
# 10.1: Temperature on a disk

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# Bessel functions and two-dimensional problems

## Temperature on a disk

Let us now turn to a different two-dimensional problem. A circular disk is prepared in such a way that its initial temperature is radially symmetric, $u(\rho,\phi,t=0) = f(\rho).$ Then it is placed between two perfect insulators and its circumference is connected to a freezer that keeps it at $$0^\circ~\rm C$$, as sketched in Fig. [fig:Tspherical].

A circular plate, insulated from above and below.

Since the initial conditions do not depend on $$\phi$$, we expect the solution to be radially symmetric as well, $$u(\rho,t)$$, which satisfies the equation \begin{aligned} \pdt u &=& k\left[\pdrhor u + \frac{1}{\rho} \pdrho u \right],\nonumber\\ u(c,t) &=&0,\nonumber\\ u(\rho,0) &=& f(\rho).\end{aligned}

The initial temperature in the disk.

Once again we separate variables, $$u(\rho,t)=R(\rho)T(t)$$, which leads to the equation

$\frac{1}{k} \frac{T'}{T} = \frac{R''+\frac{1}{\rho}R'}{R} = -\lambda.$ This corresponds to the two equations

\begin{aligned} \rho^2 R''+\rho R' + \lambda \rho^2 R &=& 0,\quad R(c)=0m\nonumber\\ T'+\lambda k T = 0.\end{aligned}

The radial equation (which has a regular singular point at $$\rho=0$$) is closely related to one of the most important equation of mathematical physics, Bessel’s equation. This equation can be reached from the substitution $$\rho = x /\sqrt{\lambda}$$, so that with $$R(r)=X(x)$$ we get the equation

$x^2 \frac{d^2}{dx^2} X(x) + x \frac{d}{dx} X(x) + x^2 X(x) = 0, \qquad X(\sqrt \lambda c) =0.$

## Bessel’s equation

Bessel’s equation of order $$\nu$$ is given by $x^2 y'' + x y' + (x^2-\nu^2) y = 0.$ Clearly $$x=0$$ is a regular singular point, so we can solve by Frobenius’ method. The indicial equation is obtained from the lowest power after the substitution $$y=x^\gamma$$, and is

$\gamma^2-\nu^2=0$

So a generalized series solution gives two independent solutions if $$\nu \neq \half n$$. Now let us solve the problem and explicitly substitute the power series,

$y = x^\nu \sum_n a_n x^n.$

From Bessel’s equation we find

$\sum_n(n+\nu)(n+\nu-1) a_\nu x^{m+\nu} +\sum_n(n+\nu)a_\nu x^{m+\nu} +\sum_n(x^2-\nu^2)a_\nu = 0$

$[(m+\nu)^2-\nu^2] a_m= -a_{m-2}$ or $a_m= -\frac{1}{m(m+2\nu)}a_{m-2}.$

If we take $$\nu=n>0$$, we have

$a_m= -\frac{1}{m(m+2n)}a_{m-2}.$

This can be solved by iteration,

\begin{aligned} a_{2k} &=& -\frac{1}{4}\frac{1}{k(k+n)}a_{2(k-1)}\nonumber\\ &=& \left(\frac{1}{4}\right)^2\frac{1}{k(k-1)(k+n)(k+n-1)}a_{2(k-2)} \nonumber\\ &=& \left(-\frac{1}{4}\right)^k\frac{n!}{k!(k+n)!}a_{0}.\end{aligned}

f we choose1 $$a_0 = \frac{1}{n!2^n}$$ we find the Bessel function of order $$n$$

[J_n(x) = \sum_{k=0}^\infty \frac{(-1)^k}{k!(k+n)!} \left(\frac{x}{2}\right)^{2k+n}.\]

There is another second independent solution (which should have a logarithm in it) with goes to infinity at $$x=0$$.

A plot of the first three Bessel functions $$J_n$$ and $$Y_n$$.

The general solution of Bessel’s equation of order $$n$$ is a linear combination of $$J$$ and $$Y$$, $y(x) = A J_n(x)+B Y_n(x).$

## Gamma function

For $$\nu$$ not an integer the recursion relation for the Bessel function generates something very similar to factorials. These quantities are most easily expressed in something called a Gamma-function, defined as

$\Gamma(\nu) = \int_0^\infty e^{-t}t^{\nu-1} dt,\;\;\;\nu>0.$

Some special properties of $$\Gamma$$ function now follow immediately:

\begin{aligned} \Gamma(1) & = & \int_0^\infty e^{-t}dt = \left.-e^{-1}\right|^\infty_0 =1-e^{-\infty}=1\nonumber\\ \Gamma(\nu) & = & \int_0^\infty e^{-t} t^{\nu-1} dt = -\int_0^\infty \frac{de^{-t}}{dt} t^{\nu-1} dt\nonumber\\ &=& \left.- e^{-t} t^{\nu-1} \right|^\infty_0 +(\nu-1) \int_0^\infty e^{-t} t^{\nu-2} dt \end{aligned}

he first term is zero, and we obtain $\Gamma(\nu) = (\nu-1)\Gamma(\nu-1)$ From this we conclude that

$\Gamma(2)=1\cdot1=1,\;\Gamma(3)=2\cdot1\cdot1=2,\; \Gamma(4)=3\cdot2\cdot1\cdot1=2,\;\Gamma(n) = (n-1)!.$

hus for integer argument the $$\Gamma$$ function is nothing but a factorial, but it also defined for other arguments. This is the sense in which $$\Gamma$$ generalises the factorial to non-integer arguments. One should realize that once one knows the $$\Gamma$$ function between the values of its argument of, say, 1 and 2, one can evaluate any value of the $$\Gamma$$ function through recursion. Given that $$\Gamma(1.65) = 0.9001168163$$ we find

$\Gamma(3.65) = 2.65\times1.65\times0.9001168163 = 3.935760779 .$

Evaluate $$\Gamma(3)$$, $$\Gamma(11)$$, $$\Gamma(2.65)$$.
$$2!=2$$, $$10!=3628800$$, $$1.65\times0.9001168163=1.485192746$$.
We also would like to determine the $$\Gamma$$ function for $$\nu<1$$. One can invert the recursion relation to read $\Gamma(\nu-1) = \frac{\Gamma(\nu)}{\nu-1},$ $$\Gamma(0.7) = \Gamma(1.7)/0.7=0.909/0.7=1.30$$.

What is $$\Gamma(\nu)$$ for $$\nu<0$$? Let us repeat the recursion derived above and find $\Gamma(-1.3) = \frac{\Gamma(-0.3)}{-1.3} = \frac{\Gamma(0.7)}{-1.3\times -0.3} = \frac{\Gamma(1.7)}{0.7\times-0.3\times -1.3}=3.33\;.$ This works for any value of the argument that is not an integer. If the argument is integer we get into problems. Look at $$\Gamma(0)$$. For small positive $$\epsilon$$ $\Gamma(\pm\epsilon)= \frac{\Gamma(1\pm\epsilon)}{\pm\epsilon} =\pm \frac{1}{\epsilon}\rightarrow \pm \infty.$ Thus $$\Gamma(n)$$ is not defined for $$n\leq0$$. This can be easily seen in the graph of the $$\Gamma$$ function, Fig. [fig:Gamma].

A graph of the $$\Gamma$$ function (solid line). The inverse $$1/\Gamma$$ is also included (dashed line). Note that this last function is not discontinuous.

Finally, in physical problems one often uses $$\Gamma(1/2)$$, $\Gamma(\half) = \int_0^\infty e^{-t}t^{-1/2} dt = 2\int_0^\infty e^{-t}dt^{1/2} = 2\int_0^\infty e^{-x^2}dx.$ This can be evaluated by a very smart trick, we first evaluate $$\Gamma(1/2)^2$$ using polar coordinates \begin{aligned} \Gamma(\half)^2 &=& 4 \int_0^\infty e^{-x^2}dx\int_0^\infty e^{-y^2}dy \nonumber\\ &=& 4 \int_0^\infty \int_0^{\pi/2} e^{-\rho^2}\rho d\rho d\phi = \pi.\end{aligned} (See the discussion of polar coordinates in Sec. [sec:polc].) We thus find $\Gamma(1/2)=\sqrt{\pi},\;\;\Gamma(3/2)=\half\sqrt{\pi},\;\;{\rm etc.}$

## Bessel functions of general order

The recurrence relation for the Bessel function of general order $$\pm\nu$$ can now be solved by using the gamma function, $a_{m} = -\frac{1}{m(m\pm 2\nu)} a_{m-2}$ has the solutions ($$x > 0$$) \begin{aligned} J_{\nu}(x) &=& \sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!\Gamma(\nu+k+1)} \left(\frac{x}{2}\right)^{\nu+2k}, \\ J_{-\nu}(x) &=& \sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!\Gamma(-\nu+k+1)} \left(\frac{x}{2}\right)^{-\nu+2k}.\end{aligned} The general solution to Bessel’s equation of order $$\nu$$ is thus $y(x) = A J_{\nu}(x)+BJ_{-\nu}(x),$ for any non-integer value of $$\nu$$. This also holds for half-integer values (no logs).

## Properties of Bessel functions

Bessel functions have many interesting properties: \begin{aligned} J_{0}(0) &= 1,\\ J_{\nu}(x) &= 0\quad\text{(if \nu>0),}\\ J_{-n}(x) &= (-1)^{n }J_{n}(x),\\ \frac{d}{dx} \left[x^{-\nu}J_{\nu}(x) \right] &= -x^{-\nu}J_{\nu+1}(x),\\ \frac{d}{dx} \left[x^{\nu}J_{\nu}(x) \right] &= x^{\nu}J_{\nu-1}(x),\\ \frac{d}{dx} \left[J_{\nu}(x) \right] &=\half\left[J_{\nu-1}(x)-J_{\nu+1}(x)\right],\\ x J_{\nu+1}(x) &= 2 \nu J_{\nu}(x) -x J_{\nu-1}(x),\\ \int x^{-\nu}J_{\nu+1}(x)\,dx &= -x^{-\nu}J_{\nu}(x)+C,\\ \int x^{\nu}J_{\nu-1}(x)\,dx &= x^{\nu}J_{\nu}(x)+C.\end{aligned}

Let me prove a few of these. First notice from the definition that $$J_{n}(x)$$ is even or odd if $$n$$ is even or odd,

$J_{n}(x) = \sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!(n+k)!} \left(\frac{x}{2}\right)^{n+2k}.$

Substituting $$x=0$$ in the definition of the Bessel function gives $$0$$ if $$\nu >0$$, since in that case we have the sum of positive powers of $$0$$, which are all equally zero.

Let’s look at $$J_{-n}$$:

\begin{aligned} J_{-n}(x) &=& \sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!\Gamma(-n+k+1)!} \left(\frac{x}{2}\right)^{n+2k}\nonumber\\ &=& \sum_{k=n}^{\infty}\frac{(-1)^{k}}{k!\Gamma(-n+k+1)!} \left(\frac{x}{2}\right)^{-n+2k}\nonumber\\ &=& \sum_{l=0}^{\infty}\frac{(-1)^{l+n}}{(l+n)!l!} \left(\frac{x}{2}\right)^{n+2l}\nonumber\\ &=& (-1)^{n} J_{n}(x).\end{aligned}

Here we have used the fact that since $$\Gamma(-l) = \pm \infty$$, $$1/\Gamma(-l) = 0$$ [this can also be proven by defining a recurrence relation for $$1/\Gamma(l)$$]. Furthermore we changed summation variables to $$l=-n+k$$.

The next one:

\begin{aligned} \frac{d}{dx} \left[x^{-\nu}J_{\nu}(x) \right] &=& 2^{-\nu}\frac{d}{dx} \left\{ \sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!\Gamma(\nu+k+1)} \left(\frac{x}{2}\right)^{2k} \right\} \nonumber\\&=& 2^{-\nu} \sum_{k=1}^{\infty}\frac{(-1)^{k}}{(k-1)!\Gamma(\nu+k+1)} \left(\frac{x}{2}\right)^{2k-1} \nonumber\\&=& -2^{-\nu} \sum_{l=0}^{\infty}\frac{(-1)^{l}}{(l)!\Gamma(\nu+l+2)} \left(\frac{x}{2}\right)^{2l+1} \nonumber\\&=& -2^{-\nu} \sum_{l=0}^{\infty}\frac{(-1)^{l}}{(l)!\Gamma(\nu+1+l+1)} \left(\frac{x}{2}\right)^{2l+1} \nonumber\\&=& -x^{-\nu} \sum_{l=0}^{\infty}\frac{(-1)^{l}}{(l)!\Gamma(\nu+1+l+1)} \left(\frac{x}{2}\right)^{2l+\nu+1} \nonumber\\&=& -x^{-\nu}J_{\nu+1}(x).\end{aligned} Similarly \begin{aligned} \frac{d}{dx} \left[x^{\nu}J_{\nu}(x) \right] &=&x^{\nu}J_{\nu-1}(x).\end{aligned}

The next relation can be obtained by evaluating the derivatives in the two equations above, and solving for $$J_{\nu}(x)$$:

\begin{aligned} x^{-\nu}J'_{\nu}(x)-\nu x^{-\nu-1}J_{\nu}(x)&=& -x^{-\nu}J_{\nu+1}(x),\\ x^{\nu}J_{\nu}(x)+\nu x^{\nu-1}J_{\nu}(x)&=&x^{\nu}J_{\nu-1}(x).\end{aligned}

ultiply the first equation by $$x^{\nu}$$ and the second one by $$x^{-\nu}$$ and add:

\begin{aligned} -2\nu \frac{1}{x}J_{\nu}(x) = -J_{\nu+1}(x)+J_{\nu-1}(x).\end{aligned}

After rearrangement of terms this leads to the desired expression.

Eliminating $$J_{\nu}$$ between the equations gives (same multiplication, take difference instead) \begin{aligned} 2 J'_{\nu}(x) &=&J_{\nu+1}(x)+J_{\nu-1}(x).\end{aligned}

Integrating the differential relations leads to the integral relations.

Bessel function are an inexhaustible subject – there are always more useful properties than one knows. In mathematical physics one often uses specialist books.

## Sturm-Liouville theory

In the end we shall want to write a solution to an equation as a series of Bessel functions. In order to do that we shall need to understand about orthogonality of Bessel function – just as sines and cosines were orthogonal. This is most easily done by developing a mathematical tool called Sturm-Liouville theory. It starts from an equation in the so-called self-adjoint form

$[r(x) y'(x)]' + [p(x)+\lambda s(x)] y(x) = 0 \label{eq:selfadj}$

where $$\lambda$$ is a number, and $$r(x)$$ and $$s(x)$$ are greater than 0 on $$[a,b]$$. We apply the boundary conditions

\begin{aligned} a_1 y(a)+ a_2 y'(a)&=&0,\nonumber\\ b_1 y(b)+ b_2 y'(b)&=&0,\end{aligned}

with $$a_1$$ and $$a_2$$ not both zero, and $$b_1$$ and $$b_2$$ similar.

If there is a solution to (\ref{eq:selfadj}) then $$\lambda$$ is real.

Assume that $$\lambda$$ is a complex number ($$\lambda = \alpha + i \beta$$) with solution $$\Phi$$. By complex conjugation we find that

\begin{aligned} [r(x) \Phi'(x)]' + [p(x)+\lambda s(x)] \Phi(x) &=& 0\nonumber\\ {}[r(x) (\Phi^*)'(x)]' + [p(x)+\lambda^* s(x)] (\Phi^*)(x) &=& 0\end{aligned}

where $$*$$ note complex conjugation. Multiply the first equation by $$\Phi^*(x)$$ and the second by $$\Phi(x)$$, and subtract the two equations: $(\lambda^*-\lambda)s(x) \Phi^*(x)\Phi(x)=\Phi(x)[r(x) (\Phi^*)'(x)]'-\Phi^*(x)[r(x) \Phi'(x)]'.$ Now integrate over $$x$$ from $$a$$ to $$b$$ and find

$(\lambda^*-\lambda)\int_a^bs(x) \Phi^*(x)\Phi(x)\,dx = \int_a^b\Phi(x)[r(x) (\Phi^*)'(x)]'-\Phi^*(x)[r(x) \Phi'(x)]'\,dx$

The second part can be integrated by parts, and we find

\begin{aligned} (\lambda^*-\lambda)\int_a^b s(x) \Phi^*(x)\Phi(x)\,dx &=& \left[\Phi'(x) r(x) (\Phi^*)'(x)-\Phi^*(x)r(x) \Phi'(x)\right|^b_a \nonumber\\ &=& r(b)\left[\Phi'(b) (\Phi^*)'(b)-\Phi^*(b)\Phi'(b)\right] \nonumber\\&& -r(a)\left[\Phi'(a) (\Phi^*)'(a)-\Phi^*(a)\Phi'(a)\right] \nonumber\\ &=&0,\end{aligned}

here the last step can be done using the boundary conditions. Since both $$\Phi^*(x)\Phi(x)$$ and $$s(x)$$ are greater than zero we conclude that $$\int_a^bs(x) \Phi^*(x)\Phi(x)\,dx > 0$$, which can now be divided out of the equation to lead to $$\lambda=\lambda^*$$.

Let $$\Phi_n$$ and $$\Phi_m$$ be two solutions for different values of $$\lambda$$, $$\lambda_n\neq \lambda_m$$, then $\int_a^b s(x) \Phi_n(x) \Phi_m(x)\,dx = 0.$

The proof is to a large extent identical to the one above: multiply the equation for $$\Phi_n(x)$$ by $$\Phi_m(x)$$ and vice-versa. Subtract and find $(\lambda_n-\lambda_m)\int_a^b s(x) \Phi_m(x)\Phi_n(x)\,dx = 0$ which leads us to conclude that $\int_a^b s(x) \Phi_n(x) \Phi_m(x)\,dx = 0.$

Under the conditions set out above

1. There exists a real infinite set of eigenvalues $$\lambda_0,\ldots,\lambda_n, \ldots$$ with $$\lim_{n\rightarrow \infty} = \infty$$.
2. If $$\Phi_n$$ is the eigenfunction corresponding to $$\lambda_n$$, it has exactly $$n$$ zeroes in $$[a,b]$$. No proof shall be given.

Clearly the Bessel equation is of self-adjoint form: rewrite $x^2 y'' + xy' + (x^2-\nu^2) y = 0$ as (divide by $$x$$) $[x y']' + (x-\frac{\nu^2}{x}) y = 0$ We cannot identify $$\nu$$ with $$\lambda$$, and we do not have positive weight functions. It can be proven from properties of the equation that the Bessel functions have an infinite number of zeroes on the interval $$[0,\infty)$$. A small list of these:

$\begin{array}{lclllll} J_0 &:& 2.42&5.52 &8.65&11.79&\ldots\\ J_{1/2} &:& \pi & 2 \pi & 3\pi & 4\pi & \ldots\\ J_{8} &:& 11.20 & 16.04 & 19.60 & 22.90 & \ldots \end{array}$

## Our initial problem and Bessel functions

We started the discussion from the problem of the temperature on a circular disk, solved in polar coordinates, Since the initial conditions do not depend on $$\phi$$, we expect the solution to be radially symmetric as well, $$u(\rho,t)$$, which satisfies the equation

\begin{aligned} \pdt u &=& k\left[\pdrhor u + \frac{1}{\rho} \pdrho u \right],\nonumber\\ u(c,t) &=&0,\nonumber\\ u(\rho,0) &=& f(\rho).\end{aligned}

With $$u(\rho,t)=R(\rho)T(t)$$ we found the equations

\begin{aligned} \rho^2 R''+\rho R' + \lambda \rho^2 R &=& 0\;\;\;\;R(c)=0\nonumber\\ T'+\lambda k T = 0.\end{aligned}

The equation for $$R$$ is clearly self-adjoint, it can be written as

$[\rho R']' + \lambda \rho R = 0$ So how does the equation for $$R$$ relate to Bessel’s equation? Let us make the change of variables $$x= \sqrt{\lambda} \rho$$. We find $\frac{d}{d\rho} = \sqrt{\lambda} \frac{d}{dx},$ and we can remove a common factor $$\sqrt{\lambda}$$ to obtain ($$X(x)=R(\rho)$$)

$[x X']' + x X = 0,$

which is Bessel’s equation of order $$0$$, i.e.,

$R(\rho) = J_0(\rho \sqrt{\lambda}).$

The boundary condition $$R(c)=0$$ shows that $c \sqrt{\lambda_n} = x_n,$

where $$x_n$$ are the points where $$J_0(x)=0$$. We thus conclude $R_n(\rho) = J_0(\rho \sqrt{\lambda_n}).$ the first five solutions $$R_n$$ (for $$c=1$$) are shown in Fig. [fig:Rn].

From Sturm-Liouville theory we conclude that $\int_0^\infty \rho d\rho \,R_n(\rho)R_m(\rho)=0\;\;{\rm if\ }n \neq m.$ Together with the solution for the $$T$$ equation, $T_n(t)= \exp(-\lambda_n k t)$ we find a Fourier-Bessel series type solution $u(\rho,t) = \sum_{n=1}^\infty A_n J_0(\rho\sqrt{\lambda_n})\exp(-\lambda_n k t),$ with $$\lambda_n= (x_n/c)^2$$.

In order to understand how to determine the coefficients $$A_n$$ from the initial condition $$u(\rho,0)=f(\rho)$$ we need to study Fourier-Bessel series in a little more detail.

## Fourier-Bessel series

So how can we determine in general the coefficients in the Fourier-Bessel series

$f(\rho) = \sum_{j=1}^\infty C_j J_\nu(\alpha_j \rho)?$

The corresponding self-adjoint version of Bessel’s equation is easily found to be (with $$R_j(\rho) = J_\nu(\alpha_j\rho)$$)

$(\rho R_j')' + (\alpha_j^2\rho - \frac{\nu^2}{\rho}) R_j = 0.$

Where we assume that $$f$$ and $$R$$ satisfy the boundary condition

\begin{aligned} b_1 f(c) + b_2 f'(c) &=& 0 \nonumber\\ b_1 R_j(c) + b_2 R_j'(c) & = & 0 \end{aligned}

From Sturm-Liouville theory we do know that $\int_0^c \rho J_\nu(\alpha_i \rho) J_\nu(\alpha_j \rho) = 0 \;\;{\rm if\ }i \neq j,$ but we shall also need the values when $$i=j$$!

Let us use the self-adjoint form of the equation, and multiply with $$2 \rho R'$$, and integrate over $$\rho$$ from $$0$$ to $$c$$,

$\int_0^c \left[(\rho R_j')' + (\alpha_j^2\rho - \frac{\nu^2}{\rho}) R_j \right]2 \rho R_j' d\rho = 0.$

This can be brought to the form (integrate the first term by parts, bring the other two terms to the right-hand side)

\begin{aligned} \int_0^c \frac{d}{d\rho}\left(\rho R_j'\right)^2 d\rho &=& 2 \nu^2 \int_0^c R_j R_j' d\rho -2\alpha_j^2 \int_0^c \rho^2 R_j R_j' d\rho\\ \left.\left(\rho R_j'\right)^2\right|^c_0 &=& \left.\nu^2 R_j^2\right|^c_0 - 2\alpha_j^2 \int_0^c \rho^2 R_j R_j' d\rho.\end{aligned}

The last integral can now be done by parts: \begin{aligned} 2 \int_0^c\rho^2 R_j R_j' d\rho &=& -2 \int_0^c\rho R_j^2 d\rho +\left. \rho R_j^2 \right|^c_0.\end{aligned}

So we finally conclude that

$2 \alpha_j^2 \int_0^c\rho R_j^2 d\rho = \left[\left(\alpha_j^2\rho^2 - \nu^2 \right) R_j^2+\left(\rho R_j'\right)^2 \right|^c_0.$

In order to make life not too complicated we shall only look at boundary conditions where $$f(c)=R(c)=0$$. The other cases (mixed or purely $$f'(c)=0$$) go very similar! Using the fact that $$R_j(r) = J_\nu(\alpha_j \rho)$$, we find $R'_j=\alpha_j J'_\nu(\alpha_j \rho).$ We conclude that

\begin{aligned} 2 \alpha_j^2 \int_0^c\rho^2 R_j^2 d\rho &=& \left[ \left(\rho \alpha_j J'_\nu(\alpha_j\rho)\right)^2 \right|^c_0 \nonumber\\ &=& {c^2}\alpha_j^2 \left(J'_\nu(\alpha_j c)\right)^2 \nonumber\\ &=& {c^2}\alpha_j^2 \left(\frac{\nu}{\alpha_j c} J_\nu(\alpha_j c) -J_{\nu+1}(\alpha_j c)\right)^2 \nonumber\\ &=& {c^2}\alpha_j^2 \left( J_{\nu+1}(\alpha_j c)\right)^2 ,\end{aligned}

We thus finally have the result $\int_0^c\rho^2 R_j^2 d\rho=\frac{c^2}{2} J^2_{\nu+1}(\alpha_j c).$

Consider the function

$f(x) = \left\{\begin{array}{ll} x^3 &\;\;0<x<10\\ 0&\;\;x>10 \end{array}\right.$

Expand this function in a Fourier-Bessel series using $$J_3$$. From our definitions we find that

$f(x) = \sum_{j=1}^\infty A_j J_3(\alpha_j x),$

with

\begin{aligned} A_j &=& \frac{2}{100 J_4(10\alpha_j)^2} \int_0^{10} x^3 J_3(\alpha_j x) dx\nonumber\\ &=&\frac{2}{100 J_4(10\alpha_j)^2} \frac{1}{\alpha^5_j} \int_0^{10\alpha_j} s^4 J_3(s) ds\nonumber\\ &=&\frac{2}{100 J_4(10\alpha_j)^2} \frac{1}{\alpha^5_j} (10 \alpha_j)^4 J_4(10\alpha_j ) ds\nonumber\\ &=& \frac{200}{\alpha_j J_4(10 \alpha_j)}.\end{aligned}

sing $$\alpha_j=\ldots$$, we find that the first five values of $$A_j$$ are $$1050.95,-821.503,703.991,-627.577,572.301$$. The first five partial sums are plotted in Fig. [fig:xcube].

## Back to our initial problem

After all that we have learned, we know that in order to determine the solution of the initial problem in Sec. 1 we would have to calculate the integrals $A_j = \frac{2}{c^2 J_1^2(c \alpha_j)}\int_0^c f(\rho) J_0(\alpha_j \rho) \rho d\rho$

1. This can be done since Bessel’s equation is linear, i.e., if $$g(x)$$ is a solution $$C g(x)$$ is also a solution.