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Math 509 Final Exam Jerry L. Kazdan
May 1, 2007 9:00 –11:
Directions This exam has three parts, Part A has 4 shorter problems (5 points each), Part B has 5 traditional problems (10 points each). Closed book, no calculators – but you may use one 3′′^ × 5 ′′^ card with notes.
Part A: Shorter Problems (4 problems, 5 points each).
A–1. Give an example of a sequence of continuous functions fn(x), 0 ≤ x ≤ 1, with fn(x) → 0 (pointwise) for all x ∈ [0, 1], but
0 |fn(x)|^ dx^ → ∞. A sketch is adequate.
Solution:
1/n 1
n^
A–2. In L 2 (− 1 , 1) with the standard inner product, show that any even function is orthogonal to any odd function (of course assume that the functions are integrable).
Solution: Let h(x) = f (x)g(x), where f (x) is even and g(x) odd. Then h(x) is odd. To show:
− 1 h(x)^ dx^ = 0. This is clear geometrically. The computation is also easy:
I :=
− 1
h(x) dx =
− 1
h(x) dx +
0
h(x) dx = −
− 1
h(−x) dx +
0
h(x) dx.
But making the change of variable t = −x we see that
− 1
h(−x) dx = −
0
h(t) dt.
Thus I = 0.
A–3. Prove that the series
∑^ ∞
1
(−1)k^ sin kx 1 + k^2
converges absolutely and uniformly for all real x.
Solution: Note that
(−1)k^ sin kx 1 + k^2
∣ ≤^
1 + k^2
for all x. Since
1 + k^2
converges, by the Weierstrass M-test the original series converges absolutely and uniformly for all x.
A–4. Let u(x, y, t) be a solution of the heat equation
∂u ∂t
∂^2 u ∂x^2
∂^2 u ∂y^2
for (x, y) in a bounded
domain D ∈ R^2 with the outer normal derivative ∇u · N = 0 on the boundary of D (here N is the unit outer normal vector field on the boundary).
If Q(t) :=
D
u(x, y, t) dx dy , show that
dQ dt
= 0 and hence that Q(t) = Q(0).
Solution: By Green’s theorem
dQ dt
D
ut(x, y, t) dx dy =
D
∆u(x, y, t) dx dy =
∂D
∇u · N ds = 0.
Part B: Traditional Problems (5 problems, 10 points each)
B–1. The following equations define a map F : (x, y, z) 7 → (u, v, w):
u(x, y, z) = x + xyz^2 v(x, y, z) = xz^2 + y w(x, y, z) = 2x + cz + z^3
Clearly F : (1, 1 , 0) 7 → (1, 1 , 2). Write p = (1, 1 , 0) and q = (1, 1 , 2). a) Compute the derivative F ′(p). Solution:
F ′(p) =
2 0 c
b) For which value(s) of the constant c can the system of equations: can be solved for x, y , z as smooth functions of u, v , w near p? Justify your assertion(s). Solution: By the inverse function theorem, the map is invertible (as a smooth map) if and only if F ′(p) is invertible. This is clearly true only for c 6 = 0. c) If c is one of these “good” values, let G : (u, v, w) 7 → (x, y, z) be the map inverse to F. Compute the derivative G′(q) and use it to compute ∂y(u, v, w)/∂v at q. Solution: By the inverse function theorem,
G′(q) = [F ′(p)]−^1 =
− 2 /c 0 1 /c
Thus, ∂y(u, v, w)/∂v|q = 1.
B–5. Let ϕk(x), x ∈ R, be a sequence of smooth functions with the following properties
i). ϕk(x) ≥ 0 for |x| < 1 /k , ϕk(x) = 0 for |x| ≥ 1 /k , ii).
R ϕk(x)^ dx^ = 1. For a continuous function f (x) with f (x) = 0 for x outside a compact set K , define
fk(x) :=
R
f (y)ϕk(x − y) dy.
Show that limn→∞ fk(x) = f (x), and that this convergence is uniform.
Solution: fk(x) − f (x) =
R
[f (x − y) − f (x)]ϕk(y) dy.
Since f (x) = 0 for x outside a compact set, it is uniformly continuous. Thus, for any > 0 there is a δ > 0 so that if |y| < δ then |f (x − y) − f (x)| < . Consequently, if 1/k < δ , then
|fk(x) − f (x)| ≤
|y|≤ 1 /k
|f (x − y) − f (x)|ϕk(y) dy ≤
|y|≤ 1 /k
ϕk(y) dy ≤ .
This holds for all x so the convergence is uniform.
