Herglotz'generalized variational principle and contact type Hamilton ...

Herglotz’ generalized variational principle and contact type Hamilton-Jacobi equations

arXiv:1804.03411v1 [math.AP] 10 Apr 2018

Piermarco Cannarsa, Wei Cheng, Kaizhi Wang and Jun Yan

Abstract We develop an approach for the analysis of fundamental solutions to Hamilton-Jacobi equations of contact type based on a generalized variational principle proposed by Gustav Herglotz. We also give a quantitative Lipschitz estimate on the associated minimizers.

1 Introduction The so called generalized variational principle was proposed by Gustav Herglotz in 1930 (see [30] and [31]). It generalizes classical variational principle by defining the functional, whose extrema are sought, by a differential equation. More precisely, the functional u is defined in an implicit way by an ordinary differential equation u(s) ˙ = F(s, ξ (s), ξ˙ (s), u(s)),

s ∈ [0,t],

(1)

Piermarco Cannarsa Dipartimento di Matematica, Università di Roma “Tor Vergata”, Via della Ricerca Scientifica 1, 00133 Roma, Italy, e-mail: [email protected] Wei Cheng Department of Mathematics, [email protected]

Nanjing

University,

Nanjing

210093,

China,

e-mail:

Kaizhi Wang School of Mathematical Sciences, Shanghai Jiao Tong University, Shanghai 200240, China, e-mail: [email protected] Jun Yan School of Mathematical Sciences, Fudan University and Shanghai Key Laboratory for Contemporary Applied Mathematics, Shanghai 200433, China, e-mail: [email protected]

1

2


with u(t) = u0 ∈ R, for t > 0, a function F ∈ C2 (R×Rn ×Rn ×R, R) and a piecewise C1 curve ξ : [0,t] → Rn . Here, u = u[ξ , s] can be regarded as a functional, on a space of paths ξ (·). The generalized variational principle of Herglotz is as follows: Let the functional u = u[ξ ,t] be defined by (1) with ξ in the space of piecewise C1 functions on [0,t], and η be an arbitrary piecewise C1 function such that η (0) = η (t) = 0. Then the value of the functional u[ξ ,t] is an extremal for the function ξ such that the variation ddε u[ξ + εη ,t] = 0. Herglotz reached the idea of the generalized variational principle through his work on contact transformations and their connections with Hamiltonian systems and Poisson brackets. His work was motivated by ideas from S. Lie, C. Carathéodory and other researchers. An important reference on the generalized variational principle is the monograph [29]. The variational principle of Herglotz is important for many reasons: – The solutions of the equations (1) determine a family of contact transformations, see [29, 11, 20, 27]; – The generalized variational principle gives a variational description of energynonconservative processes even when F in (1) is independent of t. – If F has the form F = −λ u + L(x, v), then the relevant problems are closely connected to the Hamilton-Jacobi equations with discount factors (see, for instance, [18, 17, 9, 33, 34, 36, 28, 35]). – Even for a energy-nonconservative process which can be described with the generalized variational principle, one can systematically derive conserved quantities as Noether’s theorems such as [25, 26]; – The generalized variational principle provides a link between the mathematical structure of control and optimal control theories and contact transformation (see [24]); – There are some interesting connections between contact transformations and equilibrium thermodynamics (see, for instance, [38]). In this note, we will clarify more connections between the generalized variational principle of Herglotz and Hamilton-Jacobi theory motivated by recent works in [40, 41] under a set of Tonelli-like conditions. We will begin with generalized variational principle of Herglotz in the frame of Lagrangian formalism different from the methods used in [40, 41]. Throughout this paper, let L : Rn × R × Rn be a function of class Ck such that the following standing assumptions are satisefied: (L1) Lvv (x, r, v) > 0 for all (x, r, v) ∈ Rn × R × Rn. (L2) For each r ∈ R, there exist two superlinear nondecreasing function θ r , θr : [0, +∞) → [0, +∞), θr (0) = 0 and cr > 0, such that

θ r (|v|) > L(x, r, v) > θr (|v|) − cr ,

(x, v) ∈ Rn × Rn .

(L3) There exists K > 0 such that |Lr (x, r, v)| 6 K,

(x, r, v) ∈ Rn × R × Rn.

Herglotz’ generalized variational principle

3

It is natural to introduce the associated Hamiltonian H(x, r, p) = sup {hp, vi − L(x, r, v)},

(x, r, p) ∈ Rn × R × (Rn)∗ .

v∈Rn

Let x, y ∈ Rn , t > 0 and u0 ∈ R. Set t Γx,y = {ξ ∈ W 1,1 ([0,t], Rn ) : ξ (0) = x, ξ (t) = y}.

We consider a variational problem Minimize u0 + inf

Z t 0

L(ξ (s), uξ (s), ξ˙ (s)) ds,

(2)

t such that the Carath´ where the infimum is taken over all ξ ∈ Γx,y eodory equation

u˙ξ (s) = L(ξ (s), uξ (s), ξ˙ (s)),

a.e. s ∈ [0,t],

(3)

admits an absolutely continuous solution uξ with initial condition uξ (0) = u0 . It is already known that the variational problem (2) with subsidiary conditions (3) is closely connected to the Hamilton-Jacobi equations in the form H(x, u(x), Du(x)) = c.

(4)

The readers can refer to [27] for a systematic approach of Hamilton-Jacobi equations in the form (4) especially in the context of contact geometry. In [40, 41], a weak KAM type theory on equations (4) was developed on compact manifolds under the aforementioned Tonelli-like conditions. Problem (2) is understood as an implicit variational principle ([40]) and, by introducing the positive and negative Lax-Oleinik semi-groups, an existence result for weak KAM type solutions of (4) was obtained provided c in the right side of equation (4) belongs to the set of critical values ([41]). The same approach adapts to the evolutionary equations in the form Dt u + H(x, u, Dx u) = 0. (5) Unlike the methods used in [40, 41], in this note, our approach of the equations (4) and (5) is based on the the variational problem (2) under subsidiary conditions (3). We give all the details of such a Tonelli-like theory and its connection to viscosity solutions of (4) and (5) . In view of Proposition 1 below, the infimum in (2) can be achieved. Suppose that t is a minimizer for (2) where u is uniquely determined by (3) with initial ξ ∈ Γx,y ξ condition uξ (0) = u0 . Then we call such ξ an extremal. Due to Proposition 1 below, each extremal ξ and associated uξ are of class C2 and satisfy the Herglotz equation (Generalized Euler-Lagrange equation by Herglotz) d Lv (ξ (s), uξ (s), ξ˙ (s)) ds = Lx (ξ (s), uξ (s), ξ˙ (s)) + Lu (ξ (s), uξ (s), ξ˙ (s))Lv (ξ (s), uξ (s), ξ˙ (s)).

(6)

4


Moreover, let p(s) = Lv (ξ (s), uξ (s), ξ˙ (s)) be the so called dual arc. Then p is also of class C2 and we conclude that (ξ , p, uξ ) satisfies the following Lie equation  ˙  ξ (s) = H p (ξ (s), uξ (s), p(s)); p(s) ˙ = −Hx (ξ (s), uξ (s), p(s)) − Hu (ξ (s), uξ (s), p(s))p(s);   u˙ξ (s) = p(s) · ξ˙ (s) − H(ξ (s), uξ (s), p(s)),

(7)

where the reader will recognize the classical system of characteristics for (4). The paper is organized as follows: In Section 2, we afford a detailed and rigorous treatment of (2) under subsidiary conditions (3). In Section 3, we study the regularity of the minimizers and deduce the Herglotz equation (6) and Lie equation (7) as well. In Section 4, we show that the two approaches between [40, 41] and ours are equivalent.

2 Existence of minimizers in Herglotz’ variational principle Fix x0 , x ∈ Rn , t > 0 and u0 ∈ R. Let ξ ∈ Γxt0 ,x , we consider the Carathéodory equation ( u˙ξ (s) = L(ξ (s), uξ (s), ξ˙ (s)), a.e. s ∈ [0,t], (8) uξ (0) = u0 . We define the action functional J(ξ ) :=

Z t 0

L(ξ (s), uξ (s), ξ˙ (s)) ds,

(9)

where ξ ∈ Γxt0 ,x and uξ is defined in (8) by Proposition 5 in Appendix. Our purpose is to minimize J(ξ ) over A = {ξ ∈ Γxt0 ,x : (8) admits an absolutely continuous solution uξ }. Notice that A 6= ∅ because it contains all piecewise C1 curves connecting x to y. It is not hard to check that, for each a ∈ R, A = {ξ ∈ Γxt0 ,x : the function s 7→ L(ξ (s), a, ξ˙ (s)) belongs to L1 ([0,t])}. For the following estimate, we define L0 (x, v) := L(x, 0, v). Lemma 1. Let x0 , x ∈ Rn , t > 0, u0 ∈ R. Given ξ ∈ Γxt0 ,x such that (8) admits an absolutely continuous solution, then we have that |uξ (s)| 6 exp(Ks)(|u0 | + c0s) if uξ (s) < 0. In particular, we have

(10)


5

uξ (s) > − exp(Ks)(|u0 | + c0 s),

s ∈ [0,t].

(11)

Proof. Let x0 , x ∈ Rn , t > 0, u0 ∈ R and ξ ∈ A . Suppose that uξ (s0 ) < 0, s0 ∈ (0,t]. We define E = {s ∈ [0, s0 ) : uξ (s) > 0} and a=

(

0 sup E

E = ∅, E 6= ∅.

Then, we have that uξ (s) 6 0 for all s ∈ [a, s0 ] and uξ (a) = 0 if E 6= ∅. Now, for any s ∈ [a, s0 ] we have that −|uξ (s)| = uξ (s) = uξ (a) + > − |uξ (a)| + > − |uξ (a)| +

Z s Za s a

Z s a

L(ξ (τ ), uξ (τ ), ξ˙ (τ )) d τ

L0 (ξ (τ ), ξ˙ (τ )) d τ − K

Z s a

|uξ (τ )| d τ d τ

θ0 (|ξ˙ (τ )|) d τ − c0(s − a) − K

> − |uξ (a)| − c0s − K

Z s a

Z s a

|uξ (τ )| d τ

|uξ (τ )| d τ .

Then, we have that |uξ (s)| 6 (|u0 | + c0 s) + K

Z s a

|uξ (τ )| d τ ,

s ∈ [a, s0 ].

Then Gronwall inequality implies |uξ (s)| 6 exp(K(s − a))(|u0| + c0 s) 6 exp(Ks)(|u0 | + c0 s),

s ∈ [a, s0 ].

If E = ∅, then a = 0 and the proof is the same. This leads to (10) and (11).

⊓ ⊔

In view to Lemma 1, we conclude that infξ ∈A J(ξ ) is bounded below. Now, for any ε > 0, set Aε = {ξ ∈ A : inf J(η ) + ε > uξ (t) − u0}. η ∈A

Lemma 2. Suppose x0 ∈ Rn , t, R > 0, u0 ∈ R and |x− x0 | 6 R. Let ε > 0 and ξ ∈ Aε . Then we have that uξ (t) − u0 6 t(κ (R/t) + K|u0|) exp(Kt) + ε , with κ (r) = θ 0 (r) + 2c0 . Moreover, there exist two nondecreasing and superlinear functions F, G : [0, +∞) → [0, +∞) such that |uξ (t)| 6 tF(R/t) + G(t)|u0| + ε ,

(12)

where F(r) = max{κ (r), c0 exp(Kr)} and G(r) = max{rK exp(Kr) + 1, exp(Kr)}.

6


Proof. Suppose x0 ∈ Rn , t, R > 0, u0 ∈ R and |x − x0 | 6 R. Let ε > 0 and ξ ∈ Aε . First, notice that (x, v) ∈ Rn × Rn.

|L0 (x, v)| 6 L0 (x, v) + 2c0 6 θ 0 (|v|) + 2c0,

(13)

Set κ (r) = θ 0 (r) + 2c0 . Define ξ0 (s) = x0 + s(x − x0 )/t for any s ∈ [0,t], then ξ0 ∈ A . Then, for any s ∈ [0,t], we have that |uξ0 (s) − u0 | 6

Z s 0

|L0 (ξ0 , ξ˙0 )| d τ + K

6t κ (R/t) + K

Z s 0

Z s 0

|uξ0 | d τ

|uξ0 − u0| d τ + tK|u0|.

Due to Gronwall inequality, we obtain |uξ0 (s) − u0 | 6 t(κ (R/t) + K|u0|) exp(Kt),

s ∈ [0,t]. ⊓ ⊔

Together with Lemma 1, this completes the proof.

Lemma 3. Suppose x0 ∈ Rn , t, R > 0, u0 ∈ R and |x − x0 | 6 R. Let ε > 0 and ξ ∈ Aε . Then there exist two continuous functions F1 , F2 : [0, +∞) × [0, +∞) → [0, +∞) depending on R, with Fi (r1 , ·) being nondecreasing and superlinear and Fi (·, r2 ) being nondecreasing for any r1 , r2 > 0, i = 1, 2, such that |uξ (s)| 6 tF1 (t, R/t) + C1(t)(ε + |u0 |), and

Z t 0

s ∈ [0,t]

(14)

|L(ξ , uξ , ξ˙ )| d τ 6 tF2 (t, R/t) + C2(t)(ε + |u0 |),

(15)

where Ci (t) > 0 for i = 1, 2. Proof. Suppose x0 ∈ Rn , t, R > 0, u0 ∈ R and |x − x0| 6 R. Let ε > 0 and ξ ∈ Aε . If uξ (t) > 0, we define E+ = {s ∈ [0,t] : uξ (s) > uξ (t)}. If E+ = ∅, then we have that uξ (s) 6 uξ (t) for all s ∈ [0,t]. Now, we suppose that E+ 6= ∅. It is known that E+ is the union of a countable family of open intervals {(ai , bi )} which are mutually disjoint. For any τ ∈ E+ , there exists an open interval (a, b), a component of E+ containing s, such that uξ (τ ) > uξ (t) > 0 for all τ ∈ (a, b) and uξ (b) = uξ (t). Therefore, for almost all s ∈ [a, b], we have that u˙ξ (s) = L(ξ (s), uξ (s), ξ˙ (s)) > L0 (ξ (s), ξ˙ (s)) − Kuξ (s). Invoking condition (L2), it follows that, for all s ∈ [a, b], eKb uξ (b) − eKsuξ (s) > Thus we obtain that

Z b s

eK τ L0 (ξ (τ ), ξ˙ (τ )) d τ > −c0 (b − s)eKb


uξ (s) 6 c0 (b − s)e

K(b−s)

7

+e

K(b−s)

uξ (t)

6 c0te + e [(t κ (R/t) + K|u0|)eKt + ε + |u0|] s ∈ [0,t], Kt

Kt

(16)

=tF1 (t, R/t) + G1(t)|u0 | + ε , where F1 (r1 , r2 ) := eKr1 (c0 + κ (r2 )) and G1 (r) = eKr (KeKr + 1). If uξ (t) < 0, define vξ (s) = uξ (s) − uξ (t), then vξ (s) satisfies the Carathéodory eqaution v˙ξ (s) = L(ξ (s), vξ (s) + uξ (t), ξ˙ (s)),

s ∈ [0,t]

with initial condition vξ (0) = u0 − uξ (t). Similarly, We define F+ = {s ∈ [0,t] : vξ (s) > vξ (t)}. If F+ = ∅, then we have that vξ (s) 6 vξ (t) = 0 for all s ∈ [0,t]. Now, we suppose that F+ 6= ∅ and F+ is the union of a countable family of open intervals {(ci , di )} which are mutually disjoint. For any s ∈ F+ , there exists an open interval, say (c, d), such that vξ (s) > vξ (t) = 0 for all s ∈ (c, d) and vξ (d) = vξ (t). Therefore, for almost all s ∈ [c, d], we have that v˙ξ (s) > L0 (ξ (s), ξ˙ (s)) − Kvξ (s) − K|uξ (t)|. It follows that, for all s ∈ [c, d], eKd vξ (d) − eKs vξ (s) >

Z d s

eK τ L0 (ξ (s), ξ˙ (s)) d τ − Kt|uξ (t)|eKt

> − c0teKd − Kt|uξ (t)|eKt , and this gives rise to vξ (s) 6 c0teK(d−s) + K|uξ (t)|teK(t−s) + eK(d−s)vξ (d) 6 c0teKt + Kt|uξ (t)|eKt , since vξ (d) = 0. It follows that, for all s ∈ [0,t], uξ (s) 6 c0teKt + KteKt |uξ (t)| + uξ (t) 6 c0teKt + (KteKt + 1)|uξ (t)| 6 c0teKt + (KteKt + 1)(tF2(R/t) + G2(t)|u0 | + ε )

(17)

with F2 and G2 determined by Lemma 2. By combining (16) and (17) and setting F3 (r1 , r2 ) = max{F1 (r1 , r2 ), c0 eKr1 + F2(r2 )(Kr1 eKr1 + 1)}, C1 (r) = max{G1 (r), G2 (t)(Kr1 eKr1 + 1)},

C2 (r) = max{C1 (r), eKr c0 },

we conclude that uξ (s) 6tF3(t, R/t) + C1(t)(|u0 | + ε ),

(18)

|uξ (s)| 6tF3(t, R/t) + C2(t)(|u0 | + ε ).

(19)

This leads the proof of (14) together with Lemma 1.

8


Now, by (13), Lemma 2 and (19), we have that Z s 0

|L0 (ξ , ξ˙ )|d τ 6

Z s 0

(L0 (ξ , ξ˙ ) + 2c0) d τ 6 2c0 s + uξ (s) − u0 + K

6 2c0t + tF2(t, R/t) + C2(t)(|u0 | + ε ) + |u0|

Z s 0

|uξ | d τ

+ t 2 KF2 (t, R/t) + tKC2(t)(|u0 | + ε ) 6tF4 (t, R/t) + C3(t)(|u0 | + ε ). Therefore, (15) follows from the estimate below Z t 0

|L(ξ , uξ , ξ˙ )| d τ 6

Z t 0

|L0 (ξ , ξ˙ )| d τ + K

Z t 0

|uξ | d τ

6tF4 (t, R/t) + C3(t)(|u0 | + ε ) + tK(tF3(t, R/t) + C2(t)(|u0 | + ε )) =tF5 (t, R/t) + C4(t)(|u0 | + ε ). ⊓ ⊔

This completes our proof.

Remark 1. Now, fix any ε ∈ (0, 1) and any ξ ∈ Aε ⊂ A1 . The definition of (9) can be replaced by Lφ = L(x, φ (u), v) with φ : R → R a bounded nondecreasing smooth function such that φ (u) = u for |u| 6 tF1 (t, R/t) +C1 (t)(1 + |u0 |) and φ (u) ≡ u∗ for |u| > tF1 (t, R/t) +C1 (t)(1 + |u0 |) + 1, where F1 (t, R/t) and C1 (t) are determined by (14) in Lemma 3 and F1 and C1 are both independent of ε . Therefore, to minimize J defined in (9), we can suppose that supξ ∈A1 |L(ξ (s), u, ξ˙ (s))| is bounded by an integrable function f ∈ L1 ([0,t]). In this situation, (8) is indeed a Carathéodory equation which admits a unique solution by Proposition 5. Corollary 1. For any ε ∈ (0, 1), the set {uξ : ξ ∈ Aε } is relatively compact in C0 ([0,t], R). Proof. Suppose x0 ∈ Rn , t, R > 0, u0 ∈ R and |x − x0 | 6 R. For any ε ∈ (0, 1) and ξ ∈ Aε . Recall that uξ is the unique solution of (8) by Remark 1, it follows {u˙ξ }ξ ∈Aε is equi-integrable which implies {uξ }ξ ∈Aε is equi-continuous. The boundedness of {uξ }ξ ∈Aε follows from Lemma 3. Invoking Ascoli-Arzela theorem, we get our conclusion. ⊓ ⊔ Lemma 4. Suppose x0 ∈ Rn , t, R > 0, u0 ∈ R and |x − x0 | 6 R. Let ε ∈ (0, 1) and ξ ∈ Aε . Then there exist a continuous function F = Fu0 : [0, +∞) × [0, +∞) → [0, +∞), F(r1 , ·) is nondecreasing and superlinear and F(·, r2 ) is nondecreasing for any r1 , r2 > 0, such that Z t

|ξ˙ (s)| ds 6 tF(t, R/t) + ε .

0

Moreover, the family {ξ˙ }ξ ∈Aε is equi-integrable. Proof. Let ε > 0 and ξ ∈ Aε . Then, by (L2) we obtain


9

uξ (t) − u0 =

Z t 0

>

Z t 0

>

Z t 0

L(ξ (s).uξ (s), ξ˙ (s)) ds >

Z t 0

{L(ξ (s), 0, ξ˙ (s)) − K|uξ (s)|} ds (20)

{θ0 (|ξ˙ (s)|) − c0 − K|uξ (s)|} ds {|ξ˙ (s)| − K|uξ (s)| − (c0 + θ0∗ (1))} ds.

In view of Lemma 2, Lemma 3 and (20), we obtain that Z t

|ξ˙ (s)| ds 6

0

Z t 0

K|uξ (s)| ds + t(c0 + θ0∗(1)) + uξ (t) − u0

6tK(tF1 (t, R/t) + C1(t)(ε + |u0|)) + t(c0 + θ0∗ (1)) + tF2(t, R/t) + ε := tF3 (t, R/t) + ε .

Now we turn to proof that the equi-integrability of the family {ξ˙ }ξ ∈Aε . Since θ0 is a superlinear function, then for any α > 0 there exists Cα > 0 such that r 6 θ0 (r)/α for r > Cα . Thus, for any measurable subset E ⊂ [0,t], invoking (L2), (L3) and Lemma 3, we have that Z

E∩{|ξ˙ |>Cα }

Z

Z

1 1 θ0 (|ξ˙ |)ds 6 (L0 (ξ , ξ˙ ) + c0)ds α E∩{|ξ˙ |>Cα } α E∩{|ξ˙ |>Cα } Z 1 6 (L(ξ , uξ , ξ˙ ) + K|uξ | + c0)ds α E∩{|ξ˙ |>Cα } 1 6 (uξ (t) − u0 + K(tF1(t, R/t) + C1(t)(ε + |u0 |)) + tc0) α 1 6 (tF2 (t, R/t) + 1 + K(tF1(t, R/t) + C1(t)(1 + |u0|)) + tc0 ) α 1 := F4 (t, R/t). α

|ξ˙ |ds 6

Therefore, we conclude that Z

|ξ˙ |ds 6 E

Z

E∩{|ξ˙ |>Cα }

|ξ˙ |ds +

Z

E∩{|ξ˙ |6Cα }

|ξ˙ |ds 6

1 F4 (t, R/t) + |E|Cα . α

Then, the equi-integrability of the family {ξ˙ }ξ ∈Aε follows since the right-hand side can be made arbitrarily small by choosing α large and |E| small, and this proves our claim. Proposition 1. The functional

Γxt0 ,x ∋ ξ 7→ J(ξ ) =

Z t 0

L(ξ (s), uξ (s), ξ˙ (s)) ds,

where uξ is determined by (8) with initial condition uξ (0) = u0 , admits a minimizer in Γxt0 ,x .

10


Proof. Fix x0 , x ∈ Rn , t > 0 and u0 ∈ R. Consider any minimizing sequence {ξk } for J, that is, a sequence such that J(ξk ) → inf{J(ξ ) : ξ ∈ Γxt0 ,x } as k → ∞. We want to show that this sequence admits a cluster point which is the required minimizer. Notice there exists an associated sequence {uξk } given by (8) in the definition of J(ξk ). The idea of the proof is standard but a little bit different. First, notice that Lemma 4 implies that the sequence of derivatives {ξ˙k } is equiintegrable. Since the sequence {ξ˙k } is equi-integrable, by the Dunford-Pettis Theorem there exists a subsequence, which we still denote by {ξ˙k }, and a function η ∗ ∈ L1 ([0,t], Rn ) such that ξ˙k ⇀ η ∗ in the weak-L1 topology. The equi-integrability of {ξ˙k } implies that the sequence {ξk } is equi-continuous and uniformly bounded. Invoking Ascoli-Arzela theorem, we can also assume that the sequence {ξk } converges uniformly to some absolutely continuous function ξ∞ ∈ Γxt0 ,x . For any test function ϕ ∈ C01 ([0,t], Rn ), Z t

ϕη ∗ ds = lim

Z t

k→∞ 0

0

ϕ ξ˙k ds = − lim

Z t

k→∞ 0

˙ k ds = − ϕξ

Z t 0

˙ ∞ ds. ϕξ

By the fundamental lemma in calculus of variation (see, for instance, [10, Lemma 6.1.1]), we can conclude that ξ˙∞ = η ∗ almost everywhere. Similarly, Corollary 1 implies {uξk } is relatively compact in C0 ([0,t], R). Therefore {uξk } converges uniformly to uξ∞ by taking subsequence if necessary. We recall a classical result (see, for instance, [3, Theorem 3.6] or [2, Section 3.4]) on the sequentially lower semicontinuous property on the functional L1 ([0,t], Rm ) × L1 ([0,t], Rn ) ∋ (α , β ) 7→ F(α , β ) :=

Z t

L(α (s), β (s)) ds.

0

One has that if (i) L is lower semicontinuous; (ii) L(α , ·) is convex on Rn , then the functional F is sequentially lower semicontinuous on the space L1 ([0,t], Rm ) × L1 ([0,t], Rn ) endowed with the strong topology on L1 ([0,t], Rm ) and the weak topology on L1 ([0,t], Rn ). Now, let L(αξk (s), βξk (s)) := L(ξk (s), uξk (s), ξ˙k (s)) with αξk (s) = (ξk (s), uξk (s)) and βξk (s) = ξ˙k (s), then we have lim inf k→∞

Z t 0

L(ξk (s), uξk (s), ξ˙k (s)) ds >

Z t 0

L(ξ∞ (s), uξ∞ (s), ξ˙∞ (s)) ds.

Therefore, ξ∞ ∈ Γxt0 ,x is a minimizer of J and this completes the proof of the existence result. Corollary 2. Suppose x0 ∈ Rn , t, R > 0, u0 ∈ R and |x − x0 | 6 R. If ξ ∈ Γxt0 ,x is a minimizer for (9), then, there exists a continuous function F = Fu0 ,R : [0, +∞) × [0, +∞) → [0, +∞), F(r1 , ·) is nondecreasing and superlinear and F(·, r2 ) is nondecreasing for any r1 , r2 > 0, such that Z t 0

|ξ˙ (s)| ds 6 tF(t, R/t).


11

Moreover, we conclude that ess inf |ξ˙ (s)| 6 F(t, R/t), s∈[0,t]

sup |ξ (s) − x0| 6 tF(t, R/t). s∈[0,t]

Proof. The first assertion is direct from Lemma 4. The last two inequality follows from the relations 1 ess inf |ξ˙ (s)| 6 t s∈[0,t]

Z t

|ξ˙ (s)| ds,

0

and |ξ (s) − x0 | 6

Z t

|ξ˙ (s)| ds,

0

together with the first assertion.

3 Necessary conditions and regularity of minimizers 3.1 Lipschitz estimate of minimizers To obtain the regularity properties of a minimizers ξ of (9), we need an a priori Lipschitz estimate of ξ . For such an estimate, a key point is to verify the Erdmann condition, which is standard for classical autonomous Tonelli Lagrangians. Our proof is a modification of the original one by Francis Clarke (see, [14] or [15]). Proposition 2. Suppose x0 ∈ Rn , t, R > 0, u0 ∈ R and |x − x0 | 6 R. If ξ ∈ Γxt0 ,x is a minimizer for (9), then, there exists a function F = Fu0 ,R : [0, +∞) × [0, +∞) → [0, +∞), F(r1 , ·) is nondecreasing and superlinear and F(·, r2 ) is nondecreasing for any r1 , r2 > 0, such that ess sup |ξ˙ (s)| 6 F(t, R/t). s∈[0,t]

Proof. Let ξ ∈ Γxt0 ,x be a minimizer of (9) with uξ determined by (8) and uξ (0) = u0 . R Let α : [0,t] → [1/2, 3/2] be a measurable function satisfying 0t α (s) ds = t (the set of all such functions α is denoted by Ω ), we define

τ (s) =

Z s

α (r) dr,

s ∈ [0,t],

0

then τ : [0,t] → [0,t] is a bi-Lipschitz map and its inverse s(τ ) satisfies s′ (τ ) =

1 , α (s(τ ))

a.e. τ ∈ [0,t].

Now we define a reparametrization η by η (τ ) = ξ (s(τ )). It follows that η˙ (τ ) = ξ˙ (s(τ ))/α (s(τ )). Let uη be the unique solution of (8) with initial condition uη (0) = u0 , then we have

12


J(ξ ) 6 J(η ) = =

Z t 0

Z t 0

L(η (τ ), uη (τ ), η˙ (τ )) d τ

L(ξ (s), uξ ,α (s), ξ˙ (s)/α (s))α (s) ds

where uξ ,α solves u˙ξ ,α (s) = L(ξ (s), uξ ,α (s), ξ˙ (s)/α (s))α (s). Notice that Z s

L(ξ , uξ ,α , ξ˙ /α )α − L(ξ , uξ , ξ˙ ) dr Z s Z s ˙ ˙ L(ξ , uξ , ξ /α )α − L(ξ , uξ , ξ ) dr, 6K α |uξ ,α − uξ | dr +

uξ ,α (s) − uξ (s) =

0

0

0

+

where f+ = max{ f , 0} for any measurable function f . Invoking Gronwall inequality, we obtain Z t ˙ ˙ L(ξ , uξ , ξ /α )α − L(ξ , uξ , ξ ) dr, |uξ ,α (s) − uξ (s)| 6 exp(K1t) 0

+

where K1 = 32 K > Kkα kL∞ . Therefore, we have that J(ξ ) 6

Z t

L(ξ , uξ , ξ˙ /α )α dr +

Z t

K1 |uξ ,α (s) − uξ (s)| ds Z t Z t L(ξ , uξ , ξ˙ /α )α ds + tK1 exp(K1t) 6 L(ξ , uξ , ξ˙ /α )α − L(ξ , uξ , ξ˙ ) ds. 0

0

0

0

+

For all α ∈ [1/2, 3/2], we define

Φ (s, α ) = L(ξ (s), uξ (s), ξ˙ (s)/α )α + tK1 exp(K1t) L(ξ (s), uξ (s), ξ˙ (s)/α )α − L(ξ (s), uξ (s), ξ˙ (s)) and

Λ (α ) :=

Z t

Φ (s, α (s)) ds.

0

It is clear that J(ξ ) = Λ (1) 6 Λ (α ),

α ∈ Ω.

For almost all s, by continuity, there exists δ (s) ∈ (0, 1/2] such that −1 6 Φ (s, α ) − Φ (s, 1) 6 1,

∀α ∈ [1 − δ (s), 1 + δ (s)].

+


13

By measurable selection theorem1, we can assume δ (·) is measurable. Let S ⊂ L∞ ([0,t], R) be the set of the functions α : [0,t] → [1/2, 3/2] such that α (s) ∈ [1 − δ (s), 1 + δ (s)]. It is obvious that S is convex. Now, we can formulate our problem by minimizing Λ (α ) over S ⊂ L∞ ([0,t], R) (21) where S satisfies the equality constraint h(α ) =

Z t

(α (s) − 1) ds = 0.

0

We remark that Λ is convex in S (we take Λ (α ) = +∞ if α does not lie in S), and α ∗ ≡ 1 solves (21). The next step is to write the Lagrange multiplier rule for the problem (21) and its solution. By Kuhn-Tucker Theorem (see [15, Theorem 9.4]), We obtain a nonzero vector (λ1 , λ2 ) in R2 (with λ1 = 0 or 1) such that

λ1Λ (α ) + λ2 h(α ) > λ1Λ (α ∗ ),

∀α ∈ S.

It is clear that λ1 = 1. Indeed, if λ1 = 0, then one can take α ∈ S such that h(α ) < 0 which is absurd. Therefore, we have, for any α in S, the inequality Z t 0

Φ (s, α (s)) + λ2 α (s) ds >

Z t 0

Φ (s, 1) + λ2 ds.

Invoking [15, Theorem 6.31], we deduce that, for almost every s, the function

α 7→ Φ (s, α ) + λ2 α attains a minimum over the interval [1 − δ (s), 1 + δ (s)] at the interior point α = 1. For such a value of s, let E(s) = L(ξ (s), uξ (s), ξ˙ (s)) − hLv (ξ (s), uξ (s), ξ˙ (s)), ξ˙ (s)i. Therefore, by the calculus of the subdifferentials of the convex function Φ (s, α ) (see, for instance, [32, Corollary 4.3.2]), there exists µ (s) ∈ [0, 1] such that −λ2 = E(s) + tK1 exp(K1 t)µ (s)E(s). Thus, we conclude the Erdmann condition |E(s)| 6

|λ 2 | 6 |λ2 |, 1 + tK1 exp(K1 t)µ (s)

a.e. s ∈ [0,t].

(22)

Finally, let s be such that ξ˙ (s) exists, and such that (22) holds. By convexity, we have that

1 Since δ (s) ∈ (0, 1/2], we can not apply the standard measurable selection theorem (see, for instance, [15, Corollary 6.23]) directly. But, one can apply the theorem to an increasing sequence of close-valued measurable set-valued maps.

14


L(ξ (s), uξ (s), ξ˙ (s)/(1 + |ξ˙ (s)|)) − L(ξ (s), uξ (s), ξ˙ (s)) > ((1 + |ξ˙ (s)|)−1 − 1) · hLv (ξ (s), u (s), ξ˙ (s)), ξ˙ (s)i ξ

> ((1 + |ξ˙ (s)|)−1 − 1) · (L(ξ (s), uξ (s), ξ˙ (s)) + |λ2 |). It follows that L(ξ (s), uξ (s), ξ˙ (s)) 6 L(ξ (s), uξ (s), ξ˙ (s)/(1 + |ξ˙ (s)|))(1 + |ξ˙ (s)|) + |λ2 ||ξ˙ (s)|. Let C = sups∈[0,t],|v|61 L(ξ (s), uξ (s), v), then C is finite (since Corollary 2 and Lemma 3) and we have that L(ξ (s), uξ (s), ξ˙ (s)) 6 C + (C + |λ2 |)|ξ˙ (s)|. Therefore, invoking Lemma 3, we obtain that (C + |λ2| + 1)|ξ˙ (s)| − (θ0∗ (C + |λ2| + 1) + c0) 6 θ0 (|ξ˙ (s)|) − c0 6 L(ξ (s), 0, ξ˙ (s)) 6 L(ξ (s), uξ (s), ξ˙ (s)) + K|uξ (s)| 6C + (C + |λ2 |)|ξ˙ (s)| + (|u0 | + F1(t, R/t))K. This leads to |ξ˙ (s)| 6 (θ0∗ (C + |λ2| + 1) + c0) + C + (|u0 | + F1(t, R/t))K := F2 (t, R/t), which completes the proof.

3.2 Regularity of minimizers - Herglotz equations - Lie equations Let ξ ∈ Γxt0 ,x be a minimizer of (9) where uξ is determined uniquely by (8). For any t , we denote ξ (s) = ξ (s) + λ η (s). It is λ ∈ R and any Lipschitz function η ∈ Γ0,0 λ clear that ξλ ∈ Γxt0 ,x and J(ξ ) 6 J(ξλ ). Let uξλ be the associated unique solution of (8) with respect to ξλ and the initial condition u0 . Notice that

∂ ∂ J(ξλ )|λ =0 = u (t)|λ =0 = 0. ∂λ ∂ λ ξλ Now for any s ∈ [0,t] we set

∆λ (s) = and

Z uξλ (s) − uξ (s) 1 s = L(ξλ , uξλ , ξ˙λ ) − L(ξ , uξ , ξ˙ ) d τ , λ λ 0


15

Z 1 s

L(ξλ , uξλ , ξ˙λ (s)) − L(ξλ , uξλ , ξ˙ ) d τ , λ 0 Z 1 s L(ξλ , uξλ , ξ˙ ) − L(ξ , uξλ , ξ˙ ) d τ . f2λ (s) = λ 0 f1λ (s) =

Then f1λ and f2λ are all absolutely continuous functions on [0,t], and it follows

∆λ (s) = f1λ (s) + f2λ (s) +

1 λ

Z s 0

where cλ (τ ) = L u

Z 1 0

cλ · (u − u ) d τ , L ξλ ξ u

s ∈ [0,t],

Lu (ξ (τ ), uξ (τ ) + θ (uξλ (τ ) − uξ (τ )), ξ˙ (τ )) d θ ,

τ ∈ [0,t].

Thus, we conclude that for almost all s ∈ [0,t], the following Carathéodory equation holds: cλ (s) · ∆ (s) ∆˙ λ (s) = f˙1λ (s) + f˙2λ (s) + L (23) λ u

with initial condition ∆λ (t) = aλ . Notice that limλ →0 ∆λ (t) exists and limλ →0 ∆λ (t) = limλ →0 aλ = ∂∂λ uξλ (t)|λ =0 since ξ is a minimizer of J. It is not difficult to solve (23), we obtain that Rs c λ

∆λ (s) = aλ e t

Lu (r) dr

Rs c λ

+e t

Lu (r) dr

·

Z s

e−

t

Rr c λ t

Lu (τ ) d τ

· ( f1λ (r) + f2λ (r) dr.

Since (ξλ (s), ξ˙λ (s), uξλ (s)) tends (ξ (s), ξ˙ (s), uξ (s)) as λ → 0 for almost all s ∈ [0,t], together with Proposition 2 and Corollary 1, it follows that, for all s ∈ [0,t], we have f (s) :=

Rs ∂ uξλ (s)|λ =0 = e t h(r) dr · ∂λ

Z s t

e−

Rr t

h(τ ) d τ

· g(r) dr,

f (t) = 0,

(24)

where g = Lx · η + Lv · η˙ and h = Lu which are both measurable and bounded. This implies that equality (24) becomes Z s R Z t Rs r h(r) dr τ ) d τ − h( g(s) + h(s) · e t · 0= e t · g(r) dr ds 0 t t Z s R Z t Rs r g(s) ds + e t h(r) dr · e− t h(τ ) d τ · g(r) dr = 0 t −

Z t 0

R0

=e t It follows that

0

e

Rs

Rs t h(r) dr · e− t h(r) dr · g(s)

h(r) dr

·

Z

0

ds R t − ts h(r) dr e · g(s) ds .

16


0=

Z t 0

e−

Rs t

h(r) dr

· g(s) ds =

Z t

e−

0

Rs t

h(r) dr

· (Lx · η + Lv · η˙ )(s) ds.

Invoking the fundamental lemma in calculus of variation (see, for instance, Lemma 6.1.1 in [10]), we obtain that, for almost all s ∈ [0,t], Rs d − Rts h(r) dr e Lv (ξ (s), uξ (s), ξ˙ (s)) = e− t h(r) dr Lx (ξ (s), uξ (s), ξ˙ (s)). ds

This leads to the so called Herglotz equation (for almost all s ∈ [0,t]) d Lv (ξ (s), uξ (s), ξ˙ (s)) ds = Lx (ξ (s), uξ (s), ξ˙ (s)) + Lu (ξ (s), uξ (s), ξ˙ (s))Lv (ξ (s), uξ (s), ξ˙ (s)).

(25)

Since L is of C2 and L(x, u, ·) is strictly convex, then by the standard argument as in [10, Section 6.2], we conclude that: Theorem 1. Under our standing assumptions, we have the following regularity properties for any minimizer ξ for (9): (1) Both ξ and uξ are of class C2 and ξ satisfies Herglotz equation (25) for all s ∈ [0,t] where uξ is the unique solution of (8); (2) Let p(s) = Lv (ξ (s), uξ (s), ξ˙ (s)) be the dual arc. Then p is also of class C2 and we conclude that (ξ , p, uξ ) satisfies Lie equation (7). Remark 2. If L is only of class C1 satisfying (L1), (L2) and (L3), then all the results in previous sections still hold true. Indeed, we only use the the C1 regularity property of L. More precisely, Problem (9) under the subsidiary condition (8) admits a t which is of class C1 . Moreover, L (ξ (·), ξ˙ (·)) is absolutely conminimizer ξ ∈ Γx,y v tinuous on [0,t] and ξ together with uξ uniquely determined by (8) satisfy Herglotz equation (25) for almost all s ∈ [0,t]. Proof. We first need to show that ξ is of class C1 . Let N be the set of zero Lebesgue measure where ξ˙ does not exist. For t¯ ∈ [0,t], choose a sequence {tk } ∈ [0, T ] \ N such that tk → t¯. Then ξ˙ (tk ) → v¯ for some v¯ ∈ Rn (up to subsequences) and Lv (ξ (t¯), uξ (t¯), ξ˙ (t¯)) = lim Lv (ξ (tk ), uξ (tk ), ξ˙ (tk )) =

Z t 0

k→∞

{Lx (ξ (s), uξ (s), ξ˙ (s)) + Lu (ξ (s), uξ (s), ξ˙ (s))Lv (ξ (s), uξ (s), ξ˙ (s))}ds

by (25). From the strict convexity of L it follows that the map v 7→ Lv (ξ (s), uξ (s), v) is a diffeomorphism. This implies that v¯ is uniquely determined, i.e., lim

[0,t]\N∋s→t¯

ξ˙ (s) = v. ¯

Now, by Lemma 6.2.6 in [10], ξ˙ (t¯) exists and lim[0,t]\N∋s→t¯ ξ˙ (s) = ξ˙ (t¯). It follows ξ is of class C1 . In view of (8), uξ is also of class C1 .


17

In view of (6), by setting F(s) =

Z s 0

{Lx (ξ , uξ , ξ˙ ) + Lu (ξ , uξ , ξ˙ )Lv (ξ , uξ , ξ˙ )} d τ ,

we have that {Lv (ξ (s), uξ (s), v) − F(s)}|v=ξ˙ (s) = Lv (ξ (0), uξ (0), ξ˙ (0)). Then, the implicit function theorem implies ξ˙ is of class C1 since both F and Lv are of class C1 . Therefore we conclude that ξ is of class C2 and uξ is of class C2 by (8). The rest part of the proof is standard and we omit it.

4 Concluding remarks 4.1 Equivalence of Herglotz’ variational principle and the implicit variational principle In the recent work [40, 41], the authors introduce an implicit variational principle on closed manifolds which is essentially equivalent to Hergolotz’ principle. Proposition 3 ([40]). Let M be a C2 closed manifold and let L : T M → R satisfy conditions (L1)-(L3) for M instead of Rn here. Given any x0 ∈ M and u0 ∈ R, there exists a continuous function hx0 ,u0 (t, x) defined on (0, +∞) × M satisfying hx0 ,u0 (t, x) = u0 + inf ξ

Z t 0

L(ξ (s), hx0 ,u0 (s, ξ (s)), ξ˙ (s)) ds,

where ξ is taken over all the Lipschitz continuous curve on M connecting ξ (0) = x0 and ξ (t) = x. Moreover, let ξ be any curve achieving the infimum together with the curves p and u defined by u(s) = hx0 ,u0 (s, ξ (s)),

p(s) = Lv (ξ (s), u(s), ξ˙ (s)).

Then (ξ , p, uξ ) is a solution of (7) with conditions ξ (0) = x0 , ξ (t) = x and lims→0+ u(s) = u0 . Proposition 4. Let x0 , x ∈ Rn , t > 0 and u0 ∈ R. For any ξ ∈ Γxt0 ,x being a minimizer of (9), we denote by uξ (s, u0 ) the unique solution of (8) with uξ (0, u0 ) = u0 . Then, for any 0 < t ′ < t, the restriction of ξ on [0,t ′ ] is a minimizer for A(t ′ , x0 , x, u0 ) := u0 + inf

Z t′ 0

L(ξ (s), uξ (s), ξ˙ (s)) ds

18


with uξ the unique solution of (8) restricted on [0,t ′ ]. Moreover, A(s, x0 , ξ (s), u0 ) = uξ (s, u0 ),

∀s ∈ [0,t],

(26)

and A(s1 + s2 , x0 , ξ (s1 + s2 ), u0 ) = A(s2 , ξ (s1 ), ξ (s1 + s2 ), uξ (s1 )) for any s1 , s2 > 0 and s1 + s2 6 t. In particular, if hx0 ,u0 is from Proposition 3, then uξ (s) = hx0 ,u0 (s, ξ (s)),

s ∈ [0,t].

Proof. Suppose x0 , x ∈ Rn , t > 0 and u0 ∈ R. Let ξ ∈ Γxt0 ,x be a minimizer of (9) and uξ (s) = uξ (s; u0 ) be the unique solution of (8) with uξ (0) = u0 . ′ ′ be the restriction of ξ on Now, let 0 < t ′ < t. Let ξ1 ∈ Γx,t ξ (t ′ ) and ξ2 ∈ Γξt−t (t ′ ),y [0,t ′ ] and [t ′ ,t] respectively. Then, we have that uξ (t ′ ; u0 ) = u0 + uξ (t; u0 ) − uξ (t ′ ; u0 ) =

Z t t′

Z t′ 0

L(ξ1 (s).uξ1 (s), ξ˙1 (s)) ds,

L(ξ2 (s).uξ2 (s), ξ˙2 (s)) ds.

Then both ξ1 and ξ2 are minimal curve for (9) restricted on [0,t ′ ] and [t ′ ,t] respectively by summing up the equalities above and the assumption that ξ is a minimizer of (9). In particular, (26) follows. The next assertion is direct from the relation uξ (s1 + s2 ; u0 ) = uξ (s2 ; uξ (s1 )),

∀s1 , s2 > 0, s1 + s2 6 t,

since uξ solves (8). The last assertion is a direct application of Gronwall’s inequality.

4.2 Further remarks Comparing to the method used in [40, 41], one can see more from our approach as follows: – We can derive the generalized Euler-Lagrange equations in a modern and rigorous way which does not appear in both [40, 41]; – There should be an extension of the main results of this paper under much more general conditions (like Osgood type conditions) to guarantee the existence and uniqueness of the solutions of the associated Carathéodory equation (8). – Along this line, the quatitative semiconcavity and convexity estimate of the associated fundamental solutions have been obtained in [7] recently, which is useful for the intrinsic study of the global propagation of singularities of the viscosity solutions of (4) and (5) ([8, 4, 5, 6]); – When the Lagrangian has the form L(x, v) − λ u, by solving the associated Carathéodory equation (8) directly, one gets the representation formula for the


19

associated viscosity solutions immediately ([18, 39, 42]). The representation formula bridges the PDE aspects of the problem with the dynamical ones; – Consider a family of Lagrangians in the form {L(x, v) + ∑ki=1 ai j ui }, a problem of Herglotz’ variational principle in the vector form is closely connected to certain stochastic model of weakly coupled Hamilton-Jacobi equations (see, for instance, [19, 22, 37]). Acknowledgements This work is partly supported by Natural Scientific Foundation of China (Grant No. 11631006, No. 11771283 and No.11790272), and the National Group for Mathematical Analysis, Probability and Applications (GNAMPA) of the Italian Istituto Nazionale di Alta Matematica “Francesco Severi”. The authors are grateful to Qinbo Chen, Cui Chen and Kai Zhao for helpful discussions. This work was motivated when the first two authors visited Fudan University in June 2017.

Appendix Let Ω ⊂ Rn+1 be an open set. A function f : Ω ⊂ R × Rn → Rn is said to satisfy Carathéodory condition if - for any x ∈ Rn , f (·, x) is measurable; - for any t ∈ R, f (t, ·) is continuous; - for each compact set U of Ω , there is an integrable function mU (t) such that | f (t, x)| 6 mU (t),

(t, x) ∈ U.

The classical problem of following Carathéodory equation x(t) ˙ = f (t, x(t)),

a.e.,t ∈ I

(27)

is to find an absolutely continuous function x defined on a real interval I such that (t, x(t)) ∈ Ω for t ∈ I and satisfies (27). Proposition 5 (Carathéodory). If Ω is an open set in Rn+1 and f satisfies the Carathéodory conditions on Ω , then, for any (t0 , x0 ) in Ω , there is a solution of (27) through (t0 , x0 ). Moreover, if the function f (t, x) is also locally Lipschitzian in x with a measurable Lipschitz function, then the uniqueness property of the solution remains valid. For the proof of Proposition 5 and more results related to Carathéodory equation (27), the readers can refer to [16, 23].

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