GLOBAL SOLUTIONS FOR A SEMILINEAR 2D KLEIN-GORDON EQUATION WITH EXPONENTIAL TYPE NONLINEARITY

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1 GLOBAL SOLUTIONS FOR A SEMILINEAR D KLEIN-GORDON EQUATION WITH EXPONENTIAL TYPE NONLINEARITY S. Ibrahim Department of Mathematics & Statistics, McMaster University, 180 main Street West, Hamilton, Ontario, L8S 4L8 Canada. slim.ibrahim@mcmaster.ca M. Majdoub Faculté des Sciences de Tunis, Département de Mathématiques Campus universitaire 1060, Tunis, Tunisia. mohamed.majdoub@fst.rnu.tn N. Masmoudi Department of Mathematics, Courant Institute of Mathematical Sciences New York University, 51 Mercer St. New York, NY 1001, U.S.A. masmoudi@cims.nyu.edu Submitted by:...) Abstract. We prove the existence and uniqueness of global solutions for a Cauchy problem associated to a semilinear Klein-Gordon equation in two space dimension. Our result is based on an interpolation estimate with a sharp constant obtained by a standard variational method. Accepted for publication:... AMS Subject Classifications:...,...,... 1

2 S. Ibrahim, M. Majdoub, and N. Masmoudi 1. Introduction In this paper, we study the well-posedness of the Cauchy problem associated to the semilinear wave equation E α ) t )u + ueαu = 0, where u = ut, x) is a real-valued function of t, x) R R, t = t and is the Laplace operator acting on space variables. We denote by = t the wave operator. The initial data f, g) u0, x) = fx) and t u0, x) = gx) 1.1) are in the energy space H 1 R ) L R ) and α is a nonnegative real number. H 1 is the usual Sobolev space endowed with the norm f H 1 = f L + f L. Before going any further, we shall first recall a few historic facts about this problem. First, we recall that in space dimensions d 3, the defocusing semilinear wave equation with power p reads u + u p 1 u = 0, 1.) where p is a real number with p > 1. This problem has been widely investigated and there is a large literature dealing with the local and global solvability of 1.) in the scale of the Sobolev spaces H s [11, 1, 17, 0,, 3, 6, 9, 36, 37, 38, 4] the uniqueness in suitable subspaces of finite energy solutions, [7] and the references therein ) and the asymptotics of the solutions as t goes to infinity scattering theory) [3, 5, 9, 10, 16, 13, 14, 31, 33, 35]. For a detailed bibliography, see [44]. Second, it is well-known that the Cauchy problem 1.) is locally well-posed in the usual Sobolev space H s R d ) if s > d, or when 1 s < d and p d s. The difference between the conditions for the solvability among s < d, s = d, and s > d basically comes from the Sobolev embedding H s L p if p d d s, Hs L p if p < and H s L p if p, respectively see [9, 30] and the references therein ). Finally, for the global solvability in the energy space H 1 L, there are mainly three cases. The first case is when p < p c where p c = d+ d, this is the subcritical case. In this case, Ginibre and Velo [1], showed that the problem 1.) with initial data in H 1 L has a unique global solution in the space CR, H 1 R d )) C 1 R, L R d )). If the exponent p is critical, which means p = p c ), this problem was first

3 GLOBAL SOLUTIONS FOR A SEMILINEAR D KLEIN-GORDON... 3 solved by Struwe [4] in the radial case, then by Grillakis [17] in the general case, and Shatah-Struwe [36], [37] in other dimensions. See also [19] for the case of variable metric. Notice that the proof is based on the socalled Strichartz inequalities for the solutions of the linear operator See [15, 35, 39, 41, 40, 43]). Finally in the case p > p c, the well-posedness in the energy space is an open problem except for some partial results for example see [4, 7, 5]). See also [6] for Shrödinger equations. In dimension d =, any polynomial nonlinearity is subcritiacl with respect to the H 1 norm, so an exponential nonlinearity seems to be a natural critical nonlinearity. Now we return to the equation E α ). Multiplying E α ) by t u and integrating on R, we formally obtain the following conservation law Eu, t) := t ut, ) L + ut, ) L + R e αu 1 dx = Eu, t = 0). 1.3) α A priori, one can estimate the nonlinear part of the energy using the following Moser-Trudinger type inequalitiessee []). Proposition 1. Let α 0, 4π). A constant c α exists such that R exp αux) ) 1 ) dx c α u L 1.4) for all u in H 1 R ) such that u L R ) 1. Moreover, if α 4π, then 1.4) is false. Remark 1. We point out that α = 4π becomes admissible in 1.4) if we require u H 1 R ) 1 rather than u L R ) 1. Precisely, we have sup exp 4πux) ) 1 ) dx < + u H 1 R ) 1 R and this is false for α > 4π. See [34] for more details. Remark. Notice that if u is a solution of E α ) then for any β > 0, u β is a solution of E αβ ) and obviously one may choose β such that αβ > 4π. Hence, it makes sense to take the initial data in a ball of the energy space. Remark 3. There are two points of view to deal with the Cauchy problem associated to the equation E α ). The first one is to fix the initial data in the unit ball of the energy space and distinguish the cases α < 4π, α = 4π and α > 4π. The second one is to fix α = 4π and discuss the size of the initial

4 4 S. Ibrahim, M. Majdoub, and N. Masmoudi data in the energy space. Actually these two points of view are equivalent. We will choose the second one. In all what follows, we suppose α = 4π. Definition 1. The Cauchy problem E 4π )-1.1) is said to be subcritical if E 0 := g L + f exp 4πf ) 1 L + dx < 1. R 4π It is critical if E 0 = 1 and supercritical if E 0 > 1. To establish an energy estimate, one has to consider the nonlinearity as a source term in E 4π ), so we need to estimate it in the L 1 t L x) norm. To do so, we use 1.4) combined with the so-called Strichartz estimate See [15]). Proposition Strichartz estimate). v L 4 T C 1/4 R )) [ C t v0) L R ) + v0) H 1 R ) + v + v L 1 T L R )) 1.5) But the problem with taking the L x norm is to double 4π and therefore, we loose any control of that term using only 1.4). The following estimate is an L logarithmic inequality which enables us to establish the link between e 4πu 1 L 1 T L R )) and dispersion properties of solutions of the linear Klein- Gordon equation. Proposition 3. For any real λ > π, for any real µ > 0, a constant C λ,µ exists such that, for any function u C 1/4 R ) H 1 R ), one has u L λ u µ log C λ,µ + u ) C 1/4 1.6) u µ where u µ := u L + µ u L. Recall that C 1/4 R ) denotes the space of 1/4-Hölder continuous functions endowed with the norm u C 1/4 R ) := u ux) uy) L R ) + sup x y x y 1/4. A sketch of the proof of this Proposition is discussed in the Appendix. We refer to [18] for more details. We just point out that the condition λ > π in 1.6) is optimal. Our first result is the following local in time) existence Theorem. ]

5 GLOBAL SOLUTIONS FOR A SEMILINEAR D KLEIN-GORDON... 5 Theorem 1. Assume that f L R ) < 1. Then, there exists a time T > 0 and a unique solution u to the problem E 4π )-1.1) in the space C T H 1 R )) C 1 T L R )). Moreover, u L 4 T C1/4 R )) and satisfies, for all 0 t < T, Eu, t) = Eu, 0). Here and below C T X) denotes C[0, T ); X). The proof of this Theorem is based on the combination of the three propositions given above. We derive the local wellposedness using a classical fixed point argument. Remark 4. The assumption E 0 1 in particular implies that f L R ) < 1 and consequently we have the short time existence of solutions in both subcritical and critical case. So it makes sense to deal with global existence in these cases. Remark 5. In a recent work, A. Attallah [1] proved a local existence result of solutions to E α ) for α < 4π, under more restrictive assumptions on the initial data by assuming that f = 0 and g is radially symmetric with compact support. This result is based on L t L x) estimate of the nonlinear term. Such an estimate seems to be unreasonable in the general case. As an immediate consequence of Theorem 1 we have the following global existence result. Theorem Subcritical case). Assume that E 0 < 1, then the problem E 4π )-1.1) has a unique global solution u in the class CR, H 1 R )) C 1 R, L R )). Moreover, u L 4 loc R, C1/4 R )) and satisfies the energy identity. The reasons of Definition 1 is the following. If u denotes the solution given by Theorem 1, where T < is the largest time of existence, then the conservation of the total energy gives us, in the subcritical setting, a uniform bound of ut, ) L R ) away from 1 and therefore the solution can be continued in time. Contrary to the critical case, where we loose this uniform control and therefore the total mass of energy can be concentrated in the ut, ) L R ) part. By establishing some local in space-time) identities, we show that such concentration cannot hold in the critical case and therefore we have the following theorem.

6 6 S. Ibrahim, M. Majdoub, and N. Masmoudi Theorem 3 Critical case). Assume that E 0 = 1, then the problem E 4π )-1.1) has a unique global solution u in the class CR, H 1 R )) CR, L R )). Moreover, u L 4 loc R, C1/4 R )) and satisfies the energy identity. Remark 6. It would be desirable to show that the global solution u is in L 4 R, C 1/4 R )) globally in ) time, at least when we replace the nonlinear term by u e 4πu 1 4πu. This question as well as some scattering results will be dealt with in a forthcoming paper. When the initial data are more regular, we can easily prove that the solution remains regular. More precisely, we have the following theorem. Theorem 4. Assume that f, g) H s R ) H s 1 R ) with s > 1 and f L R ) < 1. Then, the solution u given in Theorem 1 is in the space C T H s R )) C 1 T Hs 1 R )). To the best of the authors s knowledge, Theorem and Theorem 3 are the only results for global solutions of such D problems with exponential growth nonlinearities. In [9], Nakumura and Ozawa proved, under an assumption of smallness of the initial data, the existence of global solutions. The structure of the paper is as follows. In the next section, we present some notations which will be used in the sequel. Section 3 is devoted to the proof of Theorem 1 about local existence. The fourth section deals with global existence. Proofs of Theorems -3 are presented there. In the appendix, we give a sketch of the proof of Proposition 3 and some variants of the L logarithmic inequality.. Notations For Q R R and S < T, we define Q T S := {z = t, x) Q such that S t T },.1) the truncated part of Q between the times S and T. For z 0 = t 0, x 0 ) R R, we define Kz 0 ) := {z = t, x) R R such that x x 0 t 0 t} the backward cone of vertex z 0, and for fixed t Dt, z 0 ) := {x R such that x x 0 t 0 t}

7 GLOBAL SOLUTIONS FOR A SEMILINEAR D KLEIN-GORDON... 7 its section at the time t. The mantle of the cone Kz 0 ), denoted by Mz 0 ), is defined by Mz 0 ) := {z = t, x) R R such that x x 0 = t 0 t}. For x R and r R\{0}, we denote by Bx, r) the ball of R centered at x and of radius r. If x = 0, we use the notation Br) instead of Bx, r). In particular, we remark that Dt, z 0 ) = Bx 0, t 0 t). For any function u = ut, x), we define the energy density, the local energy, the flux density and the flux of u by eu, t u)t, x) := t u) + x u + e4πu 1, 4π Eu, Dt, z 0 )) := Dt,z 0 ) eu, t u)t, x) dx d z0 u) := 1 [ t u x x + xu + e4πu 4π Fluxu, MS T z 0)) := M T S z 0) 1 d z0 u) dσ, respectively. If no ambiguity can occur, we denote eu) = eu, t u). Let v 0 denote the solution of the free Klein-Gordon equation with the same initial data f, g), namely v 0 + v 0 = 0, v 0 0, x) = fx) and t v 0 0, x) = gx)..) Recall that for any real number µ > 0 and any function w in H 1 R ) w µ := w L + µ w L. In all what follows, we denote by E 0 the initial energy, namely { E 0 := fx) + gx) + e4πfx) 1 } dx. 4π R Finally, we mention that, C will be used to denote a constant which may vary from line to line. ],

8 8 S. Ibrahim, M. Majdoub, and N. Masmoudi 3. Local Existence In this section we shall prove the local wellposedness for the equation E 4π ) Theorem 1). To show the local small time) existence of solutions of E 4π ), we use a standard fixed point argument. We introduce, for any nonnegative time T, the following complete metric space E T = C[0, T ], H 1 R )) C 1 [0, T ], L R )) L 4 [0, T ], C 1/4 R )) endowed with the norm u T := sup 0 t T [ ut,.) H 1 + t ut,.) L ] + u L 4 [0,T ],C 1/4 ). Proof of Theorem 1. Let us start by proving the existence. For a positive time T and a positive real number δ, we denote by E T δ) the ball in E T of radius δ and centered at the origin. On the ball E T δ), we define the map Φ by where v Φv) := ṽ, 3.1) ṽ + ṽ = v + v 0 ) e 4πv+v 0) 1 ), ṽ0, x) = t ṽ0, x) = 0, 3.) and v 0 is defined by.). Now the problem is to show that, if δ and T are small enough, the map Φ is well defined from E T δ) into itself and it is a contraction. In order to show that the map Φ is well defined, we need to estimate the term v + v 0 ) e 4πv+v 0) 1 ) L 1 T L. Indeed, by the Energy estimate, we x) know that ṽ T C v + v 0 ) e 4πv+v 0) 1 ) L 1 T L ). x By the Hölder inequality and the Sobolev embedding, we have ) v + v 0 ) e 4πv+v 0) 1 dx v + v0 R L e 4πv+v 0) 1 ) +/ε L 1+ε C v + v 0 H 1 e 4π v+v 0 L ) e 4πv+v 0) 1 L 1+ε where ε is a nonnegative real number which will be chosen later. The first term on the right hand side of the last inequality can be easily estimated ) v + v 0 H δ + v 1 0 H 1 ) δ + f H + g 1 L := A.

9 GLOBAL SOLUTIONS FOR A SEMILINEAR D KLEIN-GORDON... 9 On the other hand, using Proposition 3, we can write e 4π v+v 0 L exp 4πλ v + v 0 µ log C λ,µ + v + v )) 0 C 1/4 v + v 0 µ for any λ > π, µ > 0, and C λ,µ given by Proposition 3. In addition, it is clear that e 4πv+v 0 ) 1 ) ) 1+ε dx e 4π1+ε)v+v 0) 1 dx. R R Now, since f L < 1, we choose a real number µ > 0 such that f µ < 1. By continuity in time of v 0, there exists a nonnegative time T 0 such that for all t in [0, T 0 ] we have v 0 t,.) µ 1 η 3.3) where η := 1 ) 1 f µ. Using the fact that v + v0 µ δ + 1 η, it follows that e 4π v+v 0 L C λ,µ + v + v ) 4πλδ+ 1 η) 0 C 1/4 δ + 1 η and, in view of Proposition 1 ) 1+ε e 4πv+v 0) 1 dx R R exp 4π1 + ε)δ + 1 η) v + v 0 ) ) ) δ + 1 η) 1 Cε, δ, η) v + v 0 L Cε, δ, η)a provided that 4π1 + ε)δ + 1 η) < 4π which is possible since 1 + ε)δ + 1 η) 1 η < 1 as ε, δ 0. Therefore, for any 0 < T T 0, we obtain T 0 v + v 0 ) e 4πv+v 0) 1) L dt C T 0 C λ,µ + v + v ) β/ 0 C 1/4 δ + dt 1 η where β := 4πλδ + 1 η). Since 4πλδ + 1 η) 81 η) < 8 as δ and λ go to /π, we can choose δ and λ such that β < 8. With this choice we get T v + v 0 ) e 4πv+v 0) 1) L dt CT 1 β/8 C λ,µ + v + v 0 C 1/4 δ + β/ 1 η 0 L 4 T dx

10 10 S. Ibrahim, M. Majdoub, and N. Masmoudi CT 1 β/8 C λ,µ T 1/4 + v + v ) β/ 0 L 4 T C 1/4 ) δ + 1 η CT 1 β/8 C λ,µ T 1/4 + δ + f H 1 + g ) β/ L δ +. 1 η From this inequality it follows immediately that, if T is small enough then Φ goes from E T δ) to E T δ). To prove that Φ is a contraction at least for T small), we consider two elements v 1 and v in E T δ) and define v = v 1 v, ṽ = ṽ 1 ṽ, v = 1 θ)v 0 +v 1 )+θv 0 +v ) with 0 θ 1. We can write v 1 + v 0 )e 4πv 1+v 0 ) 1) v + v 0 )e 4πv +v 0 ) 1) = v [ 1 + 8π v )e 4π v 1 ] for some choice of 0 θt, x) 1. By the energy estimate and the Strichartz inequality we have Φv 1 ) Φv ) C v [ 1 + 8π v )e 4π v 1 ] L T 1 T L ). x Notice that T 0 v [ 1 + 8π v )e 4π v 1 ] L dt C ε T for any ε > 0 and that by convexity 0 v e 4π1+ε) v 1 ) L dt, e 4π1+ε)[1 θ)v 0+v 1 )+θv 0 +v )] 1 θ)e 4π1+ε)v 0+v 1 ) + θe 4π1+ε)v 0+v ) and e 4π1+ε)v 0+v 1 ) + e 4π1+ε)v 0+v ), v µ δ. So arguing as before and setting β = 4π1 + ε)λδ + 1 η), we obtain Φv 1 ) Φv ) T CT 1 β/8 C λ,µ T 1/4 + δ + f H 1 + g ) β/ L δ + v 1 v T. 1 η If the parameters ε > 0, λ > /π, µ > 0 and δ > 0 are suitably chosen, then β < 8 and therefore for T small enough, Φ is a contraction map. Notice that this also proves uniqueness in the existence class, namely in v 0 + E T δ). To prove the conservation of the energy, we multiply the equation by t u and then integrate by parts. All the computations are justified since the exponential term is in L 1 T L x ).

11 GLOBAL SOLUTIONS FOR A SEMILINEAR D KLEIN-GORDON Now we shall prove the uniqueness in the energy space. Since the uniqueness holds in v 0 + E T δ), it suffices to show that, if u = v 0 + v solves E 4π )-1.1) with v C T H 1 R )) CT 1 L R )), then necessary v E T δ) at least for T small). a Let ε > 0 and a > 1 be such that a 1ε and a 1 are small enough. There exists 0 < µ 1 and T 1 > 0 such that 3.3) holds and sup vt,.) H 1 + t vt,.) L ) ε. 0 t T 1 Using Strichartz inequality and the fact that v satisfies v + v = v + v 0 ) e 4πv+v 0) 1 ), v0, x) = t v0, x) = 0, 3.4) we just need to estimate the right hand side of 3.4) in L 1 T L x ). Arguing as in the proof of the existence part, and using the following observations v + v 0 ) av0 + a a 1 v, e x+y 1 = e x 1)e y 1) + e x 1) + e y 1), we have to estimate the following three terms: For all t ]0, T 1 [ I 1 t) = e 4π1+ε) a a 1 v t,x) 1) dx, R I t) = e 4π1+ε)av 0 t,x) 1) dx, R ) I 3 t) = e 4π1+ε)av 0 t,x) 1 e 4π1+ε) a a 1 v t,x) 1) dx. R For the first term, we use Remark 1 to obtain I 1 t) C. For the second term, we write I t) e 4π1+ε)a v 0t, ) L R C 1 + v 0 t, ) C 1/4) β ) e 4π1+ε)av 0 t,x) 1 dx where β = 4π1 + ε)a1 η)λ. Finally, for the third term, we have ) ) 1/a I 3 t) e 4π1+ε)a v 0t) L e 4π1+ε)a v0 t,x) 1 dx C 1 + v 0 t) C 1/4) β. e 8π1+ε) a ) a 1) v 1 ) 1 1/a dx

12 1 S. Ibrahim, M. Majdoub, and N. Masmoudi This finishes the proof of the uniqueness in the energy space. 4. Global Existence We start this section by the following remark about the existence time. This remark will be very important to extend the solution. Remark 7. In Theorem 1, the time of existence T depends on f and g. However, in the case f L R ) + f L R ) + g L R ) < 1, this time of existence depends only on η = 1 f L R ) f L R ) g L R ). Indeed, in this case one can take µ = 1 and T 0 = +. Then, one can see easily that the choice of δ, ε, β and T depends only on µ, η and T Subcritical Case. Recall that by subcitical we mean that E 0 < 1, where { E 0 = fx) + gx) + e4πfx) 1 } dx. 4π R Let P) denote the following Cauchy problem u + ue 4πu = 0 P) u0, x) = fx) H 1 R ) t u0, x) = gx) L R ). Since the assumption E 0 < 1 implies f L < 1, it follows that the problem P) has an unique maximal solution u in the space E T where 0 < T + is the lifespan of u. We want to show that T = + which means that our solution is global in time. Assume that T <, then by the energy identity 1.3), we deduce sup ut,.) L R ) E 0 < 1. t 0,T ) Let 0 < s < T and consider the following Cauchy problem v + ve 4πv = 0 vs, x) = us, x) t vs, x) = t us, x). By a fixed point argument, as in the proof of Theorem 1, we can see that there exists a nonnegative τ and an unique solution v to our problem on the interval [s, s + τ]. Notice that τ does not depend on s see Remark 7).

13 GLOBAL SOLUTIONS FOR A SEMILINEAR D KLEIN-GORDON Choosing s close to T such that T s < τ we can prolong the solution u after the time T which is a contradiction. 4.. Critical Case. We consider now the critical case, namely E 0 = 1, and we want to prove a global existence result as in the previous section. The situation here is more delicate and the arguments used in the subcritical case do not apply here. Let us briefly explain what the major difficulty is. Since E 0 = 1 and by the conservation of the total energy 1.3), it is possible at least formally) that a concentration phenomena holds, namely lim sup t T ut,.) L = 1 where u is the maximal solution and T < + is the lifespan of u. In this case, we can not apply the previous argument to prolong the solution. The actual proof is based on proving that the concentration phenomenon does not happen. Proof of Theorem 3. Let u be the maximal solution of P) defined on [0, T ). Assume that T is finite and let us prove a contradiction. It is easy to see that sup ut,.) L = 1. 0 t<t Otherwise, there exists a real number 0 < η < 1 such that sup ut,.) L 1 η 0 t<t and hence we can prolong our solution arguing exactly in the same way as in the subcritical case. The rest of the proof is divided into several steps. First Step: We have the following proposition. Proposition 4. The maximal solution u satisfies: lim sup ut) L R ) = 1, 4.1) t T ut) t T 0 in L R ). 4.) Proof of Proposition 4: By the energy identity 1.3), we get { 0 t < T, ut, x) + t ut, x) + e4πut,x) 1 } dx = 1. 4π R

14 14 S. Ibrahim, M. Majdoub, and N. Masmoudi So, lim sup t T ut) L R ) 1. Suppose that lim sup t T ut) L R ) = L < 1. Then, a time t 0 exists such that 0 < t 0 < T and t 0 < t < T = ut) L R ) L + 1. On the other hand, by continuity, sup 0 t t 0 ut) L R ) = ut 1 ) L R ), 0 t 1 t 0. Since ut 1 ) L R ) < 1, we obtain sup ut) L R ) < 1 which is a 0 t<t contradiction. This concludes the proof of 4.1). To prove 4.), recall that u C T H 1 R )) CT 1 L R )). So, we can write t ut) us) L = t uτ) dτ s t s. It follows that ut)) admits a limit point ū in L R ) as t T, and therefore It remains to show that ū = 0. By 1.3) we have L R ut) L 1 = t ut) e 4πu 1 L dx. 4π Take the lim sup in both sides of this equality as t T, we get Fatou Lemma implies 0 = lim inf t T tut) L lim inf t T R e 4πū lim inf t T e4πu 1) = 0 and hence ū = 0. This ends the proof of Proposition π Second Step: Now, the idea is to show that, in fact, the concentration phenomena holds in the section of some backward cone of vertex z := T, x ). Since the equation E 4π ) is invariant under time translation, we can assume without loss of generality that T = 0 and the initial time is t = 1. This assumption will be made in all the sequel. We have the following. dx.

15 GLOBAL SOLUTIONS FOR A SEMILINEAR D KLEIN-GORDON Proposition 5. There exists a point x in R such that for all r > 0, we have lim sup ut) dx = ) t 0 x x r Proof of Proposition 5. By contradiction assume that for any x R, there exists two positive real numbers r x and η x such that lim sup eu)t, y) dy 1 η x. 4.4) t 0 y x r x In fact, one can choose these two numbers to be uniform with respect to x because otherwise, taking r n = η n = 1/n with n N, a sequence x n ) would exist such that lim sup eu)t, y) dy 1 1/n. t 0 y x n 1/n Now since the measure eu)t, y)dy is tight, one can extract a subsequence x ϕn) ) converging to a point x in R. So, for any r > 0 and n N lim sup eu)t, y)dy 1 /n t 0 y x r and therefore lim sup eu)t, y)dy = 1 t 0 y x r which contradicts our assumption. Now the idea is to show that in such a situation the solution u can be continued after the blow-up time T = 0. Let x R. Define the cut-off function ϕ x by 0 ϕ 1, ϕ x 1 in Bx, r/) and ϕ x 0 in Bx, r) c. Obviously, from 4.4) and Proposition 4, we have lim sup eϕ x u, ϕ x t u)t) dy 1 η. t 0 y x r Now choose a time t 1 > T r/8 such that eϕ x u, ϕ x t u)t 1 ) dy 1 η/. y x r From the local theory Theorem 1), one can solve E 4π ) with the initial data ϕ x ut 1, ), ϕ x ut 1, )) globally in time. By finite speed of propagation, we deduce that u can be continued in the backward light cone of vertex

16 16 S. Ibrahim, M. Majdoub, and N. Masmoudi x, t 1 + r/). Hence u is continued at least till the time T + r/8 which is a contradiction. A consequence of the Proposition 5 is Corollary 1. With the notations of Proposition 5 we have the following ut) dx = ) lim t 0 t < 0, x x t x x t eut)) dx = ) Proof of Corollary 1. Assume that x = 0. The proof of 4.5) is straightforward. Indeed, suppose that 4.5) is false. Then, there exists a sequence of negative real number t n ) tending to zero when n goes to infinity such that n N, ut n ) dx 1 η for some 0 < η < 1. x t n Then, arguing as in the proof of the previous proposition, we can prolong our solution which yields a contradiction. To prove 4.6), fix ε > 0. By 4.5), there exists a time t ε < 0 such that ut) dx 1 ε for t ε t < 0. By using the finite speed of x t propagation, we deduce that t < 0, x t eu)t) dx 1 ε. Letting ε go to zero, we obtain the desired result. End of the Proof of Theorem 3. Let S < T < 0, K T S be the truncated backward cone of vertex 0, 0) and M T S be its mantle. Multiplying the equation E 4π ) by t u and u yields t eu) div x t u u) = 0 4.7) t t u u) div x u u) + u t u + u e 4πu = 0 4.8) Integrating the conservation laws 4.7) and 4.8) over the backward truncated cone KS T, we obtain the following two identities: eut )) dx eus)) dx = 1 { t u x BT ) BS) MS T x + u + e4πu 1 } dσ. 4π 4.9)

17 GLOBAL SOLUTIONS FOR A SEMILINEAR D KLEIN-GORDON BT ) t ut ) ut ) dx + BS) K T S t us) us) dx + 1 M T S t u + u. x ) u dσ x { u t u + u e 4πu} dx dt = ) Notice that these identities are obtained by using an approximation argument as in [37]. Using 4.6), we deduce first that { t u x x + u + e4πu 1 } dσ = ) 4π M T S Since ut) 0 strongly in L and ut) L 1 as t goes to zero, it follows from the energy identity 1.3) that t ut) 0 strongly in L R ). Letting T go to zero in 4.10) and using 4.11), we get { t us) us) dx+ u t u +u e 4πu} dx dt = ) BS) Multiplying 4.1) by 1 S we deduce t us) us) S dx 1 u dx dt 1 S S BS) It is clear that and Moreover, K 0 S K 0 S K 0 S 1 u dx dt 1 S KS 0 as S 0 1 t u dx dt 0 S KS 0 as S 0. us) S = 1 S S 0 t uτ) dτ so, us) S ) is bounded in L R ) and hence t us) us) S dx 0 as S 0. BS) t u dx dt. 4.13) Taking the limit S 0 in 4.13), we get 0 1 which yields a contradiction. The proof of Theorem 3 is then completely achieved.

18 18 S. Ibrahim, M. Majdoub, and N. Masmoudi We end this section by giving a rapidly proof of Theorem 4. Proof of Theorem 4. We treat the case s = the general case is similar except some technical complications). Let E c v, t) = t vt,.) L + vt,.) H 1 where v = u and denotes any partial space derivative. By the energy estimate, we get t E c v, t) vu t, ) L + ve 4π u 1)t, ) L + vu e 4π u 1)t, ) L Using the fact that CE c v, t)c + e π ut, ) L ). e π ut, ) L C 1 + ut, ) C 1/4) β, β < 4, and the Gronwal s Lemma, we conclude the proof. Remark 8. Thanks to the injection H s R ) L R ), s > 1, and the following simple fact e 4πaw 1 = e 4πw 1)e 4πa 1)w + + e 4πw + 1), a N, the Cauchy problem E 4π )-1.1) is locally well-posed in H s R ) H s 1 R ) without any assumption on f L R ). 5. APPENDIX In this Appendix, we give a sketch of the proof of Proposition 3. We refer to [18] for more details and related inequalities. We start by showing the following inequality in the unit ball of R. Lemma 1. For any real number λ > π, a constant C λ exists such that, for any function u H0 1B1)) C 1/4 B1)), we have u L λ u L logc λ + Nu)), 5.1) where Nu) := u C 1/4 u. L Proof. Without loss of generality, we can normalize u L to be equal to 1. Let H0,rad 1 B1)) be the set of all decreasing and radially symmetric functions in H0 1B1)). For any parameter D 1, we denote by K D the closed convex subset of H0,rad 1 B1)) defined by K D = { u H 1 0,rad B1)); ur) 1 Dr1/4, r [0, 1] }.

19 GLOBAL SOLUTIONS FOR A SEMILINEAR D KLEIN-GORDON Consider the following problem of minimizing I[u] := u L B1)), among all the functions belonging to the set K D. This is a variational problem with obstacle. It is well known see for example, Kinderlehrer- Stampacchia [4] and L. E. Evans [8]) that it has a unique minimizer u which is in the Sobolev space W, B1)). Hence the following radially symmetric set O := {x B 1 : u x) > 1 D x 1/4 }. is open and u is harmonic in O. On the other hand, note that all harmonic functions on R can only coincide in a unique tangent point with the function r 1 Dr 1/4. Note that because of the boundary condition at r = 1, u cannot start to be harmonic near r = 0. Therefore there exists, a unique a ]0, 1[ such that u r) = 1 Dr 1/4 if r [0, a] 5.) u r) = 1 Da 1/4 ) log r if r [a, 1], log a 5.3) satisfying also the tangent condition a 1/4 = 1 Da1/4 loga 1/4 ). 5.4) In particular, note that u L = 1, u C 1/4 = D, and u L = π 4 D a 1/ π 1 Da1/4 ) loga). 5.5) loga) Substituting D from 5.4) into 5.5) and denoting by x := a 1/4 ]0, 1[, we get the following and Therefore where u L = π u C 1/4 = 1/ logx) 1 logx)) 1 x1 logx)). u L logc λ + Nu )) := Hx), 1/ logx) Hx) = πα 1 logx)) log 1 ) C λ + x. πα1/ logx))

20 0 S. Ibrahim, M. Majdoub, and N. Masmoudi Taking C λ = e in Hx), we see that Hx) goes to πα as x goes to 0. Hence, for any λ > 1 πα, there exists x λ > 0 such that λhx) 1, for any 0 < x < x λ and C λ e. Now, if x [x λ, 1], choosing the constant C λ > e big enough such that 1/ 1 logx λ )) logc λ) 1. we see that λhx) 1. Hence, by this choice of C λ, we see that λhx) 1 for all 0 < x 1. This achieves the proof of the lemma. Remark 9. Note that the limiting case λ = π is not allowed in 5.1) see [18]). Proof of Proposition 3. From the above result we derive the Proposition 3. Indeed, let u be a function in H 1 R ) C 1/4 R ), λ > π and 0 < µ 1. Fix a radially symmetric function ϕ in C0 B 4) satisfying 0 ϕ 1, ϕ 1 for r near 0, r ϕ 1 and ϕ 1. Define ϕ µ by ϕ µ x) = ϕ µ x ). Without loss of generality, we can assume that u L = u0). Note that in particular one has ϕ µ u C u 1/4 C 1/4 ϕ µ u) L u L + µ 4 u L + ϕ µ u ϕ µ udx. R Integrating by parts, ϕ µ u ϕ µ udx = 1 R ϕ µ R u dx = µ ϕ µ 8 R x) u dx. Hence, ϕ µ u) L u L + µ u L. Applying the result of Lemma 1 and using the fact that for any constant C > 0, the function x x logc λ + C x ) is increasing, the proof of Proposition 3 is achieved. Acknowledgments: S. I. would like to thank the Fields Institute & McMaster University for their support. He was supported by the NSERC grant M.M. is supported by the Laboratoire d EDP of the Faculty of Sciences of Tunis. N.M. is partially supported by an NSF grant and by an Alfred P. Sloan Fellowship.

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