Math 6230 Homework #3 Solutions 1. For the differential equation ...

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Solving for G ′ (x) gives G ′ (x) = 2 − 2xG(x) x2 . (2) − 3G(x) 2 There are two ways to obtain G ′′ (x): either differentiate (1), solve for G ′′ (x), and plug in (2), or differentiate (2) and then plug in (2). I’ll use the first method since I don’t much care for the quotient rule. We obtain x 2 G ′′ (x) + 4xG ′ (x) + 2G(x) − 6G(x)G ′ (x) 2 − 3G(x) 2 G ′′ (x) = 0, and solving for G ′′ (x) gives Then plugging in G ′ (x) from (2) gives G ′′ (x) = G ′′ (x) = 6G(x)G′ (x) 2 − 2G(x) − 4xG ′ (x) x2 − 3G(x) 2 . 6G(x) 2−2xG(x) x2−3G(x) 2 2 2−2xG(x) − 2G(x) − 4x x2−3G(x) 2 x 2 − 3G(x) 2 5. The images here are the lower-case and upper-case alphabet in Helvetica font. Classify the lower-case letters by homeomorphism type. For which letters is the uppercase version not homeomorphic to the lower-case version? Solution: The equivalence classes for the lower-case letters are {a, b, d, g, p, q}, {c, l, s, v, w, z}, {e}, {f, t, x}, {h, n, r, u, y}, {i, j}, {k, m}, {o}. The only one that’s not quite obvious is that ‘k’ is homeomorphic to ‘m’, but if you think about it carefully you’ll see the correspondence. The lower-case letters that are homeomorphic to their upper-case versions are: {a, c, k, l, o, s, t, v, w, x, y, z}. 4 .
6. Consider the parabolic coordinates (ϕ, U) given by formula (6.2.4) with open set U given by the complement of the negative vertical ray. Let fpara(τ, σ) = τ 2 be defined on the set R×(0, ∞). Is there a smooth function f : M → R such that fpara = f ◦ϕ −1 ? Solution: Let ψ : M → R 2 be the Euclidean coordinates (x, y) on the plane. Then the transition map is, by equation (6.2.4), (x, y) = φ ◦ ϕ −1 (τ, σ) = στ, 1 2 (τ 2 − σ 2 ) . (3) Now define f : U → R by f(p) = fpara◦ϕ(p) = fpara(τ, σ) if (τ, σ) = ϕ(p). The question is whether we can complete f to all of M, and in order to determine this, we need to compute f in another coordinate chart. Define frect : φ[U] → R by frect(x, y) = f ◦ φ −1 (x, y). We know that φ[U] consists of all (x, y) except the ray L = {(0, y) | y ≤ 0} (in other words, the negative vertical ray expressed in coordinates). We compute frect(x, y) = f ◦ φ −1 (x, y) = fpara ◦ ϕ ◦ φ −1 (x, y). Now the function (τ, σ) = ϕ ◦ φ−1 (x, y) is just the inverse of the function (3), defined on L, with values τ > 0 and σ ∈ R. To compute it, just solve (3) to get ⎛ (τ, σ) = ⎝ y + x2 + y2 x , y + x2 + y2 ⎞ ⎠ . Since fpara(τ, σ) = τ 2 , we conclude that frect(x, y) = τ(x, y) 2 = y + x 2 + y 2 on the complement of L, i.e., as long as x = 0, or x = 0 and y > 0. The question is now whether frect extends to all of R 2 ; if it does, then f = frect ◦ φ is defined on all of M since φ is defined on all of M. Unfortunately frect is not smooth in any neighborhood of the origin; we have f(0, y) = y + |y| = 0 y ≤ 0 2y y > 0 , so that fy(0, 0) does not exist. Hence although there is a continuous f : M → R, it cannot be smooth in coordinate charts. 7. Recall that a topological space M is second-countable if there is a countable collection {Ωn : n ∈ N} of open sets such that every other open set U ⊂ M is equal to the union of all the sets Ωn that it contains. (a) Prove that R is second-countable. (Hint: the rationals are countable.) Solution: Consider the set of all open intervals (a, b) such that a, b ∈ Q; this is a subset of Q × Q, so it is a countable collection. Let U ⊂ R be any open set. Let 5
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Page 3: Figure 1: Some u-v coordinate lines
Page 7: • If Ωi is a collection of open

Math 6230 Homework #3 Solutions 1. For the differential equation ...

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