MATH3271

FW03

Solutions to term test 1

The total number of marks on this paper is 60 but grades will be calculated out of 50.


Note the following properties of periodic functions.


If f(x) is T-periodic then it is nT-periodic where n is a positive integer.

If f(x) and g(x) are both T-periodic then so is their sum, difference, product and quotient.

is independent of a.

1. (6 marks)

(a) What is the period of the function cos(x+p) + cos(3x)?

cos(x+p) = -cos(x) is 2p -periodic

cos(3x) is 2p/3 -periodic and hence 2p -periodic.

Therefore cos(x+p) + cos(3x) is 2p -periodic.


This function is even since cosine is an even function.


(b) What is the period of the function cos(2px)sin(5px)?

cos(2px) has period 1and hence is 2-periodic

sin(5px) has period 2/5 and hence is 2-periodic

Therefore cos(2px)sin(5px) is 2-periodic.


This function is odd since cosine is even and sine id odd.


2. (4 marks)

(a) Write down the appropriate boundary conditions for the heat equation for a bar of length L when one end is kept at 0 degrees while the other end is allowed to radiate.

Boundary condition for the end kept at zero degrees is u(x,0) = 0

Boundary condition for the radiating end is

(b) Explain briefly what is meant by the Gibbs phenomenon with regard to Fourier series.

A finite Fourier series overshoots a function near a singularity.




3. (9 marks) Suppose that f(x) is a 2p-periodic function with Fourier coefficients a0, an and bn, n = 1, ¥. If the function F(x) = f(x+p) has Fourier coefficients A0, An and Bn n = 1, ¥, show that A0 = a0, An = (-1)nan.

From the formula sheet

where we have made the substitution s = x+p and used the property of the integral of a periodic function stated above to justify the change of limits. Similarly

using the same substitution as above and the fact that

cos(np(s-p)/p) = cos( nps/p -np) = (-1)n cos( nps/p)


4. (8 marks) Find the half-range sine expansion for cos(2px) for 0 < x < 1.

The half-range sine expansion is given by

where

Thus b2k = 0 while

Note that when n = 2 the second term in the integral above is zero but this gives the same result.

5. (7 marks) Define completely the function f*(x) and G(x) in d'Alembert's solution of the one-dimensional wave equation for 0 < x < 1 with initial conditions

u(x, 0) = 2x + x2

ut(x, 0) = cos(3px).

f* is the odd extension of 2x + x2 and must be periodic of period 2. Thus

f*(x) = 2x + x2 if 0 < x <1; 2x - x2 if -1 < x < 0 and f*(x+2) otherwise.

Similarly g*(x) = cos(3px) if 0 < x <1; cos(3px) if -1 < x < 0 and g*(x+2) otherwise.

Now taking a = 1 in the definition of G we have

and G(x) = G(x+2) otherwise.


6. (13 marks) Use separation of variables to derive the solution of the one-dimensional wave equation

subject to the boundary conditions u(0, t) = 0, u(p, t) = 0

and the initial conditions u(x, 0) = 3 sin(2x), ut(x, 0) = 5 sin(3x).

Give explicit expressions for the Fourier coefficients in the solution.


See the text for the derivation of the separated solution. Now for L = p

so that applying the initial conditions we have

Thus an = 0 if n ¹ 2 while a2 = 3. Also

Thus bn = 0 if n¹ 3 while b3 = 5/3c giving u(x,t) = 3 cos(2ct) sin(2x) +(5/3c) sin(3ct) sin(3x).

7. (13 marks) Use separation of variables to derive the solution of the two-dimensional heat equation

for a rectangular plate if the edges are kept at zero degrees and the initial temperature distribution is given by f(x, y) = 12 sin(2px/a)sin(4py/b), 0 < x < a, 0 < y < b.


See the text for the derivation of the separated solution

where

The initial condition gives

Thus Amn = 0 unless m = 2 and n = 4 when A24 = 12 giving

u(x, y, t) = 12 sin(2p x/a) sin(4p y/b) exp(-l242 t)