MATH 132, FALL 2009HOMEWORK 10 SOLUTIONS

1. (Sec. 4.4, ex. 2) If F (x, y) = f(x, y, g(x, y)) and f and g are C2, then (thinking of f as a functionof x, y and z):

Fx = fx + fzgx, Fy = fy + fzgy.

Differentiating again, we obtain:

Fxx = fxx + fxzgx + (fzx + fzzgx)gx + fzgxx,

and similarly for Fxy and Fyy. �

2. (Sec. 4.4, ex. 5) If f(x, y) = φ(x− cy) + ψ(x+ cy), then:

fx = φ′(x− cy) + ψ′(x+ cy), fy = −cφ′(x− cy) + cψ′(x+ cy),

hence

fxx = φ′′(x− cy) + ψ′′(x+ cy), fyy = c2φ′′(x− cy) + c2ψ′′(x+ cy).

It follows that fyy = c2fxx. �

3. (Sec. 4.4, ex. 7) If F (x, y) = e−byf(x, y), then

Fx = e−byfx, so Fxx = e−byfxx,

and Fy = −be−byf + e−byfy = e−by(fy − bf). If fy = fxx + bf , then

Fy = e−byfxx = Fxx,

as claimed. �

4. (Sec. 4.5, ex. 1) (a) If g(t) = t + x0, then g is a translation, hence a bijection, whose range inEn. Its inverse is g−1(s) = s− x0, Dg(t) = I, and Jg(t) = 1, for all t.

(b) If g(s, t) = (s+ 2t, s− t), then g is a linear map whose matrix in the standard basis is

A = Dg =[1 21 −1

].

Since Jg = −3 6= 0, g is a bijection onto E2 whose inverse is the multiplication by the matrix

A−1 =13

[1 21 −1

]=

13A.

(c) If g(s, t) = (s2−s−2, 3t), then g is not 1–1 (g(0, 0) = g(1, 0)), its range is the set (−94 ,∞)×E1,

Dg(s, t) =[2s− 1 0

0 3

],

so Jg(s, t) = 3(2s− 1), which is nonzero if s 6= 1/2.

(d) If g(s, t) = (s2 − t2, st), then g is clearly not 1–1,

Dg(s, t) =[2s −2tt s

],

1

2

so Jg(s, t) = 2s2 + 2t2, which is nonzero away from the origin.

(e) If g(s, t) = (log(st), (s2 + t2)−1) on ∆ = {(s, t) : 0 < s < t}, then g is not 1–1 (g(−s,−t) =g(s, t)),

Dg(s, t) =

[1s

1t

− 2s(s2+t2)−2 − 2t

(s2+t2)−2

],

soJg(s, t) = − 2t

s(s2 + t2)2+

2s(s2 + t2)2

6= 0

on ∆. �

5. (Sec. 4.5, ex. 6) Observe that the given assumptionn∑

i,j=1

∂jgi(t)hihj > 0,

for all h = (h1, . . . , hn) 6= 0 and all t ∈ ∆, is equivalent to the statement: “For all t ∈ ∆, Dg(t)is a positive definite matrix”, i.e.,

[Dg(t)h] · h > 0,

for all h 6= 0.Now suppose g(t1) = g(t2), for some t1 6= t2 in ∆. Let h = t2 − t1 and define a function

f : ∆ → E1 byf(t) = [g(t)− g(t1)] · h.

Observe that by the product rule (for dot products)

∇f(t) = Dg(t)h. (1)

Let us apply the Mean Value Theorem to f at t1 and t2. We obtain a point ξ on the line segmentjoining t1 and t2 such that

f(t2)− f(t1) = df(ξ)h = ∇f(ξ) · h.

The left-hand side equals zero, as is easy to check, whereas by (1), the right-hand side equals

∇f(ξ) · h = [Dg(ξ)h] · h.

Since h 6= 0, it follows that the right-hand side is positive, which is a contradiction. Therefore,g is 1–1. �