Math2221 Test 2B S2, 2014

Time allowed: 50 minutes

1. Solve the boundary-value problem

2x2u + 3xu 3u = 14x2 for 1 < x < 4, with u(1) = 9 and u(4) = 29.

2. Find the general solution of

u 4u + 4u =e2x

x.

You may assume the following formulae for the method of variation ofparameters applied to the ODE u + p(x)u + q(x)u = f(x):

u = u1v1 + u2v2, v 1 = fu2

W, v 2 =

fu1

W.

3. For the differential operator

Lu = xu ex2u + 4x2u

find the coefficients b0(x), b1(x), b2(x) of the adjoint operator

Lv = b2v + b1v + b0v.

4. The Bessel function of order n {0, 1, 2, . . .} is defined by the series

Jn(z) =

k=0

(1)k(z/2)2k+n

k!(n+ k)!.

Find the coefficient Cn in the identity

zJn1(z) + zJn+1(z) = CnJn(z)

for z C and n {1, 2, 3, . . .}.

Questions continue over the page.

1

5. Let f,g be an inner product with associated norm f =f, f. Findf,gwhen f and g are such that

f = 1, g = 3, 3f+ g = 4.

6. Consider the ODE

3z2u + z(1 z)u u = 0.

(i) Find the indicial equation for r and the recurrence relation for Ak if

u =

k=0

Akzk+r.

(ii) Which root of the indicial equation yields a solution u that is boundedat z = 0?

(iii) For this root, find A1 and A2 given that A0 = 1.

7. Show that if the boundary-value problem

u + u = f(x) for 0 < x < 3pi/4,u = 0 at x = 0,

u + u = 0 at x = 3pi/4,

has a solution u, then the function fmust satisfy 3pi/40

f(x) sin xdx = 0.

2

Solutions

1. [8 marks] Let Lu = 2x2u + 3xu 3u and observe that

Lxr =[2r(r 1) + 3r 3

]xr = (2r2 + r 3)xr = (2r+ 3)(r 1)xr,

so the general solution of the homogeneous equation Lu = 0 is

uH = Ax3/2 + Bx.

Also, Lx2 = 7x2 so a particular solution of Lu = 14x2 is uP = 2x2. Hence,the general solution of Lu = 14x2 is

u = uH + uP = Ax3/2 + Bx+ 2x2.

Since u(1) = A+B+2 and u(4) = A/8+4B+32, the boundary conditionsgive

A+ B+ 2 = 9,A/8 + 4B+ 32 = 29,

and thusA+ B = 7,

A/8 + 4B = 3.

Therefore A = 8 and B = 1 and the solution is

u = 8x3/2 x+ 2x2.

2. [8 marks] Since 2 4 + 4 = ( 2)2 the functions u1 = e2x andu2 = xe

2x are linearly independent solutions of the homogeneous equa-tion. Their Wronskian is

W =

e2x xe2x2e2x e2x + 2xe2x = e4x

so applying the formulae with f(x) = e2x/xwe have

v 1 = (e2x/x) xe2x

e4x= 1 and v 2 =

(e2x/x) e2x

e4x= x1.

Thus, v1 = x and v2 = log |x|, implying that

uP = u1v1 + u2v2 = e2x(x) + xe2x log |x| = e2x(x+ log |x|)

is a particular solution, and hence the general solution (with C = B 1) is

u = uH + uP = Ae2x + Bxe2x + uP = e

2x(A+ Cx+ x log |x|).3

3. [3 marks] The definition of L gives

Lv = (xv) + (ex2v) + 4x2v = (2v + xv ) + (2xex

2v+ ex

2v ) + 4x2v

= xv +(2 + ex

2)v +

(4x2 2xex

2)v,

showing that

b2 = x, b1 = 2 + ex2, b0 = 2x(2x ex

2).

4. [6 marks] Since

zJn1(z) = z

k=0

(1)k(z/2)2k+n1

k!(n 1 + k)!=

k=0

(1)kz2k+n22kn+1

k!(n 1 + k)!

=

k=0

(1)k(z/2)2k+n

k!(n+ k)!2(n+ k)

and

zJn+1(z) = z

k=0

(1)k(z/2)2k+n+1

k!(n+ 1 + k)!=

k=0

(1)kz2k+n+222kn1

k!(n+ 1 + k)!

=

k=1

(1)k1z2k+n22kn+1

(k 1)!(n+ k)!=

k=1

(1)k(z/2)2k+n

k!(n+ k)!(2k),

we have

zJn+1(z) + zJn1(z) = 2n(z/2)n

n!+

k=1

(1)k(z/2)2k+n

k!(n+ k)![2(n+ k) 2k

]= 2n

k=0

(1)k(z/2)2k+n

k!(n+ k)!= 2nJn(z),

so C = 2n. The identity holds for any integer n, not just n > 1, if we agreethat 1/k! = 0 for k {1,2,3, . . .}.

5. [4 marks] Since

3f+g2 = 3f+g, 3f+g = 3f2+23f,g+g2 = 9f2+6f,g+g2

we have16 = 9 + 6f,g+ 9

so 6f,g = 2 and thus f,g = 1/3.

4

6.

(i) [7 marks]

3z2u + z(1 z)u u = (3z2u + zu u) z2u

=

k=0

[3(k+ r)(k+ r 1) + (k+ r) 1

]Akz

k+r

k=0

(k+ r)Akzk+r+1

=

k=0

[3(k+ r) + 1](k+ r 1)Akzk+r k=1

(k 1 + r)Ak1zk+r

= (3r+ 1)(r 1)A0zr

+

k=1

[(3k+ 3r+ 1)(k+ r 1)Ak (k+ r 1)Ak1

]zk+r,

so the indicial equation is (3r + 1)(r 1) = 0 and the recurrencerelation is

Ak =Ak1

3k+ 3r+ 1for all k > 1.

(ii) [1 mark] The roots of the indicial equation are r1 = 1 and r2 = 1/3.Since r1 > 0 but r2 < 0, we need to choose r = r1 to obtain a solutionthat is bounded at z = 0.

(iii) [3 marks] With r = 1,

Ak =Ak1

3k+ 4for all k > 1.

Thus, putting A0 = 1 gives

A1 =A0

7=

17

and A2 =A1

10=

170

,

so

u = z

(1 +

z

7+z2

70+

).

7. [5 marks] First consider the homogeneous problem

v + v = 0 for 0 < x < 3pi/4,v = 0 at x = 0,

v + v = 0 at x = 3pi/4.(1)

5

The general solution of v +v = 0 is v = A cos x+B sin x and since v(0) = Awe see that A = 0. Thus, v = B sin x and v = B cos x, implying that whenx = 3pi/4,

v + v = B cos3pi4+ B sin

3pi4= B

2

2+ B

2

2= 0

for every B. Thus, v = B sin x is the general solution of the homogeneousproblem (1). The operator Lu = u + u is formally self-adjoint so

Lu, v = u,Lv for all u, v D,

where

f,g = 3pi/4

0f(x)g(x)dx

andD, the domain of L, is defined to be the set of allC2 functions satisfyingthe homogeneous boundary conditions. Hence, if a solution u exists thenfor any constant B,

f, v = Lu, v = u,Lv = u, 0 = 0 when v = B sin x,

and in particular, with B = 1, 3pi/40

f(x) sin xdx = 0.

6