MTH 3326
Kyle Busse
Problem Set Week 1
Section 1.1
1, 2, 4, 7
1. (a) Let u := u(x, t). Solve ut + ux = 0 subject to the condition u(x, 0) = e|x| .
It is known that solutions of the form u(x, t) = f (x t) satisfy the given differential equation.
We also ha
Answers Section 2.1
4. (a) = 2
(b)
8 1/4
|f | =
9
6. (a) Let Lu = u , then the eigenvalue problem is
Lu = u,
u(0) = 0,
u(1) = 0.
(b) = 0 is not an eigenvalue.
(c) There are no negative eigenvalues.
(d) n = (n)2 , yn (x) = sin(nx), n = 1, 2, 3, .
7. (a) Le
Answers Section 2.5
2. (a) The first line is the standard heat equation. The second and third lines are
Robin boundary conditions both with ambient temperature zero and constant of proportionality K=1. This means they both specify physically realistic con
Answers Section 2.2
3. (a) The rst line is the heat equation. u(0, t) = 0 means the left endpoint is kept at
zero degrees. ux ( , t) = 0 means the right endpoint is insulated. Finally, u(x, 0) = f (x)
gives the initial temperature at each point of the rod
Test 1
February 16, 2016
Dr. Kirsten
1.) (8 points) Consider the PDE in polar coordinates for u := u(r, ) given by
1
1
urr + ur + 2 u = 0.
r
r
Show that u(r, ) = ln r and u(r, ) = r cos are both solutions.
2.) (3 points each) In the context of the 1D heat
Test 1
February 16, 2016
Dr. Kirsten
1.) (8 points) Consider the PDE in polar coordinates for u := u(r, ) given by
1
1
urr + ur + 2 u = 0.
r
r
Show that u(r, ) = ln r and u(r, ) = r cos are both solutions.
We compute for u(r, ) = ln r that
1
ur = ,
r
urr
MTH 3326
Kyle Busse
Problem Set Week 6
Section 2.4 2, 3, 6 (a)(d), 8, 9
Section 2.5 2, 3
Section 2.4
2. Let 1 and 2 be eigenfunctions of the problem 00 + = 0, corresponding to eigenvalues 1 and 2 , respectively. Now, suppose that both 1 and 2 satisfy the
MTH 3326
Kyle Busse
Problem Set Week 7
Section 2.6 3, 4
Section 3.3 2, 6, 7
Section 2.6
3.
(a) As usual, this is the 1D heat equation: ut = kuxx on x (0, `) and t > 0. The left endpoint is fixed at a
temperature of 100, so there is a Dirichlet condition u
MTH 3326
Kyle Busse
Problem Set Week 3
Section 1.3
Section 1.4
Section 2.1
1, 2, 3, 6
1, 7, 9, 12, 15
1, 3, 7, 11
Section 1.3
1. If the left endpoint is insulated, we have that the time derivative of the heat at that point is zero. That
is, ut (0, t) = 0.
MTH 3326
Kyle Busse
Problem Set Week 9
Section 4.1 1, 4, 7
Section 4.2 1, 4, 5
Section 4.3 3, 4
Section 4.1
1. (a) The lefthand side of Equation 4.5 is given, with S being a SturmLiouville operator and with
u, v C 2 [a, b], as uS[v] vS[u]. If we let S tak
MTH 3326
Kyle Busse
Problem Set Week 3
Section 2.2
3, 4, 6
Section 2.2
3. (a) From a physical standpoint, the equation ut = kuxx (the heat equation for a rod of length `) means
that the time derivative of the temperature at a point along a pipe is proport
MTH 3326
Kyle Busse
Problem Set Week 7
Section 5.2
Section 5.3
Section 5.4
2
3, 4
4, 5, 7
Section 5.2
2. (a) In threespace, the Divergence Theorem can be stated as follows: Let M be a compact subset
of R3 with a piecewise smooth boundary M , and let F : K
MTH 3326
Kyle Busse
Problem Set Week 13
Section 6.1 2, 3, 4
2. (a) We seek to solve the polar form of Laplaces equation for (, ) and r (, ), with
u(, ) = f (). Letting u(r, ) = R(r)(), the differential equation we wish to solve renders the
following two o
MTH 3326
Kyle Busse
Problem Set Week 5
Section 2.3
2, 6, 7, 12
2. We desire to develop a formula for kn in the equation
u(x, t) =
X
p
1
k0 +
kn cos ( n x)en kt ,
2
n=1
where n = (n/`)2 , subject to the initial condition that u(x, 0) = f (x). We have that
MTH 3326
Kyle Busse
Problem Set Week 7
Section 3.4
1, 5, 9, 10
Section 3.4
1. (a) See the plot below:
(b) Indeed, the limit of f (x) at the left endpoint as approached from the right and at the right
endpoint as approached from the left are finite (both a
Answers Section 2.3
8. (a) From Exercise 2.2.3 we know
u(x, t) =
cn sin
n x en kt ,
n=1
where n = [(2n 1)/(2 )]2 . From the initial condition one nds
cn =
2
f (x) sin
n x dx
0
(c)
lim u(x, t) = 0
t