2004 final solutions

7/24/2019 2004 Final Solutions

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MATH 300 Fall 2004

Advanced Boundary Value Problems I

Solutions to Sample Final Exam

Friday December 3, 2004

Department of Mathematical and Statistical Sciences

University of Alberta

Question 1. Given the functionf(x) = cos ax, 0

x < a

find the Fourier sine series for f .

Solution:

Writingf(x) = cos ax n=1

bnsinna x,the coefficientsbn in the Fourier sine series are computed as follows:

bn= 2

a

a0

cos ax sinna x dx=

1

a

a0

sin (n+1)a x+ sin

(n1)a x

dx

= 1

1n+ 1

cos (n+1)a x

a

0

+

1

1n 1cos

(n1)a x

a

0

= 1

(n+ 1)

(1)n

+ 1

+ 1

(n 1)

(1)n

+ 1

=1 + (

1)n

1

n+ 1+ 1

n 1

.

Therefore,

bn=

4n

(n2 1) if n is even

0 if n is odd, n 3.Ifn = 1,

b1 = 2

a

a0

sin ax cosax dx=

1

asin2 ax

a

0

= 0.

The Fourier sine series for f is therefore

cos ax 8n=1

n4n2 1sin

2na x.

for 0 x < a.


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Question 2. Let

f(x) =

cos x |x| < ,

0 |x| > .

(a) Find the Fourier integral off .

(b) For which values ofx does the integral converge to f(x)?

(c) Evaluate the integral

0

sin cos x

1 2 d

for < x < .Solution:

(a) The function

f(x) =

cos x |x| <

0 |x| > is even, piecewise smooth, and is continuous at every x (,) except at x =, therefore fromDirichlets theorem the Fourier integral representation offconverges to f(x) for all x = , and

f(x)

0

A()cos x d,

where

A() = 2

0

f(x)cos xdx= 2

0

cos x cos xdx

= 1

0 cos(+ 1)x+ cos( 1)x dx=

1

sin(+ 1)x

+ 1

0

+sin( 1)x

1

0

= 1

sin(+ 1)

+ 1 +

1

sin( 1) 1

=2

sin

1 2 .

The Fourier integral representation off is therefore

f(x) 2

0

sin cos x

1

2 d.

(b) From Dirichlets theorem, the integral converges tof(x) for allx = ,and converges to 12 forx = .(c) Therefore, we have

0

sin cos x

1 2 d=

2 cos x for |x| < ,

0 for |x| > ,4 for x= .


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Question 3. Let Fc denote the Fourier cosine transform and Fs denote the Fourier sine transform. Assumethat f(x) and xf(x) are both integrable.

(a) Show that

Fc(xf(x)) =

d

dFs(f(x)).

(b) Show that

Fs(xf(x)) = ddFc(f(x)).

Solution:

(a) From the definition of the Fourier sine transform, we have

d

dFs(f(x)) = d

d

2

0

f(t)sin tdt

,

and differentiating under the integral sign,

d

dFs(f(x)) =

2

0

f(t) d

d(sin t) dt

=

2

0

tf(t)cos tdt

= Fc(xf(x)),

and therefored

dFs(f(x)) = Fc(xf(x))

as required.(b) From the definition of the Fourier cosine transform, we have

d

dFc(f(x)) = d

d

2

0

f(t)cos tdt

,

and differentiating under the integral sign,

d

dFc(f(x)) =

2

0

f(t) d

d(cos t) dt

=

2

0

tf(t)sin tdt

= Fs(xf(x)),

and therefored

dFc(f(x)) = Fs(xf(x))

as required.


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Question 4. Chebyshevs differential equation reads

(1 x2)y xy +y = 0, 1< x


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(c) In the integral 11

Tm(x)Tn(x)

(1 x2)1/2 dx

make the substitution x = cos t,so that

dx= sin t dt= (1 cos2t)1/2 dt= (1 x2)1/2 dt

that is,

dt= 1(1 x2)1/2 dx.

Therefore, 11

Tm(x)Tn(x)

(1 x2)1/2 dx= 0

cos mt cos ntdt= 0

if m= n, and the Chebyshev polynomials are orthogonal on the interval [1, 1] with respect to the

weight function w(x) =

1

(1 x2)1/2 .

Question 5. Solve the following initial value problem for the damped wave equation

2u

t2 + 2

u

t +u=

2u

x2

u(x, 0) = 1

1 +x2,

u

t(x, 0) = 1.

Hint:Do not use separation, instead consider

w(

x, t) =

et

u

(x, t

).Solution: Note that u(x, t) = et w(x, t), so that

2u

x2 =et

2w

x2

andu

t = etw+et w

t

and2u

t2 =etw 2et w

t +et

2w

t2 .

Therefore,2u

t2 + 2

u

t +u= et

2w

t2,

while2u

x2 =et

2w

x2

and ifu is a solution to the original partial differential equation, then w is a solution to the equation

et

2w

t2

2w

x2

= 0,


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and since et = 0, then w satisfies the initial value problem

2w

t2 =

2w

x2, < x < , t >0,

w(x, 0) = 11 +x2

,

w

t(x, 0) = 1.

From DAlemberts equation to the wave equation, we have (since c= 1)

w(x, t) = 1

2

1

1 + (x+t)2+

1

1 + (x t)2

+1

2

x+txt

1 ds,

so that

u(x, t) = et

2 1

1 + (x+t)2+

1

1 + (x

t)2 +tet,for < x < , t 0.

2004 final solutions

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