Circular Membrane with 
Fixed Boundary (NonSimplified)

Statement of the problem

In this file I will consider the following wave equation 
  
     (((∂^2)u)/(∂(t^2)))=(c^2) ((((∂^2)u)/(∂(x^2)))+(((∂^2)u)/(∂(y^2)))),  (x,y) ∈ D,   0 ≤ t < ∞.  
  
Here D is the unit circle in x y-plane (or, rather, r θ-plane): 

  D = {(x,y) ∈ (ℝ^2) |  (x^2)+(y^2)  ≤ 1}={(r,θ) ∈ (ℝ^2) |  r  ≤ 1,  0⩽θ⩽ 2π} 
 
and  u(x,y,t)  is the function describing the shape of the membrane at time  t .  The equation is subject to the boundary condition  u(x,y,0)= 0   whenever  (x^2)+(y^(2 ))= 1.

Clearly it is easier to deal with this equation in polar coordinates.

Let 
 v(r,θ,t) = u(r Cos[θ],r Sin[θ],t).   
Then 
((∂v)/(∂r))=((∂u)/(∂x))Cos[θ]+((∂u)/(∂y))Sin[θ]    and  ((∂v)/(∂θ))=-((∂u)/(∂x))r Sin[θ]+((∂u)/(∂y))r Cos[θ] .  
Solving for  ((∂u)/(∂x)) and  ((∂u)/(∂y))  we get 
((∂u)/(∂x))=((∂v)/(∂r))  Cos[θ]-(1/r)((∂v)/(∂θ))Sin[θ]  and  ((∂u)/(∂y))=((∂v)/(∂r)) Sin[θ]+(1/r)((∂v)/(∂θ)) Cos[θ] . 
Taking the second partial derivatives of   v(r,θ,t) we get  
(((∂^2)v)/(∂(r^2)))=(((∂^2)u)/(∂(x^2)))(Cos^2)[θ]+(((∂^2)u)/(∂y∂x))Sin[2θ]+(((∂^2)u)/(∂(y^2)))(Sin^2)[θ]  
and  
(((∂^2)v)/(∂(θ^2)))=(((∂^2)u)/(∂(x^2)))(r^2) (Sin^2)[θ]-(((∂^2)u)/(∂y∂x))(r^2) Sin[2θ] + (((∂^2)u)/(∂(y^2)))(r^2)(Cos^2)[θ]
			-((∂u)/(∂x))r Cos[θ] -((∂u)/(∂y))r Sin[θ] .  
The last equation simplifies to 

 (1/(r^2))((((∂^2)v)/(∂(θ^2)))+r ((∂v)/(∂r)))=(((∂^2)u)/(∂(x^2)))(Sin^2)[θ]-(((∂^2)u)/(∂y∂x))Sin[2θ] + (((∂^2)u)/(∂(y^2)))(Cos^2)[θ].  

Adding the last equation and the equation 
 (((∂^2)v)/(∂(r^2)))=(((∂^2)u)/(∂(x^2)))(Cos^2)[θ]+(((∂^2)u)/(∂y∂x))Sin[2θ]+(((∂^2)u)/(∂(y^2)))(Sin^2)[θ] ,
we get 

(((∂^2)u)/(∂(x^2)))+(((∂^2)u)/(∂(y^2)))= (1/(r^2))(((∂^2)v)/(∂(θ^2)))+(((∂^2)v)/(∂(r^2))) +(1/r)((∂v)/(∂r)) .

Thus, the two dimensional wave equation in polar coordinates becomes 

    (((∂^2)v)/(∂(t^2)))=(c^2) ((1/(r^2))(((∂^2)v)/(∂(θ^2)))+(((∂^2)v)/(∂(r^2))) +(1/r)((∂v)/(∂r))),  
     0 ≤ r ≤ 1, 0 ≤ θ <2π,   0 ≤ t < ∞,
	v(1,θ,t) = 0,  0 ≤ θ <2π,   0 ≤ t < ∞ , 
	v(r,θ,0) = f(r,θ), ((∂v)/(∂t))(r,θ,0)=g(r,θ).

End of this section

The separation of variables

As we discussed in class to solve this problem by separation of variables  we look for the solution of the form    v(r,θ,t) = A(r,θ) B(t).  The partial differential equation becomes: 

A(r,θ) B''(t) = (c^2) ((1/(r^2))(Subscript[A, θθ])(r,θ) +(Subscript[A, r r])(r,θ) + (1/r)(Subscript[A, r])(r,θ))B(t),
   0 ≤ r ≤ 1, 0 ≤ θ <2π,   0 ≤ t < ∞,

or equivalently,

((B''(t))/((c^2)B(t)))=(((1/(r^2))(Subscript[A, θθ])(r,θ) +(Subscript[A, r r])(r,θ) + (1/r)(Subscript[A, r])(r,θ))/(A(r,θ)))=-λ, 
       0 ≤ r ≤ 1, 0 ≤ θ <2π,   0 ≤ t < ∞.   (1)

The first set of boundary conditions leads to 

A(1,θ) = 0    and a natural condition that 

A(r,0) = A(r,2π)   and  (Subscript[A, θ])(r,0) = (Subscript[A, θ])(r,2π). 

Since the independent variables in (1) are separated we have the following equations:  For the unknown function A :

(1/(r^2))(Subscript[A, θθ])(r,θ) +(Subscript[A, r r])(r,θ) + (1/r)(Subscript[A, r])(r,θ)=-λ A(r,θ),  
       0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π,   
            A(1,θ) = 0 , 
           A(r,0) = A(r,2π),  
           (Subscript[A, θ])(r,0) = (Subscript[A, θ])(r,2π).       
                                                                                                                        (2)

The above boxed problem is called a boundary eigenvalue problem.  
For the unknown function B we get:

-B''(t) = λ (c^2) B(t),   0 ≤ t < ∞ .

Note that the unknown quantity   λ  in the equation for  B  is green since we have to solve the boundary eigenvalue problem (2) for both A  and   λ .

The problem (2) is still too complicated.  To solve it we separate variables r  and θ.  Look for the solutions in the form:  A(r,θ) = X(r)*Y(θ).  Then (2) becomes:

(1/(r^2))Y''(θ) X(r) +Y(θ)X''(r) + (1/r)X'(r)Y(θ)=-λ X(r)Y(θ),  
       0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π,   
            X(1) = 0 , |X(r)| <+∞ 
           Y(0) = Y(2π),  
           Y'(0) = Y'(2π).       
                                                                                                                        (2a)

and separating variables leads to

((Y''(θ))/(Y(θ)))=-λ (r^2)-(r^2)((X''(r))/(X(r) )) - r((X'(r))/(X(r) ))=-ν,  
       0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π,   
            X(1) = 0 , |X(r)| <+∞ 
           Y(0) = Y(2π),  
           Y'(0) = Y'(2π).       
                                                                                                                        (2b)

These are two separate boundary eigenvalue problems with unknown functions of one variable.

-Y''(θ)=νY(θ),  
             0 ≤ θ ≤ 2π,          
           Y(0) = Y(2π),  
           Y'(0) = Y'(2π).       
                                                                                                                        (2Y)

-(r^2)X''(r) - r X'(r)=(λ (r^2)-ν ) X(r), 
           0 ≤ r ≤ 1,   
            X(1) = 0 , |X(r)| <+∞ 
                                                                                                                        (2X)

We already know the solutions for the problem (2Y): 

(Subscript[ν, m]) = (m^2), m=  0, 1, 2, 3, ... 

for m=0 the eigenfunction is (Subscript[Y, 0])(θ) = 1, 
for m⩾1 the eigenfunctions are (Subscript[Y, n])(θ) = Cos[m θ]  and (Subscript[Z, n])(θ) = Sin[m θ] .

Now for each (Subscript[ν, m])  found in (2Y) we solve the equation (2X).  Rewrite (2X) as

-(r^2)X''(r) - r X'(r)+(m^2) X(r)=λ (r^2) X(r),  m=0,1,2,3,...
              0 ≤ r ≤ 1,   
            X(1) = 0 , |X(r)| <+∞ 
                                                                                                                        (2X)

This is a singular Sturm-Liouville problem.  To rewrite it in the proper S-L form  we multiply through by (1/r):

-(d/(d r))(r ((d )/(d r))X(r)) +((m^2)/r) X(r)=λ r X(r), 
              0 < r ≤ 1,   
            X(1) = 0 , |X(r)| <+∞ 
                                                                                                                        (2X SL)

End of this section.

Solving the boundary eigenvalue problem for the function A and the eigenvalue parameter  λ :

Since we have already solved the equation for Y we only need to solve

-(d/(d r))(r ((d )/(d r))X(r)) +((m^2)/r) X(r)=λ r X(r), 
              0 < r ≤ 1,   m=0,1,2,3, ... ,
            X(1) = 0 , |X(r)| <+∞ 
                                                                                                                        (2XSL)

For m=0 this problem was solved in StuLio4b.nb. For m=1,2,3,4,5, ... this problem was solved in StuLio5.nb.

Conclusion :

Here we will unify what we did in StuLio4b.nb and StuLio5.nb. These problems have only positive eigenvalues λ = (ν^2). We have found the square roots ν of the eigenvalues of the problems (2XSL) for m=0,1,2,3,4,5 by solving the following  equations:  

det[[m]]=0     where     m=0,1,2,3,4,5   and

det={BesselJ[0,μ],BesselJ[1,μ],BesselJ[2,μ],BesselJ[3,μ],BesselJ[4,μ],BesselJ[5,μ]}

Manipulate[Plot[Evaluate[det[[n]]],{μ,0,30}],{n,1,6,1}]

Using guesses we found the approximations for the square roots of the eigenvalues:

guess={{2,6,9,12,15,18,21,24,27,31},{4,7,10,13,16,20,23,26,29,31},{5,8,12,15,18,21,24,27,30.5,34},{6,10,13,16,19,23,26,29,32,35},{8,11,14,18,21,24,27,30,33,36},{9,12,16,19,22,25,28,31,34,37}};

(tevs=Table[Table[FindRoot[det[[k]]==0,{μ,guess[[k]][[n]]},MaxIterations->30,WorkingPrecision->20][[1]][[2]],{n,1,10}],{k,1,6}])//TableForm

tevs[[2]][[3]]

plco={{RGBColor[0,0,0]},{RGBColor[1,0,0]},{RGBColor[0,1,0]},{RGBColor[0,0,1]},{RGBColor[0,1,1]},{RGBColor[1,0,1]}};
Manipulate[
	Plot[
		Evaluate[det[[k]]],{μ,0,40},PlotRange->{-1,1},PlotStyle->plco[[k]],
Ticks->{
			Table[{N[tevs[[k]][[n]],3],N[tevs[[k]][[n]],3],.3},{n,1,10}],Automatic}
],{k,1,6,1}]

The eigenfunctions are obtained by substituting ν from the table tevs into solutions:

sol={BesselJ[0,r μ],BesselJ[1,r μ],BesselJ[2,r μ],BesselJ[3,r μ],BesselJ[4,r μ],BesselJ[5,r μ]}

X[r_,k_,n_]:=(sol[[k]]/.μ->tevs[[k]][[n]])

X[r,1,1]

Table[X[r,k,n],{k,1,6},{n,1,10}]//TableForm

ns=N[Table[Table[
				NIntegrate[Expand[X[r,k,n]* X[r,k,n]*r],{r,0,1},MaxRecursion->20,
PrecisionGoal->12,WorkingPrecision->10,AccuracyGoal->12],{n,1,10}],{k,1,6}],3]

Combining these with the solutions found for Y:

We already know the solutions for the problem (2Y): 

(Subscript[ν, m]) = (m^2), m=  0, 1, 2, 3, ... 

for m=0 the eigenfunction is (Subscript[Y, 0])(θ) = 1, 
for m⩾1 the eigenfunctions are (Subscript[Y, n])(θ) = Cos[m θ]  and (Subscript[Z, n])(θ) = Sin[m θ] .

Thus the eigenvalues for the problem (2) for the function A(r,θ) are given in   tevs   which is 6 by 6 table.  The eigenfunction corresponding to the eigenvalue   

tevs[[1]][[n]]  is  1*Y[r,1,n],  n=1,2,3,4,5,6   (this corresponds to m=0) 

and the eigenfunctions corresponding to the eigenvalue 

tevs[[k]][[n]]  are  

 Cos[(k-1) θ]*Y[r,k,n]  and   Sin[(k-1) θ]*Y[r,k,n]. 
Define (note that in the rest of the notebook we will use the right hand side, that is the explicit formulas for the solution and not abbreviated version given by the left hand side):

(Subscript[A, Cos])[r_,θ_,k_,n_]:=Cos[(k-1)θ]*BesselJ[k-1,tevs[[k]][[n]]r];
(Subscript[A, Sin])[r_,θ_,k_,n_]:=Sin[(k-1)θ]*BesselJ[k-1,tevs[[k]][[n]]r];

Lets explore these functions:

(Subscript[A, Cos])[r,θ,2,1]

Natural modes of vibration

Table[
	ParametricPlot3D[
		{r Cos[θ],r Sin[θ],Cos[(k-1)θ]*BesselJ[k-1,tevs[[k]][[n]]r]},{r,0,1},{θ,0,2 π},PlotRange->{{-1,1},{-1,1},{-1,1}},PlotPoints->{60,60},
	BoxRatios->{1,1,1}],{k,1,6},{n,1,6}];

Table[
	ParametricPlot3D[
		{r Cos[θ],r Sin[θ],Sin[(k-1)θ]*BesselJ[k-1,tevs[[k]][[n]]r]},{r,0,1},{θ,0,2 π},PlotRange->{{-1,1},{-1,1},{-1,1}},PlotPoints->{40,40},
	BoxRatios->{1,1,1}],{k,2,3},{n,1,3}]

Table[
	ParametricPlot3D[
		{r Cos[θ],r Sin[θ],Sin[(k-1)θ]*BesselJ[k-1,tevs[[k]][[n]]r]},{r,0,1},{θ,0,2 π},PlotRange->{{-1,1},{-1,1},{-1,1}},PlotPoints->{40,40},
	BoxRatios->{1,1,1}],{k,4,4},{n,1,2}]

These functions are orthogonal on the unit disk. This follows from the orthogonality of the functions X[r,k,n] and Sin[(k-1)θ] and Cos[(k-1)θ].  The double integral of (Subscript[A, Cos])[r,θ,1,n]*(Subscript[A, Cos])[r,θ,1,n]*r equals 2π ((∫_0)^1)X[r,1,n]*X[r,1,n]r dr and these integrals we have in the table   ns.   Also for (Subscript[k, 1]),(Subscript[k, 2]) > 1 we have:

((∫_0)^(2π))((∫_0)^1)Cos[((Subscript[k, 1])-1)θ]*BesselJ[(Subscript[k, 1])-1,tevs[[(Subscript[k, 1])]][[(Subscript[n, 1])]]r]*Cos[((Subscript[k, 2])-1)θ]*BesselJ[(Subscript[k, 2])-1,tevs[[(Subscript[k, 2])]][[(Subscript[n, 2])]]r]rdrdθ 
		=
	 (((∫_0)^(2π))Cos[((Subscript[k, 1])-1)θ]*Cos[((Subscript[k, 2])-1)θ]dθ)*
			(((∫_0)^1)BesselJ[(Subscript[k, 1])-1,tevs[[(Subscript[k, 1])]][[(Subscript[n, 1])]]r]*
					BesselJ[(Subscript[k, 2])-1,tevs[[(Subscript[k, 2])]][[(Subscript[n, 2])]]r]rdr )

((∫_0)^(2π))((∫_0)^1)Sin[((Subscript[k, 1])-1)θ]*BesselJ[(Subscript[k, 1])-1,tevs[[(Subscript[k, 1])]][[(Subscript[n, 1])]]r]*Sin[((Subscript[k, 2])-1)θ]*BesselJ[(Subscript[k, 2])-1,tevs[[(Subscript[k, 2])]][[(Subscript[n, 2])]]r]rdrdθ 
		=
	 (((∫_0)^(2π))Sin[((Subscript[k, 1])-1)θ]*Sin[((Subscript[k, 2])-1)θ]dθ)*
			(((∫_0)^1)BesselJ[(Subscript[k, 1])-1,tevs[[(Subscript[k, 1])]][[(Subscript[n, 1])]]r]*
					BesselJ[(Subscript[k, 2])-1,tevs[[(Subscript[k, 2])]][[(Subscript[n, 2])]]r]rdr )

((∫_0)^(2π))((∫_0)^1)Cos[((Subscript[k, 1])-1)θ]*BesselJ[(Subscript[k, 1])-1,tevs[[(Subscript[k, 1])]][[(Subscript[n, 1])]]r]*Sin[((Subscript[k, 2])-1)θ]*BesselJ[(Subscript[k, 2])-1,tevs[[(Subscript[k, 2])]][[(Subscript[n, 2])]]r]rdrdθ 
		=
	 (((∫_0)^(2π))Cos[((Subscript[k, 1])-1)θ]*Sin[((Subscript[k, 2])-1)θ]dθ)*
			(((∫_0)^1)BesselJ[(Subscript[k, 1])-1,tevs[[(Subscript[k, 1])]][[(Subscript[n, 1])]]r]*
					BesselJ[(Subscript[k, 2])-1,tevs[[(Subscript[k, 2])]][[(Subscript[n, 2])]]r]rdr )  = 0 .

Since we know the orthogonality relationship for Sin[(k-1)θ], Cos[(k-1)θ] , k=1,2,... and for   BesselJ[k-1,tevs[[k]][[n]]r]  for a fixed k and n=1,2, ...  we derive the orthogonality relationships for the products.  Specifically for k > 1 we have :

((∫_0)^(2π))((∫_0)^1)(((Cos[(k-1)θ]*BesselJ[k-1,tevs[[k]][[n]]r]))^2)rdrdθ 
		=
	 (((∫_0)^(2π))(((Cos[(k-1)θ]))^2)dθ)*
			(((∫_0)^1)(((BesselJ[k-1,tevs[[k]][[n]]r]))^2)rdr ) 
= π ns[[k]][[n]]

and

((∫_0)^(2π))((∫_0)^1)(((Sin[(k-1)θ]*BesselJ[k-1,tevs[[k]][[n]]r]))^2)rdrdθ 
		=
	 (((∫_0)^(2π))(((Sin[(k-1)θ]))^2)dθ)*
			(((∫_0)^1)(((BesselJ[k-1,tevs[[k]][[n]]r]))^2)rdr ) 
= π ns[[k]][[n]]

End of this section

Solving the equation for the function B:

In the equation for  B  the quantity λ, and we found the first thirty-six values for  ν= (Sqrt[λ]) in the table tevs. To find  B  we have to solve:

-B''(t) = λ (c^2) B(t),   0 ≤ t < ∞ .

Thus, the solutions that we find for  B   that correspond to the eigenvalue are
        Cos[c ν t]  and   Sin[c ν t].

Define:

c=1

(Subscript[B, p])[t_,k_,n_]:=Cos[c ν t]/.ν->tevs[[k]][[n]];
(Subscript[B, v])[t_,k_,n_]:=Sin[c ν t]/.ν->tevs[[k]][[n]];

End of this section

End of this section

Few solutions of the problem:

Finally, we summarize our investigations by concluding that we have found "few" solutions:  for each ν in the table (which can be as large as we want), we found one solutions for A(r,θ) and two solutions for B(t): 

(Subscript[A, Cos])[r,θ,k,n]*(Subscript[B, p])[t,k,n]     and     (Subscript[A, Cos])[r,θ,k,n]*(Subscript[B, v])[t,n] ,  
 k=1, 2, 3, ... ,  n = 1, 2,... .

and 

(Subscript[A, Sin])[r,θ,k,n]*(Subscript[B, p])[t,k,n]     and     (Subscript[A, Sin])[r,θ,k,n]*(Subscript[B, v])[t,n] ,  
 k= 2, 3, ... ,  n = 1, 2,... .

Here k,n  relates to the eigenvalue in the table of the squareroots of eigenvalues.  To plot these functions we have to use parametric plots of surfaces.

I prefer to see the explicit formulas for these  "few" solutions:  for each ν in the table tevs (which can be as large as we want), we found the following solutions:

BesselJ[0,tevs[[1]][[n]]r] * Cos[c tevs[[1]][[n]] t]  

and    

BesselJ[0,tevs[[1]][[n]]r] * Sin[c tevs[[1]][[n]] t]  ,  
 n = 1, 2,... .

Cos[(k-1)θ]*BesselJ[k-1,tevs[[k]][[n]]r] * Cos[c tevs[[k]][[n]] t]  

and    

Cos[(k-1)θ]*BesselJ[0,tevs[[k]][[n]]r] * Sin[c tevs[[k]][[n]] t]  ,  
 k=2,3, ...  ,  n = 1, 2,...

and

Sin[(k-1)θ]*BesselJ[k-1,tevs[[k]][[n]]r] * Cos[c tevs[[k]][[n]] t]  

and    

Sin[(k-1)θ]*BesselJ[0,tevs[[k]][[n]]r] * Sin[c tevs[[k]][[n]] t]  ,  
 k=2,3, ...  ,  n = 1, 2,...

Lets explore these functions.

k=3;
n=2;
	Cos[(k-1)θ]*BesselJ[k-1,tevs[[k]][[n]]r] * Cos[c tevs[[k]][[n]] t]

k=2;n=3;c=1;
Table[
	ParametricPlot3D[{r Cos[θ],r Sin[θ],Cos[(k-1)θ]*BesselJ[k-1,tevs[[k]][[n]]r] * Cos[c tevs[[k]][[n]] t] },{r,0,1},{θ,0,2π},
			PlotRange->{{-1,1},{-1,1},{-1.2,1.2}},PlotPoints->{30,30},BoxRatios->{1,1,.6}],{t,0,((2π)/(c tevs[[k]][[n]])),((2π*(1/25))/(c tevs[[k]][[n]]))}];

End of this section

From Few Many (FFM) and the solution of the problem:

From these "few" solutions the superposition principle allows us to create "many" solutions:  

       u(r,θ,t) =
(∑_(k=1)^∞)(∑_(n=1)^∞)(Subscript[a, Cos,k,n ])(Subscript[A, Cos])[r,θ,k,n]* (Subscript[B, p])[t,k,n ] 
	*	    + (∑_(k=1)^∞)(∑_(n=1)^∞)(Subscript[b, Cos,k,n ]) (Subscript[A, Cos])[r,θ,k,n]*(Subscript[B, v])[t,k,n]
	+      + (∑_(k=2)^∞)(∑_(n=1)^∞)(Subscript[a, Sin,k,n ]) (Subscript[A, Sin])[r,θ,k,n]*(Subscript[B, p])[t,k,n]
		      +(∑_(k=2)^∞)(∑_(n=1)^∞)(Subscript[b, Sin,k,n ]) (Subscript[A, Cos])[r,θ,k,n]*(Subscript[B, v])[t,k,n]
+
where (Subscript[a, Cos,n ]), (Subscript[b, Cos,n ]), (Subscript[a, Sin,n ]) and (Subscript[b, Sin,n ]) are arbitrary constants.

Again, I prefer to see these formulas in terms of the specific functions that are involved here.  Instead of the item tevs[[k]][[n]] of the table we can write (Sqrt[(Subscript[λ, m,n])]), where (Subscript[λ, m,n]) is the n-th eigenvalue corresponding to the eigenvalue problem

-(d/(d r))(r ((d )/(d r))X(r)) +((m^2)/r) X(r)=λ r X(r), 
              0 < r ≤ 1,   m=0,1,2,3, ... ,
            X(1) = 0 , |X(r)| <+∞ 
                                                                                                                        (2XSL)

The corresponding eigenfunctions of this problem are  

BesselJ[m,(Sqrt[(Subscript[λ, m,n])]) r].

Thus, many solutions of our problem are given by 

       u(r,θ,t) =
(∑_(m=0)^∞)(∑_(n=1)^∞)(Subscript[a, m,n]) Cos[m θ]*BesselJ[m,(Sqrt[(Subscript[λ, m,n])])r]* Cos[c (Sqrt[(Subscript[λ, m,n])])t]
 	*	    + 
 (∑_(m=0)^∞)(∑_(n=1)^∞)(Subscript[b, m,n]) Cos[m θ]*BesselJ[m,(Sqrt[(Subscript[λ, m,n])])r]* Sin[c (Sqrt[(Subscript[λ, m,n])])t] 
			             +      
              (∑_(m=1)^∞)(∑_(n=1)^∞)(Subscript[α, m,n]) Sin[m θ]*BesselJ[m,(Sqrt[(Subscript[λ, m,n])])r]* Cos[c (Sqrt[(Subscript[λ, m,n])])t] 
			                       +  		+                       (∑_(m=1)^∞)(∑_(n=1)^∞)(Subscript[β, m,n]) Sin[m θ]*BesselJ[m,(Sqrt[(Subscript[λ, m,n])])r]* Sin[c (Sqrt[(Subscript[λ, m,n])])t]

where (Subscript[a, m,n]), (Subscript[b, m,n]), (Subscript[α, m,n]) and (Subscript[β, m,n])  are arbitrary constants.

We can determine these constants such that 

f(r,θ) = u(r,θ,0) 

and 

g(r,θ) = (∂/(∂t))u(r,θ,0).

Using the orthogonality of 

  Cos[m θ]*BesselJ[m,(Sqrt[(Subscript[λ, m,n])])r] , and  Sin[m θ]*BesselJ[m,(Sqrt[(Subscript[λ, m,n])])r]
 the unit disk with the weight we can calculate 
for m=1,2,..., n=1,2, ... ,  :

(Subscript[a, 0,n ])=  (1/(2π))*(((∫_0^(2π))(∫_0^1)f(r,θ) BesselJ[0,(Sqrt[(Subscript[λ, m,n])])r] r dr  dθ)/((∫_0^1)(((BesselJ[0,(Sqrt[(Subscript[λ, m,n])])r]))^2) r dr))

and

(Subscript[a, m,n ])=  (1/π)*(((∫_0^(2π))(∫_0^1)f(r,θ) Cos[m θ] BesselJ[m,(Sqrt[(Subscript[λ, m,n])])r] r dr  dθ)/((∫_0^1)(((BesselJ[m,(Sqrt[(Subscript[λ, m,n])])r]))^2) r dr))

and    

 (Subscript[b, m,n ])=  (1/(π(Sqrt[(Subscript[λ, m,n])])))*(((∫_0^(2π))(∫_0^1)g(r,θ) Cos[m θ] BesselJ[m,(Sqrt[(Subscript[λ, m,n])])r] r dr  dθ)/((∫_0^1)(((BesselJ[m,(Sqrt[(Subscript[λ, m,n])])r]))^2) r dr))

The formulas for (Subscript[α, m,n]) and (Subscript[β, m,n]) are similar.

End of this section.

Mathematica implementation of the solution:

g[r_,θ_]=((E^(-100(((r-.5))^2))))*Sin[(E^(-50 (((θ-π))^2)))*(π/2)]

ParametricPlot3D[
	{r Cos[θ],r Sin[θ],g[r,θ]},{r,0,1},{θ,0,2π},
	PlotRange->{{-1,1},{-1,1},{-1,1}},
PlotPoints->{20,100}]

Off[NIntegrate::slwcon];
TbCosr=Table[Table[
				NIntegrate[Evaluate[Expand[((E^(-100(((r-.5))^2))))*
							 BesselJ[k-1,tevs[[k]][[n]] r]*r]],{r,0,1},MaxRecursion->20,
PrecisionGoal->12,WorkingPrecision->10,AccuracyGoal->12]
			,{n,1,6}],{k,1,6}]

TbCosθ=Table[NIntegrate[Evaluate[Expand[Sin[(E^(-50 (((θ-π))^2)))*(π/2)]* Cos[(k-1)θ]]],{θ,0,2π},MaxRecursion->20,
PrecisionGoal->12,WorkingPrecision->20,AccuracyGoal->12]
			,{k,1,6}]

TbCos0=Re[Table[(2/(π ns[[1]][[n]])) *(1/( tevs[[1]][[n]]))*
			TbCosr[[1]][[n]]*TbCosθ[[1]]
			,{n,1,6}]]

TbCos1=Table[Table[(1/(π ns[[k]][[n]])) *(1/( tevs[[k]][[n]]))*
			TbCosr[[k]][[n]]*TbCosθ[[k]]
			,{n,1,6}],{k,2,6}]

TbCos=Insert[TbCos1,TbCos0,1]

TbSinθ=Table[NIntegrate[Evaluate[Expand[Sin[(E^(-50 (((θ-π))^2)))*(π/2)]* Sin[(k-1)θ]]],{θ,0,2π},MaxRecursion->20,
PrecisionGoal->12,WorkingPrecision->10,AccuracyGoal->12]
			,{k,2,6}]

TbSin=Table[Table[(1/(π ns[[k]][[n]])) *(1/( tevs[[k]][[n]]))*TbCosr[[k]][[n]]*TbSinθ[[k-1]]
			,{n,1,6}],{k,2,6}]

M=6;c=1;
u[r_,θ_,t_]=(∑_(k=1)^M)(∑_(n=1)^M)TbCos[[k]][[n]]*(Subscript[A, Cos])[r,θ,k,n]*(Subscript[B, v])[t,k,n]

u[.5,.5,1]

ParametricPlot3D[{r Cos[θ],r Sin[θ],Evaluate[(D[u[r,θ,t],t]/.t->0)]},{r,0,1},{θ,0,2π},
	PlotPoints->{20,60},
	PlotRange->{{-1,1},{-1,1},{-1,1}}]

Manipulate[
	ParametricPlot3D[{r Cos[θ],r Sin[θ],u[r,θ,t]},{r,0,1},{θ,0,2π},
			PlotRange->{{-1,1},{-1,1},{-.07,.07}},PlotPoints->{20,90},BoxRatios->{1,1,.6}],{t,0,((2π)/2.40),((2π)/2.40)*(1/40)}]

End of this section.