What is the most general form for the wave equation ? Is it $\frac{\partial^2 \Psi}{\partial t^2}-v^2\nabla^2\Psi=0$?

For instance, can $\frac{\partial^2 \Psi}{\partial t^2}-v^2\nabla^2\Psi=cte$ be a wave equation? If yes, what is the solution in that case.


3 Answers 3


Mason handled the distinction between inhomogeneous and homogeneous differential equations, but if one is speaking of the most general possible form of the wave equation, it is,

$$\square\phi^{i_1\dots i_m}_{j_1 \dots j_n}(x) = f^{i_1 \dots i_m}_{j_1 \dots j_n}(x)$$

where both fields are rank $(m,n)$ tensors, acted upon by the Laplace-Beltrami operator $\square = \nabla^a \nabla_a$ whose action on the tensors depends on both the metric and their rank. For a scalar field with metric $\eta_{\mu \nu}$, it reduces to the most familiar form of the wave equation, $(\partial^2_t - \nabla^2)\phi = f$. (The above can also be recast in the language of differential forms.)

However, in a way this does not cover all the possibilities. For example, in general relativity, for a perturbation $h_{ab}$ of the metric, the first order change in the curvature is,

$$\delta R_{ab} \propto \Delta_L h_{ab} = \square h_{ab} -2 \nabla_{(a} \nabla^c \bar{h}_{b)c} -2 R_{d(a} h^d_{b)} +2 R_{acbd}h^{cd}$$

which is understood as the curved space 'wave operator' in the literature because it certainly admits wave solutions but is clearly not equivalent to the wave equation above as it contains other terms involving curvature tensors. Thus, the 'most general form' of the wave equation isn't something we can really write down, unless your idea of it is strictly $(\partial^2_t - \nabla^2)\phi = f$.


I'm not sure what you mean by $cte$, but I'm assuming it's some constant but I may be misinterpreting

We often talk about two classes of differential equation, homogeneous and inhomogeneous. This distinction is the root of your question, \begin{equation} \frac{1}{v^2}(\partial_t)^2 f(\vec{r},t) - \nabla ^2 f(\vec{r},t) = 0 \end{equation} is the homogeneous form of the wave equation, whereas \begin{equation} \frac{1}{v^2}(\partial_t)^2 f(\vec{r},t) - \nabla ^2 f(\vec{r},t) = u(\vec{r},t) \end{equation} is the inhomogeneous wave equation ($u(\vec{r},t)$ can also be constant if we like). This arises all over the place. One example is that electromagnetic radiation in the presence of charges and currents is governed by the inhomogeneous wave equation, the homogeneous form is only valid when $\rho=0$ and $\vec{J}=0$. Depending on who you ask, I think most people would still say the the inhomogeneous wave equation is a wave equation, but that's up to taste as it's solutions can end up having a very different character to the homogeneous ones.

In general there's not much I can say about these solutions since they'll depend heavily on the form of $u$, though I'm sure some googling will give you plenty of examples.

  • $\begingroup$ Perfect. And what about damped wave equation ? What is its form ? $\endgroup$
    – Sofiane
    Commented Oct 19, 2016 at 7:18

enter image description here


from http://www.math.psu.edu/tseng/class/Math251/Notes-PDE%20pt4.pdf


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