Show that f solves the so called wave equation

Question

Task

$\text{Let } \; c \in \mathbb{R} \; \text{ be a given parameter, with } \; c > 0$

$\text{ Show that } \; f: (\mathbb{R}^3 \setminus \{ \vec{0} \}) \times \mathbb{R} \to \mathbb{R} \; \text{ with:}$

$$ f(x,y,z,t) := \frac {\cos(\|(x,y,z)\|_2 -ct)}{\|(x,y,z)\|_2} $$

$\text{ solves the so called }$ wave equation:

$$ \frac {\partial^²f}{\partial x^2}(x,y,z,t) + \frac {\partial^²f}{\partial y^2}(x,y,z,t) + \frac {\partial^²f}{\partial z^2}(x,y,z,t) - \frac {1}{c^2} \frac {\partial^²f}{\partial t^2}(x,y,z,t) = 0 $$

$\text{for all } \; (x,y,z,t) \in (\mathbb{R}^3 \setminus \{ \vec{0} \}) \times \mathbb{R}$

This task seems to require more thinking and knowledge than actual calculations, since calculating all those by hand and adding them up will only lead to a giant unsolvable equation IMHO.

There is a extra hint given to support this:

You may use following rule:

$ \text{Let } \; g \in \mathscr{C}^2(\mathbb{R}), r: \mathbb{R}^3 \setminus \{ \vec{0} \} \to \mathbb{R} \text{ with } r(\vec{x}) := \|\vec{x}\|_2 \; \text { and } \; \vec{x} \in \mathbb{R}^3 \setminus \{ \vec{0} \} \; \text{ then: }$

$$ \frac {\partial^2 (g \circ r)}{\partial x_i^2}(\vec{x}) = \frac {g'(r(\vec{x}))}{r(\vec{x})} + \left( g''(r(\vec{x})) - \frac {g'(r(\vec{x}))}{r(\vec{x})} \right) \frac {x_i^2}{r^2(\vec{x})} \quad \text {for $\quad$ i = 1,2,3} $$

My Efforts

I tried to simply calculate the partial derivates for x,y,z and t and put them into the equation, but even with the use of the hint I ended up with a giant mess.

The partial derivates for t should be fairly easy:

$$ \frac {\partial^2 f}{\partial t^2} = -\frac {c^2\cos(\|(x,y,z)\|_2-ct)}{\|(x,y,z)\|_2} $$

But I struggle to make use of the hint, after a couple of tries I came up with this:

$$ \text{Let } \; g(x) = \frac{\cos(x-ct)}{x} \; \text{ and } \; \vec{x} = (x,y,z) $$

$$ => (g \circ r)(\vec{x}) = f(\vec{x},t) = f(x,y,z,t) $$

$$ => \frac {\partial^2 f}{\partial x_i^2} (\vec{x},t) = \frac {\partial^2 (g \circ r)}{\partial x_i^2} (\vec{x})$$ $$ = -\frac{\|\vec{x}\|_2 \sin(\|\vec{x}\|_2 - ct) + \cos(\|\vec{x}\|_2 - ct)}{\|\vec{x}\|_2} + \left( \frac{2\|\vec{x}\|_2 \sin(\|\vec{x}\|_2 - ct) + (2 - \|\vec{x}\|_4) \cos(\|\vec{x}\|_2 - ct)}{\|\vec{x}\|_6} + \frac{\|\vec{x}\|_2 \sin(\|\vec{x}\|_2 - ct) + \cos(\|\vec{x}\|_2 - ct)}{\|\vec{x}\|_2} \right) * \frac{x_i^2}{\|\vec{x}\|_4} $$

After tons of simplifications (since the derivates for the $x_i$'s are almost the same) e.g.:

$$ \|\vec{x}\|_2 = \|(x,y,z)\|_2 = \left( \sqrt{x^2+y^2+z^2} \right)^2 = x^2 + y^2 + z^2 $$

$$ ... \frac{x^2}{\|\vec{x}\|_4} + ... \frac{y^2}{\|\vec{x}\|_4} + ... \frac{z^2}{\|\vec{x}\|_4} = \frac{\|\vec{x}\|_2}{\|\vec{x}\|_4} = \frac{1}{\|\vec{x}\|_2} $$

I came up with this equation:

$$ -3 * \left( \frac{\|\vec{x}\|_2 \sin(\|\vec{x}\|_2 - ct) + \cos(\|\vec{x}\|_2 - ct)}{\|\vec{x}\|_2} \right) + \frac{2\|\vec{x}\|_2 \sin(\|\vec{x}\|_2 - ct) + (2 - \|\vec{x}\|_4) \cos(\|\vec{x}\|_2 - ct)}{\|\vec{x}\|_8} + \frac{\|\vec{x}\|_2 \sin(\|\vec{x}\|_2 - ct) + \cos(\|\vec{x}\|_2 - ct)}{\|\vec{x}\|_4} + \frac{\cos(\|\vec{x}\|_2 - ct)}{\|\vec{x}\|_2} = 0 $$

The solution I can think of right now would be to set $\|\vec{x}\|_2$ = $1$:

$$ -3 * ( \sin(1 - ct) + \cos(1 - ct) ) + 2 \sin(1 - ct) + \cos(1 - ct) + \sin(1 - ct) + \cos(1 - ct) + \cos(1 - ct) $$ $$ = -3\sin(1 - ct) -3\cos(1 - ct) + 3 \sin(1 - ct) + 3\cos(1 - ct) = 0 $$

But that seems more like a stupid hack instead of a solution, since this does not solve the equation for every $(x,y,z) \in \mathbb{R}^3$ but only for those where $\|\vec{x}\|_2$ = $1$. But it solves the equation for every $t \in \mathbb{R}$

Question

What is that fact, relation or obvious simplification I'm missing here? Since this is a quite unusual amount of work that needs to be put into calculating those per hand.

Answer 1

sorry about the mistakes, i hope this may be more useful - if $g= \frac{\cos(r-ct)}r$ we have $$ \frac{\partial g}{\partial x} = -\frac{x}{r^3}\cos(r-ct) -\frac{x}{r^2}\sin(r-ct) $$ and $$ \begin{align} \frac{\partial^2 g}{\partial x^2} = & -\frac{\cos(r-ct)}{r^3} -\frac{\sin(r-ct)}{r^2} \\ & +\frac{x^2}{r} \left(\frac{3\cos(r-ct)}{r^4} +\frac{2\sin(r-ct)}{r^3} \right) \\ & - \frac{x^2}{r} \left(\frac{-\cos(r-ct)}{r^3} +\frac{\cos(r-ct)}{r^2} \right) \end{align} $$ which gives $$ \frac{\partial^2 g}{\partial x^2}+\frac{\partial^2 g}{\partial y^2}+\frac{\partial^2 g}{\partial z^2} = -\frac{\cos(r-ct}{r} $$ but we also have: $$ \frac{\partial^2 g}{\partial t^2} = -c^2 \frac{\cos(r-ct)}{r} $$ so $$ \frac{\partial^2 g}{\partial x^2}+\frac{\partial^2 g}{\partial y^2}+\frac{\partial^2 g}{\partial z^2} -\frac1{c^2}\frac{\partial^2 g}{\partial t^2} = 0 $$