Final Exam Solution Fall 2003 - Advanced Engineering Mathematics | MATH 401, Exams of Mathematics

Material Type: Exam; Professor: Fulling; Class: ADV ENGINEERING MATH; Subject: MATHEMATICS; University: Texas A&M University; Term: Unknown 1989;

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Math. 401 (Fulling) 16 December 2003
Final Examination Solutions
1. (20 pts.) Classify each of the following equations as linear homogeneous, linear inhomoge-
neous, or nonlinear. Also, classify each as elliptic, parabolic, or hyperbolic.
(a) 2u
∂x2+2u
∂y2=3u
2
.Ans.: nonlinear, elliptic
(b) ∂u
∂t =2u
∂x2x2u.Ans.: linear homogeneous, parabolic
2. (30 pts.) Find the eigenfunctions for
y00 +λy =0,y(0) = 0,y
0
(π)=2y(π),
find the eigenvalues approximately by sketching a certain graph, and show how to expand
an arbitrary function f(x) on the interval 0 <x<πas a series in these eigenfunctions.
On general principles we know that λwill be positive, λ=ω2, at least in most cases, to give oscillatory
eigenfunctions. In that case, since y(0) = 0, we have
y(x)=sin(ωx).
The second boundary condition yields ωcos(ωπ)=2sin(ωπ), or
tan(ωπ)=ω
2.
0.5 1.0 1.5 2.0 2.5 3.0
1.5
1.0
0.5
ω
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Sketching the graphs (and recalling that 1=tan3π
4=tan7π
4=···)weseethatω
1.9, ω21.8,
ω32.7, ..., and, in general, ωnis slightly greater than (and increasingly close to) n1
2(the
location of the vertical asymptote of tan(πω)).
Assume for the moment that there are no negative or zero eigenvlues. Let yn(x)=sin(ω
n
x).
Then
f(x)=
X
n=1
cnyn(x),
where, by the orthogonality theorem for Sturm–Liouville problems,
cn=Rπ
0yn(x)f(x)dx
Rπ
0yn(x)2dx .
pf3
pf4
pf5

Partial preview of the text

Download Final Exam Solution Fall 2003 - Advanced Engineering Mathematics | MATH 401 and more Exams Mathematics in PDF only on Docsity!

Math. 401 (Fulling) 16 December 2003

Final Examination – Solutions

1. (20 pts.) Classify each of the following equations as linear homogeneous, linear inhomoge-

neous, or nonlinear. Also, classify each as elliptic, parabolic, or hyperbolic.

(a)

u

∂x

u

∂y

= 3u

. Ans.: nonlinear, elliptic

(b)

∂u

∂t

u

∂x

− x

u. Ans.: linear homogeneous, parabolic

2. (30 pts.) Find the eigenfunctions for

y

+ λy = 0, y(0) = 0, y

(π) = − 2 y(π),

find the eigenvalues approximately by sketching a certain graph, and show how to expand

an arbitrary function f (x) on the interval 0 < x < π as a series in these eigenfunctions.

On general principles we know that λ will be positive, λ = ω

, at least in most cases, to give oscillatory

eigenfunctions. In that case, since y(0) = 0, we have

y(x) = sin(ωx).

The second boundary condition yields ω cos(ωπ) = −2 sin(ωπ), or

tan(ωπ) = −

Sketching the graphs (and recalling that −1 = tan

= tan

= · · ·) we see that ω

≈ 2 .7,... , and, in general, ω

n

is slightly greater than (and increasingly close to) n −

(the

location of the vertical asymptote of tan(πω)).

Assume for the moment that there are no negative or zero eigenvlues. Let y

n

(x) = sin(ω

n

x).

Then

f (x) =

n=

c

n

y

n

(x),

where, by the orthogonality theorem for Sturm–Liouville problems,

c

n

y

n

(x)f (x) dx

y

n

(x)

dx

Alternatively, define normalized eigenfunctions

φ

n

(x) =

y

n

(x)

y

n

(x)

dx

,

and then

f (x) =

n=

c

n

φ

n

(x), c

n

=

φ

n

(x)f (x) dx.

Hidden extra credit:

(1) Evaluate the integral:

sin

n

x) dx =

1

2

(1 − cos(2ω

n

x)) dx

=

1

2

[

x −

1

2 ω

n

sin(2ω

n

x)

]

=

1

2

(

π −

1

2 ω

n

sin(2ω

n

π)

)

.

As n → ∞ this approaches

, as it should.

(2) Check that there are no zero or negative eigenvalues. This can be done by integration by parts (the

Rayleigh quotient is positive), or directly like this: For λ = 0 we have y(x) = Ax + B = Ax (by

the first boundary condition), and now the second boundary condition forces A = 0. For λ = −κ

we have similarly y(x) = C sinh(κx) and the second boundary condition can’t be satisfied (for

C nonzero). (If the second boundary condition did not contain a minus sign, there could be a

negative eigenvalue.)

  1. (30 pts.) Match each region and boundary condition with the expected form of the solution

of Laplace’s equation,

2

u

∂x

2

2

u

∂y

2

= 0. (It should not be necessary to do extensive

calculations. Use your background knowledge and common sense.)

Problems: (In the rectangular cases, the origin of Cartesian coordinates is at the lower

left corner. In the cylindrical case (f), x is the angular variable and y is the variable along the

axis of the cylinder with y = 0 at the left end.)

(a)

↓ u = 0

↑ u = 0

u = f →

→ ∞

(b)

↓ u = f

↑ u = 0

u = 0 →

→ ∞

(c)

↑ u = 0

u = f →

(d)

↑ u = f

∂u

∂x

= 0 →

The zeroth-order problem is thus

y

  • 4y

= 0, y

(0) = 0, y

(0) = 1,

whose solution is

y

(t) =

1

2

sin(2t).

The first-order problem then becomes

y

  • 4y

= −y

= − cos(2t). y

(0) = 0, y

(0) = 0.

The forcing term is on resonance, so y

will contain a secular term (proportional to t sin(2t)) that

grows with t. This problem will get worse in higher orders, so wherever we stop the perturbation

series, it will give an approximation that is not uniform in t (is not accurate except for rather small t).

(b) Solve the problem by the method of two time scales. (Find a one-term approximation

including the first-order correction.)

See Question 1 in Peter Howard’s Practice Exam 2, Version A:

http://www.math.tamu.edu/~phoward/m401/pexam2asol.pdf

  1. (20 pts.) By the method of your choice, solve the wave equation

2

u

∂t

2

=

2

u

∂x

2

(−∞ < x < ∞, −∞ < t < ∞), u(0, x) = h(x),

∂u

∂t

(0, x) = j(x).

(If you know the relevant formulas, it is not necessary to rederive them.)

First note that the conventional notation is h = f , j = g.

D’Alembert method: u(t, x) = B(x − t) + C(x + t). Matching the initial conditions yields

B(z) =

[h(z) + G(z)], C(z) =

[h(z) − G(z)],

where G is any antiderivative of j; equivalently,

u(t, x) =

[h(x − t) + h(x + t)] +

x+t

x−t

j(z) dz.

Fourier transform method: See solution to Homework 14, Exercise 4. Separation-of-variables

method is the same as the transform method except for starting out in a more wordy way.

u(t, x) =

1

2 π

ˆ

h(ω)e

iωx

cos(ωt) +

1

2 π

1

ω

ˆ(ω)e

iωx

sin ωt

if

ˆ

h(ω) =

e

−iωx

h(x) dx, etc.

  1. (30 pts.) Solve by perturbation theory

x

3

  • x − 27 = 0.

Find all real roots up through order 

2

and all complex roots up through order .

Let x ∼ x

  • x
  • 

x

. Then

x

∼ x

  • 3x

(x

  • 

x

) + 3x

(x

  • · · ·)
  • O(

)

= x

  • (3x

x

) + 

(3x

x

  • 3x

x

) + O(

).

Therefore, the equation is

0 ∼ x

  • (3x

x

) + 

(3x

x

  • 3x

x

) + x

  • 

x

− 27.

So the equation of order 

is x

= 27, or x

=

27 e

2 πin/ 3

for n = 0, 1 , 2. (In fact, n = −1 is

equivalent to n = 2 and leads to a numerical result faster.) That is, the real root is x

= 3, and there

are two complex roots,

x

= 3

[

cos

2 π

3

± i sin

2 π

3

]

=

3

2

(− 1 ± i

3).

The equations of orders 

and 

are

0 = 3x

x

  • x

, 0 = 3(x

x

  • x

x

) + x

.

For the real case, with x

= 9, we find x

= − 3 /27 = −

, then

0 = 3

(

3

9

  • 9x

)

1

9

= 27x

,

so x

= 0.

For the complex case,

x

= −

x

3 x

= −

1

3

2

3

1

− 1 ± i

3

= −

2

9

− 1 ∓ i

3

1 + 3

=

1

18

(1 ± i

3).

Summary:

x ∼ 3 −



9

  • O(

) is the real root,

x ∼

3

2

(− 1 ± i



18

(1 ± i

    • O(

) are the complex roots.

By substituting these into the original equation one can check that that equation is satisfied through

the expected order in each case. (I did that, but I will not type it out.)