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I have a function $u(r,t)$ that satisfies $$\frac{\partial^2u}{\partial t^2} - c^2 \Delta u = 0.$$

I'm looking at a solution of a problem with this setup and it states that the above equation can be written as

$$\frac{\partial^2u}{\partial t^2} + c^2Au = 0$$ with $$A = -\Delta = -\frac{1}{r}\frac{d}{dr}(r\frac{d}{dr}).$$ I have two questions:

  1. The operator $\Delta$ is defined as $\sum \frac{\partial^2}{\partial x_i^2}$ but I never see the term $\frac{\partial^2}{\partial t^2}$ in $\Delta u$ when looking at solutions. Why is this?

  2. I get $$Au = -\frac{1}{r}\frac{d}{dr}(r\frac{du}{dr}) = -\frac{1}{r}(ru')' = -u'' - \frac{u'}{r}$$ which is NOT equal to $-\Delta u = \frac{\partial }{\partial r^2}$ or $-\Delta u = \frac{\partial }{\partial r^2} + \frac{\partial }{\partial r^2}.$

What is going on?

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  • 3
    $\begingroup$ 1. The Laplacian is a sum of second order derivatives on the spatial dimensions only 2. The Laplacian $A$ that you have is in polar coordinates, not Cartesian. See the polar form of the Laplacian here. Note that in your problem, $u = u(r,t)$ is not dependent on $\theta$, so the second order derivatives in $\theta$ in the polar Laplacian don't appear, which means $$\Delta u = u_{rr} + \frac{u_{r}}{r}$$ which is what you have written. $\endgroup$ – Mattos Aug 6 '18 at 16:36
  • $\begingroup$ Okay thanks. Just one question: so the laplacian is always on spatial dimensions? If my function is dependent on time, it's not in the laplacian? @Mattos $\endgroup$ – Heuristics Aug 6 '18 at 19:31
  • $\begingroup$ $\frac{\partial^2 u}{\partial t^2}$ is the Laplacian in time. $\endgroup$ – DisintegratingByParts Aug 10 '18 at 3:50

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