Propagation of solutions of the Porous Medium Equation with reaction and their travelling wave behaviour
aa r X i v : . [ m a t h . A P ] M a y Propagation of solutions of the PorousMedium Equation with reaction and theirtravelling wave behaviour byAlejandro G´arrizSeptember 11, 2018
Abstract
We consider reaction-diffusion equations of porous medium type, with dif-ferent kind of reaction terms, and nonnegative bounded initial data. For all thereaction terms under consideration there are initial data for which the solutionconverges to 1 uniformly in compact sets for large times. We will character-ize for which reaction terms this happens for all nontrivial nonnegative initialdata, and for which ones there are also solutions converging uniformly to 0.Problems in this family have a unique (up to translations) travelling wave witha finite front and we will see how its speed gives the asymptotic velocity of allthe solutions with compactly supported initial data. We will also prove in theone-dimensional case that solutions with bounded compactly supported initialdata converging to 1 do so approaching a translation of this unique travelingwave. We will prove a similar result for non-compactly supported initial datain a certain class.
Mathematics Subject Classification.
Keywords and phrases.
Reaction-diffusion, porous medium equation, travelling wavebehaviour. 1
Introduction and main results
The aim of this paper it to characterize the large time behaviour of solutions to theCauchy problem(1.1) u t = ∆ u m + h ( u ) in Q := R N × R + , u ( · ,
0) = u ≥ R N , where m >
1. The nonlinearity h is assumed to be in C ( R + ) and to fulfill, for some a ∈ [0 , h (0) = 0 , h ( u ) ≤ u ∈ [0 , a ] , h ( u ) > u ∈ ( a, , h ( u ) < u > . Let us also ask our reaction term to be “exponentially stable” in u = 1, that is,(1.3) h ′ (1) < . We are mainly interested in the case of bounded and compactly supported initialfunctions u , though in the one-dimensional scenario we will deal also with a differentclass of initial data.If a = 0, we are in the so-called monostable case, which includes the Pearl-Verhulst(or logistic) reaction nonlinearity h ( u ) = u (1 − u ). The case a > h ( u ) <
0, and thecombustion ones, when h ( u ) = 0 for u ≤ a (in this case a is the ignition temperature).Problem (1.1)–(1.2) may be used, for example, to describe the growth and prop-agation of a spatially distributed biological population whose tendency to dispersedepends on the population density; see for instance [12, 13]. It can also be regardedas a generalization of the semilinear case m = 1. When a = 0, this semilinear problemwas introduced to study spreading questions in population genetics independently byFisher [10], and by Kolmogorov, Petrovsky and Piscounov [17]; see also the work ofSanchez-Gardu˜no and Maini [23] for more applications. When a >
0, it was proposedby Zel’dovich to study combustion problems [26]; see also [16]. Finally, in the fieldof astronomy, our equation was proposed by Newman and Sagan as a model for thepropagation of intergalactic civilizations [18].The equation in (1.1) is degenerate when u = 0, and does not have in general classicalsolutions. A function u is a weak solution to problem (1.1) if u, ∇ u m , h ( u ) ∈ L ( Q ),and(1.4) Z R N u ϕ (0) + Z t Z R N (cid:0) uϕ t − ∇ u m · ∇ ϕ + h ( u ) ϕ (cid:1) = 0for each ϕ ∈ C ∞ c ( Q ). If the equality in (1.4) is replaced by “ ≤ ” (respectively, by “ ≥ ”)for ϕ ≥
0, we have a subsolution (respectively, a supersolution). Problem (1.1) has aunique bounded weak solution when the initial function is bounded. This solution iscontinuous in Q , and satisfies in addition that ∇ u m ∈ L ( Q ). There is moreover acomparison principle between bounded subsolutions and supersolutions [21, 19].2he large time behaviour will be given in terms of travelling wave solutions (abrevi-ated TW in what follows). By this we mean a solution of the form ¯ u c ( x, t ) = V c ( x − ct )for some speed c and profile V c (depending on c ), which should satisfy V c ∈ C ( R ),( V mc ) ′ , h ( V ) ∈ L ( R ), and Z R (( V mc ) ′ ϕ ′ + c V c ϕ ′ − h ( V c ) ϕ ) = 0for all ϕ ∈ C ∞ c ( R ).Two wave profiles are said to be indistinct if one is a translation of the other one.A monotonic travelling wave solution is called a change of phase type or a wavefrontfrom 1 to 0 if it connects the two equilibrium states 1 and 0, that is,lim ξ →−∞ V c ( ξ ) = 1 , lim ξ →∞ V c ( ξ ) = 0 . If the reaction (1.2) falls in the case a = 0, there exists a minimal speed c ∗ = c ∗ ( m, h ) > c ≥ c ∗ equation (1.1) has an unique distinct monotonicchange of phase type TW profile V c . There are no admissible TW solutions for c < c ∗ though.On the other hand if a > c ∗ for which the equation hasa wavefront. In this case the sign of c ∗ matches the sign of Z h ( u ) u m − du . Theasymptotic behaviour when c ∗ ≤ Z h ( u ) u m − du > . The TW with speed c ∗ satisfies V c ∗ < ξ suchthat V c ∗ ( ξ ) = 0 for all ξ ≥ ξ and V c ∗ ( ξ ) > ξ < ξ . Moreover, V ′ c ∗ < ξ < ξ , V mc ∗ ∈ C ( R ) and ( V mc ∗ ) ′ ( ξ ) → ξ → ±∞ .Finally, V c ∗ satisfies that(1.6) lim ξ → ξ − m V m − c ∗ ( ξ ) V ′ c ∗ ( ξ ) = lim ξ → ξ − (cid:18) mm − V m − c ∗ ( ξ ) (cid:19) ′ = − c ∗ . This is a hint of the importance of the pressure variable p = mm − u m − .From now on, V c ∗ will denote the TW profile with ξ = 0. Further information aboutthese results and many others can be found in [11] and the references therein.In the first part of this work we will generalize most of the results for m = 1 presentedby Aronson and Weinberger in [1] to the case of density dependent diffusion, i.e., m >
1. Given an initial data we will study whether the solution converges to 1, asituation that, following the literature, we will denote as spreading , or to 0, which wewill call vanishing . Our first result shows that for all the reaction nonlinearities thatwe are considering there are initial data for which spreading happens.3 heorem 1.1.
There exists a three parameter family of continuous, bounded andcompactly supported functions v ( x ; x , η, ρ ) (see Section 4 for a precise description)such that if u ( x, ≥ v ( x ; x , η, ρ ) for some x ∈ R N , η, ρ > , then u converges to 1 uniformly on compact sets. It turns out that for certain monostable nonlinearities h nontrivial solutions alwaysspread, independently of the mass of the initial datum. Following [1], this is namedas the hair-trigger effect . Theorem 1.2.
Let h satisfy (1.2) with a = 0 and lim inf u → h ( u ) u m +2 /N > . If u , then u converges to 1 uniformly on compact sets. Let us remark that if h ( u ) ≤ ku p , with p ≥ m + 2 /N and k a positive constant,comparison with the problem with reaction ku p shows that for certain small initialdata our solution asymptotically vanishes; see for instance [22].Now we ask ourselves how fast is the spreading. If we move from a point y ∈ R N ina certain direction with a slow speed c in the limit we will see only the value u = 1,while if c is too fast we will surpass the free boundary of our solution, and thus wewill only see the value u = 0. Too slow will translate to c < c ∗ , and too fast to c > c ∗ .In this sense, the speed c ∗ is called the critical speed of the problem. Theorem 1.3.
If the initial datum is bounded and compactly supported, given any y ∈ R N and c > c ∗ there is a value T such that u ( y, t ) = 0 for | y − y | ≥ ct, t ≥ T. In addition, if spreading happens, for any c ∈ (0 , c ∗ ) we have lim t →∞ min | y − y |≤ ct u ( y, t ) = 1 . This question has been recently analyzed in [2, 3] for a wider class of diffusionoperators, though with less general reaction terms. In particular, these papers require h to be concave in the region where it is positive, an assumption that we do not need.The results of this first part are presented in Section 4, after two sections of a moretechnical nature. In Section 2 we obtain an important estimate for the gradient of thepressure that plays a key role in the study of the asymptotic profile using Bernstein’smethod instead of the more common nowadays estimates of semiconcavity; see [20, 7].This is done in this way due to the wide class of reaction functions we are allowinghere. In Section 3 we construct two families of sub- and supersolutions that play aey role in this paper. 4he second part of the paper is devoted to study the asymptotic behaviour of solu-tions when they converge to 1. We manage to give uniform convergence in dimension N = 1 in moving coordinates to a travelling wave profile. Without loss of generality,we may assume that u ∈ C ( R ).We start by considering a class of initial data with unbounded support for whichthe analysis is similar, though slightly simpler, than the one for the class of boundedand compactly supported initial data.Thanks to (1.3) we know that there is a value δ > h ′ ( u ) < u ∈ (1 − δ, δ ). Let A be the class of nonnegative continuous functions u suchthat u ( x ) ≡ x ≥ x for some x ∈ R , lim inf x →−∞ u ( x ) ∈ (1 − δ, δ ).Solutions with initial data in this class have a right free boundary , ζ ( t ) ≡ inf { r ∈ R : u ( x, t ) = 0 for all x ≥ r } , and converge to the travelling wave V c ∗ both in shape and speed. Theorem 1.4.
Let u be a weak solution to (1.1) with u ∈ A , and let ζ be thefunction giving its right free boundary. Then there exists a ξ ∈ R such that (1.7) lim t →∞ sup x ∈ R | u ( x, t ) − V c ∗ ( x − c ∗ t − ξ ) | = 0 , lim t →∞ ζ ( t ) − c ∗ t = ξ . This theorem will be proved in Section 5. The case h ( u ) = u p (1 − u ) with p ∈ [1 , m ]was already considered in [4]; see also [7].We remark that one can weaken the hypothesis lim inf x →−∞ u ( x ) ∈ (1 − δ, δ ) incertain cases. For example, if the nonlinearity h is such that we have the hair-triggereffect, one only needs that lim inf x →−∞ u ( x ) >
0. Conditions leading to weakerassumptions are presented in Corollary 5.1.We arrive to our main result, a description of the large time behaviour for spreadingsolutions with bounded and compactly supported initial data in dimension N = 1.Let us describe the left and right free boundary as ζ + ( t ) ≡ inf { r ∈ R : u ( x, t ) = 0 for all x ≥ r } ,ζ − ( t ) ≡ sup { r ∈ R : u ( x, t ) = 0 for all x ≤ r } . Theorem 1.5.
Let u be a solution of Problem (1.1) corresponding to a bounded andcompactly supported initial data u such that u converges to 1 uniformly in compactsets, and let ζ ± be its left and right free boundaries. There exist constants ξ ± ∈ R such that lim t →∞ sup x ∈ R + | u ( x + c ∗ t, t ) − V c ∗ ( x − ξ + ) | = 0 , lim t →∞ ζ + ( t ) − c ∗ t = ξ + , (1.8) lim t →∞ sup x ∈ R − | u ( x − c ∗ t, t ) − V c ∗ ( ξ − − x ) | = 0 , lim t →∞ ζ − ( t ) + c ∗ t = ξ − . (1.9)This will be proved in Section 6. The case of higher dimensions, which involveslogarithmic corrections, has been recently studied for the case of the Fisher-KPP5eaction term h ( u ) = u (1 − u ) in [7]. Such corrections are also expected in higherdimensions for other nonlinearities. This will be considered elsewhere.In the semilinear case, m = 1, many similar results exist. When a = 0, apart fromthe seminal work of KPP in [17], Stokes introduced the terminology of pulled and pushed cases ( c ∗ = p h ′ (0) and c ∗ > p h ′ (0) respectively) and proved a similarresult for the pushed case; see [24] and also, for more results in the pushed case, thework of Uchiyama [25]. For the pulled case we mention the work of Bramson [5].When a >
0, Kanel [15] and Fife and McLeod [9] proved also similar results. Thispaper encompass most of these results for the case m > ( u m ) ′ In order to identify the asymptotic limits of our solutions as wavefronts, we need tocheck, following [11], that these limits satisfy V ∈ C ( R ) , ( V m ) ′ ∈ L loc ( R ) . To this aim we need an estimate for the flux ( u m ) ′ of bounded solutions dependingonly on their size. Such bound will follow from a bound for the derivative of thepressure. In many cases we can get this bound from a semiconvexity result for thepressure, in essence, ∆ p + Ψ( p ) ≥ , where p is the pressure related to our solution and Ψ is a function related to our reac-tion h ; see [20, 7]. This approach is fine for reactions like u (1 − u ), but unfortunatelywe need a different tool due to the many different reactions we are encompassing, sincesome of them cannot be treated in this way. We will follow the so called Bernstein’smethod, which requires some effort. Lemma 2.1.
Let u be a solution of equation (1.1) and suppose that u is essentiallybounded. The corresponding pressure p = mu m − / ( m − satisfies ( p x ) ( x, t ) ≤ m (cid:18) t + H (cid:19) || p ( x, || ∞ , for some constant H depending only on k u k ∞ .Proof. First, we define u ε ( x, t ) as the solution of the aproximated problem u ( x, t ) = ∆ u m + h ( u ) , u ( · ,
0) = u ε ( · )where { u ε } is a family of regular positive functions that converge uniformly in com-pacts to u such that 0 < ε ≤ u ε ≤ M . We do so to be able to apply the MaximumPrinciple. It is easy to see that ε ≤ u ε ( x, t ) ≤ M . Its corresponding pressure satisfies p t = ( m − p ∆ p + p x + h ( α ( p )) α ′ ( p )6here α ( p ) = (( m − p/m ) / ( m − is the density depending on the pressure. We omitthe dependence on ε for the sake of simplicity.For a fixed T > S T = R × (0 , T ] and N = sup x,t ∈ S T p ( x, t ). Now we choose aregular function θ defined in [0,1] onto [0 , N ] that is strictly increasing, concave andsuch that ( θ ′′ /θ ′ ) ′ ≤
0, but keep in mind that a precise description will be specifiedlater. If we define p = θ ( w ) our equation becomes w t = ( m − θ ∆ w + (cid:18) ( m − θ θ ′′ θ ′ + θ ′ (cid:19) w x + h ( α ( θ ( w ))) α ′ θ ′ . We take now a cutoff function ζ ∈ C ( ¯ S T ) ∩ C ∞ ( S T ) such that 0 ≤ ζ ≤ ζ ≡ t = 0 or | x | ≥ c > x , multiply it by w x ζ and consider a point ( x , t ) ∈ S T where the function z = w x ζ reaches a maximum.In this point we have that z t ≥ , z x = 0 , z xx ≤ z ≡ z t = 2 w xt w x ζ + 2 w x ζ ζ t ≥ ⇒ w xt w x ζ ≥ − w x ζ ζ t ,z x = 2 w xx w x ζ + 2 w x ζ ζ x = 0 ⇒ w xx = − w x ζ x ζ ,z xx = 2 { w xxx w x ζ + w xx ζ + 4 w xx w x ζ x ζ + w x ζ x + ζ xx w x ζ } ≤ ⇒⇒ w xxx w x ζ ≤ w x ζ x − ζ xx w x ζ . Thus (cid:16) − mθ ′′ − ( m − θ (cid:0) θ ′′ θ ′ (cid:1) ′ (cid:17) ζ w x ≤ − (cid:0) ( m + 1) θ ′ + 2( m − θ θ ′′ θ ′ (cid:1) ζ ζ x w x + (cid:0) ζ ζ t + 2( m − θζ x − ( m − θζ ζ xx + ζ ∂∂w (cid:0) hα ′ θ ′ (cid:1)(cid:1) w x . It is time now to focus on the last derivative that appears in the previous equation.One can check that ∂∂w (cid:18) h ( α ) α ′ θ ′ (cid:19) = h ′ + h ∂∂w (cid:18) α ′ θ ′ (cid:19) = h ′ − h · α ′′ ( θ ′ ) + h · α ′ θ ′′ ( α ′ ) ( θ ′ ) . All the quantities involved here are bounded, so the only possible problem arises when α ′ →
0, but bear in mind that θ ′ is strictly positive and α ( θ ) = C θ / ( m − , α ′ = C α − m , α ′′ = C α − m for C , C , C ∈ R , and thus even when α ′ → H , independent of ε but dependent on h and thus on k u k ∞ , such that ∂∂x (cid:18) h ( α ) α θ θ w (cid:19) w x ≤ H. Let us define a = max | ζ t | , a = max | ζ x | , a = max | ζ xx | , b , , , such that0 < N b ≤ θ ′ ≤ N b , θ ′′ ≤ − N b , | θ ′′ /θ ′ | ≤ b . Then we have that(
N mb ) ζ w x ≤ ( N ( m + 1) b + 2( m − N b ) a ζ | w x | +( a + 2( m − N a + ( m − N a + H ) w x , or, in other words, ζ w x ≤ c ( H ) w x + c ζ | w x | . Since for all δ > c ζ | w x | ≤ δζ w x + c w x / δ (to see this divide by w x and study the resulting parabola), we arrive to(1 − δ ) ζ w x ≤ (cid:18) c ( H ) + c δ (cid:19) w x and thus, for all ( x, t ) ∈ S T (2.1) z ( x, t ) ≤ max z ≤ − δ (cid:18) c + c δ (cid:19) . This bound depends on the arbitrary functions ζ and θ , so we fix a point ( x , t ) ∈ S T and procced to choose them in an appropriate way. We begin with ζ and choose ζ n ( x, t ) = tt ψ (cid:18) x − x n (cid:19) where ψ is a compactly supported regular function satisfying 0 ≤ ψ ≤ , ψ =1 if | x | ≤ ψ = 0 if | x | ≥
2. This way a n = 1 /t and a n , a n → n → ∞ , sopassing to the limit in (2.1) and substituting c and c for their values we get w x ( x , t ) = z ( x , t ) ≤ N mb t + HN mb which means, since x , t were arbitrary, that p x ( x, t ) ≤ N b mb t + N b Hmb = N b mb (cid:18) t + H (cid:19) . On the other hand, we can take θ ( w ) = N w ( a − bw ) with a ≥ b and a ≥ b + 1, andminimizing with respect to a and b we get a = 2 , b = 1. If we recover the subindex ε we arrive to ( p ε ) x ( x, t ) ≤ Nm (cid:18) t + H (cid:19) and with this we finish the proof by letting ε go to 0.With this result is easy to prove the following proposition. Proposition 2.1
Let u be a solution of equation (1.1) with u ∈ L ∞ ( R ). Then forevery positive time t > u m ) ′ is continuous whenever u remains bounded.8 Building sub- and supersolutions
In this section we present some key results that will be needed later following andenhancing the ideas in [4]. They concern the construction of sub- and supersolutionsfor our problem approaching a TW profile, that we will use all along the rest of thepaper. More specifically we study functions of the form(3.1) w ( x, t ) = f ( t ) V c ∗ ( x − g ( t ))where this V c ∗ is the unique finite travelling wave solution of our equation. Thefunctions f and g will be asked to solve a system of ODEs in order to guarantee theconvergence of w to the travelling wave profile. The reader must see that f representsthe height of w while g gives the location of its front.When we are building the subsolution we want the function f to grow to 1 veryslowly so it doesn’t exceeds our solution u , and we want the function g to increasenot to fast for small times so w doesn’t surpass u through the boundary while thesolution is “starting to travel”. For big times though, we need g ′ ( t ) → c ∗ if ourintuition is true.Similarly, when we think about the supersolution we can allow g to grow this timefast at the start while maintaining its behaviour in the limit. This speed also gives abit more freedom when defining f , as we will see, but again it is better to make it goslowly to 1 to ensure that w is above u .Summarizing, our system will be ( f ′ ( t ) = ϕ ( f ) , f (0) = f ∈ (1 − δ, δ ) ,g ′ ( t ) = c ∗ f m − − kϕ ( f ) /f, g (0) = g , where k > δ is such that h ′ ( u ) < u ∈ (1 − δ, δ ). This interval must exist due to the continuity of h ′ in u = 1.The function ϕ : [1 − δ, δ ] → R is taken such that: • ϕ (1) = 0 , ϕ ′ (1) < , ϕ ( z ) > − δ,
1) and ϕ ( z ) < , δ ]. • ϕ is continuously differentiable (hence locally Lipschitz) in its domain. • sup f ∈ [1 − δ, δ ] | ϕ ′ ( f ) | ≤ H (1 − δ ) m , where H = inf θ ∈ [1 − δ, δ ] | h ′ ( θ ) | .Based on what we said before, ϕ needs to be a very flat function, almost zero. Lemma 3.1.
Consider the following system of ODEs: (3.2) ( f ′ ( t ) = ϕ ( f ) , f (0) = f ∈ (1 − δ, δ ) ,g ′ ( t ) = c ∗ f m − − kϕ ( f ) /f, g (0) = g . The following properties are satisfied. t →∞ f ( t ) = 1 . (b) If − δ < f < then f is strictly monotone increasing, and hence g ( t ) − c ∗ t isstrictly monotone decreasing. (c) If δ > f > then f is strictly monotone decreasing, and hence g ( t ) − c ∗ t isstrictly monotone increasing. (d) lim t →∞ g ( t ) = ∞ . (e) lim t →∞ g ′ ( t ) = c ∗ . (f) There exists ξ ∈ R such that lim t →∞ ( g ( t ) − c ∗ t ) = ξ .Proof. Since (a)-(e), existence and uniqueness are obvious, we only have to consider(f). If f = 1, then the result is trivial with ξ = g . If not, then f ′ ( t ) = 0 for all t > g ′ ( t ) − c ∗ f ′ ( t ) = c ∗ ( f m − ( t ) − − ϕ ( f ( t )) · ( f ( t )) − ϕ ( f ( t )) . Using L’Hopital’s rule and the fact that f → g ′ ( t ) − c ∗ f ′ ( t ) → c ∗ ( m − ϕ ) ′ (1) − k = K when t → ∞ . Thus g ′ ( t ) − c ∗ is close to kf ′ ( t ) for all large t so there must exist aconstant K ′ such that(3.3) | g ′ ( t ) − c ∗ | ≤ K ′ | f ′ ( t ) | for all t > . Since f ′ has a sign, g ′ ( t ) − c ∗ is integrable in R + , which means that g ( t ) − c ∗ t goesto some ξ when t grows.Let us now check that with the above choices w is indeed a sub- or a supersolution. Lemma 3.2.
Let w as in (3.1) and f and g as in (3.2) . Then:(i) If − δ < f < then w is a subsolution of (1.1) .(ii) If δ > f > then w is a supersolution of (1.1) .In both cases there exists ξ ∈ R such that lim t →∞ w ( ξ + c ∗ t, t ) = V c ∗ ( ξ − ξ ) uniformly with respect to ξ ∈ R . roof. We start by defining the operator L u = u t − ∆( u m ) − h ( u ) . Since V c ∗ vanishes when x ≥ g ( t ) and the flux for the profile V c ∗ is continuous,it is enough to study the sign of this operator when x < g ( t ) to see if w fills therequirements to be a sub/super-solution. Replacing in the equation it is easy toarrive to(3.4) L w = ϕ ( f )( V c ∗ + kV ′ c ∗ ) + f m h ( V c ∗ ) − h ( f V c ∗ ) . Remark:
We have made explicit use of the fact that ∆( f V c ∗ ) m = f m ∆ V mc ∗ . This issomething that, for example, we can not use in the more general case of the filtrationequation, i.e., replacing u m by a monotone increasing function in the equation.We want to prove that L w ≤
0, which thanks to (3.4) is equivalent to (i),(ii) . (i) Now we focus in the case 1 − δ < f <
1. Let us remark that we can rewriteequation (3.4) and L w ≤ k V ′ c ∗ V c ∗ + f m h ( V c ∗ ) − h ( f V c ∗ ) ϕ ( f ) V c ∗ ≤ . We can add and substract f m h ( f V c ∗ ) in the last term to get F k := 1 + k V ′ c ∗ V c ∗ + f m ( h ( V c ∗ ) − h ( f V c ∗ )) ϕ ( f ) V c ∗ + ( f m − h ( f V c ∗ ) ϕ ( f ) V c ∗ ≤ . We will check if this is true and study separatedly the cases where V c ∗ is close to 0 orclose to 1.If we stay where V c ∗ ∈ ((1 − δ ) /f , F k ≤ f m (1 − f ) h ′ ( θ ) ϕ ( f ) − ϕ (1) ≤ f m h ′ ( θ ) H (1 − δ ) m ≤ . On the other hand if V c ∗ ∈ [0 , (1 − δ ) /f ] we can apply again the Mean Value Theoremin both the third and fourth term to see that both those terms are bounded, let ussay by C , and we get F k ≤ k V ′ c ∗ V c ∗ + C ≤ k > k for a certain k >
0. Here we have made use of the facts that V ′ c ∗ < V ′ c ∗ /V c ∗ → −∞ as V c ∗ →
0, as mentioned in (1.6), to ensure that said k must exist. This way we finish the part regarding the subsolution when 1 − δ < f < (ii) When 1 + δ > f > L w ≥ ϕ ( f ) < h ( V c ∗ ) toarrive to F ′ k := 1 + k V ′ c ∗ V + ( f m − h ( V c ∗ ) ϕ ( f ) V c ∗ + h ( V c ∗ ) − h ( f V c ∗ ) ϕ ( f ) V c ∗ ≤ . V c ∗ ∈ (1 − δ,
1) then we can, by similar techniques from before, arrive to F ′ k ≤ − f ) h ′ ( θ ) ϕ ( f ) − ϕ (1) ≤ h ′ ( θ ) H (1 − δ ) m ≤ . If V c ∗ ∈ [0 , − δ ], again by similar techniques, we arrive to1 + k V ′ c ∗ V c ∗ + C ≤ k > k for a certain k >
0. We have finished the construction ofthe supersolutions and for our system of ODEs it is enough to take k > max( k , k ).The rest follows from the continuity of V c ∗ and the existence of a limit for g ( t ) − c ∗ t .lim t →∞ w ( ξ + c ∗ t, t ) = lim t →∞ f ( t ) V c ∗ ( ξ + c ∗ t − g ( t )) = lim t →∞ V c ∗ ( ξ + c ∗ t − g ( t )) = V c ∗ ( ξ − ξ ) . We focus now on initial data that are positive, bounded, piece-wise continuous andcompactly supported. We will refer to this as compactly supported initial data.In this section we will see which conditions on the initial datum and the reactionterm h guarantee that the solution will converge to the value u = 1. This part willadvance in parallel to the famous work of Aronson and Weinberger [1], in which theydevelop a similar task for the case m = 1, the Heat Equation. Many arguments thatwork out for the Heat Equation are similar (if not simpler, since we don’t have to careabout the behaviour at infinity) for the PME. The sketch of this part is the following:First we will study monostable reactions, a = 0, that are “below the Fujita exponent p F = m + 2 /N ”, for whatever this means right now. In this case the solution willpropagate no matter the initial datum given it is positive somewhere. This was called,in the semilinear case, the hair-trigger effect by Aronson and Weinberger (Chapter 3in [1]).After that we will study at the same time reaction terms with a = 0 “above theFujita exponent” and reaction terms with a >
0. In this case the initial datum willbe important for it will have to be big enough for us to stablish comparison with acertain subsolution (Chapter 5 in [1]).To be able to construct such subsolution we will need to study in advance thetravelling wave solutions in the phase-plane of the equation (Chapter 4 in [1]).
We recall that we are working with a = 0. Keeping it simple, this term must presenta power-like behaviour near the origin, i.e., h ( u ) ∼ u p when u ∼ .
12f this exponent p is less or equal than the so called Fujita exponent p F , which inthe porous medium equation in dimension N reads p F = m + 2 /N, then we will see that our solution converges to 1, regardless of the initial mass of u .This case is the main focus of this part.On the other hand, if p > p F then it is known that there exist solutions with positiveinitial data that vanish in finite time, as well as others that grow to 1; the phenomenais more complicated. We will later see this case.The idea goes as follows: we have to show that our solution does not vanish andafter that we can prove that it actually converges to 1 uniformly in compact sets.Since we will work in parallel with [1], we will only sketch the ideas of the proofswhen we consider it necessary. Lemma 4.1.
Suppose that our reaction term h , apart from the previous hypotheses,satisfies lim inf u → h ( u ) u m +2 /N > . Then lim sup t →∞ (cid:26) sup x ∈ R u ( x, t ) (cid:27) = 1 . The proof of this statement is similar to the one in [1]. Note that this is not enough toobtain convergence to 1 in compact sets, since our solution can behave like a “spike”or converge to 1 in a set that travels to infinity.The following lemma is based in the work of Kanel’ [16], and we will present just asketch of the proof, commenting the principal differences with the one in [1].
Lemma 4.2.
For a fixed δ > define the function q δ ( r ) = δ (1 − r ) if ≤ r ≤ ,q δ ( r ) = 0 if r > . Let v ( x, t ) denotes the solution of the equation v t = ∆( v m ) + ϕ ( v ) , v ( x,
0) = q δ ( | x | ) , with < δ < min( b, (3 N/k ) N/ ) . Suppose that ϕ ( v ) satisfies a = 0 and ϕ ( v ) = k v m +2 /N f or v ∈ [0 , b ] where k is a positive constant and b ∈ (0 , . Then lim t →∞ v (0 , t ) = 1 . roof. We divide this sketch of the proof in two parts.
Part 1 - First, since the initial data is radially simmetric, it is easy through nowadayswell known techniques to show that0 ≤ v ( x, t ) ≤ v (0 , t ) ≤ . Part 2 - Now the hard part is to show that v (0 , t ) is, after enough time, a monotonicfunction of time, thus its limit exists. To see this, we set z = e − lt v t , l = max w ∈ [0 , ( ϕ ) ′ ( w ) . This is done in order to control the maximal change that v t can experiment. Thiswould happen with a solution with no diffusion, a flat solution, which will give raiseto the ODE v t = ϕ ( v ) , and then ϕ ( v ) is aproximated via the Mean Value Theorem by lv .If this is so, then z satisfies the equation z t = ∆( mv m − z ) + (( ϕ ) ′ ( v ) − l ) z with an initial data z ( x,
0) = Z ( | x | ) = m δ m (1 − | x | ) m − { m − | x | + 4 | x | (1 − | x | ) − N (1 − | x | ) + kδ /N − | x | ) /N +3 } if | x | ≤ , | x | > . Note that this equation is formally parabolic in the set of positivity of v , hence wecan use the Maximum Principles.It is also easy to see that since δ < (3 N/k ) N/ , there exists an r δ ∈ (0 ,
1) such that Z ( r ) is an increasing function in [0 , r δ ], Z ( r ) < , r δ ) and Z ( r ) ≥ r δ , ∞ ).Thus, the set S ≡ { t > z (0 , t ) < } contains the point t = 0. We shall show that either S = [0 , t ) or S = [0 , ∞ ) forsome t > v (0 , t ). Part 2.1 - First we show that z ≡ t -interval,but this can be easily done by comparing v with a solution of the equation w t = ∆( w m ) + λw for some finite λ >
0, thanks to the finite derivative of ϕ in the origin. The superso-lution w has finite speed of propagation, and so does v and then z . Part 2.1 - Second, we have to show that S is an interval. This is the most difficultpart of the proof, and it is based in both a compactness argument due to the Strong14aximum Principle and an argument involving continuous paths joining the points { t = 0 , | x | = 0 } and { t = t ′ > , | x | = 0 } which needs the continuity of v and z . Thiscontinuity is given in [6, 27].The rest of this argument can be found with detail in [1], so we will not reproduceit in its whole here since it works in a similar way for us. Part 3 - Having that either S = [0 , t ) or S = [0 , ∞ ), we see that after some time z and v t have a fixed sign, so v (0 , t ) is ultimately a monotonic function of time and η ∗ = lim t →∞ v (0 , t )exists. Now if η ∗ ∈ [0 ,
1) then for all η ∈ ( η ∗ ,
1) there exists a t n such that v (0 , t ) < η for all t > t n , which according to Part 1 means that0 ≤ v ( x, t ) ≤ v (0 , t ) < η for all t > t n which is a contradiction with Lemma 4.1, so η ∗ = 1 and the lemma is proved.At this point we have enough to prove convergence of our solution to 1 in compactsets, the main result of this part. Theorem 4.1.
Suppose that our reaction term h , apart from the previous hypotheses,satisfies lim inf u → h ( u ) u m +2 /N > . Then if u is a solution of (1.1) with u we have that lim inf t →∞ u ( x, t ) = 1 uniformly in bounded subsets of R N . The proof is again analogous to the one in [1], one only has to be a bit carefulabout the bounded support of the solution and the comparison stablished in its setof positivity. Notice that this is not the result presented in the introduction, since welack the information about the speed of propagation c ∗ . The reader will see how wewill be able to add this part once we are finished with this section.Now we would like to study what happens above the Fujita exponent or with re-actions with a >
0. It is well known that even for monostable reactions there aresolutions that vanish in this case. It depends on the “size” of the initial datum. Forexample, an initial datum u ( x ) ≤ a will certainly vanish.But prior to any results in this direction, we need to study the phase-plane of thetravelling waves of the equation. 15 .2 Phase-plane analysis We have already stablished in Section 1 the existence of wavefront solutions via thework of Gilding and Kersner, but this is not the point in this section. Our point is toprove the existence of finite travelling waves that do not connect 1 and 0, but q and0, for a certain value q > a .We do so in order to be able to construct the forementioned three-parameter familyof subsolutions. We advance that the solutions we are seeking in this section will bethe “legs” that connect the flat part of the subsolution and the ground state 0.We look for solutions u ( x, t ) = q ( x · µ − ct ) , ξ = x · µ − ct where µ is an arbitraryunit vector. If we define p ( ξ ) = ( q m ) ′ ( ξ )then our equation transforms into(4.1) ( q m ) ′′ ( ξ ) + cq ′ ( ξ ) + h ( q ( ξ )) = 0and we get the system q ′ = p/mq m − , p ′ = − cp/mq m − − h ( q ) . But, defining f ( q ) = mq m − h ( q ) and dξ = mq m − dτ we arrive to the less singularsystem q ′ = p, p ′ = − cp − f ( q ) . We must work then in the plane ( q, p ).Now we can see that this is the same system that Aronson and Weinberger studiedin [1]. Thus the proofs will work out in a similar way. In fact, a study about theequivalence between these two phase-planes (and hence between their correspondingequations) was made by Engler in [8].
Remark:
There is no need to worry about the change of variable dξ = mq m − dτ .It is just a regularization of the, in some sense, former singular axis q = 0. The readermay argue that this change of variable is creating an “illusory” compact support fortrajectories through the mentioned axis, but this is not the case. All the trajectoriesthat we will need will have finite support. Let us show some calculations supportingthis claim.One can integrate in dξ = mq m − dτ to obtain ξ − ξ = Z τ τ mq m − ( τ ) dτ and see that ξ is going to remain finite whenever q and τ are finite. This alwayshappen in our trajectories.Another point of view comes from integrating by separation of variables in theequation q ′ = p/mq m − to get ξ − ξ = m Z q q q m − p dq.
16s we can see, when q → p goes to a negative value along our trajectories,and thus the integral above is finite, and so is the support of the solution describedby the trajectory. It can be proved, in fact, that even the trajectory through (0 , V c ∗ mentionedin the Introduction, which corresponds to a trajectory connecting the singular points(1 ,
0) and (0 , c = 0then a trajectory through (0 ,
0) has to satisfy p Z q mu m − h ( u ) du = 0and this is only possible if this integral is negative for all small values of q . There-fore, there is a chance that there are no trajectories through the mentioned point, itdepends on h .Also since the linearization of the system in (0 ,
0) presents the eigenvalue λ = 0, wecan’t use this tool to study this point.Finally, we comment that the null-cline { p ′ = 0 } is a curve connecting (0 ,
0) and(1 ,
0) with formula p ( q ) = − f ( q ) /c, and again it can be checked that p ′ (0) = 0.Now for a given c ≥ , ν > p c ( q ; ν ) be the only trajectory that connects thepoints (0 , − ν ) and ( q c,ν ,
0) with q c,ν ∈ (0 , q c,ν < p c ( q ; ν ) = 0 for q ∈ [ q c,ν , < ν < µ implies p c ( q ; µ ) ≤ p c ( q ; ν ) ≤ q ∈ [0 ,
1] we candefine p c ( q ) = lim ν → p c ( q ; ν ) . Let S = { ( q, p ) : 0 < q < , p < } and T c = S ∩ { ( q, p ) : 0 < q < , p = p c ( q ) } . Note that perhaps T c = ∅ .Define also q c ∈ (0 ,
1] as the value such that p c ( q ) < , q c ) and p c ( q c ) = 0 inthe case q c = 1. It follows from the monotone convergence theorem that p c ( q ) is asolution of (4.1) and T c is a curve through (0,0). If necessary we define p c ( q ) = 0 for q ∈ [ q c ,
1] if q c < Lemma 4.3.
For each c > the curve T c is a trajectory of the system in S through(0,0) and it is extremal in the sense that no other trajectory through (0,0) has pointsin S below T c . Moreover p c ( q ) ≥ (1 /c ) min u ∈ [0 , f ( u ) − cq n [0 , and thre exists a ρ c ∈ (0 , such that p c ( q ) ≤ − cq in q ∈ [0 , ρ c ] . Lemma 4.4.
Suppose that c > σ where σ ≡ sup u ∈ [0 , f ( u ) /u < ∞ . Then q c = 1 and p c (1) < . In view of these results, the quantity c ∗ = inf c> { q c = 1 , p c (1) < } is well defined and satisfies 0 ≤ c ∗ ≤ σ . Lemma 4.5. If max q ∈ [0 , Z q f ( u ) du > then c ∗ > . Remark:
Here we see the reason behind the integral condition in (1.5), but onecan ask now what happens when the quantity defined above is negative. We recallthe work of Y. Hosono [14]. In it we see that the sign of c ∗ actually matches the signof s ( m, h ) = R h ( u ) u m − du . When s ( m, h ) = 0 a stationary wave appears, but thiscase has many similarities to the case c ∗ > s ( m, h ) < Lemma 4.6. If ≤ c < d and T c is not empty, then p d ( q ) < p c ( q ) in (0 , q c ] . Also,if d > then lim c → d p c ( q ) = p d ( q ) . Thus, the family { T c } is continuous in the parameter c whenever c > . Lemma 4.7. If c = c ∗ then there exists an unique trajectory connecting the points (0 , and (1 , in the system, and thus this c ∗ and the travelling wave V c ∗ describedby such trajectory are the same as the ones presented in Section 1. The following is the final result in this part, but first let us define the quantity γ c = (cid:26) S through (0 , ,q c if the extremal trajectory T c in S through (0 ,
0) exists.
Lemma 4.8. If c ∈ (0 , c ∗ ) then γ c ∈ [0 , and for every η ∈ ( γ c , the onlytrajectory through ( η, leaves S through the mentioned point and another point inthe negative p -axis. .3 Over the Fujita exponent We are ready now to prove the propagation of solutions whose initial datum is bigenough, even when the reaction term is above the Fujita exponent or has a > c ∈ (0 , c ∗ ) , η ∈ ( γ c ,
1) and ρ > ( N − /c , thefunction v ( | x | ) = η, | x | ≤ ρ,q ( | x | − ρ ) , ρ < | x | ≤ ρ + b, , | x | ≥ ρ + b, where q ( ξ ) is the trajectory in the previous phase-plane that connects the point q (0) = ( η,
0) with the point q ( b ) = (0 , − ν ), for some b, ν > v as the solution for our Cauchy problem with initial datum v ( x,
0) = v ( | x | ). Lemma 4.9.
Under these conditions, we have that lim t →∞ v ( x, t ) = 1 uniformly in compact subsets of R N , and v ( x, t ) ≥ η for | x | ≤ ρ + (cid:18) c − N − ρ (cid:19) t and t ≥ . Proof.
Choose an arbitray c ∈ (0 , c − ( N − /ρ ) and define W ( x, t ) = v ( | x | − c t ) . Then we have that W t ( x, t ) = − c v ′ ( | x | − c t ) , ∆( W m ( x, t )) = ∂ rr ( v m ( r − c t )) + N − | x | ∂ r ( v m ( r − c t ))and h ( W ( x, t )) = h ( v ( | x | − c t )) . Thus we see that W t − ∆ W m − h ( W ) = − h ( η ) for | x | ≤ ρ + c t,q ′ ( | x | − c t − ρ )( c − ( N − / | x | − c ) for ρ + c t < | x | ≤ ρ + b + c t, | x | > ρ + b + c t. From the definition of q we see that the only point where our function is not continu-ously differentiable is in | x | = ρ + b + c t but this and the fact that W t − ∆ W m − h ( W ) ≤
19 are enough to conclude that W is a weak subsolution for our problem, since itcan be defined as the supremum of two subsolutions, the function q and 0. Since W ( x, t ) ≤ v ( x, t ) for x ∈ R N , t ≥ v ( x, h ) ≥ W ( x, h ) ≥ W ( x,
0) = v ( x,
0) for any h > v is an increasing function of time that is bounded above by the value 1.Thus, v ( x, t ) converges uniformly on compact sets to a certain τ ( x ) ≥ η . At thispoint it is easy to prove that necessarily τ ( x ) = 1, and the lemma is proved.We can now assert the main result of this section, which is the following. Theorem 4.2.
Let u be a solution of (1.1) such that u ( x, ≥ v ( x ; x , η, ρ ) for a certain v of the three-parameter family of functions defined in the previouslemma. Then, for any y ∈ R N and c ∈ [0 , c ∗ )lim t →∞ min | y − y |≤ ct u ( y, t ) = 1 , anf for any c > c ∗ lim t →∞ u ( y, t ) = 0 for | y − y | ≥ ct. Proof.
The first assertion comes from the previous lemma as we see in [1]. Thesecond one comes from a comparison with a supersolution that we build from theones presented in Section 3 , w ( x, t ) = f ( t ) V c ∗ ( x − g ( t )). This travelling wave isunidimensional, but we can take, for any unit vector ν the supersolution defined as w ν ( z, t ) = w ( x · ν, t ).But one has to keep in mind that we can’t choose f as big as we want, it has tosatisfy 1 < f < δ and in principle our solution can surpass this threshold. This isnot an issue though, since we can compare with a flat supersolution with initial data¯ u ( x, ≡ sup { u ( x, } .As we saw in Section 3, this supersolution w ν travels with speed c ∗ and has a finitefront, and thus, if we move with speed greater than c ∗ in the direction ν we will onlysee the value 0, but this ν was arbitrary, thus we are finished.The results presented in this section give convergence on compact sets to the value 1,and also show that the speed of propagation of the solution is c ∗ . But notice that weare lacking the uniform convergence in the whole space.In other words, we don’t know yet how the solution behaves near the front, i.e.,near the two free boundaries that appear in our problem. Keep in mind that ourintial data may have several disjoint sets of positivity with the corresponding freeboundaries, but after a long enough time this sets will merge into one.We will give uniform convergence in the last section.20 Convergence for initial values of class A . Dimen-sion N = 1 The aim of this section is to prove convergence towards a travelling wave for solutionswith initial data in the class A , Theorem 1.4.Adapting the work of Bir´o [4] to our equation, we will construct a sub- and asupersolution for this problem (Lemma 3.2) which will have certain needed properties(Lemma 3.1) and converge to the profile V c ∗ . Then via a special class of solutions,the eternal solutions (see next subsection) and a non-degeneracy result (Lemma 5.2)we will see that for certain sequences t k → ∞ the solution converges to V c ∗ both inshape and support. Then we will see that in fact, this is true for any sequence t → ∞ thanks to a stability result, Lemma 5.1.Notice that we say nothing about the case lim inf x →−∞ u ( x ) ≤ a , since there might bepoints were u ( x, t ) > a that compensate for the lack of mass at infinity. In theprevious section we gave a sufficient condition for the solution to grow to 1 even if itis of compact support, attending only to the concentration of the mass.In the next lemma we show stability for the solutions to the equation. Essentially,if they start close to V c ∗ , then they remain close to it in some sense. Lemma 5.1.
Let ˜ u ( ξ, t ) be a nonnegative continuous solution of the equation (1.1) such that ˜ u ( ξ, T ) ≡ for ξ ≥ ξ , | ξ | < δ and | ˜ u ( ξ, T ) − V c ∗ ( ξ ) | < δ for small δ > andfixed T > . Then there exist ε ( δ ) → as δ → , σ ( ε ) ≥ and σ ( ε ) ≥ such thatfor all t > T (1 − ε ) V c ∗ ( ξ − ξ + σ ( ε )) < ˜ u ( ξ, t ) < (1 + ε ) V c ∗ ( ξ − ξ − σ ( ε )) with ≤ σ ( ε ) ≤ Kε + log 11 − ε ≤ Kε + ε − ε , ≤ σ ( ε ) ≤ Kε + log(1 + ε ) ≤ Kε + ε, where this K is as in formula (3.3) in the proof of Lemma 3.1. This lemma is a combination of other two that can be found in the work of Bir´o butfor his particular reaction term. Since the proof of both is quite similar to the oneneeded in our case (in fact the proof of what he calls Lemma 2.4 is the same), werefer the interested reader to that paper [4].Right now it may seem that we have enough with the previous lemma and the sub-and supersolutions of section 2 for proving the stability of the convergence of oursolution, but note that in order to use Lemma 5.1 we need not only the two functions˜ u and V c ∗ to be close, but also their supports, which in principle may not happen.To see this, imagine that our solution converges uniformly to V c ∗ , i.e. to 0 for ξ ≥ ξ ∗ but there is a “thin tail” in this set that becomes smaller and smaller butwith constant (or oscillating) support. In this case we cannot say that the limit of21he supports of our solution matches the support of the limit. If this happens we willsay that the solution degenerates.Note that this may only happen in the set x ≥ ξ ∗ , since in the complement theuniform convergence of the solution to V c ∗ impose the convergence of the supports.Our goal now is to be arbitrarily close to a solution of our problem, both in densityand support. In order to fulfill this, we need two things. First, we need to prove thatalong a certain time sequence our solution does not degenerate. Second we need twoauxiliary Cauchy-Dirichlet problems. Lemma 5.2.
For every r > we have that σ ( r ) = lim sup t →∞ u ( ζ ( t ) − r, t ) > Proof.
Let us first find a supersolution of our equation. Since h ′ (0) < ∞ there mustexist a constant k such that ku ≥ h ( u ). The problem with rection term ku can betransformed via the change of variables w ( x, t ) = e − kt v ( x, t ) , τ = e k ( m − t /k ( m − τ . It is well known that this equation hasa family of explicit travelling waves, and if one writes down their formula and tracesback the change of variables it can be seen that the function v ( x, t ) = (cid:20) m − m γ ( t ) C (cid:18) C k ( m − γ m − ( t ) − x + C (cid:19) + (cid:21) m − , γ ( t ) = e kt is a weak supersolution of our equation, with arbitrary C > C ∈ R . Its freeboundary satisfies ρ ( t ) = C + C e k ( m − t k ( m − v x ( x , t ) < x < ρ ( t ) and that 0 ≤ v ( x, t ) ≤ v ( x, t + h ) for any positive h .We start now our argument and we will use a contradiction argument. Suppose thatthere exists a δ such that σ ( δ ) = 0. Then by the eventual monotonicity of u we havethat(5.1) lim t →∞ max [ ζ ( t ) − δ,ζ ( t )] u ( · , t ) = 0 . Set δ ∗ = min { δ/ , c ∗ / } . By (5.1) there must exist a pair of values C , C and a bigenough time t such that(5.2) 0 ≤ u ( x, t ) < v ( ζ ( t ) , t ) for all x ∈ [ ζ ( t ) − δ, ζ ( t )] , t ≥ t . and ρ ( t ) = ζ ( t ) + δ ∗ , ρ ( t + 1) < ζ ( t ) + 2 δ ∗ .
22e claim that ζ ( t ) ≤ ρ ( t ) for t ∈ [ t , t + 1], and if this is so then we will have ζ ( t + 1) − ζ ( t ) < ρ ( t + 1) − ζ ( t ) < δ ∗ ≤ c ∗ / t , and thus the average speed of ζ ( t ) can be no larger than 2 c ∗ / c ∗ is the asymptotic speed of propagationof our solution. Therefore, we only have to prove our claim.Again, let us assume that it is not true. Then since ρ ( t ) = ζ ( t ) + δ ∗ > ζ ( t ) and ζ is continuous there must exist a t ∈ ( t , t + 1) such that ρ ( t + 1) > ζ ( t ) > ρ ( t ),and thus for all t ∈ [ t , t ] we have that ζ ( t ) − ζ ( t ) + δ ∗ < ρ ( t + 1) − ζ ( t ) + δ ∗ < δ ∗ ≤ δ and thanks to (5.2), for such t we have that(5.3) u ( ζ ( t ) − δ ∗ , t ) < v ( ζ ( t ) , t ) < v ( ζ ( t ) − δ ∗ , t ) ≤ v ( ζ ( t ) − δ ∗ , t ) . Choose
R > max { ρ ( t + 1 , ζ ( t + 1)) } , which means that R > ζ ( t ) , ρ ( t ) for t ∈ [ t , t ],and compare u and v in [ ζ ( t ) − δ ∗ , R ] × [ t , t ]. By (5.2), (5.3) and the definitionof R we have that u ≤ v on the parabolic boundary, and thus in the whole region.In particular u ( ρ ( t ) , t ) ≤ v ( ρ ( t ) , t ) = 0. On the other hand ζ ( t ) > ρ ( t ), whichimplies that u ( ρ ( t ) , t ) >
0. This contradiction proves our claim and the lemma.This proof tries to be a simplified version of the one found in [7] for h ( u ) = u (1 − u ).In this following subsection we introduce the useful concept of eternal solutions fol-lowing the ideas in [7], trying to give a simplified version of them. We are going to study solutions of the following equation:(5.4) U t = ∆( U m ) + c ∗ U ′ + h ( U )defined in R , not only in R × { t ≥ } . From now on this solutions will be denotedwith U . Such solution will be called a eternal solution . They are a useful tool tostudy the asymptotic behaviour; in some sense they allow us to pass to the limit “intwo times”. First a solution u of problem (1.1) converges to an eternal solution U ,and then this U converges uniformly to a profile V c ∗ (we will see that it actually is aprofile V c ∗ ).Let us study then a good property present in the eternal solutions. This property iswhat make them useful for us, and is the following. Theorem 5.1.
Let ξ = x − c ∗ t and U be a nonincreasing (in the variable ξ ) weaksolution to (5.5) U t = ∆( U m ) + c ∗ U ′ + h ( U ) , ( ξ, t ) ∈ R uch that V c ∗ ( ξ + C ) ≤ U ( ξ, t ) ≤ V c ∗ ( ξ − C ∗ ) for some C > . Then there exists a constant ξ ∗ ∈ [ − C, C ] such that U ( ξ, t ) = V c ∗ ( ξ − ξ ∗ ) . This is, indeed, a very powerful characterization of eternal solutions. We are sayingthat every solution U that is nonincreasing and is trapped between two profiles is,actually, a profile. Thus, if u converges to an eternal solution that satisfies thehypotheses of Theorem 5.1, it actually converges to a stationary profile.Let us define R ∗ = sup { R : U ( ξ, t ) ≥ V c ∗ ( ξ − R ) for all ( ξ, t ) ∈ R } ,R ∗ = inf { R : U ( ξ, t ) ≤ V c ∗ ( ξ − R ) for all ( ξ, t ) ∈ R } . This pair of values R ∗ , R ∗ is well defined and finite due to the hypothesis of ourtheorem, and in order to prove it is enough to prove that indeed R ∗ = R ∗ . We willdirect our efforts in this direction through the following lemmata. Let us define thefree boundary of the function U as Ψ( t ). Lemma 5.3.
There exists a sequence { s n } ⊂ R such that U ( ξ, t + s n ) → V c ∗ ( ξ − R ∗ ) , Ψ( t + s n ) → R ∗ as n → ∞ uniformly for ( ξ, t ) in compact subsets of R .Proof. We define M ( ξ ) = inf t ∈ R | V c ∗ ( ξ − R ∗ ) − U ( ξ, t ) | . Step 1.
The first step of the proof is proving that for all ξ ∈ R , M ( ξ ) = 0, which wededuce by contradiction. Suppose that there exists a ξ such that M ( ξ ) = 2 ε > R ∗ by improvingthe upper bound.We first improve it in ( −∞ , ξ ]. Let us define the auxiliary Cauchy-Dirichlet problem W t = ∆( W m ) + c ∗ W ′ + h ( W ) for ξ ≤ ξ , t > ,W ( ξ,
0) = 1 for ξ ≤ ξ ,W ( ξ , t ) = V c ∗ ( ξ − R ∗ ) − ε for t > . It can be proved that this W is a monotone non-increasing function both in time andspace and, in fact,lim t →∞ W ( ξ, t ) = V c ∗ ( ξ − ( R ∗ − δ )) , for a certain δ > . On the other hand, by the comparison principle, we have that for each s ∈ R U ( ξ, s ) < W ( ξ, , U ( ξ , s + t ) < W ( ξ , t ) for t ≥ U ( ξ, s + t ) ≤ W ( ξ, t ) for all s ∈ R , t > , ξ ≤ ξ , which is equivalent to U ( ξ, t ) < W ( ξ, t − s ) for all s ∈ R , t > s, ξ ≤ ξ . Thus, making s → −∞ , we get U ( ξ, t ) ≤ V c ∗ ( ξ − ( R ∗ − δ )) , for ξ ≤ ξ , t ∈ R . An equivalent argument, this time with the Dirichlet problem W t = ∆( W m ) + c ∗ W ′ + h ( W ) for ξ ∈ [ ξ , R ∗ ] , t > ,W ( ξ,
0) = 1 for ξ ∈ [ ξ , R ∗ ] ,W ( R ∗ , t ) = 0 for t > ,W ( ξ , t ) = V c ∗ ( ξ − R ∗ ) − ε for t > U ( ξ, t ) ≤ V c ∗ ( ξ − ( R ∗ − δ )) , for ξ ∈ [ ξ , R ∗ ] , t ∈ R and the same δ, and thus U ( ξ, t ) ≤ V c ∗ ( ξ − ( R ∗ − δ )) , for ξ ∈ R , t ∈ R , contradicting the definition of R ∗ , and hence for all ξ ∈ R , M ( ξ ) = 0, in particularfor a certain ξ fixed.Note that we have made use of the fact that s ∈ R , instead of the more common s >
0. This means that we are making full usage of the characteristics of eternalsolutions to prove that we can actually improve our barrier for all times t ∈ R . Step 2.
Two possibilities arise now. If U ( ξ , t ) = V c ∗ ( ξ − R ∗ ) for a finite time t then the Strong Maximum Principle tells us that indeed U ( ξ, t ) ≡ V c ∗ ( ξ − R ∗ ) for all( ξ, t ) ∈ R , and thus the conclusion of the theorem comes trivially.Suppose on the contrary that for a certain unbounded sequence s n we have that U ( ξ , s n ) → V c ∗ ( ξ − R ∗ ) and define U n ( ξ, t ) ≡ U ( ξ, t + s n ) . The regularity results in [6, 27] and the equiboundedness of U n allow us to use theAscoli-Arzel´a Theorem to conclude that along a certain subsequence U n converges to¯ U , a solution of equation (5.5) with ¯ U ( ξ, t ) ≤ V c ∗ ( ξ − R ∗ ) and ¯ U ( ξ ,
0) = V c ∗ ( ξ − R ∗ ).Hence ¯ U ≡ V c ∗ and the theorem is proved because in this case convergence of solutionsimpose the convergence of the free boundary Ψ( t ).25otice that the convergence is in compacts of R . We do not get this kind ofconvergence in the next lemma, but it is not necessary. We need convergence incompacts of R ∋ ξ along a sequence of times; in particular convergence in compactsof R is a stronger result. Lemma 5.4.
There exists a sequence { ˜ s n } ⊂ R such that U ( ξ, ˜ s n ) → V c ∗ ( ξ − R ∗ ) , Ψ(˜ s n ) → R ∗ as n → ∞ uniformly for ξ in compact subsets of R .Proof. We can use again the proof of the previous lemma adapted to a lower barrier.In particular, it is enough to employ this time auxiliary Cauchy-Dirichlet problemswith initial data W ( ξ,
0) = 0 to conclude, as in the previous lemma, that there existsa sequence s j such that U ( ξ, t + s j ) → V c ∗ ( ξ − R ∗ ) as j → ∞ . uniformly for ( ψ, t ) in compact subsets of R .But note that in this case this is not enough to conclude the convergence of thesupports due to de aforementioned possible degeneration of the solutions. We needagain a non-degeneracy result similar to Lemma 5.2. This time up to a sequence ˜ s n that we will build ourselves.Fix a δ > k ∈ N satisfying k > R ∗ − R ∗ ) /c ∗ . Wewant to see that there exists a t = t ( δ ) ∈ [0 , k ] and a j = j ( δ ) such that | Ψ( t + s j ) − R ∗ | < δ. For this, we suppose the opposite, that for every t ∈ [0 , k ] and j that quantity isbigger or equal than δ , and we will arrive to a contradiction. Once we have this, it isenough to take δ n = 1 /n and ˜ s n = t ( δ n ) + s j ( δ n ) to prove our statement.We define ˜ u ( x, t ) = U ( x − c ∗ t, t ) and ˜ ζ ( t ) = Ψ( t ) + c ∗ t as its free boundary. Clearly,˜ u is a solution to ˜ u t = ∆˜ u m + h (˜ u ) in R .So fix δ > k > R ∗ − R ∗ ) /c ∗ and δ ∗ < min( δ, c ∗ ) /
3. Since U ( ξ, t + ˜ s n ) → V c ∗ ( ξ − R ∗ ) uniformly locally in R there must exist a j big enough such that0 < ˜ u ( ξ + c ∗ ( t + s j ) , t + s j ) < k for t ∈ [0 , k ] , ξ ∈ [ R ∗ , Ψ( t + s j )] ⊃ [ R ∗ , R ∗ + δ ] . Now define v as the slow travelling wave supersolution of Lemma 5.2 and ρ ( t ) as itsfree boundary, and see that because of the previous inequality there must exist forall t ∈ [0 , k −
1] two constants C and C such that(5.6) 0 ≤ ˜ u ( x, t + s j ) ≤ v ( ˜ ζ ( t + s j ) , t + s j ) for t ∈ [0 , k ] , x ≥ ˜ ζ ( t + s j ) − δ ∗ and ρ ( t + s j ) = ˜ ζ ( t + s j ) + δ ∗ , ρ ( t + 1 + s j ) < ˜ ζ ( t + s j ) + 2 δ ∗ .
26e will compare ˜ u ( x, t + s j ) and v ( x, t + s j ) in [ ˜ ζ ( t + s j ) − δ ∗ , R ] × [ t , t ] (rememberthe definitions of R and t from Lemma 5.2).At the initial time t = t there are two possibilities. If x ∈ [ ˜ ζ ( t + s j ) − δ ∗ , ˜ ζ ( t + s j )],then, since v is nonincreasing in space we have from (5.6) that ˜ u ( x, t + s j ) ≤ v ( x, t + s j ). If x ∈ [ ˜ ζ ( t + s j ) , R ] we finish by recalling that here u ≡ x = R both functions are 0, and when x = ˜ ζ ( t + s j ) − δ ∗ we have from the properties of v that˜ u ( ˜ ζ ( t + s j ) − δ ∗ , t + s j ) ≤ v ( ˜ ζ ( t + s j ) , t + s j ) ≤ v ( ˜ ζ ( t + s j ) − δ ∗ , t + s j ) . So we can compare.At this point we follow the proof of Lemma 5.2. We have again that ˜ ζ ( t + s j ) ≤ ρ ( t + s j ) for t ∈ [ t , t + 1], in particular, this and the fact that ρ ( t + 1 + s j ) < ˜ ζ ( t + s j ) + 2 δ ∗ provides˜ ζ ( t + 1 + s j ) − ˜ ζ ( t + s j ) < δ ∗ < c ∗ . But this can be done for every t ∈ [0 , k − t = k − , k − , ..., ζ ( k + s j ) − ˜ ζ ( k − s j ) < c ∗ , ˜ ζ ( k − s j ) − ˜ ζ ( k − s j ) < c ∗ ,..., ˜ ζ (1 + s j ) − ˜ ζ ( s j ) < c ∗ , and a telescoping series tells us that˜ ζ ( k + s j ) − ˜ ζ ( s j ) < c ∗ k. Using that ˜ ζ ( t ) = Ψ( t ) + c ∗ t and R ∗ < Ψ( t ) < R ∗ we get k < R ∗ − R ∗ ) /c ∗ arriving to a contradiction with k > R ∗ − R ∗ ) /c ∗ . Thus our t ( δ ) and j ( δ ) exist andwe can construct our sequence ˜ s n . Lemma 5.5. R ∗ = R ∗ .Proof. Thanks to Lemma 5.3 and the fact that both V c ∗ ( ξ − R ) and U ( ξ, t ) convergeuniformly to 1 when ξ → −∞ we know that for every δ > n ( δ ) largesuch that(1 − δ ) V c ∗ ( ξ − R + δ ) ≤ U ( ξ, s n ) ≤ (1 + δ ) V c ∗ ( ξ − R − δ ) for ξ ∈ R . − δ ) V c ∗ ( ξ − R + σ ( δ )) ≤ U ( ξ, s n ) ≤ (1 + δ ) V c ∗ ( ξ − R − σ ( δ )) for ξ ∈ R . with 0 ≤ σ ( ε ) ≤ Kε + log 11 − ε ≤ Kε + ε − ε , ≤ σ ( ε ) ≤ Kε + log(1 + ε ) ≤ Kε + ε, where K is independant of δ . It follows that R ∗ − σ ( δ ) ≤ Ψ( s n ) ≤ R ∗ + σ ( δ ) , and similarly R ∗ − σ ( δ ) ≤ Ψ(˜ s m ) ≤ R ∗ + σ ( δ ) . Thus for any t > s n , ˜ s m we have that R ∗ − σ ( δ ) ≤ Ψ( t ) ≤ R ∗ + σ ( δ ), which impliesthat 0 ≤ R ∗ − R ∗ ≤ σ ( δ ) + σ ( δ )and we get our result simply by making δ tend to 0. Having all we need, let us prove our main result for this section. The ideas now aresimple enough thanks to the previous work. Basically, what we are going to do is totrap our solution u between sub- and supersolutions and take the limit in time alonga non degenerating sequence. This limit will satisfy the conditions of Theorem 5.1,and thus it will be a profile V c ∗ . We finish by using the stability result, Lemma 5.1. Proof.
We start by setting our sub- and supersolutions below and above u ,in otherwords, there must exist w ( ξ, t ) and w ( ξ, t ) coming from Lemma 3.2 such that w ( ξ, t ) ≤ u ( ξ, t ) ≤ w ( ξ, t )and these w , converge to different profiles V c ∗ .On the other hand, thanks to Lemma 5.1 it is enough to prove that there exists asequence t k and a ξ ∈ R such thatlim k →∞ u ( ξ, t k ) = V c ∗ ( ξ − ξ )uniformly in compact sets of R withlim k →∞ ζ ( t k ) = ξ . To do so, if we prove that there exists a time sequence { t k } → ∞ without degeneracyfor every ξ < ζ ( t ), i.e.,(5.7) lim inf k →∞ u ( ζ ( t k ) − r, t k ) > r > , { t k j } such that u convergesalong it to a eternal solution without degeneracy, thanks to the regularity of thesolution, Alexandrov’s reflexion principle and the behaviour in the limit of w and w . The details are left to the reader and can be found in [7]. Let us prove then (5.7).First, we see that thanks to Lemma 5.2 and our identification of eternal solutionswe have that for each r > C < t n such that(5.8) lim n →∞ u ( ζ ( t n ) − r, t n ) = V c ∗ ( r + C − r ) in C loc ( R ) ∋ r. Now we find our sequence t k via a diagonal argument. Take a sequence r k → k > C k < t kn such that as n → ∞ , t kn → ∞ and u ( ζ ( t kn ) − r, t kn ) → V c ∗ ( r k + C k − r ) uniformly in r ∈ [0 , k ] . Since V c ∗ ( r k + C k − r ) > V c ∗ ( r k − r ) for r ∈ [ r k , k ] there must exist a n k so large that t kn k > k and u ( ζ ( t kn k ) − r, t kn k ) > V c ∗ ( r k − r ) for r ∈ [ r k , k ] . Define t k = t kn k . Then clearlylim inf k →∞ u ( ζ ( t k ) − r, t k ) ≥ lim k →∞ V c ∗ ( r k − r ) = V c ∗ ( − r ) > . So the statement and the theorem are proved.As a final remark, one can think if the hypothesis u ∈ A can be weakened. Infact, suppose only that lim inf x →−∞ u > −∞ .First, if the reaction term h is under the hair-trigger effect , then it is trivial to puta monotone non-increasing in space function below u that is going to grow to 1,hence pushing our solution into class A . We shall refer to this as Condition (1), andit depends on h .Second, there is a chance that u is not in class A but it is big enough at −∞ forus to apply the same ideas of Lemma 4.9 but with a non-symmetric function v ( | x | ) = η, x ≤ ρ,q ( | x | − ρ ) , ρ < x ≤ ρ + b, , x ≥ ρ + b, and push again u into A . We shall refer to this as Condition (2), and it depends onthe mass of u near −∞ . Corollary 5.1.
Let u be non-negative, bounded, piecewise continuous, u ( x ) ≡ for all x ≥ x , x ∈ R and satisfying either Condition (1) or Condition (2). Thenthe same results from Theorem 1.4 hold. Convergence for initial values of compact sup-port. Dimension N = 1 In this last section we study the uniform convergence of solutions of equation (1.1)with compact support in a similar way as we did in the previous section for initial datain the class A , but let us first see a preliminary result about the speed of convergenceto 1 that we will need. Lemma 6.1.
Let u be a solution of equation (1.1) of compact support such that u converges to 1 uniformly in compact sets. Then there exist ˆ c ∈ (0 , c ∗ ) , δ ∈ (0 , and M, T ∗ > such that u ( x, t ) ≤ M e − δt for all | x | ≥ and t ≥ T ∗ , (6.1) u ( x, t ) ≥ − M e − δt for all | x | ∈ [0 , ˆ c t ] and t ≥ T ∗ . (6.2)A proof for a harder lemma that this one can be found in [7], but their prove worksfine in our case. Let us show now the main proof of this section. Proof.
At this point, the proof is quite straitforward. We focus only in R + , since inthe other side is similar. What we want to do is to trap our solution between a sub-and a supersolution of the form w i = f i ( t ) V c ∗ ( x − g i ( t )), the ones that we have alreadystudied. Once this is done, we only have to pass to the moving frame ξ = x − c ∗ t andrepeat the argument showed in the proof of Theorem 1.4.Thanks to the previous lemma, we focus on times greater than a T such that forall t ≥ T − M e − δt ≤ u (0 , t ) ≤ M e − δt for a fixed M > , δ >
0. To place this w i below and above the solution in time T is easy, but keep in mind that we also have to order the functions in the parabolicfrontier { x ≡ } × { t ≥ T } . In order to do so we only need f ( t ) V c ∗ ( − g ( t )) ≤ − M e − δt , M e − δt ≤ f ( t ) V c ∗ ( − g ( t )) , but since V c ∗ is less than 1 ( f ( t ) V c ∗ ( − g ( t )) < f ( t )) and strictly monotone decreasing( f ( t ) V c ∗ ( − g ( t )) ≥ f ( t ) V c ∗ ( − g ( T )), this is the same as asking f ( t ) ≤ − M e − δt , M e − δt ≤ Cf ( t ) , where C = V c ∗ ( − g ( T )) is a positive constant. We will explain how to get the firstinequality. The second one is similar.We can choose f ( T ) < − M e − δT , and thus we only have to check that f ′ ( t ) ≤ δM e − δt . But remember that f ′ = ϕ ( f ). If we choose, for a certain k > ϕ ( f ) = k (1 − f ) , f ′ ( t ) = ke − kt we can arrive to f ′ ( t ) ≤ δM e − δt by chosing the right k , and keep in mind that inorder to find a proper subsolution we were allowed in Lemma 3.2 to make ϕ as small30s we wanted, so it is enough to pick ϕ ( f ) ≤ k (1 − f ) for a small enough k . One canrepeat this argument for the supersolution (this time with a big enough k ), apply thetechnique of Theorem 1.4 to get convergence in Ω and then repeat in Ω to finishthe proof. Acknowledgments
The author would like to thank Fernando Quir´os, from Universidad Aut´onoma deMadrid, for his help and advice in many discussions about the topic.
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