# Boundary conditions - unique solution

aaaa202
I recently solved a differential equation with the solution:

f(x) = Aexp(ikx) + Bexp(-ikx)

with the periodic boundary condition f(x+L)=f(x). This condition leads to:

Aexp(ikx)exp(ikL) + Bexp(-ikx)exp(-ikL) = Aexp(ikx) + Bexp(-ikx) (1)

Now the way I figured out the constants A and B was that I said, that (1) must be true in particular for x=0 and x=π/2k which gave some equations to solve for A and B.

But something appears weird to me with this method of using two arbitrary values for x in (1) to determine A and B. Namely how do I know that I should also get this value for A and B if I use two different values of x? I do realize you could maybe show this by a calculation letting x1 and x2 be arbitrary but from a bigger perspective: Is it always true that the solution of a differential equation is uniquely specified by a periodic boundary condition? Maybe I'm just confused about something simple - it's late at night..

## Answers and Replies

Mentor
Do you require f(x) to be real, or are you also interested in cases where f(x) is complex?

Chet

aaaa202
I have no requirement on f being real.

Mentor
Actually, your periodicity requirement puts a restriction on k, not A and B: k = 2nπ/L, where n is any integer.

Chet