**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- r^{2})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 + ρ^{2}sin(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 J_{0}(x) OR
a zero of J_{0}′(x) with a similar definition for λ_{n}.

From the lecture notes we showed
that

_{}

where y_{k}(x) = J_{0}(λ_{k}x)
and y_{m}(x) = J_{0}(λ_{m}x). 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 J_{0}(x) OR a zero of J_{0}′(x) (or a zero of
any linear combination of J_{0}(x) and J_{0}′(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'' =
μ^{2}Z then the equation for Р(ρ)
becomes ρР'' + Р'
μ^{2}ρР = 0 which has solution

Ρ(ρ) = A J_{0}(μρ)
+ B Y_{0}(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) = J_{0}(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 J_{0}′(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 J_{0}′ are just the zeros of J_{1}.

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_{mn}ρ)cos(mφ)
and J_{m}(α_{mn}ρ)sin(mφ) are eigenfunctions of
the Helmholz equation with eigenvalues λ_{mn} = α_{mn}
since a = 1 the solution of Poissons 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) 4x^{2}y′′
- 14xy′ - (18 x)y = 0

Here p(x) = -7/2x
and q(x) = -(18 x)/4x^{2}.

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

Since p_{0}
= -7/2 and q_{0} = -9/2 the indicial equation is s(s-1) 7s/2 9/2 =
0 which has roots

s_{1 }=_{ }(9 +
3√17)/4 and s_{2 }=_{ }(9 - 3√17)/4. Thus s_{1} s_{2} =
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)/x^{2}.

p and q are singular at x = 0 but xp(x) = 1
and x^{2}q(x) = -(1 + x) which have power series about x = 0. Thus x = 0 is a regular singular point.

Since p_{0}
= 1 and q_{0} = -1 the indicial equation is s(s-1) + s - 1 = 0 which
has roots

s_{1} =_{ }1 and s_{2}
=_{ }-1. Thus s_{1} s_{2}
= 2 which is an integer so we have Frobenius case III.

(iii) y′′ + (1
- x^{2})y′ + xy = 0

Here p(x) = (1 -
x^{2}) 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 + x^{2}
+
) for |x| < 1 and x^{2}q(x) = -x/(1 - x) = -x(1 + x + x^{2}
+
) which are power series about x = 0.
Thus x = 0 is a regular singular point.

Since p_{0}
= 1 and q_{0} = 0 the indicial equation is s(s-1) + s = 0 which has roots

s_{1 }=_{ }0 and s_{2}
=_{ }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 x^{2}q(x)
= x/4 which have power series about x = 0.
Thus x = 0 is a regular singular point.

Since p_{0}
= 0 and q_{0} = 0 the indicial equation is s(s-1) = 0 which has roots

s_{1} =_{ }1 and s_{2}
=_{ }0. Thus s_{1} s_{2}
= 1 which is an integer so we have Frobenius case III.

(vi) x^{3}y′′
+ x^{2}y′ - y= 0

Here p(x) = x^{2}/x^{3}
= 1/x and q(x) = -1/x^{3}.

p and q are singular at x = 0 and xp(x) = 1
and x^{2}q(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).