# Deducing a property for any function $f$ using the wave equation

I am given the wave equation in spherical coordinates for a wave function only depending on $r$: $$\frac{1}{v^2} \frac{\partial^2 \xi(r,t)}{\partial t^2} = \left( \frac{\partial^2}{\partial r^2} + \frac{2}{r} \frac{\partial}{\partial r} \right) \xi(r,t).$$ Also, I know that this equation can be solved using the ansatz $\xi(r,t) = A(r) f(kr \pm \omega t)$. Now I have to show that for any (twice differentiable) function $f$,

$$A(r) = \frac{C}{r}$$

with some constant $C$ must hold.

To prove this, I used the given ansatz and computed the partial derivatives of the wave function and plugged them into the wave equation. Doing this, I end up with the condition

$$A''(r) f(kr\pm\omega t) + A'(r)[2(f'(kr\pm \omega t)k + \frac{1}{r} f(kr\pm \omega t))] + A(r) \frac{2k}{r} f'(kr \pm \omega t) = 0,$$

where $A', A''$ and $f'$ denote the derivatives of those functions.

My question is: How do I proceed?

Obviously, this equation holds for $f = 0$. If we have $f = m$ for some $m \in \mathbb{R} \setminus \{0\}$, the condition will simplify to the second order differential equation $$A''(r) + \frac{2 A'(r)}{r} = 0.$$ Using substitution ($B(r) := A'(r)$), we find $A(r) = - \frac{c_1}{r} + c_2$ and since $A(r) \to 0$ as $r \to \infty$, we can conclude $c_2 = 0$.

But how can I similarly (?) show $A(r) = \frac{C}{r}$ for any function $f$?

Thank you very much in advance for any help.

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There is a difference between the statement

for any $f$, $A(r)$ must equal $C/r$

and

were $A(r)$ to hold for any $f$, $A(r)$ must equal $C/r$

The first statement $A(r)$ is allowed to depend on $f$. The second statement $A(r)$ cannot.

The first statement is false. Take $\omega = 0$ and set $A(r) f(kr) = 1$, then for any non-vanishing $f$ this particular $A(r)$ will work.

To show the second statement is true, you are mostly done. From the equation you derived by plugging in the ansatz, you have that, grouping terms by $f$,

$$\left(A''(r) + A'(r)\frac{2}{r}\right) f(kr\pm\omega t) +\left( A'(r) + A(r) \frac{1}{r}\right)2k f'(kr \pm \omega t) = 0$$

For this to be true for any $f$, the "coefficients" inside the parentheses must vanish identically (consider (a) the constant function $f\equiv 1$ and (b) the function $f$ defined so that on an interval $[a,b]$, $f(x) = x$). So you get

$$A'' + 2A'/r = 0 \qquad A' + A/r = 0$$

or equivalently

$$(r^2 A')' = 0 \qquad (rA)' = 0$$

which gives you precisely that $rA = C$.

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I don't quite get the difference between the two statements. I understand "for any f, A(r) must equal C/r" as "we have the same constant no matter which function $f$ we are plugging into the ansatz", whereas you say that this means A(r) depends on f. This confuses me, could you explain this part more in detail (not that I misunderstand the problem...), please? – Huy Feb 27 '12 at 13:45
it is the issue of quantifiers. The first statement, in the context that you have $\xi = Af$ a solution, usually means "$\forall f$, if $\xi = Af$ is a solution then $A = C/r$." The second statement says "if $A$ is such that $\forall f$, $\xi = Af$ is a solution, then $A = C/r$." Another way of stating the difference is that what you wrote sounds more like "I know a solution can be written as $\xi = Af$. Then I choose an $f$, and solve for $A$, then $A$ must be $C/r$." That statement is false!!! ... – Willie Wong Feb 27 '12 at 14:10
What you need to do is that you need to "assume that there is a function $A$ such that for any function $f$, $\xi = Af$ is a solution, then the function $A = C/r$." It is this freedom in choosing any function $f$ that allows you to separate the reduced wave equation and treat the two coefficient terms separately. – Willie Wong Feb 27 '12 at 14:13