Laplace and Poisson's equation.

[yungman]

Homework Statement

I want to verify for $v=\ln( r)$, that

a)$\nabla^2 v=0$ for $v$ on a disk center at $(x_0,y_0)$. Therefore $v$ is Harmonic.

b)$\nabla^2 v=\frac{1}{r^2}$ for $v$ on a sphere center at $(x_0,y_0,z_0)$. Therefore $v$ is not Harmonic.

The Attempt at a Solution

a) For circular disk, $v=\frac{1}{2}\ln[(x-x_0)^2+(y-y_0)^2]$
$$\nabla v=\frac{\hat x (x-x_0)+\hat y(y-y_0)}{[(x-x_0)^2+(y-y_0)^2]^2}$$
$$\nabla^2 v=\nabla\cdot\nabla v=\frac{[(y-y_0)^2-(x-x_0)^2]+[(x-x_0)^2-(y-y_0)^2]}{[(x-x_0)^2+(y-y_0)^2]^2}$$=0
b)For a sphere, $v=\frac{1}{2}\ln[(x-x_0)^2+(y-y_0)^2+(z-z_0)^2]$
$$\nabla v=\frac{\hat x (x-x_0)+\hat y(y-y_0)+\hat z(z-z_0)^2}{[(x-x_0)^2+(y-y_0)^2+(z-z_0)^2]^2}$$
$$\nabla^2 v=\nabla\cdot\nabla v=\frac{[(y-y_0)^2+(z-z_0)^2-(x-x_0)^2]+[(x-x_0)^2+(z-z_0)^2-(y-y_0)^2]+[(x-x_0)^2+(y-y_0)^2-(z-z_0)^2]}{[(x-x_0)^2+(y-y_0)^2+(z-z_0)^2]^2]}$$
$$\Rightarrow\;\nabla^2 v=\frac{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}{[(x-x_0)^2+(y-y_0)^2+(z-z_0)^2]^2]}=\frac{1}{r^2}$$So $v$ is Harmonic only for a two dimension disk, not in three dimension sphere.

 

[mighty2000]

This is because the Laplace operator, or the Laplacian, is defined differently for two and three dimensions. In two dimensions, the Laplacian is equal to the sum of the second partial derivatives with respect to the two variables, while in three dimensions, it is equal to the sum of the second partial derivatives with respect to the three variables. Therefore, in three dimensions, the Laplacian of v is not equal to zero, making v not Harmonic.