Traveling Wave Solutions To The Allen-Cahn Equations With Fractional Laplacians

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1 University of Connecticut Doctoral Dissertations University of Connecticut Graduate School Traveling Wave Solutions To The Allen-Cahn Equations With Fractional Laplacians Mingfeng Zhao Follow this and additional works at: ecommended Citation Zhao, Mingfeng, "Traveling Wave Solutions To The Allen-Cahn Equations With Fractional Laplacians" (14). Doctoral Dissertations. 4.

2 Traveling Wave Solutions to the Allen-Cahn Equations with Fractional Laplacians Mingfeng Zhao, Ph.D. University of Connecticut, 14 ABSTACT For any fixed s (, 1), we consider the following problem: ( ) s u(x) µu (x) = f(u(x)), x, u(x) 1, x, x ± u(x) = ±1. We proved that for any bistable nonlinearity f, then there exist a unique speed µ and unique function u up to translation such that (µ, u) is the solution to the previous problem. We use a continuation argument to show the existence of solution, in which a key ingredient is the estimation of the speed µ in terms of the potential function f. In the meantime, we prove some qualitative properties of the solution u: monotonicity, polynomial decays at infinity, Hamiltonian identity and Modica type estimate, and nondegeneracy. Moreover, we proved that for any balanced bistable nonlinearity f and any nonlinearity g C (), then for small ɛ >, the traveling speed µ ɛ corresponding to f + ɛg linearly depends on the parameter ɛ.

3 Traveling Wave Solutions to the Allen-Cahn Equations with Fractional Laplacians Mingfeng Zhao M.S. Mathematics, University of Connecticut, Storrs, USA, 1 B.S. Mathematics, Hunan University, Changsha, China, 9 A Dissertation Submitted in Partial Fulfillment of the equirements for the Degree of Doctor of Philosophy at the University of Connecticut 14

4 Copyright by Mingfeng Zhao 14

5 APPOVAL PAGE Doctor of Philosophy Dissertation Traveling Wave Solutions to the Allen-Cahn Equations with Fractional Laplacians Presented by Mingfeng Zhao, B.S. Math., M.S. Math. Major Advisor Changfeng Gui Associate Advisor Yung-Sze Choi Associate Advisor Xiaodong Yan University of Connecticut 14 ii

6 ACKNOWLEDGMENTS I would like to thank my supervisor, Dr. Changfeng Gui, for his patient guidance, encouragement and strong inspiration through all these years. Every time when I got frustrated with some problems, he could always give me some suggestions which help me out. This dissertation can not be finished without his supervision. Special thanks to my committee, Dr. Sze-Yung Choi and Dr. Xiaodong Yan for their direction, dedication, and invaluable advice. They provided me a lot of help with my research and graduation. I would also like to express my deep appreciation to all of my course teachers at UConn: Dr. Patrick McKenna, Dr. Damin Wu, Dr. Álvaro Lozano-obledo, Dr. Lan-Hsuan Huang, Dr. Keith Conrad, Dr. Sarah Glaz, Dr. Alexander Teplyaev, Dr. Maria Gordina, Dr. Luke ogers, Dr. Iddo Ben-Ari, Dr. Andrew Haas, Dr. alf Schiffler, Dr. ichard Bass, Dr. Jeffrey L. Tollefson, Dr. Erin Terwilleger and others. I thank all the wonderful staff in the Department of Mathematics for always being so helpful and friendly. People here are genuinely nice and are ready to help you out any time. I wish to thank my parents and two elder sisters. They were always supporting me and encouraging me with their best wishes through the past 7 years. Last but not the least, I would like to thank my friends for providing support and friendship that I needed. iii

7 Contents Ch. 1. Introduction Traveling Wave with Standard Laplacian Traveling Waves with Fractional Laplacians Ch.. Qualitative Properties of Traveling Wave Solutions 9.1 Monotonicity of Profile, and Uniqueness of Speed and Profile Polynomial Decays at Infinity Hamiltonian Identity and Modica Type Estimate Nondegeneracy Ch. 3. Estimates of Traveling Speeds Laplacian Case Fractional Laplacian Case Ch. 4. Existence of Traveling Wave Solution Uniform Decays at Infinity, and Linear Dependence of Speed Existence of Traveling Wave Solution Bibliography 73 iv

8 Chapter 1 Introduction 1.1 Traveling Wave with Standard Laplacian Front propagation is a natural phenomenon which has appeared in phase transition, chemical reaction, combustion, biological spreading, etc. The mechanism of front propagation is often the competing effects of diffusion and reaction. Traveling wave solutions are typical profiles of physical states near the propagating front, and are therefore of great importance in the study of reaction diffusion processes. There has been a tremendous amount of literature on traveling wave solutions in mathematics as well as in various branches of applied sciences ( see [1, 3, 7, 8, 6, 1, 8, 9, 35, 6] and references therein). Traveling wave solutions are essential building blocks in various phase field models and play an important role in pattern formation and phase separation (see [4,, 7] etc. for the classical model and [44, 55, 56] for nonlocal models with fractional Laplacians). Nonlocal phase transition models and related traveling wave solutions have been studied in [5, 3, 63] and references therein, where 1

9 the kernels of convolution in the nonlocal operators are bounded, and in [31, 3, 47] where the kernels are periodic. In the study of front propagation, traditionally the diffusion process is quite standard and normal, in the sense that the concerned particles or objects are engaged in a Brownian motion with a uniformly changed random variable. The resulting diffusion effect on the physical state, mathematically is represented by the Laplacian of this function. Therefore, the difference of various reaction diffusion systems relies on the nonlinear reaction effect which varies in combustion, chemical reaction, phase transition, biological pattern formation, etc. In general, a typical reaction diffusion system is in the form of u t (t, x) x u(t, x) = f(u(t, x)), t >, x n, (1.1.1) where f is a nonlinear function. If the front of a solution u in large time propagates at a constant speed, the solution is typically close to a profile depending on the distance away from the traveling front. We, therefore, study traveling wave solutions of one spatial variable, although more complicated traveling waves solutions do exist (see, e.g., [7, 8, 37, 39, 4, 41, 4, 43, 5, 53, 59, 6, 65, 64] and references therein). We say that u C ( ), is a traveling wave solution to (1.1.1) if u has the special form u(t, x) = g(x µt) for all (x, t). The constant µ is called the speed of the traveling wave u, and the function g is called the profile of the traveling wave. We are interested in traveling front where g connects the two states, say 1 and 1. The sliding method implies that such g strictly increases. The study of traveling wave

10 3 solution is reduced to the study of solution to the following system: g (x) µg (x) = f(g(x)), x, g (x) >, x, x ± g(x) = ±1. (1.1.) For the nonlinear function f, there are three interesting cases: Fisher-KPP or Monostable Model: f(t) > = f(±1), t ( 1, 1), f ( 1) >, f (1) <. Combustion Model: There exists some t ( 1, 1) such that f(t) =, t [ 1, t ], f(t) > = f(1), t (t, 1), f (1) <. Bistable or the Allen-Cahn Model: There exists some t ( 1, 1) such that f(t) < = f(±1) = f(t ), t ( 1, t ), f(t) >, t (t, 1), f (±1) <. (1.1.3) For the combustion and Fisher-KPP models, and 1 are usually used as the its with only 1 being stable (monostable model). The concerned states in the Allen-Cahn model are often represented by 1 and 1 with both states being stable

11 4 (bistable model). In the Allen-Cahn model, there is also another nodal point t of f in ( 1, 1); this nodal point may represents an unstable state which is not the concerned state, since otherwise the equation may be regarded as a Fisher-KPP equation by restricting u in (t, 1). Using phase plane analysis, one can show that the traveling wave solutions (g, µ) always exist for the above three types of non-linearities. More precisely, for the combustion and bistable models, there exists a unique pair (g, µ) as the traveling wave solution to (1.1.). For the Fisher-KPP model, there exists a maximal speed µ such that for any speed µ µ, there exists a traveling wave solution (g, µ) to (1.1.). Moreover, for the bistable non-linearity, the traveling wave solution to (1.1.) decays exponentially at infinity, i.e., g (x) e ν x as x, as a consequence, we get 1 g(x) e ν x as x and 1 + g(x) e ν x as x for some positive constant ν >. In addition to the method of phase analysis, one can also use the sub-super solution method and variational method to prove the same results (see [61, 66] and references therein). ecently, there have been a fast increasing number of studies on front propagation of reaction diffusion systems with an anomalous diffusion such as super diffusion, which plays important roles in various physical, chemical, biological and geological processes. (See, e.g., [61] for a brief summary and references therein.) Mathematically, such a super diffusion is related to Lévy process and may be modeled by a fractional Laplace operator ( ) s u with < s < 1, the definition of fractional Laplacians will be given in Section 1..

12 5 1. Traveling Waves with Fractional Laplacians The fractional Laplacian is often defined by Fourier transformation, for any < s < 1 and u C c ( n ), the fractional Laplacian ( ) s u is defined as the inverse Fourier transform of (π x ) s û(x), i.e., [48]) that equivalently we have ( )s u(x) = (π x ) s û(x). It is well known (see ( ) s u(x) = C n,s P.V. u(x) u(y) dz, x n, (1..1) x y n n+s where C n,s = ss Γ( n+s ) π n Γ(1 s). The integral definition (1..1) of fractional Laplacian can be used for more general functions, in particular, for u C ( n ). The fractional Laplacian can also be defined as a Dirichlet-to-Neumann map. First, the n-dimensional fractional Poisson kernel P n,s is defined as P n,s (x, y) = Γ ( ) n+s π n Γ(s) y s [ x + y ] n+s, (x, y) n+1 +. For any u C ( n ) L ( n ), let u(x, y) = P n,s (, y) u(x) for all (x, y) n+1 + be the s-harmonic extension of u. A direct computation shows that u satisfies div [ y 1 s u(x, y) ] =, (x, y) n+1 +, y u(x, y) = u(y), x n, y y1 s u y (x, y) = d s ( ) s u(x), x n, (1..) where d s = 1 s Γ(1 s). This fact is well-known for s = 1 and recently proved for all s Γ(s) (, 1) by Caffarelli and Silvestre in [1]. The s-harmonic extension method allows us to use local techniques for the local differential operator L s to study nonlocal integro-

13 6 differential operator ( ) s, which will be used in Section.3 to get a Hamiltonian identity for traveling wave solution. There have been a fast growing number of studies on the front propagation of the reaction diffusion system with an anomalous diffusion such as super diffusion, which plays an important role in various physical, chemical, biological and geological processes. Mathematically, such a super diffusion is related to the Lévy process and may be modeled by a fractional Laplacian operator ( ) s u with < s < 1, and the reaction diffusion equation with respect to such super diffusion can be given by u t + ( ) s u = f(u). We are interested in traveling wave solutions in one dimensional spatial variable. Namely, we consider solutions u(x, t) in the form of u(x µt) for some constant µ, equivalently, we will study the following problem: ( ) s u(x) µu (x) = f(u(x)), x, u (x) >, x, x ± u(x) = ±1. (1..3) Traveling wave solutions for reaction diffusion equations with the fractional Laplacians have been studied in the last few years (see [15,, 5, 51, 61]). It is very interesting to see that the study of (1..3) really depends on the nonlinearity f and parameter s. When f is a Fisher-KPP non-linearity, it is known that the front propagation speed could be very fast depending on the initial values (see [9]). The Fisher-KPP equation with a fractional Laplacian displays a very different behavior, due to the super diffusion process involved. It was discovered numerically in [5, 4, 49] that the front propagation can accelerate exponentially in time. This phenomenon is

14 7 rigorously studied and proved in [16] that for any s (, 1), the position of all level sets of solution u to reaction diffusion equation u t + ( ) s u = f(u) with u 1 moves exponentially fast in time t. Since a traveling wave front propagates linearly in t, it is an immediate consequence that there is no traveling wave solution for the Fisher-KPP equation with a fractional Laplacian. For the combustion model, the results are much more interesting. Via the subsuper solution method, it is shown in [5] that when f is a combustive non-linearity, if s (1/, 1), there exists a unique pair (µ, u) as solution to (1..3), and the solution u decays algebraically at. On the other hand, by the s-harmonic extension, it is shown in [38] that if s (, 1/], there does not exist a traveling wave solution to (1..3) for the combustion model. Moreover, they showed the nonexistence of traveling fonts for more general non-linearities including the Fisher-KPP case. For the bistable model, there are two types of bistable non-linearities: balanced and unbalanced. When the bistable non-linearity is balanced, i.e., the associated double well potential G(u) = u f(t) dt has two wells with equal depths G(1) = 1 G( 1) =, a traveling wave solution with one spatial variable for the balanced Allen-Cahn equation is indeed a standing wave, i.e., the speed µ must be zero. Such a solution is sometimes called a layer solution as it describes a transition layer structure near the interface between two physical states. The existence of the standing wave solution to (1..3) was proved in [17, 19] as a stationary point of the functional E s (u) = [ 1 ( ) s/ u ] + G(u) dx, Indeed, by using the s-harmonic extension and variational method, they proved the following result:

15 8 Theorem 1..1 (Cabré and Solà-Morales[19], Cabré and Sire [17]). For any < s < 1 and balanced bistable nonlinearity f C (), i.e., f satisfies (1.1.3) and 1 1 f(t) dt =, then there exists a unique function u C () up to translation as the solution to (1..3) with µ =. Moreover, u (x) > for all x and there exists some constant C > which only depends on s and f such that C 1 x 1+s u (x) C, x > 1. x 1+s It is shown that a multidimensional standing wave solution must be one dimensional (without counting the dimension of extension) for certain low dimensions and s (see also [13, 14]). These results are analogues of a well studied phenomenon for the classical Allen-Cahn equation and usually referred to as the De Giorgi conjecture (see [, 6, 34, 33, 36, 54]). For unbalanced bistable nonlinearity, it s natural to ask whether or not there is a traveling wave solution to the unbalanced fractional Allen-Cahn equation where G( 1) G(1). In [61], when f is a piecewise linear bistable nonlinearity, exact traveling wave solutions for fractional Allen-Cahn equations are computed for s [ 1, 1). Our main result, Theorem 4..1 in Chapter 4, proves the same statement as Theorem 1..1 for all Allen-Cahn equations with a fractional Laplacian, i.e., we can remove the restriction that 1 f(t) dt =. So the existence and nonexistence problems re- 1 lated to the traveling wave solution with a fractional Laplacian are completely solved. Several traveling waves problems similar to the fractional diffusion reaction models are investigated recently (see [15, ]).

16 Chapter Qualitative Properties of Traveling Wave Solutions In this chapter, we will always assume that < s < 1, f C () satisfies f (±1) <, and G(t) = t f(u) du. Let s consider the following problem: 1 ( ) s u(x) µu (x) = f(u(x)), x, u(x) 1, x, x ± u(x) = ±1. (..1) In the following sections, we will prove some useful qualitative properties of traveling wave solution to the Fractional Allen-Cahn equation (..1): monotonicity of profile u, uniqueness of speed µ and profile u, polynomial decays at infinity of u, Hamiltonian identity and Modica type estimate, and nondegeneracy. These properties will play important role in the proof of the existence of traveling wave solution, see more details in Chapter 4. Many of these properties are similar to those in [17, 19], 9

17 1 where the standing wave solutions with µ = are considered..1 Monotonicity of Profile, and Uniqueness of Speed and Profile In order to study traveling wave solutions, first we state a slight variation of a maximum principle in [18] to include an advection term. Lemma.1.1. Let c C(), d L () and v C () satisfy ( ) s v(x) + c(x)v (x) + d(x)v(x), x, x v(x) =. Assume that there exists some subset H (may be empty set) such that v(x), x H and d(x), x / H. Then either v(x) > in or v(x) in. Proof. We consider the following two cases: Case I: v(x) in. If v(x) in, we are done. If v(x) > in, we are done. If v(x) in and there exists some x such that v(x ) =, that is, x is a global minimum point of v, we get v (x ) = and ( ) s v(x ) <, which implies that ( ) s v(x ) + c(x )v (x ) + d(x )v(x ) = ( ) s v(x ) <. We get a contradiction. Case III: v(x 1 ) < for some x 1. Since x v(x) =, we know that v can not be a constant function in and there exists some x such that

18 11 v(x ) = inf x v(x) <. Hence we get v (x ) = and ( ) s v(x ) <. Since v(x ) <, we know x / H and d(x ), which implies that ( ) s v(x ) + c(x )v (x ) + d(x )v(x ) = ( ) s v(x ) + d(x )v(x ) <, which contradicts with the assumption. In summary, we know that either v(x) > in or v(x) in. According to Lemma.1.1, we can use the sliding method which is introduced by Berestycki and Nirenberg in [1] to prove the monotonicity of profile, and the uniqueness of speed and profile. Proposition.1. (Monotonicity). Let µ and u C () be a solution to (..1) with µ. Then 1 < u(x) < 1 and u (x) > for all x. Proof. First, we prove that 1 < u(x) < 1 in. Otherwise, since u(x) 1 in, we know that there exists some x 1 such that u(x 1 ) = 1 or 1. If u(x 1 ) = 1 = sup u(x), since u(x) = ±1, we have u(x) 1 in, which implies that x y ± u (x 1 ) = and ( ) s u(x 1 ) >. Hence we have ( ) s u(x 1 ) µu (x 1 ) f(u(x 1 ) = ( ) s u(x 1 ) >, contradiction. If u(x 1 ) = 1 = inf x u(x), since x ± u(x) = ±1, we have u(x) 1 in, which implies that u (x 1 ) = and ( ) s u(x 1 ) <. Hence we have ( ) s u(x 1 ) µu (x 1 ) f(u(x 1 ) = ( ) s u(x 1 ) <, contradiction. Hence we know that 1 < u(x) < 1, x. (.1.1) Now let s prove the monotonicity of u. For any t >, we define u t (x) = u(t + x),

19 1 w t (x) = u(t + x) u(x) for all x, and H t = {{x : w t (x) > }. It s easy to see that w t satisfies ( ) s w t (x) µ(w t ) (x) + d t (x)w t (x) =, x, w t (x), x, x w t (x) =. where f(ut (x)) f(u(x)), if w t (x), d t (x) = u t (x) u(x), if w t (x) =. By the Mean Value Theorem, we know that d t (x) f L ([ 1,1]) for all t and all x. Since f (±1) <, then there exists some < τ < 1 such that f (t) <, [ 1, τ] [τ, 1]. (.1.) Claim I: There exists some large T > such that for all t T, we have w t (x) > in. Since u(x) = ±1, there exists some large Y > such that 1 < u(x) < τ x ± for all x Y, and τ < u(x) < 1 for all x Y. By (.1.1), we have A := max x [ Y,Y ] u(x) < 1. Since u(x) = ±1, there exists some large Y 1 > such that u(x) > A for all x ± x Y 1. Now let T = Y 1 + Y >, then for all t T and fix, and all x [ Y, Y ], we have x+t Y +Y 1 +Y = Y 1, which implies that u t (x) = u(t+x) > A u(x),

20 13 in particular, we get [ Y, Y ] H t. On the other hand, for all x / H t, we have u t (x) u(x), and x / [ Y, Y ]. By the definition of Y, we know that 1 < u(x) < τ if x < Y, and τ < u(x) < 1 if x > Y. If x < Y, since u t (x) u(x), then 1 < u t (x) u(x) < τ. If x > Y, we have t + x > x > y, so τ < u t (x) = u(t + x) u(x) < 1. In both cases, we know that both u t (x) and u(x) lie in either [ 1, τ] or [τ, 1]. Since u t (x) u(x), by (.1.), we know that d t (x) for all x / H t. Since w t (x) > in [ Y, Y ], by Lemma.1.1, we know that w t (x) > for all x. Claim II: If for some fixed t >, w t (x) > in, then there exists some small ɛ t (, t) such that for all h < ɛ t, we have w t+h (x) > in. Since u t (x), u(x) ±1, as x ±, and < τ < 1, then there exists some Y > Y 1 such that for all x Y, we have u t (x) 1 + τ, and u(x) 1 + τ. In particular, we have K t := { x : u t (x) < 1 + τ } 1 + τ, or u(x) < [ Y, Y ]. Since x ± u(x) = ±1, then there exists x such that u(x ) =, in particular, x K t and K t. Since w t (x) > in, we have A := inf x K t w t (x) >.

21 14 Since u t C (), so u t are Lipschitz continuous on, in particular, u t is uniformly continuous on. Since < τ < 1, then there exists some small t > ɛ t > such that u t (y) u t (z) < min { A, 1 τ }, for all y z < ɛ t. For all h < ɛ t, we have w t+µ (x) = u t (x + µ) u t (x) < min { A, 1 τ }, x. For all x K t, by the definition of A, we get w t (x) A, which implies that w t+µ (x) = w t+µ (y) + w t (y) A + A = A >, which implies that K t H t+h. On the other hand, for any x / H t+h, we have x / K t, u(x) 1+τ and u t+h (x) u(x). If u(x) 1+τ, by (.1.1), we have 1 < ut+h (x) u(x) 1+τ < τ. If u(x) 1+τ, since wt (x) >, we have u t (x) > u(x) > 1+τ, which implies that u t+h (x) = u t+h (x) u t (x) + u t (x) > 1 τ + 1+τ = τ. In both cases, we know that both u t+h (x) and u(x) lie in either [ 1, τ] or [τ, 1]. Since u t+h (x) u(x), by (.1.), we know that d t+h (x) for all x / H t+h. Since w t+h (x) > in K t, by Lemma.1.1, we know that w t (x) > for all x. Claim III: For all t >, then w t (x) > in. Let S = { t > : w t (x) >, in }. By Claim I, [T, ) S, so S. By Claim II, S is open. If there exists

22 15 some sequence {t n } n=1 S such that t n t >, as n. For all n 1, since w tn (x) > in, by taking n, we get w t (x) in. If there exists some x such that w t (x ) =, by Lemma.1.1, we have w t (x) in, that is, u(t + x) u(x) in, which implies that u is a periodic function in, which contradicts with u(x) = ±1. So we must have x ± wt (x) > in, that is, t S. Hence, S is closed with respect to (, ). In summary, S is a nonempty open and closed subset of (, ). Since (, ) is connected, then S = (, ), that is, for all t >, then w t (x) > in. Claim IV: u (x) > in. By Claim III, we see that u is a strictly increasing function in, so u (x) in. If u (x) > in, we are done. If there exists some x such that u (x ) =, let g(x) = u (x) in, then ( ) s g(x) µg (x) = f (u(x))g(x) and g(x) for all x. Since u is strictly increasing, then g(x) in. Since g(x ) = u (x ) = inf g(x), we have x g (x ) = and ( ) s g(x ) <, which implies that ( ) s g(x ) µg (x ) f (u(x ))g(x ) = ( ) s g(x ) <, contradiction. Therefore, we must have u (x) > in. By using the same argument as Proposition.1. with small modifications, we can show the uniqueness of speed and profile. Since the proof is very similar, we omit it. Proposition.1.3 (Uniqueness). For i = 1,, let µ i and u i C () be a solution to (..1) with µ i, respectively. Then µ 1 = µ and there exists some a such that u 1 (x) u (x + a) in.

23 16. Polynomial Decays at Infinity Let s quote three functions in [17]: for any t > and all x, let p t (x) = 1 π x cos(xr)e trs dr, v t (x) = 1 + p t (r) dr, and ϕ t (x) = v t(x). Theorem 3.1 in [17] says that p t, v t C (), and there exists some f t C ([ 1, 1]) which is an odd function in [ 1, 1] and satisfies (1.1.3) and f t(±1) = 1, such that t v t is a layer solution in with nonlinearity f t, that is, v t is a solution to (..1) with f = f t and µ =. Moreover, we have x x 1+s v t(x) = 4tsΓ(s) sin(πs) >. (..1) π By (..1), the equation which ϕ t satisfies has ϕ t, let s estimate ϕ t = v t. Lemma..1. There exists some constant C > such that ϕ t(x) tc, x. x +s Proof. First, let s estimate ϕ 1(x). For all x \{}, we have ϕ 1(x) = p t(x) = π = sign x π = sign x π = sign x πx r sin(xr)e rs dr r sin( x r)e rs dr z z sin ze ( x ) s x z sin ze ( x ) z s dz. 1 x dz, Let z = x r

24 17 So we know that ϕ 1 is an odd function in \{}, and for any x >, using integration by parts, we have z sin ze ( z x) s = ze ( x) z s cos z = cos ze ( z x) s dz [ cos z e ( x) z s ze ( ( x) z s z ) ] s 1 1 s x x dz + s z s cos ze ( x) z s dz x s = e ( x) z s sin z sin ze ( ( x) z s z ) s 1 1 s { s dz + z s e ( x) z s sin z x x x s [ sin z sz s 1 e ( x) z s z s e ( ( x) z s z ) ] } s 1 1 s dz x x s + 4s = z s 1 sin ze ( x) z s dz + 4s z 4s 1 sin ze ( x) z s dz. x s x 4s dz So we get ϕ 4s(1 + s) 1(y) = π + 8s π 1 x +4s 1 x +s z s 1 sin ze ( z x) s dz z 4s 1 sin ze ( z x) s dz, x >. By taking κ =, 4 in Lemma 3.4 in [17], respectively, we can get y y z s 1 sin ze ( z y) s dz = Γ(s) sin(sπ), z 4s 1 sin ze ( z y) s dz = Γ(4s) sin(sπ). Hence, we have x x+s ϕ 4s(1 + s) 1(x) = Γ(s) sin(sπ). π

25 18 Since ϕ 1 is an odd function in \{}, from the above it, we know that there exists some C > such that ϕ 1(x) C, x. x +s ( ) ) By the formula (3.18) in [17], v t (x) = v 1 t 1 s x in, then ϕ t(x) = t 1 s ϕ 1 (t 1 s x in, which implies that ϕ t(x) = t 1 s ( ) ϕ 1 t 1 s x C t 1 s t 1 s x +s = tc, x. x +s Now we can prove the following asymptotic behaviors of traveling wave solutions. Proposition... Let µ and u C () be a solution to (..1) with µ. Then there exists some constant C > which only depends on s, µ and f such that C 1 y 1+s u (y) C, y 1. y 1+s As a consequence, we have u L p () for any 1 p, and C 1 1 u(y) C y s y, y > 1 and C u(y) C s y s, y < 1. y s

26 19 Proof. By the same compactness argument as Lemma 4.8 in [18], we know that x u (x) =. First, let s find the upper bound of u. For any δ >, let w δ,t (x) = δϕ t (x) u (x) in, by (..1), we get ( ) s w δ,t (x) µw δ,t(x) + 3 t w δ,t(x) [ ] [ ] 3 = δϕ t (x) t + f t(v t (x)) u (x) t + f (u(x)) [ ] ϕt (x) + δ µϕ t t(x). Since f t(±1) = 1 t <, f (±1) < and there exists some large T > and 1 > such that v t(y) = y ± y ± u(y) = ±1, then T + f T (v T (x)) > and 3 T + f (u(x)) <, x 1. By Proposition.1., we get [ ] [ ] 3 δϕ T (y) + f T T (v T (x)) u (x) + f (u(x)) >, x 1. (..) T By Lemma..1, we have ϕ T (x) T C, x. x +s By (..1), there exists some C 1 > and > 1 > such that 1 T ϕ T (x) C 1 x 1+s, x.

27 So we get 1 T ϕ T (x) µϕ T (x) = C 1 x µ T C 1+s x +s C 1 x +s [ x µ T C C 1 ], x. Let 3 = max {, µ T } C + 1, we have C 1 1 T ϕ T (x) µϕ T ) (x) >, x 3. In summary, we know that for all δ >, we have ( ) s w δ,t (x) µw δ,t (y) + 3 T w δ,t (x) >, x 3. Since ϕ T (x) > in, then there exists some large δ > such that for all δ δ, we have w δ,t (x) = δϕ T (x) u (x) 1 for all x Hence w δ,t satisfies ( ) s w δ,t (x) µw δ,t (x) + 3 T w δ,t (x) >, x 3, w δ,t (x) 1 >, x 3 + 1, w δ,t (x) =. x Claim I: w δ,t (x) in. If Claim I is not true, since x w δ,t (x) =, then there exists some x such that w δ,t (x ) = inf x w δ,t (x) <, in particular, w δ,t (x ) =. Since w δ,t (x) > for all x 3 + 1, we get x > 3, which implies that ( ) s w δ,t (x ) <. So we

28 1 obtain ( ) s w δ,t (x ) µw δ,t (x ) + 3 T w δ,t (x ) = ( ) s w δ,t (x ) + 3 T w δ,t (x ) <, we get a contradiction. By Claim I, we get u (x) δϕ T (x) = δv T (x) in. By (..1), we know that there exists some constant A > such that u (x) A, x 1. x 1+s For lower bound of u, we follow the same idea as the upper bound. For any δ >, we may define w δ,t (x) = w δ,t (x), x. By (..1), we have ( ) s w δ,t (x) µ w δ,t(x) + 1 4t w δ,t(x), [ ] [ ] 1 1 = δϕ t (y) t + f t(v t (x)) + u (x) 4t + f (u(x)) [ ] 1 + δ 4t ϕ t(y) + µ(ϕ t ) (y), x. x ± Since u (x) > and ϕ T1 (x) > in, f (±1) <, f t(±1) = 1 t and v t(x) = x ± u(x) = ±1, then there exists some small T 1 > and large 4 > 3 > such that for all δ >, we have 1 T 1 + f T 1 (v T1 (x)) < and 1 4T 1 + f (u(x)) >, x 4. Now look at 1 ( ) ϕ T1 (x) + µϕ T 4T 1 (x), by (..1), Claim I and v t (x) = v 1 t 1 s x, 1

29 there exist some constant C 3 > and C 4 > such that for all x 4, we have 1 ϕ T1 (x) + µϕ T 4T 1 (x) C x µ C 3 1+s y C [ 4 y 4 µ C ] 3. +s 4 x +s C 4 Taking 5 = max { 4, 4 µ C } 3 + 1, then we have C 4 1 4T 1 ϕ T1 (x) + µ(ϕ T1 ) (x) >, x 5. In summary, we know that for all δ >, we have ( ) s w δ,t1 (x) µ w δ,t 1 (x) + 1 4T 1 w δ,t1 (x) >, x 5. Since u (x) > in, then there exists some small δ 1 > such that for all < δ δ 1, we have w δ,t1 (x) = δϕ T1 (x) u (x) <, x Then w δ,t1 satisfies ( ) s w δ,t1 (x) µ w δ,t 1 (x) + 1 4T 1 w δ,t1 (x) >, x > 5, w δ,t1 (x) >, x < 5 + 1, w δ,t 1 (x) =. x By the same argument as Claim I, we can conclude that w δ,t1 (x) in, that is, u (x) δϕ T1 (x) = δv T 1 (x) in. By (..1), we know that there exists some B >

30 3 such that u (x) B, x 1. x 1+s emark..3. Let 1 p, by Proposition.., we know that u L p () for all < s < 1, but f(u( )) L p () if and only if 1 p < s < 1, that is, ( )s u L p () if and only if 1 p < s < 1..3 Hamiltonian Identity and Modica Type Estimate In this section, we always assume that µ, u C () is a solution to (..1) with µ, and u is the s-harmonic extension of u. Since u is the s-harmonic extension in + of u, then u C ( +) ( ) C + and satisfies div [y 1 s u(x, y)] =, (x, y) +, y y1 s u y (x, y) = d s [µu x (x, ) + f(u(x, ))], x, u(x, ) = u(x), x. Following [18], we show similar Hamiltonian identity and Modica-type estimate for traveling wave solution. First, let s quote a Lemma from [18]. Lemma.3.1 (Lemma 5.1 in [18]). The integral and differentiable with respect to x. Moreover, we have t 1 s u(x, t) t 1 s dt is finite x u(x, t) t 1 s dt =.

31 4 Proposition.3. (Hamiltonian Identity). For all x, we have 1 [u x(x, t) u y(x, t)]t 1 s dt = d s [ µ x ] [u x (r, )] dr + G(u(x, )), and µ [u x (r, )] dr = µ u (x) dx = G(1). Proof. By Lemma.3.1, we can define Then x v(x) = 1 v(x) = and [u x(x, t) u y(x, t)]t 1 s dt, x. v (x) = [u x (x, t)u xx (x, t) u y (x, t)u xy (x, t)]t 1 s dt, x. Since div [y 1 s u(x, y)] = in +, we get u xx (x, y) = y s 1 D y [y 1 s u y (x, y)], (x, y) +. Use the integration by parts, we can get v (x) = = [ ux (x, t)t s 1 D y [t 1 s u y (x, t)] u y (x, t)u xy (x, t) ] t 1 s dt u x (x, t)d x [t 1 s u y (x, t)] dt = t t 1 s u y (x, t)u x (x, t) = d s u x (x, ) [µu x (x, ) + f(u(x, ))] [ = d s µu x(x, ) + d ] G(u(x, )), x. dx u y (x, t)u xy (x, t)t 1 s dt

32 5 Then there exists some constant C > such that x ] v(x) = d s [ µ u x(r, ) dr + G(u(x, )) + C, x. Take x in the above identity, since G( 1) =. We complete the proof. v(x) =, we know that C = x emark.3.3. A direct consequence of Proposition.3. is that µ has the same sign as G(1) = 1 f(t) dt. In particular, u is standing wave (i.e., µ = ) if and only if 1 G(1) = G( 1) =. For the second identity µ u (x) dx = G(1) in Propositoin.3., we have another direct approach. Multiply u (x) on the both sides of (..1), we know that it suffices to show that ( ) s u(x)u (x) dx =. (.3.1) Proof of (.3.1). By the Dominated Convergence Theorem and changing of variables, we have ( ) s u(x)u (x) dx = C n,s [u(x + y) + u(x y) u(x)]u (x) dydx n y n n+s = C n,s [u(x + y) + u(x y) u(x)]u ɛ (x) y n n+s = C n,s C n,s ɛ n ɛ n y >ɛ y >ɛ y >ɛ [u(x + y) u(x)]u (x) y n+s [u(x y) u(x)]u (x) y n+s dydx dydx dydx

33 6 = C n,s C n,s = C 1,s = C 1,s ɛ ɛ n ɛ n ɛ [ y >ɛ y >ɛ x y >ɛ x y >ɛ [u(y) u(x)]u (x) x y n+s dydx [u(y) u(x)]u (x) x y n+s dydx [u(x) u(y)]u (x) dydx x y 1+s u(x)u (x) dydx x y 1+s x y >ɛ ] u(y)u (x) dydx. x y 1+s u(x)u (x) For dydx, by Fubini-Tonelli s Theorem, and apply the integration by parts, we x y >ɛ x y 1+s have x y >ɛ u(x)u (x) dydx = x y 1+s = = z >ɛ z >ɛ z >ɛ u(x)u (x) dydx Let z = y x z 1+s [ ] u(x)u (x) dx dz [ ] 1 u(x) dz 1 z 1+s 1 z 1+s u(y)u (x) For x y >ɛ x y dydx, let J ɛ(z) = 1+s an even function in, and Since x ± J ɛ u(x) = x ± J ɛ (z) dz = =, Since x u(x) = ±1. z >ɛ 1 z 1+s 1 n \B 1 ()(z) in, then J ɛ is 1 1 dz = z 1+s sɛ. s u(x) = ±1, by the Dominated Convergence Theorem, we have x ± J ɛ (y)u(x y) dz = ± J ɛ (y) dz = ± 1 sɛ, s

34 7 which implies that J ɛ u(x) u(x) = 1 x ± sɛ. s Use the integration by parts, we have x y >ɛ u(y)u (x) dydx = J x y 1+s ɛ u(x)u (x)dx = J ɛ u(x)u(x) u(x)j ɛ u (x) dx [ ] = u(x) J ɛ (x y)u (y) dy dx = J ɛ (y x)u(x)u (y) dydx Since J is even = J ɛ (x y)u(y)u (x) dxdy By changing the positions of x and y u(y)u (x) = dydx. x y >ɛ x y 1+s So we know that x y >ɛ u(y)u (x) dydx =. x y 1+s In summary, we have x y >ɛ u(x)u (x) u(y)u (x) dydx = dydx =, x y 1+s x y >ɛ x y 1+s which implies that ( ) s u(x)u (x) dx =.

35 8 Proposition.3.4 (Modica Type Estimate). For all (x, y) +, we have 1 y x ] [u x(x, t) u y(x, t)]t 1 s dt < d s [ µ u x(r, ) dr + G(u(x, )). As a consequence, we have x G(u(x)) > µ u (r) dr, x. (.3.) Proof. Consider the function v(x, y) = 1 y [u x(x, t) u y(x, t)]t 1 s dt, (x, y) +. By Lemma.3.1, we know that x v(x, y) =, uniformly in y, (.3.3) which implies that v L ( +). It s easy to see that v x (x, y) = y [u x (x, t)u xx (x, t) u y (x, t)u xy (x, t)]t 1 s dt, v y (x, y) = 1 [u x(x, y) u y(x, y)]y 1 s. Since div [y 1 s u(x, y)] =, we get u xx (x, y) = y s 1 D y [y 1 s u y (x, y)], (x, y) +.

36 9 Use integration by parts, we obtain v x (x, y) = y = y [ ux (x, t)t s 1 D y [t 1 s u y (x, y)] u y (x, t)u xy (x, t) ] t 1 s dt u x (x, t)d y [t 1 s u y (x, t)] dt y = u x (x, ) t t 1 s u y (x, t) y 1 s u y (x, y)u x (x, y) u y (x, t)u xy (x, t)t 1 s dt = u x (x, )d s [µu x (x, ) + f(u(x, ))] y 1 s u y (x, y)u x (x, y) [ = d s µ[u x (x, )] + d ] G(u(x, )) y 1 s u y (x, y)u x (x, y). (.3.4) dx Consider the function x ] w(x, y) = d s [ µ [u x (r, )] dr + G(u(x, )) v(x, y), (x, y) +. By Proposition.3., we have y w(x, y) = for all x, and x [ x ] µ [u x (r, )] dr + G(u(x, )) =. By (.3.3), we know that w(x, y) uniformly in y, as x. Since v L ( +), u(x, ) = u(x) 1 in, and u x (r, ) dr = u (r) dr <, we get w L ( +). Claim I: For all (x, y) +, we have div [y 1 s w(x, y)] = (s 1)y 1 4s u x(x, y), div [y s 1 w(x, y)] = (s 1)y 1 u y(x, y). (.3.5)

37 3 In fact, for w(x, y), we have [ w x (x, y) = d s µ[u x (x, )] + d ] G(u(x, )) v x (x, y) dx = y 1 s u x (x, y) u y (x, y) By (.3.4) w y (x, y) = v y (x, y) = y1 s [u y(x, y) u x(x, y)]. For D w(x, y), we have w xx (x, y) = y 1 s [u xx (x, y)u y (x, y) + u x (x, y)u xy (x, y)] w yy (x, y) = = (1 s)y s + y1 s [u y(x, y) u x(x, y)] [u y(x, y)u yy (x, y) u x (x, y)u xy (x, y)] (1 s)y s [u y(x, y) u x(x, y)] +y 1 s [u y (x, y)u yy (x, y) u x (x, y)u xy (x, y)] w(x, y) = y 1 s [u xx (x, y)u y (x, y) + u x (x, y)u xy (x, y)] (1 s)y s + [u y(x, y) u x(x, y)] +y 1 s [u y (x, y)u yy (x, y) u x (x, y)u xy (x, y)] = y 1 s u y (x, y) u(x, y) + (1 s)y s [u y(x, y) u x(x, y)]. Since div [y 1 s u(x, y)] =, we know that u(x, y) = (s 1)y 1 u y (x, y). For div [y 1 s w(x, y)], we have div [y 1 s w(x, y)] = y 1 s w(x, y) + (1 s)y s w y (x, y)

38 31 [ ] = y 1 s y 1 s (1 s)y s u y (x, y) u(x, y) + [u y(x, y) u x(x, y)] +(1 s)y s y1 s [u y(x, y) u x(x, y)] = y 4s u y (x, y) u(x, y) + (1 s)y 1 4s u y(x, y) + (s 1)y 1 4s u x(x, y) = (s 1)y 1 4s u y(x, y) + (1 s)y 1 4s u y(x, y) + (s 1)y 1 4s u x(x, y) = (s 1)y 1 4s u x(x, y). For div [y s 1 w(x, y)], we have div [y s 1 w(x, y)] = y s 1 w(x, y) + (s 1)y s w y (x, y) [ ] = y s 1 y 1 s (1 s)y s u y (x, y) u(x, y) + [u y(x, y) u x(x, y)] +(s 1)y s y1 s = u y (x, y) u(x, y) = (s 1)y 1 u y (x, y). [u y(x, y) u x(x, y)] Claim II: w is not a constant function in +. If Claim II is not true, that is, w(x, y) w(, ) in +. Since v(x, ) in, we have x ] w(, ) d s [ µ u x(r, ) dr + G(u(x, )), x. This implies that µ u x (x, ) = d dx G(u(x, )) = f(u(x, ))u x(x, ), x.

39 3 Since u x (x, ) = u (x) > and u(x, ) = u(x) in, then we get µu (x) = f(u(x)) in, which implies that ( ) s u(x) = in. Since u L (), by Liouville s theorem for the fractional harmonic functions, then u is a constant function in, which contradicts with x ± u(x) = ±1. Claim III: w(x, y) > A := inf w(z) for all (x, y) +. z + If Claim III is not true, then there exists some (x, y ) + such that w(x, y ) = A = inf (x,y) + w(x, y). Since u (x) > in, then u x (x, y) > for all (x, y) +. It s easy to see that w x (x, y) = y 1 s u y (x, y)u x (x, y) in +, then we have u y (x, y) = y s 1 wx(x, y) u x (x, y). By (.3.5), we get div [y s 1 w(x, y)] = (s 1)y s uy(x, y) u x (x, y) w x(x, y), in +. Hence w satisfies an locally uniformly elliptic equation in +. By the strong maximum principle and Claim II, we know that (x, y ) +, that is, (x, ) is a boundary minimum point of w in +, in particular, w x (x, ) =, that is, = f(u(x, ))u x (x, ) µu x(x, ). Since u x (x, y) > in +, then µu x (x, ) + f(u(x, )) =, which implies that y y1 s u y (x, y) =. (.3.6) We consider the following two cases:

40 33 Case I: < s 1. By (.3.5), then div [y1 s w(x, y)] in +. Since w(x, y) > w(x, ) for all (x, y) +, by Hopf Lemma, Proposition 4.11 in [18], we know y 1 s u y (x, y) <, which contradicts with (.3.6). y Case II: 1 > s > 1. By the extension, we know that u y(x, y) = in. Since y w(x, y) > w(x, ) in +, then inf y y s 1 w y (x, y). On the other hand, it s easy to see that w y (x, y) = 1 [u y(x, y) u x(x, y)]y 1 s in +, which implies that inf y [u x(x, ) u y(x, y)] = u x(x, ) sup y = u x(x, ) >. u y(x, ) Therefore, we get a contradiction. In summary, we know that for all < s < 1, w(x, y) > A in +. Since w(x, y) uniformly in y, as x, and w(x, y), as y, then w(x, y) =. By Claim III, we must have A and w(x, y) > in (x,y) +, that is, for all (x, y) +, we have 1 y x ] [u x(x, t) [u y(x, t)]t 1 s dt < d s [ µ u x(r, ) dr + G(u(x, )). emark.3.5. Since x ± u(x) = ±1, by (.3.), we know that nonnegative speed

41 34 µ implies that G(t) > in ( 1, 1), and x x ( ) s u(r)u (r) dr = G(u(x)) + µ u (r) dr <, x. (.3.7).4 Nondegeneracy The following proposition the nondegeneracy of traveling wave solution, which allows to use the implicit function theorem near the traveling wave solution. Proposition.4.1. Let µ and u C () be a solution to (..1) with µ. Assume h and φ C () satisfy ( ) s φ(x) µφ (x) f (u(x))φ(y) + hu (x) =, x, φ() =, x φ(x) =. Then h = and φ(x) in. Proof. By Proposition.1., we know that v(x) := u (x) > in. We consider the following two cases: Case I: h. For any δ >, let w δ (x) = v(x) δφ(x) in, then w δ satisfies ( ) s w δ (x) µw δ (x) f (u(x))w δ (y) = δhu (x), x, w δ () >, w δ(x) =. x Since f (±1) < and x u(x) = ±1, we can find some fixed large > such

42 35 that f (u(x)) > for all x. Since v(x) > in, there exists some fixed small ɛ > such that < δ ɛ and all x + 1, we have w δ (x) >. By Lemma.1.1, we know that w δ (x) > in. Hence, we can define Λ = sup {ɛ > : w δ (x) > in, δ (, ɛ)} ɛ. If Λ =, that is, for all δ >, we have w δ (x) > in, which implies that φ(x) in. Since φ() =. By Proposition.1.1, we know that φ(y) in. Hence h. If Λ <, then w Λ (x) in. Since φ() =, v(x) > in, by Lemma.1.1, we know w Λ (x) > in. eplace v by w Λ in the previous argument, we can find a larger Λ > Λ such that for all δ (, Λ ), we have w δ (y) > in, which contradicts with the definition of Λ. Hence, in this case, we have h = and φ(x) in. Case II: h. In this case, let k = h and ψ(x) = φ(x) in, then it s easy to see that k and ψ satisfies x ψ(x) = and ( ) s ψ(x) µψ (x) f (u(x))ψ(x) + ku (x) =, x. Applying the result of Case I, we know that k = and ψ(x) in, which implies that, h = and φ(x) in. In summary, we can conclude that h = and φ(x) in.

43 Chapter 3 Estimates of Traveling Speeds In this chapter, we will estimate the traveling speed in terms of potential function f, which is crucial in the proof of the existence of traveling waves. In the chapter, we always assume that < s < 1, f is a bistable nonlinearity, that is, f satisfies (1.1.3), and G(t) = t f(u) du Laplacian Case In this section, let s get some motivations from the case of usual Laplacian. As I mentioned in Section 1.1 (see [61, 66] for more details), there exists a unique pair (µ, φ) up to translation such that there exists some constant ν >, we have φ (x) µφ (x) = f(φ(x)), x, φ (x) e ν x, as x, x ± φ(x) = ±1. 36

44 37 If µ <, consider ψ(x) = φ( x) in, then ψ (x) = φ ( x) in and x ± ψ(x) = ±1. Let f(t) = f( t) in, it s easy to see that ψ satisfies ψ (x) ( µ)ψ (x) = f(ψ(x)), x In summary, without loss of generality, we can assume the speed µ, and consider the following problem: φ (x) µφ (x) = f(φ(x)), φ (x) e ν x, as x, φ (x) >, x ± x φ(x) = ±1, µ x (3.1.1) Multiply φ (x) on both sides of the first equation in (3.1.1) and integrate from to y, we get Since y y y φ (x)φ (x) dx µ [φ (x)] dx = f(φ(x))φ (x) dx. x φ (x) = and dg(φ(x)) dx = f(φ(x))φ (x), we get 1 y [φ (y)] + µ [φ (x)] dx = G(φ(y)), y. (3.1.) Since x φ (x) = and x following Hamiltonian identity: φ(x) = 1, by taking y in (3.1.), we get the µ [φ (x)] dx = G(1) (3.1.3)

45 38 Since µ, we have the Modica type estimate: 1 [φ (y)] G(φ(y)), y, (3.1.4) which implies that G(t) > in ( 1, 1). Now we are ready to estimate the speed µ. If G(1) =, by (3.1.3), µ =, there is nothing to do. In the following, we assume G(1) >, that is, µ >, so we only need to find the upper bound of µ. In the following, we introduce three different approaches. Since f satisfies (1.1.3), we know that G is strictly increasing on [ 1, t ], and strictly decreasing on [t, 1], in particular, there exists some t 1 ( 1, t ) such that < G(1) < G(t), t (t 1, 1), and G(t ) = max t [ 1,1] G(t). By translation, without loss of generality, we can assume that φ() = t. Take y = in (3.1.), we have 1 [φ ()] = G(t) µ G(t) µ [φ (x)] dx [φ (x)] dx Since µ > = G(t) G(1), By (3.1.3). By (3.1.3), we have 1 [φ ()] G(t) G(1). (3.1.5)

46 39 Proposition For any t (t 1, 1), we have < µ < f L ([ 1,1]). [G(t) G(1)] Proof. Since x φ (x) =, by (3.1.5), we know that there exists some x such that φ (x ) = sup x φ (x) φ () [G(t) G(1)], which implies that φ (x ) =. By (3.1.1), we get µ = f(φ(x )) φ (x ) f C([ 1,1]) φ () f C([ 1,1]) [G(t) G(1)]. emark The approach of Proposition involves the Hamiltonian identity (3.1.5) and the relation between φ (x) and φ (x) which is very special for the Laplacian. Proposition For any t (t 1, 1), we have < µ B + B + 4AC, A Where A = G(t) G(1), B = f(t) G(t), and C = f L ([ 1,t]) G(1).

47 4 In particular, we have < µ f L ([ 1,t ]) G(1) G(t ) G(1). Proof. Multiply φ (x) on (3.1.1) and integrate over (, ), use the integration by parts, then [φ (x)] dx = = µ [φ ()]. φ (x)φ (x) dx f(φ(x))φ (x) dx f(φ(x))φ (x) dx Since µ >, by (3.1.5), we obtain µ[g(t) G(1)] µ [φ ()] f(φ(x))φ (x) dx Let s estimate f(φ(x))φ (x) dx. Use the integration by parts, since φ(x) = x 1 and f( 1) =, we obtain f(φ(x))φ (x) dx = f(φ(x))φ (x) + = f(t)φ () + φ (x)f (φ(x))φ (x) dx f (φ(x))[φ (x)] dx For f(t)φ (), by (3.1.4), we get f(t)φ () f(t) G(t)

48 41 For f (φ(x))[φ (x)] dx, then f (φ(x)) [φ (x)] dx f L ([ 1,t]) [φ (x)] dx f L ([ 1,t]) [φ (x)] dx = f L ([ 1,t]) G(1), By (3.1.4). µ In summary, we have µ[g(t) G(1)] f(t) G(t) + f L ([ 1,t]) G(1) µ. Since µ >, we get [G(t) G(1)]µ f(t) G(t)]µ f L ([ 1,t]) G(1) Let A = G(t) G(1), B = f(t) G(t) and C = f L ([ 1,t]) G(1), then we have Aµ Bµ C, which implies that µ B + B + 4AC A The third approach is quite similar to Proposition 3.1.3, but it gives different estimate.

49 4 Proposition For any t (t 1, 1), we have < µ B + B + 4AC, A Where A = [G(t) G(1)], B = f(t) G(1) G(t), and C = f L ([ 1,t]) [G(1)]. In particular, we have < µ G(1) f L ([ 1,t ]) G(t ) G(1). Proof. Multiply φ (x) on the both sides of (3.1.1) and integrate over (, ), use the integration by parts, then [φ (x)] dx = µ = µ φ (x) = µ [φ ()] φ (x)φ (x) dx f(φ(x))φ (x) dx f(φ(x))φ (x) dx f(φ(x))φ (x) dx < f(φ(x))φ (x) dx, Since µ, φ () >. By the proof of Proposition 3.1.3, we have f(φ(x))φ (x) dx f(t) G(t) + f L ([ 1,t]) G(1) µ.

50 43 So we get [φ (x)] dx f(t) G(t) + f L ([ 1,t]) G(1) µ. By (3.1.5), use the Cauchy-Schwarz s inequality, we get G(t) G(1) 1 [φ ()] = [ [ φ (x)φ (x) dx Since x φ (x) = ] 1 [ ] 1 [φ (x)] dx [φ (x)] dx [φ (x)] dx ] 1 [ [φ (x)] dx ] 1 [ f(t) G(t) + f L ([ 1,t]) G(1) µ ] 1 [ G(1) µ ] 1, By (3.1.3). Since µ >, we get [G(t) G(1)]µ f(t) G(1) G(t)]µ f L ([ 1,t]) [G(1)]. Let A = [G(t) G(1)], B = f(t) G(1) G(t) and C = f L ([ 1,t]) [G(1)], then we have Aµ Bµ C, which implies that µ B + B + 4AC. A emark The approaches of Proposition and Proposition involve

51 44 the Hamiltonian identity (3.1.5) Modica type estimate (3.1.4),. These two approaches will be used in Section 3. to estimate the speeds with fractional Laplacians. emark For the lower bound of µ, by the Hamiltonian identity (3.1.3) and Modica type estimate (3.1.4), since φ (x) > in and x ± φ(x) = ±1, we have µ = G(1) φ (x) dx G(1) G(φ(x)) φ (x) dx = G(1) 1. 1 G(t) dt 3. Fractional Laplacian Case By the discussion for the Laplacian case in Section 3.1, in this section, we can assume that f is a unbalanced bistable nonlinearity such that 1 f(t) dt >. So we only 1 need to consider the following problem: ( ) s u(x) µu (x) = f(u(x)), x, u (x) >, x, x ± u(x) = ±1, µ >. (3..1) Since µ >, by (.3.), we have G(1) >. Since f satisfies (1.1.3), we know that G is strictly increasing on [ 1, t ], and strictly decreasing on [t, 1], in particular, there exists some t 1 ( 1, t ) such that < G(1) < G(t), t (t 1, 1), and G(t ) = max t [ 1,1] G(t). For any t (t 1, 1), by translation, without of generality, we can assume u() = t. Multiply u (x) on the first equation of (3..1) and integrate over (, ), use the

52 45 integration by parts, we have ( ) s u(x)u (x) dx = f(u(x))u (x) dx µ u (x) dx = G(u(x)) µ u (x) dx = G(t) µ u (x) dx G(t) µ u (x) dx. By Proposition.3., we have G(t) G(1) ( ) s u(x)u (x) dx. (3..) emark The formula (3..) is the counterpart of (3.1.5) in the Laplacian case. Notice that in the Laplacian case, we have [ φ (x)]φ (x) dx = 1 [φ ()]. Now we are ready to estimate the speed, but we need to separate to three parts: < s < 1/ (supercritcal), s = 1/ (critical) and 1/ < s < 1 (subcritical). Proposition 3... Let < s < 1, then for all t (t 1, 1), we have G(t) G(1) C 1,s s [t + 1] + C 1,s s 1 s 1 s G(1), >. µ

53 46 In particular, we know that < µ [ 4C 1,s 1 s ( ) 1 s t s ] 1 s 1 G(1). G(t) G(1) Proof. For any >, then ( ) s u(x) u(y) u(x) = C 1,s P.V. dy x y 1+s u(x) u(y) = C 1,s dy Since < s < 1 x y 1+s u(x + z) u(x) = C 1,s dz Let y = x + z z 1+s u(x + z) u(x) u(x + z) u(x) = C 1,s dz + C z 1+s 1,s z 1+s z < z dz For C 1,s z u(x + z) u(x) z 1+s dz, since u(x) 1 in, then For C 1,s C 1,s C 1,s z < z < z u(x + z) u(x) 1 dz C z 1+s 1,s dz z z 1+s 1 = 4C 1,s dz z1+s u(x + z) u(x) z 1+s = C 1,s s. s dz, since u (x) > in, then u(x + z) u(x) dz = C z 1+s 1,s 1 C 1,s z < 1 u (x + tz)z dt z 1+s z < u (x + tz) dz 1 dzdt z s

54 47 Hence we know that ( ) s u(x) C 1,s s 1 s + C 1,s z < u (x + tz) 1 dzdt, z x s Since u (x) > in, multiply u (x) on the both sides of the above inequality, and integrate over (, ), we get [ C1,s ( ) s u(x)u (x) dx 1 s + C 1,s s z < = C 1,s s 1 [t + 1] + C 1,s s z < z s ] u 1 (x + tz) dzdt u (x) dx z s [ 1 ] u (x + tz) u (x) dxdt dz For u (x + tz) u (x) dx, by Cauchy-Schwarz s inequality, then u (x + tz) u (x) dx = [ [u (x + tz)] dx [ ] 1 [u (x + tz)] dx [u (x)] dx ] 1 [ [ = G(1), By Proposition.3.. µ [u (x)] dx [u (x)] dx ] 1 ] 1 So we get C 1,s C 1,s z < z < 1 z s 1 z s [ 1 [ 1 ] G(1) µ dt ] u (x + tz) u (x) dxdt dz dz

55 48 = C 1,s G(1) µ 1 dz zs = C 1,s 1 s 1 s G(1), Since s < 1. µ In summary, we get ( ) s u(x)u (x) dx C 1,s s s [t + 1] + C 1,s 1 s 1 s G(1) µ, > By (3..), we get G(t) G(1) ( ) s u(x)u (x) dx C 1,s s [t + 1] + C 1,s s 1 s 1 s G(1), >. µ Let s take C 1,s s [t + 1] = C 1,s s 1 s 1 s G(1) µ[t + 1], that is, =, then we µ sg(1) obtain G(t) G(1) 4C 1,s 1 s 1 s G(1) µ = 4C ( 1,s µ[t + 1] 1 s sg(1) = 4C 1,s 1 s ) 1 s G(1) µ ( ) 1 s ( ) s t + 1 µ, s G(1) which implies that [ ( ) ] 1 1 s s 4C 1,s t + 1 µ 1 s 1 G(1). s G(t) G(1)

56 49 emark When < s < 1/, the point-wise estimate of ( ) s u(x) is the most important part in the proof of Proposition 3.. For the case s = 1/, the following lemma tells us the energies of u x and u y in x direction are the same, which is crucial in the estimate of speed µ. Lemma Let s = 1, and u be the 1/-harmonic extension in + of u. Then we have [u x (x, y)] dx = [u y (x, y)] dx, y. Proof. Since u y (x, ) = ( ) 1 u(x) = µu (x)+f(u(x)) in, u L () and emark..3, we know that u y (, ) L (). Since u is harmonic in +, then u y (, y) L () for all y. In particular, we can consider the function ψ(y) = [u y(x, y) u x(x, y)]dx, y. Differentiating ψ and using integration by parts, we get ψ (y) = = = u y (x, y)u yy (x, y) dx u y (x, y)u yy (x, y) dx + =, y. u y (x, y) u(x, y) dx u x (x, y)u xy (x, y) dx u y (x, y)u xx (x, y) dx Claim I: u(x, y) y dx =.

57 5 Let P (x, y) = P 1, 1 (x, y) = 1 π y x + y in +, we get P (, y) L () = 1 π = 1 πy y [x + y ] dx, as x. By Young s inequality, we have u x (, y) L () = P (, y) u L () P (, y) L () u L 1 (), as y. Look at u y (x, y), by Proposition.., we know that there exists some constant C 1 > such that u y (x, ) = ( ) 1 u(x) = µu (x) + f(u(x)) C 1, x. 1 + x Since u y is harmonic in +, so there exists some constant C > C 1 > such that u y (x, y) C 1 + y, (x, y) +. Since 1 z = 1 x + iy = x iy x, we know that x + y +. Let v(x, y) = C x x + y and y are harmonic in x + y C(x + y + 1) x + (y + 1), we know that v is harmonic in and ±u y (x, ) C(1 + x) 1 + x = v(x, ) for all x and ±u y (, y) C 1 + y = v(, y) for all

58 51 x. By the weak maximum principle, we know that ±u y (x, y) v(x, y) = C(x + y + 1), x, y. x + (y + 1) C(y + 1 x) Let w(x, y) = x + (y + 1), we get w is harmonic in and ±u y (x, ) C 1 1 x C(1 x) = w(x, ) for all x and ±u 1 + x y (, y) C = w(, y) for all y. By 1 + y the weak maximum principle, we know that ±u y (x, y) w(x, y) = C(y + 1 x), y, x. x + (y + 1) In summary, we have u y (x, y) C(y x ) x + (y + 1) C (x, y + 1), (x, y) +. Notice that 1 (x, y + 1) dx 1 x + y dx = π y, as y. So we know that u y (, y) L (), as y. In summary, we have u(x, y) dx =. y By Claim I, we get ψ(y) in +, which completes the proof. emark Lemma 3..4 tells us that [( ) 1 u(x) dx is controlled by u (x) dx,

59 5 in fact, they are equal. Proposition Let s = 1, then for any t (t 1, 1), we have < µ G(1) G(t) G(1) G(1) G(t ) G(1). Proof. Let u be the 1/-harmonic extension of u, by Cauchy-Schwarz s inequality, then we know that ( ) 1 u(x)u (x) dx = = = u y (x, ) u (x) dx [ [u y (x, )] dx [ ] 1 [u y (x, )] dx [ ] 1 [u x (x, )] dx [u (x)] dx ] 1 [ [ [ [u (x)] dx [u (x)] dx [u (x)] dx = G(1), By Proposition.3.. µ ] 1 ] 1 ] 1 By Lemma 3..4 So we get µ G(1) ( ) 1 u(x)u (x) dx G(1), By (3..). G(t) G(1)