Locally Lipschitzian Guiding Function Method for ODEs.

Transcription

1 Locally Lipschitzian Guiding Function Method for ODEs. Marta Lewicka International School for Advanced Studies, SISSA, via Beirut 2-4, 3414 Trieste, Italy. 1 Introduction Let f : [, T ] R n R n ; we investigate the problem of existence of T- periodic solutions to the first order differential equation with f in the right hand side. Namely, we seek for solutions to the problem: x (t) = f(t, x(t)) x() = x(t ), (1) by generalizing the well known guiding function method. Such an approach can be found in several works, however, under some heavier assumptions. For example, in [M] f is assumed to be locally lipschitzian and the guiding function to be C 1. In [GP] f needs to be of the Caratheodory type only, while the corresponding guiding function must be still C 1. In fact, in [GP] the more general, multivalued problem x (t) ϕ(t, x(t)) x() = x(t ). (1 ) is under consideration. In our paper we get rid of the assumption of the guiding function to be C 1. In fact, in Theorem 1, which is motivated by [GP], we need it to be locally lipschitzian only (Theorem 1). Our second main result (Theorem 2) characterizes a class of guiding functions, satisfying the conditions of Theorem 1. This result is an extension of the theorem on the index of coercive potentials [K] (a remarkable reformulation of which was done in [M]). 1

2 2 PRELIMINARIES 2 2 Preliminaries In this section we review some of the standard facts and definitions. Let X, Y be topological spaces. We say that X is an R δ set, whenever it is homeomorphic with an intersection of a decreasing sequence of compact, metric ANRs. A multivalued mapping ϕ : X Y (we will always suppose that a multivalued mapping has nonempty values) is called upper semicontinuous (u.s.c.) if, for every open set U Y, the set ϕ 1 (U) = x X : ϕ(x) U} is open in X. If X is a space with measure, we say that ϕ is measurable if ϕ 1 + (U) = x X : ϕ(x) U } is measurable for any open set U Y. We say that ϕ is admissible if X and Y are compact metric ANRs and ϕ is u.s.c. with R δ values. A map ϕ : X X is called decomposable if it has a decomposition: ϕ D ϕ : X = X 1 ϕ 2 X1 X2... ϕn X n = X ϕ = ϕ n... ϕ 2 ϕ 1, (2) where each ϕ i is admissible. Let A be an open subset of X such that a decomposable map ϕ (with a decomposition (2)) has no fixed points on its boundary, that is x ϕ(x) for every x A. In such a case it is possible to define a fixed point index Ind X (D ϕ, A) Z having the following properties: (i) (existence) If Ind X (D ϕ, A) then ϕ has a fixed point in A. (ii) (additivity) Let A i, (1 i n) be open, pairwise disjoint subsets of A. Suppose, that ϕ has no fixed points in A \ n i=1 A i. Then the indices Ind X (D ϕ, A i ), 1 i n, are well defined and Ind X (D ϕ, A) = n i=1 Ind X (D ϕ, A i ). (iii) (homotopy invariance) Let ψ : X X be a decomposable mapping, with a decomposition: ψ D ψ : X = X 1 ψ 2 X1 X2... ψn X n = X, ψ = ψ n... ψ 2 ψ 1. Suppose that the decompositions D ϕ and D ψ are homotopic, that is there exists a decomposable homotopy χ : X [, 1] X, having a decomposition: D χ : X [, 1] = X [, 1] χ 1 X 1 [, 1]... χ n 1 X n 1 [, 1] χn X n = X, χ = χ n χ n 1... χ 1, where, for 1 i n, there exist admissible χ i : X i 1 X i such that the following conditions are fulfilled: χ i (, ) = ϕ i, χ i (, 1) = ψ i, χ i (, λ) = χ i (, λ) λ} (for i n) and: x A λ [, 1] x χ(x, λ).

3 2 PRELIMINARIES 3 Then Ind X (D ψ, A) is well defined and: Ind X (D ψ, A) = Ind X (D ϕ, A). (iv) (contraction) Suppose, that in the decomposition of D ϕ we have X n 1 = Y X and the mapping ϕ n is the inclusion: ϕ n = i : Y X. Then ϕ Y has a decomposition D ϕ Y : Y = X ϕ 1 Y X 1 ϕ 2 X2... ϕ n 1 X n 1 = Y. Moreover, if ϕ Y has no fixed points in (A Y ), then Ind Y (D ϕ Y, A Y ) is well defined and Ind Y (D ϕ Y, A Y ) = Ind X (D ϕ, A). (v) (units) If ϕ is constant, that is ϕ(x) = B X for every x X, then Ind X (D ϕ, A) = where D ϕ : X = X ϕ X1 = X. 1 for A B for A B =, Notice, that the fixed point index is defined for a decomposition of a multivalued map, not for the map itself. However, when it is clear which decomposition we mean, we will simply write Ind X (ϕ, A). For the above definitions and other properties of Ind we refer to [BK] and [AGL]. Let ϕ : [, T ] R n R n be a Caratheodory multifunction, that is measurable in the first variable for every x R n and u.s.c. in the second variable for almost every t [, T ] we say that ϕ has integrably bounded growth (with the bounding function µ) if there exists a function µ L 1 ([, T ], R) such that y µ(t)(1+ x ) for every x R n, t [, T ] and y ϕ(t, x). The Poincaré operator for the differential inclusion (1 ) is a multivalued mapping S ϕ : R n C([, T ], R n ), given by S ϕ (x ) = x : [, T ] R n, x is absolutely continuous, x (t) ϕ(t, x(t)) for almost every t [, T ], x() = x }. We say that x : [, T ] R n is a solution of the inclusion x (t) ϕ(t, x(t)), whenever x S ϕ (x()). It is known that if ϕ is of the Caratheodory type, has compact and convex values and has integrably bounded growth, then S ϕ is u.s.c. with R δ values. Let ψ : [, T ] R n [, 1] R n be a multivalued mapping with compact, convex values, such that for every (x, λ) R n [, 1] the mapping ψ(, x, λ) is measurable and that for almost every t [, T ] the map ψ(t,, ) is u.s.c.; assume that there exists µ L 1 ([, T ], R) such that for every λ [, 1] the multifunction ψ(,, λ) has integrably bounded growth with µ as its bounding function. Then the map R n [, 1] (x, λ) S ψ(,,λ) (x) is u.s.c. with R δ values.

4 3 AUXILIARY RESULTS 4 For the statements above, see [AC], [G], [AGL]. From now on, by V will be assumed to be a locally lipschitzian function from R n to R. Let us define Ω V = x R n, V is not differentiable in x}. The Rademacher Theorem states that Ω V has measure. By the generalized gradient of V at point x R n, denoted by V (x ), we mean the convex hull of the set of all limits lim V (x i ) where x i } i i=1 is any sequence of points in R n \ Ω V, converging to x. The following properties of gradient are known: Ω V (i) Let S R n be of measure. Then V (x ) can be obtained by replacing with Ω V S in the above definition. (ii) The multivalued mapping V : R n R n is u.s.c. with compact, convex values. (iii) (mean value theorem) Let x, y R n. Then there exists a point u in a segment (x, y) and a point w V (x) such, that V (x) V (y) = w, y x. The reader is refered to [C] for more more material on this topic. Given an euclidean space, by B(ε) we will denote an open ball of the radius ε, centered in the origin. 3 Auxiliary Results Let us recall the following lemma [GP]: Lemma Assume ϕ : [, T ] R n R n has integrably bounded growth. Fix r >, there exists r > such that, for every solution of the problem x (t) ϕ(t, x(t)) x() > r, we have x(t) > r for every t [, T ]. Moreover, r depends only on r and on the function µ in the definition of integrably bounded growth for the multivalued map ϕ. Lemma 1 Let Ω R n have measure. Suppose we are given a Lebesgue measurable function z : [, T ] R n. Then, for every ε > there exists x B(ε) R n such that z(t) + x Ω for almost every t [, T ].

5 3 AUXILIARY RESULTS 5 Proof Consider the characteristic function of Ω, χ Ω : R n R and a measurable function F : [, T ] B(ε) R n, given by F (t, x) = z(t) + x. Without loss of generality we may assume that Ω is Borel. Then the composition χ Ω F is measurable and from Fubini Theorem we get B(ε) ( T χ Ω F dt)dx = T ( χ Ω F dx)dt =. Hence for almost every x B(ε), T (χ Ω F )(x, t)dt =. B(ε) The following fact is straightforward: Lemma 2 Let z : [, T ] R n be absolutely continuous. Then the composition V z is absolutely continuous and for every t [, T ] such that z is differentiable at t and V is differentiable at z(t), V z is differentiable at t and (V z) (t) = V (z(t)), z (t). The following lemma is of a basic importance for our later considerations: Lemma 3 Let z be as in the previous lemma. Suppose that for almost every t [, T ] we have: v V (z(t)) v, z (t) >. (3) Then (V z) (t) > for almost every t [, T ]. In particular, z(t ) z(). Proof By Lemma 1 there exists a sequence x k } k=1 convergent to R n such that: t [, T ] \ A k > z k (t) = z(t) + x k Ω V, where A [, T ] is a set of measure, containing the points in which z fails to be differentiable. We may also suppose that (3) is valid for every t [, T ]\A. It is easily seen that the absolutely continuous functions V z k are equibounded and that (V z k ) are uniformly dominated by an integrable function t C z (t) with C a positive constant (this follows from Lemma 2). Therefore (see e.g. [AC]) we can extract a subsequence V z ki } i=1 such that the derivatives (V z ki ) converge weakly in L 1 ([, T ], R) to (V z). By Mazur Lemma there exists a sequence of convex combinations w i } i=1, w i = λ i j (V z kj ), convergent to (V z) in L 1 ([, T ], R). Thus, without loss of generality, we have lim i w i (t) = (V z) (t) for every t [, T ] \ A. Fix t [, T ]\A. By (3) there exists a real α t > such that v, z (t) > 2α t for every v V (z(t)). By Lemma 2 and uppersemicontiniuity of V, there exists a number k t such that, for every k k t, we have (V z k ) (t) = V (z(t) + x k ), z (t) v + u, z (t) : v V (z(t)) and u α t / z (t) } j=i

6 3 AUXILIARY RESULTS 6 and, consequently, (V z k ) (t) > α t for every k k t. Thus, w i (t) > α t for i large enough and we conclude that (V z) (t) α t >, which proves the first statement of the lemma. In particular, V z(t ) V z() = T (V z) (τ)dτ >, so z(t ) z(). Definition 1 V is said to be a direct potential (with a constant r > ) if x R n \ B(r ) v, w V (x) v, w Note that the above definition will not change if we replace by >. Our definition agrees with the classical one for C 1 functions presented, for example, in [M]. The following lemma defines the index of a direct potential: Lemma 4 Let V be a direct potential (with a constant r ). Set ϕ : R n R n as follows: ϕ = Id R n V. We define Ind(V ) = Ind B(2r ) (r 2r ϕ, B(r )), where r 2r is the radial retraction of R n onto B(2r ). (Here ϕ and r 2r are admissible, so the decomposition for defining the index is taken naturally as the composition r 2r ϕ.) Then Ind B(2R) (r 2R ϕ, B(R)) = Ind(V ) for every R r. Proof The proof is straightforward and follows from homotopy invariance, additivity and contraction properties of the fixed point index. Let V be as in the previous lemma. Now consider a multivalued mapping W V : R n R n, given W V (x) = conv(r 1 V (x)), where r 1 : R n R n is the radial retraction onto B(1) R n and conv stands for the convex hull. It is not hard to see that W V is u.s.c., bounded by 1 and has compact, convex values. The following lemma is analogous to a result obtained in [GP]: Lemma 5 Set a number T >. There exists R > such that for every r > R there is t r (, T ] such that every solution of the problem: x (t) W V (x(t)) x() = r (4) has the following properties: (i) t (, t r ] v V (x()) x(t) x(), v >, (ii) t (, T ] x(t) x().

7 4 MAIN RESULTS 7 Proof Lemma gives the existence of a number R > r such that for every solution of the problem: x (t) W V (x(t)) x() > R we have x(t) > r for every t [, T ]. Since V is u.s.c. with compact values and V is a direct potential, we obviously have: R > r ε R > x, y B(R) \ B(r ) R n : x y < ε R w V (x) v V (y) w, v >. (5) Fix r > R and let t r ε r+t, t r (, T ]. Let x be a solution of (4). We have x(t) r + t x (τ) dτ r + T for every t [, T ] and x(t) x() t x (τ) dτ t r ε r+t for every t (, t r ]. Hence, by (5): t (, t r ] w V (x(t)) v V (x()) w, v >. Combining the above with the following evident remark: we obtain and, finally, y R n w W V (y) α 1 αw V (y), (6) t (, t r ] w W V (x(t)) v V (x()) w, v > t (, t r ] v V (x()) x(t) x(), v = t x (τ), v dτ >. Moreover, by (6), we have v, x (t) > for almost every t [, T ] and every v V (x(t)), which, by Lemma 3 completes the proof of (ii). 4 Main Results When V is in C 1, the following definition reduces to the corresponding one in [GP]. Definition 2 Let f : [, T ] R n R n, V be a direct potential with a constant r. V is called a guiding function for f, whenever x R n \ B(r ) w V (x) t [, T ] f(t, x), w.

8 4 MAIN RESULTS 8 Here comes our first main result. Theorem 1 Let f : [, T ] R n R n be a Caratheodory function with integrably bounded growth. Suppose that f has a guiding function V (with a constant r ) such that Ind(V ). Then the problem (1) has at least one solution. Proof Consider the following family of differential inclusions, with κ [, 1] : z (t) κw V (z(t)) + (1 κ)f(t, z(t)). By Lemma, there exists R > r such that for any z : [, T ] R n that is a solution of the above problem for some κ [, 1] and satisfies z() R, we have z(t) > r for every t [, T ]. Let t R be as in Lemma 5. Take the decomposable homotopy H : B(2R) [, 1] B(2R), given by H(x, λ) = r 2R ((1 λ) (x V (x)) + λ (2x e tr S WV (x))), where r 2R is, as usual, the radial retraction of R n onto B(2R) and e tr : C([, T ], R n ) R n is by definition e tr (z) = z(t R ) (the explicit formula for the decomposition of H is not complicated but long, so we omit it). We will show that H has no fixed points in B(R). Conversely, suppose that there exists x H(x, λ), with x B(R) and λ [, 1]. Then v V (x) z S WV (x) = (1 λ)(x v) + λ(2x z(t R )) x 2 = (1 λ) 2 v 2 +λ 2 z(t R ) z() 2 +2(1 λ)λ v, z(t R ) z(), which contradicts Lemma 5. By the above formula, Lemma 4 and the homotopy invariance of the fixed point index we have Ind(V ) = Ind B(2R) (r 2R (2Id B(2R) e tr S WV ), B(R)). (7) (the decompositions of the multivalued maps in the formula are induced by the decomposition of the homotopy H). Consider G : [, T ] R n [, 1] R n, given by G(t, x, λ) = k(λ)w V (x) + (1 k(λ))f(t, x), where k : [, 1] R is by definition k(λ) = 1 for λ [, 1/2) 2 2λ for λ [1/2, 1].

9 4 MAIN RESULTS 9 It is easy to check that G has the properties of the map ψ, introduced in paragraph 1, thus the following homotopy is decomposable K : B(2R) [, 1] B(2R), K(x, λ) = r 2R (2x e h(λ) S G(,,λ) (x)), where h : [, 1] R is given by h(λ) = 2(T tr )λ + t R for λ [, 1/2) T for λ [1/2, 1]. Now suppose that the problem (1) has no solutions. We will first show that K has no fixed points in B(R). Conversely, suppose that x K(x, λ) for a point x B(R) and λ [, 1]. If λ [, 1/2), then z(h(λ)) = z() for a function z S WV (x), that contradicts Lemma 5 (ii). If λ [1/2, 1), then z(t ) = z() for a function z S G(,,λ) (x). We have z (t), v = k(λ) w t,v, v + (1 k(λ)) f(t, z(t)), v for almost every t [, T ], every v V (z(t)) and a point w t,v W V (z(t)). Now f(t, z(t)), v > because V is a guiding function for (1) and w t,v, v > from (6) and the fact that V is a direct potential. Consequently z (t), v > for almost every t [, T ] and every v V (z(t)), which contradicts Lemma 3. The case λ = 1 is already excluded by our assumption that (1) has no solutions. Now from the above and homotopy invariance of the fixed point index we have Ind B(2R) (r 2R (2Id B(2R) e tr S WV ), B(R)) = Ind B(2R) (r 2R (2Id B(2R) e T S f ), B(R)). Recalling (7) Ind B(2R) (r 2R (2Id B(2R) e T S f ), B(R)) = Ind(V ). By existence property of the fixed point index we obtain a fixed point of the mapping r 2R (2Id B(2R) e T S f ), namely x B(R) such that x 2x e T S f (x) which is equivalent to x e T S f (x). This means that (1) has a solution (with the initial value x), that is against our contradictory assumption and thus proves the theorem. Our next result gives a condition for a direct potential to have nonzero index, as it is required in Theorem 1. Theorem 2 Let V be a direct potential (with a constant r > ). Suppose that V is coercive, that is V (x) = +. Then Ind(V ) = 1. lim x

10 4 MAIN RESULTS 1 Proof For every γ R denote by A γ the open and bounded subset of R n, given by V 1 ((, γ)). Take the numbers α, β > α and R > r > r such that: B(r ) A α A α B(r) B(r) A β A β B(R). Let β α T = 2 min v, w : v V (x), w W V (x), x A β \ A α } The above formula makes sence in view of (6) (V is a direct potential) and A β \ A α R n \ B(r ). We will consider the following differential inclusion: and its solutions on the interval [, T ]. We have divided the later proof into five steps. z (t) W V (z(t)) (8) Step 1 We will prove that for every γ α the set A γ is positively invariant, that is x A γ z S WV (x) t [, T ] z(t) A γ. The composition V z is absolutely continuous, while (6) implies that for almost every t [, T ] such that z(t) R n \ B(r ) we have v, z (t) < for every v V (z(t)). Using the same method as in the proof of Lemma 3, we obtain (V z) (t) < for almost every t [, T ] such that z(t) R n \ B(r ). If there exist t, t 1 [, T ], that z(t ) A γ and z(t) A γ for every t (t, t 1 ], then V z(t ) = γ and V z(t) > γ for t (t, t 1 ], hence V z(t 1 ) = V z(t ) + t 1 (V z) (τ)dτ < γ, a contradiction. t Step 2 Consider the mapping S 1 : B(2R) (, 1] C([, T ], R n ) S 1 (x, λ) = w : [, T ] R n : w(t) = z(λt) x λ with z S WV (x)}. S 1 is u.s.c. and has R δ values, because S 1 (x, λ) = 1 λ (S λw V (x) x). Moreover w (t) 1 for every w S 1 (B(2R) (, 1]) and for almost every t [, T ]. Hence, by Ascoli-Arzelá Theorem, the set S 1 (B(2R) (, 1]) is relatively compact in C([, T ], R n ). Let S 2 : B(2R) [, 1] C([, T ], R n ), S 2 (x, λ) = S 1 (x, λ) for λ w = lim w k : w k S 1 (x k, λ k ) with x n x, λ k >, λ k } k for λ =. It is readily verified that S 2 is u.s.c. and has compact values.

11 4 MAIN RESULTS 11 Finally, let S : B(2R) [, 1] C([, T ], R n ) S(x, λ) = S 2 (x, λ) for λ conv S 2 (x, ) for λ =. It is not hard to see that S is u.s.c. with R δ values. Step 3 Consider the decomposable homotopy K : B(2R) [, 1] B(2R), K(x, λ) = r 2R (λ(e T S(x, ) + x) + (1 λ)(x V (x))), where r 2R and e T are as in the proof of Theorem 1. We will show that K has no fixed points in A β. First, let us remark that: x A β α x > ε x > y B(x, ε x ) w W V (y) v V (x) w, v > α x. Now let w S 2 (x, ) and v V (x) for a point x A β. By the definition z of S 2, w(t ) = lim k (λ k T ) x k k λ k, where lim x k = x, lim λ k = and λ k >, k k z k S WV (x k ). For k large enough λ k < ε x /(2T ) and x k x < ε x /2, thus, for any t [, λ k T ] and hence z k (t) x x k x + z k (t) x k ε x z k(λ k T ) x k, v = 1 λ k λ k λ k T z k(τ), v dτ > 1 λ k λ k T α x dτ = α x T. This implies w(t ), v α x T >. Now suppose that x K(x, λ) for an x A β and λ [, 1], hence λe T S(x, ) (1 λ) V (x). By the definition of S(, ) there exist w 1,..., w k } k S 2 (x, ), λ 1,..., λ k } R +, λ i = 1 and v V (x) such that i=1 = λ k λ i w i (T ) (1 λ)v 2 i=1 = λ 2 k λ i w i (T ) 2 +(1 λ) 2 v 2 +2λ(1 λ) k λ i w i (T ), v >, i=1 a contradiction. Homotopy invariance and additivity of the fixed point index now yield i=1 Ind(V ) = Ind B(2R) (r 2R (Id B(2R) V ), B(R)) = Ind B(2R) (r 2R (Id B(2R) V ), A β ) = Ind B(2R) (r 2R (e T S(, ) + Id B(2R) ), A β ).

12 4 MAIN RESULTS 12 Step 4 Let H : B(2R) [, 1] B(2R) H(x, λ) = r 2R (e T S(x, λ) + x), H is a decomposable homotopy without fixed points in A β. To show this last statement, it is enough to prove that x H(x, λ) for any x A β and λ (, 1]. Suppose the contrary; then x = z(λt ) for a point x A β and z S WV (x). The sets A β and A α A β are positively invariant by Step 1, hence z(t) A β \A α A β \B(r ) for every t [, λt ]. As in Step 1, we obtain (V z) (t) < for almost every t [, λt ] and, finally = V z(λt ) V (x) = λt (V z) (τ)dτ <, a contradiction. The homotopy invariance of the fixed point index forces: Ind B(2R) (r 2R (e T S(, ) + Id B(2R) ), A β ) = Ind B(2R) (r 2R e T S WV, A β ). Step 5 Take the decomposable homotopy L : B(2R) [, 1] B(2R) L(x, λ) = r 2R (λ e T S WV (x)). If we had x A β, z S WV (x) and λ [, 1) such that x = λz(t ), it would follow = x λz(t ) > x z(t ) and z(t ) R n \ B(r) R n \ A α. Thus V z(t ) > α and z(t) A α for every t [, T ]. Hence: β α > V (x) V z(t ) = T (V z) (τ)dτ. (9) Using the same technique as in the proof of Lemma 3 it can be easily seen that, for almost every t [, T ], z (t) W V (z(t)) and there exists x t k} k=1, convergent to R n such that (V z) (t) is a limit of a sequence wk} t k=1, where each wk t is a convex combination of some numbers in the set V (z(t) + x t i), z (t), i = k, k + 1,...}. Set a number k, for i large enough we have V (z(t) + x i ), z (t) v, z (t) : v V (z(t))+b(1/k)}. Consequently, (V z) (t) v, z (t) : v V (z(t)) + B(1/k)} and (V z) (t) v, w : v V (z(t)), w W V (z(t))} for almost every t [, T ]. Recalling (9), we obtain: β α > T min v, w : v V (x), w W V (x), x A β \ A α } = 2(β α), a contradiction. In this way we have shown that L has no fixed points in A β. By the homotopy invariance and units properties of the fixed point index, Ind B(2R) (r 2R e T S WV, A β ) = Ind B(2R) (, A β ) = 1 and, finally, by Steps 3 5 we obtain Ind(V ) = 1.

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