I'm trying to find the time dependent solution to the heat equation in a very long cylinder (the problem only depends on radius, r, and time, t) with the initial value $T(r,0)=T_1$ and the Dirichlet conditions $T(R,t)=T_2$ and $T(0,t)=T_0$, where $R$ is the radius of the cylinder. The PDE is

$$\partial_tT-a\nabla^2T=0 \, .$$

I started to use the separation of variables method and correctly got the general solution

$$T(r,t)=\sum_{n=1}^\infty A_n J_0(k_nr)e^{-ak_n^2t}$$

where $J_0(k_nr)$ is a Bessel function.

I don't understand how to decide the coefficient $A_n$. My textbook says that I should first use the initial value, which in this case would remove the exponential term, leaving

$$T(r,0)=T_1-T_2=\sum_{n=1}^\infty A_nJ_0(k_nr) \, .$$

Now my book says I should multiply with the function inside the sum and change $n$ to an $m$ and then integrate, which would give me

$$\int_0^R J_0(k_mr)(T_1-T_2)dr=\sum A_n\int_0^R J_0(k_nr)J_0(k_mr)dr \, .$$

However the correct answer to this problem says that you should multiply with $J_0(k_mr)r$ instead of $J_0(k_mr)$ and then puts $m=n$.

My questions are therefore:

  • Where does the extra $r$ come from?
  • Why can you assume $m=n$?

I feel like I don't understand the theory behind finding these coefficients in series so I hope someone can explain the basics.

  • 6
    $\begingroup$ This is one of the better homework questions I've seen on here in a while. $\endgroup$
    – Sean
    Mar 11, 2015 at 11:39
  • 1
    $\begingroup$ I wonder if there's a way this can be linked in the help pages as an example of a (high level) well posed homework type question. $\endgroup$
    – DanielSank
    Aug 4, 2015 at 16:24

1 Answer 1


In one way you might say that the $r$ is coming from the fact that you are in cylindrical coordinates, but more importantly, you want to get rid of the infinite sum over $n$. You do so by using a orthogonality relation of the Bessel functions. Here one uses

$$\int_0^1 x J_\alpha(u_{\alpha,m}x)J_\alpha((u_{\alpha,n}x)\mathrm{d}x = \frac{\delta_{m,n}}{2}[J_{\alpha+1}(u_{\alpha,m})]^2$$

where $u_{\alpha,m}$ is the $m$'s zero of $J_\alpha(x)$. Check the wiki. So, you are getting rid of the infinite sum by multiplying with "orthogonal" functions. You do the same in a Fourier series where you would use

$$\int_0^T \sin(\omega_n t) * ... \mathrm{d}t $$

with $\omega_n= 2\pi n/T$ and you know that

$$\int_0^T \sin(\omega_n t) \sin(\omega_m t) \mathrm{d}t \propto\delta_{m,n} \, .$$

It's the same thing. Hence, the trick is actually that this integral requires $m=n$ due to the orthogonality; all other contributions are automatically zero.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.