Nonlnear Dffer. Equ. Appl. 21 (2014), 589 620 c 2013 Sprnger Basel 1021-9722/14/040589-32 publshed onlne December 31, 2013 DOI 10.1007/s00030-013-0259-5 Nonlnear Dfferental Equatons and Applcatons NoDEA Watng tme effect for moton by postve second dervatves and applcatons Qng Lu Abstract. We dscuss the watng tme effect for the evoluton of a planar graph governed by ts postve part of second dervatve. For any smooth perodc functon whch contans fntely many convex peces n one perod, we show that the watng tme s contnuous by usng comparson arguments. Moreover, we show that the convex parts keep expandng n sze n a strct manner, whch answers an open queston posed by Kohn and Serfaty (Commun Pure Appl Math 59:344 407, 2006) nthsspecal case. The results on watng tme effect are also appled to the statonary problem of mean curvature type on an unbounded nonconvex doman for our study of ts game-theoretc nterpretaton. Mathematcs Subject Classfcaton (1991). 49L25, 35J93, 35K93, 49N90. Keywords. Vscosty solutons, Determnstc games, Curvature flow equatons. 1. Introducton In ths paper, we study the moton of a functon drven by the postve part of ts second dervatve. The smplest example s as follows. Suppose u(x, t) s a functon defned on R (0, ) and satsfes { ut (u xx ) + =0 nr (0, ), (1.1) u(x, 0) = u 0 (x) n R, where a + denotes max{a, 0} for any a R. It s a specal case of second-order Hamlton Jacob Bellman equatons related to optmzaton problem of the controlled Brownan moton. In spte of the seemngly smple structure, (1.1) s actually a fully nonlnear, degenerate ellptc equaton of non-dvergence form. But the standard vscosty soluton theory stll apples. We get the exstence and unqueness of solutons wthout much dffculty; see [6, 11] and [12].
590 Q. Lu NoDEA We focus our attenton to the geometrc propertes of the vscosty soluton. Notce that the equaton wll be turned nto the usual heat equaton provded that the soluton u(x, t) s known to be convex n the varable x; n partcular, when the ntal data u 0 s convex, then the convexty s preserved durng the whole evoluton, as proved n [13] and [18] etc. On the other hand, f u 0 s concave, then t s clear that u 0 tself s the unque statonary soluton of (1.1). A general problem remans ncomplete about the stuaton when the convex and concave peces both exst ntally. In ths case, a reasonable result should be that the convex part of the curve moves mmedately as n the former case whle the nonconvex part stays at the ntal poston for a whle before startng to evolve, whch s the so called watng tme effect. One may actually defne the watng tme for any pont on the graph of u 0,whchwe denote by T 0. Then our predcton above amounts to sayng that T 0 (z) =0 when z s on the convex part of u 0 and T 0 (z) > 0 when z s on the concave part. In ths paper, we ntend to gve rgorous and precse descrptons of the watng tme, especally ts contnuty n space. The key turns out to be the nvestgaton of the moton of nflecton ponts of u 0. 1.1. Moton by the postve part of second dervatve In general the dstrbuton and structure of the nflecton ponts of u 0 could be very complcated but n order to smplfy our ntroducton and grasp the essence of our problem, we pck a qute specal but typcal nonconvex ntal graph. We frst assume that u 0 s perodc and of C 2 class. Let I =(a, b) be one perod of u 0 such that u 0 (a) =u 0 (b) = max R u 0. For the moment, we add a condton of unque convex pece, sayng that u 0 s dvded by only two nflecton ponts at x = α and x = β(α <β) so that u 0 s (strctly) convex on (α, β) and (strctly) concave on I\[α, β]. See Fg. 1. Under the assumptons above, we consder the followng general one-dmensonal problem: { u t h(u x )(u xx ) + =0 nr (0, ), u(x, 0) = u 0 (x) n R, (1.2) where h s a gven functon satsfyng (A1) h C(R) and mn p R h(p) > 0 for any R>0. Ths assumpton s used to exclude the degeneracy caused by h. It s clear that (1.1) s ts specal case. We now state the smplest verson of our man results on the watng tme. Theorem 1.1. Assume (A1). Letu be the soluton of (1.2) wth perodc ntal condton u 0 of class C 2.Letα t and β t stand for the x-coordnates of the only two nflecton ponts of u(x, t) n a perod [a, b] at each tme t>0 wth α 0 = α and β 0 = β. Then the followng statements hold: (1) α t and β t are contnuous n t; (2) α t s strctly decreasng and β t s strctly ncreasng n t; (3) α t a and β t b as t.
Vol. 21 (2014) Watng tme for moton by postve second dervatves 591 Fgure 1. The ntal graph y = u 0 (x) Snce (α t,β t ) conssts of the locatons of all convex ponts of u(x, t) at t, the above consequence amounts to sayng that the convex parts of u keep expandng n a strct manner. As a result, we easly obtan the contnuty of the watng tme T 0 wth respect to x. In Sect. 3, we gve a more general verson of Theorem 1.1. The key ngredent s the nvestgaton of the moton of all nflecton ponts. Note that n ths general settng the nflecton ponts may collde each other durng the evoluton. 1.2. Motvaton: the game nterpretaton of mean curvature flow Let us ntroduce the motvaton of the above analyss about the watng tme. We frst remark that the equaton { ut (uxx)+ 1+u =0 nr (0, ), 2 x (1.3) u(x, 0) = u 0 (x) n R s another specal case of (1.2). Ths equaton s the graph formulaton of the socalled moton by postve curvature, whose correspondng level-set formulaton s as follows: ( ) U U t U dv =0 nr 2 (0, ), (1.4) (E1) U + U(x, 0) = U 0 (x) n R 2, (1.5) where U 0 s a defnng functon of the ntal curve. Ths equaton s a lttle dfferent from the normal mean curvature flow equaton, whch s ndependently studed n [4] and [8]. We refer to [12] for detals on the well-posedness of (E1) n the framework of vscosty solutons. It also has applcatons n mage processng [21]. We remark that watng tme effect was studed for other geometrc motons. For example, one may refer [5] for ths phenomenon for Gauss curvature flows. Our equaton s clearly dfferent from thers.
592 Q. Lu NoDEA Our watng tme results partally answered an open problem proposed by Kohn and Serfaty [16] about (E1). They asked whether or not for any nonconvex curve n the plane there exsts a free boundary separatng the (movng) convex part and the (statonary) concave part. They also asked whether t s true that the concave part decreases monotoncally n sze and any part of the curve never stops once t starts to move. It s known that n general the answers are negatve f no regularty condtons more than Lpschtz contnuty are assumed; see the example gven by G. Barles and F. Da Lo n [16, Appendx C.3]. We gve affrmatve answers to all of these questons for any smooth perodc graph wth fntely many concave peces. Another closely related motvaton s to study the determnstc game nterpretaton for the statonary counterpart of (E1) also proposed by Kohn and Serfaty [16]. For a bounded doman Ω R 2, they ntroduced a famly of ext tme games wth a parameter ɛ>0, whose value functons T ɛ (z) converge, under several condtons on Ω, to the soluton T of the statonary problem of mean curvature type: (E2) ( ) T T dv 1 = 0 n Ω, (1.6) T T =0 on Ω, (1.7) whch s sometmes called normalzed 1-Laplace equaton. Ths approxmaton gves an nterestng representaton of the solutons to the equaton. See also [17] for the determnstc game approach to general ellptc and parabolc equatons and [22 25] for a stochastc tug-of-war game approach to the p-laplace equaton wth p>1. Related extensons of ths new method to the Hesenberg group are recently addressed n [9, 10]. More precsely, t s proved n [16] that the relaxed sem-lmts T and T are respectvely a subsoluton and a supersoluton of (E2) and t suffces to use the comparson prncple to conclude the convergence T ɛ T as ɛ 0. The usual comparson theorem to guarantee the unqueness of contnuous solutons of ths Drchlet problem s known only when Ω s convex [8] and the desred convergence above follows easly n ths case. It s however less complete when the convexty of Ω s dropped. Wthout any assumptons on the convexty or regularty of Ω, the solutons can easly become dscontnuous, as s shown agan n the example of G. Barles and F. Da Lo n [16, Appendx C.3]. The Drchlet boundary condton may not be realzed n the strct sense and therefore the usual comparson prncple does not hold. The best one can expect n the nonconvex case seems to be a unque result for possbly dscontnuous solutons by showng the so-called weak comparson prncple for solutons wth boundary condton nterpreted n the vscosty sense; namely, f U and V are respectvely a subsoluton and a supersoluton wth the boundary condton (1.7) n the vscosty sense, then U V and U V n Ω, where W and W are respectvely the lower and upper semcontnuous envelopes of any bounded functon W. Applyng ths weaker comparson result, one may obtan the game approxmaton for the possbly dscontnuous soluton and the convergence s certanly n a weaker sense.
Vol. 21 (2014) Watng tme for moton by postve second dervatves 593 1.3. Applcaton of results for watng tme The above weak approach does work but t requres Ω to be star-shaped [16]. Wthout the star-shapedness, even the weak comparson prncple s not necessarly true. An example nvolvng Ω of fgure-eght type s provded n [19], where a necessary condton, related to the fattenng phenomenon, for the weak comparson prncple s gven as well. In ths work, we wll show that T ɛ may actually converge to a contnuous soluton of (1.6) n a nonconvex (unbounded) doman. We prove the convergence for the case when Ω s represented by the graph of a functon u 0 satsfyng the perodcty and havng fntely many smooth concave parts n each perod, as was descrbed prevously. In ths specal case, we prove that the soluton wll become contnuous. Ths mprovement s obtaned manly due to the mproved regularty of the boundary Ω. Our method s to pose stronger boundary condtons whch the lmt of T ɛ satsfes. The man dfference from [16] les n the followng two aspects. 1. We stress that under our assumptons on Ω, the doman Ω s unbounded. The problem on solvablty of (E2) n an unbounded doman seems to be new to our best knowledge. We overcome ths dffculty by lnkng the value functons T ɛ wth the game values approxmatng the soluton of (E1). We are able to determne an effectve doman. In fact, we show that Γ:={(x, y) R 2 : y = max u 0 } dvdes Ω nto a half plane, n whch T ɛ =, and nfntely many congruent bounded regons, where T ɛ s locally bounded. Let Ω e be one of the regon. We also show that T ɛ (z) when z Ω e satsfes dst(z,γ) 0. Ths means that a sngular Drchlet boundary condton on Γ appears n the lmt PDE problem. Sngular boundary problems are studed only for second-order semlnear equatons by Lasry and Lons [20] and for frst-order Hamlton Jacob equatons n [1, 3, 7] wth applcatons to large tme asymptotcs n [14, 15]. Our sngular boundary problem s dfferent from all these works, snce our equaton s quaslnear and degenerate ellptc. 2. As the boundary of Ω e s composed of Γ and Ω. We need to gve another boundary condton on Ω. It turns out that the lmtng boundary value of T ɛ as ɛ 0 s nothng else but the watng tme T 0 of (1.3). Here we apply our contnuty results about the watng tme. The game nterpretaton of (E1) wth U 0 beng a defnng functon of Ω s used agan to connect (1.3) and (E2). Wth the preparaton above, we can characterze the lmt of T ɛ as the unque contnuous soluton T of the followng problem wth mxed boundary condtons: (E3) ( ) T T dv 1=0 T nω e, T (z) =T 0 (z) for all z Ω, (1.8) T (z) as dst(z,γ) 0, (1.9)
594 Q. Lu NoDEA where T 0 s the watng tme we obtaned prevously. Note that (1.8) snow nterpreted n the strct sense and therefore the usual comparson theorem follows easly. We prove that T ɛ T unformly n any compact subset of Ω e. By usng the weak approach as n [16], we may show that the soluton T of (E3) s also a soluton of ( ) T T dv 1=0 nω e, T (E4) T (z) = 0 for all z Ω, (1.10) T (z) as dst(z,γ) 0, (1.11) where (1.10) s fulflled n the vscosty sense. Ths ndcates that f one can follow [16] to get a unque weak soluton of (E4), then t must be contnuous and satsfy (E3). We are not able to show the unqueness of weak solutons but a comparson prncple s gven to show that the game-related contnuous soluton T s the bggest among all of the weak solutons n the effectve doman. Indeed, our comparson prncple, whose proof s smlar to the argument used for state constrant problems [26], states that any upper semcontnuous subsoluton s not above any contnuous supersoluton. The unqueness of weak solutons wll be completed under current assumptons f we can show a symmetrc comparson result, whch says that any contnuous subsoluton s not larger than any lower semcontnuous supersoluton. We gve a very smple example, revealng that ths s mpossble wthout makng any extra assumptons on the accessblty of the supersoluton. We fnally remark that t s possble to extend our arguments to a bounded doman wth analogous regularty by applyng the analyss of watng tme locally. Ths paper s organzed n the followng way. In Sect. 2 we brefly ntroduce some basc propertes of the Eq. (1.2). In Sect. 3 we study n detal the watng tme effect and show ts contnuty. We prove a general verson of Theorem 1.1. The applcaton to the statonary level-set equaton of mean curvature type s presented n Sect. 4. We establsh the assocated games and show the convergence of game values to the unque contnuous soluton of the statonary problem. Some dscussons ncludng a comparson prncple are also presented for the maxmalty of the contnuous soluton among all weak solutons. Notaton. In ths artcle, we use the followng notatons. For any z R n and r>0, we use B r (z) to denote the open ball n R n centered at z wth radus r. For any z 1,z 2 R 2, we denote by z 1 z 2 the lne segment between z 1 and z 2,.e., z 1 z 2 := {kz 1 +(1 k)z 2 : k [0, 1]}. For any functon u : R n+1 R of class C 2,weuseu x,u t and u xx j to denote the partal dervatves of u,.e., u x := u/ x, u t := u/ t and = 2 u/ x x j for all, j =1, 2,...,n. We denote the gradent of u n u xx j
Vol. 21 (2014) Watng tme for moton by postve second dervatves 595 space by u := ( u/ x 1, u/ x 2,..., u/ x n ). We also wrte u to represent the dervatve du/dx provded that u s a functon of one varable x. 2. Moton by postve second dervatves Let us begn wth a bref revew of some basc results about the Eq. (1.2), whch gves a general descrpton of the moton of a graph by ts postve second dervatve n one dmenson. Wth the condton (A1), we fnd that (1.2) s a fully nonlnear degenerate parabolc equaton. For any classcal soluton u C 2 (R (0, )), u satsfes u t h(u x )u xx = 0 when u xx 0 but u t = 0 when u xx < 0, whch reveals that (1.2) s actually a combnaton of a parabolc second-order equaton and a (trval) frst order equaton and the type change occurs at the nflecton ponts of u. It s also clear that (1.3) s a specal case of (1.2). In spte of the nonlnearty, we can stll get a unque soluton n the framework of vscosty soluton theory. Theorem 2.1. Assume (A1). Letu 0 be a Lpschtz contnuous functon on R. Then there exsts a unque vscosty soluton u of (1.2). Moreover, u s Lpschtz contnuous n space. We refer to [6] and [12] for the defnton of vscosty solutons and the proof for Theorem 2.1. The soluton of (1.2) obvously enjoys monotoncty n tme. Lemma 2.2. Assume (A1). Letu 0 be a Lpschtz contnuous functon on R. Let u be the soluton of (1.2). Thenu(x, t) u(x, s) for any x R and t s 0. Snce u satsfes u t 0 n R (0, ) n the vscosty sense, we can easly show the monotoncty by followng [2, Lemma 5.15] or [15, Lemma 4.4]. Remark 2.1. It s easly seen that Lemma 2.2 can be extended to Cauchy- Drchlet problems. The proof s analogous and therefore omtted here as well. We assume the followng perodcty n space for smplcty. (A2) u 0 s a perodc Lpschtz contnuous functon on R. LetI := (a, b) denote one of ts open perods satsfyng u 0 (a) =u 0 (b) = max u 0(x) x R and u 0 <M n I.
596 Q. Lu NoDEA We present all our analyss n the nterval I. Denote by Ω 0 the epgraph of u 0 n I,.e., Ω 0 := {(x, y) R 2 : y>u 0 (x),x I}. (2.1) Let M := max x R u 0 (x) andm = mn x R u 0 (x). Note that the constant M s a supersoluton, whch, by comparson prncple, mples that the soluton u satsfes u(a, t) =u(b, t) =M for all t 0. The large-tme behavor of (1.3) s qute smple for ths specfc ntal value. Proposton 2.3. (Large-tme behavor) Assume (A1) and (A2). Let u be the soluton of (1.2). Thenu(x, t) M = max x R u 0 (x) unformly for all x R as t. Proof. One may frst take the relaxed lmts of u as t : u(x) = lmsup t u(x, t) and u(x) = lmnf u(x, t). t Snce the constant M s a supersoluton and by Lemma 2.2 u s nondecreasng n t, wehaveu 0 (x) u(x, t) M, whch mples that u 0 u u M n R. In partcular, we have u(a) = u(a) = u(b) = u(b) = M. Hence by the standard stablty theory for vscosty solutons, we may prove u n I s a supersoluton of { h(ux )(u xx ) + =0 ni =(a, b) (2.2) u(a) =u(b) =M. We assert that u = M n I. Indeed, f there exsts x 0 I such that u(x 0 ) <M, then we may fnd a smooth (quadratc) functon φ n [a, b] satsfyng φ(a) = φ(b) =M, u(x 0 ) <φ(x 0 ) <Mand φ xx > 0nI. It s obvous that there exsts x I such that mn (u φ)(x) =u(x) φ(x) < 0. x [a,b] By the defnton of vscosty supersolutons, we get h(φ x (x))(φ xx (x)) + 0, whch yelds that φ xx (x) 0. Ths contradcts the fact that φ xx > 0nI. It s now clear that u = u = M, whch mples the unform convergence of u(x, t) to the constant soluton of (2.2) ast. 3. Watng tme effect A very specal property of the moton by postve second dervatves (1.2) s ts watng tme before movng. We dvde the perod I nto a convex part, a concave part and the remanng part. In order to further smplfy our exposton and show the essence of our argument, let us make the followng assumptons. (A3) Let I 0 = m =1 {α 0,β0}, where m Z and α0,β 0 I such that (a) α0 <β0 <α0 +1 <β0 +1 for all =1, 2,...,m 1;
Vol. 21 (2014) Watng tme for moton by postve second dervatves 597 (b) u 0 s convex n I + = m =1 (α 0,β0); (c) u 0 s of class C 1 n I := I\(I + I 0 ) wth u 0 strctly decreasng n each nterval of I ; (d) For any α0,β 0 I 0, k α, u 0 (x) u 0 (α 0 := lm 0) u 0 (x) u 0 (α x α 0 x α0 lm 0) x α 0 + x α0 and k β, u 0 (x) u 0 (β 0 := lm 0) u 0 (x) u 0 (β x β0 + x β0 lm 0) x β0 x β0 ; (e) k α, 0 <k β, 0 for any =1, 2,...,m. For our convenence of notaton, we let α0 m+1 = b and β0 0 = a. The assumptons (a) (c) gve a partton of the graph by convexty. Note that we do not explctly assume any regularty, more than Lpschtz contnuty, of the convex part I + but we assume some relaton between the left and rght dervatves at the nflecton ponts n (d), whch, together wth (a) (c), mples that u 0 s semconvex. The assumpton (e) roughly states that each convex pece (α0,β 0) contans strct convex ponts. It s easy to see that a smooth graph wth fntely many convex and concave peces fulflls (A3). Snce α0 and β0 correspond to the nflecton ponts n the smooth case, we stll call them nflecton ponts n our general settng (A3) as well. 3.1. Exstence of watng tme We frst show the general exstence of watng tme effect. Smlar results for (E1) are shown n [16, Theorem 1.7] by a game approach. Lemma 3.1. (Exstence of watng tme) Assume (A1) (A3). Letu be the soluton of (1.2). For any x I, there exsts τ x > 0 dependng on x such that u(x, t) =u 0 (x) for all t [0,τ x ]. Proof. Fx an arbtrary x 0 I and take w(x, s) =Bx + C wth B,C R properly chosen so that there exst x 1,x 2 I satsfyng x 1 < x 0 < x 2, w(x 0, 0) = u 0 (x 0 )and w(x, 0) >u 0 (x) for all x [x 1,x 0 ) (x 0,x 2 ]. By contnuty of the soluton u(x, t), there exsts τ>0such that u(x 1,s) < w(x 1, 0) and u(x 2,s) <w(x 2, 0) for any s [0,τ]. Hence, w s a supersoluton of v s h(v x )(v xx ) + =0 n(x 1,x 2 ) (0,τ), v(x 1,s)=u(x 1,s)andv(x 2,s)=u(x 2,s) for all s [0,τ], (3.1) v(x, 0) = u 0 (x) for all x [x 1,x 2 ]. The comparson theorem yelds that u(x 0,s) w(x 0,s) u 0 (x 0 ) and therefore u(x 0,s)=u 0 (x 0 ) for any s [0,τ] by the monotoncty of u n tme derved n Lemma 2.2.
598 Q. Lu NoDEA For any z 0 =(x 0,u 0 (x 0 )) R 2, defne T 0 (z 0 ):=sup{τ : u(x 0,t)=u 0 (x 0 ) for any t τ} (3.2) We call T 0 (z 0 )thewatng tme of the curve y = u 0 (x) atz 0. Note that under the assumptons (A1) (A3), for any x I such that u 0 (x) <M= lm t u(x, t), we have T 0 (x) < by Proposton 2.3. 3.2. Short tme convexty of movng peces In what follows, we nvestgate the behavor of nflecton ponts when tme t s small. Defne for any =0, 1, 2,...,m αt +1 := sup{x : u(x, t) =u 0 (x), β0 x α0 +1 }, (3.3) βt := nf{x : u(x, t) =u 0 (x), β0 x α0 +1 }. Lemma 3.1 asserts that the defntons above are vald and βt <αt +1 when t>0 s suffcently small. For each =0, 1,...,m,let τ := sup{t [0, ) :βt <αt +1 }. It s therefore clear that τ > 0 for all =0, 1,...,m. Denote τ := mn τ =sup{t [0, ) :βt <αt +1 for all =1, 2,...,m}, (3.4) =0,1,...,m whch gves a more precse bound n t for whch the defntons n (3.3) are vald. Moreover, Lemma 2.2 mples the monotoncty of αt +1 and βt;thats, αt +1 αs +1 and βt βs for all 0 s t<τ. We may also defne ατ +1 and β τ by lettng t = τ n (3.3). In fact, α +1 τ due to the contnuty of the soluton u. = lm t τ α+1 t and β τ = lm t τ β t Remark 3.1. We remark that τ can be. When τ =, wemusthavethe followng stuaton: for any =0, 1,...,m, there exsts x [β0,α 0 +1 ] such that u 0 (x )=M = max x I u 0 (x). Indeed, suppose by contradcton that there exsts some such that u 0 (x) <M ɛ wth ɛ>0 for all x [β0,α 0 +1 ]. Then by (3.3) and the fact that τ =, we obtan that (βt,α t +1 ) [β0,α 0] for all t>0, whch yelds the exstence of a certan x(t) (βt,α t +1 ) such that u(x(t),t) = u 0 (x(t)) < M + ɛ. By takng a subsequence, we may let x(t) x I as t. By Proposton 2.3, we end up wth u 0 (x )=M, whch s a contradcton. We next study the moton of all nflecton ponts when t<τ. Remark 3.2. Under the assumptons (A1) (A3), we may easly obtan for any t<τ and any =0, 1,...,m u(x, s) =u 0 (x) for all x (βt,α t +1 )ands [0,t). (3.5) Indeed, from the defnton of αt +1 and βt and the contnuty of u, wehave u(αt +1,t)=u 0 (αt +1 )andu(βt,t)=u 0 (βt), whch by Lemma 2.2 mples that u stays statonary at αt +1 and βt before tme t; n other words, u(αt +1,s)=
Vol. 21 (2014) Watng tme for moton by postve second dervatves 599 u 0 (αt +1 )andu(βt,s)=u 0 (βt) for all 0 s t. Notng that u(a, s) =u(b, s) = u 0 (a) =u 0 (b) =M for all s 0, we fnd that u(x, s) s actually a soluton of u s h(u x )(u xx ) + =0 n(βt,α t +1 ) (0,t), u(βt,s)=u 0 (βt)andu(α t +1,s)=u 0 (αt +1 ) for all s [0,t], u(x, 0) = u 0 (x) for all x [βt,α t +1 ]. We are thus led to (3.5) due to the fact that u 0 s the unque soluton of the above problem. Theorem 3.2. (Contnuty n tme of nflecton ponts) Assume (A1) (A3). Let u be the soluton of (1.2). Letα t,β t be defned as n (3.3) for all =1, 2,...,m and t [0,τ ),whereτ s gven n (3.4). Thenα t,β t :[0,τ ) I are both contnuous functons. Proof. Let us only show the contnuty for βt wth any fxed =1, 2,...,m.We fx t 0 [0,τ ). Our goal s to show lm t t0 βt = βt 0. Notce that βt 0 <αt +1 0, snce t<τ. 1. We frst show the rght contnuty,.e., lm t t β t = βt 0. 0+ Indeed, by the monotoncty of βt n t, we have lm nf t t0+ β t β t0.on the other hand, by Remark 3.2, u(x, t 0 )=u 0 (x) for all x (βt 0,αt +1 0 ). Set ũ(x, t) =u(x, t + t 0 ). Then ũ s the soluton of (1.2) wth ntal data u(x, t 0 ). Usng Lemma 3.1 for ũ and any x (βt 0,αt +1 0 ), we may take τ = t t 0 > 0 small such that ũ(x, s) =u(x, t 0 ) for all s [0,τ], whch yelds that u(x, t) =u(x, τ + t 0 )=u 0 (x). We therefore get βt <xas long as t>t 0 s suffcently close to t 0. If follows that lm sup t t0+ βt βt 0. 2. We next consder the left contnuty. Thanks to the monotoncty of αt +1 and βt n t, we may assume by contradcton that there exsts an ncreasng sequence t n t 0 as n such that lm n β t n <βt 0 δ 0 for some δ 0 > 0 and lm n α+1 t n αt +1 0. It follows from Remark 3.2 that u(x, t n )=u 0 (x) for all x [βt n,αt +1 n ), whch, by contnuty of u, mples that u(x, t n ) = u 0 (x) and therefore u(x, t 0 )=u 0 (x) for all x [βt 0 δ 0,αt +1 0 ). Ths contradcts the defnton of βt +1 0. Remark 3.3. The second half of our proof above can be used to show that there exsts =0, 1, 2,...,msuch that ατ +1 = β τ. Indeed, there s =0, 1, 2,...,m such that τ = τ.fxsuch. By defnton, we easly get βτ α+1 τ. Both sdes are actually equal, for otherwse we may utlze the exstence result of
600 Q. Lu NoDEA watng tme, Lemma 3.1, for the ponts between them and the ntal tme t = τ to derve a contradcton that τ >τ. Lemma 3.3. (Support functons at nflecton ponts) Assume (A1) (A3). Let u be the soluton of (1.2). For any =1,...,m and t [0,τ ),lety = L α, t (x) and y = L β, 1 t (x) be the equatons of tangent lnes to u 0 respectvely at αt and βt 1.Then u(x, t) max{l α, t (x),l β, t (x)} for all x [αt,β t]. (3.6) Proof. The statement obvously holds when t = 0 due to (A3). We only show u(,t) L β, t ( ) n[αt,β t]fort>0. The concluson follows by usng a symmetrc argument to prove u(,t) L α, t ( ) n[αt,β t]. Suppose by contradcton that there exsts t>0 such that there s x 1 [αt,β t] satsfyng u(x 1,t) <L β, t (x 1 ), whch yelds that u(x 1,s) <L β, t (x 1 )for any 0 s t. Then we may take y 1 (u(x 1,t),L β, t (x 1 )). Take another tangent lne y = L (x) tou 0 passng through (x 1,y 1 ). We assume that they are tangent at x = x 0. Then we must have x 0 <βt.wepck x 2 from the nterval (βt,α t +1 ), whch s nonempty due to t<τ, and denote y 2 = L (x 2 ). It s obvous that u(x 2,s)=u 0 (x 2 ) <y 2 for all s [0,t]. It s now clear that y = L (x) s a supersoluton of the Cauchy-Drchlet problem (3.1) wth τ = t. Snce u s the soluton of the same equaton, by comparson theorem for (3.1), we have u(x, s) L (x) for all x [x 1,x 2 ]and s [0,t], and n partcular, u(x 0,t) L (x). Ths mples that u(x 0,t)= u 0 (x 0 ). On the other hand, by defnton of βt,wegetu(x 0,t) >u 0 (x 0 ), whch s a contradcton. Remark 3.4. The concluson n Lemma 3.3 holds for t = τ as well. We only need to send the lmt n (3.6) ast τ and the contnuty of u yelds u(x, t) max{l α, τ (x),lβ, τ (x)} for all x [α τ,β τ ]. For any =1, 2,...,m and t (0,τ ), denote Ω t := {(x, y) R 2 : y>u(x, t), α t <x<β t}. (3.7) An mmedate consequence of Lemma 3.3 s as follows. Lemma 3.4. Assume (A1) (A3). For any =1, 2,...,m and t (0,τ ),letω 0 and Ω t be defned as n (2.1) and (3.7) respectvely. Then for any z 1,z 2 Ω t, we have z 1 z 2 Ω 0. Proof. Set S := {(x, y) R 2 : y L α, t (x) L β, t (x),y u 0 (x) andαt x βt}. Then by defnton we have S Ω 0. It s also easly seen that S s convex. On the other hand, Lemma 3.3 and Lemma 2.2 mples that Ω t S. Hence the convex hull of Ω t s contaned n Ω 0.
Vol. 21 (2014) Watng tme for moton by postve second dervatves 601 Lemma 3.5. (Convexty of movng graph peces) Assume (A1) (A3). Letu be the soluton of (1.2). For any =1, 2,...,m and t (0,τ ),letαt and βt be defned as n (3.3). Then u(kx 1 +(1 k)x 2,t) ku(x 1,t)+(1 k)u(x 2,t) for any k [0, 1], provded that x 1,x 2 [αt,β t]. Proof. Set for any x [x 1,x 2 ]ands [0,t], v(x, s) :=u(x 2,s) x x 1 + u(x 1,s) x 2 x. x 2 x 1 x 2 x 1 It follows from Lemma 3.4 that v(x, 0) u 0 (x). It s also obvous that v(x 1,s)=u(x 1,s)andv(x 2,s)=u(x 2,s) for all s [0,t]. Moreover, due to Lemma 2.2, we obtan v s (x, s) 0 n the vscosty sense for any x [x 1,x 2 ] and s (0,t). Snce v s lnear n space varable, the observatons above yeld that v s a supersoluton of (3.1) wth λ = t. Therefore by comparson prncple, we have v(x, t) u(x, t) for all x [x 1,x 2 ]. Our concluson mmedately follows. Remark 3.5. We remark that the statement n Lemma 3.5 s true for t = τ by contnuty of the soluton u. 3.3. Local bounds for the moton of nflecton ponts Our frst goal s to show the strct expanson of each convex pece for all t [0,τ ). To ths end, we need to gve bounds for αt and βt for every =1, 2,...,m. Roughly speakng, we wll fnd an nterval (a,b ) such that (αt,β t) (a,b ) [β0 1,α0 +1 ] and the slopes of the tangents to u 0 n (a,α0] are greater than those n [βt,b )fort τ. To be more precse, let us follow the procedure below to get (a,b ) for any fxed =1, 2,...,m. Step 1. We start wth the followng obstacle problem: mn f C([β0 1,α0 +1 ]) subject to { f 0n(β0 1,α0 +1 ) n the vscosty sense, f u 0 n (β0 1,α0 +1 (3.8) ). Owng to (A3), we can easly fnd a smooth concave functon f C 2 ([β0 1, α0 +1 ]) satsfyng f u 0 n (β0 1,α0 +1 ), f(β0 1 )=u 0 (β0 1 )andf(α0 +1 )= u 0 (α0 +1 ). Applyng Perron s method, we can get a unque soluton u 0 C([β0 1,α0 +1 ]) of (3.8), whch s concave n (β0 1,α0 +1 ) and satsfes that u 0(β0 1 )=u 0 (β0 1 )andu 0(α0 +1 )=u 0 (α0 +1 ). In addton, u 0 s harmonc at x (β0 1,α0 +1 ) where u 0 (x) u 0 (x). Step 2. Take a =sup{ξ (β0 1,α0 +1 ):u 0(x) =u 0 (x) for all x (β0 1,ξ)}, (3.9) b =nf{ξ (β0 1,α0 +1 ):u 0(x) =u 0 (x) for all x (ξ,α0 +1 )}. (3.10) It s obvous that a αt <βt b by defntons. By the assumpton (A3)(e), we have a <α0 and b >β0. It follows that u 0 >u 0 n (a,b ), whch mples that
602 Q. Lu NoDEA Fgure 2. Plot of u 0, a and b u 0 s a straght lne on (a,b ) wth slope k := u 0(b ) u 0 (a ). (3.11) b a See Fg. 2 for a sketch of a, b and u 0. Proposton 3.6. (Slope bounds for convex peces) Assume (A1) (A3). Let a,b be gven as n (3.9) (3.10) and u 0 be the soluton of (3.8) for every =1, 2,...,m.Then lm x a u + 0(x) lm x b u 0(x). (3.12) Remark 3.6. The nequalty (3.12) reduces to u 0(a ) u 0(b )fa and b are known to appear n (β0 1,α0 +1 ). We keep the weak form (3.12) because t s possble that a = β0 1 or b = α0 +1 and we do not make any further assumptons on the regularty of u 0 at the nflecton ponts. Proof. Recall that u 0 s of class C 1 n (a,α0) (β0,b ). The relaton (3.12) follows easly, snce the graph of u 0 s a straght lne segment above u 0 n [a,b ] wth u 0(a )=u 0 (a )andu 0(b )=u 0 (b ). The functon u 0 s mportant n that t plays the role of the asymptotc profle of (1.2) ast provded that each convex pece evolves wthout beng nfluenced by the others. Lemma 3.7. (Local large-tme behavor) Assume (A1) (A3). Leta,b be gven as n (3.9) (3.10) and u 0 be the soluton of (3.8) for every =1, 2,...,m.Let v(x, t) be the soluton of v t h(v x )(v xx ) + =0 n (a,b ) (0,λ), v(a,t)=u 0 (a ) and v(b,t)=u 0 (b ) for all t [0,λ), (3.13) v(x, 0) = u 0 (x) for all x [a,b ]
Vol. 21 (2014) Watng tme for moton by postve second dervatves 603 wth λ =. Thenv(x, t) u 0(x) unformly for x [a,b ] as t. Moreover, v(x, t) < u 0(x, t) for all (x, t) (a,b ) [0, ). (3.14) Proof. Suppose that the slope of the affne functon u 0 s k. Set ṽ(x, t) :=v(x, t) k(x a ). Then ṽ s the soluton of ṽ t h(ṽ x + k)(ṽ xx ) + =0 n(a,b ) (0, ), ṽ(a,t)=u 0 (a ) and ṽ(b,t)=u 0 (a ) for all t [0, ), ṽ(x, 0) = u 0 (x) k(x a ) for all x [a,b ]. (3.15) Notcng that u 0 (a ) s the unque statonary soluton of the equaton above, we may apply the same argument as n the proof of Proposton 2.3 to get that ṽ(x, t) u 0 (a ) unformly for all x [a,b ]ast, whch amounts to sayng that v(,t) u 0( ) n[a,b ]. We next show (3.14). We frst show that for any t 0, mn ṽ(x, t) <u 0(a ). (3.16) x (a,b ) Let us construct a supersoluton of (1.2). Let W 0 (x) =u 0 (a )+Asn(ωx + d) wth A, ω and d properly chosen such that [a,b ]saperodofw 0 wth W 0 (a )=W 0 (b ) = max W 0 R and W 0 (x) >u 0 (x) k(x a ) for all x [a,b ]. We solve the heat equaton W t CW xx =0 nr (0, ) (3.17) wth C>0andW (x, 0) = W 0 (x). The unque soluton of the Cauchy problem s W (x, t) =u 0 (a )+Ae Cω2t sn(ωx+d). We thus have W (a,t)=w(b,t) > u 0 (a ) for all t 0. In addton, W (x, t) sconvexnxwhenever W (x, t) < u 0 (a ). Snce W u 0 (a ) s clearly another soluton of (3.17), W (x, t) = mn{w (x, t),u 0 (a )} s a supersoluton of (3.17). We next clam that W restrcted n [a,b ] [0, ) s also a supersoluton of (3.15) provded that C>0slarge enough. Indeed, the boundary condtons are clearly satsfed. Also, by calculaton, we have sup W x (x, t) Aω for any C>0. (x,t) R (0, ) Now lettng C max p Aω h(p + k), we have W t h(w x )W xx W t CW xx provded that W xx 0. It follows that W (x, t) s a supersoluton of (3.15) n the regon where W < u 0 (a ). Besdes, t s easy to see that W (x, t) satsfes the defnton of supersolutons for (x, t) (a,b ) (0, ) wth W (x, t) >u 0 (a ) because W s just a constant around those ponts. Our last verfcaton s for those (x, t) satsfyng W (x, t) =u 0 (a ) but n fact W cannot be tested from below there.
604 Q. Lu NoDEA By comparson prncple for (3.15), we obtan ṽ(x, t) W (x, t) and (3.16) follows mmedately. Notcng that u 0(x) > v(x, t) n(a,α t] [β t,b ) due to (A3)(c) and the convexty of u n [α t,β t], we easly deduce (3.14) from (3.16). The argument above can be adapted to the proof of the followng general result. Theorem 3.8. Suppose v s the vscosty soluton of v t h(v x )(v xx ) + =0 n (a,b ) (0,T), v(x 1,t)=u 0 (x 1 ) and v(x 2,t)=u 0 (x 2 ) for all t [0, ), v(x, 0) = u 0 (x) for all x [x 1,x 2 ] wth u 0 (x) < u 0 (x) = (x 2 x)u 0 (x 1 )+(x x 1 )u 0 (x 2 ) for any x (x 1,x 2 ). x 2 x 1 Then v(x, t) u 0 (x) as t unformly n x [x 1,x 2 ]. Moreover, v(x, t) < u(x) for every (x, t) (x 1,x 2 ) [0, ). Lemma 3.7 asserts that an ndependent convex pece wll eventually become a straght lne and ths asymptotc state s realzed only at t =. We next utlze ths fact to deduce the followng key result. For any t (0,τ ], we set k α, t := lm x α t u(x, t) u(α t,t) x α t and k β, u(x, t) u 0 (β t := lm t) x βt + x βt. (3.18) Lemma 3.9. (Strct upper bounds of convex peces) Assume (A1) (A3). Letu be the soluton of (1.2). Letαt,β t be defned as n (3.3) and u 0 be defned as n (3.8). Then u(x, t) < u 0(x) for any (x, t) (αt,β t) [0,τ ) (3.19) and k α, t for any =1, 2,...,m and t [0,τ ),wherek α, t <k β, t, (3.20) and k β, t s gven n (3.18). Proof. We frst note that (3.19) s a straghtforward consequence of Lemma 3.7, snce u restrcted for t [0,τ ) s the soluton of (3.13) wth λ = τ.toshow (3.20), let us frst dscuss the case that τ =. As mentoned n Remark 3.1, there are x [β0,α 0 +1 ] such that u 0 (x ) = M = max x I u 0 (x) foreach = 0, 1,...,m. Lemma 2.2 yelds that u(x,t) = M for all t 0, whch mples that αt,β t [x,x +1 ] for any t 0andx 1 a <b x wth = 1, 2,...,m. We next show that n fact αt,β t (x 1,x ). If t s not the case, say αt = x 1 for some fxed t 0and =0, 1,...,m, then by Lemma 3.3, the graph of u(x, t) s above the tangent lne to the graph of u 0 at αt = x 0, whch s exactly y = M, for all x [αt,β t]. It follows that u(x, t) M
Vol. 21 (2014) Watng tme for moton by postve second dervatves 605 for all x [a,b ], whch contradcts Lemma 3.7. By the strct concavty of u 0 n [x 1,α 0) and n (β 0,x ], we mmedately get k α, t < 0=u 0(x 1 )=u 0(x ) <k β, t for all t 0. We then dscuss the case τ <. Let us show that ether αt >a or βt <b for all t [0,τ ). We agan argue by contradcton. Suppose on the contrary that αt = a and βt = b. As n the proof of Proposton 3.6, the slope k of u 0 n (a,b ) s equal to ether u 0(a )oru 0(b ), whch gves u(x, t) u 0 n (a,b ) by Lemma 3.3. Ths s a contradcton to Lemma 3.7, snce u can be vewed as the soluton of (3.13) for any when t<τ. When αt >a and βt <b,tseasytoget(3.20), agan due to the strct concavty of u 0 n [a,α0) (β0,b ]. Suppose αt > a and βt = b. Then b <α +1 0 for otherwse t contradcts the fact that t<t. Ths means that k β, t = u 0(b ). On the other hand, the concavty of u 0 yelds k α, t < lm x a+ u 0(x). Therefore (3.20) follows easly from Proposton 3.6. The other case that αt = a and βt <b s smlarly treated. Remark 3.7. Wth slght modfcaton n the proof above, we may also prove that u(,τ ) < u 0( ) n (ατ,β τ ) (3.21) and k α, τ <kβ, τ (3.22) for all = 1, 2,...,m provded that τ <. Indeed, Lemma 3.3 and Lemma 3.7 stll enable us to deduce that ether ατ > a or βτ < b n ths case. The proof s exactly the same as that of Theorem 3.20 f ατ >a and βτ <b. We therefore assume, for nstance, that ατ >a and βτ = b. We are not able to exclude the possblty that b = α0 +1 ths tme but we may nstead use Lemma 2.2 and (A3)(d) to show that k β, τ kα,+1 0 = lm x b u 0(x), whch, combned wth Proposton 3.6, mples (3.22). The strct upper boundedness (3.21) s also easly shown, snce u(ατ,τ ) < u 0(ατ )andu(,τ )sconvexn [ατ,β τ ], as mentoned n Remark 3.5. Remark 3.8. By Lemma 3.9 and Remark 3.7, we must have αt 1 >βt 0 a and βt m <αt m+1 = b ether for all t [0, ) fτ = or for all t [0,τ ]f τ <. 3.4. Short tme behavor of the graph Lemma 3.10. (Immedate move of a corner) Assume (A1). Assume that u 0 C(R) s perodc. Let u be the soluton of (1.2) wth ntal data u 0.Iftheres x 0 R such that u 0 (x) u 0 (x 0 ) u 0 (x) u 0 (x 0 ) < lm < lm <, (3.23) x x 0 x x 0 x x 0+ x x 0 then u(x 0,t) >u 0 (x 0 ) for any t>0.
606 Q. Lu NoDEA Proof. Assume by contradcton that there s t 0 > 0 such that u(x 0,t)=u 0 (x 0 ) for all t [0,t 0 ]. Snce u(x, t) s nondecreasng n t, by the assumptons gven, t s easy to construct a smooth perodc functon w : R [0,δ] such that (1) w(x, t) =w 0 (x) for all x R and t [0,t 0 ]; (2) w 0 (x) u 0 (x) for all x R and w 0 (x 0 )=u 0 (x 0 ); (3) w 0 (x 0 ) > 0. Then t follows that w satsfes w t (x 0,t) h(w x (x 0,t))(w xx (x 0,t)) + < 0 for all t (0,t 0 ), whch contradcts the fact that u(x, t) s a soluton of (1.2) asw(x, t) touches u(x, t) from below at the pont (x 0,t) wth t arbtrarly taken from (0,t 0 ). The argument n the proof above can also be generalzed to show a smlar stuaton durng the evoluton. Lemma 3.11. If u(x, t) s a contnuous and space-perodc soluton of (1.2) satsfyng u(x, t 0 ) u(x 0,t 0 ) u(x, t 0 ) u(x 0,t 0 ) < lm < lm < x x 0 x x 0 x x 0+ x x 0 for some x 0 R and t 0 0. Thenu(x 0,t) >u(x 0,t 0 ) for all t>t 0. Let ũ(x, t) =u(x, t + t 0 ). Then ũ s the soluton of (1.3) wth ntal data ũ(x, 0) = u(x, t 0 ). We reach the concluson by applyng Lemma 3.10. Theorem 3.12. (Nonstop movng of convex peces) Assume (A1) (A3). Letu be the soluton of (1.2) and τ be gven as n (3.4). Thenu(x, t) >u(x, s) for all x [α s,β s] and 0 s<t<τ. Proof. Suppose by contradcton that there are s 0, δ>0andx 0 [α s,β s] such that u(x 0,t)=u(x 0,s) for all t [s, s + δ]. By Lemma 3.5, wehave u(x, s) u(x 0,s) u(x, s) u(x 0,s) < lm lm <. x x 0 x x 0 x x 0+ x x 0 We may derve a contradcton mmedately by usng Lemma 3.11 f u(x, s) u(x 0,s) u(x, s) u(x 0,s) lm < lm. x x 0 x x 0 x x 0+ x x 0 We therefore only dscuss the case that u(x, s) u(x 0,s) u(x, s) u(x 0,s) lm = lm, x x 0 x x 0 x x 0+ x x 0 whch means u(,s) s dfferentable at x 0. Let us denote by ˆk the quantty above,.e., the dervatve at x 0. By Lemma 3.3 and Lemma 3.5, wehave, we must have ether ˆk >ks α, or ˆk <ks β,. Let us assume the latter wthout loss of generalty. Note that n ths case, we may take a smooth perodc functon w 0 whch fulflls the followng condtons: u 0(α s) ˆk u 0(β s). Snce Lemma 3.9 states that k α, s <k β, s
Vol. 21 (2014) Watng tme for moton by postve second dervatves 607 (1) Its graph s symmetrc about (x 0,u 0 (x 0 )),.e., w 0 (x) =2u 0 (x 0 ) w 0 (2x 0 x). (2) It serves as a one-sded lower bound n the sense that w 0 (x) =ˆkx for all x [x 0,x 0 + δ 1 ) and wth some δ 1 > 0andw 0 (x) <u(x, s) for all x (β s, ). (3) Its frst dervatve s nondecreasng n [x 0,β s]and w 0(x) k>ˆk for all x (βs, ). Set K =sup x w 0(x) and c = mn p K h(p). It s clear that c>0due to (A1). Let w be the unque smooth soluton of a heat equaton { wt cw xx =0 nr (s, ), (3.24) w(x, s) =w 0 (x) n R. We then utlze the symmetry to get w(x 0,t)=u 0 (x 0 )andw xx (x 0,t)=0for all t s. We adopt the strong maxmum prncple for w x to obtan w x (x 0,t) > w 0(x 0 )fort sand therefore w x (x 0,t) > ˆk u(x, t) u(x 0,t) = lm for all t>s. (3.25) x x 0+ x x 0 Moreover, the maxmum prncple for w xx mples that w xx 0n(x 0, ) [s, s + δ). Besdes, there exsts μ (0,δ) small such that w(βs,t) u(βs,t)for all t [s, s + μ]. It s not dffcult to see that w also satsfes w t h(w x )(w xx ) + 0 n (x 0,βs) (s, s + μ), w(x, s) u 0 (x) for all x [x 0,βs], (3.26) w(x 0,t) u(x 0,t)andw(βs,t) u(βs,t) for all t [s, s + μ]. Indeed, snce the ntal and boundary condtons are satsfed, as proved above, we only need to show that w fulflls the frst nequalty. Note that the maxmum prncple for (3.24) gves w x (x, t) <K for all x R and t 0. Snce w stays convex n space n (x 0, ) (s, s + μ). Then a drect calculaton shows that w t = cw xx h(w x )(w xx ) + n (x 0,βs) (s, s + μ), whch completes the verfcaton that w fulflls (3.26). As a result, u(x, t) w(x, t) for all x [x 0,βs]andt [s, s + μ), whch n turn mples that u(x, t) u(x 0,t) lm w x (x 0,t) > x x 0+ x x ˆk u(x, t) u(x 0,t) = lm 0 x x 0 x x 0 for any t (s, s+μ). In partcular, the relaton above holds for t = s+μ. Snce μ<δ, Lemma 3.11 now comes nto play for us to obtan u(x 0,s+δ) >u(x 0,s), whch s a contradcton. We conclude the proof by pontng out that a symmetrc verson of the argument above apples to the case when ˆk >k α, s.
608 Q. Lu NoDEA Corollary 3.13. (Short tme behavor of nflecton ponts) Assume (A1) (A3). Then for any =1, 2,...,m and 0 s<t<τ, we have αt <αs and βt >βs. Proof. We only show βt >βs. By Lemma 2.2, t s not dffcult to fnd that every movng pece, whch s defned on [αt,β t], s expandng; n other words, we have αt αs and βt βs for all t>s 0. It suffces to show that βt βs but ths s a straghtforward consequence of Theorem 3.12 and the contnuty of the soluton u. 3.5. Global behavor of the graph In the prevous secton we have gven a rgorous analyss for the moton of the graph and nflecton ponts before the frst collson tme τ. We are nterested n the evoluton after τ n what follows. Our basc dea s to regard the soluton u(x, t) as a new ntal condton every tme when two nflecton ponts ht each other and apply the precedng results repeatedly. To be more precse, we nvestgate the shape of the graph of u(x, t) at t = τ < below. Lemma 3.14. Assume (A1) (A3). Letu be the soluton of (1.2) and τ be defned as n (3.4). Assume τ <. Then there exsts I0 = m =1 {α 0, β 0}, where m Z wth 1 m <mand α 0, β 0 I such that (1) α 0 < β 0 < α +1 0 < β +1 0 for all =1, 2,...,m 1; (2) u(,τ ) s convex n I+ := m =1 (α 0, β 0); (3) u(,τ ) s of class C 1 n I := I\(I+ I0 ) wth x u x (x, τ ) strctly decreasng n each nterval of I ; (4) For any α 0, β 0 I0, k α, u 0 (x) u 0 (α 0 := lm 0) x α 0 + x α 0 k β, u 0 (x) u 0 (β 0 := lm 0) x β 0 x β 0 u 0 (x) u 0 (α lm 0) x α 0 x α 0 u 0 (x) u 0 (β lm 0) x β 0 + x β. 0 (5) k α, 0 < k β, 0 for any =1, 2,...,m. Proof. It follows from Remark 3.3 that there exsts =0, 1,...,m such that β τ = α+1 τ. (3.27) Let us take the least, denoted by 1, satsfyng (3.27) for our explanaton. Assume that j := 1 + j 1 also satsfes (3.27) for all j =1, 2,...,l wth l m. We frst clam that 1 > 0and l < m.if 1 = 0, then u 0(a) = u 0(βτ 0 )=u 0(ατ 1 ) = 0, whch, by Remark 3.4 and Remark 3.5, yelds that u 0(βτ 1 ) = 0 too. Ths s a contradcton to Remark 3.7. Weget l <mfor the same reason. We therefore let α 0 = ατ for all =1, 2,..., 1, β 0 = βτ for all =1, 2,..., 1 1andβ 1 0 = β l+1 τ. We contnue relabelng the nflecton ponts n ths way f l +1<mand get α 0 and β 0 for =1, 2,...,m wth and
Vol. 21 (2014) Watng tme for moton by postve second dervatves 609 m <m. Ths gves the partton n (1). Also, t s clear that (3) holds, snce I I and u(x, τ )=u 0 (x) for all x I. Usng Remark 3.4, we conclude (4) wth ease. We next show (2) and (5). By Remark 3.5 and Remark 3.7, we may assert that u(,τ ) s convex n each nterval (α τ,β τ )and k α, τ <kβ, τ (3.28) for all =1, 2,...,m. Snce (α 0, β 0)=(ατ,β τ ) for all =1, 2,..., 1 1, the convexty and the slope relaton stll hold for these ntervals. We thus need to prove them n the nterval (α 1 0, β1 0 )= l+1 j=1 (αj 0,βj 0 ). We adopt Remark 3.4 and obtan u(,τ ) L β,j τ n [αj τ,βj τ ] and u(,τ ) L α,j+1 τ n [αj+1 τ,βj+1 τ ] for all j =1, 2,...,l, from whch the convexty follows easly, snce L β,j L α,j+1 τ. Fnally, we combne the relaton kβ,j τ = kα,j+1 τ for j =1, 2,...,l wth (3.28) and get τ = k α,1 0 = k α,1 τ <kβ,1+l τ = kα,1 0. The proof of (2) and (5) for the other nflecton ponts α 0 and β 0 s analogous. Lemma 3.14 amounts to sayng that u(,τ ) satsfes the ntal condton (A3). Snce t obvously satsfes (A2) as well, we may dscuss the graph moton after t = τ by repeatng the analyss presented n Sects. 3.2, 3.3 and 3.4 for only fnte tmes. In order to state our man theorem, we defne for any t 0 I t := m =0(βt,α t +1 ), I0 t = I I t and I+ t = I\(I t t 0). Theorem 3.15. (Long term nonstop movng of convex peces) Assume (A1) (A3). Letu be the soluton of (1.2) and αs and βs be defned for any s 0 as n (3.3). Thenu(x, t) >u(x, s) for all t>s 0 and x I+ s I0. s Theorem 3.16. (Contnuty of watng tme) Assume (A1) (A3). Then the watng tme T 0 (z) of the curve y = u 0 (x) at z =(x, u 0 (x)) s contnuous wth respect to x I. Moreover, T 0 (x) as [a, b] x x 0 when x 0 = a or b. Proof. To smplfy our notaton n the proof, we vew T 0 as a functon of x,.e., T 0 (x) :=T 0 (x, u 0 (x)). It s clear that T 0 (x) = 0 for all x I +. Let us prove the contnuty of T 0 for any x 0 I\I +. Wthout loss, we assume x 0 [β 0,b). We frst show the contnuty of T 0 at any fxed x I such that T 0 (x) [0,τ ). We have three dfferent cases to dscuss. 1. T 0 (x) = 0. Then we have x I + I 0 due to Theorem 3.1. In other words, there exsts =1, 2,...,m such that x [α0,β 0]. By Corollary 3.13, for any δ>0, we have [α0,β 0] (αδ,β δ ). We are thus allowed to take r>0 small such that (x r, x + r) (αδ,β δ ), whch means that T 0(y) <tfor all y (x r, x + r).
610 Q. Lu NoDEA 2. T 0 (x) =s (0,τ ). Then x a and x b due to Remark 3.8. We clam there exsts = 1, 2,...,m such that x = αs or x = βs. Indeed, x / m+1 =1 (β 1 s,αs), for otherwse T 0 (x) >s, as a result of Theorem 3.1 appled to the soluton of (1.2) wth ntal data u(x, s). Also, x / m =1 (α s,βs), snce, by Corollary 3.13, T 0 (x) <sfor every x n the set. For any δ>0 small, we use Corollary 3.13 agan to get r>0 such that (x r, x + r) (αs+δ,β s+δ )\(α s δ,β s δ ), whch mples that T 0(y) (s δ, s + δ) for any y (x r, x + r). We extend our argument to the remanng case when T 0 (x) τ by consderng the problem wth updated ntal condtons. In addton, for the endponts a and b, then we may apply Theorem 3.8 wth v = u, x 1 = a and x 2 = b to conclude that T 0 (x) as x = a or b. 4. Applcaton to statonary level-set equatons In ths secton, we turn to nvestgate the statonary level-set equaton (E2). We attempt to obtan a game-theoretc nterpretaton for ths equaton when Ω s represented by the graph of a perodc smooth functon wth multple concave parts. Our games are played n an unbounded doman. Let the step sze of space be ɛ>0. Suppose that the game starts from z R 2 and there are two players, Player I and Player II. The followng set of game rules s repeated durng the game. Game rules: Inthe-th round ( =1, 2,...,N) wth poston η R 2, (1) Player I frst chooses a vector v R 2 wth v 1. (2) Player II has a rght to support or reverse Player I s choce; that s, Player II may take b = ±1. (3) Once the decsons above have been made, we update the poston, lettng t be η + 2ɛb v. There are two ways to proceed and end ths game: 1. (Fnte horzon problem:) One way s to set a fxed endng tme t 0. To be more precse, we take N =[t/ɛ 2 ]. Let y(t; z,b,v) be the game state after N rounds. We are then able to get U 0 (y(t; z,b,v)), whch n fact depends on the ntal poston z, duraton t, step sze ɛ and certanly the decsons b, v of both players. Suppose Player I wants to maxmze U 0 (y(t; z,b,v)) and Player II wants to mnmze t. The value functon s thus defned as U ɛ (z,t) = max v 1 mn b 1 max v 2 mn b 2 max mn U 0 (y(t; z,b,v)) (4.1) v N b N 2. (Tme optmal problem:) The other way s to consder the frst ext tme from Ω; namely, the game wll end only when the poston frst leaves Ω. The endng round N tself depends on the strateges of both players. We let τ ɛ (z; b, v) :=Nɛ 2 denote the frst tme of ext from Ω. If Player I attempts to mnmze τ ɛ and Player II tres to maxmze τ ɛ, then we may defne value functon for ths game as
Vol. 21 (2014) Watng tme for moton by postve second dervatves 611 T ɛ (z) = mn v 1 max b 1 mn max τ ɛ (z; b, v). (4.2) v 2 b 2 It s clear that T ɛ (z) =0fz/ Ω. We adopt the conventon that T ɛ (z) = f the game startng from x cannot be ended. The frst type of game provdes us wth the approxmaton and, as a byproduct, the exstence of solutons of (E1). Notce that the unqueness follows from the comparson prncple; see [12]. Theorem 4.1. [16] Let U ɛ be the value functon defned n (4.1). IfU 0 s bounded and unformly contnuous, then U ɛ U locally unformly as ɛ 0, whereu s the vscosty soluton of (E1). The lnk between the level set and graph formulatons can be easly seen from the followng. Proposton 4.2. Suppose that u 0 : R R s a unformly contnuous functon and u s the unque soluton of (1.3). LetU(x, y, t) =u(x, t) y for all x, y R and t 0. ThenU s the unque soluton of (E1) satsfyng U 0 (x, y) =u 0 (x) y. Proof. The proof s straghtforward. Note that the ntal condton s clearly satsfed. We now show that U s a subsoluton of (eq: E1a). Suppose there are (x 0,y 0 ) R 2, t 0 > 0andφ C (R 2 (0, )) such that max (U φ) =(U φ)(x 0,y 0,t 0 ). R 2 (0, ) Snce U(x, y, t) =u(x, t) y, y u(x 0,t 0 ) y φ(x 0,y,t 0 ) attans a maxmum at y 0. Then t s obvous that φ y (x 0,y 0,t 0 )= 1. (4.3) On the other hand, (x, t) u(x, t) y 0 φ(x, y 0,t) attans a maxmum at (x 0,t 0 ). Then by the defnton of subsolutons of (1.3), we have φ t (x 0,t 0 ) (φ xx(x 0,t 0 )) + 1+φ 2 x(x 0,t 0 ) 0, whch, together wth (4.3), yelds ( ) φ φ t φ dv 0 at (x 0,y 0,t 0 ). φ + The proof for supersoluton verfcaton s smlar. We now turn our attenton to the game wth ext tme. The dynamc programmng prncple for ths game s T ɛ (z) = mn max T ɛ (z + 2ɛbv)+ɛ 2. (4.4) v 1 b=±1 for all z such that for any v 1, there exsts b = ±1 satsfyng z + 2ɛbv Ω. We let T ɛ (z) =ɛ 2 f there s v 0 1 such that z + 2ɛbv / Ω for both b = ±1. The value functon T ɛ s supposed to converge, as ɛ 0, to a soluton of the correspondng statonary problem lke (E2) n an analogous manner. It however turns out to be qute dffcult n general. As was descrbed above, the
612 Q. Lu NoDEA man dffculty s the loss of comparson theorems due to the lack of (strct) boundary data. We am to overcome ths dffculty and present a game-based convergence theorem by usng our results n the prevous secton on the watng tme. To ths end, we assume (A4) the boundary Ω :={(x, y) R 2 : y = u 0 (x)} and Ω = {(x, y) R 2 : y>u 0 (x)}, where u 0 : R R s a perodc functon fulfllng (A2) (A3). Let us defne the relaxed lmts of T ɛ for any z Ω T (z) = lmsup T ɛ (z) and T (z) = lmnf T ɛ (z) ɛ 0 ɛ 0 A natural queston arsng from ths defnton s whether T ɛ (z) (also T and T ) s fnte for all z Ω. The answer s negatve because of the unboundedness of the doman. Indeed, let Γ = {(x, y) R 2 : y = M(= max I u 0 )}. Then Γ dvdes the doman Ω nto a half plane and countably many congruent parts. Takng one of the congruent parts, we denote Ω e = {(x, y) I R : u 0 (x) <y<m}, (4.5) whch s connected due to (A4). Roughly speakng, T ɛ (z) s fnte when z Ω e but s nfnte when z / Ω e. In order to prove ths, we need the connecton between tme-dependent games and the assocated statonary games. Lemma 4.3. (Game connectons) Assume (A4). LetU(z,t) be the unque soluton of (E1) wth a choce of U 0 satsfyng {z R 2 : U 0 (z) < 0} =Ω. Let T ɛ be the value functon defned as n (4.2). Then for any z Ω and t 0, (1) U(z,t) > 0 mples t T (z); (2) U(z,t) < 0 mples t T (z). Proof. Fx z 0 Ωandt 0arbtrarly. (1) Our goal s to show T (z 0 ) t under the assumpton that U(z 0,t) > 0. We may take δ>0such that U ɛ (z,t) >δfor all z B δ (z 0 ) owng to Theorem 4.1. It means that for any z B δ (z 0 ) there exsts a strategy v z of Player I such that U 0 (y(t; z,v z,b)) >δ,.e., y(t; z,v z,b) / Ωfor any b. Note that here we abuse the notaton, usng v and b to denote strateges, whch represent the sequences of choces of both players. As a result, by defnton we have T ɛ (z) <tfor all z B δ (z 0 ), whch mples that T (z 0 ) t. (2) We now prove T (z 0 ) t provded that U(z 0,t) < 0. Applyng Theorem 4.1 as above, we have δ > 0 such that U ɛ (z,t) < δ for all z B δ (z 0 ). We thus can ensure the exstence of a strategy b z of Player II satsfyng U 0 (y(t; z,v,b z )) < δ for any v; n other words, y(t; z,v,b z ) Ω. We clam that y(s; z,v,b z ) Ω for all s t. If t s not true,.e., there exsts s 0 t such that y(s 0 ; z,v,b z ) / Ω. Then Player I may keep choosng the trval opton v = 0 for the rest of the fnte horzontal game. Ths strategy of Player I yelds y(t; z,v,b z ) / Ω, whch s clearly a contradcton. Hence,