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I want to solve the equation below

$$\partial_t F(r,t)= \frac{a}{r^{d-1}}\partial_r\big(r^{d-1} \partial_r F(r,t)\big)$$

where $r$ denotes the radius in spherical coordinates, and $a$ is a constant. The initial and boundary conditions are:

$$F(r,t=0)=\delta(r-R_0)$$

and

$$\partial_r F(r,t)|_{r=R_0}=0$$

Does anyone have any idea of if it is solvable or not?

Rodrigo de Azevedo
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Zhi Ruo
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2 Answers2

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Let $F(r,t) = R(r)T(t)$ then

$$ \frac{T'}{aT} = \frac{\frac{1}{r}(rR')'}{R} = -\lambda $$

and

\begin{cases} T' + \lambda aT = 0 \\ (rR')' + \lambda rR = 0 \end{cases}

The radial equation is

$$ r^2R'' + rR' + \lambda r^2R = 0 $$

of which the solution is in the form of Bessel functions

$$ R(r) = J_0(\sqrt{\lambda}r) $$

The boundary condition requires

$$ R'(r_0) = -\sqrt{\lambda}J_1(\sqrt{\lambda}r_0) = 0 $$

Then $\sqrt{\lambda_n}r_0 = \alpha_n$ where $\alpha_n$ are the zeroes of $J_1(x)$. There are no closed forms for these zeroes, but numerical approximations can be found. Rewriting the solution

$$ R_n(r) = J_0\left(\frac{\alpha_n r}{r_0}\right) $$

The general solution is then

$$ F(x,t) = \sum_{n=1}^\infty c_n J_0\left(\frac{\alpha_n r}{r_0}\right) e^{-a(\alpha_n/r_0)^2t}$$

Next, I will prove that the eigen-solutions $R_n$ are orthogonal with respect to a weighting function. By definition we have

$$ (rR_n')' = -\lambda_n rR_n $$

and all solutions satisfy $R_n'(r_0) = 0$

Through integration by parts

\begin{align} \int_0^{r_0} (rR_n')'R_m dr &= rR_n'R_m\Big\vert_0^{r_0} -\int_0^{r_0} rR_n'R_m' dr = -\int_0^{r_0} rR_n'R_m' dr \\ \int_0^{r_0} (rR_m')'R_n dr &= rR_m'R_n\Big\vert_0^{r_0} -\int_0^{r_0} rR_n'R_m' dr = -\int_0^{r_0} rR_m'R_n' dr \end{align}

Using the definition

$$ \int_0^{r_0} \big[(rR_n')'R_m - (rR_m')R_n \big] dr = -(\lambda_n-\lambda_m)\int_0^{r_0} rR_nR_m dr = 0 $$

which follows that

$$ \int_0^{r_0} rR_nR_m dr = 0 $$

for $n\ne m$

Finally, the initial condition

$$ F(x,0) = \delta(r-s) = \sum_{n=1}^\infty c_n J_0 \left(\frac{\alpha_n r}{r_0}\right) $$

where $\delta(r)$ is the polar Dirac-Delta and $0<s<r_0$

Using the proven orthogonality

\begin{align} \int_0^{r_0} r\delta(r-s) J_0\left(\frac{\alpha_m r}{r_0}\right) &= \int_0^{r_0}\sum_{n=1}^\infty c_n rJ_0 \left(\frac{\alpha_n r}{r_0}\right)J_0\left(\frac{\alpha_m r}{r_0}\right) dr \\ J_0\left(\frac{\alpha_m s}{r_0}\right) &= c_m \int_0^{r_0} rJ_0^2\left(\frac{\alpha_m r}{r_0}\right) dr \end{align}

Edit 3/5: In spherical coordinates the Laplacian turns out to be

$$ \nabla^2 F = \frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial F}{\partial r}\right) $$

Following this, we have the radial equation

$$ (r^2R')' + \lambda r^2 R = 0 $$

This has a solution in terms of spherical Bessel functions. Once again, we only take the one that's finite at $r=0$

$$ R(r) = j_0(\sqrt{\lambda}r) = \frac{\sin(\sqrt{\lambda}r)}{r} $$

The B.C. gives

$$ R'(r_0) = 0 \implies \sqrt{\lambda}r_0\cos(\sqrt{\lambda}r_0) - \sin(\sqrt{\lambda}r_0) = 0 \implies \sqrt{\lambda}r_0 = \tan(\sqrt{\lambda}r_0) $$

Let $\alpha_n$ be the solutions to the equation $x=\tan(x)$ (again ignoring $\alpha_0=0$), you have the general solution

$$ F(x,t) = \sum_{n=1}^\infty \frac{c_n}{r} \sin\left(\frac{\alpha_n r}{r_0}\right)e^{-a(\alpha_n/r_0)^2t}$$

In spherical the weighting factor is $r^2$ and not $r$, so we have

$$ c_n = \frac{\int_0^{r_0} F(x,0) R_n(r)\ r^2 dr}{\int_0^{r_0} R_n(r)\ r^2 dr} = \frac{\int_0^{r_0}F(x,0)\sin\left(\frac{\alpha_n r}{r_0}\right)\ r dr}{\int_0^{r_0}\sin^2\left(\frac{\alpha_n r}{r_0}\right)\ dr} $$

Dylan
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  • I'm using the definition of the Dirac-delta in polar coordinates: $$ \int_\Bbb R r\delta(r-s)f(r) dr = f(s) $$ or $\delta(x,y) = \frac{1}{r}\delta(r,\phi)$. If your initial condition is just the regular delta, then adjust accordingly. But I don't think that's the case. – Dylan Mar 01 '19 at 15:47
  • I thought your title was typo so I didn't say anything, but yes you did. $\frac{1}{r}\partial_r(r\partial_r)$ the Laplacian in cylindrical/polar coordinates. The Laplacian in spherical is $\frac{1}{r^2}\partial_r(r^2\partial_r)$. Fortunately the eigensolutions to the spherical problem are a bit easier. It's $R(r) = \frac{\sin(\sqrt{\lambda r})}{r}$ – Dylan Mar 05 '19 at 03:05
  • What do you mean? There are no close form for $\alpha_n$, you have to find them numerically. – Dylan Mar 05 '19 at 11:00
  • I'm confused what you're asking? I already gave an expression for $c_m$. If you want to know what it looks like in spherical, I'll add to my answer later. – Dylan Mar 05 '19 at 11:04
  • if you integrate the initial condition inside the sphere, you will find a finite result. But if you integrate over your solution you will get zero! So I think something is not correct – Zhi Ruo Mar 05 '19 at 11:05
  • Let us continue this discussion in chat. – Dylan Mar 05 '19 at 11:07
  • Do you have any idea about this question? https://math.stackexchange.com/questions/3184706/solving-a-pde-outside-a-sphere – Zhi Ruo Apr 12 '19 at 09:50
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By variable separation, taking $F(r,t)=G(r)T(t)$, you get $$ T(t)=Ae^{-k^2t} \qquad G(r)=C_1J_0\left(k\frac{r}{a}\right)+C_2Y_0\left(k\frac{r}{a}\right). $$ We remove the singularity in the origin by taking $C_2=0$. Then, by the second boundary condition, we get $$ J_1\left(k\frac{R_0}{a}\right)=0 $$ That has an infinite set of solutions $u_n=\frac{R_0}{a}k_n$. So, the solution can be written in the form $$ F(r,t)=\sum_{n=0}^\infty B_n e^{-\frac{u_n^2a^2}{R_0^2}t}J_0\left(u_n\frac{r}{R_0}\right) $$ given the Fourier-Bessel series $$ \delta(r-R_0)=\sum_{n=0}^\infty B_n J_0\left(u_n\frac{r}{R_0}\right). $$

Jon
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