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

AI Thread Summary
Negative values for n are not permissible because they lead to solutions that contradict boundary conditions, specifically the requirement that V approaches zero as x approaches infinity. The sine function's odd property means that solutions for negative n would merely replicate those for positive n, albeit with different constants. However, introducing negative n would reintegrate terms that should be excluded based on the boundary conditions. Consequently, allowing n to be negative would violate the established conditions of the problem. Thus, maintaining n as a non-negative integer is essential for the integrity of the solution.
laser1
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Griffiths Pg 133 4th Edition
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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
 
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Because ##\sin (-x) = - \sin(x)##, the solutions for negative ##n## simply repeat the solution for positive ##n##.
 
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PeroK said:
Because ##\sin (-x) = - \sin(x)##, the solutions for negative ##n## simply repeat the solutionf for positive ##n##.
But with different constant, though?
 
laser1 said:
But with different constant, though?
The constant ##C## is arbitrary.
 
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.
 
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Gavran said:
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
 
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