How can the wave equation be rearranged to include r?

[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?

 

[Orodruin]

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.

 

[Botttom]

Yes, this works!

Is it possible to find solutions using the second formula?

Thanks!

 

[DrSuage]

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

 

[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?

 

[DrSuage]

The product rule should give you this result

 

[Botttom]

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

 

[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.

 

[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 don't get the result.

 

[DrSuage]

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

 

[Botttom]

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

 

[DrSuage]

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

 

[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.

 

[Botttom]

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

 

[Strum]

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

 

[Botttom]

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