The Hamilton-Jacobi Method

Nel documento Weak KAM Theorem in Lagrangian Dynamics Seventh Preliminary Version (pagine 11-21)

A natural problem in dynamics is the search for subsets invari-ant by the flow φs. Within the framework which concerns us the Hamilton-Jacobi method makes it possible to find such invariant subsets.

To explain this method, it is better to think of the Hamiltonian H as a function on cotangent bundle TM . Indeed, under the assumptions (1) and (2) above, we see that the Legendre transform L : T M → TM , defined by

L(x, v) = (x,∂L

∂v(x, v)),

is a diffeomorphism of T M onto TM . We can then regard H as a function on TM defined by

H(x, p) = p(v) − L(x, v), where p = ∂L

∂v(x, v).

As the Legendre transform L is a diffeomorphism, we can use it to transport the flow φt: T M → T M to a flow φt : TM → TM defined by φt = LφtL−1.

Theorem 0.1.1 (Hamilton-Jacobi). Let ω be a closed 1-form on M . If H is constant on the graph Graph(ω) = {(x, ωn) | x ∈ M }, then this graph is invariant by φt.

We can then ask the following question:

Given a fixed closed 1-form ω0, does there exist ω another closed 1-form cohomologous with ω0 such that H is constant on the graph of ω?

The answer is in general negative if we require ω to be contin-uous. It is sometimes positive, this can be a way to formulate the Kolmogorov-Arnold-Moser theorem, see [Bos86].

However, there are always solutions in a generalized sense. In order to explain this phenomenon, we will first show how to reduce the problem to the 0 cohomology class. If ω0 is a fixed closed 1-form, let us consider the Lagrangian Lω0 = L − ω0, defined by

Lω0(x, v) = L(x, v) − ω0,x(v).

Since ω0is closed, if we consider only curves γ with the same fixed endpoints, the map γ 7→ R

γω0 is locally constant. It follows that Lω0 and L have the same extremal curves. Thus they have also the same Euler-Lagrange flow. The Hamiltonian Hω0 associated with Lω0 verifies

Hω0(x, p) = H(x, ω0,x+ p).

By changing the Lagrangian in this way we see that we have only to consider the case ω0 = 0.

We can then try to solve the following problem:

Does there exist a constant c ∈ R and a differentiable function u : M → R such that H(x, dxu) = c, for all x ∈ M ?

There is an “integrated” version of this question using the semi-group Tt: C0(M, R) → C0(M, R), defined for t ≥ 0 by

Ttu(x) = inf{L(γ) + u(γ(0)) | γ : [0, t] → M, γ(t) = x}.

It can be checked that Tt+t = Tt◦ Tt, and thus Tt is a (non-linear) semigroup on C0(M, R).

A C1 function u : M → R, and a constant c ∈ R satisfy H(x, dxu) = c, for all x ∈ M , if and only if Ttu = u − ct, for each t ≥ 0.

Theorem 0.1.2 (Weak KAM). We can always find a Lipschitz function u : M → R and a constant c ∈ R such that Ttu = u − ct, for all t ≥ 0.

xiii The case M = Tn, in a slightly different form (viscosity solu-tions) is due to P.L. Lions, G. Papanicolaou and S.R.S. Varadha-ran 87, see [LPV87, Theorem 1, page 6]. This general version was obtained by the author in 96, see [Fat97b, Th´eor`eme1, page 1044].

Carlsson, Haurie and Leizarowitz also obtained a version of this theorem in 1992, see [CHL91, Theorem 5.9, page 115].

As u is a Lipschitz function, it is differentiable almost every-where by Rademacher’s Theorem. It can be shown that H(x, dxu) = c at each point where u is differentiable. Moreover, for such a func-tion u we can find, for each x ∈ M , a C1 curve γx:] − ∞, 0] → M , with γx(0) = x, which is a solution of the multivalued vector field

“ gradLu”(x) defined on M by

“ gradLu”(y) = L−1(y, dyu).

These trajectories of gradLu are minimizing extremal curves.

The accumulation points of their speed curves in T M for t → −∞

define a compact subset of T M invariant under the Euler-Lagrange flow ϕt. This is an instance of the so-called Aubry and Mather sets found for twist maps independently by Aubry and Mather in 1982 and in this full generality by Mather in 1988.

We can of course vary the cohomology class replacing L by Lω and thus obtain other extremal curves whose speed curves define compact sets in T M invariant under φt. The study of these extremal curves is important for the understanding of this type of Lagrangian Dynamical Systems.

Chapter 1

Convex Functions:

Legendre and Fenchel

Besides some generalities in the first two sections, our main goal in this chapter is to look at the Legendre and Fenchel transforms.

This is now standard material in Convex Analysis, Optimization, and Calculus of Variations. We have departed from the usual viewpoint in Convex Analysis by not allowing our convex functions to take the value +∞. We think that this helps to grasp things on a first introduction; moreover, in our applications all functions have finite values. In the same way, we have not considered lower semi-continuous functions, since we will be mainly working with convex functions on finite dimensional spaces.

We will suppose known the theory of convex functions of one real variable, see for example [RV73, Chapter 1]or [Bou76, Chapitre 1].

1.1 Convex Functions: General Facts

Definition 1.1.1 (Convex Function). Let U be a convex set in the vector space E. A function f : U → R is said to be convex if it satisfies the following condition

∀x, y ∈ U, ∀t ∈ [0, 1], f (tx + (1 − t)y) ≤ tf (x) + (1 − t)f (y).

1

The function f is said to be strictly convex if it satisfies the fol-lowing stronger condition

∀x 6= y ∈ U, ∀t ∈]0, 1[, f (tx + (1 − t)y) < tf (x) + (1 − t)f (y).

It results from the definition that f : U → R is convex if and only if for every line D ⊂ E the restriction of f on D ∩ U is a convex function (of one real variable).

Proposition 1.1.2. (i) An affine function is convex (an affine function is a function which is the sum of a linear function and a constant).

(ii) If (fi)i ∈ I is a family of convex functions : U → R, and supi∈Ifi(x) < +∞ for each x ∈ U then supi∈Ifi is convex.

(iii) Let U be an open convex subset of the normed space E.

If f : U → R is convex and twice differentiable at x ∈ U , then D2f (x) is non-negative definite as a quadratic form. Conversely, if g : U → R admits a second derivative D2g(x) at every point x ∈ U , with D2g(x) non-negative (resp. positive) definite as a quadratic form, then g is (resp. strictly) convex.

Properties (i) and (ii) are immediate from the definitions. The property (iii) results from the case of the functions of a real vari-able by looking at the restrictions to each line of E.

Definition 1.1.3 (C2-Strictly Convex Function). Let be a in the vector space E. A function f : U → R, defined on the convex subset U of the normed vector space E, is said to be C2-strictly convex if it is C2, and its the second derivative D2f (x) is positive definite as a quadratic form, for each x ∈ U .

Exercise 1.1.4. Let U be an open convex subset of the normed space E, and let f : U → R be a convex function.

a) Show that f is not strictly convex if and only if there exists a pair of distinct points x, y ∈ U such that f is affine on the segment [x, y] = {tx + (1 − t)y | t ∈ [0, 1].

b) If f is twice differentiable at every x ∈ U , show that it is strictly convex if and only if for every unit vector v ∈ E the set {x ∈ U | D2f (x)(v, v) = 0} does not contain a non trivial segment parallel to v. In particular, if D2f (x) is non-negative definite as a quadratic form at every point x ∈ U , and the set of

3 points x where D2f (x) is not of maximal rank does not contain a non-trivial segment then f is strictly convex.

Theorem 1.1.5. Suppose that U is an open convex subset of the topological vector space E. Let f : U → R be a convex function. If there exists an open non-empty subset V ⊂ U with supx∈V f (x) < +∞, then f is continuous on U.

Proof. Let us first show that for all x ∈ U , there exists an open neighborhood Vx of x, with Vx ⊂ U and supy∈Vxf (y) < +∞. In-deed, if x /∈ V , we choose z0 ∈ V . The intersection of the open set U and the line containing x and z0 is an open segment contain-ing the compact segment [x, z0]. We choose in this intersection, a point y near to x and such that y /∈ [x, z0], thus x ∈]y, z0[, see fig-ure 1.1. It follows that there exists t with 0 < t0 < 1 and such that x = t0y + (1 − t0)z0. The map H : E → E, z 7→ x = t0y + (1 − t0)z sends z0to x, is a homeomorphism of E, and, by convexity of U , it maps U into itself. The image of V by H is an open neighborhood Vx of x contained in U . Observe now that any point x of Vx can be written as the form x = t0y + (1 − t0)z with z ∈ V for the same t0 ∈]0, 1[ as above, thus

f (x) = f (t0y + (1 − t0)z)

≤ t0f (y) + (1 − t0)f (z)

≤ t0f (y) + (1 − t0) sup

z∈V

f (z) < +∞.

This proves that f is bounded above on Vx.

Let us now show that f is continuous at x ∈ U . We can suppose by translation that x = 0. Let V0 be an open subset of U containing 0 and such that supy∈V0f (y) = M < +∞. Since E is a topological vector space, we can find an open set ˜V0 containing 0, and such that t ˜V0 ⊂ V0, for all t ∈ R with |t| ≤ 1. Let us suppose that y ∈ ǫ ˜V0∩ (−ǫ ˜V0), with ǫ ≤ 1. We can write y = ǫz+ and y = −ǫz, with z+, z ∈ ˜V0 (of course z = −z+, but this is irrelevant in our argument). As y = (1 − ǫ)0 + ǫz+, we obtain f (y) ≤ (1 − ǫ)f (0) + ǫf (z+), hence

∀y ∈ ǫ ˜V0∩ (−ǫ ˜V0), f (y) − f (0) ≤ ǫ(M − f (0)).

U

V Vx

z0

x y

Figure 1.1:

We can also write 0 = 1+ǫ1 y + 1+ǫǫ z, hence f (0) ≤ 1+ǫ1 f (y) +

ǫ

1+ǫf (z) which gives (1 + ǫ)f (0) ≤ f (y) + ǫf (z) ≤ f (y) + ǫM . Consequently

∀y ∈ ǫ ˜V0∩ (−ǫ ˜V0), f (y) − f (0) ≥ −ǫM + ǫf (0).

Gathering the two inequalities we obtain

∀y ∈ ǫ ˜V0∩ (−ǫ ˜V0), |f (y) − f (0)| ≤ ǫ(M − f (0)).

Corollary 1.1.6. A convex function f : U → R defined on an open convex subset U of Rnis continuous.

Proof. Let us consider n+1 affinely independent points x0, · · · , xn∈ U . The convex hull σ of x0, · · · , xn has a non-empty interior. By convexity, the map f is bounded by maxni=0f (xi) on σ.

Most books treating convex functions from the point of view of Convex Analysis do emphasize the role of lower semi-continuous convex functions. When dealing with finite valued functions, the following exercise shows that this is not really necessary.

5 Exercise 1.1.7. Let U be an open subset of the Banach space E.

If f : U → R is convex, and lower semi-continuous show that it is in fact continuous. [Indication: Consider the sequence of subsets Cn = {x ∈ U | f (x) ≤ n}, n ∈ N. Show that one of these subsets has non-empty interior.]

We recall that a function f : X → Y , between the metric spaces X, Y , is said to be locally Lipschitz if, for each x ∈ X, there exists a neighborhood Vx of x in X on which the restriction f|Vx is Lipschitz.

Theorem 1.1.8. Let E be a normed space and U ⊂ E an open convex subset. Any convex continuous function f : U → R is a locally Lipschitz function.

Proof. In fact, this follows from the end of the proof of Theorem 1.1.5. We now give a direct slightly modified proof.

We fix x ∈ U . Since f is continuous, there exists r ∈]0, +∞[

and M < +∞ such that sup

y∈ ¯B(x,r)

|f (y)| ≤ M.

We have used the usual notation ¯B(x, r) to mean the closed ball of center x and radius r.

Let us fix y, y ∈ ¯B(x, r/2). We call z the intersection point of the boundary ∂B(x, r) = {x ∈ E | kx− xk = r} of the closed ball ¯B(x, r) with the line connecting y and y such that y is in the segment [z, y], see figure 1.2. We of course have kz−yk ≥ r/2. We write y = tz + (1 − t)y, with t ∈ [0, 1[, from which it follows that y−y= t(z−y). By taking the norms and by using kz−yk ≥ r/2, we see that

t ≤ ky− yk2 r.

The convexity of f gives us f (y) ≤ tf (z)+ (1− t)f (y), from which we obtain the inequality f (y) − f (y) ≤ t(f (z) − f (y)). It results that

f (y) − f (y) ≤ 2tM ≤ 4M

r ky − yk, and by symmetry

∀y, y ∈ ¯B(x, r/2), |f (y) − f (y)| ≤ 4M

r ky − yk.

z

y

y

Figure 1.2:

Corollary 1.1.9. If f : U → R is convex with U ⊂ Rn open and convex, then f is a locally Lipschitz function.

We recall Rademacher’s Theorem, see [EG92, Theorem 2, page 81]or [Smi83, Theorem 5.1, page 388].

Theorem 1.1.10 (Rademacher). A locally Lipschitz function de-fined on open subset of Rn and with values in Rm is Lebesgue almost everywhere differentiable.

Corollary 1.1.11. A convex function f : U → R, where U is open convex of Rn, is Lebesgue almost everywhere differentiable.

It is possible to give a proof of this Corollary not relying on Rademacher’s Theorem, see [RV73, Theorem D, page 116]. We conclude this section with a very useful lemma.

Lemma 1.1.12. Let f : V → R be a convex function defined on an open subset V of a topological vector space.

(a) A local minimum for f is a global minimum.

7 (b) If f is strictly convex, then f admits at most one minimum.

Proof. (a) Let x0 be a local minimum. For y ∈ U and t ∈ [0, 1]

and close to 1 we have

f (x0) ≤ f (tx0+ (1 − t)y) ≤ tf (x0) + (1 − t)f (y),

thus (1 − t)f (x0) ≤ (1 − t)f (y) for t close to 1. It follows that f (y) ≥ f (x0).

(b) It results from the convexity of f that the subset {x | f (x) ≤ λ} is convex. If λ = inf f , we have {x | f (x) = inf f } = {x | f (x) ≤ inf f }. If f is strictly convex this convex set cannot contain more than one point.

Nel documento Weak KAM Theorem in Lagrangian Dynamics Seventh Preliminary Version (pagine 11-21)