Solving the heat equation in an infinite cylinder has resulted in the following expression for the general solution $$u(t,\rho,\theta)=\sum_{n,m=1}^\infty u_{n,m}(t,\rho,\theta)=\sum_{n,m=1}^\infty e^{-a^2\lambda^2_{n,m}t}J_m\left(\lambda_{n,m}\rho\right)(a_{n,m}\cos{n\theta}+b_{n,m}\sin{n\theta})$$ where $J_m$ is the $m$-th Bessel function of the first kind, and $\lambda_{n,m}$ has to do with its $n$-th zero. I want to find an expression for the coefficients so that this solution verifies the initial condicion of the problem, which I've written as $$u(0,\rho,\theta)=f(\rho,\theta)$$ plugging that in the solution yields the equality $$\sum_{n,m=1}^\infty J_m\left(\lambda_{n,m}\rho\right) (a_{n,m}\cos{n\theta}+b_{n,m}\sin{n\theta})=f(\rho,\theta)$$ And I'm thinking on using the orthogonality of such functions to find the expression of the coefficients $a_{n,m},b_{n,m}$. I know the orthogonality of the trigonometric functions, $\sin$ and $\cos$.

Regarding Bessel functions, though, I've looked up and so far, I've only been able to find the orthogonality of Bessel functions with the same index, that is $$J_m(\lambda_{n,m}\rho)\bot J_m(\lambda_{n',m}\rho)$$ with $n\neq n'$ (with respect to the weight $\rho$). However, that is not enough to be able to find the coefficients, as ideally I would need that $$J_m(\lambda_{n,m}\rho)\bot J_{m'}(\lambda_{n',m'}\rho)$$ with $m\neq m'$ and $n\neq n'$, with respect to some weight.

Any idea how to proceed?

  • $\begingroup$ I haven't done cylindrical solution since a while ago, but I think there's something wrong with your solution. Shouldn't the solution be $\sum J_m(\lambda_{m,n} \rho)(\sin(m\theta) + \cos(m\theta))$ ? $\endgroup$ – MGirard Jan 17 '17 at 22:27
  • $\begingroup$ The exponential term comes from the time part of the heat equation, which in cylindrical coordinates is $$u_t=u_{\rho\rho}+\frac{1}{\rho}u_\rho+\frac{1}{\rho^2}u_{\theta\theta}$$ so when you separate variables you end up with a linear ODE of first order and constant coefficients in the time function, which yields the exponential factor. $\endgroup$ – F.Webber Jan 17 '17 at 22:34
  • $\begingroup$ I just threw away the time dependance and other coefficients. I meant that the order of the bessel function should match the angular order. That would also provide the orthogonality. $\endgroup$ – MGirard Jan 17 '17 at 22:52
  • $\begingroup$ Oh, you're completely right, what a mistake. I still can't see how to derive the coefficients though, since that's a double series, if I multiply both sides of the equality by, say, $J_m(\lambda_{n,m}\rho)$, then by integrating I'd get rid of all the $J_m(\lambda_{n,m}\rho)J_m(\lambda_{k,m}\rho)$ terms, but not the $J_m(\lambda_{n,m}\rho)J_l(\lambda_{k,l}\rho)$ ones (with different orders) that will also appear. $\endgroup$ – F.Webber Jan 17 '17 at 23:04
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    $\begingroup$ When you integrate over $\theta$, you can use the orthogonality of the sines to get the different order to be orthogonal $\endgroup$ – MGirard Jan 17 '17 at 23:18

If you're going to solve on an infinite cylinder $0 \le \rho < \infty$, then you have to use continuous sums of the eigenfunctions, instead of discrete. That means Hankel transforms, which are analogous to Fourier transforms. If you want to expand a function $f \in L^2_{\rho}[0,\infty)$ as $$ f(\rho) = \int_{0}^{\infty}c(k)J_{\nu}(k\rho)kdk, $$ then the coefficient function $c(k)$ in the expansion must be $$ c(k) = \int_{0}^{\infty}f(\rho)J_{\nu}(k\rho)\rho d\rho. $$ In your case, you can't do this with discrete $\lambda_{m,n}$. The correct solution is $$ u(t,\rho,\theta)=\sum_{n=0}^{\infty}\int_{0}^{\infty}e^{-a^2k^2 t}\{c_{n}(k)\cos(n\theta)+d_{n}(k)\sin(n\theta)\}J_{n}(k\rho)kdk, $$ where the coefficient funcions $c_n(k)$, $d_n(k)$ are determined by the initial conditions. To find coefficients $c_n$ and $d_n$ such that $$ u(0,\rho,\theta) = f(\rho,\theta) $$ requires two inversions. For example, for $n > 0$, $$ c_n(k) = \int_{0}^{\infty}\left(\frac{1}{\pi}\int_{0}^{2\pi}f(\rho',\theta')\cos(n\theta')d\theta'\right)J_n(k\rho')\rho'd\rho' \\ = \frac{1}{\pi}\int_{0}^{2\pi}\int_{0}^{\infty}f(\rho',\theta')J_n(\rho'k)\cos(n\theta')\rho'd\rho'd\theta' $$

  • $\begingroup$ Oh, it seems I'll really have to have a deeper look into this. Thanks for explaining the required steps. $\endgroup$ – F.Webber Jan 19 '17 at 20:38

TL;DR You can recover your $a$'s and $b$'s by taking a "spatial" Fourier transform of $f$ followed by computing the "angular" Fourier series of the result

Start with $$ f(\rho,\theta)=\sum_{n=0}^{\infty}\int_{0}^{\infty}kdkA_n(k) J_n(k)e^{in\theta}J_n(k\rho), $$ where I've switched to using a complex $A_n$ rather than your real $a_n$ and $b_n$.

We'll take the Fourier transform of $f$ in $x$ and $y$ where $x=\rho\cos\theta$ and $y=\rho\sin\theta$. We'll also define polar coordinates in the Fourier domain by $u=w\cos\phi$ and $v=w\sin\phi$. \begin{align*} \hat{f}(u,v)&=\int_{-\infty}^\infty dx\int_{-\infty}^\infty dy \sum_{n=0}^{\infty}\int_{0}^{\infty}kdkA_n(k)J_n(k\rho)e^{in\theta}J_{n}(k\rho)e^{-i(ux+vy)} \\ &=\int_0^{2\pi}d\theta\int_0^\infty 2\pi \rho d\rho \sum_{n=0}^{\infty}\int_{0}^{\infty}kdkA_n(k)J_n(k\rho)e^{in\theta}J_{n}(k\rho)e^{-i\rho(u\cos\theta+v\sin\theta)}\\ &=\int_0^\infty 2\pi \rho d\rho \sum_{n=0}^{\infty}A_n(k)\int_{0}^{\infty}kdk J_{n}(k\rho)\int_0^{2\pi}d\theta e^{in\theta}e^{-i\rho(u\cos\theta+v\sin\theta)}\\ &=\int_0^\infty 2\pi \rho d\rho \sum_{n=0}^{\infty}A_n(k)\int_{0}^{\infty}kdk J_{n}(k\rho)\int_0^{2\pi}d\theta e^{in\theta}e^{-i\rho(w\cos(\theta-\phi))}\\ &=\int_0^\infty 2\pi \rho d\rho \sum_{n=0}^{\infty}A_n(k)\int_{0}^{\infty}kdk J_{n}(k\rho)\int_0^{2\pi}d\theta e^{in(\theta+\phi)}e^{-i\rho(w\cos(\theta))}\\ &=\sum_{n=0}^{\infty}e^{in\phi}\int_{0}^{\infty}kdk A_n(k)\int_0^\infty \rho d\rho J_{n}(k\rho)J_n(w\rho)\\ &=\sum_{n=0}^\infty e^{in\phi}\int_{0}^{\infty}kdk A_n(k)\delta(k-w)\\ &=\sum_{n=0}^\infty e^{in\phi} A_n(w)\\ \end{align*} Where at multiple steps I used a number of identities I found on Wikipedia.

You can now extract the $A_n(w)$ from the last expression by considering it as a Fourier series in $\theta$.

The surprising thing (to me) is that the final result doesn't explicitly mention Bessel functions at all. . It's particularly convenient if you're using a computer algebra package as they can have good specialised Fourier transform routines and can be a bit lacklustre dealing with Bessel functions.

(If equations at Wikipedia were numbered I'd indicate where I used each one. I apologise in advance for any factors of $2\pi$ I left out so please check my work.)


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