# Rearranging wave equation

1. Jun 20, 2015

### Botttom

Hello!

The wave equation given: ${1\over{c^2}} {\partial^2 \phi\over{\partial t^2}} = \Delta \phi$ with $r = \sqrt{x^2+y^2+z^2}$ needs to be rearranged, so that ${1\over{c^2}} {\partial^2 ( r \phi) \over{\partial t^2}} = {\partial^2 (r \phi) \over{\partial r^2}}$.

Are there any tricks to obtain this result?

2. Jun 20, 2015

### Orodruin

Staff Emeritus
This is only possible if you have a wave which does not depend on the angular coordinates in spherical coordinates, i.e., a spherical wave. Then you need express the Laplace operator in spherical coordinates.

3. Jun 20, 2015

### Botttom

Yes, this works!

Is it possible to find solutions using the second formula?

Thanks!

4. Jun 20, 2015

### DrSuage

it sure is but you're gonna need to specify some initial conditions otherwise your solution can be any periodic function whatsoever that solves the wave equation...

5. Jun 21, 2015

### Botttom

I would like to show, that if $\phi$ is a solution to the equation than each partial derivative of $\phi$ is also a solution.
I am failing to show that just by plugging the derivative in. How can i do that?

6. Jun 23, 2015

### DrSuage

The product rule should give you this result

7. Jun 23, 2015

### Botttom

Do you mean product rule in this formula>?
${1\over{c^2}} {\partial^2 ( r \phi) \over{\partial t^2}} = {\partial^2 (r \phi) \over{\partial r^2}}$.

8. Jun 23, 2015

### DrSuage

could you link the exact wording of the question?
without actually doing it my intuition suggests that to check that a partial derivative of phi solves the equation, make the substitution $$\phi \rightarrow \frac{\partial\phi}{\partial r}$$
or $$\phi \rightarrow \frac{\partial\phi}{\partial t}$$
and then apply the product rule.

Last edited: Jun 23, 2015
9. Jun 23, 2015

### Botttom

The question is to show, that if $\phi$ solves the equation ${1\over{c^2}} {\partial^2 ( r \phi) \over{\partial t^2}} = {\partial^2 (r \phi) \over{\partial r^2}}$
than so should every partial derivative $$\frac{\partial\phi}{\partial x }, \frac{\partial\phi}{\partial y }, \frac{\partial\phi}{\partial z }, \frac{\partial\phi}{\partial t }.$$
And it seems that the product rule has to be applied here for more than 20 times, but i still dont get the result.

10. Jun 24, 2015

### DrSuage

maybe write the general solution for the product $r\phi$ then take derivatives?

11. Jun 25, 2015

### Botttom

Yes, but the general solution is not known in this case

12. Jun 26, 2015

### DrSuage

The general solution is $\phi = \frac{Au(r-ct)}{r} + \frac{Bv(r+ct)}{r}$

13. Jun 26, 2015

### Strum

Perhaps I have misunderstood the problem but anyway. Since r is independent the partial derivative of $\phi$ with time is obviously a solution.
$\partial^{2}_{t}(r\partial_{t}\phi) = \partial_{t}(\partial_{t}^{2} (r\phi)) = \partial^{2}_{r}(r\partial_{t}\phi) = \partial_{t}(\partial_{r}^{2} (r\phi))$. Now do the same with the other partial derivatives. I think we get something along the lines of $\partial_{t}^{2}(\cos(\theta)\phi) = \partial_{r}^{2}(\cos(\theta)\phi)$ in 2 dimensions.

14. Jun 27, 2015

### Botttom

Well , yes the part with the time derivative is obvious. The other part is not.
Why would there be a cosine(theta) ?

15. Jun 28, 2015

### Strum

It would be from $\partial_{x}r = \cos(\phi)$. Isn't that correct?

16. Jun 28, 2015

### Botttom

oh you mean $${\partial r\over {\partial x}} = {x \over r} = {r \cos(\theta) \over {r}}$$