Stationary mean-field games Diogo A. Gomes

Transcription

1 Stationary mean-field games Diogo A. Gomes

2 We consider is the periodic stationary MFG, { ɛ u + Du V (x) = g(m) + H ɛ m div(mdu) = 0, (1) where the unknowns are u : T d R, m : T d R, with m 0 and m = 1, and H R.

3 We suppose that V : T d R is C, g : R + R (or g : R + 0 R), C in the set m > 0, satisfying T d g(m) C T d mg(m). We say that (u, m, H) or (u, m) is a classical solution of, respectively, (1) if u and m are C, m > 0, (u, m) solves (1).

4 Maximum principle bounds Bounds for H Proposition Let u be a classical solution of (1). Suppose that g 0. Then, H sup T d V.

5 Maximum principle bounds Because u is periodic, it achieves a minimum at a point, x 0. At this point, Du(x 0 ) = 0 and u 0. Consequently, Hence, H sup V. V (x 0 ) H + g(m) H.

6 First-order estimates Proposition There exists a constant, C, such that, for any classical solution, (u, m, H), of (1), we have Td Du 2 2 (1 + m) + 1 g(m)mdx C. 2

7 First-order estimates Multiply the Hamilton-Jacobi equation by (m 1) and Fokker Planck equation by u, adding them, and integrating by parts gives Td Du 2 (1 + m) + mg(m)dx = V (m 1) + g(m)dx. 2 T d Using the assumption on g, we obtain the result.

8 First-order estimates Corollary Let (u, m, H) be a classical solution of (1). Suppose that g 0. Then, there exists a constant, C, not depending on the particular solution, such that H C.

9 First-order estimates We have: Du 2 2 L 1. From the assumptions and the preceding estimate, g(m) L 1. Therefore, integrating the Hamilton-Jacobi equation, we obtain the bound for H.

10 Integral Bernstein estimate Bernstein estimates Here, we examine the Hamilton-Jacobi equation, u(x) + Du(x) V (x) = H, with V L p. Our goal is to bound the norm of Du in L q for some q > 1.

11 Integral Bernstein estimate Lemma Let u C 3 and v = Du 2. Suppose that V C 1. Then, there exist, c, C > 0, which do not depend on u or V such that, for every p > 1, v p vdx T d 4pc ( (p + 1) 2 v T d ) d 2 (p+1)d ( d d 2 dx C v p+2 dx T d ) p+1 p+2 and 2 DV Du v p dx 1 D 2 u T d 2 T d 2 v p dx + C p T V d 2 v p dx.

12 Integral Bernstein estimate By integration by parts, we have the identity v p vdx = pv p 1 Dv 2 4p dx = T d T d (p + 1) 2 Dv p dx. T d Next, we use Sobolev s inequality to obtain Dv p dx+ T d T d v p+1 dx c moreover, from Young s inequality, v p+1 2 ( 2 = c 2 ( ) p+1 v p+1 dx v p+2 p+2 dx, T d T d v (p+1)d d 2 T d ) d 2 d dx,

13 Integral Bernstein estimate For the second inequality, we integrate again by parts to get DV Du v p dx = T d V u v p dx + p T d V v p 1 Dv Dudx. T d Next, we apply a weighted Cauchy inequality to each of the terms in the prior identity to get V u v p dx 1 D 2 u 2 v p dx + C V 2 v p dx. T d 8 T d T d Next, because v = Du 2, we have Dv = 2D 2 udu, hence p V v p 1 Dv Dudx 2p V v p D 2 u dx 1 D 2 u T d T d 8 T d 2 v p dx + C p T V d 2 v p dx. Using the two preceding bounds, we get the second estimate.

14 Integral Bernstein estimate Bernstein estimate Theorem Let u be C 3 and V C 1. Then, for any p > 1, there exists a constant, C p > 0, that depends only on H, such that ) Du 2d(p+1) C p (1 + V L d 2 (T d 2d(1+p). ) L d+2p (T d ) Note that γ p = 2d(1+p) d+2p d when p and that γ p is increasing when d > 2.

15 Integral Bernstein estimate We set v = Du 2. Differentiating the Hamilton-Jacobi equation Thus, v = 2 = 2 u xi = 1 2 v x i + V xi. d ( ) 2 d uxi x j 2 u xi u xi (2) i,j=1 i,j=1 i=1 d ( ) 2 d ( ) 1 uxi x j 2 u xi 2 v x i + V xi. i=1

16 Integral Bernstein estimate By multiplying (2) by v p and integrating over T d, we have v p vdx + 2 D 2 u 2 v p dx T d T d = Du Dv v p dx 2 DV Du v p dx. T d T d The Lemma provides bounds for the first term on the left-hand side and the last term on the right-hand side. For δ > 0, there exists a constant, C δ > 0, such that Du Dv v p dx δ D 2 u 2 v p dx + T d T d for every p > 1. C δ p + 1 T d v p+2 dx,

17 Integral Bernstein estimate Now, we claim that for any large enough p > 1, there exists C p > 0 that does not depend on u, such that ( ) (d 2) ( ) 1 v d(p+1) d(p+1) d 2 dx Cp V 2βp βp dx + Cp, (3) T d T d where β p is the conjugate exponent of d(p+1) (d 2)p. Further, β p d 2 when p.

18 Integral Bernstein estimate To prove the previous claim, we use the lemma to get c p ( v d(p+1) d 2 T d ) d 2 d dx + 2 D 2 u 2 v p dx T d ) p+1 c p v (T p+2 p+2 dx Du Dv v p dx 2 DV Du v p dx, d T d T d where c p := 4p C (p+1) 2, for some constant C.

19 Integral Bernstein estimate From the second estimate in the Lemma, and Young s inequality, (z p+1 p+2 ɛz + C p,ɛ ), we have c p ( v d(p+1) d 2 T d ) d 2 d dx C p T d V 2 v p dx + [ ( ( Cδ p δ )] 2 + δ D 2 u 2 v p dx T ) d v p+2 dx + C p,δ. T d

20 Integral Bernstein estimate The key idea is now to use the Hamilton-Jacobi equation again: D 2 u 2 v p dx 1 u 2 v p dx = 1 v T d d T d d T d 2 + V H 2 v p dx 1 v 2 v p dx 1 V 2 v p dx 1 3d T d d T d d C c v p+2 dx C V 2 v p dx C p, T d T d where the second inequality follows from (a b c) a2 b 2 c 2. v p dx

21 Integral Bernstein estimate For a small δ and a large enough p, the preceding inequalities give Hence, ( ) d 2 v d(p+1) d d 2 dx C p V T T 2 v p dx + C p d d ( ) (d 2)p ( ) 1 d(p+1) C p dx V 2βp βp dx + Cp. Td v d(p+1) d 2 T d ( ) (d 2) ( ) 1 v d(p+1) d(p+1) d 2 dx Cp V 2βp βp dx + Cp. T d T d This last estimate gives (3), and the theorem follows.

22 The Bernstein method for MFGs We fix a C 2 potential, V : T d R, and look for a solution, (u, m, H), of the MFG u(x) + Du(x) V (x) = H + m α, m div (mdu) = 0, (4) udx = 0, mdx = 1, with u, m : T d R and H R.

23 The Bernstein method for MFGs Theorem Let (u, m, H) solve (4) and 0 < α 1 d 1. Suppose that u, m C 2 (T d ). Then, for every q > 1, there exists a constant, C q > 0, that depends only on V L 1+ α 1, such that (T d ) Du L q (T d ) C q.

24 The Bernstein method for MFGs We use Bernstein s estimate with V replaced by V (x) m α. By the previous results, we have H C, m α L 1+ 1 α (T d ) C. Because d α, γ p α, Bernstein estimate gives Du L q (T d ) C q for every q > 1.

25 Bootstrapping regularity Proposition Let (u, m, H) solve (4) with u and m in C (T d ), and let m > 0. Then, there exists a constant, C > 0, such that Hence, m, 1 m L. ln m W 1,q (T d ) C.

26 Bootstrapping regularity Standard elliptic regularity theory applied to Hamilton-Jacobi equation yields u W 2,q (T d ) C q, for every q > 1. Therefore, Morrey s Embedding Theorem implies that u C 1,β (T d ), for some β (0, 1).

27 Bootstrapping regularity Next, set w = 2 ln m. Straightforward computations show that w satisfies w Dw 2 Du Dw + 2 div(du) = 0. Integrating, we conclude that Dw L 2.

28 Bootstrapping regularity The Bernstein estimate gives Dw L 2d(p+1) d 2 (T d ) C p C + Du Dw L 2d(1+p) + 2d(1+p) d+2p (T d ) div(du) L d+2p (T d ). Since 2d(p+1) d 2 > 2d(1+p) d+2p, we get Dw Lq for any q 1.

29 Bootstrapping regularity Hence, ln m is a Hölder continuous function. Because T d mdx = 1, m is bounded from above and from below. Consequently, ln m L q (T d ) is a priori bounded by some universal constant that depends only on q.

30 Bootstrapping regularity Proposition Let (u, m, H) solve (4) with u and m in C (T d ), and let m > 0. For any k 1 and q > 1, there exists a constant, C k,q > 0, such that D u k D, k m C k,q. L q (T d ) L q (T d )

31 Bootstrapping regularity The preceding results give u W 2,a (T d ) C a for every 1 < a <. Also, Proposition 7 gives ln m W 1,a (T d ) C for any 1 < a <. By differentiating the first equation in (4), we obtain D x u = D x g(m) D 2 udu. (5) Finally, we observe that the right-hand side of (5) is bounded in L a (T d ). Thus, u W 3,a (T d ) C 3,a, which leads to m W 2,a (T d ) C 2,a. The proof proceeds by iterating this procedure up to order k.

32 Continuation method stationary problems Here, we illustrate the continuation method by proving the existence of smooth solutions of u + Du V (x) = H + g(m), m div (Dum) = 0, T u = 0, d T m = 1. d (6)

33 Continuation method stationary problems First, for 0 λ 1, we consider the family of problems m λ div(du λ m λ ) = 0, u λ Du λ 2 2 λv + H λ + g(m λ ) = 0, T u d λ = 0, T m d λ = 1. (7)

34 Continuation method stationary problems Set { } H k (T d, R) = f H k (T d, R) : f dx = 0 T d and consider F k = H k (T d, R) H k (T d, R) R,, which is a Hilber space with norm w 2 F k = ψ 2 Ḣ k (T d,r) + f 2 H k (T d,r) + h 2 for w = (ψ, f, h) F k. H+(T k d, R), for k > d 2 is the set of (everywhere) positive functions in H k (T d, R). For any k > d 2, let F+ k = H k (T d, R) H+(T k d, R) R. A classical solution is a tuple, (u λ, m λ, H λ ) F+. k k 0

35 Continuation method stationary problems Theorem Assume that g, V C (T d ) with g (z) > 0 for z (0, + ) and that we have the a priori estimate for any solution of (7): H u λ W k,p (T d ) + m λ W k,p (T d ) C k,p. L (T d ) m λ Then, there exists a classical solution to (6).

36 Continuation method stationary problems For large enough k, define E : R F k + F k 2 by E(λ, u, m, H) = m div(dum) u Du 2 2 λv + H + g(m) T d m + 1 Our system equivalent to E(λ, v λ ) = 0, where v λ = (u λ, m λ, H λ )..

37 Continuation method stationary problems The partial derivative of E in the second variable at v λ = (u λ, m λ, H λ ), L λ = D 2 E(λ, v λ ): F k F k 2, is L λ (w)(x) = f (x) div(du λ f (x) + m λ Dψ) ψ(x) Du λ Dψ + g (m λ (x))f (x) + h T d f, where w = (ψ, f, h) F k. In principle, L λ is a linear map on F k for a large enough k. However, it is easy to see that it admits a unique extension to F k for any k > 1.

38 Continuation method stationary problems Let Λ := { λ 0 λ 1, (7) has a classical solution (u λ, m λ, H λ ) }. Note that 0 Λ as (u 0, m 0, H 0 ) (0, 1, g(1)) is a solution to (7) for λ = 0. Our goal is to prove Λ = [0, 1]. The a priori bounds in the statement mean that Λ is a closed set. To prove that Λ is open, we s apply the implicit function theorem.

39 Continuation method stationary problems Let F = F 1. For w 1, w 2 F with smooth components, set B λ [w 1, w 2 ] = w 2 L λ (w 1 ). T d For smooth w 1, w 2, B λ [w 1, w 2 ] = [m λ Dψ 1 Dψ 2 + f 1 Du λ Dψ 2 f 2 Du λ Dψ 1 T d + g (m λ )f 1 f 2 + Df 1 Dψ 2 Df 2 Dψ 1 + h 1 f 2 h 2 f 1 ]. This last expression defines a bilinear form B λ : F F R.

40 Continuation method stationary problems Claim B λ is bounded, i.e., B λ [w 1, w 2 ] C w 1 F w 2 F. To prove the claim, we use Holder s inequality on each summand.

41 Continuation method stationary problems Claim There exists a linear bounded mapping, A: F F, such that B λ [w 1, w 2 ] = (Aw 1, w 2 ) F. This claim follows from Claim 10 and the Riesz Representation Theorem.

42 Continuation method stationary problems Claim There exists a positive constant, c, such that Aw F c w F for all w F. If the previous claim were false, then there would exist a sequence, w n F, with w n F = 1 such that Aw n 0.

43 Continuation method stationary problems Let w n = (ψ n, f n, h n ). Then, T d m λ Dψ n 2 + g (m λ )f 2 n = B λ [w n, w n ] 0. (8) By combining the a priori estimates on 1 m λ with the fact that g is strictly increasing and smooth, we have g (m λ ) > δ > 0.

44 Continuation method stationary problems Then, (8) implies that ψ n 0 in H 1 0 and f n 0 in L 2. Taking ˇw n = (f n f n, 0, 0) F, we get T d [ Df n 2 + m λ Dψ n Df n + f n Du λ Df n ] = B[w n, ˇw n ] = (Aw n, ˇw n ), Therefore, 1 ) 2 Df n 2 L 2 (T d ) ( Dψ C n 2 L 2 (T d ) + f n 2 L 2 (T d ) (Aw n, ˇw n ) 0, where the constant, C, depends only on u λ. Because Dψ n, f n 0 in L 2, we have f n 0 in H 1 (T d ).

45 Continuation method stationary problems Finally, we take w = (0, 1, 0). Accordingly, we get T d [ Du λ Dψ n + g (m λ )f n ] + h n = B[w n, w] = (Aw n, w) 0. Because Dψ n, f n 0 in L 2, we have h n 0. Hence, w n F 0, which contradicts w n F = 1.

46 Continuation method stationary problems Claim R(A) is closed in F. This claim follows from the preceding one.

47 Continuation method stationary problems Claim R(A) = F. By contradiction, suppose that R(A) F. Then, because R(A) is closed in F, there exists a vector, w 0, with w R(A). Let w = (ψ, f, h). Then, 0 = (Aw, w) = B λ [w, w] θ Dψ 2 + δ f 2. T d Therefore, ψ = 0 and f = 0. Next, we choose w = (0, 1, 0). Similarly, we have h = B λ [ w, w] = (A w, w) = 0. Thus, w = 0, and, consequently, R(A) = F.

48 Continuation method stationary problems I Claim For any w 0 F 0, there exists a unique w F such that B λ [w, w] = (w 0, w) F 0 for all w F. Consequently, w is the unique weak solution of the equation L λ (w) = w 0. Moreover, w F 2 and L λ (w) = w 0 in the sense of F 2.

49 Continuation method stationary problems Consider the functional w (w 0, w) F 0 on F. By the Riesz Representation Theorem, there exists ω F such that (w 0, w) F 0 = (ω, w) F. Taking w = A 1 ω, we get B[w, w] = (Aw, w) F = (ω, w) F = (w 0, w) F 0. Therefore, f is a weak solution to and ψ is a weak solution to f div(m λ Dψ + fdu λ ) = ψ 0 ψ Du λ Dψ + g (m λ )f + h = f 0.

50 Continuation method stationary problems Standard results from the regularity theory for elliptic equations combined with bootstrapping arguments give w = (ψ, f, h) F 2. Thus, L λ (w) = w 0. Consequently, L λ is a bijective operator from F 2 to F 0. Then, L λ is injective as an operator from F k to F k 2 for any k 2. To prove that it is also surjective, take any w 0 F k 2. Then, there exists w F 2 such that L λ (w) = w 0. Finally elliptic regularity and bootstrapping imply that w F k. Hence, L λ : F k F k 2 is surjective and, therefore, also bijective.

51 Continuation method stationary problems Claim L λ is an isomorphism from F k to F k 2 for any k 2. Because L λ : F k F k 2 is bijective, we just need to check that it is also bounded. The boundedness follows directly from bounds on u λ and m λ and the smoothness of V and g.

52 Continuation method stationary problems Claim The set Λ is open.

53 Continuation method stationary problems We choose k > d/2 + 1 so that H k 1 (T d, R) is an algebra. For each λ 0 Λ, the partial derivative, L = D 2 E(λ 0, v λ0 ): F k F k 2, is an isomorphism. By the Implicit Function Theorem, there exists a unique solution v λ F k + to E(λ, v λ ) = 0, in some neighborhood, U, of λ 0. Finally, because H k 1 (T d, R) is an algebra, bootstrapping yields that u λ and m λ are smooth. Therefore, v λ is a classical solution to (6). Hence, U Λ, which proves that Λ is open.

54 Continuation method stationary problems We have proven that Λ is both open and closed; hence, Λ = [0, 1]. This argument ends the proof of the theorem.