# Why does the inhomogenous wave equation solution change form with a different number of spatial dimensions?

Assume $$c = 1$$ for what follows.

For the general inhomogenous wave equation in one spatial dimension $$\left(\frac{\partial^2}{\partial t^2} - \frac{\partial^2}{\partial x^2}\right)\phi = v(x, t),$$ the article The Wave Equation with a Source gives (where we ignore the homogenous part of the solution, e.g. by setting initial conditions to $$0$$): $$\phi(x, t) = \frac{1}{2} \int_0^t \int_{x - (t - s)}^{x + (t - s)} v(y, s)\ dy \ ds.$$

For the general inhomogenous wave equation in three spatial dimensions $$\left(\frac{\partial^2}{\partial t^2} - \nabla^2\right)\phi = v({\bf r}, t),$$ the article Solution of Inhomogeneous Wave Equation gives $$\phi({\bf r}, t) = \int \frac{ v({\bf r}', t - \vert{\bf r} -{\bf r}'\vert)} {4\pi \vert{\bf r} - {\bf r}'\vert}\ dV'.$$

Why does the integrand in the three-dimensional case depend inversely on $$\vert{\bf r} - {\bf r}'\vert$$, while the integrand in the one-dimensional case doesn't?

• The first equation with the denominator in the integrand is for 3D waves. The second one without a denominator is for 1D waves. The dimension of space will change the formula.
– LPZ
Commented Apr 16, 2023 at 18:49
• But conceptually it doesn't make sense to me why a one dimensional wave equation would have a solution that behaves completely differently from a three dimensional wave equation --- what characteristic of three dimensions that does not exist in one dimension makes the behavior depend inversely on distance? Thanks for the comment. Commented Apr 16, 2023 at 18:53
• You can already see the issue with the Green’s function of the Laplacian. In fact, the power law in $r$ can be deduced simply by dimensional analysis. Note that if you take a source that is translational invariant along $y,z$ in the 3D case, you recover the 1D equation.
– LPZ
Commented Apr 16, 2023 at 18:56
• Stackexchange likes nice tidy question-answer pairs. Since your question as written has been answered, I suggest that you either ask another question, or (probably much better) edit this question to reflect what you really want to know (starting with the title, something like "why does the inhomogeneous solution change with change in dimension"). Commented Apr 16, 2023 at 18:57
• To lpz: Thanks! To TimWescott: Got it, I'll try fixing it when I have time. Commented Apr 16, 2023 at 19:22

The wave equation is linear, so the inhomogeneous problem: $$(\partial_t^2-\Delta)\phi=v$$ can be solved by a convolution with the Green’s function: $$\phi=G*v\\ (\partial_t^2-\Delta)G=\delta(x,t)$$ Your question can therefore be reformulated as explaining the spatial dimension $$D$$ dependence of power law of $$G$$.
This can be partially tackled by dimensional analysis. Setting $$c=1$$, $$x,t$$ have dimension $$1$$. The delta function has dimension $$-(D+1)$$. A partial derivative has dimension $$-1$$ so the d’Alembertian has dimension $$-2$$. Therefore, $$G$$ has dimension $$1-D$$. You therefore expect that: $$G\sim r^{1-D}$$ This gives the correct scaling in 1D and 2D, but not in 3D.
This is because you have an extra Dirac delta factor $$\delta(r-t)$$ which has dimension $$-1$$. Correcting this you do get for $$D=3$$: $$G\sim r^{-1}$$ while still being consistent with dimensional analysis. This extra Dirac delta factor has a simple interpretation: it is Huygens’ principle.