Week4 5 self study - Problems PDF

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MATH0056: Mathematical Methods 4

Self-study exercises for Weeks 4 and 5 with solutions

1. Exam 2019 Q3. (Self-study for week 4 material) This question gives an alternative proof of the recurrence relation for Legendre polynomials without the use of generating functions. (a) Let ak,n denote the coefficient in xk in the monomial expansion of Pn (x), i.e. Pn (x) = an,n xn + an−1,n xn−1 + · · · + a0,n . Using Rodrigues’ formula, show that the leading coefficient an,n is given by an,n =

(2n)! 2n (n!)2

.

(b) For each n ≥ 0, show that the polynomial xPn (x) of degree n + 1 can be written as n+1 Pn+1 (x) + Q(x), xPn (x) = 2n + 1 where Q(x) is some polynomial of degree at most n. (c) To determine Q(x), show that Q(x) is orthogonal to Pj (x) for all j ≤ n − 2. Conclude that there exists constants A and B such that Q(x) = APn (x) + BPn−1 (x) Using the normalisation condition Pn (1) = 1 for all n ≥ 0, and the fact that R1 n . Deduce from this the x|Pn (x)|2 dx = 0, show that A = 0 and B = 2n+1 −1 recurrence relation (n + 1)Pn+1 (x) = (2n + 1)xPn (x) − nPn−1 (x) 2. Self-study exercise for week 5 material (a) Solve the eigenvalue problems Ly = −

d2 y = λy dx2

for two following sets of boundary conditions (with L > 0 constant) i. y(0) = 0 and y(L) = 0 ii. y(0) = 0 and y ′ (L)=0

This example illustrates how changes in the boundary conditions can influence the eigenvalues/functions. (b) Consider the differential operator L defined by   dy 1 d x Ly = , x dx dx which is related to Bessel’s equation. Show that L is self-adjoint on the space of smooth functions defined on (0, 1) that vanish at x = 1 with respect to the inner-product h·, ·ix defined by Z 1 xf (x)g(x)dx. hf, gix = 0

SOLUTIONS 1. Exam 2019 Q3 (a) We first expand (x2 − 1)n using the Binomial formula n   X n (−1)n−k x2k , (x − 1) = k 2

n

k=0

and therefore deduce that n   1 dn X n (−1)n−k x2k Pn (x) = n 2 n! dxn k=0 k n   1 X n = n (−1)n−k (2k)(2k − 1)(2k − 2) . . . (2k − n + 1)x2k−n . 2 n! k=0 k The term in the sum that leads to the highest coefficient is for k = n, i.e.   (2n)! n 1 (2n)! 1 = n an,n = n (−1)0 (2n)(2n − 1) . . . (n + 1) = n . 2 n! n 2 n! n! 2 (n!)2 (b) To show that Q(x) is a polynomial of degree n, we check the coefficient associated to xn+1 in the monomial expansion of Q(x). In particular, we get an,n −

n+1 (2n)! n + 1 (2n + 2)! an+1,n+1 = n − 2 2n + 1 2n+1 (n + 1)!2 (2n + 1) 2 (n!) (2n)! n + 1 2(n + 1)(2n + 1)(2n)! = n = 0. − 2 2n+1 (n + 1)!2 2n + 1 2 (n!)

Therefore Q(x) is a polynomial of degree at most n. R1 (c) For every j ≤ n − 2, −1 Pj xPn dx = 0 since xPj is then a poly of degree at most n − 1 and Pn is orthogonal to all polynomials of degree less than n. Similarly Pn+1 is orthogonal to all polys of degree at most n, so Q is orthogonal to all polys of degree at most n − 2. It follows then that Q(x) = APn (x) + BPn−1 (x) for some A and B. Inserting Pn (1) = 1 we see that 1=

n+1 + A + B, 2n + 1

and we have 2n + 1 A= 2 n+1 = so B = 1 − 2n+1 relation for Pn+1 (x).

Z

1

2n + 1 Q(x)Pn (x)dx = 2 −1

n . 2n+1

Z

1

xPn (x)2 dx = 0, −1

Re-arranging the above then yields the recurrence

2. (a) Applying the results from the handout on the sign of eigenvalues, we find that the eigenvalues for each these problems are necessarily non-negative, i.e. λ ≥ 0. In both cases, we obtain from √ the equation and the condition y(0) = 0 solutions of the form y(x) = A sin( λx), which are nontrivial only for λ > 0. The boundary condition at x = L then determines the eigenvalues and eigenfunctions as follows: √ i. The condition y(L) = 0 implies λL = kπ, k ≥ 1 integer. Therefore we get eigenvalues and eigenfunctions  2 p kπ , yk (x) = sin( λk x) λk = L √ √ ii. The condition y ′ (L) = 0 implies cos( λL) = 0 so λL = (k + 1/2) π for k ≥ 1 integer, therefore   p (2k + 1)π 2 , yk (x) = sin( λk x). λk = 2 This shows how in general the eigenvalues/functions are therefore dependant on the boundary conditions as well as the differential operator. (b) We consider two smooth functions f and g that vanish at x = 1 i.e. f (1) = g(1) = 0, and compute     Z 1 Z 1 d df df 1 d hLf, gix = x gdx x x gdx = dx dx x dx 0 dx 0 Z 1 Z 1 d ′ ′ ′ 1 ′ 1 xf g dx = − [xfg ] 0 + = [xf g]0 − f (xg ′ ) dx | {z } | {z } dx 0 0 =0 as g(1)=0

=

Z

0

1

xf

=0 as f (1)=0

1 d (xg ′ ) dx = hf, Lgix x dx

so L is then self-adjoint with respect to h·, ·ix for smooth functions f and g that vanish at x = 1....