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MA259 Multivariable Calculus Assignment 3
发布时间:2022-11-17
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MA259 Multivariable Calculus
Assignment 3
2022
1. (a) Suppose that f : Rn 一 R and r : (a, b) 一 Rn are continuously differentiable and that r solves the
system of ODEs
d
dt
Prove that if [α, β] 仁 (a, b), then f (r(α)) ≥ f (r(β)) with equality if, and only if, Af (r(α)) = 0 and r(t) = r(α) A t e [α, β]. (See (4.18) in ?4.6.2 in the notes.)
Remark. Solutions to (1) generate the so-called gradient flow of f .
(b) Let f : R 一 R be differentiable and define u : R2 一 R by u(x, y) := xf (xy + x). Show that u
satisfies
∂u ∂u
∂x ∂y
(See ?4.6.4 in the notes.)
(c) Mean Value theorem for a scalar-valued function of several variables on a convex set.
Suppose that U is a convex open subset of Rn and that f e Cī (U). Prove that, for any x, y e U, 3 θ e (0, 1) such that
f (y) - f (x) = ╱ Af ((1 - θ)x + θy) ← . (y - x).
2. Consider f e Cī (B, Rk ) for which 3 α > 0 such that
|Df (0)h| ≥ α|h| A h e Rn .
Set A := Df (0) and define F : B 一 Rk by
F (x) := f (x) - Ax.
(a) Calculate DF (0).
(b) Use the continuity of DF (x) and the Mean Value Inequality to prove that 3 δ > 0 such that
|F (x) - F (y)| ≤ α|x - y| A x, y e B6 .
(c) Deduce that
|f (x) - f (y)| ≥ α|x - y| A x, y e B6 .
In particular, f is injective on B6 . (Compare with Question 6 in Examples Sheet 3.)
Slogan: If f is continuously differentiable at x and Df is injective at x, then f is injective near x.
3. Define f : Rn / {0} 一 Rn by
x
f (x) :=
(a) Calculate the n 人 n Jacobian matrix ∂f (x).
Hint. Write f (x) as the column vector ╱ , . . . ,
←T and compute ∂jfi(x). The Kronecker symbol
δij for the n 人 n identity matrix can be useful in this computation. You are allowed to quote the calculations in Example 4.6.4 in the notes.
(b) For each x e Rn / {0}, let Πx be the hyperplane orthogonal to x:
Πx := {v e Rn : x . v = 0}.
Prove that Πx is an eigenspace of ∂f (x), whose dimension and corresponding eigenvalue you should state explicitly.
(c) Find the kernel of ∂f (x) and calculate the rank of ∂f (x).
(d) Optional challenge, not for credit. Observe that f is the projection of x onto Sn一 ī along the ray joining 0 and x. Use this geometric interpretation of f to interpret the results in parts (b) and
(c) geometrically.
4. Direct verification of Green’s theorem for some special regions . Mainly first year review.
Given f e Cī ([0, 1]) such that f (0) = k > 0, f′(t) < 0 A t e (0, 1), f (1) = 0, let C be the oriented closed curve obtained by going from 0 to 1 along the x-axis, up along the graph of f from (1, 0) to (0, k) and down from k to 0 along the y-axis.
(a) Recall from First Year Analysis that f′(t) < 0 A t e (0, 1) implies the existence of the inverse function f一 ī : [0, k] 一 [0, 1], which is also differ-
entiable and (f一 ī )′(y) = 1
Given g e C(R2), use the change of variables (x = f一 ī (y) or, equiva- lently, y = f (x)) formula to show that
\〇ī g(x, f (x)) f′(x) dx = - \〇k g(f一 ī (y), y) dy .
y
1 x
For the rest of this question, let v : R2 一 R2 , v (x, y) = ╱a(x, y), b(x, y) ←, be a continuously differen- tiable vector field.
(b) Write down 尸C v . dr as a sum of three integrals involving a, b, f and f′ . One of these integrals
尸
(c) Let Ω be the region enclosed by C. Use the Fundamental Theorem of Calculus (as was done for a rectangle in ?6.1 of the notes) and the result in part (a) to verify that 尸C v . dr = …Ω curl v dx dy , where curl v := -
.
Remark. By piecing together regions like that in this question along their vertical and horizontal segments, one can verify Green’s Theorem for a large class of regions that need not have a straight segment in their boundary.