NONLOCAL p-laplace EQUATIONS DEPENDING ON THE L p NORM OF THE GRADIENT MICHEL CHIPOT AND TETIANA SAVITSKA

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1 NONLOCAL -LAPLACE EQUATIONS DEPENDING ON THE L NORM OF THE GRADIENT MICHEL CHIPOT AND TETIANA SAVITSKA Abstract. We are studying a class of nonlinear nonlocal diffusion roblems associated with a -Lalace-tye oerator, where a nonlocal quantity is resent in the diffusion coefficient. We address the issues of existence and uniqueness for the arabolic setting. Then we study the asymtotic behaviour of the solution for large time. For this urose we introduce and investigate in details the associated stationary roblem. Moreover, since the solutions of the stationary roblem are also critical oints of some energy functional, we make a classification of its critical oints. 1. Introduction We consider the roblem of finding u = ux, t) weak solution to u t a u ) u u = f in, T ), 1.1) u = on Γ, T ), u, ) = u in, where is a bounded oen set of R n, n 1 with Lischitz boundary Γ. We assume 1.) a is continuous, aξ) >, ξ R. By we denote the L )-norm, 1 < < + and we assume 1.3) f = fx) W 1,q ) := ) W 1,, u W 1, ) L ), q = 1. For notions on Sobolev saces we refer to [5], [14], [15]. During the last decades many mathematicians have been studying roblems associated with the -Lalace oerator, which aears in a variety of hysical fields see for instance [1], []). In articular a lot of attention has been devoted to nonlocal roblems. One of the justification of such models lies in the fact that in reality the measurements are not made ointwise but through some local average. Some interesting features of nonlocal roblems and more motivation are described in [4], [5], [7], [8], [9] and in the references therein. The ellitic roblems with our tye of nonlocality have been studied in [1], [13] and the stability issues for a local case were considered in [3]. Furthermore, the roblem 1.1) was examined for = in [1] and [11]. We now describe the results obtained in this aer. In sections and 3 we study the existence and uniqueness of a weak solution of roblem 1.1). Next in section 4 we investigate the corresonding stationary roblem and show that deending on 1

2 MICHEL CHIPOT AND TETIANA SAVITSKA the function a it can have from a unique u to a continuum of solutions. In articular, since stationary solutions are also critical oints of some energy functional see.8)) we rove the existence of its global minimizer. The main results of this aer are contained in section 5, where we give the classification of the critical oints of the energy functional define by.8), assuming that the function a satisfies just 1.) and 1 < < +. We also resent an algorithm for finding a global minimizer or global minimizers it may be not unique) of the energy.8). Finally in the last section we study the asymtotic behaviour of the solution of roblem 1.1) as time goes to infinity. We rove that the solution of roblem 1.1) converges to a stationary solution, which is a global minimizer of.8), in case of uniqueness of such a stationary oint. Moreover, we also resent some local stability result for the case of uniqueness of a global minimizer.. Existence Theorem.1. Let the assumtions above hold and assume that there exist two constants λ, Λ such that.1) < λ aξ) Λ, ξ R and that.) f L q ). Then, for any T > there exists u solution to u L, T ; W 1, )) C[, T ]; L )),.3) u t L q, T ; W 1,q )), u, ) = u, u t, v + a u ) u u vdx = f, v v W 1, ) in D, T ), where, denotes the airing between L q ) and L ), ut) = u, t), D, T ) is the sace of distributions on, T ). Proof. Consider λ 1,..., λ n,... a basis in W 1, ) L ) smooth and that without loss of generality, we will suose orthonormal in L ). If u = β i λ i consider i n u n t) = γ i t)λ i i=1

3 NONLOCAL -LAPLACE EQUATIONS 3 solution to.4) u nvdx + a u n ) u n u n vdx = f, v v [λ 1,..., λ n ], u n ) = n β i λ i, i=1 where [λ 1,..., λ n ] is the sace sanned by λ 1,..., λ n. Taking v = λ j and using the fact that the λ i s are orthonormal we see that.4) is equivalent to the Cauchy roblem γ j t) = a n ) n n γ i t) λ i γ i t) λ i γ i t) λ i λ j dx i=1 i=1 i=1.5) + f, λ j, j = 1,... n, γ j ) = β j, j = 1,... n. Since the right hand side of the first equation above is continuous in γ i this Cauchy roblem ossesses a solution. Moreover, using the formulation.4) and taking v = u n we see that u nu n dx + a u n ) u n dx = f, u n, which imlies using.1), Poincaré s and Young s inequalities 1 d dt u n + λ u n dx C f q u n ε u n + C ε f q q. Choosing for instance ε = λ we arrive to 1 d dt u n + λ u n dx C ε f q q. After an integration in t this leads to 1.6) u nt) + λ t t u n dxdt C ε f q qdt + 1 u n). In articular we see that u n t) remains bounded in time and thus the solution to.4) or.5) is global in time is just a norm in [λ 1,..., λ n ], where all the norms are equivalent). Remark that u n remains bounded in time uniformly. To see that taking v = u n in.4) we get.7) u n dx + a u n ) u n u n u ndx = f, u n. Introducing.8) Eu) = 1 A ) u dx f, u

4 4 MICHEL CHIPOT AND TETIANA SAVITSKA with.9) Az) = z we see that.7) can be written.1) t Eu n ) = as)ds, u n dx. Thus Eu n ) decreases in time and is bounded from above for every t. The bound for u n follows then from the estimate.11) Eu n ) λ u n C f q u n. From.6),.11) we deduce that u n is bounded in L, T ; W 1, )) L, T ; L ) W 1, )). Furthermore, from the first equation in.4) and Hölder s inequality we derive easily ) 1 ) 1 ) 1 u nvdx Λ u n 1)q q dx v dx + C f q v dx, i.e. and u n 1,q Λ u n 1 + C f q, u n is bounded in L q, T ; W 1,q )) L q, T ; W 1,q ) + L )) indeendently of n, W 1,q ) + L ) denotes a dual sace to W 1, ) L q ). We have that W 1, ) L ) W 1,q ) + L ), where the first embedding is comact see [14]). Hence by Aubin-Lions lemma the embedding of W := {v L, T ; W 1, )), v L q, T ; W 1,q ))} in L, T ; L )) is comact. Thus we can find a subsequence of n such that In fact, u n u in L, T ; W 1, )), u n u in L, T ; L )), 1 a u n ) a in L, T ) - weak*, u n T ) ut ) in L ), u n u n χ in L q, T ; W 1,q ))..1) u n L, T ; L )) = L Q T ), Q T =, T ). Indeed, integrating.1) from to T we derive T.13) u n dx = Eu n )) Eu n T )). Using the Young inequality in.11) we get Eu) λ u C f q q λ q q λ u = 1 q ) q C f q, λ 1

5 NONLOCAL -LAPLACE EQUATIONS 5 hence Eu n ) is bounded from below indeendently of n. Thus from.13) we obtain.1). The fact that u C[, T ], L )) follows by the standard argument see [15]). By rescaling the time in the following way, setting.14) αt) = t a u, s) )ds, we reduce solving the roblem 1.1) to solving the roblem see [1]): w t w f w = a w in, αt )), ).15) w = on Γ, αt )), where w, ) = u in, wx, αt)) = ux, t). Relacing in.14) u by u n, we can also write the first equation of.4) as.16) u nvdx + u n f, v u n vdx = a u n ). Now assing to the limit in.16) one has in the distributional sense in Q T.17) u t χ = a f therefore u t L q, T ; W 1,q ))). Taking v = u n in.16) we obtain 1 d dt u n + u n dx = f, u n a u n ) and by integration on, T ) we get.18) u n dxdt = Q T T f, u n a u n ) dt + u n) u nt ). Since u n u in L f Q T ), a u n ) a f in L q Q T ) and using the fact that u n T ) ut ) from.18) we get lim n.19) lim n Thus from the inequality Q T u n dxdt T a f, u dt + u ut ). Q T u n u n v v) u n v)dxdt we derive by taking the lim for any v L, T ; W 1, )).) T a f, u dt + u ut ) T + χ, v dt v v u v)dxdt. Q T

6 6 MICHEL CHIPOT AND TETIANA SAVITSKA By integrating.17) after having multilied by u we get and integrating over, T ) we obtain.1) T χ, u dt = 1 d dt u χ, u = a f, u T Thus combining.),.1) we have T a f, u dt + u ut ). χ + v v, u v dt v L, T ; W 1, )). Taking v = u δw, δ >, we see T χ + u δw) u δw), w dt w L, T ; W 1, )). Letting δ we get easily T χ + u u, w dt = w L, T ; W 1, )) and the equation.17) reads u t u u = a f. Going back to.19),.1) we derive lim n u n dxdt Q T u dxdt Q T and u n u in L Q T ) strongly. In other words T u n u) dxdt. U to a subsequence we have u n u) dxdt a.e. t, i.e. this imlies and then 1 a u n ) 1 a u ) u n u a.e. t, since the sequence is bounded this convergence take also lace in any L, T ) and a = roof. ) lim u n dxdt n Q T a.e. t 1 a u, which comletes the ) 3. Uniqueness For the reader convenience we start this section by formulating some auxiliary lemmas, used throughout the aer.

7 NONLOCAL -LAPLACE EQUATIONS 7 Lemma 3.1. see [6]) Let 1 < < +. There exist ositive constants c, C such that for every ξ, η R n where c N ξ, η) ξ ξ η η) ξ η) C N ξ, η), N ξ, η) = { ξ + η } ξ η, a dot denotes the Euclidean roduct in R n. Lemma 3.. Let a, b be non negative numbers. Then Proof. We can suose a > b. Then a b = 1 a b) a b a b {a + b} 1. d dt b + ta b) dt = 1 1 ta + 1 t)b 1 dt a b) 1 b + ta b) b + ta b) a b)dt b + ta b) 1 { a + b } 1 dt = a b {a + b} 1. Theorem 3.1. If in addition to the assumtions of Theorem.1 for some L it holds that 3.1) aξ) aξ ) L ξ ξ ξ, ξ R and then the solution to.3) is unique. 3.) f L ), Proof. Let u 1, u be two weak solutions to u L, T ; W 1, )), u t L q, T ; W 1,q )), u t u u = f a u ). By subtraction we obtain u 1 u ) t u ) 1 u 1 u ) 1 u = a u 1 ) 1 a u f. ) Multilying by u 1 u, integrating over and using.1), 3.1) we get 3.3) 1 d dt u 1 u + u1 u 1 u ) u u1 u )dx L u1 λ u fu 1 u )dx

8 8 MICHEL CHIPOT AND TETIANA SAVITSKA From Lemma 3. and the Hölder inequality we derive 3.4) u1 u = u 1 u )dx = u1 + u ) 1 u1 u ) dx u1 + u ) u 1 + u ) u 1 u dx 1 u 1 u ) dx u1 + u ) ) 1 dx u1 + u ) u1 u ) dx ) 1. From Lemma 3.1 we obtain u 1 u 1 u u ) u 1 u )dx c u1 + u ) u1 u ) dx. Combining 3.3) and the two inequalities above leads to 1 d dt u 1 u + c u1 + u ) u1 u ) dx L λ fu 1 u )dx u1 + u ) ) 1 dx u1 + u ) ) 1 u1 u ) dx c u1 + u ) u1 u ) dx + Ct) u 1 u dx. In the last inequality above we use Young s inequality. Note that C L 1, T )). Therefore, we have 1 d dt u 1 u Ct) u 1 u dx. The uniqueness follows then from Gronwall s inequality. Theorem 3.. Let the assumtions 1.), 1.3) hold and if in addition the function a is such that 3.5) s as )s 1 is nondeacreasing, then the solution to.3) is unique. Proof. Let u 1, u be two solutions to.3), then taking v = u 1 u and by subtraction one has 1 d 3.6) dt u 1 u + a u 1 ) u 1 u 1 a u ) u u ) u 1 u )dx =.

9 NONLOCAL -LAPLACE EQUATIONS 9 By exanding the integral term I one gets I = a u 1 ) u 1 a u 1 ) u 1 u 1 u + a u ) u a u ) u u u 1 )dx. Recall that a u i ), i = 1, are indeendent of x and can be ulled out of the integrals. Using Hölder s inequality we see u 1 u 1 u dx u u 1 1, u u u 1 dx u 1 u 1. Then using 3.5) it comes ) I a u 1 ) u 1 u 1 1 u ) + a u ) u u 1 u 1 ) = a u 1 ) u 1 1 a u ) u 1 u 1 u ). Hence, 3.6) imlies therefore the result follows. d dt u 1 u, Remark 3.1. Note that 3.5) holds in articular for a nondecreasing. 4. The stationary roblem In this section we consider the associated stationary roblem to the roblem 1.1), that is the following roblem a u ) u u = f in, 4.1) u = on Γ. We will assume here that f W 1,q ). In a weak form u is a weak solution to u W 1, ), 4.) a u ) u u vdx = f, v v W 1, ). Here and after, denotes the airing between W 1,q ) and W 1, ). In order to solve the stationary roblem we introduce ϕ the solution to 4.3) ϕ W 1, ), ϕ ϕ vdx = f, v v W 1, ). It is known that for f W 1,q ) 4.3) admits a unique solution [6].

10 1 MICHEL CHIPOT AND TETIANA SAVITSKA Theorem 4.1. Suose that 1.) holds, 1 < < +. Then for f W 1,q ), the maing u u is one-to-one maing from the set of solutions to 4.) onto the set of solutions in R of the equation 4.4) aµ) 1 µ = ϕ. Proof. Let u be a solution to the stationary roblem, then 4.5) a u ) u u vdx = f, v = ϕ ϕ vdx v W 1, ), which imlies 4.6) a u ) 1 1 u = ϕ, from where follows 4.7) a u ) 1 u = ϕ. Hence u is a solution to 4.4). Let now µ be a solution to 4.4), u denotes the solution to u W 1, ), 4.8) aµ) u u vdx = f, v v W 1, ), then aµ) 1 1 u = ϕ. Therefore, we get aµ) 1 u = ϕ = aµ) 1 µ u = µ and u is a solution to 4.). Now to show the injectivity we have u 1 = u a u 1 ) = a u ) u 1 = u, due to the uniqueness of the solution of 4.8). Remark 4.1. The stationary oints are determined by the solutions to 4.9) aµ) = ϕ 1 µ 1 1. Thus it can haen that there is one solution, several, infinitely many solutions or no solution just in case where a is not bounded away from ). It deends on the function a, see Figure 4.1. The solutions of the roblem 4.) can be also found as critical oints of the energy Eu), defined by 4.1) Eu) = 1 ) A u dx f, v, where 4.11) Az) = and z as)ds 4.1) E u) = a u ) u u f.

11 NONLOCAL -LAPLACE EQUATIONS 11 y y = ϕ 1 µ 1 1 y y = ϕ 1 µ 1 1 y = aµ) y = aµ) µ µ µ 1 µ µ 3 µ y a) Unique solution y = ϕ 1 µ 1 1 b) Several solutions y y = ϕ 1 µ 1 1 c) No solution y = aµ) µ y = aµ) d) Infinitely many solutions µ Figure 4.1 If u is a critical oint of E on W 1, ) then u is a solution to 4.). Indeed, if u is a critical oint then for arbitrary v W 1, ) it holds d dδ Eu + δv) δ= = a u + δv) ) Thus u + δv) u + δv) v f, v ) a u ) u u v f, v =, v W 1, ), namely u is a solution to 4.) and a stationary oint. δ= =. Theorem 4.. Let.1) holds, f W 1,q ), then Eu) admits a global minimizer on W 1, ). Proof. To rove this theorem we will use the direct method of calculus of variations. We claim that E is coercive and bounded from below. Indeed, Hölder s and Poincaré s inequalities imly therefore f, u f 1,q u, 4.13) Eu) = 1 A u ) f, v λ u f 1,q u.

12 1 MICHEL CHIPOT AND TETIANA SAVITSKA Since > 1 the coerciveness follows. Now coming back to 4.13) and using Young s inequality we obtain 4.14) Eu) λ u f 1,q) q λ λ q q u = 1 ) q f 1,q. q Thus E is also bounded from below. Let u n W 1, ) be a minimizing sequence of E. From 4.13) it follows that u n is bounded in W 1, ). Hence for some u W 1, ) we have u n u in W 1, ). Next we show that E is weakly lower semicontinuous on W 1, ). In fact, it holds that lim u n u n the norm is weakly lower semicontinuous). Considering a subsequence u nk such that lim u n = lim u n k n k and due to the fact that u nk is a minimizing sequence we see inf Eu) = lim Eu nk ) = 1 lim unk as)ds f, u W 1, ) k 1 u λ 1 as)ds f, u = Eu ), which imlies u is a minimizer of E on W 1, ). Therefore, the result follows. Note that the minimizer might be not unique. 5. Remarks on the stationary oints Suose first we are in case of Figure 5., then we have: Theorem 5.1. Let u 1 be the stationary oint corresonding to µ 1 such that 5.1) aµ) < ϕ 1 µ 1 1 µ µ, µ 1 ), 5.) aµ) > ϕ 1 µ 1 1 µ µ 1, µ). Then u 1 is a local minimizer for E. More recisely one has: Eu 1 ) < Eu) u u 1, u µ, µ). Proof. Recall that by Theorem 4.1 we have that 5.3) µ 1 = u 1, u 1 = ϕ. aµ 1 ) 1 1

13 NONLOCAL -LAPLACE EQUATIONS 13 y y = ϕ 1 µ 1 1 y = aµ) µ µ 1 µ µ Figure 5. i) Suose u > µ 1. Then from.8), 5.) we have 5.4) Eu) Eu 1 ) = 1 u u 1 > 1 ϕ 1 as)ds f, u + f, u 1 u u 1 s 1 1 ds f, u + f, u 1 = ϕ 1 u ϕ 1 u 1 f, u + f, u 1. From 4.3) and using Hölder s inequality we see 5.5) f, u = ϕ ϕ udx u dx ) 1 ) 1 ϕ q 1) q dx = u ϕ 1, where q = 1. Now by 4.4) and 5.3) we obtain 5.6) f, u 1 = ϕ ϕ u 1 dx = ϕ ϕ ϕ dx = ϕ u 1 aµ 1 ) 1 = ϕ 1 u 1. 1 ϕ Hence, combining 5.4) 5.6) we derive Eu) > Eu 1 ) for u > µ 1. ii) Suose now u < µ 1. Then as above we get Eu) Eu 1 ) = 1 u1 u as)ds f, u + f, u 1,

14 14 MICHEL CHIPOT AND TETIANA SAVITSKA and by 5.1), 5.5), 5.6) we can conclude 5.7) Eu) Eu 1 ) > 1 ϕ 1 u1 u s 1 1 ds f, u + f, u 1 = ϕ 1 u 1 + ϕ 1 u f, u + f, u 1. Thus we have Remark 5.1. If one assumes Eu) > Eu 1 ) for u µ, µ), u u 1. aµ) ϕ 1 µ 1 1 µ µ 1, aµ) ϕ 1 µ 1 1 µ µ 1. Then one gets only Eu) Eu 1 ). Thus E can osses infinitely many global minimizers see Figure 4.1d). Lemma 5.1. Let u be the stationary oint corresonding to µ such that 5.8) aµ) > ϕ 1 µ 1 1 µ µ, µ ), 5.9) aµ) < ϕ 1 µ 1 1 µ µ, µ) see Figure 5.3). Then u is a oint of local maximum for E in the direction of ϕ, where ϕ is the solution of the roblem 4.3). More recisely one has: for every δ such that δ Eu ) > Eu + δϕ), 1 aµ ) 1 1, u + δϕ) µ, µ). y y = ϕ 1 µ 1 1 y = aµ) µ µ µ µ Figure 5.3

15 NONLOCAL -LAPLACE EQUATIONS 15 Proof. As above by Theorem 4.1 we have that 5.1) µ = u, u = ϕ aµ ) 1 1 i) Let us first assume that u + δϕ) > µ. Then from.8), 4.3), 5.9) we have 5.11) Eu + δϕ) Eu ) = 1 if 1 aµ ) 1 1 < 1 ϕ 1 u+δϕ) u u+δϕ) u as)ds δ f, ϕ s 1 1 ds δ ϕ = ϕ 1 ) u + δϕ) u δ ϕ ) = ϕ 1 1 aµ ) δ ϕ + δ. Thus it holds that. ϕ aµ ) 1 1 Eu + δϕ) < Eu ) for u + δϕ) > µ. δ ϕ =, ii) Suose now u + δϕ) < µ. Then similarly, from.8), 4.3), 5.8) we get 5.1) Eu + δϕ) Eu ) = 1 as in art i). Hence, < ϕ 1 u u +δϕ) as)ds δ f, ϕ u u + δϕ) ) δ ϕ = Eu + δϕ) < Eu ) for u + δϕ) < µ. Lemma 5.. Let u be a solution to the roblem 4.). Suose that 1.) holds and that ψ W 1, ), ψ is such that 5.13) f, ψ =. Then 5.14) Eu + ψ) > Eu), i.e. u is a oint of minimum for E in any direction direction of the hyerlane defined by 5.13). Proof. Let us consider ψ which satisfies 5.13). Then for u + ψ) > u from.1) we have Eu + ψ) Eu) = 1 u+ψ) u as)ds >.

16 16 MICHEL CHIPOT AND TETIANA SAVITSKA Hence it remains to rove that u + ψ) > u. Due to 5.13) and since a > we get u u ψdx =. Then we see u + ψ) u = = = d u + sψ) dxds ds u + sψ) u + sψ) ψdxds From Lemma 3.1 we have u + sψ) u + sψ) u u ) sψ) u + sψ) u + sψ) u u ) ψdxds. c u + sψ) + u ) sψ). This shows that u + ψ) u. If the equality holds then u + sψ) + u ) ψ = a.e. x, s, 1). This imlies that for u = we have ψ = and for u = as well. Thus ψ =, which contradicts our assumtions. This comletes the roof of the theorem. Theorem 5.. Let f, 1.) holds, u be a solution to 4.) such that 5.8), 5.9) hold see Figure 5.3, u corresonds to µ ). Then u is a saddle oint for the energy.8). Proof. The statement of the theorem is a consequence of Lemmas 5.1 and 5.. Theorem 5.3. Let u 1 and u be two solutions of the roblem 4.) corresonding to the solutions µ 1 < µ of the equation 4.4) resectively. Let 5.15) ys) = ϕ 1 s 1 1, then one has Proof. From.8) one has Eu 1 ) Eu ) = 1 Eu 1 ) Eu ) = 1 u1 u µ µ 1 as) ys))ds. Due to the definition of u i, i = 1, see 5.3)) we get 1 µi ys)ds = 1 ϕ 1 ui see 5.6)). Hence, the result follows. s 1 1 ds as)ds f, u 1 + f, u. = ϕ 1 u i = f, u i, i = 1,.

17 NONLOCAL -LAPLACE EQUATIONS 17 Corollary 5.1. Let u 1 and u be two solutions of the roblem 4.) corresonding to the solutions µ 1 < µ of the equation 4.4). If we assume that 5.16) aµ) > yµ) for µ 1 < µ < µ 5.17) res. aµ) < yµ), aµ) = yµ)), then Eu 1 ) < Eu ) res. Eu 1 ) > Eu ), Eu 1 ) = Eu )). Corollary 5.. Let u i, i = 1,, 3 be solutions of the roblem 4.) corresonding to solutions µ i, i = 1,, 3 of the equation 4.4) such that 5.18) aµ) > yµ) for µ 1 < µ < µ ; 5.19) aµ) < yµ), µ < µ < µ 3. Then we can conclude the following where 5.) A ij := see Figure 5.4). A 1 > A 3 Eu 3 ) > Eu 1 ); A 1 < A 3 Eu 1 ) > Eu 3 ); A 1 = A 3 Eu 1 ) = Eu 3 ), µj µ i as) ys))ds, i = 1,, j =, 3. y y = ϕ 1 µ 1 1 A 1 A 3 µ 1 µ µ 3 y = aµ) µ Figure 5.4. Several solutions

18 18 MICHEL CHIPOT AND TETIANA SAVITSKA Corollary 5.3. The absolute minimum of E corresonds to a oint µ such that 5.1) 5.) µ µ as) ys))ds, µ > µ, µ is a stationary oint, µ µ as) ys))ds, µ < µ, µ is a stationary oint. Now we give the algorithm of finding solutions of 4.), which are global minimizers for Eu) defined by.8). First we introduce some notation. We label the solutions to 4.4) as µ 1 < µ <... < µ N with the convention that we choose only one oint µ i in the interval µ, µ) i when the solutions consist of one interval µ, µ) i see Figure 5.5). We denote by {u} 1, {u},... {u} N the sets of solutions of 4.), corresonding to µ 1 < µ <... < µ N solutions of 4.4). Then due to our convention we see that {u} i can consist of one oint or infinitely many oints. y y = ϕ 1 µ 1 1 Stationary oints y = aµ) µ 1 µ µ i µ i µ i µ j µ Figure 5.5. Infinitely many solutions Corollary 5.4. Let µ i, i I := {1,..., N} be solutions of 4.4) with the convention above. Suose that k, l I are such that 5.3) A kl = max A µj ij = max i,j I i,j I as) ys))ds, µ i where 5.4) A ij := µj µ i as) ys))ds, i j, i, j I. Then we can conclude as follows: A kl > the set {u} k consists of global minimizers for Eu), A kl < the set {u} l consists of global minimizers for Eu), A kl = the sets {u} i, i I consist of global minimizers for Eu).

19 NONLOCAL -LAPLACE EQUATIONS 19 Proof. By Corollary 5.1 for arbitrary u {u} i, i I it holds that Eu) = E i, i I. Since the set I is finite there exist min E i, which we denote by E min. Then from i I Theorem 5.3 and from 5.3) for arbitrary i I we have E i E min max E i E j = 1 i,j I max A ij = 1 i,j I A kl = 1 = A kl, if A kl E l E k, if A kl > 1 = E k E l, if A kl < A kl, if A kl <, if A kl =. Let us first assume that A kl >. Then since i is arbitrary it holds also for i = l and we have E l E min E l E k, i.e. E k E min, therefore E k = E min. The result for A kl < will follow by taking i = k. If now A kl =, then for all i I we get E min = E i, which comletes the roof of the Corollary. Remark 5.. Corollary 5.4 is valid also in case when I = + if there exists max A ij. Note that in case of the stationary roblem 4.) having infinitely many i,j I solutions, the energy.8) can have a unique, several or infinitely many global minimizers. We start with a lemma: 6. Asymtotic behaviour Lemma 6.1. Let u be a weak solution to 1.1) and suose that.1) holds. There exists a sequence t k such that where u is a stationary oint. u k = u, t k ) u in W 1, ) as t k +, Proof. Taking v = u t in.3) we obtain a u ) u u u t dx f, u t = t u, Then from 4.13), 4.14) follows that t Eu) = t u Eut)). 6.1) Eut)) E, From above we get u is uniformly bounded in t. Eut)) Eus)) = s t t u ξ)dξ < +, s t u ξ)dξ,

20 which imlies for a sequence t k t u, t k ) in L ). MICHEL CHIPOT AND TETIANA SAVITSKA From the equation in.3) with u k = u, t k ) we obtain 6.) t u, t k )u k dx + a u k ) u k dx = f, u k. U to a subsequence it holds that Passing to the limit in 6.) we see u, t k ) = u k u in W 1, ), u k u in L ), u k l, g k = u k u k g in L q )) n. 6.3) al )l = f, u. Next taking v W 1, ) L ) we get t u, t k )vdx + a u k ) Passing to the limit we derive 6.4) al ) u k u k vdx = f, v. g vdx = f, v. Since a > combining 6.3), 6.4) we can conclude that l = g u dx. We claim that u k u strongly in W 1, ). Indeed, for there exists a constant C > such that χ k = uk u k u ) u uk u )dx C u k u ) dx. Develoing χ k = u k dx This imlies g k u dx l g u dx u u u k dx + u dx u dx + u dx =. l = lim k u k = u, g = u u. Hence u is a stationary oint. To show that u k u in W 1, ) strongly in case 1 < < it is enough to notice that by Lemma 3.1 one has c u k u ) u k + u ) dx χk.

21 NONLOCAL -LAPLACE EQUATIONS 1 Writing u k u ) dx = u k u ) u k + u ) ) uk + u ) ) dx and using Hölder s inequality with, it comes u k u ) dx u k u ) u k + u ) ) dx This comletes the roof of the lemma. uk + u ) ) dx Cχ k. Corollary 6.1. Suose that E admits a unique global minimizer u u is also a solution to the roblem 4.)) and that the initial value u of.3) satisfies Eu ) < Eu i ) for any stationary oint u i u. Then Proof. Recall that we have Then by Lemma 6.1 and 6.1) we get u, t) u in W 1, ) as t +. Eu) Eu ) < Eu i ), u i u. Eut)) Eu ), where u is the global minimizer of E and a solution of the roblem 4.). Due to the fact that ut) is uniformly bounded in W 1, ) for some subsequence we have u, t k ) v in W 1, ). Then by the weak lower semicontinuity of E see the roof of Theorem 4.) we obtain Eu ) = lim t k Eut k)) Ev ). Since u is a unique global minimizer of E, then it holds that Eu ) < Ev ) for u v, hence u = v. This holds for every subsequence and the convergence is in fact strong see Lemma 6.1), therefore the result follows. Acknowledgements: The research leading to these results has received funding from Lithuanian-Swiss cooeration rogramme to reduce economic and social disarities within the enlarged Euroean Union under roject agreement No CH- 3-SMM-1/. The research of the first author was suorted also by the Swiss National Science Foundation under the contracts # /1 and The research of the second author was suorted by the Euroean Initial Training Network FIRST under the grant agreement n. PITN-GA Finally the aer was comleted during a visit of the first author at the Newton Institute in Cambridge. The nice atmoshere of the institute is greatly acknowledged.

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