Asymptotic Behavior of a Nonlocal KPP Equation with an Almost Periodic Nonlinearity
aa r X i v : . [ m a t h . A P ] J a n ASYMPTOTIC BEHAVIOR OF A NONLOCAL KPP EQUATION WITH ANALMOST PERIODIC NONLINEARITY
YAN ZHANG
Abstract.
We consider a space-inhomogeneous Kolmogorov-Petrovskii-Piskunov (KPP) equationwith a nonlocal diffusion and an almost-periodic nonlinearity. By employing and adapting thetheory of homogenization, we show that solutions of this equation asymptotically converge to itsstationary states in regions of space separated by a front that is determined by a Hamilton-Jacobivariational inequality. Introduction
The aim of this paper is to analyze the large space/long time asymptotic behavior of the nonlocalreaction-diffusion equation(1.1) u t ( x, t ) − Z J ( y )[ u ( x − y, t ) − u ( x, t )] dy − f ( x, u ) = 0 , where J is a continuous, nonnegative, compactly supported, and symmetric kernel, and f is amonostable (KPP) type nonlinearity in u for which the canonical example is f ( u ) = u (1 − u ). Tostudy the asymptotic behavior of (1.1), we introduce the “hyperbolic” scaling ( x, t ) ( ǫ − x, ǫ − t ) . As ǫ →
0, the time scaling reproduces long-time behavior of (1.1), while the space scaling reproducesin bounded sets behavior for large space variables. The new unknown is now given by u ǫ ( x, t ) := u ( ǫ − x, ǫ − t ) . We introduce an initial condition u ǫ ( · ,
0) = u ( · ), and we can easily see that u ǫ satisfies(1.2) u ǫt ( x, t ) − ǫ Z J ( y )[ u ǫ ( x − ǫy, t ) dy − u ǫ ( x, t )] dy − ǫ f (cid:16) xǫ , u ǫ (cid:17) = 0 in R n × (0 , ∞ ) ,u ǫ ( x,
0) = u ( x ) . The behavior of u ǫ as ǫ → u ǫ , it is necessary to make assumptions about theoscillatory behavior of f , and in this paper we assume that f is an almost-periodic function in the xǫ variable. Our main result, Theorem 3.1, states that as ǫ → u ǫ respectively converges to thetwo equilibria of f , which for simplicity we take to be constant, in the two regions { φ < } andint( { φ = 0 } ), where φ is the solution of the Hamilton-Jacobi variational inequality(1.3) max( φ t + H ( Dφ ) , φ ) = 0 in R n × (0 , ∞ ) φ = (cid:26) G × { }−∞ on R n \ G × { } .G is the support of u , and H ( p ) is an “effective Hamiltonian” resulting from the homogenizationof (1.2). This behavior was shown for a nonlocal equation very close to (1.1) that models thepropagation of an invasive species in ecology by Perthame and Souganidis in [24], and similarasymptotic behavior was found for a non-local Lotka-Volterra equation by Barles, Mirrahimi, andPerthame in [8]. Date : September 2, 2015.2010
Mathematics Subject Classification.
Key words and phrases.
Nonlocal KPP equation, asymptotic behavior, homogenization. ecause the behavior of the solutions of (1.1) consists of two equilibrium states joined togetherby a transition layer near the interface defined by (1.3), and the effective Hamiltonian H ( p ) canbe interpreted as the propagation speed of this interface, our work is connected with the well-studied areas of traveling wave solutions of the KPP equation and the speed of their associatedtraveling fronts. Recent articles concerning these aspects of nonlocal KPP equations include thoseby Coville, D´avila, and Mart´ınez, who in [10] and [11] studied (1.1) in the case where f is periodicin x . They showed that there exists a critical speed which is the lowest speed for which there existsa pulsating front solution of (1.1). The existence of traveling wave solutions and of a critical speedwas considered for a non-local KPP equation similar to (1.1) by Berestycki, Nadin, Perthame, andRyzhik in [9]. Lim and Zlatos in [21] gave conditions on the inhomogeneity of f in order to proveexistence or non-existence of transition fronts for (1.1), where they also studied the range of speedsfor which transition fronts exist.The local version of (1.1), i.e. the equation where the integral term is replaced by a uniformlyelliptic second-order operator, has been studied extensively. Its rescaled form reads(1.4) u ǫt − ǫa ij (cid:16) x, xǫ (cid:17) u ǫij + ǫ − f (cid:16) x, xǫ , u ǫ (cid:17) = 0 . It was originally studied in the 1930’s by Fisher in [16] and by Kolmogorov, Petrovskii, andPiskunov in [20]. Freidlin in [17] studied the behavior of (1.4) using probabilistic methods forthe xǫ -independent problem. Evans and Souganidis in [14] extended [17] and introduced a differentapproach based on PDE methods which has proven to be more flexible. The asymptotic behavior of u ǫ in the presence of periodic space-time oscillation was analyzed by Majda and Souganidis in [23].Our work is an extension of [14] and [23] to the nonlocal case. There is also a vast literature dealingwith the long-time behavior of (1.4), going back to the work of Aronson and Weinberger [5].Due to the presence of the oscillatory variable xǫ in (1.2), the theory of homogenization plays acrucial part in the analysis of this equation as ǫ →
0. The study of homogenization of Hamilton-Jacobi equations in periodic settings began with the work of Lions, Papanicolaou, and Varadhan[22], and homogenization for “viscous” Hamilton-Jacobi equations was studied by Evans [15] andMajda and Souganidis [23]. The fundamental tool in the periodic setting is the fact that it is possibleto solve the macroscopic problem, or “cell problem.” Homogenization in the almost-periodic casewas established by Ishii [18], who used the almost periodic structure to construct approximatecorrectors.Arisawa in [3] and [4] studied the periodic homogenization of integro-differential equations withL´evy operators, equations that are similar in structure to the ones we consider, and we employthe general ideas of her work. She considered the “ergodic problem”, which is the same as the cellproblem, and proved that approximate correctors exist by considering the limit along a subsequenceof a family of functions that satisfy an approximated cell problem and showing that the limitingequation satisfies a strong maximum principle. Since such a limiting equation and strong maximumprinciple are not available in our case, we will use more direct techniques based on an analysis ofthe nonlocal term to prove the existence of the approximate corrector, and we show that almost-periodicity provides enough of a “compactness” criterion in order to make these techniques work.The paper is organized as follows. In Section 2, we make precise our assumptions. In Section 3,we state our main result, Theorem 3.1, and we give a heuristic justification for it. In Section 4, wegive the proof of homogenization, and in Section 5, we finish the proof of Theorem 3.1 using thehomogenization result. 2.
Assumptions
We assume that f : R n +1 → R is smooth and has bounded derivatives. In particular, it satisfies(2.1) sup x ∈ R n , | u |≤ L {| D x,u f ( x, u ) | + | D x,u f ( x, u ) |} < ∞ for each L > , x,u denotes derivatives with respect to x and u ; in the rest of this paper D denotes derivativeswith respect to the space variable x . We also assume that f is of KPP type i.e. monostable in the u variable. That is, for every x ∈ R n , f satisfies(2.2) (cid:26) f ( x, u ) < u ∈ ( −∞ , ∪ (1 , ∞ ) ,f ( x, u ) > u ∈ (0 , , and(2.3) c ( x ) := ∂f∂u ( x,
0) = sup u> u − f ( x, u ) ≥ κ > . Because (2.1) implies that f ( x, u ) is smooth and has locally in u and globally in x bounded firstand second derivatives in both variables, we can see that c ( x ) is smooth, bounded, and Lipschitzcontinuous. Define K := max( k c ( x ) k ∞ , k Dc k ∞ ).Concerning the kernel J , we assume that(2.4) J is compactly supported in a set O ⊂ B (0 , ¯ r ) , J ≥ ,J ∈ C ( R n ) , J ( x ) = J ( − x ) for all x ∈ R n , R R n J ( y ) dy = ¯ J < ∞ , There exists r > J ( y ) ≥ A > B (0 , r ) . Concerning the initial condition u , we assume that(2.5) u ∈ C ( R n ) , ≤ u ≤ , and G := spt( u ) = { x | u ( x ) = 0 } is compact . We assume that the nonlinearity is almost-periodic, that is, we assume that the family(2.6) { c ( · + z ) : z ∈ R n } is relatively compact in BUC( R n ) . Note that the typical assumption of 1-periodicity is a specific case of almost-periodicity.3.
Main Result, Heuristic Derivation
We now state our main result.
Theorem 3.1.
Assume (2.1) - (2.6) . Then there exists a continuous function H : R n → R such thatas ǫ → , u ǫ → in { φ < } and u ǫ → in int { φ = 0 } locally uniformly, where φ is the uniquesolution of (1.3) . Next we explain in a heuristic way the origin of the variational inequality and why it controlsthe asymptotic behavior of the u ǫ . Following the work for local KPP equations mentioned in theintroduction, we now use the classical Hopf-Cole transformation(3.1) u ǫ = exp( ǫ − φ ǫ ) . It is immediate that for t = 0, φ ǫ = −∞ on R n \ G and φ ǫ → G as ǫ → . The interestingpart of the transformation comes into play for t >
0. We can see via straightforward calculationsthat φ ǫ solves φ ǫt ( x, t ) + ¯ J − Z J ( y ) exp (cid:18) φ ǫ ( x − ǫy, t ) − φ ǫ ( x, t ) ǫ (cid:19) dy − u ǫ f (cid:16) xǫ , u ǫ (cid:17) = 0 , an equation which can be analyzed using homogenization techniques. We assume that φ ǫ admitsthe asymptotic expansion φ ǫ ( x, t ) = φ ( x, t ) + ǫv ( xǫ ) + O ( ǫ ) . Writing z = xǫ and performing a formalcomputation, we obtain φ t + ¯ J − Z J ( y ) exp (cid:18) φ ( x − ǫy, t ) − φ ( x, t ) ǫ + v ( z − y ) − v ( z )) (cid:19) dy − u ǫ f ( z, u ǫ ) = 0 . Formally, we can say that as ǫ → ǫ − ( φ ( x − ǫy, t ) − φ ( x, t )) → − y · Dφ ( x, t ) . In addition, if u ǫ → ǫ →
0, then(3.2) ( u ǫ ) − f ( z, u ǫ ) → ∂f∂u ( z,
0) = c ( z ) . riting p = Dφ ( x, t ), we see that oscillatory behavior disappears in the limit as ǫ → H ( p ) and a function v that solves(3.3) ¯ J − Z J ( y ) exp( − y · p ) exp( v ( z − y ) − v ( z )) dy − c ( z ) = H ( p ) , which is a typical macroscopic problem or “cell problem” from homogenization theory. The issueis to find H ( p ), referred to as the effective Hamiltonian, so that (3.3) admits a solution v , typicallyreferred to as a “corrector,” with appropriate behavior at infinity i.e. strict sublinearity, so that H ( p ) is unique. If an effective Hamiltonian and a corresponding corrector can be found, then wesee that φ ǫ converges to a function φ that satisfies φ t + H ( Dφ ) = 0 , provided that we also ensurethat φ < u ǫ → φ shouldsatisfy the Hamilton-Jacobi variational inequalitymax( φ t + H ( Dφ ) , φ ) = 0 , which combined with the initial condition at t = 0 is precisely (1.3). Then (3.1), the fact that φ ǫ → φ , and an additional argument, found in Section 5, to show that u ǫ → { φ = 0 } imply that u ǫ satisfies the behavior described by Theorem 3.1.4. Proof of Homogenization
We proceed to prove Theorem 3.1 rigorously. Our primary result in this section is the homoge-nization of (4.1), that is, we show that solutions φ ǫ of(4.1) φ ǫt + ¯ J − Z J ( y ) exp (cid:18) φ ǫ ( x − ǫy, t ) − φ ǫ ( x, t ) ǫ (cid:19) dy − f (cid:0) xǫ , exp( ǫ − φ ǫ ) (cid:1) exp( ǫ − φ ǫ ) = 0 on R n × (0 , ∞ ) φ ǫ = ǫ log( u ) on int( G ) × { } φ ǫ ( x, t ) → −∞ as t → x ∈ R n \ G converge locally uniformly to φ , the solution of the homogenized equation (1.3). Theorem 4.1.
Under the assumptions (2.1) - (2.6) , φ ǫ converges locally uniformly to φ as ǫ → on R n × (0 , ∞ ) . Our first objective is to find H ( p ) such that the cell problem (3.3) admits “approximate correc-tors” v + , v − that satisfy (4.6) and (4.7) respectively, as the existence of approximate correctors issufficient to prove homogenization. The proofs in the almost-periodic and periodic cases are verysimilar, so we will present the proof in the almost periodic case and explain how the proof differsin the periodic case.We start by making the typical approximation to the cell problem (see [22]) and consider thefollowing equation in R n for λ > λv λ ( z ) + ¯ J − Z J ( y ) exp( − y · p ) exp( v λ ( z − y ) − v λ ( z )) dy − c ( z ) = 0 . First we need to show that this problem is well-posed. The proof follows along similar lines of othercomparison proofs (see [1], [2], [6], [7], [12], [19]).
Proposition 4.2.
Let u ( z ) ∈ USC( R n ) be a bounded subsolution of (4.2) , and let v ( z ) ∈ LSC( R n ) be a bounded supersolution of (4.2) . Then u ≤ v in R n . In addition, there exists a unique boundedcontinuous solution of (4.2) .Proof. We first prove that comparison holds. Assume for a contradiction that M := sup z ∈ R n u ( z ) − v ( z ) > . Then define M δ := max z ∈ R n u ( z ) − v ( z ) − δ | z | . ote that this quantity is positive for δ sufficiently small. Because of our assumption that u, v arebounded, there exists a point z δ such that M δ = u ( z δ ) − v ( z δ ) − δ | z δ | . We can deduce that(4.3) lim δ → M δ = M, lim δ → δ | z δ | = 0 . Because u, v are respectively a subsolution and a supersolution of (4.2), we obtain λu ( z δ ) + ¯ J − Z J ( y ) exp( − y · p ) exp( u ( z δ − y ) − u ( z δ )) dy − c ( z δ ) ≤ λv ( z δ ) + ¯ J − Z J ( y ) exp( − y · p ) exp( v ( z δ − y ) − v ( z δ )) dy − c ( z δ ) ≥ . Subtracting the second inequality from the first, we have(4.4) λ ( u ( z δ ) − v ( z δ )) − Z J ( y ) exp( − y · p )[exp( u ( z δ − y ) − u ( z δ )) − exp( v ( z δ − y ) − v ( z δ ))] dy ≤ . We know that for any y ∈ R n ,(4.5) u ( z δ − y ) − u ( z δ ) ≤ v ( z δ − y ) − v ( z δ ) − δ [2 z δ · y − | y | ] . (4.3) implies that as δ →
0, for any y ∈ R n , δ [2 z δ · y − | y | ] →
0. Therefore, because u, v arebounded, and y is contained in a ball B (0 , ¯ r ) in the integral term of (4.4), we can apply (4.5) to(4.4) and take the limit δ → λ ( u ( z δ ) − v ( z δ )) ≤ o δ (1) , but this is a contradiction because theleft hand side is uniformly positive by (4.3). Therefore, M ≤
0, which was what we wanted to show,and this completes the proof of comparison/uniqueness. The existence of a bounded continuoussolution given a comparison principle is a consequence of Perron’s method, and follows in the sameway as the analogous result in [1]. (cid:3)
The next proposition, which shows that there exist approximate correctors to the cell problem,is the main objective of this section. It is similar to the analogous one found in [3]. In that workArisawa considers the “ergodic problem”, which is essentially the statement of Proposition 4.4, fora different nonlocal equation, a periodic integro-differential equation containing a L`evy operator,that bears resemblance to (4.2). In our proof, we will employ some techniques from [3] along withsome new ones involving an analysis of the nonlocal term of (4.2).For the almost periodic setting, we introduce the concept of “uniform almost periodicity”.
Definition 4.3.
The collection of functions { f s } s ∈ I for I an arbitrary index set is uniformlyalmost periodic in s if given any sequence { x j } ∈ R n there exists a subsequence, also called { x j } for convenience, such that for any ǫ > N such that for all s ∈ I and all j, k ≥ N , k f s ( x j + · ) − f s ( x k + · ) k ∞ < ǫ. This definition means that for { f s } the almost periodicity condition (2.6) holds uniformly in s .This concept will be used to give sufficient “compactness” for the almost periodic setting in orderto apply the techniques that are applicable to the periodic setting. Proposition 4.4.
Assume (2.1) - (2.4) and (2.6) . For any p ∈ R n , there exists a unique H ( p ) suchthat for each ν > there exist bounded, Lipschitz continuous functions v + , v − solving ¯ J − Z J ( y ) exp( − y · p ) exp( v + ( z − y ) − v + ( z )) dy − c ( z ) ≤ H ( p ) + ν (4.6) ¯ J − Z J ( y ) exp( − y · p ) exp( v − ( z − y ) − v − ( z )) dy − c ( z ) ≥ H ( p ) − ν, . (4.7) In addition, (4.8) lim λ ↓ λv λ ( z ) = − H ( p ) niformly in R n , where v λ is the solution to (4.2) .Proof. We first show that if there exists a constant H ( p ) such that for every ν >
0, there existfunctions v + , v − satisfying Proposition 4.4, then H ( p ) is unique. This argument was originallyfound in [22]. Lemma 4.5.
Suppose that there exists H ( p ) such that for any ν > there exist bounded, Lipschitzfunctions v + , v − satisfying (4.6) and (4.7) respectively. Then H ( p ) is unique.Proof. Suppose for a contradiction that there exists
A < B such that for any ν > v + ν , v − ν that satisfy the following for all z ∈ R n :¯ J − Z J ( y ) exp( − y · p ) exp( v + ν ( z − y ) − v + ν ( z )) dy − c ( z ) ≤ A + ν ¯ J − Z J ( y ) exp( − y · p ) exp( v − ν ( z − y ) − v − ν ( z )) dy − c ( z ) ≥ B − ν. Fix a sufficiently small ν such that B − ν > A + 2 ν . Because v + ν and v − ν are bounded, we can adda constant to v + ν to ensure that v + ν > v − ν on R n . In addition, for ǫ sufficiently small,(4.9) ǫv − ν + ¯ J − Z J ( y ) exp( − y · p ) exp( v − ν ( z − y ) − v − ν ( z )) dy − c ( z ) ≥ B − ν> A + 2 ν ≥ ǫv + ν + ¯ J − Z J ( y ) exp( − y · p ) exp( v + ν ( z − y ) − v + ν ( z )) dy − c ( z )holds for all z ∈ R n . Now comparison, which can be applied for (4.9) because v + ν and v − ν arebounded and Lipschitz continuous, now implies that v − ν ≥ v + ν . This is a contradiction, and thus B = A and so H is unique. (cid:3) Now we proceed with proving that there exists such a constant H ( p ). Let z ∈ R n be fixed, anddefine w λ ( z ) := v λ ( z ) − v λ ( z ) and C λ := λ k w λ k ∞ . We know that C λ is bounded due to comparisonfor (4.2) between λw λ and constant functions depending on sup R n | c ( · ) | . We claim that if C λ → λ →
0, then the proposition follows. This is true because k λv λ ( z ) − λv λ ( z ) k ∞ = λ k w λ k ∞ → . Because λv λ ( z ) is bounded in λ , there exists a subsequence such that we can define H ( p ) :=lim λ → − λv λ ( z ) , such that (4.8) holds. Now we can see that upon taking λ sufficiently small sothat k v λ − H ( p ) k ∞ < ν , v λ satisfies (4.6) and (4.7). Then Lemma 4.5 allows us to finish the proofof Proposition 4.4 in this case.Therefore, suppose for a contradiction that lim inf λ → C λ >
0. Since C λ is uniformly bounded, wecan extract a sequence λ n → n →∞ C λ n = C ′ > . We will subsequently call thissubsequence λ for convenience. Now define˜ w λ ( z ) = w λ ( z ) k w λ k ∞ . Then we have that upon writing ˜ c ( z ) = c ( z ) − λv λ ( z ), ˜ w λ satisfies(4.10) λ ˜ w λ + λC λ ¯ J − λC λ Z J ( y ) exp( − y · p ) exp (cid:18) C λ λ ( ˜ w λ ( z − y ) − ˜ w λ ( z )) (cid:19) dy − λC λ ˜ c ( z ) = 0 . Our objective is to show that ˜ w λ converges uniformly to zero. Assume for a contradiction thatthere exist sequences λ j , z j such that λ j → w λ j ( z j ) → δ = 0 , and suppose without loss ofgenerality that δ is positive. We claim that there exists a point ˆ z such that for all j sufficientlylarge,(4.11) ˜ w λ j (ˆ z ) ≥ δ . n the case where c ( z ) is periodic, ˜ w λ is also periodic, and then a point ˆ z satisfying (4.11) can befound by compactness because the sequence { z j } can be taken to lie in the unit cube. In the casewhere c ( z ) is almost-periodic, we use the fact that the family { ˜ w λ } λ ≤ is uniformly almost periodicin λ in the sense of Definition 4.3. This is true because by separated z dependence and comparisonfor (4.10), we know that for any x , x ∈ R n and any λ ≤
1, there exists a uniform constant C suchthat | ˜ w λ ( · + x ) − ˜ w λ ( · + x ) | ≤ C | ˜ c ( · + x ) − ˜ c ( · + x ) | , and now because ˜ c ( z ) is uniformly almost periodic in λ , which follows from the fact that c ( z ) is analmost periodic function, we have that the family { ˜ w λ } λ ≤ is uniformly almost periodic.Because { ˜ w λ } λ ≤ is uniformly almost periodic, we can extract a subsequence of { z j } , also called { z j } , and take N sufficiently large so that(4.12) | ˜ w λ ( z j + z ) − ˜ w λ ( z k + z ) | ≤ δ j, k ≥ N , all z ∈ R n , and for all λ ≤
1. Now if we fix k ≥ N , then for j sufficiently large,the fact that ˜ w λ ( z j ) → δ and (4.12) applied with z = 0 implies ˜ w λ j ( z k ) ≥ δ , so z k is a point ˆ z thatsatisfies (4.11).Now we use (4.11) and ˜ w λ ( z ) = 0 to reach a contradiction. Note that since the integrand ofthe nonlocal term is always nonnegative, we can restrict our integration domain to suitable subsetswhen seeking lower bounds. We consider (4.10) as λ j →
0. We have that λ ˜ w λ → k ˜ w λ k ∞ = 1, and there exists a constant C < ∞ such that (cid:13)(cid:13)(cid:13)(cid:13) λ ˜ c ( z ) C λ (cid:13)(cid:13)(cid:13)(cid:13) ∞ = (cid:13)(cid:13)(cid:13)(cid:13) C λ ( λc ( z ) − λ v λ (0)) (cid:13)(cid:13)(cid:13)(cid:13) ∞ ≤ C because C λ → C ′ > λ →
0. Therefore we consider the nonlocal second term W λ ( z ) := − λC λ Z J ( y ) exp( − y · p ) exp (cid:18) C λ λ ( ˜ w λ ( z − y ) − ˜ w λ ( z )) (cid:19) dy. If we consider the line between ˆ z and z and cover it with M := | ˆ z − z | r balls of radius r , thenbecause ˜ w λ j increases by at least δ on that line from z to ˆ z , then there exists x j ∈ R n such thatosc B ( x j , r ) ˜ w λ j ≥ δ := δ M ,
Write A j = B ( x j , r ), and consider x j,min , x j,max ∈ A j to be the points respectively where ˜ w λ j isminimized and maximized over A j . Then we know that ˜ w λ j ( x j,max ) − ˜ w λ j ( x j,min ) ≥ δ . In addition,we have due to comparison for (4.10) and the separated z dependence, ˜ w λ is Lipschitz continuouswith constant K = KC ′ for all λ sufficiently small. This gives us(4.13) ˜ w λ j ( x j,max − y ) − ˜ w λ j ( x j,min ) ≥ δ y ∈ B (0 , r ) , r := δ K . Finally, to obtain a contradiction, we consider W λ j ( z ) with z = x j,min , and we define A := B ( x j,min − x j,max , min( r , r − | x j,min − x j,max | )) . A is contained in B (0 , r ) by construction, andits radius is positive because | x j,min − x j,max | < r . In addition, for y ∈ A , (4.13) implies that˜ w λ j ( x j,min − y ) − ˜ w λ j ( x j,min ) ≥ δ . This gives us W λ j ( x j,min ) ≥ λ j C λ j Z A J ( y ) exp( − y · p ) exp (cid:18) C λ j λ j ( ˜ w λ j ( x j,min − y ) − ˜ w λ j ( x j,min )) (cid:19) dy ≥ C λ j Z A J ( y ) exp( − y · p ) exp (cid:18) C λ j δ λ j (cid:19) dy ≥ C λ j exp (cid:18) C λ j (cid:19) , here C , C > C λ j exp( C λ j ) is unbounded as λ j →
0, which means that W λ is unbounded, a contradiction to (4.10). Therefore, ˜ w λ converges uniformly to zero as λ → k ˜ w λ k ∞ = 1 for all λ by construction, a contradiction. Therefore, C λ →
0, and we have theexistence of H satisfying (4.8). In addition, given ν >
0, we also have the existence of bounded,Lipschitz continuous v + and v − satisfying (4.6) and (4.7) respectively, because we can simply take v + = v − = v λ for λ sufficiently small depending on ν . This concludes the proof of Proposition4.4 because this means that any convergent subsequence (in the uniform metric) of λv λ ( · ) mustconverge to − H ( p ), and so the full sequence λv λ ( x ) converges uniformly to − H ( p ), which is uniqueby Lemma 4.5. (cid:3) Remark . Arisawa in [3] proves that the analogue to ˜ w λ in her (periodic) setting convergesuniformly to zero by using the uniform equicontinuity of ˜ w λ to find a limiting function ˜ w alonga subsequence as λ → w λ → w solves an equation that has a strong maximum principle. The techniques demonstrated in theabove proof overcome the fact that in our case taking λ → u ǫ , the solution of (1.2), and φ ǫ , the solution of (4.1). Note that because φ ǫ is given by(3.1), since we know that (1.2) is well posed (see [1]), and in particular that a comparison principleholds, we know that a comparison principle holds for (4.1) as well. Lemma 4.7.
Assume that f satisfies (2.1)-(2.3), J satisfies (2.4), u satisfies (2.5), and ǫ < .Then(1) ≤ u ǫ ≤ on R n × [0 , ∞ ) .(2) For each compact subset Q of [ R n × (0 , ∞ )] ∪ [int( G ) × { } ] , there exists a constant C ( Q ) ,independent of ǫ , such that (4.14) | φ ǫ | ≤ C ( Q ) on Q. Proof.
The first part is a consequence of comparison for (1.2) and the fact that the constantfunctions 0 and 1 are respectively a subsolution and a supersolution of (1.2). To prove the secondpart, note that it suffices to show a lower bound because u ǫ ≤ φ ǫ ≤
0. To do this,we adapt the argument from Lemma 2.1 of [14]. First, we can assume without loss of generalitythat there exists an
R > B (0 , R ) ⊂ int( G ) and inf B (0 ,R ) u > . We first show that φ ǫ is bounded from below on B (0 , R ) × (0 , ∞ ). To this end, define the function ϕ : R n × [0 , ∞ ) → R ∪ {−∞} by ϕ ( x, t ) := ( | x | − R − αt − β if ( x, t ) ∈ B (0 , R ) × [0 , ∞ ) −∞ if ( x, t ) ∈ R n \ B (0 , R ) × [0 , ∞ ) , where α, β are positive constants to be chosen. We can now compute ϕ ,t + ¯ J − Z J ( y ) exp (cid:18) ϕ ( x − ǫy, t ) − ϕ ( x, t ) ǫ (cid:19) dy − f (cid:0) xǫ , exp( ǫ − ϕ ) (cid:1) exp( ǫ − ϕ ) ≤ − α + ¯ J − Z J ( y ) exp (cid:18) ϕ ( x − ǫy, t ) − ϕ ( x, t ) ǫ (cid:19) dy − κ ≤ − α + ¯ J − κ, where the second inequality follows due to (2.3). Therefore, upon taking α sufficiently large, wecan make the right hand side negative on B (0 , R ) × (0 , ∞ ). If we take β = log(inf B (0 ,R ) u ) , thenwe have that ϕ ≤ φ ǫ on B (0 , R ) c × (0 , ∞ ) ∪ B (0 , R ) × { } . Now comparison for (4.1) implies that ϕ ≤ φ ǫ on B (0 , R ) × (0 , ∞ ), which means that(4.15) | φ ǫ | ≤ C on B (cid:18) , R (cid:19) × (0 , ∞ ) e now provide a lower bound for the points ( x, t ) ∈ R n × (0 , ∞ ) such that | x | > R . Define ϕ ( x, t ) := ( − γ | x | t − σt − τ if t > −∞ if t = 0 , where γ, σ, τ are positive constants to be determined. We can compute ϕ ,t + ¯ J − Z J ( y ) exp (cid:18) ϕ ( x − ǫy, t ) − ϕ ( x, t ) ǫ (cid:19) dy − f (cid:0) xǫ , exp( ǫ − ϕ ) (cid:1) exp( ǫ − ϕ ) ≤ − σ + γ | x | t − Z J ( y ) exp (cid:18) − γ ( − x · y ) + ǫ | y | ) t (cid:19) dy − κ = − σ − (cid:18)Z J ( y ) exp (cid:18) γ ( x · y ) − γǫ | y | ) t (cid:19) dy − γ | x | t (cid:19) − κ = − σ − Z J ( y ) exp (cid:18) γ (( x − ǫy ) · y ) t (cid:19) dy − (cid:18) √ γ | x | t (cid:19) ! − κ ≤ R n \ B (0 , R × (0 , ∞ )] . We justify the last inequality. It suffices to show that(4.16) Z J ( y ) exp (cid:18) γ (( x − ǫy ) · y ) t (cid:19) dy − (cid:18) √ γ | x | t (cid:19) ≤ . By (2.4) there exists r > Z B (0 , R ) \ B (0 ,r ) J ( y ) = C > , so by the symmetry of J , we know that there is a positive constant C such that Z B (0 , R ) \ B (0 ,r ) J ( y )(( x − ǫy ) · y ) t dy ≥ C | x | t . Therefore, if we take γ sufficiently large, we have that (4.16) is satisfied, and thus ϕ is a subsolutionof (4.1). Select τ to be larger than the constant from (4.15), and define ˆ φ ǫ ( x, t ) := φ ǫ ( x, t + ξ ) for ξ >
0. Then we have that ϕ ≤ ˆ φ ǫ on (cid:20) B (cid:18) , R (cid:19) × [0 , ∞ ) (cid:21) ∪ (cid:20) B (cid:18) , R (cid:19) c × { } (cid:21) . This means that we can apply comparison to conclude that ϕ ≤ ˆ φ ǫ on B (0 , R ) c × (0 , ∞ ), andtaking ξ → (cid:3) We will now use the perturbed test function method (see [15], [4]) to prove Theorem 4.1.
Proof of Theorem 4.1.
For each ( x, t ) ∈ R n × (0 , ∞ ), we define(4.17) φ ∗ ( x, t ) = lim sup ǫ → , ( x ′ ,s ) → ( x,t ) φ ǫ ( x ′ , s ) , φ ∗ ( x, t ) = lim inf ǫ → , ( x ′ ,s ) → ( x,t ) φ ǫ ( x ′ , s )to be the half-relaxed upper and lower limits (see [12]); note that the local uniform bounds on φ ǫ from Lemma 4.7 implies that φ ∗ ( x, t ) , φ ∗ ( x, t ) ∈ R for all ( x, t ) ∈ R n × (0 , ∞ ). We first show that φ ∗ is a solution of(4.18) max( φ ∗ t + H ( Dφ ∗ ) , φ ∗ ) ≤ R n × (0 , ∞ ) . Because φ ǫ ≤ φ ∗ t + H ( Dφ ∗ ) ≤ R n × (0 , ∞ ) n the viscosity sense. Take a smooth test function ϕ and a point ( x , t ) ∈ R n × (0 , ∞ ) suchthat ( x, t ) φ ∗ ( x, t ) − ϕ ( x, t ) has a strict global maximum at ( x , t ) (see [12]). Assume fora contradiction that ϕ t ( x , t ) + H ( Dϕ ( x , t )) = θ > . Set p := Dϕ ( x , t ), and define theperturbed test function(4.20) ϕ ǫ ( x, t ) := ϕ ( x, t ) + ǫv − (cid:16) xǫ ; p (cid:17) where v − is given by Proposition 4.4 for ν sufficiently small to be determined. We claim that for r, ǫ sufficiently small, the following holds in the viscosity sense:(4.21) ϕ ǫt + ¯ J − Z J ( y ) exp (cid:18) ϕ ǫ ( x − ǫy, t ) − ϕ ǫ ( x, t ) ǫ (cid:19) dy − c (cid:16) xǫ (cid:17) ≥ θ B ( x , r ) × ( t − r, t + r ) . To show (4.21), select another smooth test function ψ and a point ( x , t ) ∈ B ( x , r ) × ( t − r, t + r ) such that ( x, t ) ( ϕ ǫ − ψ )( x, t ) has a global minimum at ( x , t ). If we define η ( z, t ) := ǫ − ( ψ ( ǫz, t ) − ϕ ( ǫz, t )) , then we know that( z, t ) v − ( z ) − η ( z, t ) has a global minimum at (cid:16) x ǫ , t (cid:17) . In particular, we know that ϕ t ( x , t ) = ψ t ( x , t ), because v doesn’t depend on t . Now because v − is a viscosity solution of (4.7), η satisfies¯ J − Z J ( y ) exp( − y · p )) exp (cid:16) η (cid:16) x ǫ − y, t (cid:17) − η (cid:16) x ǫ , t (cid:17)(cid:17) dy − c (cid:16) x ǫ (cid:17) ≥ H ( p ) − ν = − ϕ t + θ − ν. We can write η (cid:16) x ǫ − y, t (cid:17) − η (cid:16) x ǫ , t (cid:17) = ψ ( x − ǫy, t ) − ψ ( x , t ) ǫ − ϕ ( x − ǫy, t ) − ϕ ( x , t ) ǫ , and as ǫ →
0, because ϕ is smooth, ǫ − ( ϕ ( x − ǫy, t ) − ϕ ( x , t )) → − y · Dϕ ( x , t ) , so then for r, ǫ, ν sufficiently small, we get¯ J − Z J ( y ) exp (cid:18) ψ ( x − ǫy, t ) − ψ ( x , t ) ǫ (cid:19) dy − c (cid:16) x ǫ (cid:17) ≥ H ( p ) − θ − ψ t + θ , which is exactly (4.21). Here we have used the fact that v − is Lipschitz continuous. On the otherhand, applying (2.3) to (4.1) yields that(4.22) φ ǫt + ¯ J − Z J ( y ) exp (cid:18) φ ǫ ( x − ǫy, t ) − φ ǫ ( x, t ) ǫ (cid:19) dy − c (cid:16) xǫ (cid:17) ≤ R n × (0 , ∞ ). Now because we have (4.21) and (4.22), we can usecomparison to conclude that for ǫ sufficiently small,max B ( x ,r ) × [ t − r,t + r ] ( φ ǫ − ϕ ǫ ) = max D ( φ ǫ − ϕ ǫ ) , where D = { B ( x , r ) c × [ t − r, t + r ] } ∪ { B ( x , r ) × { t − r }} . Upon taking ǫ →
0, this identitycontradicts our initial assumption that φ ∗ − ϕ has a strict global maximum at ( x , t ).It now remains to verify the initial condition; that is, we would like to show that(4.23) φ ∗ = (cid:26) G × { }−∞ on R n \ G × { } . This follows in the same way as in [23]. It is clear by (3.1) that for x ∈ G , φ ǫ ( x , → ǫ →
0, and so φ ∗ = 0 on G × { } . So it remains to show that φ ∗ = −∞ on R n \ G × { } . First weprove a preliminary claim. Fix µ > ζ satisfying ζ = 0 on G , ζ > R n \ G , ≤ ζ ≤ . We claim that(4.24) min( φ ∗ t + H ( Dφ ∗ ) , φ ∗ + µζ ) ≤ R n × { } olds in the viscosity sense. Suppose that ϕ is a smooth test function and φ ∗ − ϕ has a strict localmaximum at some ( x , ∈ R n × { } . If x ∈ G , then ζ ( x ) = 0 and (4.24) holds. Otherwise,suppose that x ∈ R n \ G , and that φ ∗ ( x , > − µζ ( x ) > −∞ . By definition of φ ∗ , there existpoints ( x ǫ , t ǫ ) such that ( x ǫ , t ǫ ) → ( x ,
0) and φ ǫ ( x ǫ , t ǫ ) → φ ∗ ( x , φ ǫ ( x,
0) = −∞ for all x near x , the points ( x ǫ , t ǫ ) must lie in R n × (0 , ∞ ), and so we can repeat the precedinghomogenization argument to show that ϕ ∗ t + H ( Dϕ ∗ ) ≤ x , x ∈ R n \ G , and suppose for a contradiction that φ ∗ ( x , > −∞ . Fix δ > ϕ δ ( x, t ) = δ − | x − x | + γt, for γ to be selected (in terms of δ ) below. Since φ ∗ isupper semicontinuous and bounded above, we know that φ ∗ − ϕ δ has a maximum at some point( x δ , t δ ) ∈ R n × [0 , ∞ ). This implies that(4.25) − δ − | x δ − x | ≥ φ ∗ ( x δ , t δ ) − ( δ − | x δ − x | + γt δ ) ≥ φ ∗ ( x , > −∞ . Now if t δ >
0, then we know that ϕ δt ( x δ , t δ ) + H ( Dϕ δ ( x δ , t δ )) ≤ , which means that(4.26) γ + H (cid:18) − x δ − x ) δ (cid:19) ≤ , but this is a contradiction upon taking γ = γ ( δ ) sufficiently large by (4.25). Therefore, t δ = 0. Nowif φ ∗ ( x , > − µζ ( x ), then since x δ → x by (4.25), then this means that φ ∗ ( x δ , > − µζ ( x δ )for δ sufficiently small. Therefore, by (4.24), we get (4.26), which is once again a contradiction.Therefore, we must have that φ ∗ ( x , ≤ − µζ ( x ). However, since ζ ( x ) > µ is arbitrary,then (4.25) cannot hold and we have another contradiction. This finishes our proof of (4.23).We can prove in a similar fashion that φ ∗ is a supersolution of (1.3), using v + instead of v − forthe perturbed test function (4.20); the proof differs at the point where we deduce an analogousstatement to (4.22). Due to the nature of the variational inequality (1.3), it suffices to prove that φ ∗ is a supersolution of (1.3) on { φ ∗ < } . In this case, instead of using (2.3), we use the factthat ( u ǫ ) − f ( z, u ǫ ) → c ( z ) on { φ ∗ < } , which follows from 3.1. Note that if φ ∗ = 0, then thepreceding statement would not hold, (2.3) would not give the correct inequality to prove that φ ∗ isa supersolution of φ t + H ( p ) = 0 if we attempted to duplicate the proof from the subsolution case.It is precisely at this point that the variational inequality (1.3) for φ is necessary.The proof also deviates from the subsolution case when we show that the initial condition (4.23)holds for φ ∗ . Because in this case we know that φ ∗ = −∞ on ( R n \ G ) × { } since φ ǫ = −∞ on that set, we need to show that φ ∗ = 0 on G × { } . Instead of (4.24), in this case we showthat max( φ ∗ ,t − H ( Dφ ∗ ) , φ ∗ ) ≥ G × { } , and in the proof we change the definition of ϕ δ to ϕ δ = − δ − | x − x | − γt. Because H satisfies (4.27), the result of [13] can be applied, whichmeans that comparison holds for (1.3), and so φ ∗ = φ ∗ = φ . This implies that φ ǫ converges locallyuniformly to φ , which was what we wanted. (cid:3) We now discuss the properties of the effective Hamiltonian H . In the case of a homogeneousnonlinearity f i.e. c ( z ) ≡ c , constant functions are correctors, so we can write the form of H ( p ) tobe H ( p ) = ¯ J − c − Z J ( y ) exp( − y · p ) dy. In particular, we can see that in this situation, the effective Hamiltonian is concave, negativelycoercive, and continuous in p , and we now prove that these properties of H ( p ) hold in general. Proposition 4.8.
The effective Hamiltonian H has the following properties:(1) p H ( p ) is concave.(2) There exist positive constants K , K , K , K > , C , C such that H ( p ) satisfies (4.27) − K exp( K | p | ) − C ≥ H ( p ) ≥ − K exp( K | p | ) − C for all p ∈ R n . In particular, this implies that H is uniformly and negatively coercive.
3) There exist constants C , C such that for all p , p ∈ R n , (4.28) | H ( p ) − H ( p ) | ≤ C exp( C (1 + | p | + | p | )) | p − p | . Proof.
Select p , p ∈ R d , and for each λ >
0, write v λi := v λ ( z, ω ; p i ). Set p := p + p and˜ v λ := ( v λ + v λ ). It is clear that by convexity of the exponential we have that ˜ v λ satisfies λ ˜ v λ + ¯ J − Z J ( y ) exp( − y · p ) exp(˜ v λ ( z − y ) − ˜ v λ ( z )) dy − c ( z ) ≥ . By comparison we have that v λ ( y ; p ) ≤ ˜ v λ ( y ). Multiplying this inequality by − λ and taking λ → H ( p ) ≥ ( H ( p ) + H ( p )) , which means that H is concave.To prove (4.27), we first note that since c ( z ) is bounded, we can find sufficiently large K , K , C so that − λ − ( K exp( K | p | ) + C ) is a subsolution of (4.2). This gives half of (4.27). To prove theother half, we note that since J is symmetric, there exists K , K such that Z J ( y ) exp( − y · p ) dy ≥ K exp( K | p | ) , and so this means that − λ − ( K exp( K | p | ) − C ) is a supersolution of (4.2). This gives us theother half of (4.27).To show (4.28), we prove a lemma giving a modulus of continuity estimate for v λ . Lemma 4.9.
There exist C , C > such that for each λ > , p , p ∈ R n (4.29) sup y ∈ R n | λv λ ( y ; p ) − λv λ ( y ; p ) | ≤ C exp( C (1 + | p | + | p | )) | p − p | . Proof.
Let v λ , v λ be the solution of (4.2) with p and p respectively. We claim that there existconstants C , C such that v λ ± C λ − exp( C (1+ | p | + | p | )) | p − p | are respectively a supersolutionand a subsolution of (4.2) with p , which by comparison for (4.2) implies (4.29). As the other casefollows similarly, we show here that˜ v := v λ + C λ − exp( C (1 + | p | + | p | )) | p − p | is a supersolution of (4.2) with p . Define A := C exp( C (1 + | p | + | p | )) | p − p | for notationalconvenience. We can compute λ ˜ v ( z ) + ¯ J − Z J ( y ) exp( − y · p ) exp(˜ v ( z − y ) − ˜ v ( z )) dy − c ( z )= λv λ ( z ) + A + ¯ J − Z J ( y ) exp( − y · p ) exp( v λ ( z − y ) − v λ ( z )) dy − c ( z )= λv λ ( z ) + A + ¯ J − Z J ( y ) exp( − y · ( p + ( p − p )) exp( v λ ( z − y ) − v λ ( z )) dy − c ( z ) ≥ λv λ ( z ) + A + ¯ J − Z J ( y ) exp( − y · p ) exp( v λ ( z − y ) − v λ ( z )) dy − c ( z )+ (1 − exp( K | p − p | )) Z J ( y ) exp( − y · p ) exp( v λ ( z − y ) − v λ ( z )) dy, (4.30)where K is a constant. Now we know that because v λ is a solution of (4.2), the quantity Z J ( y ) exp( − y · p ) exp( v λ ( z − y ) − v λ ( z )) dy is uniformly bounded in λ . Therefore, if we take appropriate constants C , C , then the last termof (4.30) is nonnegative, which shows that ˜ v is a supersolution of (4.2) with p , as desired.(4.28) now follows from Lemma 4.9 by taking λ → (cid:3) Note that these properties of H imply that (1.3) has a unique solution (see [12], [13]). . Convergence of u ǫ We will now prove Theorem 3.1. That is, we show that as ǫ → u ǫ , the solution of (1.2),converges locally uniformly to 0 and 1 in regions determined by φ , the solution of (1.3). This proofis based on [23] and [14]. Proof of Theorem 3.1.
We have shown that(5.1) φ ǫ → φ locally uniformly , and this combined with the Hopf-Cole transformation (3.1) implies u ǫ = exp( ǫ − φ ǫ ) = exp( ǫ − ( φ + o ǫ (1))) , so u ǫ → { φ < } . It remains to show that u ǫ → { φ = 0 } . Fix a point ( x , t ) ∈ int { φ = 0 } , and define ϕ ( x, t ) = −| x − x | − | t − t | . Then ifwe consider a domain B ( x , r ) × [ t − h, t + h ] ⊂ int { φ = 0 } for r, h sufficiently small, and define( x ǫ , t ǫ ) to be a point where φ ǫ − ϕ is minimized over this domain, (5.1) implies that ( x ǫ , t ǫ ) → ( x , t )as ǫ →
0. We now use ϕ as a test function in (4.1), which we can do for ǫ sufficiently small, to seethat at ( x ǫ , t ǫ ), ϕ t ( x ǫ , t ǫ ) + ¯ J − Z J ( y ) exp (cid:18) ϕ ( x ǫ − ǫy, t ǫ ) − ϕ ( x ǫ , t ǫ ) ǫ (cid:19) dy ≥ ( u ǫ ) − f ( z ǫ , u ǫ ) , where z ǫ := ǫ − x ǫ . We have that Z J ( y ) exp (cid:18) ϕ ( x ǫ − ǫy, t ǫ ) − ϕ ( x ǫ , t ǫ ) ǫ (cid:19) dy = Z J ( y ) exp( − y · ( x ǫ − x ) + ǫ | y | ) dy = ¯ J + o ǫ (1)as ǫ →
0. Therefore, we can conclude that as ǫ → f ( z ǫ , u ǫ ( x ǫ , t ǫ )) ≤ o (1) u ǫ ( x ǫ , t ǫ ) , but we also have by (2.3) that there exists a > z and all U ∈ [0 , f ( z, U ) ≥ − aU + c ( z ) U ≥ − aU + κU. In particular by (2.1) we can take a = 12 sup x ∈ R n ,u ∈ [0 , | f uu ( x, u ) | , and then if we consider f ( x, u ) = f ( x, u ) + au − κu , we have by (2.3) that f ( x,
0) = 0 , f u ( x, ≥ , for all x , and f uu ( x, u ) ≥ u ∈ [0 ,
1] and all x . Therefore f ( x, u ) ≥ u ∈ [0 ,
1] and all x ,and thus (5.3) is proved.(5.2) and (5.3) imply that o ǫ (1) u ǫ ( x ǫ , t ǫ ) ≥ − a ( u ǫ ( x ǫ , t ǫ )) + κu ǫ ( x ǫ , t ǫ ) . Because u ǫ is nonnegativeby Lemma 4.7, o ǫ (1) ≥ − au ǫ ( x ǫ , t ǫ ) + κ, and so(5.4) lim inf ǫ → u ǫ ( x ǫ , t ǫ ) ≥ κa > . However, since ( x ǫ , t ǫ ) is a local minimum of φ ǫ − ϕ , we have φ ǫ ( x ǫ , t ǫ ) − ϕ ( x ǫ , t ǫ ) ≤ φ ǫ ( x , t ) − ϕ ( x , t ) which means that ǫ log( u ǫ ( x ǫ , t ǫ )) − o ǫ (1) ≤ ǫ log( u ǫ ( x , t )) , and so due to (5.4), u ǫ ( x , t ) ≥ u ǫ ( x ǫ , t ǫ ) + o ǫ (1) ≥ κa > ǫ → , Therefore,(5.5) lim inf ǫ → u ǫ ≥ α > O ⊂ int { φ = 0 } for α = α ( O ).We now need to show that(5.6) lim ǫ → u ǫ = 1uniformly on compact subsets of int { φ = 0 } . First note that it suffices to consider the cylinder A = B ( x , r ) × ( t , t + h ) ⊂ int { φ = 0 } and to prove (5.6) in A ′ = B ( x , r ) × ( t + h , t + h ). We now that (5.5) holds uniformly on A ′ . Due to (2.1), (2.2), and (2.3), for any θ > β = β ( α, θ ) > f ( z, u ) ≥ β (1 − θ − u )for all u ∈ [ α ,
1] and for all z ∈ B ( x , r ). We can prove this by noting that because f ≥ u ∈ [0 , θ, β > f ( z, u ) := f ( z,u ) β − (1 − θ ) + u ≥ z ∈ B ( x , r ) and u ∈ [1 − θ, u ∈ [ α , − θ ].Then, because z lies in a compact set, then by (2.2) there exists ρ > f ( z, u ) ≥ ρ for all ( z, u ) ∈ B ( x , r ) × [ α , − θ ]. If we now take β = ρ , it is clear that (5.7) holds for( z, u ) ∈ B ( x , r ) × [ α , − θ ], thus giving us the inequality we need.Applying (5.7) to (1.2) gives us(5.8) u ǫt − ǫ Z J ( y ) ( u ǫ ( x − ǫy, t ) − u ǫ ( x, t )) dy + βǫ ( u ǫ − θ ) ≥ A. Now define f ǫ ( t ) = 1 − θ − e − δ ( t − t ǫ , g ǫ ( x, t ) = f ǫ ( t ) ψ ( x ) − ǫγ ( t − t ) , where ψ satisfies 0 ≤ ψ ≤ , ψ = 0 on B ( x , r ) c , ψ = 1 on B (cid:0) x , r (cid:1) . Using the fact that J issymmetric, we have that as ǫ → ǫ Z J ( y )( g ǫ ( x − ǫy, t ) − g ǫ ( x, t )) dy = ǫ Z J ( y ) (cid:18) g ǫ ( x + ǫy, t ) − g ǫ ( x, t ) + g ǫ ( x − ǫy, t ) ǫ (cid:19) dy ≥ − Cǫ Z J ( y ) | D g ǫ ( x, t ) | dy ≥ − Cǫ and we can compute g ǫt ( x, t ) + βǫ ( g ǫ ( x, t ) − θ ) = − ǫγ + δǫ e − δ ( t − t ǫ ψ ( x ) + βǫ ( f ǫ ( t ) ψ ( x ) − ǫγ ( t − t ) − θ ) ≤ δǫ [1 − θ − f ǫ ( t ) ψ ( x )] + βǫ ( f ǫ ( t ) ψ ( x ) − θ ) ≤ − ǫγ + β − δǫ ( f ǫ ( t ) ψ ( x ) − θ ) , and so g ǫt − ǫ Z J ( y )( g ǫ ( x − ǫy, t ) − g ǫ ( x, t )) dy + βǫ ( g ǫ − θ ) ≤ A, if we take γ > δ > g ǫ is a subsolutionof (5.8) in A . We have by construction that g ǫ ≤ R n × { t } ) ∪ ( B ( x , r ) c × ( t , t + h )), soby comparison we have that g ǫ ≤ u ǫ in A , but this means that since lim ǫ → g ǫ = 1 − θ in A ′ and θ is arbitrary, we have (5.6). This finishes the proof of Theorem 3.1. (cid:3) Acknowledgements
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