UBC Mathematics 402(101)—Assignment 5
Due in class on Friday 13 February 2015
1. Consider the following free-endpoint problem on the fixed interval [a, b]:
)
(
Z b
L(x(t), ẋ(t)) dt .
Λ[x] := ℓ(x(a), x(b)) +
min
x∈P WS[a,b]
a
Assume that both ℓ: R × R → R and L: R × R → R are convex functions. Suppose an
admissible arc x
b has the property that some arc p ∈ P WS[a, b] satisfies the following
conditions, which generalize (IEL) and (NBC) to problems where L and ℓ need not
be differentiable:
(i) [ṗ(t) p(t)] ∈ ∂L(b
x(t), x(t)),
ḃ
(ii) [p(a)
t ∈ [a, b],
− p(b)] ∈ ∂ℓ (b
x(a), x
b(b)).
Prove that x
b truly provides a global minimum.
2. Prove: If f : (0, +∞) → R is convex, then g(x) = xf (1/x) is convex on (0, ∞). (Partial
credit will be available for succeeding under the assumption f ∈ C 2 ; full marks for
treating the case f ∈ C 1 ; and bonus points for giving a proof valid for arbitrary
convex f .)
3. Consider the basic
p problem with fixed endpoints (a, A) and (b, B) and Lagrangian
L(t, x, v) = f (t) 1 + v 2 + g(t)x, where f, g ∈ C 1 [a, b] are given and f (t) > 0 for all
t ∈ [a, b]. Prove that the basic problem has a smooth global solution if and only if
there exists a C 2 function G: [a, b] → R satisfying the three conditions
Z b
G(t)
′
p
G (t) = g(t) ∀t ∈ (a, b), |G(t)| < f (t) ∀t ∈ [a, b], B − A =
dt.
f (t)2 − G(t)2
a
4. Consider the basic problem with fixed 0 < a < b and L(t, x, v) = t2 v 2 + 2βtxv + γx2 :
)
(
Z b
L(t, x(t), ẋ(t)) dt : x(a) = A, x(b) = B .
(P )
Λ[x] =
min
x∈P WS[a,b]
a
(i) Show that whenever γ − β ≥ −1/4, there is a unique admissible extremal for any
choices of A and B. (One way to do this is just to find it. The case γ −β = −1/4
will require special treatment.)
(ii) Find and sketch the region in the (β, γ)-plane corresponding to parameter vectors
that make the function (x, v) 7→ L(t, x, v) convex for all t > 0.
(iii) Explain why the basic problem above is essentially equivalent to the one obtained
by replacing the integrand with
e x, v) = L(t, x, v) + α x2 + 2txv ,
L(t,
File “hw05”, version of 06 Feb 2015, page 1.
Typeset at 16:25 February 6, 2015.
for any real constant α.
(iv) Use a shrewd choice of α = α(β, γ) in (iii) to prove that if γ − β ≥ −1/4,
problem (P ) has a unique global minimizer for any choices of A and B. Add
this half-space to the sketch initialized in part (i).
Note: The idea illustrated in this problem, of adding a suitable auxiliary function to
extend the region where the given integrand is convex, will play a significant role in
later theoretical developments.
5. For each of the two problems below, find the unique global solution. Explain carefully
how you know it has these properties.
Z 1
Z 1
2
ẋ(t) dt = 1 .
tx(t) dt : x(0) = 0, x(1) = 0,
(a) min
0
0
(b) min
Z
0
π
2
2x(t) sin t + ẋ (t) dt : x(0) = x(π) = 0,
File “hw05”, version of 06 Feb 2015, page 2.
Z
π
x(t) dt = 1 .
0
Typeset at 16:25 February 6, 2015.