General solution of 1D vs 3D wave equations

[yucheng]

TL;DR Summary

N/A

For the 1 dimensional wave equation,

$$\frac{\partial^2 u}{\partial x ^2} - \frac{1}{c^2}\frac{\partial ^2 u }{\partial t^2} = 0$$

$u$ is of the form $u(x \pm ct)$
For the 3 dimensional wave equation however,
$$\nabla ^2 u - \frac{1}{c^2}\frac{\partial ^2 u }{\partial t^2} = 0$$It appears that solutions need not be of the form $u(\vec{k} \cdot \vec{r} - \nu t)$, for instance spherical waves $u = A/r \; \mathrm{exp}[-i(kr - vt)]$

Am I right? Why is it so?

 

[Frabjous]

Conservation of energy. A spherical wave spatial expands so it’s magnitude needs to decrease.

 

[yucheng]

Conservation of energy. A spherical wave spatial expands so it’s magnitude needs to decrease.

But why doesn't it apply to the 1-dimensional case?

 

[Frabjous]

But why doesn't it apply to the 1-dimensional case?

Take a distance r from the origin. For a spherical wave, things are spread out on the surface of a sphere with surface area 4πr². On a line, things are still located at a single point.

 

[pasmith]

Summary: N/A

For the 1 dimensional wave equation,

$$\frac{\partial^2 u}{\partial x ^2} - \frac{1}{c^2}\frac{\partial ^2 u }{\partial t^2} = 0$$

$u$ is of the form $u(x \pm ct)$
For the 3 dimensional wave equation however,
$$\nabla ^2 u - \frac{1}{c^2}\frac{\partial ^2 u }{\partial t^2} = 0$$It appears that solutions need not be of the form $u(\vec{k} \cdot \vec{r} - \nu t)$, for instance spherical waves $u = A/r \; \mathrm{exp}[-i(kr - vt)]$

Am I right? Why is it so?

The change of variable $\zeta = x + ct$, $\eta = x - ct$ turns the 1D wave equation into $$\frac{\partial^2 u}{\partial \zeta\,\partial \eta} = 0$$ with general soluton $$u = f(\zeta) + g(\eta) = f(x + ct) + g(x-ct).$$ There is no equivalent of this change of variable in 2 or more spatial dimensions, so other types of solution are possible. In particular, solutions of the form $u(\mathbf{x},t) = f(\mathbf{x})e^{\pm i\omega t}$ are possible where $f$ satisfies $$\nabla^2 f + \frac{\omega^2}{c^2}f = 0.$$ This equation separates in many coordinate systems, and only in cartesians do we get solutions of the form $$f(\mathbf{x}) = \int_{\|\mathbf{k}\| = \omega/c}A(\mathbf{k})e^{i\mathbf{k}\cdot \mathbf{x}}\,dS$$ so that $$u(\mathbf{x},t) = \int_{\|\mathbf{k}\| = \omega/c}A(\mathbf{k})e^{i(\mathbf{k}\cdot \mathbf{x} \pm \omega t)} \,dS.$$ In 3D with spherical symmetry it can be shown that $u(r,t)/r$ satisfies the 1D wave equation, with general solution as above.

 

[yucheng]

1D wave equation

1D wave equation? Really?

 

[yucheng]

1D wave equation? Really?

Oops let me quote it in context...

In 3D with spherical symmetry it can be shown that u(r,t)/r satisfies the 1D wave equation

Should it really be called 1D? I mean after a change of variables from Cartesian to spherical, then taking the partial derivatives with respect to $\theta$ and $\phi$ as zero, we get an ODE, but...

 

[Frabjous]

Should it really be called 1D?

Some people do, some people don’t. You shouldn’t let semantics bother you.