MATH3271

FW03

Solutions to Assignment 4

1. (a) Solve the problem of a vibrating circular membrane of radius 1 if it is initially undisplaced with an initial velocity given by g(r,f) = (1- r²)sin(2φ). Take c = 1.

The solution for a vibrating membrane which is initially undisplaced is given by

where λ_mn = α_mn since a = 1 and c = 1. The coefficients are given by

by the orthogonality of the trig fuinctions {cos(mφ), sin(mφ)}. Similarly

Thus B_mn = 0 if m ≠ 2 and

so that

(b) Modify your MAPLE program from assignment 3 to calculate the first 10 coefficients of the series solution and plot the tenth partial sum of this series for several values of t.

See MAPLE output

2. (a) Find the steady-state temperature of a disk of radius 1 if the temperature on the boundary is given by 1 + sin(2φ).

The steady state temperature distribution is given by

where

Thus u(ρ,φ) = 1 + ρ²sin(2φ).

(b) Use MAPLE to plot the tenth partial sum of this solution.

See MAPLE output.

3. (a) From the proof of the orthogonality integral(see section 4.8 of the text) deduce that

if m ≠ n and λ_m = α_m/a where α_m is a zero of J₀(x) OR a zero of J₀′(x) with a similar definition for λ_n.

From the lecture notes we showed that

where y_k(x) = J₀(λ_kx) and y_m(x) = J₀(λ_mx). Thus the rhs is zero at the lower limit x = 0 and is also zero at the upper limt x = a if λa is a zero of J₀(x) OR a zero of J₀′(x) (or a zero of any linear combination of J₀(x) and J₀′(x)).

(b) Find the steady-state temperature of a cylinder of radius a and height h if the side is insulated and the temperature of the bottom is kept a zero degrees while the temperature distribution on the top is given by f(ρ).

To find the steady-state temperature distrobution in a cylinder we must solve

subject to the insulating boundary conditions

plus the conditions that u(ρ,z) remain finite while on the bottom u(ρ,0) = 0 and on the top

u(ρ,h) = f(ρ).

As before we use separation of variables, i.e. u(ρ,z) = Р(ρ)Z(z) which leads to

If we let Z'' = μ²Z then the equation for Р(ρ) becomes ρР'' + Р' – μ²ρР = 0 which has solution

Ρ(ρ) = A J₀(μρ) + B Y₀(μρ)

while the solution for Z is

Z(z) = C cosh(μz) + D sinh(μz)

Since Ρ(ρ) must remain finite in the cylinder we must take B = 0. Similarly the condition that u is zero when z = 0 requires C = 0. Thus a basic solution is

u(ρ,z) = J₀(μρ)sinh(μz)

Since

the insulating boundary requires that μa = β_0n or μ_0n = β_0n/a where β_0n is a zero of J₀′(x). Thus the general solution is

Apply the boundary condition on the top of the cylinder we get

This is just a Fourier-Bessel series for f(ρ) so we get

where

The last part of the above equation can be derived from the results in question 36 of section 4.8. Also it can be shown that β_0n = α_1n (see equation 4 of section 4.8) i.e.the zeros of J₀′ are just the zeros of J₁.

4.(a) Find a solution to the two-dimensional Poisson equation in polar coordinates

, 0 < ρ < 1, 0 < φ < 2π

with boundary condition u(1,φ) = 0.

Since J_m(α_mnρ)cos(mφ) and J_m(α_mnρ)sin(mφ) are eigenfunctions of the Helmholz equation with eigenvalues λ_mn = α_mn since a = 1 the solution of Poisson’s equation with rhs ρ cos(φ) is

where

using the orthogonality of {cso(mφ), sin(mφ)}. Similarly

if m = 1 and zero otherwise while

Thus

(b) Modify your MAPLE program from question 1 to calculate the first 10 coefficients in the series solution and plot the tenth partial sum of this solution.

See MAPLE output

5. For each of the following differential equations, determine if x = 0 is an ordinary point, a regualr singular point or neither. If it is a regular singula point, solve the indicial equation and state which case of the Frobeniuns method applies.

(i) 4x²y′′ - 14xy′ - (18 – x)y = 0

Here p(x) = -7/2x and q(x) = -(18 – x)/4x².

p and q are singular at x = 0 but xp(x) = -7/2 and x²q(x) = -(18 – x)/4 which have power series about x = 0. Thus x = 0 is a regular singular point.

Since p₀ = -7/2 and q₀ = -9/2 the indicial equation is s(s-1) – 7s/2 – 9/2 = 0 which has roots

s₁ = (9 + 3√17)/4 and s₂ = (9 - 3√17)/4. Thus s₁ – s₂ = 3√17/2 which is not an integer so we have Frobenius case I.

(ii) xy′′ + y′ - (1 + x)y/x = 0

Here p(x) = 1/x and q(x) = -(1 + x)/x².

p and q are singular at x = 0 but xp(x) = 1 and x²q(x) = -(1 + x) which have power series about x = 0. Thus x = 0 is a regular singular point.

Since p₀ = 1 and q₀ = -1 the indicial equation is s(s-1) + s - 1 = 0 which has roots

s₁ = 1 and s₂ = -1. Thus s₁ – s₂ = 2 which is an integer so we have Frobenius case III.

(iii) y′′ + (1 - x²)y′ + xy = 0

Here p(x) = (1 - x²) and q(x) = x.

p and q are finite at x = 0 so x = 0 is an ordinary point.

(iv) x(1 – x)y′′ + (1 – 3x)y′ - y = 0

Here p(x) = (1 – 3x)/x(1 – x) and q(x) = -1/x(1 - x).

p and q are singular at x = 0 but xp(x) = (1 – 3x)/(1 – x) = (1 – 3x)(1 + x + x² + …) for |x| < 1 and x²q(x) = -x/(1 - x) = -x(1 + x + x² + …) which are power series about x = 0. Thus x = 0 is a regular singular point.

Since p₀ = 1 and q₀ = 0 the indicial equation is s(s-1) + s = 0 which has roots

s₁ = 0 and s₂ = 0. Thus the roots are equal which is Frobenius case II.

(v) 4xy′′ + 2xy′ + y= 0

Here p(x) = 2x/4x = 1/2 and q(x) = 1/4x.

q is singular at x = 0 but xp(x) = x/2 and x²q(x) = x/4 which have power series about x = 0. Thus x = 0 is a regular singular point.

Since p₀ = 0 and q₀ = 0 the indicial equation is s(s-1) = 0 which has roots

s₁ = 1 and s₂ = 0. Thus s₁ – s₂ = 1 which is an integer so we have Frobenius case III.

(vi) x³y′′ + x²y′ - y= 0

Here p(x) = x²/x³ = 1/x and q(x) = -1/x³.

p and q are singular at x = 0 and xp(x) = 1 and x²q(x) = -1/x so that q does not have power series about x = 0. Thus x = 0 is neither an ordinary or a regular singular point (it is an irregular singular point).