Fdtxjr

I want to solve the equation below

$$partial_t F(r,t)= fracar^d-1partial_rbig(r^d-1 partial_r F(r,t)big)$$

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?

By variable separation, taking $F(r,t)=G(r)T(t)$, you get
$$
T(t)=Ae^-k^2t qquad G(r)=C_1J_0left(kfracraright)+C_2Y_0left(kfracraright).
$$
We remove the singularity in the origin by taking $C_2=0$. Then, by the second boundary condition, we get
$$
J_1left(kfracR_0aright)=0
$$
That has an infinite set of solutions $u_n=fracR_0ak_n$. So, the solution can be written in the form
$$
F(r,t)=sum_n=0^infty B_n e^-fracu_n^2a^2R_0^2tJ_0left(u_nfracrR_0right)
$$
given the Fourier-Bessel series
$$
delta(r-R_0)=sum_n=0^infty B_n J_0left(u_nfracrR_0right).
$$

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

$$ fracT'aT = fracfrac1r(rR')'R = -lambda $$

and

begincases
T' + lambda aT = 0 \
(rR')' + lambda rR = 0
endcases

The radial equation is

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

of which the solution is in the form of Bessel functions

$$ R(r) = J_0(sqrtlambdar) $$

The boundary condition requires

$$ R'(r_0) = -sqrtlambdaJ_1(sqrtlambdar_0) = 0 $$

Then $sqrtlambda_nr_0 = alpha_n$ 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

$$ R_n(r) = J_0left(fracalpha_n rr_0right) $$

The general solution is then

$$ F(x,t) = sum_n=1^infty c_n J_0left(fracalpha_n rr_0right) e^-a(alpha_n/r_0)^2t$$

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

beginalign
int_0^r_0 (rR_n')'R_m dr &= rR_n'R_mBigvert_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_nBigvert_0^r_0 -int_0^r_0 rR_n'R_m' dr = -int_0^r_0 rR_m'R_n' dr
endalign

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 $nne m$

Finally, the initial condition

$$ F(x,0) = delta(r-s) = sum_n=1^infty c_n J_0 left(fracalpha_n rr_0right) $$

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

Using the proven orthogonality

beginalign
int_0^r_0 rdelta(r-s) J_0left(fracalpha_m rr_0right) &= int_0^r_0sum_n=1^infty c_n rJ_0 left(fracalpha_n rr_0right)J_0left(fracalpha_m rr_0right) dr \
J_0left(fracalpha_m sr_0right) &= c_m int_0^r_0 rJ_0^2left(fracalpha_m rr_0right) dr
endalign

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

$$ nabla^2 F = frac1r^2fracpartial partial rleft(r^2fracpartial Fpartial rright) $$

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 $r=0$

$$ R(r) = j_0(sqrtlambdar) = fracsin(sqrtlambdar)r $$

The B.C. gives

$$ R'(r_0) = 0 implies sqrtlambdar_0cos(sqrtlambdar_0) - sin(sqrtlambdar_0) = 0 implies sqrtlambdar_0 = tan(sqrtlambdar_0) $$

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

$$ F(x,t) = sum_n=1^infty fracc_nr sinleft(fracalpha_n rr_0right)e^-a(alpha_n/r_0)^2t$$

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

$$ c_n = fracint_0^r_0 F(x,0) R_n(r) r^2 drint_0^r_0 R_n(r) r^2 dr = fracint_0^r_0F(x,0)sinleft(fracalpha_n rr_0right) r drint_0^r_0sin^2left(fracalpha_n rr_0right) dr $$