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
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
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
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 2.1
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 2.3
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