2. Use d'Alembert's formula to write down the
explicit solution to the problems given in question 1(iii). Make sure you define completely the functions
f* and G.

In this case the
odd extension of f(x) is

f*(x) = 1 – cos(2πx)
if 0 < x < 1

= cos(2πx) – 1 if -1 < x <
0

= f*(x+2) otherwise

and the odd
extension of g is

g*(x) = -(1+x)/6
if -1 < x < -0.4,

= x/4 if -0.4 < x < 0.4,

= (1-x)/6 if 0.4 < x < 1

= g*(x+2) otherwise

and

_{}

Note that G(x) is
even, i.e. G(x) = G(-x) since g*(x) is an odd function.

3.(a) Solve the one-dimensional heat equation with
the conditions given below. Take L = 1
and c = 1.

(i) u(0,t) = 20,
u(1,t) = 10, u(x,0) = 20 + 20x – 30x^{2};

From equations
(13) and (14) of section 3.5 the steady-state solution is u(x) = 20 - 10x and

_{}

Thus b_{2k}=
0 and b_{2k+1 }= 240/((2k+1) p)^{3} so that the solution is

_{}

(ii) u_{x}(0,t)
= 0, u_{x}(1,t) = 0, u(x,0) = 20 + 20x -
30x^{2};

From equations
(4) and (5) of section 3.6 we have

_{}

_{}

Thus a_{2k}
= -120/(2kπ)^{2} and a_{2k+1} = 40/((2k+1)π)^{2}
and the solution is

_{}

(b) Plot the solution to each of the above
problems for a sequence of different times.
Choose times to show the behaviour of the temperature from its initial
values to times approaching the steady state solution.

See MAPLE output.

4.(a) Prove the orthogonality relation

_{}

for a ¹ b provided tan(a) = -ka, tan(b) = -kb.

_{}

since tan(a) =
-ka, tan(b) = -kb.

(b) Use the MAPLE function fsolve to find the
first five positive solutions of tan(x) = -x.
Set the range option to ensure you get the desired root in each case.

See MAPLE output.

5.(a) Find the solution for a vibrating, square
membrane with c = 1 whose edges are of length unity and whose initial shape is
given by the function f(x,y) = sin(px) sin(2py)

The edges of the
membrane are held fixed and the membrane is initially at rest.

The coefficients
of the double Fourier sine series for f(x,y) are

_{}

From the
orthogonality properties of the sine functions this is zero unless m = 1 and n
= 2 in which case A_{12} = 1 so that the solution is u(x,t) = sin(px) sin(2py) cos(Ö5pt) since

λ_{12}
= (1 + 4)^{1/2}.

(b) Find the temperature of a square plate with c
=1 with sides of length unity if the edges are kept a zero degrees and the
original temperature is given by f(x,y) = sin(2px) sin(py)

The Fourier coefficients
are

_{}

From the
orthogonality properties of the sine functions this is zero unless m = 2 and n
= 1 in which case A_{21} = 1 so that the solution is u(x,t) = sin(2px) sin(py) exp(-5π^{2}t) since

λ_{21}
= (4 + 1)^{1/2}.

(c) Plot the solutions of part (a) for several
values of t. Use a separate plot for
each value of t.

See MAPLE output.

6. Derive the solution of the two-dimensional
heat equation for a rectangular plate if one pair of opposite edges is kept at
zero degrees while the other pair of opposite edges is insulated so that no
heat can escape.

We take the
rectangular plate to lie in 0 < x < a, 0 < y < b. Thus we have to solve the two-dimensional
heat equation subject to the boundary conditions

u(0,y,t) = 0,
u(a,y,t) = 0 (the edges x = 0 and x = a are kept at zero degrees)

u_{y}(x,0,t)
= 0 and u_{y}(x,b,t) = 0 (the edges y = 0 and y = b are insulated so
there is no temperature gradient in the y-direction)

Using separation
of variables we write u(x,y,t) = X(x)Y(y)T(t) and as in the original derivation
we get:

1) X"/X = -m^{2} with solution X
= Acos(mx) + Bsin(mx).

Applying the boundary
conditions X(0) = X(a) = 0 requires A = 0 and m = mπ/a giving the normal modes X(x) = sin(mpx/a);

2) Y"/Y =
-υ^{2} with solutions Y = Ccos(υy) + Dsin(υy)

Thus Y' = - υCsin(υy)
+ υDcos(υy)

Applying the
boundary conditions Y'(0) = Y'(b) = 0 requires D = 0 and υ = nπ/b
giving the normal modes Y(y) = cos(nπy/b);

3) T' = -c^{2}(m^{2} + υ^{2})T
with normal modes T = exp(-λ_{mn}^{2}t) where λ_{mn}^{2}
= c^{2}π^{2}[(m/a)^{2} + (n/b)^{2}].

Thus the solution
is

_{}_{}

If the initial
condition is that u(x,y,0) = f(x,y) then this gives the double Fourier series

_{}

where the
coefficients are given by

_{}

provided n
≠ 0. If n = 0 the coefficient becomes

_{}

If you chose to
insulate the edges x = 0 and x = a the derivation is similar with x and y
interchanged giving

_{}

where the
coefficients are given by

_{}

provided m ≠
0. If
m = 0 the coefficient becomes

_{}