defined everywhere, we just need that its products with "decent" functions are integrable. Or maybe I misunderstood you? – akhmeteli Jan 10 '13 at 2:44
yes, my argument is only applicable to functions. Nevertheless what sort of behaviour is to expect around $t=0$? I think that this is just a "step" of height $s(\vec r)$, then $t$-derivative gives $\delta(t) s(r)$, second derivative gives $\delta'(t)s(r)$. The second equation would give the same if we assumed additionally that $\psi(t<0,\vec r)=0$. I guess then the solution may in fact be just a discontinous function around $t=0$ (and of course a distribution by local integrability). The problem is the second equation does not gieanything special about $t<0$ – Lacek Jan 11 '13 at 9:25
@Lacek: I stand corrected. Please see the edit to my answer. – akhmeteli Jan 12 '13 at 8:51

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