Conversion to Polar Coordinates

Converting to cylindrical coordinates,

In[6]:=
CoordinatesFromCartesian[{x,y,z},Cylindrical]
SetCoordinates[Cylindrical]

Out[6]=
       2    2
{Sqrt[x  + y ], ArcTan[x, y], z}

Out[7]=
Cylindrical[r, theta, z]

the wave equation becomes

In[8]:=
waveCylindrical := 

   D[u[r,theta,t],{t,2}] == Div[c^2 Grad[u[r,theta,t]]]
waveCylindrical

Out[9]=
                           2  (0,2,0)

 (0,0,2)                  c  u       [r, theta, t]

u       [r, theta, t] == (------------------------ + 

                                     r

 

      2  (1,0,0)                 2    (2,0,0)

     c  u       [r, theta, t] + c  r u       [r, theta, t]) / r

The vibrations of a drumhead satisfy the wave equation, where u is the displacement from the horizontal equilibrium position. If the initial conditions are independent of theta, then the solution will also be indepenent of theta. This simplifies the wave equation.

In[10]:=
Clear[u]

wave := D[u[r,t],{t,2}] == Div[c^2 Grad[u[r,t]]]
wave

Out[12]=
                 2  (1,0)          2    (2,0)

 (0,2)          c  u     [r, t] + c  r u     [r, t]

u     [r, t] == -----------------------------------

                                 r

For a drum of radius one, the boundary condition at the outer edge would be u(1,t) = 0, t > 0, since the drumhead does not move at r = 1. We need another boundary condition, since the partial differential equation is second order in r. We will be able to determine an appropriate boudnary condition as we solve the differential equation. We need two initial conditions, since the equation is second order in t. Let's assume the drumhead is initially displaced from equilibrium, but with no initial velocity. u(r,0) = u0(r)

(d/dt) u(r,0) = 0

Up to Vibrating Drumhead