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[SOLVED] Circular Membrane

dwsmith

Well-known member
Feb 1, 2012
1,673
$u_{tt} = c^2\nabla^2u$ where $\nabla^2 = \frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial}{\partial r}\right) + \frac{1}{r^2}\frac{\partial^2}{\partial\theta^2}$.
Suppose $u(r,\theta,t) = G(r,\theta)e^{i\omega t}$.
Then
$$
-\omega^2Ge^{i\omega t} = c^2\nabla^2Ge^{i\omega t}
$$
which leads to the Helmholtz equation $(\nabla^2G + k^2G = 0)$ where $k^2 = \frac{\omega^2}{c^2}$.
Let $G(r,\theta) = R(r)e^{\pm im\theta}$.
Then
$$
\left[\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial}{\partial r}\right) + \frac{1}{r^2}\frac{\partial}{\partial r}\right]Re^{im\theta} + k^2Re^{im\theta} = 0
$$
which leads to
$$
r^2R'' + rR' + (k^2r^2 - m^2)R = 0\quad \text{(Bessel's equation)}.
$$
Therefore, $R(r) = \mathcal{J}_m(kr) = \sum\limits_{j = 0}^{\infty}\frac{(-1)^j}{(j!)^2(k + m)!}\left(\frac{kr}{2}\right)^{2j + m}$.
The circular membrane is of radius $a$ whose edges are fixed.
That is, $R(a) = \mathcal{J}_{mn}(k_{mn}a) = 0$.
Let $z_{mn}$ be the zeros of $J_{mn}$ and $z_{mn} = k_{mn}a\iff k_{mn} = \frac{z_{mn}}{a}$.
Then the form of the general solution is
\begin{alignat*}{3}
u(r,\theta,t) & = & \sum_{n = 1}^{\infty} \sum_{m = 0}^{ \infty} \mathcal{J}_{mn} \left(z_{mn}\frac{r}{a}\right) \left[A_{mn}\cos m\theta + B_{mn}\sin m\theta\right] \cos\left(z_{mn}\frac{ct}{a}\right)\\
& + & \sum_{n = 1}^{\infty} \sum_{m = 0}^{ \infty} \mathcal{J}_{mn} \left(z_{mn}\frac{r}{a}\right) \left[C_{mn}\cos m\theta + D_{mn}\sin m\theta\right] \sin\left(z_{mn}\frac{ct}{a}\right)
\end{alignat*}
Using the first initial condition, we have\begin{alignat*}{5}
u(r,\theta,0) & = & \sum_{n = 1}^{\infty} \sum_{m = 0}^{ \infty} \mathcal{J}_{mn} \left(z_{mn}\frac{r}{a}\right) \left[A_{mn}\cos m\theta + B_{mn}\sin m\theta\right] & = & 0
\end{alignat*}
How can I create a table of the eigenvalues for the Bessel equation in Mathematica?
Would it be
Code:
ClearAll["Global`*"]z = Table[N[BesselJZero[m, n]], {n, 1, 20}, {m, 0, 5}];
z // TableForm
{
 {2.40483, 3.83171, 5.13562, 6.38016, 7.58834, 8.77148},
 {5.52008, 7.01559, 8.41724, 9.76102, 11.0647, 12.3386},
 {8.65373, 10.1735, 11.6198, 13.0152, 14.3725, 15.7002},
 {11.7915, 13.3237, 14.796, 16.2235, 17.616, 18.9801},
 {14.9309, 16.4706, 17.9598, 19.4094, 20.8269, 22.2178},
 {18.0711, 19.6159, 21.117, 22.5827, 24.019, 25.4303},
 {21.2116, 22.7601, 24.2701, 25.7482, 27.1991, 28.6266},
 {24.3525, 25.9037, 27.4206, 28.9084, 30.371, 31.8117},
 {27.4935, 29.0468, 30.5692, 32.0649, 33.5371, 34.9888},
 {30.6346, 32.1897, 33.7165, 35.2187, 36.699, 38.1599},
 {33.7758, 35.3323, 36.8629, 38.3705, 39.8576, 41.3264},
 {36.9171, 38.4748, 40.0084, 41.5207, 43.0137, 44.4893},
 {40.0584, 41.6171, 43.1535, 44.6697, 46.1679, 47.6494},
 {43.1998, 44.7593, 46.298, 47.8178, 49.3204, 50.8072},
 {46.3412, 47.9015, 49.4422, 50.965, 52.4716, 53.963},
 {49.4826, 51.0435, 52.586, 54.1116, 55.6217, 57.1173},
 {52.6241, 54.1856, 55.7296, 57.2577, 58.7708, 60.2702},
 {55.7655, 57.3275, 58.873, 60.4032, 61.9192, 63.4221},
 {58.907, 60.4695, 62.0162, 63.5484, 65.067, 66.5729},
 {62.0485, 63.6114, 65.1593, 66.6932, 68.2142, 69.7229}}
\begin{alignat*}{5}
u(r,\theta,0) & = & 0 & & \\
u_t(r,\theta,0) & = & \delta(\mathbf{x} - \mathbf{x}_0) & = & \delta(r - r_0, \theta - \theta_0) \end{alignat*}
$$
\int_A\delta(\mathbf{x} - \mathbf{x}_0)f(r,\theta)dA = \int_0^{2\pi}\int_0^a\delta(r - r_0, \theta - \theta_0)f(r,\theta)rdrd\theta = f(\mathbf{x}_0)
$$
 
Last edited:

dwsmith

Well-known member
Feb 1, 2012
1,673
$$
\sum_{n = 1}^{\infty} \sum_{m = 0}^{ \infty} \mathcal{J}_{mn}\left(z_{mn}\frac{r}{a}\right)A_{mn}\cos m\theta = - \sum_{n = 1}^{\infty} \sum_{m = 0}^{ \infty} \mathcal{J}_{mn}\left(z_{mn}\frac{r}{a}\right)B_{mn}\sin m\theta
$$
Does this imply that $A_{mn} = B_{mn} = 0$? That is my thought.
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,123
$$
\sum_{n = 1}^{\infty} \sum_{m = 0}^{ \infty} \mathcal{J}_{mn}\left(z_{mn}\frac{r}{a}\right)A_{mn}\cos m\theta = - \sum_{n = 1}^{\infty} \sum_{m = 0}^{ \infty} \mathcal{J}_{mn}\left(z_{mn}\frac{r}{a}\right)B_{mn}\sin m\theta
$$
Does this imply that $A_{mn} = B_{mn} = 0$? That is my thought.
Putting in my two cents:

Both sums are functions of [tex]\theta[/tex] so the only time you could require that [tex]A_{mn} = B_{mn} = 0[/tex] would be when [tex]cos(m \theta ) = -sin(m \theta)[/tex].

-Dan
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Putting in my two cents:

Both sums are functions of [tex]\theta[/tex] so the only time you could require that [tex]A_{mn} = B_{mn} = 0[/tex] would be when [tex]cos(m \theta ) = -sin(m \theta)[/tex].

-Dan
Does m being an integer help?
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,123
Does m being an integer help?
Ummm...Not sure what you're trying to say there?

We have the condition that
[tex]A_{mn}~cos(m \theta) + B_{mn}~sin(m \theta ) = 0[/tex]

All I'm saying is that if we have
[tex]cos(m \theta) = - sin(m \theta )[/tex]

[tex]tan(m \theta) = -1[/tex]

etc, etc.
Only certain values of [tex]\theta[/tex] will mandate that the coefficients are zero.

-Dan
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Ummm...Not sure what you're trying to say there?

We have the condition that
[tex]A_{mn}~cos(m \theta) + B_{mn}~sin(m \theta ) = 0[/tex]

All I'm saying is that if we have
[tex]cos(m \theta) = - sin(m \theta )[/tex]

[tex]tan(m \theta) = -1[/tex]

etc, etc.
Only certain values of [tex]\theta[/tex] will mandate that the coefficients are zero.

-Dan
Then I am at a loss on solving for the coefficients.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Would it be easier perhaps to start with the second initial condition?
If this is correct(below)
$$
C_{0n} = \frac{\int_0^a \int_0^{2\pi}r \delta(r - r_0,\theta - \theta_0) \mathcal{J}_{0n} \left(z_{0n}\frac{r}{a}\right)d\theta dr}{z_{0n}ac\pi\mathcal{J}_{1}^2(z_{0n})} = \frac{\mathcal{J}_{0n} \left(z_{0n}\frac{r_0}{a}\right)}{z_{0n}ac\pi \mathcal{J}_{1}^2(z_{0n})},
$$
$$
C_{mn} = \frac{2\int_0^a \int_0^{2\pi}r \delta(r - r_0,\theta - \theta_0) \mathcal{J}_{mn} \left(z_{mn}\frac{r}{a}\right)\cos m\theta d\theta dr}{z_{mn}ac\pi\mathcal{J}_{m + 1}^2(z_{mn})} = \frac{2\mathcal{J}_{mn} \left(z_{mn}\frac{r_0}{a}\right)\cos m\theta_0}{z_{mn}ac\pi\mathcal{J}_{m + 1}^2(z_{mn})}\quad m\neq 0,
$$
and
$$
D_{mn} = \frac{2\int_0^a \int_0^{2\pi}r \delta(r - r_0,\theta - \theta_0) \mathcal{J}_{mn} \left(z_{mn}\frac{r}{a}\right) \sin m\theta d\theta dr}{z_{mn}ac\pi \mathcal{J}_{m + 1}^2(z_{mn})} = \frac{2\mathcal{J}_{mn} \left(z_{mn}\frac{r_0}{a}\right)\sin m\theta_0}{z_{mn}ac\pi\mathcal{J}_{m + 1}^2(z_{mn})}.
$$
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Since we have the double Bessel-Fourier series, A = B = 0 is the answer unless I am missing something I am sure it is 0.
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
$$
\sum_{n = 1}^{\infty} \sum_{m = 0}^{ \infty} \mathcal{J}_{mn}\left(z_{mn}\frac{r}{a}\right)A_{mn}\cos m\theta = - \sum_{n = 1}^{\infty} \sum_{m = 0}^{ \infty} \mathcal{J}_{mn}\left(z_{mn}\frac{r}{a}\right)B_{mn}\sin m\theta
$$
Does this imply that $A_{mn} = B_{mn} = 0$? That is my thought.
Hi dwsmith, :)

No. This does not imply that \(A_{mn} = B_{mn} = 0\). However \(A_{mn} = B_{mn} = 0\) is a trivial solution to,

\[\sum_{n = 1}^{\infty} \sum_{m = 0}^{ \infty} \mathcal{J}_{mn}\left(z_{mn}\frac{r}{a}\right)A_{mn}\cos m\theta = - \sum_{n = 1}^{\infty} \sum_{m = 0}^{ \infty} \mathcal{J}_{mn}\left(z_{mn}\frac{r}{a}\right)B_{mn}\sin m\theta\]

Note that if you have two infinite series \(\sum_{n=1}^{\infty}a_{n}\) and \(\sum_{n=1}^{\infty}b_{n}\) equal to one another means that the two series will converge to the same value not that \(a_{n}=b_{n}\).

\[\sum_{n=1}^{\infty}a_{n}=\sum_{n=1}^{\infty}b_{n} \not\Rightarrow a_{n}=b_{n}\]

Kind Regards,
Sudharka.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Hi dwsmith, :)

No. This does not imply that \(A_{mn} = B_{mn} = 0\). However \(A_{mn} = B_{mn} = 0\) is a trivial solution to,

\[\sum_{n = 1}^{\infty} \sum_{m = 0}^{ \infty} \mathcal{J}_{mn}\left(z_{mn}\frac{r}{a}\right)A_{mn}\cos m\theta = - \sum_{n = 1}^{\infty} \sum_{m = 0}^{ \infty} \mathcal{J}_{mn}\left(z_{mn}\frac{r}{a}\right)B_{mn}\sin m\theta\]

Note that if you have two infinite series \(\sum_{n=1}^{\infty}a_{n}\) and \(\sum_{n=1}^{\infty}b_{n}\) equal to one another means that the two series will converge to the same value not that \(a_{n}=b_{n}\).

\[\sum_{n=1}^{\infty}a_{n}=\sum_{n=1}^{\infty}b_{n} \not\Rightarrow a_{n}=b_{n}\]

Kind Regards,
Sudharka.
It is true because of
Since we have the double Bessel-Fourier series, A = B = 0 is the answer unless I am missing something I am sure it is 0.
See http://www.mathhelpboards.com/f13/integrating-delta-bessel-function-2655/ for the coefficients of the double Bessel-Fourier series.