# Insulated Heat equation

So it is well known that the heat equation for a rod with perfectly insulated ends at $$x= 0$$ and $$x = l$$ is given by: $$u_{t} = k u_{xx}$$ It has the following BCs:

$$\frac{\partial u}{\partial x}|_{(0,t)}=0 \ \hbox{ and } \frac{\partial u}{\partial x}|_{(l,t)}=0$$

and IC :

$$\frac{\partial u}{\partial x}|_{(x,0)}= h(x)$$

Just wondering how the BCs for a spherically symmetric sphere with radius $$r=a$$ that is perfectly insulated around the boundary would look like. Should it look like this? If we only consider radial heat flow, the heat equation would be:

$$u_t = k (u_{rr}+ \frac{2}{r} u_{r})$$.

By substituting $$v = ur$$, we have: $$v_t = k v_{rr}$$

with the BCs given by: $$v_{r}(0,t) = v_{r} (a,t) = 0$$

and IC given by: $$v = r h(r)$$.

• What do you mean by $v=ur$? Dec 1, 2020 at 16:18
• i added a clarification in the question Dec 3, 2020 at 2:49
• Are you sure you want $\partial_r[v](0,t)=\partial_r[v](a,t)=0$ and not $\partial_r[u](0,t)=\partial_r[u](a,t)=0$ ? The former will (I think) lead to complex, non-physical solutions. Dec 3, 2020 at 17:12
• Updated with Neumann boundary conditions on $u$. Dec 4, 2020 at 0:17
• You do not have a boundary at $r=0$. The boundary conditions will be $\partial_ru(a,t)=0$ and $|u(0,t)|$ bounded. Dec 4, 2020 at 1:11

SO the heat equation is $$\partial_t u=\lambda\nabla^2u$$ With $$\lambda$$ being some diffusivity constant. We can use the Laplacian in spherical coordinates and remove the $$\theta,\phi$$ dependency to get $$\partial_t u=\frac{\lambda}{r^2}\partial_r\left(r^2\partial_ru\right)$$ Let's assume the equation is separable. $$\frac{G'(t)}{G(t)}=\frac{\lambda}{r^2F(r)}\partial_r\left(r^2F'(r)\right)=-c$$ Let's focus on the spacial part. Make the substitution $$F(r)=f(r)/r$$ so $$F'(r)=\frac{r\cdot f'(r)-f(r)}{r^2}$$ Which leads us to $$f''(r)=\frac{-c}{\lambda}f(r)$$ $$f(r)=a_1\sin\left(\sqrt{\frac{c}{\lambda}}r\right)+a_2\cos\left(\sqrt{\frac{c}{\lambda}}r\right)$$ Imposing the boundary conditions, namely $$f(0)=f(a)=0$$ we require $$a_2=0$$ and we get constants $$c_n=\lambda \frac{n^2\pi^2}{a^2}$$ Which gives our spacial eigenfunctions $$F_n(r)=A_n\operatorname{sinc}\left(\frac{n\pi}{a}r\right)=A_nj_0\left(\frac{n\pi}{a}r\right)=A_n\sqrt{\frac{a}{2nr}}J_{1/2}\left(\frac{n\pi}{a}r\right)$$ Using Bessel functions (this will be important later)

The time component follows: $$\frac{G'(t)}{G(t)}=-c=-\lambda \frac{n^2\pi^2}{a^2}$$ $$G_n(t)=B_n\exp\left(-\lambda \frac{n^2\pi^2}{a^2}t\right)$$

So we can write our solution as (using $$A_n\sqrt{\frac{a}{2n}}\cdot B_n= C_n$$) $$u(r,t)=\sum_{n=1}^\infty C_n\frac{J_{1/2}\left(\frac{n\pi}{a}r\right)}{\sqrt{r}}\exp\left(-\lambda \frac{n^2\pi^2}{a^2}t\right)$$ Initial conditions:

We can see that $$\sqrt{r}\cdot u(r,0)=\sum_{n=1}^\infty C_nJ_{1/2}\left(\frac{n\pi}{a}r\right)$$ This is a perfect time to use Fourier-Bessel series. Let $$\alpha_{1/2,n}$$ be the $$n$$th root of $$J_{1/2}$$, which in this case is the rather simple $$\alpha_{1/2,n}=n\pi$$. So $$\sqrt{r}u_0(r)=\sum_{n=1}^\infty C_nJ_{1/2}\left(\frac{\alpha_{1/2,n}}{a}r\right)$$ The literature then tells us that $$C_n=\frac{\int_0^a r^{3/2}u_0(r)J_{1/2}\left(\frac{\alpha_{1/2,n}}{a}r\right)\mathrm{d}r}{(a/2)J_{3/2}(\alpha_{1/2,n})^2}$$

EXAMPLE: Let $$u_0(r)=(r^2-1)\log(r)$$. The first few coefficients are $$\{C_n\}=\{0.624023,0.386661,0.10893,...\}$$ Here is a plot of the initial data $$u_0(r)$$ as well as its first five partial series- And the Mathematica code used (so you can try yourself):

Subscript[u, 0][r_] := (r^2 - 1) Log[r]
c[n_] := 2*
NIntegrate[
r^(3/2) Subscript[u, 0][r] BesselJ[1/2, n*\[Pi]*r], {r, 0, 1}]/
BesselJ[3/2, n*\[Pi]]^2
Subscript[u, 0][r_, N_] :=
Sum[c[n]*(Sqrt Sin[n*\[Pi]*r])/(\[Pi]*r*Sqrt[n]), {n, 1, N}]
Plot[{Subscript[u, 0][r], Subscript[u, 0][r, 1],
Subscript[u, 0][r, 2], Subscript[u, 0][r, 3], Subscript[u, 0][r, 4],
Subscript[u, 0][r, 5]}, {r, 0.01, 1},
PlotLegends -> {"\!$$\*SubscriptBox[\(u$$, $$0$$]\)(r)",
"\!$$\*SubscriptBox[\(u$$, $$0$$]\)(r;1)",
"\!$$\*SubscriptBox[\(u$$, $$0$$]\)(r;2)",
"\!$$\*SubscriptBox[\(u$$, $$0$$]\)(r;3)",
"\!$$\*SubscriptBox[\(u$$, $$0$$]\)(r;4)",
"\!$$\*SubscriptBox[\(u$$, $$0$$]\)(r;5)"}]


Here $$u_0(r;N)$$ is the $$N$$th partial series. Another example when $$u_0(r)=\sqrt{1-r}$$ : EDIT: What if we have Neumann, not Dirichlet, boundary conditions? That is, what if we want $$\partial_r[u](0,t)=\partial_r[u](a,t)=0$$ Well, let's return to our spacial part of our solution: $$F(r)=a_1\frac{\sin\left(\sqrt{\frac{c}{\lambda}}r\right)}{r}+a_2\frac{\cos\left(\sqrt{\frac{c}{\lambda}}r\right)}{r}$$ The derivative is $$F'(r)=\frac{\cos\left(\sqrt{\frac{c}{\lambda}}r\right)\left(a_1\sqrt{\frac{c}{\lambda}}r-a_2\right)-\sin\left(\sqrt{\frac{c}{\lambda}}r\right)\left(a_1+a_2\sqrt{\frac{c}{\lambda}}r\right)}{r^2}$$ In order to get $$F'(0)=0$$ we require $$a_2=0$$. Then our expression for the derivative simplifies to $$F'(r)=a_1\frac{\sqrt{\frac{c}{\lambda}} r\cos\left(\sqrt{\frac{c}{\lambda}}r\right)-\sin\left(\sqrt{\frac{c}{\lambda}}r\right)}{r^2}$$ What we're trying to do is find values of $$c$$ such that $$F'(a)=0$$ given $$\lambda,a$$. Unfortunately, the roots of $$F(r)$$ are not easily obtainable in closed form. Letting $$\beta_{\lambda,n}$$ be the $$n$$th positive root of $$F'(r)$$ given the constant $$\lambda$$ (we can find these via numerical algorithms) we require $$\beta_n=\sqrt{\frac{c_n}{\lambda}}a\implies c_n=\lambda \left(\frac{\beta_n}{a}\right)^2$$ So our spacial eigenfunctions are $$F_n(r)=A_n\operatorname{sinc}\left(\frac{\beta_n}{a}r\right)=A_n j_0\left(\frac{\beta_n}{a}r\right)=A_n\sqrt{\frac{\pi}{2\frac{\beta_n}{a}r}}J_{1/2}\left(\frac{\beta_n}{a}r\right)\to A_n\frac{1}{\sqrt{r}}J_{1/2}\left(\frac{\beta_n}{a}r\right)$$ Where of course we redefine $$A_n$$ as we go along for convenience. After this point, we follow the same process as before with the Fourier Bessel series and whatnot - just in this case we have to compute the roots $$\beta_n$$ beforehand.

• I've just noticed you wanted $\partial_r[u](0,t)=\partial_r[u](a,t)=0$, not $u(0,t)=u(a,t)=0$. This makes things more difficult. I'll try to update the answer soon. Dec 3, 2020 at 16:40