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 λmn = αmn since a = 1 and c = 1.  The coefficients are given by

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

Thus Bmn = 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 λm = αm/a where αm is a zero of J0(x) OR a zero of J0′(x) with a similar definition for λn.

From the lecture notes we showed that

where yk(x) = J0kx) and ym(x) = J0mx).  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 = β0n or μ0n = β0n/a where β0n 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 β0n = α1n (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 Jm(αmnρ)cos(mφ) and Jmmnρ)sin(mφ) are eigenfunctions of the Helmholz equation with eigenvalues λmn = αmn 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).