J77 said:
Was giving a presentation yesterday so couldn't follow this thread but adding to Halls of Ivy...
I'll first write your stuff in latex so we get a better view:
r^2 \frac{1}{R}\frac{\partial^2 R}{\partial r^2} + r \frac{1}{R}\frac{\partial R}{\partial r} + \frac{1}{\Theta}\frac{\partial^2\Theta}{\partial\theta^2} + r^2\lambda = 0
Separation of variables gives:
\frac{1}{\Theta}\frac{\partial^2\Theta}{\partial\theta^2}=const.=-m^2
which gives:
r^2 \frac{\partial^2 R}{\partial r^2} + r \frac{\partial R}{\partial r} + (r^2\lambda-m^2)R = 0
This last term is Bessels equation: The boundary conditions (geometry of problem) give you which functions are valid - you should consider both normal and modified kind, both 1st and 2nd of.
Again, \Theta is periodic and therefore satisfies Ae^{\pm im\theta}
The above isn't the full answer but should help...
Ok guys, thanks for all the help. Here's what I got. How do you guys get the tex code, do ypu write it by hand or do you have any help software?
I didnt do it exactly as you wrote J77, but it's the same way of thinking.
\frac{1}{\Theta}\frac{\partial^2\Theta}{\partial\theta^2}=const.=-k
which I solved to T = A*exp(-i*sqrt(k)*t) + B*exp(-i*sqrt(k)*t)
T(t) = T(t+2*pi)
A*exp(-i*sqrt(k)*t) + B*exp(-i*sqrt(k)*t) =
A*exp(-i*sqrt(k)*(t+2pi)) + B*exp(-i*sqrt(k)*(t+2pi))
exp(i*sqrt(k)*t) = exp(i*sqrt(k)*(t+2*pi))
sqrt(k)*t = sqrt(k)*t + 2pi*sqrt(k) + 2pi*n
sqrt(k) = -n
The R part get the solution
R(r) = a*J_-n(sqrt(lambda)*r) + b*Y_-n(sqrt(lambda)*r)
and the T part (former theta part, looks so greasy to write theta without tex) get the solution
T(t) = A*exp(-i*n*t) + B*exp(i*n*t)
and u(r,t) = R(r)*T(t)
now the problem is how to fit the neumann condition u(1,t) = 0 to this.
And I also want to know how to do for dirichlet condition for r = 1