# Wave equation - dissipation

The book states that the wave equation assumes no dispersion and no dissipation, with dissipation defined as a loss of energy and thus a diminution of amplitude.

How can a spherical wave be described by the wave equation if it's amplitude decreases like $\frac{1}{r}$ ? It seems contradictory ...

• I think the book was talking about plane waves in which the amplitude remains constant. – Paul Mar 16 '15 at 10:55

I assume you're talking about the simple wave equation $$\ddot \phi=\phi'' \qquad (1)$$ To simplify things, let's talk about one dimensional waves. Everything still holds for higher dimensions, you just replace $\phi'$ by $\vec\nabla\phi$ everywhere. The energy density is $$u=\frac{1}{2}\left((\dot\phi)^2+(\phi')^2\right)\ .$$
Note that $u$ is the local energy density, and it is not at all constant (except for special cases such as standing waves or plane waves). However, the total energy of the system, which is given by $$U=\int u\, d^3\!x$$ is constant. This is easy to see if you multiply $(1)$ by $\dot \phi$ to get \begin{align} 0&=\dot \phi\left(\ddot\phi-\phi''\right) =\frac{\partial}{\partial t}\left(\frac{1}{2}\dot\phi\right)^2-\dot\phi\phi''\\ &=\frac{\partial}{\partial t}\left(\frac{1}{2}\dot\phi^2\right)-\frac{\partial}{\partial x}\left(\phi'\dot\phi\right)+\phi'\dot\phi' =\frac{\partial}{\partial t}\left[\frac{1}{2}\dot\phi^2+\frac{1}{2}\phi''^2\right]-\frac{\partial}{\partial x}\left(\phi'\dot\phi\right)\\ &=\frac{\partial }{\partial t}u-\frac{\partial}{\partial x}\left(\phi'\dot\phi\right) \end{align} This is known as (the) continuity equation. The quantity $J=\phi'\dot\phi$ is known as the energy flux. To see that the global energy is constant, apply Gauss' theorem to $\dot u$.
The physical meaning of the continuity equation is that energy is globally conserved, but locally it is not necessarily so. Local changes in the energy density $u$ are compensated by the divergence of the energy flux.