# Problem with solution of a PDE, Neumann functions

1. Apr 14, 2013

### nearlynothing

hello everyone

i'm in my sixth semester of undergraduate physics and currently taking a math methods of physics
class. So far we've been working with boundary value problems using PDE's.

In the textbook we're using and from which i've been reading mostly (mathematical physics by eugene butkov), the general approach to solving PDE's is first to use the method of separation of variables and then of course solving the resulting ODE's, by applying boundary conditions, etc.

The thing is, in solving the ODE's, the author often "discards" particular solutions to them by choosing the coorresponding constant coefficient in the general solution to be equal to zero, this is done whenever this particular solution is "unacceptable on physical grounds", which sounds quite reasonable until you're forced to do such a thing when some solution becomes unbounded at some point in the domain you're working with.

while solving a problem of a circular vibrating membrane, in one of the ODE's arising from the separation of variables, there's a bessel differential eq., whose general solution contains a Neumann function, which is not bounded at the origin, which is of course on the membrane.

Here the author says (i'll just paraphrase): we reject the Neumann function since it's not bounded at the origin, by setting its coefficient to be equal to zero.

my doubt is, by setting that coefficient equal to zero, in order to avoid physically unreasonable solutions, we seem to act as if that Neumann function ceases to exist, while all we're doing is multiply it by zero, which of course makes it vanish at all points, right?

but what happens then at the origin where the Neumann function is not even bounded, can we still say that, by multiplying it by zero, the result will be as well zero?

thanks in advance and sorry if my english is sometimes unclear or obscure so to speak, as it's not my native language.

2. Apr 14, 2013

### AlephZero

I'll use a simpler example, rather than Bessel functions.

Suppose the general solution of an ODE is y = Ax + B/x where A and B are arbitrary constants.

If B ≠ 0, the solution is not defined when x = 0., and it is unbounded when x is close to 0.

But y = Ax is a solution of the ODE which is defined at x = 0, and obviously y(0) = 0.

People (and textbooks) often describe this by saying "if y is bounded (or finite) when x = 0, then B = 0" but that is not quite right. They really mean the B/x term is removed completely from the solution, not that B = 0.