Helmholtz wave equation

[compxci]

If I was only considering the "elliptical arc" as 1/6 of a circle and all I was concerned with was the radial and angular dependencies (w.r.t.) and two sinusoidal sources acting in unison as the forcing term which is actually 2 sin(60t), how would this wave equation be set up and what would be the B.C.'s? Initial displacement is zero @t=0, initial velocity ~=0 @t=0.

Helmholtz's equation: Ztt = del(grad(Z))+Lambda(Z)+F(r,theta) => Z(r,theta,t) 0<r<R, 0< theta<PI, t>0
=>Z(r,theta,0) =0 0<r<R, 0<theta<PI
=>Zt(r,theta,0)~=0
periodic conditions: Z(r,0)=Z(r,PI)
Ztheta(r,0)=Ztheta(r,PI)

Where, F(r,theta) = 2sin(60t)
My problem is I'm not sure what the B.C.'s really are, it is not until milliseconds later that the first ripple is incidence upon the wave guide and this only describes the incidence wave and not it's reflection. Can anyone Please shed some light on this problem, time is running out!


Could it or would it be advisable to solve it as an Eigenfunction expansion method?? I must do the analysis first, then I can implement the code.

 

[gtoguy97]

Yes, you can solve it using an eigenfunction expansion method. This involves writing the Helmholtz equation in terms of its eigenfunctions and then solving for the coefficients which represent the radial and angular dependence of the solution. The boundary conditions would be that Z(r,theta,0)=0 and Zt(r,theta,0)~=0 for 0<r<R and 0<theta<PI. Additionally, you would need to enforce the periodic conditions, i.e. Z(r,0)=Z(r,PI) and Ztheta(r,0)=Ztheta(r,PI).