# Laplace Operator

1. Oct 7, 2005

### jet10

I have read in an electrodynamics book that
$$\int d^3r \frac{\Delta\Phi(r)}{r}=\int d^3r \Phi(r)}\Delta \frac{1}{r}$$
is possible through partial integration. But how?? Can some one help me on this by showing me the steps? Thanks

2. Oct 7, 2005

### Physics Monkey

$$\int d^3 r \nabla \cdot \left( \frac{1}{r} \nabla \Phi \right) = \int d^3 r \left( \nabla \frac{1}{r} \cdot \nabla \Phi + \frac{1}{r} \Delta \Phi \right)$$

$$\int d^3 r \nabla \cdot \left( \Phi \nabla \frac{1}{r} \right) = \int d^3 r \left( \nabla \frac{1}{r} \cdot \nabla \Phi + \Phi \Delta \frac{1}{r}\right)$$

Now the left hand side of each equation can written as a surface integral which gives zero. This gives
$$\int d^3 r \left( \nabla \frac{1}{r} \cdot \nabla \Phi + \frac{1}{r} \Delta \Phi \right) = 0$$
$$\int d^3 r \left( \nabla \frac{1}{r} \cdot \nabla \Phi + \Phi \Delta \frac{1}{r} \right) = 0$$
but since these two expressions have the same first term you can easily see that
$$\int d^3 r \frac{1}{r} \Delta \Phi = \int d^3 r \Phi \Delta \frac{1}{r}$$

3. Oct 7, 2005

### jet10

Thanks for your reply. Just one thing I am not able to see yet. How are these equal to zero?
$$\int d^3 r \nabla \cdot \left( \frac{1}{r} \nabla \Phi \right)=\oint dA \frac{1}{r} \nabla \Phi=0?$$
$$\int d^3 r \nabla \cdot \left( \Phi \nabla \frac{1}{r} \right)=\oint dA \Phi \nabla \frac{1}{r}=0?$$

Last edited: Oct 7, 2005
4. Oct 7, 2005

### Physics Monkey

You integrate on a surface at infinity, and with very weak assumptions about the fall off of $$\Phi$$ with distance, you can show those area integrals go to zero.

5. Oct 7, 2005

### jet10

That's clever. Thanks!