Is the Transformed Function a Solution to Laplace's Equation?

[merrypark3]

Homework Statement

Show that

If \phi(x,y,z) is a solution of Laplace's equation, show that
\frac{1}{r}\phi (\frac{x}{r^2} ,\frac{y}{r^2} , \frac{z}{r^2} ) is also a solution

Homework Equations

The Attempt at a Solution

let \psi= \frac{1}{r} \phi (\frac{x}{r^2} ,\frac{y}{r^2} , \frac{z}{r^2} ) is a solution.

Then in the spherical coordinate,

\psi=\frac{1}{r} \phi ( \frac{1}{r} , \theta , \varphi )

So input \psi to the spherical laplace equation.

\frac{1}{r^2}\frac{∂}{∂r} (r^2 \frac{∂\psi}{∂r}) = \frac {2}{r^4} \frac{∂\phi}{∂r} - \frac{1}{r^3} \frac{∂^2 \phi}{∂r^2}

The derivation with other angles, same to the original one, except 1/r times factor.

But the r part for the original one is

\frac{1}{r^2}\frac{∂}{∂r} (r^2 \frac{∂\phi}{∂r}) = \frac {2}{r} \frac{∂\phi}{∂r} + \frac{∂^2 \phi}{∂r^2}

What's wrong with me?

 

[gabbagabbahey]

let \psi= \frac{1}{r} \phi (\frac{x}{r^2} ,\frac{y}{r^2} , \frac{z}{r^2} ) is a solution.

Here you are assuming what you are trying to prove. That is not usually a good idea.

Instead, just compute the Laplacian of \psi= \frac{1}{r} \phi (\frac{x}{r^2} ,\frac{y}{r^2} , \frac{z}{r^2} ) and show that it is zero if \nabla^2 \phi(x,y,z)=0

Then in the spherical coordinate,

\psi=\frac{1}{r} \phi ( \frac{1}{r} , \theta , \varphi )

No. What are x, y and z in terms of spherical coordinates r, \theta and \varphi? What does that make \frac{x}{r^2}, \frac{y}{r^2}, and \frac{z}{r^2}?

 

[merrypark3]

\psi= \frac{1}{r} \phi (\frac{x}{r^2} ,\frac{y}{r^2} , \frac{z}{r^2} )= \frac{1}{r} \phi(\frac{sin \theta cos\varphi}{r}, \frac{sin \theta sin\varphi}{r},\frac{cos \theta}{r})

\frac{∂}{∂r} \phi(\frac{sin \theta cos\varphi}{r}, \frac{sin \theta sin\varphi}{r},\frac{cos \theta}{r})= -( \frac{ sin\theta cos\varphi}{r^2} \frac{∂\phi}{∂(\frac{sin \theta cos\varphi}{r})}+\frac{ sin\theta sin\varphi}{r^2} \frac{∂\phi}{∂(\frac{sin \theta sin\varphi}{r})}+\frac{ cos\theta}{r^2}\frac{∂\phi}{∂(\frac{sin \theta sin\varphi}{r})} )

Then how can I calculate the terms like \frac{∂\phi}{∂(\frac{sin \theta cos\varphi}{r})}?

 

[gabbagabbahey]

\psi= \frac{1}{r} \phi (\frac{x}{r^2} ,\frac{y}{r^2} , \frac{z}{r^2} )= \frac{1}{r} \phi(\frac{sin \theta cos\varphi}{r}, \frac{sin \theta sin\varphi}{r},\frac{cos \theta}{r})

good.

\frac{∂}{∂r} \phi(\frac{sin \theta cos\varphi}{r}, \frac{sin \theta sin\varphi}{r},\frac{cos \theta}{r})= -( \frac{ sin\theta cos\varphi}{r^2} \frac{∂\phi}{∂(\frac{sin \theta cos\varphi}{r})}+\frac{ sin\theta sin\varphi}{r^2} \frac{∂\phi}{∂(\frac{sin \theta sin\varphi}{r})}+\frac{ cos\theta}{r^2}\frac{∂\phi}{∂(\frac{sin \theta sin\varphi}{r})} )

Then how can I calculate the terms like \frac{∂\phi}{∂(\frac{sin \theta cos\varphi}{r})}?

Just use a substitution like \bar{x} \equiv \frac{\sin \theta \cos\varphi}{r}, \bar{y} \equiv \frac{\sin \theta \sin\varphi}{r}, and \bar{z} \equiv \frac{\cos \theta}{r} to make it easier to write:

\frac{\partial \phi\left( \frac{x}{r^2} ,\frac{y}{r^2} , \frac{z}{r^2} \right)}{\partial (\frac{\sin \theta \cos\varphi}{r})} = \frac{ \partial \phi( \bar{x}, \bar{y}, \bar{z})}{ \partial \bar{x}}?

In the end, you should get something like \nabla^2 \psi = \text{some factor} * \left[ \frac{ \partial^2 \phi( \bar{x}, \bar{y}, \bar{z})}{ \partial^2 \bar{x}} + \frac{ \partial^2 \phi( \bar{x}, \bar{y}, \bar{z})}{ \partial^2 \bar{y}} + \frac{ \partial^2 \phi( \bar{x}, \bar{y}, \bar{z})}{ \partial^2 \bar{z}} \right]

Aside: Changing the size on all your posts makes them difficult to read (for my eyes anyways), please stop it.

 

[steelclam]

I am stuck on this too

Hi!

I was just trying to solve this problem through this same approach (make the substitutions \overline{x}=\sin(\theta)\cos(\phi)/r, etc...). But it turns out to be *very* complicated. Do you have finished this one? Can you show me more explicit steps? Thanks a lot!