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Exercises 1

Question 1. Use d'Alembert's solution to solve to the IVPs

(a) utt = uxx, u(x, 0) = 0, ut(x, 0) = 1/(1 + x2)

                                       1   |x| ≤ 1
(b) utt = uxx, u(x, 0) =                                ut(x, 0) = 0.
                                        0   otherwise,

Sketch the solution u(x, t) at t = 0 and several times t > 0.

Question 2. By a similar approach to that used in obtaining d'Alembert's solution, find the solution to

utt = c2uxx, u(x, x) = f (x), ut(x, x) = g(x),

in which conditions are given on the line t = x.

Question 3. Verify that the two-dimensional Laplacian ? for polar coordinates (r, θ), where x = r cos θ, y = r sin θ, is

? = ∂2/∂r2 + 1/r.∂/∂r + 1/r22/∂θ2 .

Question 4. Using the result Γ( 1/2 ) = √π in the series expression for J1/2 (x), show that J1/2(x) = √2/πx. sinx and find a corresponding expression for J-1/2 (x).

Question 5. Consider the series expansion for Jν (x) in the case that ν is an integer. Prove that for integer index n,

J-n(x) = Jn(-x) = (-1)nJn(x).

Question 6. Prove (a)

d/dx(xν Jv (x)) = xv Jv-1(x),      d/dx(x-vJv(x)) = x Jν+1(x)

and (b) deduce that

J'ν (x) = 1/2 (Jν-1(x) - Jν+1(x)),  Jν (x) = x/2ν (Jν-1(x) + Jν+1(x)).

Question 7. Prove that

G(x, t) := ex/2 (t-t-1) = ∑n=-∞ Jn(x)tn.

G(x, t) is a generating function for Bessel functions of integer order.

(a) By using the invariance of G(x, t) under certain transformations of x and t, deduce the results of Q5.

(b) By an appropriate choice of t, deduce from (*) that

cos(x sin θ) = J0(x) + 2∑k=1 J2k(x) cos(2kθ),

and


sin(x sin θ) = 2∑k=0 J2k+1(x) sin((2k + 1)θ).

Question 8. Prove by induction that for all non-negative integers n,

Jn(x) = (-x)n(1/x.d/dx)n J0(x).

Question 9. You are given that for ν > -1, all zeros of Jν are real. Using Rolle's theorem and the identities in Q6 (a), prove that for ν > -1 there is a zero of Jν between and any two positive zeros of Jν+1 and vice versa. Deduce that between consecutive positive zeros of Jν there is exactly one zero of Jν+1. (It is said that the zeros of Jν and Jν+1 are interlaced.)

Question 10. Let ν > -1. You are given that xJ'ν (x) + hJν (x) has infinitely many positive zeros. Show that if λn is the nth such zero then

01rJνmr)Jνnr) dr = 1/2 δmn[1 + (h2 - ν2)/λ2n)] Jνn)2.

The Fourier-Dini expansion of f on [0, 1] is the expression

f (r) = ∑n=1AnJνnr).

Show that

An = 2λn2/ (λn2 + h2 - ν2)Jνn)2 01rf(r)Jνnr) dr.

Question 11. A semi-infinite cylinder of conducting material r ≤ a, z ≥ 0 has temperature T such that T = T1 on z = 0 and T → T0 at z → ∞. Outside the curved surface the temperature is also T0 and Newton's law of cooling gives the boundary condition

∂T/∂r = -c(T - T0), on r = a,

where c is a positive constant.

The steady temperature distribution is an axially symmetric (i.e. θ-independent) solution of Laplace's equation. Show that it is given by

T(r, z) = T0 + 2(T1 - T0) Σn=1nJ1n)J0nr/a)enz/a)/(λn2 + a2c2)J0n)2)

where λn is the nth positive zero of xJ'0(x) + acJ0(x).

Exercises 2

Consider Bessel's equation with ν = 1/2,

y'' + 1/x.y' + (1 - 1/4x2) y = 0. (*)

You should find the general solution in two ways:

Question 1. (Easier) Find and solve the second order ODE satisfied by new dependent variable z = √x.y. Hence find the general solution of (*).

Question 2. (Harder) Use the method of Frobenius to find a series solution ( Σn=0 anxn-1/2) in which the first two coefficients (a0 and a1) are arbitrary. Find a closed form expression for this solution (i.e. in terms of standard functions not infinite series).

Compare the two answers you get. (They should be the same of course.)

Exercises 3

Consider a semi-infinite conducting cylinder of radius a with the closed end at z = 0. The temperature at z = 0 is maintained at T1 and the curved sides are insulated. The temperature T → T0 as z → ∞. The steady state temperature T in the interior of the cylinder satisfies Laplace's equation. Assuming that. T does not depend on the polar angle θ, write down the boundary value problem that determines T.

Show that T(r, z) = T0 + ∑m=1 Amemz/a Jomr/a),

where λm is the mth zero of J'0 and Am are constants. Find an expression for Am in terms of T0, T1, λm and J0m) and J1m).

Exercises 4

Question 1. Use the base cases P0(x) = 1, P1(x) = x and the recurrence formula

(n + 1)Pn+1(x) = (2n + 1)xPn(x) - nPn-1(x)

for n ≥ 1, to derive the Legendre polynomials Pn(x) for 2 ≤ n ≤ 4.

Question 2. Determine all non-zero associated Legendre functions P n m (x) for m ≤ n ≤ 3.

Question 3. By considering the generating function G(x, t) = (1-2xt+t2)-1/2 = Σn=0 Pn(x)tn, prove that for all n ≥ 0, Pn(-x) = (-1)nPn(x) and Pn(1) = 1.

Question 4. For all integers n ≥ 0 the Legendre polynomial Pn(x) satisfies Legendre's equation (1 - x2)y'' - 2xy' + n(n + 1)y = 0. Use this fact to prove that for any n ≥ 0, P'n(1) = 1/2n(n + 1) and P'n(-1) = (-1)n1/2n(n + 1).

Exercises 5

Prove that (guided by the steps indicated below)

(n + 1)Pn+1 - (2n + 1)xPn(x) + nPn-1 = 0, (*)
for n ≥ 0.

Step 1 The polynomial xPn(x) has degree n + 1. Explain why it may expressed as a linear combination of Legendre polynomials Pm(x) where m is odd/even when n is even/odd.

Step 2 Further, show that xPn(x) is a linear combination of just two Legendre polynomials aPn+1(x) + bPn-1(x) where a and b are constants to be determined.

Step 3 Evaluate xPn(x) and its derivative at x = 1 and solve the resulting simultaneous equations for a and b.

Step 4 Obtain the relation (*).

Exercises 6

Consider the IVP

ut = (σx2u), u(x,0) = {x   0 < x ≤ 1

                                 {0   x > 1 
where σ > 0 is a constant and x > 0. Find the changes of variable which transform this into an IVP for the heat equation. Hence solve the IVP for u(x, t).

Exercises 7

Find the solutions of the Cauchy problems with u(x, 0) = f (x) and ut(x, 0) = g(x) where

Question (i) f(x) = 1/(1+ x2 + y2 + z2), g(x) = 0,

Question (ii) f(x) = 0, g(x) = 1/(1 + x2 + y2 + z2)

Question (iii) f(x) and g(x) are arbitrary but depend only on r.

Exercises 8

Question 1) Find the solution of the problem for u(x^, 0) = 0 and ut(x^, 0 ) = xy. The solution must be xyt as computed via the 2D formula.

Question 2) Likewise, compare the solution for u(x^, 0) = z2 ut(x^, 0) = 0 and verify that it needs u(x^, t ) = z2 + c2t2

Question 3) assume u(x^, 0) = x2 + y2 + z2 and compare the solution of obtained by means of the 3D solution with that computed radial symmetry

U(n, t) = 1/2h ((h-ct) f(h - ct) + (h + cr)f(h + ct)) as g = 0.

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