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 – 30x²;

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 - 30x²;

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 A₁₂ = 1 so that the solution is u(x,t) = sin(px) sin(2py) cos(Ö5pt) since

λ₁₂ = (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 A₂₁ = 1 so that the solution is u(x,t) = sin(2px) sin(py) exp(-5π²t) since

λ₂₁ = (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 = -m² 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 = -υ² 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' = -c²(m² + υ²)T with normal modes T = exp(-λ_mn²t) where λ_mn² = c²π²[(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