Super-linear propagation for a general, local cane toads model. (English) Zbl 1407.35112
Summary: We investigate a general, local version of the cane toads equation, models the spread of a population structured by unbounded motility. We use the thin-front limit approach of L. C. Evans and the last author [Indiana Univ. Math. J. 38, No. 1, 141–172 (1989; Zbl 0692.35014)] to obtain a characterization of the propagation in terms of both the linearized equation and a geometric front equation. In particular, we reduce the task of understanding the precise location of the front for a large class of equations to analyzing a much smaller class of Hamilton-Jacobi equations. We are then able to give an explicit formula for the front location in physical space. One advantage of our approach is that we do not use the explicit trajectories along which the population spreads, which was a basis of previous work. Our result allows for large oscillations in the motility.
MSC:
| 35K57 | Reaction-diffusion equations |
| 35F21 | Hamilton-Jacobi equations |
| 35K15 | Initial value problems for second-order parabolic equations |
Keywords:
reaction diffusion equations; long range; long time limits; motility; cane toads equation; mutation; spatial sorting; front propagationCitations:
Zbl 0692.35014References:
| [1] | The proof of Theorem1.4 The proof hinges on the locally uniform convergence of vto I guaranteed by Proposition1.3. We show how to conclude Theorem1.4assuming this proposition, which is proved in Section3. Our proof follows the general outline of [17], with the relevant modifications made to handle the technical issues arising from the boundary. Proof of Theorem1.4.We first consider the setfI > 0g. Fix any point .x0; 0; t0/ such that I.x0; 0; t0/ > 0 with t0> 0. Since vconverges to I locally uniformly as tends to zero by Proposition1.3, v.x; ; t / > ı for some ı; r > 0 and any .x; ; t /2 Br.x0; 0; t0/ when is sufficiently small. It follows that u.x; ; t / 6 expf ı=g for all sufficiently small and all .x; ; t /2 Br.x0; 0; t0/. Hence uconverges to zero uniformly on Br.x0; 0; t0/ as tends to zero. Now we consider the set IntfI D 0g. Fix .x0; 0; t0/2 IntfI D 0g. There are two cases to investigate depending on whether 0is positive or zero. First assume that 0> 0. Define a test function .x; ; t /D jtt0j2C jxx0j2C j0j2; and note that, since I 0 near .x0; 0; t0/, I has a strict local maximum at .x0; 0; t0/ on a small enough ball centered at .x0; 0; t0/. It follows that v has a maximum at some point .x; ; t/ such that .x; ; t/ converges to .x0; 0; t0/ as tends to zero. Because 0> 0, we may restrict to sufficiently small so that > 0. Then, using (5), we find, at .x; ; t/, tDxx C Dj xj2C j j26u1: An explicit computation, using only the form of and the fact that .x; ; t/ converges to .x0; 0; t0/ as tends to zero, shows that the left hand side tends to zero as tends to zero. We infer that 1 6 lim inf!0u.x; ; t/. On the other hand, recall that .x; ; t/ is the location of a minimum of uexpf =g. Hence we have that lim infu.x0; 0; t0/ > lim infu.x; ; t/ exp˚1jtt0j2C jxx0j2C j0j2 > 1: !0!0 Initially, u061 in RŒ0; 1/. The maximum principle implies that u61 in RŒ0; 1/Œ0; 1/ for all . It follows that 1 > lim sup!0u.x0; 0; t0/. As a consequence, lim sup!0u.x0; 0; t0/D lim inf!0u.x0; 0; t0/D 1, which implies that u.x0; 0; t0/ converges to 1 as tends to zero. This concludes the proof in the case that 0D 0. If 0D 0, define .x; ; t /WD jtt0j2C jxx0j2C j2j2; and let .x; ; t/ be a maximum of v. Since and vconverge to and I , respectively, as tends to zero and I has a strict local maximum at .x0; 0; t0/, it follows that .x; ; t/ converges to .x0; 0; t0/ as tends to zero. We claim that > 0 for all > 0, and we proceed by contradiction. Suppose that D 0 for any > 0. Because v has a local maximum at .x; 0; t/, v.x; 0; t/ 6 .x; 0; t/: By (5), the left hand side is 0. The right hand side is, by construction,22. This is a contradiction. It follows that > 0 for all > 0. Then (5) yields, at .x; ; t/, tD. / xx C D. /j xj2C j j26u1: As above, an explicit computation shows that the left hand side tends to zero as tends to zero. Hence, lim inf u.x; ; t/ > 1. 490C.HENDERSON,B.PERTHAME AND P.E.SOUGANIDIS By construction, .x; ; t/ is the location of a local minimum of uexpf =g. Thus, jtt0j2C jxx0j2C j2j2 lim inf u.x0; 0; t0/ > lim inf u.x; ; t/ exp>1: From the conclusion of the previous case, when 0> 0, recall that u61, which immediately yields lim sup!0u.x0; 0; t0/ 6 1. Arguing as above, we conclude that u.x0; 0; t0/ converges to 1 as tends to zero. This concludes the proof. |
| [2] | The limit of the sequence.v/>0– the proof of Proposition1.3 We proceed in three steps. In the first, we obtain uniform bounds on von compact subsets of .G0 ft D 0g/ [ .R Œ0; 1/ RC/. In the second, we take the half-relaxed limits of the sequence .v/>0to obtain vand v, and we show that they are respectively super- and sub-solutions of (6). Finally in the last step, we use comparison to show that vD vD I and conclude that v converges locally uniformly to I . 3.1An upper bound forv By the maximum principle, 0 6 u61 and so v> |
| [3] | In order to take the half-relaxed limits, we need an upper bound on vthat is uniform in . Lemma 3.1 Suppose that Assumption1.1and Assumption1.2hold. Fix any compact subsetQ of .G0 ft D 0g/ [ R Œ0; 1/ RC : There exists C D C.Q/ > 0 such that, if .x; ; t/ 2 Q, then v.x; ; t / 6 C: Further, if Q G0 Œ0; 1/, then there exists a constant C0D C0.Q/ such that, if.x; ; t /2 Q, then v.x; ; t / 6 C0:(12) Proof.We begin by noticing that, when > 0, we may ignore the boundaryf D 0g: Indeed, using the Neumann boundary condition, we may extend u, and thus v, evenly to R R RC. The parabolic regularity theory yields that vsatisfies (5) on R R RCwith D. / replaced by D.jj/; for more details see [31]. For the remainder of this proof, we abuse notation by letting u and vrefer to their even extensions. Next, we set some notation. For any R > 0 and .x0; 0/2 R R, let QR.x0; 0/WD .x0R; x0C R/ .0R; 0C R/: In the sequel, we use QRto refer to QR.0; 0/. We proceed in two steps. First, for any T; R > 0 and .x0; 0/ such thatj0j > R=2 and QR.x0; 0/ G0, we build a barrier on QR.x0; 0/ Œ0; T that yields an upper bound on v in Q3R=4.x0; 0/ Œ0; T that is uniform in . Since G0is open, it is easy to see that [[ G0DQ3R=4.x0; 0/: R2.0;1/j0j>R=2; QR.x0;0/G0 Thus, the bound we have is enough to conclude an upper bound on vthat is independent of on any compact subset of G0 Œ0; T . The second step extends this by building a barrier on sets of the SUPER-LINEAR PROPAGATION FOR A GENERAL,LOCAL CANE TOADS MODEL491 form .QL.x0; 0/n QR=2.x0; 0// ŒT1; T1C T , where R, x0, and 0are as in the first step, T > 1, and L > R. This crucially uses the bound obtained in the first step to control the portion of the parabolic boundary @QR=2 .T1; T1C T /. This provides an upper bound on vthat is independent of on compact subsets of R R RC, which finishes the proof. Our proof follows the ideas of [17] with a few key modifications, which we mention as they arise. The added complication that occurs in our proof is due to the interplay of the degeneracy of Dat D 0 and its growth at D 1. We point out that a crucial observation that saves our computations is restricting to cubes QR.x0; 0/ wherej0j > R=2, see Step Two below. # Step one:Since the equation is translation invariant in x, we may assume that x0D 0, without loss of generality. We may also assume, without loss of generality, that 0> 0, which, in turn, implies that 0> R=2. For notational ease, we translate the equation in . That is, we define .x; ; t /D v.x; C 0; t / and D0. /D D.C 0/. It follows that, tD0xx C Djxj2C jj2C 1e=D 0in R R RC:(13) An upper bound on in QRimplies the desired bound on vin QR.0; 0/. We proceed by building a barrier. Consider, for ˛, ˇ, and that are positive constants to be determined, the auxiliary function R2x2CR22in QR RC: We point out that differs from the barrier used in [17], and this difference simplifies many computations because it separates the variables. Straightforward calculations yield tD0xx C D0j xj2C j j2C 1e = 28x2282! .R2x2/2C.R2x2/3C.R22/2C.R22/3 C 42D0x2C2 .R2x2/4.R22/4 D ˛ C.R2C 3x2/C222.R2C 32/: .R2x2/3R2x2.R22/3R22 (14) We define 20 max.1C D0.0// WD 6R2; ˇWD max.x; ; 0/D maxv0.x; /; and ˛WDj0j6R: QRQR.0;0/R4 (15) Consider the second term in the last line of (14). Whenjxj 2 ŒR=2; R, we have 2x22.R=2/22 R2C 3x2 > R2C 3R2 D4R2D 0; R2x2R2.R=2/23 492C.HENDERSON,B.PERTHAME AND P.E.SOUGANIDIS where the last equality follows from the definition of (15). Whenjxj 2 Œ0; R=2, we have 2D02x22 maxD0 .R2x2/3R2x2.R2C 3x2/>.R2j0.R=2/j6R2/3.02R2/ >˛2: We see that, for all .x; /2 QR, ˛2D02x2 C.R2C 3x2/>0: 2.R2x2/3R2x2 A similar argument shows that, for all .x; /2 QR, ˛222 2.R22/3R22.R2C 32/>0: These two inequalities, applied to (14), show that tD0xx C D0j xj2C j j2C 1e =>0in QR RCI that is, is a super-solution of (13) in QR RC. Next, the choice of ˇ ensures that, on QR, .; ; 0/ > ˇ > v0: Further, the strong maximum principle implies that u> 0 on R R RC, which implies that v, and thus , is finite in C. Because D C1 on @QR RC, > on @QR RC. The maximum principle R R R implies that 0 6 6 on QR RC. In particular, there exists some CR> 0, which depends only on 0, R, D, and u0, such that, on Q3R=4 Œ0; 1/, v6 6 CR.1C t/:(16) We now establish (12). Since QR.0; 0/ G0D fu0> 0g, then ! 1 ˇDmaxv06 log: .x; /2QR.0;0/min.x; /2QR.0;0/u0.x; / We recall that minu0> 0 due to the continuity of u0. Also, it follows from their .x; /2QR.0;0/ definitions that ˛; 6 C , for some constant C depending only on D, R, and 0. We conclude that, for any .x0; 0/ and R; T > 0 such that QR.x0; 0/ G0andj0j > R=2, there exists a constant C that depends only on u0, R, D, .x0; 0/, and T such that 6 C in Q3R=4.x0; 0/ Œ0; T : Given a compact subset Q G0 Œ0; 1/, it can be covered by finitely many sets of the form Q3R=4.x0; 0/ Œ0; T where QR.x0; 0/ G0andj0j > R=2. Hence, we conclude that, for any such Q, there exists CD C.Q/ such that 6 C on Q; that is, (12) holds. # Step two:Let R and 0be as above and fix L > R and T > 1. Define .x; ; t /D v.x; C 0; tC T1/. Then, satisfies (13). In view of the bound (16), a bound on in QLn QR=2 Œ0; T , yields a bound on von QL.0; 0/ ŒT1; T1C T . To obtain such a bound, we build a barrier. Before beginning, we note that, in [17], the authors are able to construct a barrier on their analogue of QR=2 RCdirectly. This approach will not work in our setting since D0is unbounded. This is the reason that we strict to cubic annuli in physical and trait space. SUPER-LINEAR PROPAGATION FOR A GENERAL,LOCAL CANE TOADS MODEL493 Define 4T L4.1C maxD 2j0j6L0.0//C L8 ˇWDmaxDmaxvand WD: R=2.0;0/ŒT1;T1CTR2 min1;minD0.0/ j0j6R=2 (17) Since 0> R=2, it follows that the denominator of is bounded below by a positive constant independent of . Also, ˇ is bounded above depending only on u0, 0, R, and T , due to (16). Define 1 x; ; tCWD ˇ CCin QLn QR=2 Œ0; T : Tt .L2x2/t .L22/ Note that the restriction 0> R=2 has the consequence that when D0 0, jj O.1/. This is the key observation in constructing that allows us to side-step any complications stemming from the degeneracy D0. 0/D D.0/D 0. Also, note that this barrier is different from the one constructed in [17]. Indeed, since we are restricted to a compact set in physical and trait space, it is crucial that be larger than on the boundary @QL. Hence, we may not use the quadratically growing barrier from [17]. We show that is super-solution of (13). A straightforward computation yields tD0xx C D0jxj2C jj2C 1e= 28x2 t2.L2x2/t2.L22/D0t .L2x2/2Ct .L2x2/3 42x2422 t .L44/2t .L2x2/3t2.L2x2/4Ct2.L22/4(18) ” D2x2D0D0t .L2C 3x2/.L2x2/.L2x2/4 t2.L2x2/42 C22t .L2C 32/.L2x2/.L22/4#: .L22/42 Since .x; /2 QLn QR=2, we consider three cases: (1)jxj > R=2 > jj; (2) jj > R=2 > jxj; and (3)jxj; jj > R=2. Case one:Ifjxj > R=2 > jj, notice that 2x2D0D0t .L2C 3x2/.L2x2/.L2x2/4 .L2x2/42 .L2x2/42j0j6R=20.0/4T L4jmax0j6LD0.0/L28(19) .L2x2/44T L4CL28>.L2.R=2/12/44T L4CL28; 494C.HENDERSON,B.PERTHAME AND P.E.SOUGANIDIS where we used the definition of (17) in the second-to-last inequality. On the other hand, .L22/422t .L2C 32/.L22/.L222/4 >4T L4CL8: .L2.R=2/2/42 Summing these two inequalities and recalling (18) yields tD0xx C D0jxj2C jj2C 1e=>0 whenjxj > R=2 and jj 6 R=2. Case two:Ifjj > R=2 > jxj, the argument is handled in exactly the same way, so we omit it and conclude again that tD0xx C D0jxj2C jj2C 1e=>0 whenjxj 6 R=2 and jj > R=2. Case three:Ifjxj; jj > R=2, then, following the argument in (19) in case one, we see that 2x2t .L2C 3x2/.L2x2/ > 0. Hence, 2x2D0D0t .L2C 3x2/.L2x2/.L2x2/4>1: .L2x2/422 Also, arguing similarly as in (19) and using the definition of (17), we find 22t .L2C 32/.L22/.L22/4 .L22/42 .L22/424T L4L28>.L212/44T L4CL28>12: Summing these two inequalities and recalling (18) implies that tD0xx C D0jxj2C jj2C 1e=>0 whenjxj; jj > R=2. The combination of all three cases above implies that is a super-solution of (13) in .QLn QR=2/ .0; T /. By the definition of ˇ (17), it follows that > on @QR=2 Œ0; T . Also, since is finite on QLŒ0; T (see the discussion at the end of Step One) and D C1 on .QLnQR=2/ft D 0g and on @QL Œ0; T , then > on .QLn QR=2/ ft D 0g and on @QL Œ0; T . It follows that > on the parabolic boundary of .QLn QR=2/ .0; T /. The maximum principle then implies that 6 in .QLn QR=2/ .0; T /. Given the definition of and the preliminary bound on von Q3R=4(16), it follows that there exists a constant C that depends only on u0, 0, D, L, R, and T such that v6Cin QL=2.0; 0/ 2;2C T: TT Since L and T are arbitrary, this concludes the proof. SUPER-LINEAR PROPAGATION FOR A GENERAL,LOCAL CANE TOADS MODEL495 3.2The half-relaxed limits We next recall the definition of the classical half-relaxed limits vand v: v.x; ; t /Dlim supv.y; ; s/andv.x; ; t /Dlim infv.y; ; s/:(20) .y;;s/!.x;;t/;.y;;s/!.x;;t/; !0!0 The existence of these limits is guaranteed by Lemma3.1along with the fact that, as discussed in Section1, v> |
| [4] | We point out that vis lower semi-continuous while vis upper semi-continuous. Equations forvandv.Our first step is to prove that vand vsatisfy the limits that the theory of viscosity solutions suggest. The issues here are the boundary behavior and verifying the initial conditions. Lemma 3.2 The relaxed lower limitvsatisfies in the viscosity sense ( min˚.v/tC D./j.v/xj2C j.v/j2C 1; v > 0inR RC RC; max˚.v/; min˚.v/tC j.v/j2C 1; v > 0onR f0g RC;(21) and ( 0inG0; v.; ; 0/ Dc(22) 1inG0: Proof.We verify (21) first. Assume that, for some test function , v has a strict local minimum at .x0; 0; t0/2 R Œ0; 1/ RC. We may then choose kconverging to 0 and .yk; k; sk/ converging to .x0; 0; t0/ as k tends to infinity such that .yk; k; sk/ is a local minimum of vk in R Œ0; 1/ Œ0; 1/ and v.x0; t0; 0/D limk!1vk.yk; k; sk/: If .x0; 0; t0/2 R RC RC, then, for sufficiently large k, .yk; k; sk/2 R RC RC. Since vsolves (5), at .yk; k; sk/, we have, at .yk; k; sk/, kk 0 6 tkDxxk C Dj xj2C j j2C 1e =k 6tDj xj2C j j2C 1 C o.1/: Here and in the sequel, we use o.1/ to mean a quantity that tends to zero in the limit. Taking the limit as k tends to infinity and using the smoothness of yields, at .x0; 0; t0/, 0 6 tC D x2C 2 C 1: As discussed above, v>0 on R RC RC. From this and the inequality above, we conclude that min˚.v/tC D./j.v/xj2C j.v/j2C 1; v > 0in R RC RC; which finishes the proof in this case. Assume next that .x0; 0; t0/2 R f0g RC. If k> 0 for infinitely many k, the fact that vk solves (5) yields, at .yk; k; sk/, kk2 0 6 tkDxxk C DxC 2C 1ek=k k2 6tC DxC 2C 1 C o.1/: Letting k tend to infinity, we find, at .x0; 0; t0/, 0 6 tC D x2C 2 C 1: If kD 0 for all k sufficiently large, then, since vksatisfies Neumann boundary conditions, we have 496C.HENDERSON,B.PERTHAME AND P.E.SOUGANIDIS 0 6.yk; 0; sk/. Letting k tend to infinity, we find 0 6.x0; 0; t0/. In either case, we have verified that max.v/; min˚.v/tC j.v/j2C 1; vo>0onR f0g RC: Finally we need to consider the initial condition (22). Fix > 0 and any smooth function 2 C1.R Œ0; 1/I Œ0; 1/ such that jG 0 and j 0RRCnG0> 0. Then 8no <max.v/tC Dj.v/xj2C j.v/j2C 1; v>0inR Œ0; 1/ f0g; (23) :max.v/; .v/tC j.v/j2C 1; v>0inR f0g f0g: Indeed, if .x0; 0/2 G0, (23) holds since v>0 and 0 on G0. If .x0; 0/2 R RCn G0 and v.0; x0; 0/ < .x0; 0/ then, since vis finite at .x0; 0/, we argue exactly as in the second paragraph of this proof to obtain .v/tC Dj.v/xj2C j.v/j2C 1 > 0. We proceed similarly if 0D 0 using the arguments of the third paragraph of this proof. Hence, we obtain (23). It follows immediately from (12) of Lemma3.1and the definition of lim inf that vD 0 on f0g G0. If .x0; 0/2 R RCn G0, then we assume, by contradiction, that v.x0; 0; 0/ <1. Choose sufficiently large so that v.x0; 0; 0/ < .x0; 0; 0/. Let 18 1C D.0/v.x0; 0; 0/ ıD 1 CC:(24) ıı Notice that ıtends to infinity as ı tends to zero. Define the test function ı.x; ; t /WD ı1.jx x0j2C j0j2/ıt: Since vis lower semi-continuous, vıattains a minimum at some .xı; ı; tı/2 R Œ0; 1/ Œ0; 1/. Further, v.x0; 0; 0/ <C1 and ı.x; ; t / tends to infinity locally uniformly away from .x0; 0; 0/. Thus, .xı; ı; tı/ converges to .x0; 0; 0/ as ı tends to zero. As .xı; ı; tı/ is a minimum of vı, we see that jxıx0j2C jı0j26v ı.x0; 0; 0/:(25) We now collect four properties that hold when ı is small and rely the fact that .xı; ı; tı/ converges to .x0; 0; 0/ as ı tends to zero. Firstly, by (25), if tı> 0 for any ı then v.x0; 0; 0/ > 0 and, thus, v.xı; ı; tı/ > 0 if ı is sufficiently small due to the the lower semi-continuity of v. Secondly, (25), the lower semi-continuity of v, and the fact that v.x0; 0; 0/ < .x0; 0; 0/, imply that if ı is sufficiently small, 0 < v.xı; ı; tı/ < .xı; ı; tı/. Thirdly, the continuity of D implies that D.ı/ 6 2D.0/ for all ı sufficiently small. Fourthly and finally, since 0> 0, then ı> 0 if ı is sufficiently small. Fix ı0> 0 such that, if ı2 .0; ı0/ then all four properties above hold. Suppose that tı> 0 for some ı2 .0; ı0/. Using that vsatisfies (21) for tı> 0 and v.xı; ı; tı/ > 0, we have 0 6 t.xı; ı; tı/C D.ı/ x.xı; ı; tı/2C .xı; ı; tı/2C 1 4 2D.0/C 1 jxıx0j2C jı0j2(26) ı2C 1: Above we used that D.ı/ 6 2D.0/. Using now (25) in (26), we find 4 2D.0/C 1v.x0; 0; 0/ 0 6ıCC 1:(27) ı SUPER-LINEAR PROPAGATION FOR A GENERAL,LOCAL CANE TOADS MODEL497 In view of the definition of ı(24), the right hand side is negative. This yields a contradiction. If tıD 0 for all ı 2 .0; ı0/, the proof is the same as above, with (23) playing the role of (21). Indeed, as observed above, we have that v.xı; ı; tı/ < .xı; ı; tı/. Using this and that v satisfies (23), we find, at .xı; ı; tı/, tC Dj xj2C j j2C 1 > 0: Using the definition of and the choice of ı, we obtain the same contradiction as in (27). Having reached a contradiction in both cases, we conclude that v.x0; 0; 0/D C1. We now obtain the equation for v. The argument is slightly more complicated since v>0 and, hence, for the first equation must consider the cases where vis zero or positive. Lemma 3.3 The upper relaxed half limitvis a viscosity solution to 8 min˚.v/tC Dj.v/xj2C j.v/j2C 1; v 6 0inR RC RC; < (28) :min.v/; min˚.v/tC j.v/j2C 1; v60;onR f0g RC; and (0inG0; v.; ; 0/ Dc(29) 1inG0: Proof.The proof of Lemma3.3is similar to that of Lemma3.2, thus we omit some details and provide only a sketch of the proof. We first verify (28). Assume that, for some test function , v has a strict local maximum at .x0; 0; t0/2 R Œ0; 1/ RC. We may then choose kconverging to 0 and .yk; k; sk/ converging to .x0; 0; t0/ as k tends to infinity such that .yk; k; sk/ is a local maximum of vk and v.x0; t0; 0/D limvk.yk; k; sk/: k!1 To check (28), we need only consider the setfv> 0g since (28) is satisfied whenever vD 0. If t0> 0 and 0> 0, then for sufficiently large k, tk; k> 0 and, at .yk; k; sk/, kk 0 > tkDxxk C Dj xj2C j j2C 1evk=k: Since vk.yk; k; sk/ converges to v.x0; 0; t0/ > 0 as k tends to1, the last term tends to zero as k tends to infinity. In addition, the regularity of implies that, after taking the limit k to infinity, at .x0; 0; t0/, 0 > tC Dj xj2C j j2C 1: If 0D 0 we argue similarly as in Lemma3.2. We now consider the case t0D 0. Fix any point .x0; 0/2 G0. Using (12), we have that v converges to zero uniformly on any compact subset of G0 Œ0; 1/. Hence v.x0; 0; 0/D 0. c On the other hand, fix any point .x0; 0/2 G0, and notice that v.x0; 0; 0/D log.u0.x0; 0; 0//D log.0/D C1. It then follows immediately from the definition of lim sup that v.x0; 0; 0/D 1. This concludes the proof. 3.3The equality ofvandv As noted above, by construction, v6v. In addition, vand vare a super- and a sub-solution to the same equation with the same initial conditions except on the small set @G0. In this section, we show that vD v. 498C.HENDERSON,B.PERTHAME AND P.E.SOUGANIDIS Existence and uniqueness ofI .We outline the argument developed in Crandall, Lions and Souganidis [16] that yields that there exists a unique solution to (6) with initial condition (7). For any open, convex, C3set U , let CUWD f 2 C0.R Œ0; 1// W jU 0g and denote by S.t / the solution to (6) with the initial data 2 CU. The existence and uniqueness of S.t / are well-understood; see, [15]. In addition, arguments as in Section3.1give bounds on S.t / in R Œ0; 1/ RC. Let I.x; ; t /WD sup 2CS.t /: Following [16], we observe that I is the unique G0 maximal solution of (6). We note that, due to the Neumann boundary conditions, this does not follow directly from [16]. The extension is, however, straightforward. The equality ofvandv. Proof of Proposition1.3.First, we show that v>I . To this end, fix any 2 CG0. Observe that v.; ; 0/ > on R Œ0; 1/. The standard comparison principle, along with Lemma3.2, yields v>S.t / on R Œ0; 1/ RC. Since this is true for all , we find ID sup 2CS.t / 6 v G0: Next, we show that v6I . Fix ı > 0 and define GıWD f.x; / 2 G0W dist..x; /; G0c/ > ıg: Let IıD sup 2CS.t /. Fix any > 0. By Lemma3.1, we have that v.; ; / is finite on Gı R Œ0; 1/ and is zero on G0. Hence, there exists 2 CGısuch that v.; ; / 6 . From the comparison principle, it follows that, for all .x; ; t /2 R Œ0; 1/ Œ0; 1/, v.x; ; tC / 6 S.t/.x; / 6 supS.t /0.x; / D Iı.x; ; t /: 02CGı Taking to zero, we obtain v6Iıon R Œ0; 1/ RC. Further, it is easy to see1that, there exists ı, which tends to zero as ı does and depends only on G0and ı, such that Iı.; ; ı/2 CG0. We conclude that v.x; ; tC ı/ 6 Iı.x; ; tC ı/ 6 I.x; ; t / for all .x; ; t /2 R Œ0; 1/ RC. Taking ı to zero, we conclude that v6I , as desired. Hence we have that v6v6I 6 v, which implies that all three functions must be equal. In particular, we have that vconverges locally uniformly to I , finishing the proof. |
| [5] | The relationship betweenI , J , and w – Propositions1.5and1.6 We now characterize the location of the front in a more tractable manner; that is we prove Propositions1.5and1.6. We do not follow the approach of [18,19], in which the author shows directly that ID maxfJ; 0g by developing a theory for and checking a condition on the minimizing paths of J . As this condition is difficult to verify, we, instead, opt for a PDE proof based on the work in [26] using w in an intermediate step. We note that, since the Hamiltonian associated to (6), H.x; ; px; p/WD D./jpxj2C jpj2C 1, is not homogeneous, that is, it depends on , the arguments from [26] do not directly apply. We outline our proof below, and make note of the differences with [26]. In order to prove Propositions1.5and1.6, we show equivalence of the various level and superlevel sets involved and then we apply Theorem1.4. The inclusionfJ > 0g fI > 0g fw D 1g 1This is intuitively clear and can be observed in many ways. In the current manuscript, the quickest is, perhaps, using the inclusionfIı> 0g fwıD 1g seen in Section4.3, where wısatisfies (8) with G0replaced by Gı. A straightforward computation using (31) yields ısuch that wı.; ; ı/jG0 0, from which the claim follows. SUPER-LINEAR PROPAGATION FOR A GENERAL,LOCAL CANE TOADS MODEL499 follows from the maximum principle, as in [26]. To close this chain of inclusions, we requirefw D 1g fJ > 0g. This is accomplished in [26] via the Hopf-Lax formula; however, this only applies when the Hamiltonian is independent of .x; ; t / and so is not useful here. We get around this by using the fact that w is given as the solution to a variational problem similar to the one defining J . We can then compare these two functions directly. In order to follow this outline, we first show the following two key facts: that J is a sub-solution of (6) and that w can be represented by a variational problem. 4.1The equation forJ We first show that J solves (JtC D./jJxj2C jJj2C 1 D 0inR RC RC; (30) minf J; JtC jJj2C 1g 6 0onR f0g RC; from which it follows that J is a sub-solution of (6). The main difficulty is verifying the boundary condition. We note that J actually satisfies the Neumann boundary condition in , but this is not necessary for our purposes so we do not show it. Proof of(30). In AppendixA, we discuss how the classical arguments may be easily adapted to show that J solves (30) on RRCRC. The main point is that optimal trajectories in the definition of J exist and remain bounded away from the set R f0g, see AppendixA. As such, one may show that the dynamic programming principle is verified and argue as usual. Next, we show that min˚J; JtC jJj2C 1 6 0 on R f0g RC. For any test function ’, assume that J’ has a strict maximum at .x0; 0; t0/2 R f0g RCin a ball2Br.x0; 0; t0/. Without loss of generality, assume that .J /.x0; 0; t0/D 0 and r < t0. If’.x0; 0; t0/ 6 0 then we are finished. Hence, we may assume that ’.x0; 0; t0/ < 0. Fix any smooth function W Œ0; 1/ ! R such that .0/ D 0, .1/ D1, and .2/D 0, which is strictly increasing on Œ0; 1=2[ Œ1; 1/ and strictly decreasing on Œ1=2; 1. For any ; ı > 0, let ’ı;.x; ; t /WD ’.x; ; t/ C .=ı/: If ı > .2r /1, observe that ’ı;.x; ; t / > ’.x; ; t / > J.x; ; t / for all .x; ; t /2 Br.x0; 0; t0/, with equality only at .x0; 0; t0/. Define ıWD inf˚ı > 0 W if ı0> ı then ’ı0;> J on Br.x0; 0; t0/n f.x0; 0; t0/g : Then there exists .x; ; t/2 Br.x0; 0; t0/n f.x0; 0; t0/g such that ’ı;.x; ; t/D J.x; ; t/. First, we claim that =ı2 Œ1=2; 1. Since > J on Br.x0; 0; t0/n f.x0; 0; t0g and ./ > 0 for 2 .0; 1=2/ [ Œ2; 1/ it cannot be that =ı2 .0; 1=2/ [ Œ2; 1/. We now show that =ı… .1; 2/. We argue by contradiction, supposing that =ı2 .1; 2/. Let rWD =ı. By the construction of there exists l< rsuch that .l/D .r/. Let ıWD ır=l. Notice that 2Here, we define a ball as follows: for any .x; ; t /2 R Œ0; 1/ Œ0; 1/, let B R.x; ; t / WD f.y; ; s/ 2 R Œ0; 1/ Œ0; 1/ W jxyj2C jj2C jtsj2< R2g. In particular, we include only those points in the ambient space R Œ0; 1/ Œ0; 1/. 500C.HENDERSON,B.PERTHAME AND P.E.SOUGANIDIS ı > ı, which implies that ’ı;> J in Br.x0; 0; t0/n f.x0; 0; t0/g by the definition of ı. Notice also that =ıD l, which implies that .=ı/D .l/D .r/D .=ı/. Thus, we find J.x; ; t/ < ’ı;.x; ; t/D ’.x; ; t/C .=ı/ D ’.x; ; t/C .=ı/D ı;.x; ; t/D J.x; ; t/; which is a contradiction. Hence, =ı2 Œ1=2; 1, and, in particular, .=ı/ 6 0. Second we claim that .x; ; t/ converges to .x0; 0; t0/ as tends to zero. Fix any sequence k tending to zero as k tends to zero and extract a convergent sub-sequence, which we denote the same way, such that that ıkconverges to some ı02 Œ0; .2r/1 and .xk; k; tk/ converges to some .x00; 00; t00/2 Br.x0; 0; t0/ as k tends to infinity. By continuity, we observe that J.x00; 00; t00/D limJ.xk; k; tk/D lim’ık;k.xk; k; tk/D ’.x00; 00; t00/: k!1k!1 It follows that .x00; 00; t00/D .x0; 0; t0/ because J’ is negative in Br.x0; 0; t0/n f.x0; 0; t0/g. Since every sequence has a sub-sequence such that .xk; k; tk/ converges to .x0; 0; t0/ as k tends to infinity, we conclude that .x; ; t/ converges to .x0; 0; t0/ as tends to 0. We now verify (30) on Rf0gRC. Fix sufficiently small such that .x; ; t/2 Br.x0; 0; t0/. Notice that >ı=2 > 0, t> t0r > 0, which implies that .x; ; t/2 R RC RC. Also, notice that .x; ; t/ is the location of a local maximum of J’ı;. Hence, recalling that J solves (30) in R RC RC, at .x; ; t/, 0 > .’ı;/tC Dj.’ı;/xj2C j.’ı;/j2C 1 D ’tC Dj’xj2Cˇˇ’Cˇ2 ˇıˇˇC 1: Because .x; ; t/ converges to .x0; 0; t0/ as tends to zero, D.0/D 0, and ’ is smooth, we have ’t.x0; 0; t0/C j’.x0; 0; t0/j2C2’.x; ; t/ .=ı/C2j ıı2.=ı/j2C 1 6 o.1/: Recall that ’.x0; 0; t0/ < 0 by assumption. Hence, ’.x; ; t/ < 0 for sufficiently small. Using this and the fact that that .=ı/ 6 0, we take to zero to find that, ’t.x0; 0; t0/C j’.x0; 0; t0/j2C 1 6 0: This concludes the proof. 4.2A representation formula forw Recall that w satisfies (8) and (9). Following work of Lions [25], we define, for any .x; /2 RRC q and pD .px; p/2 R2, N.x; ; p/WD12px2=D. /C p2: Then let ˆ1 d .x; /; .y; / WDinfN.;P/ds: Ax;;f.y;/g;10 Without the boundary, it follows from [25, Section 3.4] that ( w.x; ; t /D infw.y; ; 0/W d .x; /; .y; / 6 tD0;if d .x; /; G0 6 t;(31) 1;otherwise. The modifications in our setting are straightforward, with the main difficulties handled similarly as in our treatment of J . As such, we omit it. SUPER-LINEAR PROPAGATION FOR A GENERAL,LOCAL CANE TOADS MODEL501 4.3The proofs of Proposition1.5and Proposition1.6 Proof of Proposition1.5and Proposition1.6.First, we claim thatfI > 0g fw D 1g. To begin, we note that w is a super-solution of (6) because q 2D. /p2xC p26D. /px2C p2C 1: Following [26], we let IWD tanh.I / and observe that I and w satisfy the same initial data. The maximum principle implies that I 6 w, which, in turn, givesfI > 0g fw > 0g D fw D 1g. Since tanh is increasing, we have thatfI > 0g D fI > 0g, and thus fI > 0g fw D 1g. We note that J is a sub-solution of (6) satisfying the same initial conditions as I . It follows that J 6 I . This implies thatfJ > 0g fI > 0g. Now we show thatfw D 1g fJ > 0g. We remark that it is known that this inclusion is not true in general for propagation problems, see the appendix of [26]. Fix .x; ; t /2 RRCRCsuch that w.x; ; t /D 1. It follows that d..x; /; G0/ > t . Suppose that, for the sake of contradiction, J.x; ; t / 6 0. Let 2 Ax;;G0;1be any minimizing trajectory in the formula for J . Using the Cauchy-Schwarz inequality and the fact that J.x; ; t / 6 0, we find t >J.x; ; t /C t D1CP22ds1=2>ˆtN.;P/ds=ˆtds1=2: 04D.2/400 ´t It follows that0N.;P/ds 6 t. Define the re-scaled trajectory Q W Œ0; 1 ! R RCbyQ.s/ D .st /. ThenQ 2 Ax;;G0;1. Using the definition of d and then changing variables¸ yields ˆ1ˆt d..x; /; G0/ 6N.Q; PQ/ds DN.;P/ds 6 t: 00 By hypothesis, d..x; /; G0/ > t , which is in contradiction to the inequality above. It follows that J.x; ; t / > 0, and, thus, thatfw > 0g fJ > 0g. Combining all inclusions above, we have thatfJ > 0g D fI > 0g D fw D 1g. From Theorem1.4, this yields the convergence of uto 0 infw D 1g and fJ > 0g in Proposition1.5and Proposition1.6, respectively. Taking the complements of these sets and recalling that I > 0, we see thatfJ 6 0g D fI D 0g D fw D 0g. In view of Theorem1.4, we have that uconverges to 1 on Intfw D 0g and IntfJ 6 0g. This completes the proof of Proposition1.5. To complete the proof of Proposition1.6, we must show thatfJ < 0g D IntfJ 6 0g. To this end, notice thatfJ < 0g is open, due to the continuity of J . This implies that fJ < 0g IntfJ 6 0g. On the other hand, fix any .x; ; t /2 IntfJ 6 0g and suppose for the sake of contradiction that J.x; ; t /D 0. There exists r > 0 such that Br.x; ; t / IntfJ 6 0g. It follows that J has a maximum at .x; ; t / in Br.x; ; t /, which implies, by using the constant function 0 as a test function, that 0C D 02C 02C 1 6 0: This is a contradiction. Hence, J.x; ; t/ < 0 and we obtain IntfJ 6 0g fJ < 0g. We conclude that fJ < 0g D IntfJ 6 0g. The proof of Proposition1.6is now complete. 502C.HENDERSON,B.PERTHAME AND P.E.SOUGANIDIS Appendices A. Brief comments aboutJ and w as a solutions of (30), (8) Due to the degeneracy of (30) at D 0 and the loss of coercivity of the quadratic form in the equation as tends to1, (30) falls outside the classical theory of Hamilton-Jacobi equations. In view of this, we include here some remarks that are meant to convince the reader that J and w have the usual properties, that is they satisfy the dynamic programming principle, solve respectively (30) and (8) in R RC RC, and their extremal paths are given by the Euler-Lagrange equations. Since the arguments are similar, in the remainder of the appendix we only discuss J . The main observation that we establish here is that extremal paths are bounded away from1 and 0. Lemma A.1 Suppose that Assumption1.1and Assumption1.2hold, and fix.x; ; t /2 R RC RC. Let2 H1..0; t /I R RC/ be a trajectory such that ˆt P1.s/2P2.s/2 C1ds 6 J.x; ; t /C t 04D.2.s//4 There existsCx;;t, depending only on.x; ; t / and D, such that, for all s2 Œ0; t, 2.s/ 6 Cx;;t. Proof.We proceed in two steps. First, by comparing withQ, the trajectory that connects .x; / and any point of G0linearly, we find Cx;;G0depending only on x, , and G0, such that J.x; ; t /C t 6 Cx;;G0t1. Secondly, we use obtain a bound on 2. Indeed, for any s2 .0; t/, we obtain ˆspsˆsP 2.r/2pp 2.s/DP2.r/dr 6 2sdr 6 2tJ.x; ; t /C t 6 2pCx;;G0: 004 This concludes the proof. It follows that, for any approximately extremal trajectory , 2is bounded. As a result, D.2/ is bounded from above and the quadratic form in the integrand of J is uniformly coercive. Hence any approximately extremal trajectory will be bounded in H1..0; t /I R RC/. Using compactness we obtain an extremal trajectory, ; however, we cannot rule out the existence of times s2 .0; t/ such that 2.s/D 0. We summarize the above observations in the following identity: let Ax;;G0;tWD f 2 H1..0; t /I R Œ0; 1// W .0/ D .x; /; .t/ 2 G0g, then ˆtP 1.s/2P2.s/2 J.x; ; t /DminC1ds:(A1) Ax;;G0;t04D.2.s//4 The difference between (10) and (A1) is that, in the latter, we allow trajectories to hit the boundary R f0g. The goal of the next lemma is to rule this out. c Lemma A.2 Suppose that Assumption1.1and Assumption1.2hold. Fix.x; ; t /2 G0 RCand let2 H1..0; t /I R Œ0; 1// be a trajectory such that ˆtP J.x; ; t /D1.s/2CP2.s/21ds: 04D.2.s//4 (i) For any ˛2 R, any non-empty maximal connected component of f2< ˛g includes either 0 ort as an endpoint. In particular, 2cannot have a strict local minimum. SUPER-LINEAR PROPAGATION FOR A GENERAL,LOCAL CANE TOADS MODEL503 (ii) There does not exist an non-empty interval Œs; s Œ0; t on which is constant. (iii) Fix any s02 Œ0; t. Then, for all s 2 .0; s0/, 2.s/ > minf2.s0/; g. Proof of(i). We proceed by contradiction, assuming that there exists s1; s22 .0; t/ with s1< s2, 2.s1/D 2.s2/D ˛, and .s1; s2/ f2< ˛g. We define a new trajectory Q.s/ D .s/1Œ0;s1[Œs2;t C .1.s/; ˛/1.s1;s2/. It is clear thatQ 2 Ax;;G0;t. By the monotonicity of D, we see that D.2.s// 6 D.Q2.s// for all s2 Œ0; t. Thus, from (A1) ˆt“PQ2PQ2# J.x; ; t / 61C21ds 04D.Q2/4 ˆP2ˆs2 D1CP22dsCP1.s/2dsC t Œ0;s1[Œs2;t 4D.2/4s14D.˛/ ˆP2ˆs2 <1CP22dsCP12CP22dsC t D J.x; ; t/: Œ0;s1[Œs2;t 4D.2/4s14D.˛/4 The strict inequality comes from the fact that 2.s/ < ˛ for all s2 .s1; s2/ and 2.s1/D 2.s2/D ˛. This is a contradiction, concluding the proof of claim (i). Proof of(ii). We proceed by contradiction. Suppose that is constant on Œs; s for 0 6 s < s 6 t . For the ease of notation, assume that sD t, but the general case is handled similarly. Define Q.s/ D .1.ss=t //; 2.ss=t //: We notice thatQ 2 Ax;;G0;t. Thus, from (A1), ˆt”PQ2PQ2#ˆ J.x; ; t /C t 61C2dsDs2t“P1.ss=t /2P2.ss=t /2#ds 04D.Q2/4t04D 2.ss=t / C4 sP1.s/2P2.s/2#sˆtP2 DdsD1CP22ds t04D 2.s/ C4t04D.2/4 s DJ.x; ; t /C t: t By assumption, s < t . Hence, J.x; ; t /C t D 0, which in turn implies that P 0. This is a c contradiction because .0/2 G0and .t /2 G0. This concludes the proof of claim (ii). Proof of(iii). We proceed by contradiction. Suppose that there exists s02 Œ0; t and s12 .0; s0/ such that 2.s1/ 6 minf.s0/; g. We assume that minf.s0/; g D .s0/, but the argument is similar in the other case. We first consider the case when 2.s0/ > 0. If mins2Œ0;s02.s/ < 2.s0/, fix any ˛2 .mins2Œ0;s02.s/; 2.s0//. Applying part (i), we obtain a contradiction sincef2< ˛g must have a connected component contained in .0; s0/ which does not contain 0 as an endpoint. It follows that 2.s1/D 2.s0/ and that 2.s/>2.s0/ for all s2 Œ0; s0. If maxŒ0;s12; maxŒs1;s02> 2.s0/, we can argue exactly as above, with the choice ˛2 .2.s0/; minfmaxŒ0;s12; maxŒs1;s02g/, to obtain a contradiction via part (i). Hence, we consider only the case that 2.s/D 2.s0/ for all s2 Œs1; s0, though the case 2.s/D 2.s0/ for all s2 Œ0; s1 follows similarly. By part (ii), it must be thatfs 2 .s1; s0/W P1.s/¤ 0g has positive measure. Fix > 0 to be determined, let T.s/D ..s2s1/j2s.s2C s1/j/, and define 504C.HENDERSON,B.PERTHAME AND P.E.SOUGANIDIS the trajectoryQ.s/ D .s/ C .0; T.s//1Œs1;s1.s/. It is clear thatQ 2 Ax;;G0;t. Using first that D. /D pand a Taylor expansion and then that 2 2.s0/ in .s1; s0/, we find, from (A1), ˆt”PQ 1.s/2PQ2.s/2# J.x; ; t /C t 6ds 04DQ2.s/ C4 ˆs0“P 1.s/21p2.s0/p 1T.s/C O.2/# DC 2ds s14D 2.s0/ ˆ”P 1.s/2P2.s/2# Cds Œ0;t nŒs1;s04D 2.s/ C4 ˆs0”P 1.s/2p2.s0/p 1T.s/# DC O.2/dsC J.x; ; t/ C t: s14D 2.s0/ Using the explicit form of Tand thatfs 2 .0; s1/W P1.s/¤ 0g has positive measure, the first term on the last line is negative when is sufficiently small. The above then simplifies to J.x; ; t / < J.x; ; t /, which is a contradiction. Under the assumption that 2.s0/ > 0, we have examined all cases and obtained a contradiction in each one. We conclude that, when 2.s0/ > 0, the claim holds. Suppose that 2.s0/D 0. By applying part (i) with ˛ tending to zero, we find 2.s/D 0 for all s2 Is1, where Is1is either Œ0; s1 or Œs1; t . Since D.2.s//D 0 for all s 2 Is1, it follows thatP2.s/D 0 for all s 2 Is1, otherwise J would be infinite. Thus, is constant on Is1, which contradicts part (ii). This concludes the proof. Since extremal trajectories remain bounded away from zero, they do not “see” the boundary. Hence the standard theory of Hamilton-Jacobi equations applies showing that J solves (30) and has all the expected properties. B. The precise location of the front What follows is a somewhat informal discussion of how to compute and prove the precise asymptotics of the front location in (1) when the initial data has compact support. We first discuss how to “guess” the asymptotics in terms of an abstract representation formula using the limiting equation (6). Second, we outline the main modifications to the work in [11] in order to prove this abstract guess. Finally, we compute an explicit value for this guess from the abstract formula. The work below is not rigorous, but it is a simple exercise to turn this discussion into a proof. Connecting the front location with the Hamilton-Jacobi equation.We make precise what we mean by “front” in this context. For a solution u of (1), we refer to the region where x > 0 and maxu.x; ; t / transitions from 1 to 0 as the front, see Figure2. As we shall see, up to lower order terms, it is enough to fix any m2 .0; 1/ and track the level set of u of height m; that is we may define the front as maxfx W 9 > 0; u.x; ; t/ D mg, cf. [11, Section 1]. We discuss, heuristically, that the front location corresponds to the location of the boundary of the zero level set of I when G0D f.0; 0/g. We do this by noting of fJ D 0g D @fI D 0g (see SUPER-LINEAR PROPAGATION FOR A GENERAL,LOCAL CANE TOADS MODEL505 Section4), and using J for all computations. Due to the fact that the assumption G0D f.0; 0/g falls outside the scope of this paper (cf. Assumption1.2), all discussion in this subsection is not rigorous; however, we discuss below how to make it rigorous (see the next subsection). Roughly, to see the connection between the solution u of (1) with the function J of (10) we proceed as follows. Fix t > 0 and let tD t1. For any .x; ; s/2 R RC Œ0; 1/, define ut.x; ; s/D uxpD.1=t/;;s!andvtD tlog ut: ttt From Proposition1.6, we expect that, as t tends to infinity, and thus ttends to zero, ( uxtpD.t /; t; stD ut.x; ; s/!1;if J.x; ; s/ < 0;(B1) 0;if J.x; ; s/ > 0; and that vtconverges to J , where J is given by (10) with the set G0to be determined. Since u.; ; 0/ has compact support, it follows that ut.; ; 0/ tends to zero locally uniformly on f.0; 0/gc and that ut.0; 0; 0/ is positive. Heuristically, we then expect vt.; ; 0/ to converge to 0 on f.0; 0/g and1 on f.0; 0/gc. This, in view of the convergence of vtto J , suggests that, in the definition of J (10), we should take G0D f.0; 0/g. Let xfWD max˚x W 9 > 0; J.x; ; 1/ D 0 D max ˚x W minJ.x; ; 1/D 0 :(B2) The second equality above is easy to check by hand. It is also easy to observe thatjxj < xfimplies that maxJ.x; ; 1/D 1 and jxj > xfimplies that maxJ.x; ; 1/ < 0. Returning to (B1) and setting sD 1, we expect that as t tends to 1, if jxj > xfthen u.xtpD.t/; t; t/ converges to 0, while ifjxj < xfthen u.xtpD.t/; t; t/ converges to 1 for some . This suggests that the front location is given by xftpD.t /C otpD.t /:(B3) How to make this rigorous.There are two approaches that one may use to make (B3) rigorous. The first is to develop the theory of “maximal solutions” to accommodate cases such as ours, where G0is not a smooth open set, but, instead, a one-point set. The second is to re-use the approach of [11], in our context. For simplicity, we discuss the second approach now. A slightly stronger assumption than Assumption1.1is that there exists a C1function FW RC! Cand a real number p > 0 such that D. /=F . / converges to 1 and @ Rlog F converges to p as tends to |
| [6] | This is satisfied by the example (2), with the choice F . /D log. C 1/. Under this hypothesis, the strategy from [11], may be repeated with the following minimal adaptation. We first discuss the proof that the front location is bounded below by (B3). Let xfbe as above and define fto be a point such that J.xf; f; 1/D minJ.xf; ; 1/D 0. The lower bound in [11] is obtained by building a sub-solution along moving, growing ellipses. The major difficulty in adapting this strategy in our context is identifying the correct trajectory for the ellipse to follow. Let .X; / be the optimal trajectory in the definition of J (10) beginning at .X.0/; .0//D .0; 0/ and ending at .X.1/; .1//D .xf; f/. Instead of using the trajectories in [11, equation (4.9)], we define, for any large time T > 0, and use the trajectory T.t /D .XT.t /; T.t // where .XT; T/ 506C.HENDERSON,B.PERTHAME AND P.E.SOUGANIDIS satisfy ˆtsF T.s/P T.t /D T .t=T /;andXT.t /DX .s=T /ds: 0.s=T /p From our assumptions, we see that F .T/ D.T/ pD.T /, so that XT.T / xfTpD.T /. In the case considered in [11], F . /D and the trajectories above are exactly those used in [11]. Once the trajectories have been determined, one may complete the proof of the lower bound exactly as in [11] with all further modifications straightforward. The reason for the additional assumptions on the regularity of F can be seen in the hypotheses of [11, Lemma 4.1]. This yields, for all m2 .0; 1/, max˚x 2 R W 9 > 0; u.x; ; t/ > m lim inf>xf: t !1tpD.t/ On the other hand, an upper bound may be easily obtained using the Hamilton-Jacobi set-up (see, for a similar argument, [12]). We note that the explicit upper bound in [11] cannot be used here since it is a particularity of the case D. /D . We conclude that, for all x > xf. limsupu.x; ; t /D 0: t !1p x>xtD.t /; >0 In particular, this implies that max˚x 2 R W 9 > 0; u.x; ; t/ D m limD xf: t !1tpD.t/ and we conclude that the front location is given by (B3), as claimed. Computingxfusing(B2)We now compute xfexplicitly. We have the Hamiltonian system for .X; ; P; Q/: XPD 2P D;PD 2Q;andPPD 0;QPD D0P2;(B4) with the boundary conditions .X.0/; .0//D .0; 0/ and .X.1/; .1// D .xf; f/. We see that PD A=2 for some constant A depending on .xf; f/. Next, we differentiate the equation for to get that RD 2 PQDA2D0./: Multiplying this by P and integrating, we find 2 .s/P2D P.0/2A2D .s/:(B5) Further, from (B4), we see that PXD AD./. Hence, we have that ˆ1XP2P2! 0D J.xf; f; 1/DC1ds 04D./4 ˆ1A2D./.0/P2A2D./! DC1ds; 044 which implies that P.0/D 2. To compute A, we need the following key observation: SUPER-LINEAR PROPAGATION FOR A GENERAL,LOCAL CANE TOADS MODEL507 Lemma B.1 Suppose that.X; / is the minimizing trajectory given above. Then P.s/ > 0 for all s2 Œ0; 1/ and P.1/D 0. Heuristically this is because any downward motion in could be used instead to increase X for the same “cost.” The details of the proof can be seen in [12]. From LemmaB.1and (B5), we find .s/.4PA2D.//1=2D 1: Using that D./ D pand integrating, we see that 1 .s/DFp1.2s/;(B6) ´s where D 22=pA2=pand Fp.s/D0.1rp/1=2dr. Using (B6) along with the condition.1/PD 0, we see that 0 D.1/P2D 4 A2p.Fp1.2//p. Re-arranging this and using the formula for , this yields 1D Fp1.2/, which can be re-written 2D Fp.1/. To compute Fp.1/, we use the easy-to-establish identities ˆ1ˆ1p rp2 pdrD1rpdr;(B7) 01rpp0 andˆˆ 1rpdrD11rp11p1r drD1ˇ 1;3;(B8) 0p0pp2 where ˇ is the beta function, see [1, Section 6.2]. Using (B7) and (B8) yields ˆ11rpC rppC 2ˆ1pp Fp.1/DpdrD1rpdrDC 2ˇ 1;3:(B9) 01rpp0p2p2 Thus, we have a formula for , which, in turn, yields a formula for AD 2p=2. Having computed A, we are now in a position to conclude. Indeed,XPD AD./ D pFp1.2s/pDA4Fp1.2s/p: Using (B7) and (B8), we find ˆ1ˆ1 xfD X.1/ DFp1.2s/pdsD4prpdrD42ˇ 1;3 A0AFp.1/01rpAFp.1/p2p2 81 D: ApC 2 The third equality comes from the change of variables rD Fp1.2s/. Plugging in for A, we have !1 818pC 2 13 2C pA2C pp2ˇp;2:(B10) We use the well-known identities ˇ.x; y/pD.x/ .y/= .xC y/,.xC 1/ D x .x/, and .1=2/D (see [1, Section 6]), to simplify the expression involving ˇ. Indeed, pC 2 13pC 22ppC 221Cp1p1Cp1 p2p2Dp23C1D2p1C11C1 D1C1 : 2p2p2p2p Combining this with (B10), yields the desired expression (11). 508C.HENDERSON,B.PERTHAME AND P.E.SOUGANIDIS Acknowledgments.CH was partially supported by the National Science Foundation Research Training Group grant DMS-1246999. BP has been supported by the French “ANR blanche” project Kibord: ANR-13-BS01-0004. PS was partially supported by the National Science Foundation grants DMS-1266383 and DMS-1600129 and the Office for Naval Research Grant N00014-17-1-2095. References |
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