# 3D and 1D spherical waves equations and solutions

In class, we showed that one particular solution to the scalar wave equation in 3D was a **plane wave**, for which the wave took an argument of the form__ __$r \pm v t$. It was called a plane wave because, at a fixed time, the function

__$U (r \pm v t)$__has the same value over a plane specified by $r\hat{u} = constant$. In this problem, you will find a different type of solution under spherical symmetry.

a) Assume we are looking for a spherically symmetric solution to the scalar wave equation like $U(r,t)/r$, where r is the radial position coordinate in spherical coordinates. Show that if $U(r. t)/r$ satisfies the scalar wave equation in 3D, $U(r, t)$ satisfies a 1D scalar wave equation.

b) For a fixed time, what are the surfaces over which this spherical wave is constant? How do these surfaces propagate as time advances?

Answers

Let $f(r,t) = U(r,t) / r$

The 3D wave equation is

$d^2f/d t^2 = v^2*\nabla^2(f)$

In spherical coordinates the Laplacian is written as

$\nabla^2 = d^2/d r^2 +(2/r)*d/dr$

In our case

$d f/d r = d/d r (U/r) = 1/r*d U/d r -U/r^2$

$d^2f/d r^2 = d/d r (1/r*d U/d r -U/r^2) = (1/r)*d^2U/d r^2 -(1/r^2)*d U/d r -$

$-(1/r^2)*d U/d r -2U/r^3$

therefore

$\nabla^2(f) = d^2f/d r^2 +(2/r)*d f/d r =$

$=(1/r)*d^2U/d r^2 -(2/r^2)*d U/d r -2U/r^3 + (2/r)*[1/r*d U/d r -U/r^2] =(1/r)*d^2U/d r^2$

But

$d^2f/d t^2 =(1/r)*d^2U/d t^2$

And therefore the 3D wave equation becomes

$(1/r)*d^2U/d t^2 =v^2*(1/r)*d^2U/d r^2$

or equivalent

**$d^2U/d t^2 =v^2*d^2U/d r^2$**

WHICH IS THE 1D WAVE EQUATION that has as solution $U(r,t)$

b) Solutions of the form

$f(r,t) = U(r,t) / r$

have __constant surfaces SPHERICAL__ (or said simpler spheres). As the wave propagates these surfaces become of bigger and bigger radius (the spheres become bigger and bigger with time).