Math. 312 17 November 2000

Test C – Solutions

1. (45 pts.) Solve the heat equation in a disk with insulated edge:

u = u(t, r, θ), t > 0, 0 ≤ r < 3, 0 ≤ θ < 2π,

\frac{∂u}{∂t} = ∇2u = \frac{∂2u}{∂r2} + \frac{1}{r} \frac{∂u}{∂r} + \frac{1}{r2} \frac{∂2u}{∂θ2} ,

\frac{∂u}{∂r}(t, 3, θ) = 0,

u(0, r, θ) = f(r, θ).

(Be as explicit as you can about the eigenvalues and eigenfunctions that arise. Sketching one of the eigenfunctions would be a good idea.)

In a word, we expect Bessel functions. But let’s go through the steps of variable separation to get there.

Consider u_{sep}(t, r, θ) = T (t)R(r)Θ(τ). Then

T′RΘ = T R′′θ + \frac{1}{r} T R′Θ + \frac{1}{r^{2}} T RΘ′′.

Thus

\frac{T′}{T} = −ω^{2} = \frac{R′′}{R} + \frac{1}{r} \frac{R′}{R} + \frac{1}{r^{2}} \frac{Θ′′}{Θ} .

(We should come back later and check that the separation constant is always negative, but for a heat equation we expect most of the modes, at least, to decay in time.) Thus T (t) = e^{−ω^{2}t}.

Multiply the remaining equation by r^{2} to separate the variables, and introduce another separation constant:

− \frac{Θ′′}{Θ} = n^{2} = r^{2} \frac{R′′}{R} + r \frac{R′}{R} + ω^{2}r^{2}.

The periodic boundary conditions on Θ ensure that n^{2} is really positive, and in fact that n is an integer. We have Θ(θ) = e^{iνθ}, ν = ±n.

Now we’re left with

\frac{R′′}{R} + \frac{1}{r} \frac{R′}{R} + (ω^{2} − \frac{n^{2}}{r^{2}} ) = 0,

a form of Bessel’s equation. Let z = ωr and Z(z) = R(r). Then d/dr = ω d/dz, so

Z′′ + \frac{1}{z} Z′ + (1 − \frac{n^{2}}{z^{2}} ) Z = 0

(where the primes now indicate differentiation with respect to z). The solutions of this are J_{n}(z) and Y_{n}(z), but because our region includes the origin and the solution must be smooth there, only J functions can appear. Thus R(r) = J_{n}(ωr). The boundary condition requires that R′(3) = 0. Therefore, 3ω must be a zero of the derivative of J_{n}. So, define y_{nj} to be the jth zero: J′(y_{nj}) = 0. Then ω_{nj} = \frac{1}{3} y_{nj} .

312C-F00 Page 2

Now, could ω^{2} possibly be negative or 0? Extra credit to anybody who realized the following: For n ≠ 0, the first zero of J′_{n} is smaller than the first zero of J_{n}, so our first eigenfunction has no nodes and there is no way that there could be another eigenfunction with an even smaller eigenvalue. (Alternative argument: The eigenfunction would have to be r^{n} (for ω = 0) or the modified Bessel function I_{n}(κr) (for ω^{2} = −κ^{2} < 0), and the derivatives of these functions have no zeros.) But for n = 0 the first zero of the derivative (other than the one at the origin) comes after the first derivative of J_{0} itself, so there probably is another eigenfunction with no nodes lurking somewhere. In fact, the eigenvalue ω^{2} is 0, and the eigenfunction is the constant function 1 (leading to u_{sep}(t, r, θ) = 1 × 1 × 1 = 1, and hence to the term A_{00} in the next equation).

We are now ready to superpose the normal mode solutions u_{sep} ; this requires summing over n and j.

u(t, r, θ) = A_{00} + ∑_{ν=−∞}^{∞} ∑_{j=1}^{∞} A_{νj}e^{−ω_{nj}^{2}t}J_{n}(ω_{nj}r)e^{iνθ},

where n = |ν| and ω_{nj} = \frac{1}{3} y_{nj} .

Finally we have to determine the coefficients from the initial data.

f(r, θ) = A_{00} + ∑_{ν=−∞}^{∞} ∑_{j=1}^{∞} A_{νj}J_{n}(ω_{nj}r)e^{iνθ}.

We can solve this either in one step, by taking inner products with the two-dimensional eigenfunctions ϕ_{νj} = e^{iνθ}J_{n}(ω_{nj}r), or in two steps, doing an ordinary Fourier series calculation and then a Fourier-Bessel series calculation. Thinking in the first way, one sees that

A_{00} = \frac{1}{9π} ∫_{0}^{2π} dθ ∫_{0}^{3} r drf(r, θ)

(the 9π is the area of the disk, which is the square of the norm of the eigenfunction 1). Let’s get the other constants the other way:

ϕ_{n}(r) ≡ \frac{1}{2π} ∫_{0}^{2π} e^{−iνθ}f(r, θ) dθ = ∑_{j=1}^{∞} A_{νj}J_{n}(ω_{nj}r),

and then

A_{νj} = \frac{∫_{0}^{3} f_{ν}(r)J_{n}(ω_{nj}r) r dr}{∫_{0}^{3} J_{n}(ω_{nj}r)^{2} r dr} .

Whew!

2. (45 pts.) Let’s study the wave equation

\frac{∂2u}{∂t2} = \frac{∂2u}{∂x2}

on the interval 0 < x < 1 with the boundary conditions

u(t, 0) = 0, \frac{∂u}{∂x}(t, 1) + βu(t, 1) = 0,

where β is a positive constant.

(a) Separate variables, find the eigenfunctions, and indicate graphically how to find the eigenvalues. Give an approximate formula for the large eigenvalues.

Let u_{sep}(t, x) = T (t)X(x). Then

\frac{T′′}{T} = −ω^{2} = \frac{X′′}{X} ,

and X must satisfy X′(1)+βX(1) = 0 and X(0) = 0. Thus X(x) = sin(ωx) and ω cos ω +β sin ω = 0. The eigenvalue equation is most conveniently written

tan ω = − \frac{ω}{β} ,

which can be graphed exactly as in the class notes. For large n the nth solution is slightly greater than (n − \frac{1}{2} )π. The positivity of β is the condition that excludes negative and zero solutions of ω^{2}.

(b) Solve the wave equation with initial data

u(0, x) = f(x), \frac{∂u}{∂t}(0, x) = g(x).

Superpose the normal-mode solutions found above, noting that there are two solutions to the time equation:

u(t, x) = ∑_{n=1}^{∞}[a_{n} sin(ω_{n}x) cos(ω_{n}t) + b_{n} sin(ω_{n}x) sin(ω_{n}t)].

Then

f(x) = ∑_{n=1}^{∞} a_{n} sin(ω_{n}x), g(x) = ∑_{n=1}^{∞} b_{n}ω_{n} sin(ω_{n}x).

Therefore,

a_{n} = \frac{∫_{0}^{1} f(x) sin(ω_{n}x) dx}{∫_{0}^{1} sin(ω_{n}x)^{2} dx} , b_{n} = \frac{∫_{0}^{1} g(x) sin(ω_{n}x) dx}{ω_{n} ∫_{0}^{1} sin(ω_{n}x)^{2} dx} .

3. (10 pts.) Answer ONE of these. (Extra credit for both.) Relatively brief and qualitative answers are expected, not complete calculations. (A) What would happen in Question 1 if the disk were replaced by an annulus (ring) with

inner boundary r = 1? Suppose that that edge is held at a constant temperature, u(t, 1, θ) = T.

First of all, if T ≠ 0 we need to subtract off a steady-state solution. It is easy to see that v(t, r, θ) = T is that solution, since it satisfies the heat equation and the other boundary condition.

Let w = u − v. Then w satisfies the same problem except that u(t, 1, θ) = 0 and u(0, r, θ) = f(r, θ) − T ≡ g(r, θ). Now we could separate variables in the same way as before, but this time the radial function R(r) = Z(ωr) would have to satisfy R(1) = 0 as well as R′(3) = 0. Since the origin is no longer in the region, there is no reason why Y_{n} can’t appear. So Z(z) = α_{n}(ω)J_{n}(z)+β_{n}(ω)Y_{n}(z), and the equations Z′(3ω) = 0 and Z(ω) = 0 determine both the ratio of α to β and the values of ω that can occur.

(B) What would happen in Question 2 if the constant β were negative? Useful information: The hyperbolic tangent function has a graph like this, with slope 1 at the origin and asymptote 1 at +∞:

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First, in the graphical solution the graph of − \frac{ω}{β} now slopes up instead of down, so the nth solution is slightly less than (n + \frac{1}{2} )π. (Actually, if β < 1 this straight line also intersects the first branch of the tangent function, so there is a zeroth solution somewhere less than \frac{π}{2} .)

Second, it is now possible to have a negative solution for ω^{2}. In that case, setting κ^{2} = −ω^{2}, we have X(x) = sinh(κx) and

tanh κ = \frac{κ}{β} .

If β > 1, this equation has one solution, as you can see by adding the diagonal straight line to the graph above. (The accompanying function of t will also be a sinh or cosh, so this mode is an instability in the system — probably making this problem physically implausible!) If β < 1 the negative eigenvalue does not exist, being replaced by the extra positive eigenvalue mentioned previously. If β = 1, zero is an eigenvalue with eigenfunction X(x) = x.