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This is a cleaned up and refined repost of my previous attempt, after I did some research on the subject.

SOLVED

The only problem here was that I was making tons of little mistakes, always watch your signs!

Task

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

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

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

$\text{ solves the homogeneous wave equation:}$

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

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

$\text{You may use following rule:}$

$ \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

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

$$ \frac {\partial^2 f}{\partial t^2}(x,y,z,t) = -\frac {c^2\cos(r-ct)}{r} \Rightarrow \frac {1}{c^2} \frac {\partial^²f}{\partial t^2}(x,y,z,t) = -\frac {\cos(r-ct)}{r} $$

$$ \Rightarrow \frac {\partial^²f}{\partial x^2}(x,y,z,t) + \frac {\partial^²f}{\partial y^2}(x,y,z,t) + \frac {\partial^²f}{\partial z^2}(x,y,z,t) = -\frac {\cos(r-ct)}{r} $$

$$ \sum_{i=1}^{3} \frac {\partial^2 f}{\partial x_i^2}(\vec{x},t) = \sum_{i=1}^{3} \frac {\partial^2 (g \circ r)}{\partial x_i^2}(\vec{x}) $$

$$ = 3 * \frac {g'(r(\vec{x}))}{r(\vec{x})} + \left( g''(r(\vec{x})) - \frac {g'(r(\vec{x}))}{r(\vec{x})} \right) \frac {1}{r(\vec{x})} $$

$$ \Rightarrow 3 * \frac {g'(r(\vec{x}))}{r(\vec{x})} + \left( \frac {g''(r(\vec{x}))}{r(\vec{x})} - \frac {g'(r(\vec{x}))}{r^2(\vec{x})} \right) = -\frac {\cos(r-ct)}{r} $$

$$ g'(r(\vec{x})) = -\frac {\sin(r-ct)}{r} - \frac {\cos(r-ct)}{r^2} $$

$$ g''(r(\vec{x})) = \frac {2\sin(r-ct)}{r^2} - \frac {\cos(r-ct)}{r} + \frac {2\cos(r-ct)}{r^3} $$

$$ \Rightarrow \sum_{i=1}^{3} \frac {\partial^2 f}{\partial x_i^2}(\vec{x},t) = -3\frac {r\sin(r-ct)+\cos(r-ct)}{r^3} + \left( \frac {2\sin(r-ct)}{r^2} - \frac {\cos(r-ct)}{r} + \frac {2\cos(r-ct)}{r^3} + \frac {r\sin(r-ct)+\cos(r-ct)}{r^3}\right) * r^2 $$

$$ = -3\frac {r\sin(r-ct)+\cos(r-ct)}{r^3} + \left( \frac {3r\sin(r-ct) - (r^2-3)\cos(r-ct)}{r^5} \right) * r^2 $$

$$ = -\frac {3r\sin(r-ct)+3\cos(r-ct)}{r^3} + \frac {3r\sin(r-ct) + 3\cos(r-ct) - r^2\cos(r-ct)}{r^3} $$

$$ = -\frac {3r\sin(r-ct)+3\cos(r-ct)}{r^3} + \frac {3r\sin(r-ct) + 3\cos(r-ct)}{r^3} - \frac{r^2\cos(r-ct)}{r^3} = -\frac {\cos(r-ct)}{r} $$

$$ \Rightarrow \sum_{i=1}^{3} \frac {\partial^2 f}{\partial x_i^2}(\vec{x},t) + \frac {\cos(r-ct)}{r} = -\frac {\cos(r-ct)}{r} + \frac {\cos(r-ct)}{r} = 0 $$

Question

Where is my mistake / How to solve this equation?

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Shawniac
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1 Answers1

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We have

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

For short

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

Now consider $$g(r)=\frac{\cos (r-ct)}{r}.$$

According to the given rule it is

$$\frac{\partial^2 g}{\partial x_i^2}=\frac{g'(r)}{r}+\left(g''(r)-\frac{g'(r)}{r}\right)\frac{x_i^2}{r^2} \\= -\frac{r\sin(c-t)+\cos(r-ct)}{r^3}+\frac{3r\sin(c-t)-(r^2-3)\cos(r-ct)}{r^5}x_i^2.$$ So

$$\sum_{i=1}^3 \frac{\partial^2 f}{\partial x_i^2}=-\frac{3r\sin(c-t)+3\cos(r-ct)}{r^3}+\frac{3r\sin(c-t)-(r^2-3)\cos(r-ct)}{r^3}\\ =-\frac{\cos(r-ct)}{r}.$$

mfl
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  • I saw some errors on my side now, I my Question to resemble what I did with your help, now I'm stuck again, could you have a look at that? – Shawniac Nov 02 '14 at 22:47
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    If I am not wrong, your second derivative should read $g''(r)= \frac {2\sin(r-ct)}{r^2} - \frac {\cos(r-ct)}{r} + \frac {2\cos(r-ct)}{r^3}$ – mfl Nov 02 '14 at 22:52
  • You are totally right again, now I'm stuck on the very end!!! Somehow you say that $\sum_{i=1}^{3} x_i^2 = r^2$ while I only see that this equals $r$. How's that? – Shawniac Nov 02 '14 at 23:19
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    Note that $r=|(x,y,z)|_2=\sqrt{x^2+y^2+z^2}$ and so $\sum_ix_i^2=r^2.$ – mfl Nov 02 '14 at 23:21
  • Oh I thought (as noted in my question), that $r = |(x,y,z)|_2 = |(x,y,z)|^2 = x^2 + y^2 + z^2$, this was the final stone on my way to solve this task, thanks a ton for your fast and great amount of help! – Shawniac Nov 02 '14 at 23:25
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    You're welcome. – mfl Nov 02 '14 at 23:27