Verifying Solution of 3-D Laplace Eq. u=1/(x^2+y^2+z^2)^2

[fk378]

Homework Statement

Verify that the function u=1/(x^2 + y^2 + z^2)^2 is a solution of the 3-dimensional Laplace equation uxx+uyy+uzz=0

The Attempt at a Solution

I know how to solve the partial derivatives, so I know that uxx=uyy=uzz for this problem. How can their sum equal 0?

 

[lzkelley]

are you sure its not ^3/2?

 

[Dick]

are you sure its not ^3/2?

You mean ^(1/2), yes? 1/r is the Green's function for the Laplace equation.

 

[fk378]

Ah, yes, the function should read:
u=1/(x^2 + y^2 + z^2)^(1/2)

Can you explain how the sum of the partial derivatives should equal zero, if their individual expressions are equal and positive?

 

[Dick]

Ah, yes, the function should read:
u=1/(x^2 + y^2 + z^2)^(1/2)

Can you explain how the sum of the partial derivatives should equal zero, if their individual expressions are equal and positive?

You said you know you to find the second derivatives. Then do it. The individual expression aren't 'equal and positive'. Tell me what is the second derivative u_xx? It has two terms which cancel when summed over x,y and z.

 

[fk378]

For u_xx I'm getting 3(x^2 + y^2 + z^2)^-(5/2)

 

[Dick]

That's not what I get. I get an 'x' in the numerator after the first derivative coming from the chain rule. When I apply the quotient rule to that to get the second derivative I get two terms.

 

[fk378]

Ok, for my u_xx I now get -(x^2 + y^2 + z^2)^(-3/2) + 3(x^2) (x^2 + y^2 + z^2)^(-5/2)

 

[Dick]

Right. Now sum over x, y and z.