- #1
member 428835
Hey PF!
I have a quick question. When I was solving a PDE via separation of variables, I was able to come up with a same format solution for ##n \geq 1## but when ##n=0## I had a different "type" of solution. This doesn't really bother me since I am dealing with a linear PDE. However, I matched my last boundary condition ##u(a, \theta) = f(\theta)## where ##u=u(r, \theta)##. I satisfied this boundary condition only for the equation when ##n \geq 1##. So why do I even bother with the ##n=0## case?
I can give more details specific to the problem if you want.
Thanks!
Josh
I have a quick question. When I was solving a PDE via separation of variables, I was able to come up with a same format solution for ##n \geq 1## but when ##n=0## I had a different "type" of solution. This doesn't really bother me since I am dealing with a linear PDE. However, I matched my last boundary condition ##u(a, \theta) = f(\theta)## where ##u=u(r, \theta)##. I satisfied this boundary condition only for the equation when ##n \geq 1##. So why do I even bother with the ##n=0## case?
I can give more details specific to the problem if you want.
Thanks!
Josh