Let Cn be an n-dimensional cross-polytope and Γp(Cn) be the smallest positive number γ such that Cn can be covered by p translates of γCn. We obtain better estimates of Γ2n(Cn) for small n and a universal upper bound of Γ2n(Cn) for all positive integers n.
Citation: Feifei Chen, Shenghua Gao, Senlin Wu. Covering cross-polytopes with smaller homothetic copies[J]. AIMS Mathematics, 2024, 9(2): 4014-4020. doi: 10.3934/math.2024195
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Let Cn be an n-dimensional cross-polytope and Γp(Cn) be the smallest positive number γ such that Cn can be covered by p translates of γCn. We obtain better estimates of Γ2n(Cn) for small n and a universal upper bound of Γ2n(Cn) for all positive integers n.
Necessary and sufficient conditions for optimality play a crucial role in solving problems in the calculus of variations. The main necessary conditions for such problems are the well-known conditions of Euler, Weierstrass, Legendre and Jacobi. In many cases, depending on the smoothness of the assumptions, Euler's necessary condition corresponds to a second order differential equation which restricts solutions to lie in a family of trajectories with certain uniformity properties. Also, the necessary condition of Jacobi cannot be applied when the extremal has corners and is not nonsingular. This is an unfortunate feature since, in general, the admissible arcs or trajectories which are candidates for solving the problem are neither nonsingular nor smooth.
On the other hand, one fundamental aspect of the theory of sufficient conditions for optimality consists in slightly strengthening the necessary conditions. Concretely, if an admissible arc satisfies the strengthened conditions of Euler, Legendre and Jacobi, then it is a strict weak minimum. Additionally, if this admissible arc also satisfies the strengthened condition of Weierstrass, then it is a strict strong minimum. Some of the techniques used to obtain sufficiency include the construction of a Mayer field on which the extremals are independent of the path with respect to an invariant integral commonly called the Hilbert integral, the existence of a symmetric solution of the matrix Riccati inequality associated with the problem, a verification function satisfying the Hamilton-Jacobi equation, a quadratic function that satisfies a Hamilton-Jacobi inequality, the nonexistence of conjugate points on the underlying half-closed time interval, or the incorporation of some convexity arguments on the functions delimiting the calculus of variations problem (see for example [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17] and references therein).
It is important to mention that the smoothness and the nonsingularity assumptions are crucial in the sufficiency theories mentioned above. In other words, the classical sufficiency theory in the calculus of variations, in general, may not give a response when the extremal under consideration is singular or has corners. In fact, as mentioned in [6], there is a gap between the set of necessary and sufficient conditions and [6,Section 3.7] is entirely devoted to the study of some problems for which the nonsingularity assumption fails. There, one finds a method which is only applicable to particular examples and may not hold in general, since "although this algorithm sheds light on the theory, it provides no panacea. Indeed there are no panaceas." Additionally, we refer the reader to [13], where the importance of the nonsingularity assumption in the classical calculus of variations sufficiency theory is fully explained.
In this paper we derive two new sufficiency results which provide sufficient conditions for strong local minima in certain classes of parametric and nonparametric calculus of variations problems of Bolza with variable or free end-points, inequality and equality nonlinear isoperimetric constrains, and nonlinear mixed pointwise inequality and equality constraints. The main novelty of our new sufficient theorems concerns their applicability to cases in which the extremals under consideration may be singular and nonsmooth, that is, the strengthened condition of Legendre and the continuity of the derivative of the proposed extremal are no longer required. More precisely, given an admissible extremal whose derivative is not continuous nor piecewise continuous but only essentially bounded, the elements comprising the new sufficiency theorems are the classical transversality condition, a crucial inequality which arises from the original algorithm used to prove the main result of the article, the necessary condition of Legendre, but not its strengthened version, the positivity of the second variation over the set of all nonnul admissible variations, and three refined Weierstrass conditions which are related to the functions delimiting the problems.
Another distinguishing characteristic of the main sufficiency result of the nonparametric calculus of variations problem presented in this paper is the fact that the initial and final end-points of the states are completely free, that is, they are not only variable end-points but they may belong to any set which is not necessarily a smooth manifold described by some functions which usually involve some type of equality or inequality conditions.
The paper is organized as follows. In Section 2 we pose the parametric calculus of variations problem we shall deal with together with some basic definitions and the statement of the main result of the article. In Section 3 we enunciate the nonparametric calculus of variations problem we shall study together with some basic definitions and a corollary which is also one of the main results of the paper. Section 4 is devoted to state two auxiliary lemmas in which the proof of the theorem is strongly based. Section 5 is dedicated to the proof of the main theorem of the article. In Section 6 we prove the lemmas given in Section 4 and, in the final section, we provide an example which shows how the sufficient theory developed in this paper widens the range of applicability of the classical calculus of variations theory.
Suppose we are given an interval T:=[t0,t1] in R, and functions l(b):Rp→R, lγ(b):Rp→R (γ=1,…,K), Ψi(b):Rp→Rn (i=0,1), L(t,x,˙x):T×Rn×Rn→R, Lγ(t,x,˙x):T×Rn×Rn→R (γ=1,…,K) and φ(t,x,˙x):T×Rn×Rn→Rs. Set
R:={(t,x,˙x)∈T×Rn×Rn∣φα(t,x,˙x)≤0(α∈R),φβ(t,x,˙x)=0(β∈S)} |
where R:={1,…,r} and S:={r+1,…,s} (r=0,1,…,s). If r=0 then R=∅ and we disregard statements involving φα. Similarly, if r=s then S=∅ and we disregard statements involving φβ.
It will be assumed throughout the paper that L, Lγ (γ=1,…,K) and φ=(φ1,…,φs) have first and second derivatives with respect to x and ˙x. Moreover, we shall assume that the functions l, lγ (γ=1,…,K) and Ψi (i=0,1) are of class C2 on Rp. Also, if we denote by c(t,x,˙x) either L(t,x,˙x), Lγ(t,x,˙x) (γ=1,…,K), φ(t,x,˙x) or any of its partial derivatives of order less than or equal to two with respect to x and ˙x, we shall assume that if C is any bounded subset of T×Rn×Rn, then |c(C)| is a bounded subset of R. Additionally, we shall assume that if {(Γq,Λq)} is any sequence in AC(T;Rn)×L1(T;Rn) such that for some U⊂T measurable and some {(Γ0,Λ0)}∈AC(T;Rn)×L∞(T;Rn), (Γq(t),Λq(t))→(Γ0(t),Λ0(t)) uniformly on U, then for all q∈N, c(t,Γq(t),Λq(t)) is measurable on U and
c(t,Γq(t),Λq(t))→c(t,Γ0(t),Λ0(t))uniformly on U. |
Note that all conditions above concerning the functions L, Lγ (γ=1,…,K) and φ, are satisfied if the functions L, Lγ (γ=1,…,K), φ and their first and second derivatives with respect to x and ˙x are continuous on T×Rn×Rn.
Set
X:=AC(T;Rn),Us:=L∞(T;Rs),A:=X×Rp. |
We shall use the notation xb to denote any element xb:=(x,b)∈A. Let B any subset of Rp which we shall call the set of parameters. The parametric calculus of variations problem we shall deal with, denoted by (P), is that of minimizing the functional
I(xb):=l(b)+∫t1t0L(t,x(t),˙x(t))dt |
over all xb∈A satisfying the constraints
{c(t,x(t),˙x(t))is integrable on T.b∈B.x(ti)=Ψi(b) for i=0,1.Ii(xb):=li(b)+∫t1t0Li(t,x(t),˙x(t))dt≤0(i=1,…,k).Ij(xb):=lj(b)+∫t1t0Lj(t,x(t),˙x(t))dt=0(j=k+1,…,K).(t,x(t),˙x(t))∈R(a.e. in T). |
Elements b=(b1,…,bp)∗ (the notation ∗ denotes transpose) in B will be called parameters, elements xb in A will be called arcs or trajectories, and a trajectory xb is admissible if it satisfies the constraints. The notation x0b0 refers to an element (x0,b0)∈A.
Let us now introduce some definitions which will be used throughout the paper.
∙ An arc x0b0 solves (P) if it is admissible and I(x0b0)≤I(xb) for all admissible arcs xb. For strong local minima, an admissible arc x0b0 is called a strong minimum of (P) if it is a minimum of I relative to the following norm
‖xb‖:=|b|+supt∈T|x(t)|=|b|+‖x‖C, |
that is, if for some ϵ>0, I(x0b0)≤I(xb) for all admissible arcs satisfying ‖xb−x0b0‖<ϵ.
∙ For all x∈X, we use the notation (˜x(t)) in order to represent (t,x(t),˙x(t)). Also, (˜x0(t)) represents (t,x0(t),˙x0(t)).
∙ Given K real numbers λ1,…,λK, for any xb admissible define the functional I0 by
I0(xb):=I(xb)+K∑γ=1λγIγ(xb)=l0(b)+∫t1t0L0(˜x(t))dt, |
where l0:Rp→R is given by
l0(b):=l(b)+K∑γ=1λγlγ(b), |
and L0:T×Rn×Rn→R is given by
L0(t,x,˙x):=L(t,x,˙x)+K∑γ=1λγLγ(t,x,˙x). |
∙ Given λ1,…,λK, for all (t,x,˙x,ρ,μ)∈T×Rn×Rn×Rn×Rs, define the Hamiltonian of the problem by
H(t,x,˙x,ρ,μ):=⟨ρ,˙x⟩−L0(t,x,˙x)−⟨μ,φ(t,x,˙x)⟩, |
where ρ∈Rn denotes the adjoint variable and μ∈Rs is the associated multiplier of the mixed constraints.
∙ Given (ρ,μ)∈X×Us, and λ1,…,λK, for all (t,x,˙x)∈T×Rn×Rn, define the following function associated to the Hamiltonian,
F0(t,x,˙x):=−H(t,x,˙x,ρ(t),μ(t))−⟨˙ρ(t),x⟩. |
∙ Given (ρ,μ)∈X×Us and λ1,…,λK, for any xb admissible define the functional J0 by
J0(xb):=⟨ρ(t1),x(t1)⟩−⟨ρ(t0),x(t0)⟩+l0(b)+∫t1t0F0(˜x(t))dt. |
∙ The notation yβ refers to any element (y,β) in A.
∙ Given (ρ,μ)∈X×Us, and λ1,…,λK, for any xb∈A with ˙x∈L∞(T;Rn) and any yβ∈A consider the first variations of J0 and Iγ (γ=1,…,K) with respect to xb over yβ which are given, respectively, by
J′0(xb;yβ):=⟨ρ(t1),y(t1)⟩−⟨ρ(t0),y(t0)⟩+l′0(b)β+∫t1t0{F0x(˜x(t))y(t)+F0˙x(˜x(t))˙y(t)}dt, |
I′γ(xb;yβ):=l′γ(b)β+∫t1t0{Lγx(˜x(t))y(t)+Lγ˙x(˜x(t))˙y(t)}dt. |
∙ For all (t,x,˙x)∈T×Rn×Rn, denote by
Ia(t,x,˙x):={α∈R∣φα(t,x,˙x)=0}, |
the set of active indices of (t,x,˙x) with respect to the mixed inequality constraints.
∙ For all xb∈A, denote by
ia(xb):={i=1,…,k∣Ii(xb)=0}, |
the set of active indices of xb with respect to the isoperimetric inequality constraints.
∙ Given xb∈A, let Y(xb) be the set of all yβ∈A with ˙y∈L2(T;Rn) satisfying
{y(ti)=Ψ′i(b)β(i=0,1),I′i(xb;yβ)≤0(i∈ia(xb)),I′j(xb;yβ)=0(j=k+1,…,K),φαx(˜x(t))y(t)+φα˙x(˜x(t))˙y(t)≤0(a.e. in T,α∈Ia(˜x(t))),φβx(˜x(t))y(t)+φβ˙x(˜x(t))˙y(t)=0(a.e. in T,β∈S). |
The set Y(xb) will be called the set of admissible variations along xb.
∙ Given (ρ,μ)∈X×Us, and λ1,…,λK, for any xb∈A with ˙x∈L∞(T;Rn) and any yβ∈A with ˙y∈L2(T;Rn), we define the second variation of J0 with respect to xb over yβ, by
J′′0(xb;yβ):=⟨l′′0(b)β,β⟩+∫t1t02Ω0(x;t,y(t),˙y(t))dt, |
where for all (t,y,˙y)∈T×Rn×Rn,
2Ω0(x;t,y,˙y):=⟨y,F0xx(˜x(t))y⟩+2⟨y,F0x˙x(˜x(t))˙y⟩+⟨˙y,F0˙x˙x(˜x(t))˙y⟩. |
∙ Given (ρ,μ)∈X×Us, λ1,…,λK and x0b0∈A, we say that x0b0 is singular, if for some τ∈T, |H˙x˙x(˜x0(τ),ρ(τ),μ(τ))|=0. It satisfies the Legendre condition if
F0˙x˙x(˜x0(t))=−H˙x˙x(˜x0(t),ρ(t),μ(t))≥0(a.e. in T) |
and the strengthened Legendre condition, if F0˙x˙x(˜x0(t))>0 (t∈T).
∙ Denote by E0 the Weierstrass excess function of F0, given by
E0(t,x,˙x,u):=F0(t,x,u)−F0(t,x,˙x)−F0˙x(t,x,˙x)(u−˙x). |
∙ Similarly, the Weierstrass excess function of Lγ (γ=1,…,K), is given by
Eγ(t,x,˙x,u):=Lγ(t,x,u)−Lγ(t,x,˙x)−Lγ˙x(t,x,˙x)(u−˙x). |
∙ For all π=(π1,…,πn)∗∈Rn, set
V(π):=(1+|π|2)1/2−1. |
∙ For all x∈X, define
D(x):=V(x(t0))+∫t1t0V(˙x(t))dt. |
∙ As we mentioned above the symbol ∗ denotes transpose.
It is well-known that, under certain normality assumptions (see for example [10]), if x0b0 is a strong minimum of (P), then there exist ρ∈X and μ∈Us with μα(t)≥0 and μα(t)φα(˜x0(t))=0 (α∈R,a.e. in T) and multipliers λ1,…,λK with λi≥0 and λiIi(x0b0)=0 (i=1,…,k) such that
˙ρ(t)=−H∗x(˜x0(t),ρ(t),μ(t)) and H∗˙x(˜x0(t),ρ(t),μ(t))=0(a.e. in T). |
The relations given above are the Euler-Lagrange equations of the constrained problem (P) and if (x0,ρ,μ) satisfies the Euler-Lagrange equations, then (x0,ρ,μ) will be called an extremal.
The following theorem is the main result of the article. Given an admissible arc x0b0 with ˙x0 neither continuous nor piecewise continuous but only essentially bounded, this theorem gives sufficient conditions assuring that x0b0 is a strong minimum of problem (P). Hypothesis (ⅰ) of Theorem 2.1 is known as the transversality condition, hypothesis (ⅱ) arises from the original proof of the theorem, condition (ⅲ) is the necessary condition of Legendre, but not its strengthened version, hypothesis (ⅳ) is the positivity of the second variation over the set of all nonnull admissible variations and finally, hypothesis (ⅴ) involves three conditions related to the Weierstrass excess functions.
2.1 Theorem: Let x0b0 be an admissible arc with ˙x0∈L∞(T;Rn). Assume that Ia(˜x0(⋅)) is piecewise constant on T, and there exist (ρ,μ)∈X×Us with μα(t)≥0 and μα(t)φα(˜x0(t))=0 (α∈R,a.e. in T), two positive numbers h,ϵ, and multipliers λ1,…,λK with λi≥0 and λiIi(x0b0)=0 (i=1,…,k) such that (x0,ρ,μ) is an extremal and the following holds:
(i) l′0(b0)+ρ∗(t1)Ψ′1(b0)−ρ∗(t0)Ψ′0(b0)=0.
(ii) ρ∗(t1)Ψ′′1(b0;β)−ρ∗(t0)Ψ′′0(b0;β)≥0for all β∈Rp.
(iii) H˙x˙x(˜x0(t),ρ(t),μ(t))≤0 (a.e. in T).
(iv) J′′0(x0b0;yβ)>0 for all nonnull yβ∈Y(x0b0).
(v) For all xb admissible with ‖x−x0‖C<ϵ,
a. E0(t,x(t),˙x0(t),˙x(t))≥0 (a.e. in T).
b. ∫t1t0E0(t,x(t),˙x0(t),˙x(t))dt≥h∫t1t0V(˙x(t)−˙x0(t))dt.
c. ∫t1t0E0(t,x(t),˙x0(t),˙x(t))dt≥h|∫t1t0Eγ(t,x(t),˙x0(t),˙x(t))dt| (γ=1,…,K).
Then for some θ1,θ2>0 and all admissible trajectories xb satisfying ‖xb−x0b0‖<θ1,
I(xb)≥I(x0b0)+θ2min{|b−b0|2,D(x−x0)}. |
In particular, x0b0 is a strong minimum of (P).
Suppose we are given an interval T:=[t0,t1] in R, two sets B0,B1⊂Rn and functions ℓ(x1,x2):Rn×Rn→R, ℓγ(x1,x2):Rn×Rn→R (γ=1,…,K), L(t,x,˙x):T×Rn×Rn→R, Lγ(t,x,˙x):T×Rn×Rn→R (γ=1,…,K) and ϕ(t,x,˙x):T×Rn×Rn→Rs. Set
ˉR:={(t,x,˙x)∈T×Rn×Rn∣ϕα(t,x,˙x)≤0(α∈R),ϕβ(t,x,˙x)=0(β∈S)} |
where R:={1,…,r} and S:={r+1,…,s} (r=0,1,…,s). If r=0 then R=∅ and we disregard statements involving φα. Similarly, if r=s then S=∅ and we disregard statements involving φβ.
It will be assumed throughout this section that L, Lγ (γ=1,…,K) and ϕ=(ϕ1,…,ϕs) have first and second derivatives with respect to x and ˙x. Moreover, we shall assume that the functions ℓ, ℓγ (γ=1,…,K) are of class C2 on Rn×Rn.
Also, if we denote by c(t,x,˙x) either L(t,x,˙x), Lγ(t,x,˙x) (γ=1,…,K), ϕ(t,x,˙x) or any of its partial derivatives of order less than or equal to two with respect to x and ˙x, we shall assume that all the assumptions posed in Section 2 in the statement of the problem are satisfied.
As in Section 2, X will denote the space of absolutely continuous functions mapping T to Rn and Us:=L∞(T;Rs) the space of essentially bounded functions mapping T to Rs.
The nonparametric calculus of variations problem we shall deal with, denoted by (ˉP), consists in minimizing the functional
J(x):=ℓ(x(t0),x(t1))+∫t1t0L(t,x(t),˙x(t))dt |
over all x∈X satisfying the constraints
{c(t,x(t),˙x(t))is integrable on T.x(ti)∈Bi for i=0,1.Ji(x):=ℓi(x(t0),x(t1))+∫t1t0Li(t,x(t),˙x(t))dt≤0(i=1,…,k).Jj(x):=ℓj(x(t0),x(t1))+∫t1t0Lj(t,x(t),˙x(t))dt=0(j=k+1,…,K).(t,x(t),˙x(t))∈ˉR(a.e. in T). |
Elements x in X will be called arcs or trajectories, and a trajectory x is admissible if it satisfies the constraints.
An arc x0 solves (ˉP) if it is admissible and J(x0)≤J(x) for all admissible arcs x. For strong minima, an admissible arc x0 is called a strong minimum of (ˉP) if it is a minimum of J relative to the norm
‖x‖:=supt∈T|x(t)|, |
that is, if for some ϵ>0, J(x0)≤J(x) for all admissible arcs satisfying ‖x−x0‖<ϵ.
Let Ψ:Rn→Rn×Rn be any function of class C2 such that B0×B1⊂Ψ(Rn). Associate the nonparametric problem (ˉP) with the parametric problem of Section 2, which we denote by (PΨ), that is, (PΨ) will be the parametric problem given in Section 2, with p=n, B=Ψ−1(B0×B1), l=ℓ∘Ψ, lγ=ℓγ∘Ψ (γ=1,…,K), L=L, Lγ=Lγ (γ=1,…,K), φ=ϕ and Ψ0,Ψ1 the components of Ψ, that is, Ψ=(Ψ0,Ψ1). Recall that the notation xb means (x,b) where b∈Rn is a parameter.
3.1 Lemma: The following is satisfied:
(i) xb is an admissible arc of (PΨ) if and only if x is an admissible arc of (ˉP) and b∈Ψ−1(x(t0),x(t1)).
(ii) If xb is an admissible arc of (PΨ), then
J(x)=I(xb). |
(iii) If x0b0 is a solution of (PΨ), then x0 is a solution of (ˉP).
Proof: Conditions (ⅰ) and (ⅱ) follow from the definitions of the problems. Now, let x an admissible arc of (ˉP) and let b∈Ψ−1(x(t0),x(t1)). By (ⅰ), x0 is an admissible arc of (ˉP) and xb is an admissible arc of (PΨ). Then by (ⅱ),
J(x0)=I(x0b0)≤I(xb)=J(x) |
which shows (ⅲ).
The following corollary which is a consequence of Theorem 2.1 and Lemma 3.1, provides a set of sufficient conditions of problem (ˉP).
3.2 Corollary: Let Ψ:Rn→Rn×Rn be any function of class C2 such that B0×B1⊂Ψ(Rn) and let (PΨ) be the parametric problem defined in the previous paragraph of Lemma 3.1. Let x0b0 be an admissible arc of (PΨ) with ˙x0∈L∞(T;Rn). Assume that Ia(˜x0(⋅)) is piecewise constant on T, and there exist (ρ,μ)∈X×Us with μα(t)≥0 and μα(t)φα(˜x0(t))=0 (α∈R,a.e. in T), two positive numbers h,ϵ, and multipliers λ1,…,λK with λi≥0 and λiIi(x0b0)=0 (i=1,…,k) such that (x0,ρ,μ) is an extremal and the following holds:
(i) l′0(b0)+ρ∗(t1)Ψ′1(b0)−ρ∗(t0)Ψ′0(b0)=0.
(ii) ρ∗(t1)Ψ′′1(b0;β)−ρ∗(t0)Ψ′′0(b0;β)≥0for all β∈Rn.
(iii) H˙x˙x(˜x0(t),ρ(t),μ(t))≤0 (a.e. in T).
(iv) J′′0(x0b0;yβ)>0 for all nonnull yβ∈Y(x0b0).
(v) For all xb admissible with ‖x−x0‖<ϵ,
a. E0(t,x(t),˙x0(t),˙x(t))≥0 (a.e. in T).
b. ∫t1t0E0(t,x(t),˙x0(t),˙x(t))dt≥h∫t1t0V(˙x(t)−˙x0(t))dt.
c. ∫t1t0E0(t,x(t),˙x0(t),˙x(t))dt≥h|∫t1t0Eγ(t,x(t),˙x0(t),˙x(t))dt| (γ=1,…,K).
Then, x0 is a strong minimum of (ˉP).
In this section we state two auxiliary results which will be used to prove Theorem 2.1. The proof of these results will be given in Section 6. As before X denotes AC(T;Rn).
In the following two lemmas, we shall assume that we are given x0∈X and {xq} a sequence in X such that
limq→∞D(xq−x0)=0 and dq:=[2D(xq−x0)]1/2>0(q∈N). |
For all q∈N and t∈T, let
yq(t):=xq(t)−x0(t)dq. |
We say that ˙xq(t)→˙x0(t) almost uniformly on T, if for any ϵ>0, there exists Uϵ⊂T measurable with m(Uϵ)<ϵ such that ˙xq(t)→˙x0(t) uniformly on T∖Uϵ.
4.1 Lemma: For some subsequence of {xq}, again denoted by {xq}, and some y0∈X with ˙y0∈L2(T;Rn), ˙xq(t)→˙x0(t) almost uniformly on T, yq(t)→y0(t) uniformly on T and {˙yq} converges weakly in L1(T;Rn) to ˙y0.
4.2 Lemma: Let U⊂T measurable, R0∈L∞(U;Rn×n) and {Rq} a sequence in L∞(U;Rn×n). If ˙xq(t)→˙x0(t) uniformly on U, Rq(t)→R0(t) uniformly on U and R0(t)≥0 (t∈U), then
lim infq→∞∫U⟨Rq(t)˙yq(t),˙yq(t)⟩dt≥∫U⟨R0(t)˙y0(t),˙y0(t)⟩dt. |
The proof of Theorem 2.1 will be divided in three Lemmas. In Lemmas 5.1, 5.2 and 5.3 below, we shall be assuming that all the hypotheses of Theorem 2.1 are satisfied. Before enunciating the lemmas, we shall introduce some definitions.
First of all, note that given x=(x1,…,xn)∗∈Rn and b=(b1,…,bp)∗∈Rp, if we define xi,bj∈Rn+p by xi:=(x1,…,xn,0,…,0)∗ and bj:=(0,…,0,b1,…,bp)∗, then
xi+bj=(x1,…,xn,b1,…,bp)∗=(xb)∈Rn+p. |
Define ˜F0:T×Rn+p×Rn→R by
˜F0(t,ξ,˙x):=l0(ξn+1,…,ξn+p)t1−t0+F0(t,ξ1,…,ξn,˙x). |
Observe that the Weierstrass excess function ˜E0:T×Rn+p×Rn×Rn→R of ˜F0 is given by
˜E0(t,ξ,˙x,u):=˜F0(t,ξ,u)−˜F0(t,ξ,˙x)−˜F0˙x(t,ξ,˙x)(u−˙x). |
It is clear that for all (t,x,˙x,u)∈T×Rn×Rn×Rn and all b∈Rp,
˜E0(t,xi+bj,˙x,u)=E0(t,x,˙x,u). |
Define
˜J0(xb):=⟨ρ(t1),x(t1)⟩−⟨ρ(t0),x(t0)⟩+∫t1t0˜F0(t,x(t)i+bj,˙x(t))dt. |
We have that J0(xb)=˜J0(xb) for all xb∈A, and
˜J0(xb)=˜J0(x0b0)+˜J′0(x0b0;xb−x0b0)+˜K0(x0b0;xb)+˜E0(x0b0;xb) | (1) |
where
˜E0(x0b0;xb):=∫t1t0˜E0(t,x(t)i+bj,˙x0(t),˙x(t))dt,˜K0(x0b0;xb):=∫t1t0{˜M0(t,x(t)i+bj)+⟨˙x(t)−˙x0(t),˜N0(t,x(t)i+bj)⟩}dt,˜J′0(x0b0;xb−x0b0):=⟨ρ(t1),x(t1)−x0(t1)⟩−⟨ρ(t0),x(t0)−x0(t0)⟩+∫t1t0{˜F0ξ(t,x0(t)i+b0j,˙x0(t))([x(t)−x0(t)]i+[b−b0]j)+˜F0˙x(t,x0(t)i+b0j,˙x0(t))(˙x(t)−˙x0(t))}dt, |
and ˜M0, ˜N0 are given by
˜M0(t,xi+bj):=˜F0(t,xi+bj,˙x0(t))−˜F0(t,x0(t)i+b0j,˙x0(t))−˜F0ξ(t,x0(t)i+b0j,˙x0(t))([x−x0(t)]i+[b−b0]j), |
˜N0(t,xi+bj):=˜F∗0˙x(t,xi+bj,˙x0(t))−˜F∗0˙x(t,x0(t)i+b0j,˙x0(t)). |
We have,
˜M0(t,xi+bj)=12⟨[x−x0(t)]i+[b−b0]j,˜P0(t,xi+bj)([x−x0(t)]i+[b−b0]j)⟩, | (2 |
˜N0(t,xi+bj)=˜Q0(t,xi+bj)([x−x0(t)]i+[b−b0]j), | (2 |
where
˜P0(t,xi+bj):=2∫10(1−λ)˜F0ξξ(t,[x0(t)+λ(x−x0(t))]i+[b0+λ(b−b0)]j,˙x0(t))dλ,˜Q0(t,xi+bj):=∫10˜F0˙xξ(t,[x0(t)+λ(x−x0(t))]i+[b0+λ(b−b0)]j,˙x0(t))dλ. |
5.1 Lemma: For some ν,κ>0 (κ≤ϵ) and any admissible arc xb satisfying ‖xb−x0b0‖<κ,
˜E0(x0b0;xb)≥h[D(x−x0)−V(x(t0)−x0(t0))],|˜K0(x0b0;xb)|≤ν‖xb−x0b0‖[1+D(x−x0)]. |
Proof: By condition (v)(b) of Theorem 2.1, given xb admissible with ‖xb−x0b0‖<ϵ,
˜E0(x0b0;xb)=∫t1t0˜E0(t,x(t)i+bj,˙x0(t),˙x(t))dt=∫t1t0E0(t,x(t),˙x0(t),˙x(t))dt≥h∫t1t0V(˙x(t)−˙x0(t))dt=h[D(x−x0)−V(x(t0)−x0(t0))]. |
On the other hand, by (2) and using [t,b] in order to denote (t,x(t)i+bj), observe that for some constants c0,c1>0, for all xb admissible with ‖xb−x0b0‖<1 and almost all t∈T,
|˜M0[t,b]+⟨˙x(t)−˙x0(t),˜N0[t,b]⟩|=|12⟨[x(t)−x0(t)]i+[b−b0]j,˜P0[t,b]([x(t)−x0(t)]i+[b−b0]j)+2˜Q∗0[t,b](˙x(t)−˙x0(t))⟩|≤12|[x(t)−x0(t)]i+[b−b0]j|⋅[|˜P0[t,b]||[x(t)−x0(t)]i+[b−b0]j|+2|˜Q∗0[t,b]||˙x(t)−˙x0(t)|]≤c0|[x(t)−x0(t)]i+[b−b0]j|⋅[|[x(t)−x0(t)]i+[b−b0]j|+|˙x(t)−˙x0(t)|]≤c0|[x(t)−x0(t)]i+[b−b0]j|⋅[|x(t)−x0(t)|+|b−b0|+|˙x(t)−˙x0(t)|]≤c0|[x(t)−x0(t)]i+[b−b0]j|⋅[‖xb−x0b0‖+|˙x(t)−˙x0(t)|]≤c0|[x(t)−x0(t)]i+[b−b0]j|⋅[1+|˙x(t)−˙x0(t)|]≤c1|[x(t)−x0(t)]i+[b−b0]j|⋅[1+|˙x(t)−˙x0(t)|2]1/2. |
Setting ν:=max{c1,c1(t1−t0)}, for all xb admissible with ‖xb−x0b0‖<1,
|˜K0(x0b0;xb)|≤c1‖xb−x0b0‖∫t1t0[1+V(˙x(t)−˙x0(t))]dt≤ν‖xb−x0b0‖[1+D(x−x0)−V(x(t0)−x0(t0))]≤ν‖xb−x0b0‖[1+D(x−x0)] |
and so the conclusion of the lemma is obtained with κ=min{ϵ,1}.
5.2 Lemma: If conclusion of Theorem 2.1 is false, then there exists a subsequence {xqbq} of admissible arcs such that
limq→∞D(xq−x0)=0 and dq:=[2D(xq−x0)]1/2>0(q∈N). |
Proof: If conclusion of Theorem 2.1 is false, then for all θ1,θ2>0, there exists an admissible arc xb such that
‖xb−x0b0‖<θ1andI(xb)<I(x0b0)+θ2min{|b−b0|2,D(x−x0)}. | (3) |
Since
μα(t)≥0 (α∈R,a.e. in T) and λi≥0 (i=1,…,k), |
if xb is admissible, then I(xb)≥J0(xb). Also, since
μα(t)φα(˜x0(t))=0 (α∈R,a.e. in T) and λiIi(x0b0)=0 (i=1,…,k), |
then I(x0b0)=J0(x0b0). Therefore (3) implies that, for all θ1,θ2>0, there exists xb admissible with
‖xb−x0b0‖<θ1 and J0(xb)<J0(x0b0)+θ2min{|b−b0|2,D(x−x0)}. |
Let κ and ν be the positive numbers given in Lemma 5.1. Thus, if conclusion of Theorem 2.1 is false, then for all q∈N, there exists xqbq admissible such that
‖xqbq−x0b0‖<min{κ,1/q},J0(xqbq)−J0(x0b0)<min{|bq−b0|2q,D(xq−x0)q}. | (4) |
Clearly, D(xq−x0)=0 if and only if xq=x0. Then, by the second relation of (4),
D(xq−x0)=0⟹bq≠b0. |
Suppose D(xq−x0)=0 for infinitely many q's. For i=0,1, we have
0=xq(ti)−x0(ti)=Ψi(bq)−Ψi(b0)=∫10Ψ′i(b0+λ[bq−b0])(bq−b0)dλ, | (5) |
0=Ψi(bq)−Ψi(b0)=Ψ′i(b0)(bq−b0)+∫10(1−λ)Ψ′′i(b0+λ[bq−b0];bq−b0)dλ. | (6) |
Denoting by (bq,b0) the line segment in Rp joining the points bq and b0, by the second relation of (4), by condition (ⅰ) of Theorem 2.1, by (6), and the mean value theorem, there exists Ξq∈(bq,b0) such that
0>J0(x0bq)−J0(x0b0)=l0(bq)−l0(b0)=l′0(b0)(bq−b0)+12⟨l′′0(Ξq)(bq−b0),bq−b0⟩=ρ∗(t0)Ψ′0(b0)(bq−b0)−ρ∗(t1)Ψ′1(b0)(bq−b0)+12⟨l′′0(Ξq)(bq−b0),bq−b0⟩=1∑i=0(−1)i+1∫10(1−λ)ρ∗(ti)Ψ′′i(b0+λ[bq−b0];bq−b0)dλ+12⟨l′′0(Ξq)(bq−b0),bq−b0⟩. | (7) |
Choose an appropriate subsequence of {(bq−b0)/|bq−b0|} (without relabeling), such that
limq→∞bq−b0|bq−b0|=β0 | (8) |
for some β0∈Rp with |β0|=1. By (5),
Ψ′i(b0)β0=0(i=0,1). | (9) |
For all (t,ξ,˙x)∈T×Rn+p×Rn and γ=1,…,K, if we set
˜Lγ(t,ξ,˙x):=lγ(ξn+1,…,ξn+p)t1−t0+Lγ(t,ξ1,…,ξn,˙x), |
and for all (t,ξ,˙x,u)∈T×Rn+p×Rn×Rn and γ=1,…,K, if we set
˜Eγ(t,ξ,˙x,u):=˜Lγ(t,ξ,u)−˜Lγ(t,ξ,˙x)−˜Lγ˙x(t,ξ,˙x)(u−˙x), |
we have that for all xb∈A and γ=1,…,K,
˜Iγ(xb)=˜Iγ(x0b0)+˜I′γ(x0b0;xb−x0b0)+˜Kγ(x0b0;xb)+˜Eγ(x0b0;xb) |
where
˜Eγ(x0b0;xb):=∫t1t0˜Eγ(t,x(t)i+bj,˙x0(t),˙x(t))dt,˜Kγ(x0b0;xb):=∫t1t0{˜Mγ(t,x(t)i+bj)+⟨˙x(t)−˙x0(t),˜Nγ(t,x(t)i+bj)⟩}dt,˜I′γ(x0b0;xb−x0b0):=∫t1t0{˜Lγξ(t,x0(t)i+b0j,˙x0(t))([x(t)−x0(t)]i+[b−b0]j)+˜Lγ˙x(t,x0(t)i+b0j,˙x0(t))(˙x(t)−˙x0(t))}dt,˜Iγ(xb):=∫t1t0˜Lγ(t,x(t)i+bj,˙x(t))dt, |
and ˜Mγ, ˜Nγ are given by
˜Mγ(t,xi+bj):=˜Lγ(t,xi+bj,˙x0(t))−˜Lγ(t,x0(t)i+b0j,˙x0(t))−˜Lγξ(t,x0(t)i+b0j,˙x0(t))([x−x0(t)]i+[b−b0]j), |
˜Nγ(t,xi+bj):=˜L∗γ˙x(t,xi+bj,˙x0(t))−˜L∗γ˙x(t,x0(t)i+b0j,˙x0(t)). |
We have
˜Mγ(t,xi+bj)=12⟨[x−x0(t)]i+[b−b0]j,˜Pγ(t,xi+bj)([x−x0(t)]i+[b−b0]j)⟩, |
˜Nγ(t,xi+bj)=˜Qγ(t,xi+bj)([x−x0(t)]i+[b−b0]j), |
where
˜Pγ(t,xi+bj):=2∫10(1−λ)˜Lγξξ(t,[x0(t)+λ(x−x0(t))]i+[b0+λ(b−b0)]j,˙x0(t))dλ,˜Qγ(t,xi+bj):=∫10˜Lγ˙xξ(t,[x0(t)+λ(x−x0(t))]i+[b0+λ(b−b0)]j,˙x0(t))dλ. |
Since x0bq and x0b0 are admissible, for all i∈ia(x0b0), we have
0≥Ii(x0bq)=Ii(x0bq)−Ii(x0b0)=˜Ii(x0bq)−˜Ii(x0b0)=˜I′i(x0b0;x0bq−x0b0)+˜Ki(x0b0;x0bq)+˜Ei(x0b0;x0bq)=I′i(x0b0;x0bq−x0b0)+˜Ki(x0b0;x0bq)=l′i(b0)(bq−b0)+˜Ki(x0b0;x0bq). |
As one readily verifies, for all γ=1,…,K,
limq→∞˜Kγ(x0b0;x0bq)|bq−b0|=0. |
Then, for i∈ia(x0b0),
0≥l′i(b0)β0=I′i(x0b0;0β0). | (10) |
On the other hand, once again since x0bq and x0b0 are admissible, for all j=k+1,…,K, we have
0=Ij(x0bq)−Ij(x0b0)=˜Ij(x0bq)−˜Ij(x0b0)=˜I′j(x0b0;x0bq−x0b0)+˜Kj(x0b0;x0bq)+˜Ej(x0b0;x0bq)=I′j(x0b0;x0bq−x0b0)+˜Kj(x0b0;x0bq)=l′j(b0)(bq−b0)+˜Kj(x0b0;x0bq). |
Then, for j=k+1,…,K,
0=l′j(b0)β0=I′j(x0b0;0β0). | (11) |
Consequently, by (9), (10) and (11), 0β0∈Y(x0b0).
By (7), (8) and condition (ⅱ) of Theorem 2.1, it follows that
0≥12[ρ∗(t1)Ψ′′1(b0;β0)−ρ∗(t0)Ψ′′0(b0;β0)+⟨l′′0(b0)β0,β0⟩]≥12⟨l′′0(b0)β0,β0⟩=12J′′0(x0b0;0β0) |
which contradicts (ⅳ) of Theorem 2.1. Therefore, we may assume that for all q∈N,
dq=[2D(xq−x0)]1/2>0. |
Since (x0,ρ,μ) is an extremal, for all q∈N,
˜J′0(x0b0;xqbq−x0b0)=⟨ρ(t1),xq(t1)−x0(t1)⟩−⟨ρ(t0),xq(t0)−x0(t0)⟩+l′0(b0)(bq−b0) |
and thus
limq→∞˜J′0(x0b0;xqbq−x0b0)=0. | (12) |
By (1), the first relation of (4) and Lemma 5.1, for all q∈N,
˜J0(xqbq)−˜J0(x0b0)=˜J′0(x0b0;xqbq−x0b0)+˜K0(x0b0;xqbq)+˜E0(x0b0;xqbq)≥˜J′0(x0b0;xqbq−x0b0)−ν‖xqbq−x0b0‖+D(xq−x0)(h−ν‖xqbq−x0b0‖)−hV(xq(t0)−x0(t0)), |
then, by (4), for all q∈N,
D(xq−x0)(h−νq−1q)<νq+hV(xq(t0)−x0(t0))−˜J′0(x0b0;xqbq−x0b0). |
By (12),
limq→∞D(xq−x0)=0. |
5.3 Lemma: If conclusion of Theorem 2.1 is false, then condition (iv) of Theorem 2.1 is false.
Proof: Let {xqbq} be the sequence of admissible arcs given in Lemma 5.2. Then,
limq→∞D(xq−x0)=0 and dq=[2D(xq−x0)]1/2>0(q∈N). |
Case (1): Suppose first that the sequence {(bq−b0)/dq} is bounded in Rp.
For all q∈N and t∈T, define
yq(t):=xq(t)−x0(t)dq and ωq(t):=yq(t)i+bq−b0dqj. |
By Lemma 4.1, for some y0∈X with ˙y0∈L2(T;Rn) and a subsequence of {xq}, again denoted by {xq}, {˙yq} converges weakly in L1(T;Rn) to ˙y0. Once again, by Lemma 4.1,
limq→∞yq(t)=y0(t)uniformly on T. | (13) |
Since the sequence {(bq−b0)/dq} is bounded in Rp, then we may assume that there exists some β0∈Rp such that
limq→∞bq−b0dq=β0. | (14) |
First, we are going to show that for i=0,1,
y0(ti)=Ψ′i(b0)β0. | (15) |
Note first that for i=0,1 and all q∈N, we have that
yq(ti)=∫10Ψ′i(b0+λ[bq−b0])(bq−b0)dqdλ. | (16) |
By (13), (14) and (16), we obtain (15). Now, we claim that
J′′0(x0b0;y0β0)≤0 and y0β0≢(0,0). | (17) |
To prove it, observe that by (2), (13) and (14),
limq→∞˜M0(t,xq(t)i+bqj)d2q=limq→∞12⟨ωq(t),˜P0(t,xq(t)i+bqj)ωq(t)⟩=12⟨y0(t)i+β0j,˜F0ξξ(t,x0(t)i+b0j,˙x0(t))[y0(t)i+β0j]⟩, |
limq→∞˜N0(t,xq(t)i+bqj)dq=limq→∞˜Q0(t,xq(t)i+bqj)ωq(t)=˜F0˙xξ(t,x0(t)i+b0j,˙x0(t))[y0(t)i+β0j] |
both uniformly on T. With this in mind and since {˙yq} converges weakly in L1(T;Rn) to ˙y0, we have
limq→∞˜K0(x0b0;xqbq)d2q=12∫t1t0{⟨y0(t)i+β0j,˜F0ξξ(t,x0(t)i+b0j,˙x0(t))[y0(t)i+β0j]⟩+2⟨˙y0(t),˜F0˙xξ(t,x0(t)i+b0j,˙x0(t))[y0(t)i+β0j]⟩}dt. | (18) |
Since (x0,ρ,μ) is an extremal and by condition (ⅰ) of Theorem 2.1, it follows that
limq→∞˜J′0(x0b0;xqbq−x0b0)d2q=limq→∞1d2q[⟨ρ(t1),xq(t1)−x0(t1)⟩−⟨ρ(t0),xq(t0)−x0(t0)⟩+l′0(b0)(bq−b0)]=limq→∞1d2q[ρ∗(t1)(Ψ1(bq)−Ψ1(b0)−Ψ′1(b0)(bq−b0))−ρ∗(t0)(Ψ0(bq)−Ψ0(b0)−Ψ′0(b0)(bq−b0))]=limq→∞1d2q∫101∑i=0(−1)i+1(1−λ)ρ∗(ti)Ψ′′i(b0+λ[bq−b0];bq−b0)dλ=12[ρ∗(t1)Ψ′′1(b0;β0)−ρ∗(t0)Ψ′′0(b0;β0)]. | (19) |
Consequently, by (1), (4), (19), and condition (ⅱ) of Theorem 2.1,
0≥limq→∞˜K0(x0b0;xqbq)d2q+lim infq→∞˜E0(x0b0;xqbq)d2q. | (20) |
Now, let us show that
lim infq→∞˜E0(x0b0;xqbq)d2q≥12∫t1t0⟨˙y0(t),˜F0˙x˙x(t,x0(t)i+b0j,˙x0(t))˙y0(t)⟩dt. | (21) |
To this end, let U a measurable subset of T such that ˙xq(t)→˙x0(t) uniformly on U. For all q∈N and t∈U, we have that
1d2q˜E0(t,xq(t)i+bqj,˙x0(t),˙xq(t))=12⟨˙yq(t),Rq(t)˙yq(t)⟩, |
where
Rq(t):=2∫10(1−λ)˜F0˙x˙x(t,xq(t)i+bqj,˙x0(t)+λ[˙xq(t)−˙x0(t)])dλ. |
Clearly,
limq→∞Rq(t)=R0(t):=˜F0˙x˙x(t,x0(t)i+b0j,˙x0(t))uniformly on U. |
By condition (ⅲ) of Theorem 2.1, we have
˜F0˙x˙x(t,x0(t)i+b0j,˙x0(t))=R0(t)≥0(t∈U). |
With this in mind, and since by (v)(a) of Theorem 2.1 for all q∈N,
E0(t,xq(t),˙x0(t),˙xq(t))≥0(a.e. in T), |
by Lemma 4.2,
lim infq→∞˜E0(x0b0;xqbq)d2q=lim infq→∞1d2q∫t1t0˜E0(t,xq(t)i+bqj,˙x0(t),˙xq(t))dt=lim infq→∞1d2q∫t1t0E0(t,xq(t),˙x0(t),˙xq(t))dt≥lim infq→∞1d2q∫UE0(t,xq(t),˙x0(t),˙xq(t))dt=lim infq→∞1d2q∫U˜E0(t,xq(t)i+bqj,˙x0(t),˙xq(t))dt=12lim infq→∞∫U⟨˙yq(t),Rq(t)˙yq(t)⟩dt≥12∫U⟨˙y0(t),R0(t)˙y0(t)⟩dt. |
As U can be chosen to differ from T by a set of an arbitrarily small measure and the function
t↦⟨˙y0(t),R0(t)˙y0(t)⟩ |
belongs to L1(T;R), this inequality holds when U=T, and this establishes (21). With this in mind, by (18) and (20), we have
0≥∫t1t0{⟨˙y0(t),˜F0˙x˙x(t,x0(t)i+b0j,˙x0(t))˙y0(t)⟩+2⟨˙y0(t),˜F0˙xξ(t,x0(t)i+b0j,˙x0(t))[y0(t)i+β0j]⟩+⟨y0(t)i+β0j,˜F0ξξ(t,x0(t)i+b0j,˙x0(t))[y0(t)i+β0j]⟩}dt=⟨l′′0(b0)β0,β0⟩+∫t1t0{⟨˙y0(t),F0˙x˙x(˜x0(t))˙y0(t)⟩+2⟨˙y0(t),F0˙xx(˜x0(t))y0(t)⟩+⟨y0(t),F0xx(˜x0(t))y0(t)⟩}dt=⟨l′′0(b0)β0,β0⟩+∫t1t02Ω0(x0;t,y0(t),˙y0(t))dt=J′′0(x0b0;y0β0). |
Now, let us show that y0β0≢(0,0). By (20), the first conclusion of Lemma 5.1, the fact that V(π)≤|π|2/2 for all π∈Rn,
0≥limq→∞˜K0(x0b0;xqbq)d2q+h2−h2lim supq→∞|xq(t0)−x0(t0)|2d2q=limq→∞˜K0(x0b0;xqbq)d2q+h2−h2lim supq→∞|Ψ0(bq)−Ψ0(b0)|2d2q=limq→∞˜K0(x0b0;xqbq)d2q+h2−h2lim supq→∞|∫10Ψ′0(b0+λ[bq−b0])(bq−b0dq)dλ|2=limq→∞˜K0(x0b0;xqbq)d2q+h2−h2|Ψ′0(b0)β0|2=limq→∞˜K0(x0b0;xqbq)d2q+h2−h2|y0(t0)|2. |
With this in mind and (18), the fact that y0β0≡(0,0) contradicts the positivity of h and this establishes (17).
Let us now show that
I′i(x0b0;y0β0)≤0(i∈ia(x0b0)). | (22) |
To this end, note that, for all γ=0,1,…,K,
limq→∞˜Mγ(t,xq(t)i+bqj)dq=limq→∞12⟨[xq(t)−x0(t)]i+[bq−b0]j,˜Pγ(t,xq(t)i+bqj)ωq(t)⟩=0, |
limq→∞˜Nγ(t,xq(t)i+bqj)=limq→∞˜Qγ(t,xq(t)i+bqj)([xq(t)−x0(t)]i+[bq−b0]j)=0, |
all uniformly on T and {˙yq} converges weakly in L1(T;Rn) to ˙y0, then for all γ=0,1,…,K,
limq→∞˜Kγ(x0b0;xqbq)dq=0. | (23) |
As in (19), we have
limq→∞˜J′0(x0b0;xqbq−x0b0)dq=limq→∞1dq1∑i=0(−1)i+1∫10(1−λ)ρ∗(ti)Ψ′′i(b0+λ[bq−b0];bq−b0)dλ=0. | (24) |
By (4), (23) and (24),
0≥lim supq→∞˜J0(xqbq)−˜J0(x0b0)dq=lim supq→∞˜E0(x0b0;xqbq)dq. |
Since ˜E0(x0b0;xqbq)≥0 (q∈N), then
limq→∞˜E0(x0b0;xqbq)dq=0. |
Therefore, by condition (v)(c) of Theorem 2.1, for all γ=1,…,K,
limq→∞˜Eγ(x0b0;xqbq)dq=0. | (25) |
As for all q∈N and i∈ia(x0b0),
0≥Ii(xqbq)=Ii(xqbq)−Ii(x0b0)=˜Ii(xqbq)−˜Ii(x0b0)=˜I′i(x0b0;xqbq−x0b0)+˜Ki(x0b0;xqbq)+˜Ei(x0b0;xqbq)=I′i(x0b0;xqbq−x0b0)+˜Ki(x0b0;xqbq)+˜Ei(x0b0;xqbq), |
then, by (23) and (25), for all i∈ia(x0b0),
0≥limq→∞I′i(x0b0;xqbq−x0b0)dq. |
Therefore, since yq(t)→y0(t) uniformly on T, {˙yq} converges weakly in L1(T;Rn) to ˙y0 and (bq−b0)/dq→β0, then for all i∈ia(x0b0),
0≥limq→∞I′i(x0b0;xqbq−x0b0)dq=I′i(x0b0;y0β0) |
which establishes (22).
Now, let us show that
I′j(x0b0;y0β0)=0(j=k+1,…,K). | (26) |
Indeed, as for all q∈N and j=k+1,…,K,
0=Ij(xqbq)−Ij(x0b0)=˜Ij(xqbq)−˜Ij(x0b0)=˜I′j(x0b0;xqbq−x0b0)+˜Kj(x0b0;xqbq)+˜Ej(x0b0;xqbq)=I′j(x0b0;xqbq−x0b0)+˜Kj(x0b0;xqbq)+˜Ej(x0b0;xqbq), |
by (23) and (25), for all j=k+1,…,K,
0=limq→∞I′j(x0b0;xqbq−x0b0)dq=I′j(x0b0;y0β0) |
which is precisely (26).
We claim that, for all α∈Ia(˜x0(t)),
φαx(˜x0(t))y0(t)+φα˙x(˜x0(t))˙y0(t)≤0(a.e. in T). | (27) |
Indeed, for all α∈R, q∈N, almost all t∈T and λ∈[0,1], define
Gαq(t;λ):=φα(t,x0(t)+λ[xq(t)−x0(t)],˙x0(t)+λ[˙xq(t)−˙x0(t)]), |
Wαq(t):=[−φα(˜xq(t))]1/2, |
Zα0(t):=−φαx(˜x0(t))y0(t)−φα˙x(˜x0(t))˙y0(t), |
where as usual (˜xq(t)):=(t,xq(t),˙xq(t)) (a.e. in T). Given t∈[t0,t1) a point of continuity of Ia(˜x0(⋅)) and α∈Ia(˜x0(t)), since Ia(˜x0(⋅)) is piecewise constant on T, there exists an interval [t,ˉt]⊂T with t<ˉt such that φα(˜x0(τ))=0 for all τ∈[t,ˉt]. Using the notation
[τ]:=(τ,x0(τ)+λ[xq(τ)−x0(τ)],˙x0(τ)+λ[˙xq(τ)−˙x0(τ)]), |
we have
0≤limq→∞∫[t,ˉt]∩U(Wαq(τ))2dqdτ=limq→∞1dq∫[t,ˉt]∩U{−φα(˜xq(τ))+φα(˜x0(τ))}dτ=−limq→∞1dq∫[t,ˉt]∩U{Gαq(τ;1)−Gαq(τ;0)}dτ=−limq→∞1dq∫[t,ˉt]∩U∫10∂∂λGαq(τ;λ)dλdτ=−limq→∞1dq∫[t,ˉt]∩U∫10{φαx[τ](xq(τ)−x0(τ))+φα˙x[τ](˙xq(τ)−˙x0(τ))}dλdτ=−limq→∞∫[t,ˉt]∩U∫10{φαx[τ]yq(τ)+φα˙x[τ]˙yq(τ)}dλdτ=∫[t,ˉt]∩U{−φαx(˜x0(τ))y0(τ)−φα˙x(˜x0(τ))˙y0(τ)}dτ=∫[t,ˉt]∩UZα0(τ)dτ. |
Since U can be chosen to differ from T by a set of an arbitrarily small measure, then
0≤∫ˉttZα0(τ)dτ. |
If Zα0(τ)<0 on a measurable set Θ such that Θ⊂[t,ˉt] and m(Θ)>0, then
0>∫Θ∩UZα0(τ)dτ=limq→∞∫Θ∩U(Wαq(τ))2dqdτ≥0 |
which is a contradiction. Therefore, Zα0(τ)≥0 a.e. in [t,ˉt] with t∈[t0,t1) an arbitrary point of continuity of Ia(˜x0(⋅)) and, therefore, Zα0(t)≥0 for almost all t∈T which shows (27).
Now, let us show that for all β∈S,
φβx(˜x0(t))y0(t)+φβ˙x(˜x0(t))˙y0(t)=0 (a.e. in T). | (28) |
Indeed, for all β∈S, q∈N, almost all t∈T and λ∈[0,1], define
Hβq(t;λ):=φβ(t,x0(t)+λ[xq(t)−x0(t)],˙x0(t)+λ[˙xq(t)−˙x0(t)]). |
For all β∈S, q∈N and almost all t∈T, we have
0=Hβq(t;1)−Hβq(t;0)=∫10∂∂λHβq(t;λ)dλ=∫10{φβx[t](xq(t)−x0(t))+φβ˙x[t](˙xq(t)−˙x0(t))}dλ. |
Therefore, for all β∈S, q∈N and almost all t∈T,
0=∫10{φβx[t]yq(t)+φβ˙x[t]˙yq(t)}dλ. | (29) |
By (29), for all t∈T and β∈S,
0=∫[t0,t]∩U{φβx(˜x0(τ))y0(τ)+φβ˙x(˜x0(τ))˙y0(τ)}dτ. |
Since, as before, U can be chosen to differ from T by a set of an arbitrarily small measure, then for all t∈T and β∈S,
0=∫tt0{φβx(˜x0(τ))y0(τ)+φβ˙x(˜x0(τ))˙y0(τ)}dτ |
and so (28) is verified. Consequently, from (15), (22), (26), (27) and (28), y0β0∈Y(x0b0). This fact together with (17) contradicts condition (ⅳ) of Theorem 2.1.
Case (2): Now, suppose that the sequence {(bq−b0)/dq} is not bounded. Then,
limq→∞|bq−b0dq|=+∞. | (30) |
Choose an appropriate subsequence of {(bq−b0)/|bq−b0|} (without relabeling), and ˉβ0∈Rp with |ˉβ0|=1, such that
limq→∞bq−b0|bq−b0|=ˉβ0. | (31) |
For all q∈N and t∈T, define
ˉωq(t):=xq(t)−x0(t)|bq−b0|i+bq−b0|bq−b0|j. |
By Lemma 4.1 and (30),
limq→∞xq(t)−x0(t)|bq−b0|=limq→∞yq(t)⋅dq|bq−b0|=y0(t)⋅0=0uniformly on T. | (32) |
For i=0,1 and all q∈N, we have
xq(ti)−x0(ti)|bq−b0|=∫10Ψ′i(b0+λ[bq−b0])(bq−b0|bq−b0|)dλ. | (33) |
By (31), (32) and (33), for i=0,1,
Ψ′i(b0)ˉβ0=0. | (34) |
Now, by (2), (31) and (32),
limq→∞˜M0(t,xq(t)i+bqj)|bq−b0|2=limq→∞12⟨ˉωq(t),˜P0(t,xq(t)i+bqj)ˉωq(t)⟩=12⟨0ˉβ0,˜F0ξξ(t,x0(t)i+b0j,˙x0(t))0ˉβ0⟩=⟨ˉβ0,l′′0(b0)ˉβ0⟩2(t1−t0), |
limq→∞˜N0(t,xq(t)i+bqj)|bq−b0|=limq→∞˜Q0(t,xq(t)i+bqj)ˉωq(t)=˜F0˙xξ(t,x0(t)i+b0j,˙x0(t))0ˉβ0=0 |
both uniformly on T. Together with Lemma 4.1, this implies that
limq→∞˜K0(x0b0;xqbq)|bq−b0|2=12⟨ˉβ0,l′′0(b0)ˉβ0⟩+limq→∞∫t1t0⟨dq|bq−b0|⋅˙yq(t),˜N0(t,xq(t)i+bqj)|bq−b0|⟩dt=12⟨ˉβ0,l′′0(b0)ˉβ0⟩. | (35) |
As in (19), we have
limq→∞˜J′0(x0b0;xqbq−x0b0)|bq−b0|2=12[ρ∗(t1)Ψ′′1(b0;ˉβ0)−ρ∗(t0)Ψ′′0(b0;ˉβ0)]. | (36) |
Even more, by (1), (4), (36), and condition (ⅱ) of Theorem 2.1,
0≥limq→∞˜K0(x0b0;xqbq)|bq−b0|2+lim infq→∞˜E0(x0b0;xqbq)|bq−b0|2. | (37) |
Consequently, since ˜E0(x0b0;xqbq)≥0 (q∈N), by (35) and (37),
0≥12⟨ˉβ0,l′′0(b0)ˉβ0⟩=12J′′0(x0b0;0ˉβ0). | (38) |
Let us now show that for all i∈ia(x0b0),
I′i(x0b0;0ˉβ0)≤0. | (39) |
To prove it, note that since
limq→∞xq(t)−x0(t)|bq−b0|=0 uniformly on T, |
and {˙yq} converges weakly in L1(T;Rn) to ˙y0, for all γ=0,1,…,K,
limq→∞˜Kγ(x0b0;xqbq)|bq−b0|=0. | (40) |
As in (24),
limq→∞˜J′0(x0b0;xqbq−x0b0)|bq−b0|=0. | (41) |
Then, by (4), (40) and (41),
0≥lim supq→∞˜J0(xqbq)−˜J0(x0b0)|bq−b0|=lim supq→∞˜E0(x0b0;xqbq)|bq−b0|≥0. |
Thus, by condition (v)(c) of Theorem 2.1, for all γ=1,…,K,
limq→∞˜Eγ(x0b0;xqbq)|bq−b0|=0. | (42) |
As for all q∈N and i∈ia(x0b0),
0≥Ii(xqbq)=Ii(xqbq)−Ii(x0b0)=˜Ii(xqbq)−˜Ii(x0b0)=˜I′i(x0b0;xqbq−x0b0)+˜Ki(x0b0;xqbq)+˜Ei(x0b0;xqbq)=I′i(x0b0;xqbq−x0b0)+˜Ki(x0b0;xqbq)+˜Ei(x0b0;xqbq), |
then, by (40) and (42), for all i∈ia(x0b0),
0≥limq→∞I′i(x0b0;xqbq−x0b0)|bq−b0|. |
Hence, since
{˙yq} converges weakly to ˙y0 in L1(T;Rn),dq|bq−b0|→0, |
limq→∞xq(t)−x0(t)|bq−b0|=0 uniformly on T, and bq−b0|bq−b0|→ˉβ0, |
for all i∈ia(x0b0),
0≥limq→∞I′i(x0b0;xqbq−x0b0)|bq−b0|=l′i(b0)ˉβ0=I′i(x0b0;0ˉβ0) |
which establishes (39). Finally, let us show that for all j=k+1,…,K,
I′j(x0b0;0ˉβ0)=0. | (43) |
Indeed, as for all q∈N and j=k+1,…,K,
0=Ij(xqbq)−Ij(x0b0)=˜Ij(xqbq)−˜Ij(x0b0)=˜I′j(x0b0;xqbq−x0b0)+˜Kj(x0b0;xqbq)+˜Ej(x0b0;xqbq)=I′j(x0b0;xqbq−x0b0)+˜Kj(x0b0;xqbq)+˜Ej(x0b0;xqbq), |
then, by (40) and (42), for all j=k+1,…,K,
0=limq→∞I′j(x0b0;xqbq−x0b0)|bq−b0|=I′j(x0b0;0ˉβ0) |
which is precisely (43). Consequently, (34), (38), (39) and (43) contradicts condition (ⅳ) of Theorem 2.1 and this completes the proof of Theorem 2.1.
Proof of Lemma 4.1: For all q∈N and almost all t∈T, define
cq:=[1+12V(xq(t0)−x0(t0))]1/2 and Wq(t):=[1+12V(˙xq(t)−˙x0(t))]1/2. |
For all q∈N, note that
|yq(t0)|2c2q+∫t1t0|˙yq(t)|2W2q(t)dt=|xq(t0)−x0(t0)|2d2q[1+12V(xq(t0)−x0(t0))]+1d2q∫t1t0|˙xq(t)−˙x0(t)|21+12V(˙xq(t)−˙x0(t))dt=|xq(t0)−x0(t0)|2D(xq−x0)[2+V(xq(t0)−x0(t0))]+1D(xq−x0)∫t1t0|˙xq(t)−˙x0(t)|22+V(˙xq(t)−˙x0(t))dt=1D(xq−x0)(V(xq(t0)−x0(t0))+∫t1t0V(˙xq(t)−˙x0(t))dt)=D(xq−x0)D(xq−x0)=1. |
Clearly limq→∞cq=1. Then, there exist some subsequence of {xq}, again denoted by {xq}, some ˉy0∈Rn and some σ0∈L2(T;Rn) such that
limq→∞yq(t0)cq=limq→∞yq(t0)=ˉy0, |
{˙yq/Wq}converges weakly in L2(T;Rn) to σ0. |
As W2q(t)≥Wq(t)≥1 for all q∈N and for almost all t∈T, we have
0≤∫t1t0[Wq(t)−1]dt≤∫t1t0[W2q(t)−1]dt=12∫t1t0V(˙xq(t)−˙x0(t))dt≤D(xq−x0). |
Thus, it follows that
limq→∞∫t1t0[Wq(t)−1]dt=limq→∞∫t1t0[W2q(t)−1]dt=0. |
Note also that
∫t1t0[Wq(t)−1]2dt=∫t1t0[W2q(t)−1]dt−2∫t1t0[Wq(t)−1]dt. |
Then for any h∈L∞(T;Rn),
limq→∞‖hWq−h‖2=0 |
and so
limq→∞∫t1t0⟨h(t),˙yq(t)⟩dt=limq→∞∫t1t0⟨h(t)Wq(t),˙yq(t)Wq(t)⟩dt=∫t1t0⟨h(t),σ0(t)⟩dt. |
Therefore, {˙yq} converges weakly in L1(T;Rn) to σ0. Hence, {˙yq} is equi-integrable on T and therefore the sequence {yq} is equi-continuous on T. Thus, if y0(t):=ˉy0+∫tt0σ0(τ)dτ, then, y0∈X with ˙y0∈L2(T;Rn) and
limq→∞yq(t)=limq→∞yq(t0)+limq→∞∫tt0˙yq(τ)dτ=y0(t)uniformly on T, |
{˙yq}converges weakly in L1(T;Rn) to ˙y0=σ0. |
Now, let us show that ˙xq(t)→˙x0(t) almost uniformly on T. For almost all t∈T, define
W(t):=[1+12V(˙x(t))]1/2. |
Observe that
∫t1t0|˙x(t)|22W2(t)dt=∫t1t0|˙x(t)|22+V(˙x(t))dt=∫t1t0V(˙x(t))dt≤D(x), |
∫t1t02W2(t)dt=2t1−2t0+∫t1t0V(˙x(t))dt≤2t1−2t0+D(x). |
From these relations, we have
‖˙x‖21≤∫t1t0|˙x(t)|22W2(t)dt∫t1t02W2(t)dt≤D(x)[2t1−2t0+D(x)]. |
Consequently, ‖˙xq−˙x0‖1→0 and so some subsequence of {˙xq} converges almost uniformly to ˙x0 on T.
Proof of Lemma 4.2: Recall the definition of Wq given in the proof of Lemma 4.1. As Wq(t)→1 uniformly on U, then for all h∈L2(U;Rn),
limq→∞∫U⟨˙yq(t),h(t)⟩dt=limq→∞∫U⟨˙yq(t)Wq(t),Wq(t)h(t)⟩dt=∫U⟨˙y0(t),h(t)⟩dt, |
that is, {˙yq} converges weakly in L2(U;Rn) to ˙y0. As R0(t)≥0 (t∈U), the function
˙y↦∫U⟨R0(t)˙y(t),˙y(t)⟩dt |
is convex in L2(U;Rn) and since this function is strongly continuous on L2(U;Rn), then this function is weakly lower semicontinuous in L2(U;Rn). Thus,
lim infq→∞∫U⟨R0(t)˙yq(t),˙yq(t)⟩dt≥∫U⟨R0(t)˙y0(t),˙y0(t)⟩dt. |
Since Rq(t)→R0(t) uniformly on U, it follows that
lim infq→∞∫U⟨Rq(t)˙yq(t),˙yq(t)⟩dt≥∫U⟨R0(t)˙y0(t),˙y0(t)⟩dt. |
In this section, we show with an example how our sufficiency theory is able to detect optimality even when the proposed extremal to be a strong minimum is singular and its derivative is only essentially bounded. It is worthwhile observing that the initial and final end-points of the states of admissible trajectories are not restricted to belong to any manifold described by any smooth function, in contrast, these boundary points must only lie in the set of real numbers, that is, these boundary points are completely free.
In Example 7.1 since no isoperimetric constraints occur l0, L0, F0, E0 and J′′0 correspond simply to l, L, F, E and J′′ respectively.
7.1 Example: Let x0:[0,1]→R be any absolutely continuous function with ˙x0∈L∞([0,1];R) and x0(0)=x0(1)=0. Consider the nonparametric problem (ˉP) of minimizing
J(x):=x2(0)−x(1)+∫10{exp(t(˙x(t)−˙x0(t)))+x(t)}dt |
over all arcs x∈X satisfying the constraints
{c(t,x(t),˙x(t))is integrable on [0,1].x(0)∈R,x(1)∈R.(t,x(t),˙x(t))∈ˉR(a.e. in [0,1]) |
where
ˉR:={(t,x,˙x)∈[0,1]×R×R∣ϕ1(t,x,˙x)≤0}, |
ϕ1(t,x,˙x):=(˙x−˙x0(t))2−exp(t(˙x−˙x0(t)))+t(˙x−˙x0(t))+1, |
X:=AC([0,1];R) and c(t,x,˙x) denotes either
L(t,x,˙x):=exp(t(˙x−˙x0(t)))+x, |
ϕ1(t,x,˙x), or any of its partial derivatives of order less than or equal to two with respect to x and ˙x.
For this problem we shall consider the data of the nonparametric problem given in Section 3 which are given by T=[0,1], n=1, r=1, s=1, k=K=0, B0=R, B1=R, ℓ(x1,x2)=x21−x2, L(t,x,˙x)=exp(t(˙x−˙x0(t)))+x, ϕ1(t,x,˙x)=(˙x−˙x0(t))2−exp(t(˙x−˙x0(t)))+t(˙x−˙x0(t))+1.
Clearly, all the assumptions posed in the statement of the problem are easily verified.
Also, it is evident that the trajectory x0 is admissible of (ˉP). Let Ψ:R→R×R be defined by Ψ(b):=(b,b). Clearly, Ψ is C2 in R and B0×B1⊂Ψ(R). The associated parametric problem of Section 2 denoted by (PΨ) has the following data, p=1, B=Ψ−1(B0×B1)=R, l=ℓ∘Ψ, L=L, φ=φ1=ϕ1 and Ψ0, Ψ1 the components of Ψ, that is, Ψ0(b)=b, Ψ1(b)=b (b∈R). Recall that the notation xb means (x,b) where b∈R is a parameter.
Observe that if we set b0:=0, then x0b0 is admissible of (PΨ). Also, clearly Ia(˜x0(⋅)) is constant on T. Let (ρ(t),μ1(t)):=(t,0) (t∈T) and note that (ρ,μ)∈X×U1, μ1(t)≥0, μ1(t)φ1(˜x0(t))=0 (a.e. in T).
Now, observe that the Hamiltonian H is given by
H(t,x,˙x,ρ(t),μ(t))=t˙x−exp(t(˙x−˙x0(t)))−x. |
Also, note that
Hx(˜x0(t),ρ(t),μ(t))=−1,H˙x(˜x0(t),ρ(t),μ(t))=0(a.e. in T), |
and so (x0,ρ,μ) satisfies the Euler-Lagrange equations which are given by
˙ρ(t)=−Hx(˜x0(t),ρ(t),μ(t)),H˙x(˜x0(t),ρ(t),μ(t))=0(a.e. in T). |
Thus, (x0,ρ,μ) is an extremal. As Ψ0(b)=b, Ψ1(b)=b, l(b)=b2−b (b∈R), then
l′(b0)+ρ(1)Ψ′1(b0)−ρ(0)Ψ′0(b0)=0 |
and so condition (ⅰ) of Corollary 3.2 is satisfied. Also, as one readily verifies,
ρ(1)Ψ′′1(b0;β)−ρ(0)Ψ′′0(b0;β)=0 for all β∈R |
and so condition (ⅱ) of Corollary 3.2 is fulfilled.
Additionally, as
H˙x˙x(˜x0(t),ρ(t),μ(t))=−t2(a.e. in T), |
it follows that (ⅲ) of Corollary 3.2 is satisfied and (x0,ρ,μ) is singular since
H˙x˙x(˜x0(0),ρ(0),μ(0))=0. |
As φ1x(˜x0(t))=φ1˙x(˜x0(t))=0 (a.e. in T), then Y(x0b0) is given by all yβ∈X×R with ˙y∈L2(T;R) satisfying y(0)=β, y(1)=β. We have,
J′′(x0b0;yβ)=2β2+∫10t2˙y2(t)dt>0 |
for all yβ∈Y(x0b0), yβ≠(0,0), and hence (ⅳ) of Corollary 3.2 is satisfied. Now, observe that for all (t,x,˙x)∈T×R×R,
F(t,x,˙x)=−H(t,x,˙x,ρ(t),μ(t))−˙ρ(t)x=−t˙x+exp(t(˙x−˙x0(t))). |
Hence for almost all t∈T, if xb is admissible,
E(t,x(t),˙x0(t),˙x(t))=F(t,x(t),˙x(t))−F(t,x(t),˙x0(t))−F˙x(t,x(t),˙x0(t))(˙x(t)−˙x0(t))=−t˙x(t)+exp(t(˙x(t)−˙x0(t)))+t˙x0(t)−1=exp(t(˙x(t)−˙x0(t)))−t(˙x(t)−˙x0(t))−1≥0 |
which implies that (v)(a) of Corollary 3.2 is verified with any ϵ>0. Finally, for all xb admissible, we have
∫10E(t,x(t),˙x0(t),˙x(t))dt=∫10{exp(t(˙x(t)−˙x0(t)))−t(˙x(t)−˙x0(t))−1}dt≥∫10(˙x(t)−˙x0(t))2dt≥2∫10V(˙x(t)−˙x0(t))dt. |
Therefore, (v)(b) of Corollary 3.2 is satisfied with any ϵ>0 and h=2. Since k=K=0, it is evident that (v)(c) of Corollary 3.2 is also verified with any ϵ>0 and h=2. By Corollary 3.2, x0 is a strong minimum of (ˉP).
The author is grateful to Dirección General de Asuntos del Personal Académico, Universidad Nacional Autónoma de México, for the support given by the project PAPIIT-IN102220.
The author declares no conflict of interest.
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