A radial heat flow problem

mark

 This is essentially problem #3 on page 207 of our text but adapted to take advantage of some of the fun tools we've developed recently. Specifically, let's try to solve the heat problem on the unit disk:

$$u_t = \Delta u, \: u(1,t) = 0, \: u(r,0) = 5 r^3 (1 - r),$$

where the problem is described in terms of polar coordinates. To do so:

1. Express your solution as an infinite series involving Bessel functions with unknown coefficients $c_n$.
2. Write out the first three terms of the series explicitly with numerically computed values of $c_1$, $c_2$, and $c_3$.
3. Plot the graph of a higher precision approximation to the radial temperature profile at time $t=0.02$.
4. Write down the temperature at the center of the disk at time $t=0.02$ to 4 digits of precision.

AbS

(1) We start by separating variables, and assume $u(r,t) = R(r)T(t)$. Our PDE then becomes

$$(RT)_t = (RT)_{rr} + \frac{1}{r}(RT)_r$$ or

$$RT' = R''T + \frac{1}{r}R'T.$$

Dividing both sides by $RT$, we get

$$\frac{T'}{T} = \frac{R''}{R} +\frac{1}{r}\frac{R'}{R} = -\lambda.$$

Thus, $\frac{T'}{T} = -\lambda$, so $T' = -\lambda T.$ Hence,

$$T = e^{-\lambda t}.$$ Meanwhile,

$$\frac{R''}{R} + \frac{1}{r}\frac{R'}{R} = -\lambda,$$ so

$$R'' + \frac{1}{r}R' = -\lambda R.$$

This can be rewritten as $$-(rR'(r))' = \lambda rR(r).$$

The solutions to this radial differential equation are the Bessel functions $$R(r) = c_1J_0\left(\sqrt{\lambda}r\right) + c_2Y_0\left(\sqrt{\lambda}r\right).$$

Since $Y_0$ is unbounded at $z=0, c_2$ must equal $0,$ so $$R(r) = c_1J_0\left(\sqrt{\lambda}r\right).$$

From our boundary conditions, we know $$R(1) = c_1J_0\left(\sqrt{\lambda}\right) = 0,$$ so

$$\sqrt{\lambda} = z_n, n=1,2,3...$$

where $z_n$ are the zeros of $J_0.$ Hence, the eigenvalues are $\lambda_n = z_n^2$ and the corresponding eigenfunctions are $R_n(r) = J_0(z_nr).$ Thus, our full solution is

$$u(r,t) = \sum_{n=1}^{\infty}c_ne^{-\lambda_n^2t}J_0(z_nr).$$

(2) By our initial conditions,

$$u(r,0) = \sum_{n=1}^{\infty}c_nJ_0(z_nr) = 5r^3(1-r).$$

Since Bessel functions satisfy the orthogonality condition,

$$c_n = \frac{\int_{0}^{1}5r^3(1-r)J_0(z_nr)rdr}{\int_{0}^{1}J_0^2(z_nr)^2rdr}.$$

Plugging an initial condition $u(r,0)$ into the "Radial heat flow" ObservableHQ page gives us the normalized Bessel function $F(r)$ for each term of the solution and its constant coefficient, which we'll denote $b_n,$ where

$$u(r,0) = \sum_{n=1}^{\infty}b_nF_n(r) = \sum_{n=1}^{\infty}b_n\frac{R_n(r)}{||R_n||}.$$

Since $u(r,0) = \sum_{n=1}^{\infty}c_nR_n(r),$ it follows that

$$c_n = \frac{b_n}{||R_n||}.$$

We plug in $u(r,0) = 5r^3(1-r)$ and find that

$$u(r,t) \approx \frac{0.17333401324973097}{0.3670927157694067}e^{-z_1^2t}J_0(z_1r) + \frac{-0.18156860394077448}{0.24060355211655088}e^{-z_2^2t}J_0(z_2r) + \frac{0.0731801393963544}{0.16437392311147672}e^{-z_3^2t}J_0(z_3r)$$

$$= 0.4721804759e^{-z_1^2t}J_0(z_1r) - 0.7546380855e^{-z_2^2t}J_0(z_2r) + 0.4452052857e^{-z_3^2t}J_0(z_3r)$$

(3)

Using the same Observable page, with N=10, we graph the function at time $t = 0.02$:

(4)

Using the final tool on the Observable page with $N=10$, we find $$u(0, 0.02) = 0.086.$$

mark

@AbS This looks great - I just have one minor complaint. When you write:

(2) By our initial conditions,

$$u(r,0) = \sum_{n=1}^{\infty}c_nJ_0(z_nr) = 5r^3(1-r).$$

I guess you really mean something like

In order to satisfy our initial conditions, we want to find values of each $c_n$ such that ...