- #1

SqueeSpleen

- 141

- 5

$$

u_{xx} +u_{yy} =0

$$

I usually have a rectangle as boundary conditions, so I use the principe of superposition and arrive to

$$

\dfrac{X''(x)}{X(x)} = - \dfrac{Y''(y)}{Y(y)} = - \lambda

$$

Clearly, lambda can be both negative and positive as there's no difference between ##X(x)## and ##Y(y)## in the sense that the problem is symmetric. Then, depending in the boundary conditions I usually chose ##\lambda## so I can use a Fourier series with real scalars in the variable in which I have my boundary conditions different from zero (is it's in both then I solve search the solution adding solutions to different problems where only one side is not homogeneous).

My question is...

I might chose ## \lambda ## of the opposite sign and I'll also arrive to a solution, but this will have terms more difficult to compute, I'm right?

I mean

If my solution is, for example

$$

\sum_{n=1}^{\infty} \left( a_{n} \sin (nx) + b_{n} \cos(nx) \right) \left( c_{n} \sinh (ny) + d_{n} \cosh (ny) \right)

$$

Should I be able to express it as

$$

\sum_{n=1}^{\infty} \left( a'_{n} \sinh (nx) + b'_{n} \cosh (nx) \right) \left( c'_{n} \sin (ny) + d'_{n} \cos (ny) \right)

$$

and the choice of ##\lambda## sign is only for convenience?

A more concrete example. Given the boundary conditions

$$ u_{x} (0,y) = u_{x} ( \pi , y) =0 \qquad 0 < y < \pi $$

$$ u_{x} (x,0) =0 \qquad u_{x} ( x, \pi ) = f(x) \qquad 0 < x < \pi $$

I chose ##\lambda## sign in order to be able to make a Fourier series in ##x##, but I guess I should be able to chose it in the other sense and make a Fourier series in ##y## and one with hyperbolic sine and hyperbolic cosine in ##x##. I've done a complex analysis course, so I don't worry if the proof uses things like

$$

\dfrac{ \sinh (iz) }{ i } = sin(z) \qquad \cosh (iz) = cos(z)

$$