Math 381 Fall 2006 Exam 1
Instructor: Sabin Cautis
Wednesday, October 11, 2006
Instructions: This is a closed book, closed notes exam. You have fifty minutes.
Do all the problems but notice there is a bonus problem which you should try
only after finishing all the other questions. Please do all your work on the paper
provided – using the extra page if you need extra space. You must show your
work to receive full credit on a problem.
Please print you name clearly here.
Print name:
Upon finishing please sign the pledge below:
On my honor I have neither given nor received any aid on this exam.
Grader’s use only:
1a.
/45
1b.
/10
2a.
/30
2b.
/15
Bonus.
/5
1. Consider the following partial differential equation
∂2u
∂u
=−
∂x2
∂t
where u = u(x, t) is a function of x and t.
(a)[45 points] Use the separation of variables method to find the general
solution to this PDE on the interval [0, L] subject to the boundary conditions u(0, t) = 0 and u(L, t) = 0. Please show all your work including all
three cases (λ < 0, λ = 0, λ > 0) for the possible eigenvalues.
We assume that u(x, y) = φ(x)h(t). Then the give PDE takes the form
h(t)
dh
d2 φ
= −φ .
2
dx
dt
Separating variables we get
1 dh
1 d2 φ
=−
=λ
2
φ dx
h dt
where λ is the eigenvalue. From the second equality we find
so h = e . For the other equality we get
cases to consider λ > 0, λ = 0 and λ < 0.
−λt
d2 φ
dx2
dh
dt
= −λ · h
= λφ and there are three
√
√
If λ > 0 the general solution is φ(x) = c1 e λx + c2 e− λx . The boundary
condition u(0, t) = 0 means that φ(0) = 0 which means
c + c2 =
0 so that
√
√ 1
the general solution must have the form φ(x) = c(e λx − e− λx ). The
boundary condition u(L, t) = 0 means φ(L) = 0 so we have
c(e
√
λL
− e−
√
λL
) = 0.
√
√
This is impossible unless c = 0 since e λL − e− λL > 0 (recall that ex is a
strictly increasing function). So if λ > 0 there are no nontrivial solutions.
If λ = 0 the general solution is φ(x) = c1 x + c2 . The condition φ(0) = 0
means c2 = 0. The condition φ(L) = 0 means c1 = 0 so again there are
no nontrivial solutions.
√
√
If λ < 0 the general solution is φ(x) = c1 sin( −λx) + c2 cos( −λx). The
condition
√ 0. The condition φ(L) = 0 means that
√ φ(0) = 0 means c2 =
c1 sin( −λL) = 0 which gives −λL = nπ for some n = 1, 2, 3, . . . . Thus
2 2
λ = − nLπ2 for some n.
We conclude that the general solution is
u(x, t) =
∞
X
bn sin(−
n=1
nπx −λt
)e .
L
(b)[10 points] What is the general solution if the boundary conditions are
changed to u(0, t) = 0 and u(L, t) = 1? (you may assume the general
solution from part (a) is given).
Denote by u(x, t) the solution from part (a). Suppose u1 and u2 are two
solutions with the boundary conditions. Then u1 − u2 is a solution with
the original boundary conditions u(0, t) = u(L, t) = 0. So all we need to
do is find one solution u0 to our PDE having the new boundary conditions
and then every solution will be of the form u0 + u(x, t). Clearly, u0 = x/L
is a solution to our PDE and it satisfies the new boundary conditions.
Thus the general solution with these new boundary conditions is
x+
∞
X
n=1
bn sin(−
nπx −λt
)e .
L
(question: what is the general solution for any boundary condition of the
form u(0, t) = a and u(L, t) = b where a, b are arbitrary constants?)
2. Let f (x) = 1.
(a) [30 points] Find the sine Fourier series of f (x) on the interval [−2, 2]
(please show your work and simplify your answer as much as possible).
P∞
The sine Fourier series is n=1 bn sin( nπx
L ) where in this case L = 2. The
coefficients bn are given by
bn
=
=
=
=
Z
nπx
2 L
)dx
f (x) sin(
L 0
L
Z 2
nπx
)dx
sin(
2
0
nπx 2 2
− cos(
) |
2 nπ 0
2
(1 − cos(nπ))
nπ
This simplifies to give 0 if n is even and
4
nπ
if n is odd.
(b) [15 points] To what function does the sine Fourier series converge on
the interval [−4, 4] (draw a precise graph). Very briefly explain why this
is the case.
Away from the points of discontinuity the sine Fourier series converges
to the odd periodic extension g(x) of f (x) = 1 on [0, 2]. At a discontinuity x = x0 , Fourier’s theorem says that the Fourier series converges to
+
−
1
2 (g(x0 ) + g(x0 )). Thus we obtain the following graph:
1
−4
−3
−2
−1
0
1
−1
2
3
4
3. [5 points BONUS]
State the divergence theorem! (either in two or three dimensions) Hint: if
you don’t remember it, think of the fundamental theorem of calculus and
guess a generalization to two or three dimensions.
We give the statement in three dimensions. Let Φ = (Φ1 , Φ2 , Φ3 ) be a
vector field (not a function) in three dimensions – i.e. you have a vector
at each point. Then the divergence theorem says that integrating the
divergence of the vector field is the same as integrating the vector field
along the boundary: more precisely if R is a region and ∂R is its boundary
then
Z
Z
Φ · n̂dS
∇ · ΦdV =
R
∂R
where n̂ is the unit normal vector to the boundary.