# Prove the Laplacian of Function g is equal to g

1. Feb 17, 2016

### zr95

1. The problem statement, all variables and given/known data

2. Relevant equations
the gradient of g is (d/dx,d/dy,d/dz)
the divergence of g is d/dx+d/dy+d/dz

3. The attempt at a solution
When I run through even using only a few terms to see if I can get the final result of it equaling g I end up with u^2 terms as coefficients and this seems to me as if it is no longer equal.

2. Feb 17, 2016

### Samy_A

1) Can you show what you get as result?
2) What does it mean that u is a unit vector?
(3) In relevant equations, that should actually be partial derivatives.)

Last edited by a moderator: Feb 17, 2016
3. Feb 17, 2016

### zr95

1) u1u2...un*e^(u1x1+u2x2+...unxn)
2) u is a unit vector as in it can be any vector with a magnitude of 1

4. Feb 17, 2016

### Samy_A

How did you get that?
Let's start with the beginning: what is $\displaystyle \frac {\partial g}{\partial x_1}$?

5. Feb 17, 2016

### zr95

A partial derivative of g with respect to x. So we take the gradient and it becomes grad(g)=(u1e^(u1x1)+u2e^u2x2....) and so on. The thing I gave previously is after taking the divergence of the gradient.

6. Feb 17, 2016

### zr95

Its fine now. In case anyone else needs this for future reference.