Harmonic Functions: Why n>2 Condition Imposed

[praharmitra]

the function f:\Re^n \rightarrow \Re = |X|^{2-n} is harmonic, but only for n > 2.

why is the condition n > 2 imposed. Isn't harmonic for all values of n??

 

[phreak]

Shouldn't the domain be \mathbb{R}^n\setminus \{ 0 \}? Otherwise, said f could not even exist at 0, and hence is clearly not harmonic.

 

[praharmitra]

ya I'm sorry that is the domain...my mistake...

but still, why shoud n>2, why can't n = 1?..

note: n stands for dimension of the space and hence is an integer >=1

 

[malawi_glenn]

oh, is this classical physics? ;-)

Try what happens if n=1, is the function harmonic then?

 

[praharmitra]

of course it is classical physics. It plays a MAJOR role in electromagnetic theory. and the potential formulation of the theory.

for n = 1, it is harmonic as laplacian(|X|) = 0 right? so it satisfies.

but then, why put the condition n > 2. I have seen it in many books

 

[malawi_glenn]

well then calculus problems are classic physics problems aswell. Just trying to make sure you get help.

Now harmonic functions are functions which are continuously differentiable twice and satisfy Laplace equation.

So play around with the criterion for twice continuously differentiable.