MAP 5336.ORV1 (Nevai)
Homework #1 Solutions
Copyright c 2012 Andrew Nevai
1. There are at least two ways to solve this problem. (a) Let D = cfw_x : |x| R and I = cfw_t : |t| T . The D is convex and f (x, t) = tx4 + et is continuous on D I. In addition, f
MAP 5336.ORV1 (Nevai): Midterm 2 Solutions
Copyright
2012 Andrew Nevai
SHOW ALL WORK FOR FULL CREDIT SIMPLIFY AND BOX FINAL ANSWERS 1. Let B = X(T ) and let X(t) be the principal fundamental matrix. () Suppose x(t + T ) = x(t) for t 0 with x(0) 0. On one
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MAP 5336.ORV1 (Nevai)
Homework #5 Solutions
z1 z2
Copyright
2012 Andrew Nevai
1. (a) We can express u + f (u)u + g(u) = 0 as z = F (z) where z = F (z) = z2 . -f (z1 )z2 - g(z1 )
=
u u
, and
2 Let k > 0. It is clear that 1 z2 0 with equality if and only i
MAP 5336.ORV1 (Nevai)
Homework #4 Solutions
Copyright
2012 Andrew Nevai
(1)
1. Assume f (t) is continuous for t 0 with 0
f (t) < and let 0 be a real constant. The equation
u + 2 + f (t) u = 0 can be written equivalently as x = A(t)x Indeed, A(t) = A0 + B
MAP 5336.ORV1 (Nevai)
1. Consider
Homework #3 Solutions
Copyright
2012 Andrew Nevai
(1)
x = A(t)x
where
A(t) =
2 cos t 0 3
Observe that A(t) is periodic with period T = 2. (a) The principal fundamental matrix of (1) is a 2 2 matrix X(t) whose columns are
MAP 5336.ORV1 (Nevai)
d d 1. Let L = (t2 - 1) dt2 - 2t dt + 2
2
Homework #2 Solutions
Copyright c 2012 Andrew Nevai
(a) Define M = t21 L. Since Lx = 0 M x = 0 (for t = 1), it follows that the space of -1 solutions to Lx = 0 is two-dimensional (at least on