Math 110A
Final Exam
Sep. 6 (7:00-9:50pm), 2014
Instructions:
1. No books, no notes, no electronic devices (except a calculator), no cheating.
2. One calculator and one cheat sheet (A4 or standard letter size) allowed.
3. Write your name, PID, and section number on front of your blue book.
4. In the upper right corner on front of your blue book, draw the following table of
grading:
1
2
3
4
5
6
7
(
)
Σ
5. Write your solutions clearly in your blue book:
(a) Carefully indicate the number and letter of each question.
(b) Present your answers in the same order they appear in the exam.
(c) Start each question on a new slide of page.
6. Show all of your work. No credit will be given for unsupported answers.
7. Write your solutions clearly and legibly. No credit will be given for illegible
solutions.
8. If any question is not clear, ask for clarification.
9. Turn in your blue book, no need to turn in this sheet.
Page 1 of 10
Math 110A
Final Exam
Sep. 6 (7:00-9:50pm), 2014
1. (1 point) Carefully read and complete the instructions at the other page of this exam sheet.
This exam sheet contains 8 problems, you need to finish problems 2-7, and choose
to finish one problem from problem 8 and problem 9.
Notice that there is a parenthesis in the table of grading under instruction 4, please put 8 if
you want us to grade problem 8, and please put 9 if you want us to grade problem 9.
Page 2 of 10
Math 110A
Final Exam
Sep. 6 (7:00-9:50pm), 2014
2. (10 points) Water passes through a hose in x direction with constant velocity V . The water
density is u(x, t). Derive a PDE for u(x, t).
Solution:
ut + V ux = 0.
This is a typical transport equation. For solution details, you can read the lecture notes on
Aug 6 or section 1.3 of the book.
Page 3 of 10
Math 110A
Final Exam
Sep. 6 (7:00-9:50pm), 2014
3. (9 points) Derive a general solution to the following transport equation
ux + y 2 uy = 0
Solution:
The characteristics
The solution is
dy
= y2
dx
1
u(x, y) = f (x + )
y
Page 4 of 10
Math 110A
Final Exam
Sep. 6 (7:00-9:50pm), 2014
4. (10 points) Assume n and m are both integers, when are the following two functions
f (x) = sin nx,
g(x) = cos mx
orthogonal in the interval (0, π)?
2 sin x cos y = sin(x + y) + sin(x − y)
Solution:
Z
π
sin nx cos mxdx =
0
• When n = 0,
Z
1
2
Z
π
(sin(n + m)x + sin(n − m)x)dx
0
π
sin nx cos mxdx = 0
0
• When n = m or n = −m,
π
Z π
Z
1 π
1
sin nx cos mxdx =
(sin 2nx)dx = − cos 2nx = 0
2 0
4n
0
0
• Otherwise,
Z π
Z
1 π
sin nx cos mxdx =
(sin(n + m)x + sin(n − m)x)dx
2 0
0
1
(−1)n+m − 1 (−1)n−m − 1
=
−
−
2
n+m
n−m
Notice that when n − m is even, (−1)n−m = 1 and (−1)n+m = 1, and hence
Z π
sin nx cos mxdx = 0
0
When n − m is odd, (−1)n−m = −1 and (−1)n+m = −1, and hence
Z π
1
2n
1
+
=
6= 0
sin nx cos mxdx =
n
+
m
n
−
m
(n
+
m)(n
− m)
0
Notice that n = m and n = −m are included in the case where n − m is even.
Hence we conclude that when n = 0 or n − m is even, we have
f (x) = sin nx,
g(x) = cos mx
orthogonal in the interval (0, π).
Page 5 of 10
Math 110A
Final Exam
Sep. 6 (7:00-9:50pm), 2014
5. (10 points)
1
2
e−x /4kt ,
for t > 0
4πkt
is known as the fundamental solution to the diffusion equation. Now use the fact that
Z ∞
√
2
e−p dp = π
S(x, t) = √
−∞
to show that
Z
∞
S(x, t)dx = 1
−∞
Solution:
Using change of coordinates, this has been shown many times in class, we also did a similar
problem
Z x
1
S(x, t)dx =
2
−∞
during the review class. Please refer to notes on Aug 29 for details.
Page 6 of 10
Math 110A
Final Exam
Sep. 6 (7:00-9:50pm), 2014
6. (10 points) Find the solution to the wave equation
utt = uxx ,
u(x, 0) = 1,
ut (x, 0) = sin x,
for 0 < x < ∞
for 0 < t < ∞
u(0, t) = 0,
Solution:
odd extension φ(x) = 1 when x > 0 and φ(x) = −1 when x < 0; ψ(x) = sin x for x ∈ R. Then
the solution is
Z
1
1 x+t
ψ(s)ds
u(x, t) = (φ(x + t) + φ(x − t)) +
2
2 x−t
• When x − t > 0,
1
1
u(x, t) = (1 + 1) +
2
2
Z
x+t
sin sds = 1 +
x−t
1
1
cos (x + t) − cos (x − t)
2
2
• When x − t < 0,
1
1
u(x, t) = (−1 + 1) +
2
2
Z
x+t
sin sds =
x−t
Page 7 of 10
1
1
cos (x + t) − cos (x − t)
2
2
Math 110A
Final Exam
Sep. 6 (7:00-9:50pm), 2014
7. (10 points) Find the Fourier sine series of f (x) = x in the interval (0, 1), using the formula
An =
2
l
Z
l
f (x) sin
0
nπx
dx
l
with the corresponding l.
Solution:
This has been done many times as well, you may refer to example 3, section 5.1, in page 109
for details.
Page 8 of 10
Math 110A
Final Exam
Sep. 6 (7:00-9:50pm), 2014
8. (10 points) Find the solution to the Laplace’s equation
in {0 < x < 1, 0 < y < 1}
∆u = 0,
u(0, y) = u(x, 0) = 0,
u(x, 1) = x,
u(1, y) = y
( Hint: there is a shortcut to solve it, the same as one of your homework problems. )
Solution:
Assume
u(x, y) = Ax2 − Ay 2 + Bxy + Cx + Dy + E = 0
Then the boundary condition u(0, y) = 0 implies
−Ay 2 + Dy + E = 0
Hence A = D = E = 0. The condition u(x, 0) = 0 implies that
Ax2 + Cx + E = 0
Hence A = C = E = 0. Then we have
u(x, y) = Bxy
Now u(x, 1) = x and u(1, y) = y implies that
Bx = x,
By = y
both implies that
B=1
Hence we obtain
u(x, y) = xy
Page 9 of 10
Math 110A
Final Exam
Sep. 6 (7:00-9:50pm), 2014
9. (10 points) Prove that the solution to the Poisson equation with Dirichlet boundary
with ~x ∈ D
∆u(~x) = f (~x),
with ~x ∈ ∂D
u(~x) = h(~x),
is unique. Where D is a domain in R2 , and ∂D is the boundary of D.
Solution:
Assume solution not unique, then exist two solutions u and v such that
with ~x ∈ D
∆u(~x) = f (~x),
with ~x ∈ ∂D
u(~x) = h(~x),
and
with ~x ∈ D
∆v(~x) = f (~x),
with ~x ∈ ∂D
v(~x) = h(~x),
Hence if we let
w =u−v
then w satisfies
with ~x ∈ D
∆w(~x) = 0,
with ~x ∈ ∂D
w(~x) = 0,
By maximum principle,
w≤0
Hence
u(~x) ≤ v(~x)
On the other hand, if we let
w =v−u
then w satisfies
with ~x ∈ D
∆w(~x) = 0,
w(~x) = 0,
with ~x ∈ ∂D
By maximum principle,
w≤0
Hence Hence
v(~x) ≤ u(~x)
Since we have both u ≤ v and v ≤ u, we must have
u=v
throughout the domain. Thus we have uniqueness.
Page 10 of 10