Practice Final
Math 112 A, 30 November, 2005
Name:
ID:
1. Show that the problem
x
(1 + x2 + y 2 ) u +
x
(1 + x2 + y 2 ) u u = 1
y
u=1
y
for x2 + y 2 < 1
for x2 + y 2 = 1
(1)
has at most one solution.
2. Let C be a piecewise continuously dierentiable close
Practice Midterm I
Math 112 A, 14 October, 2005
Name:
ID:
You may need the following:
u(x, t) =
1
1
[f (x + ct) + f (x ct)] +
2
2c
x+ct
g () dx +
x
xct
1
2c
x+c(tt)
t
0
xc(tt)
F (, t) dxdt
x
(1)
1. Consider the problem
2 u 2u
2 = sin(3x), 0 < x < 1, t
Practice Midterm II
Math 112 A, 17 November, 2005; Third Correction
Name:
ID:
You may need the following:
v dxdy =
D
v n ds,
(1)
C
and
sin2 (nx) dx =
0
.
2
(2)
1. Show that the problem
x
(1 + x2 + y 2 ) u +
x
(1 + x2 + y 2 ) u u = 1
y
u=1
y
for x2 + y 2