Answers to HW set 1
1.2/6 :
u(x, t) = g (t x/c) exp(x/c), if x < ct,
u(x, t) = 0, if x > ct
1.3/1
uxx 1/2,
u(6, T + 0.5) 64.005
1.3/2
E (t) = 2k
u2 dx
x
0
1.3/3
v := g (t) + [h(t) g (t)]x/
w = u v and wt wxx = g (t) + [g (t) h (t)]x/.
1.3/5
v :=
2k +
Answers to HW set 2
1.5/5: You can dierentiate E , integrate by parts (using the fact that since u is stationary at
the endpoints ut = 0 at the end points) and obtain
dE/dt =
0
(0 utt 0 uxx )ut dx = 0.
1.7/6: Let v = w u. Note v = 0 on the boundary. In
Answers to HW set 3
2.1/1b:
1
1 + erf
2
x 2kt
4kt
2.2/3:
exp(x + kt)
xct
u(x, t) =
s( /c) d + C
0
2.4/2:
1
4kt
exp (x y )2 /4kt exp (x + y )2 /4kt
dy
0
2.4/6:
T0 + A exp(
Amplitude is A exp(
2k
x) cos
2k
x) and phase shift is x
x t
2k
2k .
Extra #3 W
Homework 3, Due Sept. 17th, 10 pts each
1. Solve the Laplaces equation inside a 60 wedge of radius a subject to the
boundary conditions
u(r, 0) = 0;
u(r, /3) = 0,
u(a, ) = f ().
R 1 dx
2. Consider 0 1+x
2
a) Evaluate explicitly
1
b) Use Taylor series of 1