Why can't n be negative in Laplace's equation?

[laser1]

TL;DR

N/A

Griffiths Pg 133 4th Edition

Why can't n be negative? Is there a reason for this? My thought is that if n is negative, as sine is odd, the negative gets absorbed into C, a constant. Is this correct?

Would it be equally correct to let n be a negative integer?

Thank you

 

[PeroK]

Because $\sin (-x) = - \sin(x)$, the solutions for negative $n$ simply repeat the solution for positive $n$.

 

[laser1]

Because $\sin (-x) = - \sin(x)$, the solutions for negative $n$ simply repeat the solutionf for positive $n$.

But with different constant, though?

 

[PeroK]

But with different constant, though?

The constant $C$ is arbitrary.

 

[Gavran]

The solution of the problem is $$ V(x,y)=(Ae^{k x}+Be^{-k x})(C\sin(ky)+D\cos(ky)) $$ where $k>0$ and where the condition (iv) $V\to0$ as $x\to\infty$ requires that A is equal to zero.
If $n$ is negative $-k$ will be positive and the part, which has already been excluded from the solution by $A=0$, will be included into the solution again and the condition (iv) will be violated.

 

[laser1]

The solution of the problem is $$ V(x,y)=(Ae^{k x}+Be^{-k x})(C\sin(ky)+D\cos(ky)) $$ where $k>0$ and where the condition (iv) $V\to0$ as $x\to\infty$ requires that A is equal to zero.
If $n$ is negative $-k$ will be positive and the part, which has already been excluded from the solution by $A=0$, will be included into the solution again and the condition (iv) will be violated.

Thank you