# Homework Help: Verify that the function U is a solution for Laplace Equation.

1. Oct 2, 2011

### DavidAp

Verify that the function U = (x^2 + y^2 + z^2)^(-1/2) is a solution of the three-dimensional Laplace equation Uxx + Uyy + Uzz = 0.

First I solved for the partial derivative Uxx,
Ux
= 2x(-1/2)(x^2 + y^2 + z^2)^(-3/2)
= -x(x^2 + y^2 + z^2)^(-3/2)

Uxx
= -(x^2 + y^2 + z^2)^(-3/2) + -x(2x)(-3/2)(x^2 + y^2 + z^2)^(-5/2)
= 3(x^2)(x^2 + y^2 + z^2)^(-5/2) - (x^2 + y^2 + z^2)^(-3/2)

From there I saw that for finding the partial derivative Uyy & Uzz I would just have the change the variable being squared in the beginning of the function. So,

Uxx =3(x^2)(x^2 + y^2 + z^2)^(-5/2) - (x^2 + y^2 + z^2)^(-3/2)
Uyy =3(y^2)(x^2 + y^2 + z^2)^(-5/2) - (x^2 + y^2 + z^2)^(-3/2)
Uzz =3(z^2)(x^2 + y^2 + z^2)^(-5/2) - (x^2 + y^2 + z^2)^(-3/2)

Because, through my observation, everything should just stay the same. However, when added together I get,

Uxx + Uyy + Uzz
= 3(3x^2 + 3y^2 + 3z^2)[(x^2 + y^2 + z^2)^(-5/2) - (x^2 + y^2 + z^2)^(-3/2)]

As you can see this ridiculously long function is not zero which leads to my question. Why didn't everything cancel itself out? Wasn't that suppose to happen, everything cancels itself out so I can say that Laplace's Rule works and it all equals to zero? I'm confused.

Again, thank you so much for reviewing my question and not being deterred at the sheer sight of my derivation of the partial derivatives and algebra.

2. Oct 2, 2011

### snipez90

Advice: write U_xx = 3(x^2)(x^2 + y^2 + z^2)^(-5/2) - (x^2 + y^2 + z^2)^(-3/2) as a single fraction, and then proceed.

Also I'm not sure how you got what you got when you added the 3 terms together.

3. Oct 2, 2011

### Staff: Mentor

I don't see anything wrong with what you did - the only problem is that you didn't take it far enough.

Uxx = -(x2 + y2 + z2)-3/2 + 3x2(x2 + y2 + z2)-5/2

I rewrote all three expressions using positive exponents (getting fractions), and then combined the fractions.

For example,
$$U_{xx} = \frac{-(x^2 + y^2 + z^2) + 3z^2}{(x^2 + y^2 + z^2)^{5/2}}$$

If you add all three of the indicated second partials, you do in fact get zero.