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The circularly symmetric heat equation is

$\frac{\partial{u}}{\partial{t}} = k\frac{1}{r}\frac{\partial}{\partial{r}}(r\frac{\partial{u}}{\partial{r}})$

When we have the boundary conditions being

$u(a,t) = 0$ and $u(r,0) = f(r)$

I found that the solution is $u(r,t)=\sum_{n=1}^{\infty}D_nJ_0(\sqrt{\lambda_n}r)e^{-\lambda_nkt}$

where

$D_n=\frac{\int_0^af(r)J_0(\sqrt{\lambda_n}r)rdr}{\int_0^aJ_0^2(\sqrt{\lambda_n}r)rdr}$

I found this solution via separation of variables, (assuming solution of form $u(r,t)=T(t)R(r)$) and was able to reduce the resulting spatial ODE into Bessel's differential equation of order $0$.

So my question is, if instead of $u(a,t)=0$ we had that the boundary condition is $\frac{\partial{u}}{\partial{r}}(a,t)=0$ and with the same initial condition, how would we go about finding the general solution?

Note that I'm very new to the theory of PDEs and only learnt about Bessel's functions and all that good stuff last week, but am eager to learn. A description of where to implement this boundary condition and what it results in would be greatly appreciated.

Thanks.

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1 Answer 1

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The solution would still be given by $$ u(x,t)=\sum_{n=1}^{\infty}D_nJ_0(\sqrt{\lambda_n}r)e^{-\lambda_nkt}, \tag{1} $$ but the eigenvalues $\lambda_n$ would be determined by solving $$ J_0'(\sqrt{\lambda}a)=0, \tag{2} $$ instead of $J_0(\sqrt{\lambda}a)=0$. Despite the different set of eigenvalues, the functions $J_0(\sqrt{\lambda_m}r)$ and $J_0(\sqrt{\lambda_n}r)$ remain orthogonal for $m\neq n$, thanks to the identity$^{(*)}$ $$ \int_0^1 xJ_{\nu}(\alpha x)J_{\nu}(\beta x)\,dx = \frac{\alpha J_{\nu}(\beta)J_{\nu}'(\alpha)-\beta J_{\nu}(\alpha)J_{\nu}'(\beta)}{\beta^2-\alpha^2}\qquad[\alpha\neq\beta,\quad \nu>-1]. \tag{3} $$ Therefore, the coefficients $D_n$ in $(1)$ are also given by $$ D_n=\frac{\int_0^af(r)J_0(\sqrt{\lambda_n}r)rdr}{\int_0^aJ_0^2(\sqrt{\lambda_n}r)rdr}, \tag{4} $$ but with the eigenvalues given by $(2)$.


$^{(*)}$ I.S. Gradshteyn and I.M. Ryzhik, Table of Integrals, Series, and Products, 7th edition, Eq.(6.521).

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  • $\begingroup$ Thanks so much! Very clear and informative answer. $\endgroup$
    – freddie123
    Commented May 11, 2021 at 10:14

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