On asymptotic speed of solutions to level-set mean curvature flow equations with driving and source terms
OON ASYMPTOTIC SPEED OF SOLUTIONS TOLEVEL-SET MEAN CURVATURE FLOW EQUATIONSWITH DRIVING AND SOURCE TERMS
YOSHIKAZU GIGA, HIROYOSHI MITAKE, AND HUNG V. TRAN
Abstract.
We investigate a model equation in the crystal growth, which is describedby a level-set mean curvature flow equation with driving and source terms. We establishthe well-posedness of solutions, and study the asymptotic speed. Interestingly, a newtype of nonlinear phenomena in terms of asymptotic speed of solutions appears, whichis very sensitive to the shapes of source terms.
Contents
1. Introduction 21.1. Problem and Background 21.2. Main results 32. Wellposedness 63. Heuristic observation 93.1. Trotter-Kato product formula 93.2. Notations and Spherical symmetric case 104. Spherical symmetric case 114.1. The case R < n − R > n − R = n − f F ± [ E ]( t ) 227. Conclusion 268. Appendix 27References 28 Date : November 15, 2018.2010
Mathematics Subject Classification.
Key words and phrases.
Forced Mean Curvature equation, Discontinuous source term, Crystal growth,Asymptotic speed, Obstacle problem.The work of YG was partially supported by Japan Society for the Promotion of Science (JSPS)through grants KAKENHI a r X i v : . [ m a t h . A P ] D ec . Introduction
Problem and Background.
In this paper, we study a level-set forced mean cur-vature flow equation motivated by a crystal growth phenomenon described below. Thecrystal grows in both vertical and horizontal directions. The vertical direction growthis stimulated by a nucleation, and the horizontal one is given by a surface evolution.We assume further that the surface evolution is described by the mean curvature with aconstant force. Under these assumptions, the equation of interest is(C) u t − (cid:18) div (cid:18) Du | Du | (cid:19) + 1 (cid:19) | Du | = c E in R n × (0 , ∞ ) ,u ( · ,
0) = 0 on R n , (1.1)where c > E ⊂ R n are a given constant and a compact set respectively, and E ( x ) := (cid:26) x ∈ E, x (cid:54)∈ E. Here for ( x, t ) ∈ R n × [0 , ∞ ), u ( x, t ) is the height of the crystal at location x at time t .Hypotheses:1. There is a source term c E whichstimulates the nucleation.2. Each level set evolves following thelaw V = κ + 1, where κ is its meancurvature in the direction of the outernormal vector. Figure 1.
Image of the crystal growthLet us explain the background of our model from the theory of crystal growth [3]. Weconsider a perfectly flat surface of a crystal immersed in a supersaturated media. Thecrystal grows by catching adatoms. Assume that there are no dislocations on the surfaceso that no spiral growth is expected. There are several models to explain the growth ofa perfectly flat crystal surface and their theories are often called two-dimensional nucle-ation growth theory [15, 11]. They are roughly classified into single nucleation growth andmultinucleation growth. Single nucleation growth is easy to explain. A nucleation startssomewhere and it spreads across the surface at an infinity velocity and the surface be-comes flat again and waits next nucleation. Multinucleation growth model was originallyintroduced by Hillig [9] and developed in many ways especially to calculate the growthrate of the crystal surface, e.g. [21]. There are several multinucleation growth models inthe literature including ours, which is considered as a kind of the birth and spread model[15, Section 2.6] originally proposed by Nielsen [14]. n the birth and spread model, an island of a layer of width d of molecules, which lookslike a pancake, is attached on the crystal surface at the first nucleation. This pancake-likeshape spreads to the surface with a finite speed V = v ∞ ( ρ c κ + 1) , (1.2)where V and κ , respectively, denote the outer normal velocity of the lateral part of theisland, and the outward curvature of the curve bounding the set in the plane where theisland and the original crystal surface touches. The constants ρ c > v ∞ > E ⊂ R n be a set where nucleationstarts with supersaturation c . Let u denote the height of the crystal surface, and initiallythe crystal surface is perfectly flat, i.e, u ( · , ≡
0. We assume that outside E thereis no supersaturation or no way to nucleate. Within very short time d = ∆ t > d E , i.e. u ( · , d ) = d E .During next short time ∆ t this island spreads horizontally to the crystal surface with thevelocity V = v ∞ ( ρ c κ + 1) and we get a profile u ( · , d ) = d E (∆ t ) , where E ( s ) denotes thesolution of the above eikonal-curvature flow (1.2) starting from E at time s >
0. In thenext stage another nucleation starts again in E and all pancake shapes spread with thesame eikonal-curvature flow. By the repetition of this process we obtain a solution of theTrotter-Kato approximation of our original problem. Our original equation is its limit as∆ t →
0. In fact, equation (1.1) is obtained when v ∞ = 1 and ρ c = 1.In the literature, it is sometimes assumed as in [15, Section 2.6] that ρ c = 0 since ρ c is very small, and in this case, our continuum model is reduced to the Hamilton-Jacobiequation with discontinuous source term studied recently by Giga and Hamamuki [6],which justifies a proposed solution of T. P. Shulze and R. V. Kohn [17] modeling spiralgrowth phenomena consisting of a pair of screw dislocations with opposite orientation. Inour model we take the curvature effect into account so that the model is more realistic.Indeed, due to the curvature effect, an interesting physical phenomenon can be explained,which is new and intuitive. Moreover, from mathematical point of view, we will face manyserious difficulties because of the appearance of the curvature term.1.2. Main results.
Mathematically, equation (C) has two main parts. One comes fromthe nucleation described by the right hand side of (C), and the other one comes from thesurface evolution. In general, the surface evolution of each level set is not only nonlinearbut also nonhomogeneous and not monotone. This makes the interaction between the nu-cleation and the surface evolution extremely nonlinear. In a sense, in order to understand he behavior of solutions of (C), we need to understand the double nonlinear effects com-ing from the surface evolution and the interaction, which will be explained more clearlyin Sections 3 and 5.In this paper, we first establish the well-posedness of solutions to (C) as the equation isnot so standard because of the appearance of the discontinuous source term in the righthand side. In particular, we show that the maximal solution is unique. Our main goalthen is to study the asymptotic growth speed of the maximal solution u to (C), i.e.,lim t →∞ u ( x, t ) t , which describes the speed of the crystal growth. It is important noting that the growth of u can be seen in a heuristic way via the Trotter-Kato product formula in Section 3 whichhelps a lot in analyzing the double nonlinear effects.In a specific case where E = B (0 , R ) for some given R >
0, the growth is completelyunderstood and is analyzed in Section 4. Here B ( x, R ) denotes the open ball of radius R > x ∈ R n . The analysis could be done in an explicit way here as the surfaceevolution is homogeneous. More precisely, balls remain balls after the surface evolution.This also makes the interaction between the nucleation and the surface evolution simpleand clear. The governing equation (C) becomes a first order equation but it is noncoercive.As studied in [7, 8] the asymptotic speed may depend on the place when the equation isnoncoercive. See also [20] for its physical background. We prove the following theoremthrough Propositions 4.3, 4.6 and 4.8. Theorem 1.1.
Assume that E = B (0 , R ) for some R > fixed. Let u be the maximalsolution of (C) . The followings hold (i) If R < n − , then u has the formula (4.2) and is bounded on R n × [0 , ∞ ) . Inparticular, lim t →∞ u ( x, t ) t = 0 uniformly for x ∈ R n . (ii) If R > n − , then u has the formula (4.5) and lim t →∞ u ( x, t ) t = c locally uniformly for x ∈ R n . (iii) If R = n − , then u = ct B (0 ,n − . In particular, lim t →∞ u ( x, t ) t = c uniformly for x ∈ B (0 , n − , and lim t →∞ u ( x, t ) t = 0 uniformly for x ∈ R n \ B (0 , n − . Some more general results for the case of inhomogeneous source terms which are radiallysymmetric are analyzed in Section 4.5 as well. See Theorem 4.12 and Remark 1.In the general case where E is not radially symmetric, it turns out that a new type ofnonlinear phenomena appears. Very roughly speaking, if the nucleation site E is smallenough, then the solution of (C) does not grow up globally as t → ∞ . On the other hand,if the set E is big enough, then the solution of (C) grows up locally uniformly in R n as → ∞ with an asymptotic speed c , which is the rate of nucleation. If the set E is ofmiddle size in a sense, it seems that the asymptotic speed depends on its shape in a verydelicate and sensitive way.It is worthwhile emphasizing here that we find such phenomena as a fact of experimentin the crystal growth. In the experiment the set E corresponding the places where adatomsare sprayed on crystal surfaces. Such a situation is quite popular for growth of metalsin manufacturing technology of semiconductors although spreading mechanism has othereffects. See e.g., [22]. For such phenomena as well as quantum dots the curvature effectmay not be neglected in two-dimensional setting.We provide a framework to get estimates of growth rate in Sections 3 and 5. We thenchoose a representative case where E is of square shape in R to study in details in Section6. We establish the following theorem through Propositions 6.1 and 6.6. Theorem 1.2.
Assume that n = 2 and E = { ( x , x ) : | x i | ≤ d, i = 1 , } for some d > fixed. Let u be the maximal solution of (C) . The followings hold (i) If d < / √ , then u is bounded on R × [0 , ∞ ) . In particular, lim t →∞ u ( x, t ) t = 0 uniformly for x ∈ R . (ii) If d > , then lim t →∞ u ( x, t ) t = c locally uniformly for x ∈ R . (iii) If d = 1 , then lim t →∞ u ( x, t ) t = c uniformly for x ∈ B (0 , . (iv) If / √ < d < , then there exist α, β such that < α < β < c and α ≤ lim inf t →∞ u ( x, t ) t ≤ lim sup t →∞ u ( x, t ) t ≤ β locally uniformly for x ∈ R . Parts (i)–(iii) of Theorem 1.2 are obtained straightforwardly by using Theorem 1.1 andthe comparison principle. To prove part (iv) of Theorem 1.2, a first important step is tounderstand the behavior of the level sets of the top and bottom of solutions to (C). Westudy this by using a set theoretic approach (see [5, Chapter 5] for instance) in Section 5.This perspective gives us rough estimates (Theorems 5.4, 5.6) on the behavior of the heightof solutions to (C), but this is not enough to obtain the precise behavior of lim t →∞ u ( x, t ) /t .Indeed, we have not yet been able to get the precise behavior lim t →∞ u ( x, t ) /t in part (iv)of Theorem 1.2, which is the case where E is of middle size.We conclude this Introduction to give some related works studying large-time asymp-totic behavior of solutions of problems which are non-coercive or of second order problem.The list is not exhaustive at all. As an example of non-coercive Hamilton-Jacobi equationsthe instability of flatness in crystal growth is discussed in [20, 7, 8], and the turbulentflame speed is studied in the context of G-equations in [18, 19]. In [4] the large-timebehavior of solutions of mean curvature flow equations with driving force is studied. cknowledgments. The first two authors are grateful to Professor Etsuro Yokoyamaand Professor Hiroki Hibino for their kind comments on the two-dimensional nucleation.2.
Wellposedness
In this section, we consider a little bit more general equation u t − (cid:18) div (cid:18) Du | Du | (cid:19) + 1 (cid:19) | Du | = f ( x ) in R n × (0 , ∞ ) ,u ( · ,
0) = u on R n , (2.1)where f : R n → [0 , ∞ ) is a bounded function, and u : R n → [0 , ∞ ) is a continuous withsupp u , supp f ⊂ B (0 , R ) for some R > . (2.2)This is an important assumption as, for any T >
0, we only deal with compactly supportedsolutions of (2.1) and (C) on R n × [0 , T ]. Notice thatdiv (cid:18) Du | Du | (cid:19) | Du | = tr (cid:20)(cid:18) I − Du ⊗ Du | Du | (cid:19) D u (cid:21) , where I is the identity matrix of size n . Set σ ( p ) := I − ( p ⊗ p ) / | p | , and H ( p, X ) := − tr [ σ ( p ) X ] − | p | for ( p, X ) ∈ ( R n \ { } ) × S n , (2.3)where S n is the set of real symmetric matrices of size n .We first recall the definition of viscosity solutions to an equation with discontinuousfunctions, which was introduced in [10]. Definition 1 (Viscosity solutions) . Let u : R n × [0 , ∞ ) → R be a locally bounded. Wesay that u is a viscosity subsolution of (2.1) if u ∗ ( · , ≤ u on R n , and τ + H ∗ ( p, X ) ≤ f ∗ ( x ) for all ( x , t ) ∈ R n × (0 , ∞ ) , ( p, X, τ ) ∈ J + u ∗ ( x , t ) . We say that u is a viscosity supersolution of (2.1) if u ∗ ( · , ≥ u on R n , and τ + H ∗ ( p, X ) ≥ f ∗ ( x ) for all ( x , t ) ∈ R n × (0 , ∞ ) , ( p, X, τ ) ∈ J − u ∗ ( x , t ) . Here for a locally bounded function h on R m for m ∈ N , we denote the upper semi-continuous envelope (resp., lower semicontinuous envelope) by h ∗ and h ∗ defined as h ∗ ( x ) := lim δ → sup { h ( y ) : | x − y | ≤ δ } and h ∗ ( x ) := lim δ → inf { h ( y ) : | x − y | ≤ δ } ,respectively, and we write J + u ∗ ( x, t ) and J − u ∗ ( x, t ) for the super and sub semijets of u ∗ , u ∗ at ( x, t ) ∈ R n × (0 , ∞ ), respectively.We say that u is a viscosity solution of (2.1) if it is both a viscosity subsolution and aviscosity supersolution of (2.1).It is well-known that if the function f on the right hand side of (2.1) is continuouson R n , then the comparison principle and the uniqueness of solutions hold. See [5] forinstance. On the other hand, if we deal with discontinuous functions f on the right handside of (2.1), we lose the uniqueness of viscosity solutions in general. See [6] for someexamples and observations of first order Hamilton-Jacobi equations with discontinuoussource terms. We only have the comparison principle in a weak sense. The followingresult is standard in the theory of viscosity solutions, but we present it here to make thepaper self-contained. roposition 2.1 (Weak Comparison Principle) . Fix
T > . Assume that v ∈ USC ( R n × [0 , T ]) and w ∈ LSC ( R n × [0 , T ]) , which are compactly supported, i.e., v ( x, t ) = w ( x, t ) = 0 for all x ∈ R n \ B (0 , R T ) , t ∈ [0 , T ] and some R T > , (2.4) are a viscosity subsolution and a viscosity supersolution of (2.1) with f, g on the righthand side respectively, where f and g are locally bounded functions satisfying f ∗ ≤ g ∗ on R n . Then v ≤ w on R n × [0 , T ] .Proof. We argue by contradiction and suppose that max R n × [0 ,T ] ( v − w ) >
0. Then, thereexists a small constant α > ε > x,y ∈ R n t ∈ [0 ,T ] (cid:26) v ( x, t ) − w ( y, t ) − | x − y | ε − αT − t (cid:27) > . As v ( x, t ) = w ( x, t ) = 0 for all ( x, t ) ∈ ( R n \ B (0 , R T )) × [0 , T ], the maximum is attainedat ( x ε , y ε , t ε ) ∈ B (0 , R T ) × (0 , T ) and by passing to a subsequence if necessary, we canassume ( x ε , y ε , t ε ) → ( x , x , t ) as ε → x ∈ B (0 , R T ) , t ∈ [0 , T ].In view of Ishii’s lemma, for any ρ >
0, there exist ( a ε , p ε , X ε ) ∈ J , + v ( x ε , t ε ) and( b ε , p ε , Y ε ) ∈ J , − w ( y ε , t ε ) such that a ε − b ε = α ( T − t ε ) , p ε = | x ε − y ε | ( x ε − y ε ) ε , (cid:18) X ε − Y ε (cid:19) ≤ A + ρA , (2.5)where A := 1 ε | x ε − y ε | (cid:18) I − I − I I (cid:19) + 2 ε (cid:18) ( x ε − y ε ) ⊗ ( x ε − y ε ) − ( x ε − y ε ) ⊗ ( x ε − y ε ) − ( x ε − y ε ) ⊗ ( x ε − y ε ) ( x ε − y ε ) ⊗ ( x ε − y ε ) (cid:19) . The definition of viscosity solutions implies the following inequalities: a ε + H ∗ ( p ε , X ε ) ≤ f ∗ ( x ε ) , and b ε + H ∗ ( p ε , Y ε ) ≥ g ∗ ( y ε ) . (2.6)Note that (2.5) implies X ε ≤ Y ε .In the case p ε (cid:54) = 0, i.e., x ε (cid:54) = y ε , we have H ∗ ( p ε , X ε ) − H ∗ ( p ε , Y ε ) = tr [ σ ( p ε )( Y ε − X ε )] ≥ . In the case p ε = 0, we have x ε = y ε . Due to (2.5), we have A = 0, which implies X ε ≤ Y ε ≥
0. Thus, H ∗ ( p ε , X ε ) ≥ H ∗ (0 ,
0) = 0 , and H ∗ ( p ε , Y ε ) ≤ H ∗ (0 ,
0) = 0 . In both cases, H ∗ ( p ε , X ε ) − H ∗ ( p ε , Y ε ) ≥
0. Combine this with (2.6) to yield αT ≤ α ( T − t ε ) ≤ f ∗ ( x ε ) − g ∗ ( y ε ) . Let ε → ε → ( f ∗ ( x ε ) − g ∗ ( y ε )) ≤ lim sup ε → f ∗ ( x ε ) − lim inf ε → g ∗ ( y ε ) ≤ ( f ∗ − g ∗ )( x ) ≤ , which is a contradiction. (cid:3) n the case where we drop the curvature term, i.e., we consider the Hamilton–Jacobiequation with a discontinuous source term, if we additionally assume( f ∗ ) ∗ = f ∗ on R n , (2.7)then we can prove the uniqueness of viscosity solutions in the class of upper semicontinuousfunctions in the sense of [10]. This can be done by using a control approach which is ananalogue of [1]. On the other hand, as far as the authors know, there is no uniquenessresults for (2.1) under (2.7). Therefore, in this paper, we consider the maximal viscositysolutions of (2.1). To make it clear, we give its definition here. Definition 2 (Maximal viscosity solutions) . We say that u is a maximal viscosity solutionof (2.1) if u is a viscosity solution of (2.1) satisfying (2.4) and for every viscosity solution v of (2.1) satisfying (2.4), u ≥ v on R n × [0 , ∞ ). Theorem 2.2 (Existence and Uniqueness) . There exists a unique maximal viscosity so-lution u of (2.1) .Proof. Fix
T >
0. For k ∈ N and x ∈ R n , define f k ( x ) := sup y ∈ R n ( f ∗ ( y ) − k | x − y | ) . It is straightforward that f k ∈ C ( R n ) and f k ( x ) ↓ f ∗ ( x ) pointwise as k → ∞ .By the standard theory of viscosity solutions, there exists a unique viscosity solution u k ∈ C ( R n × [0 , T ]) of (2.1) with the right hand side f k , and by the comparison principle,we can easily prove u k ( x ) ↓ u ( x ) for all x ∈ R n as k → ∞ . Furthermore, in light ofassumption (2.2), there exists R T > u k ( x, t ) = u ( x, t ) = 0 for all ( x, t ) ∈ ( R n \ B (0 , R T )) × [0 , T ] , and k ∈ N . Note that in view of this monotonicity u = inf k ∈ N u k = lim sup ∗ k →∞ u k , (2.8)where lim sup ∗ is the upper half-relaxed limit. Thus, u ∈ USC ( R n × [0 , T ]).Clearly, u k is a supersolution of (2.1) for each k ∈ N , which yields immediately that u is also a supersolution of (2.1) in view of the inf-stability. Thanks to (2.8), u is asubsolution of (2.1) in view of the stability property of the upper half-relaxed limit forviscosity subsolutions. Therefore, u is a solution of (2.1). Moreover, u is compactlysupported on R n × [0 , T ].Next, we prove that u is continuous at t = 0. By [5, Lemma 4.3.4], there exists aviscosity subsolution v ∈ C ( R n × [0 , T ]) with a compact support of (2.1) with f = 0 onthe right hand side. Thus, by the comparison principle, v ( y, t ) − u ( x ) ≤ u ( y, t ) − u ( x ) ≤ u k ( y, t ) − u ( x ) for all x, y ∈ R n , t ∈ [0 , T ], which implies u ( y, t ) → u ( x ) as ( y, t ) → ( x, u is the unique maximal viscosity solution of (2.1). Indeed, takeany v to be a viscosity solution of (2.1). By the comparison principle, v ∗ ≤ u k . Let k → ∞ to deduce the desired result. (cid:3) . Heuristic observation
In this section, we give a formal argument in order to understand the behavior ofsolutions of (C). Our goal in this section is to explain intuitively with a geometric aspecthow the asymptotic average of solutions depends on the shape of E . This is basically thesame as the derivation of the problem from physics explained in Introduction.3.1. Trotter-Kato product formula.
We consider the following double-step method:(N) (cid:26) v t = c E in R n × (0 , ∞ ), v ( · ,
0) = u in R n ,and (P) w t = (cid:18) div (cid:16) Dw | Dw | (cid:17) + 1 (cid:19) | Dw | in R n × (0 , ∞ ), w ( · ,
0) = u in R n .We call (N) and (P) the nucleation problem and the propagation problem , respectively.We define the operators S ( t ) : L ∞ ( R n ) → L ∞ ( R n ), and S ( t ) : Lip ( R n ) → Lip ( R n ),respectively by S ( t )[ u ] := u + c E t, and S ( t )[ u ] := w ( · , t ) , (3.1)where w is the unique viscosity solution of (P).For x ∈ R n , τ > , i ∈ N , set U τ ( x, iτ ) := S ( τ ) (cid:0) S ( τ ) S ( τ ) (cid:1) i [ u ] . (3.2)The function U τ ( x, iτ ) is called the Trotter-Kato product formula , we can expect for x ∈ R n and t = iτ > i →∞ U τ ( x, iτ ) = u ( x, t ) locally uniformly for x ∈ R n , (3.3)under some condition, where u is the “solution” of (C). In the framework of the theoryof viscosity solutions, Barles and Souganidis in [2] first proved (3.3). Our situation hereactually does not fit into the framework of [2], as we do not have the comparison principlefor (C) because of the discontinuous source term c E on the right hand side of (C). Nev-ertheless, it is quite reasonable to assume that (3.3) holds in order to guess the behaviorof the solution u to (C).In light of this, the behavior of u ( x, t ) /t as t → ∞ can be consider as the behavior oflim t →∞ (cid:32) lim τ → iτ = t U τ ( x, iτ ) iτ (cid:33) . (3.4)The advantage of considering U τ ( x, iτ ) lies in the fact that its graph is a pyramid offinite number of steps of height cτ . The double-step method can then be described in ageometrical way as follows: (N) At each nucleation step, we drop from above an amount of cτ E crystal down tothe pyramid with the assumption that the crystals are not sticky; (P) At each propagation step, each layer of the pyramid evolves under a forced meancurvature flow ( V = κ + 1). et us emphasize that, in general, the growth of the pyramid is highly nonlinear. Thereason comes from the fact that the behavior of each layer is extremely complicated, whichwill be pointed out in Section 5 in more clearly. One particular layer can receive someamount of crystal in each nucleation step, then changes its shape in each propagation step.Of course the layers change not only in a nonlinear way but also in a nonhomogeneousway in each propagation step. Furthermore, the changes are not monotone (unlike thecase V = 1). These affect the next nucleation step seriously as the receipt of crystals ateach layer will change dramatically from time to time. More or less, this says that theproblem has double nonlinear effects.3.2. Notations and Spherical symmetric case.
In this subsection, we make the anal-ysis above clearer by a careful step by step analysis. In order to do so, we introduce no-tations below. For A ⊂ R n and t >
0, let F [ A ]( t ) be the solution to the surface evolutionequation (S) V = κ + 1 on Γ( t ) with Γ(0) = A. Fix i ∈ N and τ >
0. For j ∈ { , . . . , i } and k ∈ { , . . . , j } , we define the sets E τ ( j, k ) ⊂ R n , which are the layers of the pyramids, as follows: E τ (1 ,
1) :=
E,E τ (2 ,
1) := E ∪ F [ E τ (1 , τ ) , E τ (2 ,
2) := E ∩ F [ E τ (1 , τ ) ,E τ (3 ,
1) := E ∪ F [ E τ (2 , τ ) , E τ (3 ,
2) := ( E ∩ F [ E τ (2 , τ )) ∪ F [ E τ (2 , τ ) ,E τ (3 ,
3) := E ∩ F [ E τ (2 , τ ) , ...Let us now use this system to investigate spherical symmetric cases, i.e., E = B (0 , R ) . It is worth pointing out that the spherical symmetric cases are easy to understand becauseof the fact that balls remain balls after the evolution under the forced mean curvature flow V = κ + 1. So the changes in shapes in each propagation step are homogeneous, whichmake behaviors of the nucleation steps and the pyramids extremely clear. Our concerntherefore is only whether B (0 , R ) grows or shrinks under the propagation step or not.This leads to the distinction between the three cases: R < n − , R > n − , and R = n − . (3.5)We first consider the case where R < n −
1. As the curvature term is stronger thanthe force term, the surfaces start to shrink. Thus, E τ (1 ,
1) = B (0 , R ) ,E τ (2 ,
1) = B (0 , R ) , E τ (2 ,
2) = B (0 , R ( τ )) ,E τ (3 ,
1) = B (0 , R ) , E τ (3 ,
2) = B (0 , R ( τ )) , E τ (3 ,
3) = B (0 , R (2 τ )) , ... here R ( t ) is the solution of the ODE ˙ R ( t ) = − n − R ( t ) + 1 for t > R (0) = R . On the other hand, if R > n −
1, then the curvature term is weaker than the forceterm and the surfaces start to expand. Thus, E τ (1 ,
1) = B (0 , R ) ,E τ (2 ,
1) = B (0 , R ( τ )) , E τ (2 ,
2) = B (0 , R ) ,E τ (3 ,
1) = B (0 , R (2 τ )) , E τ (3 ,
2) = B (0 , R ( τ )) , E τ (3 ,
3) = B (0 , R ) . ...By using these observations, we can somehow understand the behavior of (3.4) in eachcases of (3.5).In the next section, we only consider the spherical case and rigorously derive the formulaof u as well as its large time average.4. Spherical symmetric case
In this section, we study the case where E = B (0 , R ) for some R > , and investigate the large time average of the maximal viscosity solution u .It is reasonable to look for radially symmetric solution u of (C) of the form u ( x, t ) = φ ( | x | , t ) = φ ( r, t ) . Then, u t = φ t , Du = φ r x | x | ,D u = φ rr x ⊗ x | x | + φ r | x | (cid:16) I − x ⊗ x | x | (cid:17) . Plugging these into (C) to reduce it to the following initial value problem, which is basi-cally a singular noncoercive Hamilton–Jacobi equation in 1 − dimension,(C (cid:48) ) φ t − ( n − φ r r − | φ r | = c [0 ,R ] ( r ) in (0 , ∞ ) × (0 , ∞ ) ,φ ( · ,
0) = 0 on [0 , ∞ ) . From the next subsequences, we consider three cases divided in (3.5). .1. The case R < n − . In order to obtain the maximal viscosity solution, we ap-proximate from above to get a decreasing sequence of supersolutions. Its limit will be themaximal viscosity solution once we prove that it is a subsolution.Fix ε > R + ε < n −
1. We first solve a boundary valueproblem of a linear ordinal differential equation:(ODE) (cid:18) − ( n − r + 1 (cid:19) ψ εr = cI ε ( r ) in (0 , R + ε ) ,ψ ε ( R + ε ) = 0 , where I ε ( r ) := r ∈ [0 , R ] , ( R + ε ) − rε for r ∈ [ R , R + ε ] , r ∈ [ R + ε, + ∞ ) . It is clear that, for 0 ≤ r ≤ R + ε , ψ ε ( r ) := (cid:90) r cI ε ( s ) − n − s + 1 ds − (cid:90) R + ε cI ε ( s ) − n − s + 1 ds. We set ψ ε ( r ) := 0 for r ≥ R + ε , and extend ψ ε to the whole R in a symmetric way (i.e.,set ψ ε ( r ) = ψ ε ( − r ) for all r ≤ u ε ∈ C ( R n × [0 , ∞ )) by u ε ( x, t ) := min { ψ ε ( | x | ) , ct } . (4.1) Figure 2.
Picture of u ε in Case 1 Lemma 4.1.
The function u ε is a viscosity supersolution of (2.1) for g ( x ) = cI ε ( | x | ) .Proof. The claim is clear for ( x, t ) ∈ B (0 , R + ε ) × (0 , ∞ ) as u ε is the minimum of twosupersolutions there. Also there is nothing to check in case | x | > R + ε as u ε = g = 0there.We only need to check carefully the case where | x | = R + ε . It is worth to mentionfirst that for any t >
0, we always have that u ε ( x, t ) = ψ ε ( | x | ) for x in a neighborhood of ∂B (0 , R + ε ) and t ∈ ( t / , t + 1). In other words, u ε does not change with respect totime in this neighborhood. Take φ ∈ C ( R n ) to be a test function such that ψ ε − φ hasa strict minimum at x ∈ ∂B (0 , R + ε ). In light of Lemma 8.1 in Appendix, for some s ≤ Dφ ( x ) = s x R + ε , and tr [ σ ( Dφ ( x )) D φ ( x )] ≤ ( n − sR + ε . hus, for s > H ( Dφ ( x ) , D φ ( x )) − cI ε ( R + ε ) = − tr [ σ ( Dφ ( x )) D φ ( x )] − | Dφ ( x ) |≥ (cid:0) ( n − − ( R + ε ) (cid:1) | s | R + ε ≥ , which implies the conclusion. (cid:3) We define ψ ( r ) := lim ε → ψ ε ( r ) = inf ε → ψ ε ( r ) for r ∈ R . Actually, ψ can be computed explicitly as following ψ ( r ) = (cid:40) c (( r + ( n −
1) log | r − n + 1 | ) − ( R + ( n −
1) log | R − n + 1 | )) for r ∈ [0 , R ] , r ∈ ( R , ∞ ) . For ( x, t ) ∈ R n × [0 , ∞ ), set v ( x, t ) := min { ψ ( | x | ) , ct } = lim ε → u ε ( x, t ) = inf ε> u ε ( x, t ) . (4.2)It is important noticing that u ε converges to v uniformly in R n × [0 , ∞ ).By the comparison principle and Lemma 4.1, it is clear that u ε ≥ u and hence v ≥ u .Furthermore, u ε is also a supersolution of (C) for all ε >
0, and so is v . We now showthat in fact u = v . In order to achieve this, we need the following result Lemma 4.2.
The function v is a subsolution of (C) .Proof. Set T := ( n −
1) log( n − − ( R +( n −
1) log | R − n +1 | ). For t ≥ T , v ( x, t ) = ψ ( | x | )for all x ∈ R n and there is nothing to check.Let us now fix ( x , t ) ∈ R n × (0 , T ) such that v ( x , t ) = ψ ( x ) = ct . Assume that v − φ has a strict maximum at ( x , t ) for some test function φ ∈ C ( R n × (0 , ∞ )) andthat φ ( x , t ) = v ( x , t ) = ct . Clearly, 0 ≤ φ t ( x , t ) ≤ c . Thanks to Lemma 8.1, forsome s ∈ [ ψ (cid:48) ( | x | ) , Dφ ( x , t ) = s x | x | , and tr [ σ ( Dφ ( x )) D φ ( x )] ≥ ( n − s | x | . Thus, for s > φ t ( x , t ) − H ( Dφ ( x ) , D φ ( x )) − c E ( x ) ≤ φ t ( x , t ) + s ( | x | − n + 1) | x | − c. (4.3)For t < t , let r ( t ) be the function in ( | x | , R ) which satisfies t = r ( t ) + ( n −
1) log( n − − r ( t )) − ( R + ( n −
1) log( n − − R )) , and set x ( t ) := r ( t ) x / | x | . Then we have φ ( x ( t ) , t ) ≥ v ( x ( t ) , t ) = ct for t ≤ t and φ ( x ( t ) , t ) = φ ( x , t ) = ct . Therefore, c ≥ ddt ( φ ( x ( t ) , t )) | t = t = φ t ( x , t ) + Dφ ( x , t ) · x (cid:48) ( t )= φ t ( x , t ) + r (cid:48) ( t ) (cid:18) Dφ ( x , t ) · x | x | (cid:19) = φ t ( x , t ) + s | x | − n + 1 | x | . We combine this and (4.3) to get the result. (cid:3) n conclusion, we obtain Proposition 4.3.
Let u be the maximal solution of (C) . Then, we have the formula (4.2) ,and thus u is bounded on R n × [0 , ∞ ) . In particular, lim t →∞ u ( x, t ) t = 0 in C ( R n ) . Proof.
Since the maximal solution u is obtained by (4.2), we have u ( x, t ) = min { ψ ( | x | ) , ct } ≤ ψ ( | x | ) for all ( x, t ) ∈ R n × [0 , ∞ ) . This immediately implies the conclusion. (cid:3)
The case R > n − . Fix ε >
0. We first look at an initial-boundary value problemof a linear partial differential equation in 1-dimension:(L) ε ϕ εt + (cid:18) − ( n − r + 1 (cid:19) ϕ εr = 0 in ( R + ε, ∞ ) × (0 , ∞ ) ϕ ε ( R + ε, t ) = ct on [0 , ∞ ) . By using the method of characteristics, we can find a solution to the above PDE ϕ ε ( r, t ) = c (cid:0) t − r − ( n −
1) log( r − ( n − R + ε + ( n −
1) log( R + ε − ( n − (cid:1) . Define u ε : R n × [0 , ∞ ) → R as u ε ( x, t ) := (cid:26) ct for all ( x, t ) ∈ B (0 , R + ε ) × [0 , ∞ )( ϕ ε ( | x | , t )) + for all ( x, t ) ∈ ( R n \ B (0 , R + ε )) × [0 , ∞ ) . Figure 3.
Picture of u ε in Case 2 Lemma 4.4.
The function u ε is a viscosity supersolution of (2.1) for g ( x ) = c B (0 ,R + ε ) .Proof. We only need to check at ( x , t ) ∈ R n × (0 , ∞ ) where u ε ( x , t ) = ϕ ε ( | x | , t ) = 0.Assume that u ε − φ has a strict minimum at ( x , t ) for some test function φ ∈ C ( R n × [0 , ∞ ) and u ε ( x , t ) = φ ( x , t ) = 0. In light of Lemma 8.1, for some s ∈ [ ϕ r ( | x | , t ) , Dφ ( x , t ) = s x | x | , and tr [ σ ( Dφ ( x , t )) D φ ( x , t ) | ] ≤ ( n − s | x | . Thus, φ t ( x , t ) + H ( Dφ ( x , t ) , D φ ( x , t )) − c B (0 ,R + ε )) ( x ) ≥ φ t ( x , t ) + s ( | x | − n + 1) | x | . (4.4) or t < t , let r ( t ) be the function in ( R + ε, | x | ) which satisfies t = r ( t ) + ( n −
1) log( r ( t ) − n + 1) − ( R + ε + ( n −
1) log( R + ε − n + 1)) , and set x ( t ) := r ( t ) x / | x | . Then we have φ ( x ( t ) , t ) ≤ u ε ( x ( t ) , t ) = 0 for t ≤ t and φ ( x ( t ) , t ) = φ ( x , t ) = 0. Therefore,0 ≤ ddt ( φ ( x ( t ) , t )) | t = t = φ t ( x , t ) + Dφ ( x , t ) · x (cid:48) ( t )= φ t ( x , t ) + r (cid:48) ( t ) (cid:18) Dφ ( x , t ) · x | x | (cid:19) = φ t ( x , t ) + s | x | − n + 1 | x | . We combine this and (4.4) to get the result. (cid:3)
For ( x, t ) ∈ R n × [0 , ∞ ), set v ( x, t ) := lim ε → u ε ( x, t ) = inf ε> u ε ( x, t ) , It is important noticing that u ε converges to v locally uniformly in R n × [0 , ∞ ). Weactually have v ( x, t ) = (cid:26) ct for all ( x, t ) ∈ B (0 , R ) × [0 , ∞ )( ϕ ( | x | , t )) + for all ( x, t ) ∈ ( R n \ B (0 , R )) × [0 , ∞ ) , (4.5)where ϕ is the solution to (L) , i.e., ϕ ( r, t ) = c (cid:0) t − r − ( n −
1) log( r − ( n − R + ( n −
1) log( R − ( n − (cid:1) . By the comparison principle and Lemma 4.4, it is clear that u ε ≥ u and hence v ≥ u .Furthermore, u ε is also a supersolution of (C) for all ε >
0, and so is v . We now showthat in fact u = v . In order to achieve this, we need the following result Lemma 4.5.
The function v is a subsolution of (C) .Proof. It is enough to test at ( x , t ) ∈ R n × (0 , ∞ ) in case | x | = R . Assume that v − φ has a strict maximum at ( x , t ) for some test function φ ∈ C ( R n × [0 , ∞ )) and v ( x , t ) = φ ( x , t ) = ct . We first note that φ t ( x , t ) = c . We use Lemma 8.1 once moreto deduce that for some s ∈ [ ϕ r ( R , t ) , Dφ ( x , t ) = s x R , and tr [ σ ( Dφ ( x , t )) D φ ( x , t )] ≥ ( n − sR . Hence, for s > φ t ( x , t ) + H ( Dφ ( x , t ) , D φ ( x , t )) − c E ( x ) ≤ s R − n + 1 R ≤ . (cid:3) In conclusion, we obtain
Proposition 4.6.
Let u be the maximal solution of (C) . Then, u has the formula (4.5) ,and lim t →∞ u ( x, t ) t = c locally uniformly for x ∈ R n . .3. The critical case R = n − . We denote by u r the maximal solution of (C) when E = B (0 , r ) when r > n −
1. By the comparison principle, we get that u ≤ u r and hence u ≤ lim r → u r = inf r> u r = ct B (0 ,n − =: v. It is clear that v is a supersolution of (C). We now show that v is in fact a subsolution of(C), which yields again that u = v = ct B (0 ,n − . Lemma 4.7.
The function v = ct B (0 ,n − is a subsolution of (C) .Proof. As usual, it is enough to test at ( x , t ) ∈ R n × (0 , ∞ ) in case | x | = R = n − v − φ has a strict maximum at ( x , t ) for some test function φ ∈ C ( R n × [0 , ∞ )) and v ( x , t ) = φ ( x , t ) = ct . We note first that φ t ( x , t ) = c . Lemma 8.1 yieldsthat for some s ≤ Dφ ( x , t ) = s x R , and tr [ σ ( Dφ ( x , t )) D φ ( x , t )] ≥ ( n − sR = s. Hence, for s > φ t ( x , t ) + H ( Dφ ( x , t ) , D φ ( x , t )) − c E ( x ) ≤ s − s = 0 . (cid:3) We conclude by the following result.
Proposition 4.8.
We have the following asymptotics lim t →∞ u ( x, t ) t = c uniformly for x ∈ B (0 , n − , and lim t →∞ u ( x, t ) t = 0 uniformly for x ∈ R n \ B (0 , n − . Some immediate consequences.
By using the above results, we get some resultsfor general compact sets E when it is either small enough or large enough. Corollary 4.9. If E ⊂ B ( y, n − for some y ∈ R n , then lim t →∞ u ( x, t ) t = 0 uniformly for x ∈ R n . If B ( y, n − ⊂ int E for some y ∈ R n , then lim t →∞ u ( x, t ) t = c locally uniformly for x ∈ R n . Corollary 4.10.
Assume that E = (cid:83) ki =1 B ( y i , r i ) for some given k ∈ N , y i ∈ R n , and < r i ≤ n − for ≤ i ≤ k . Assume further that the closed balls B ( y i , r i ) for ≤ i ≤ k are disjoint. Denote by K = { i : 1 ≤ i ≤ k, r i = n − } . Then lim t →∞ u ( x, t ) t = c uniformly for x ∈ (cid:91) i ∈ K B ( y i , r i ) and lim t →∞ u ( x, t ) t = 0 uniformly for x ∈ R n \ (cid:91) i ∈ K B ( y i , r i ) . roof. For 1 ≤ i ≤ k , denote by u i the solution of (C) corresponding to c B ( y i ,r i ) . Theimportant point is that { u i > } ⊂ B ( y i , r i ), which implies that { u i > } ∩ { u j > } = ∅ for i (cid:54) = j. Hence, u = max ≤ i ≤ k u i . The results then follow from Propositions 4.3 and 4.8. (cid:3) Optimal control interpretation and inhomogeneous f . We now consider aslightly more general version of (C) in the spherical symmetric situation. More precisely,we are concerned with (2.1) in case f ( x ) = h ( | x | ) where h : [0 , ∞ ) → [0 , ∞ ) is uppersemicontinuous and there exists R > h ⊂ [0 , R ] . It is again reasonable to look for radially symmetric solution u ( x, t ) = φ ( | x | , t ) = φ ( r, t ).Then φ satisfies (cid:40) φ t − n − r φ r − | φ r | = h ( r ) in (0 , ∞ ) × (0 , ∞ ) ,φ ( · ,
0) = 0 on [0 , ∞ ) . (4.6)Let ψ = − φ , and then ψ solves (cid:40) ψ t − n − r ψ r + | ψ r | + h ( r ) = 0 in (0 , ∞ ) × (0 , ∞ ) ,ψ ( · ,
0) = 0 on [0 , ∞ ) . (4.7)The Hamiltonian of (4.7) is ˜ H ( p, r ) = | p | − n − r p + h ( r ), which is convex and singular at r = 0. We can easily compute the corresponding Lagrangian ˜ L as˜ L ( q, r ) = (cid:40) − h ( r ) if (cid:12)(cid:12) q + n − r (cid:12)(cid:12) ≤ ∞ otherwise . Therefore, the representation formula of ψ in light of optimal control theory is ψ ( r, t ) = inf (cid:26)(cid:90) t ( − h ( γ ( s ))) ds : γ ([0 , t ]) ⊂ (0 , ∞ ) , γ ( t ) = r, (cid:12)(cid:12)(cid:12)(cid:12) γ (cid:48) ( s ) + n − γ ( s ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:27) . This yields that, for ( r, t ) ∈ (0 , ∞ ) × [0 , ∞ ), φ ( r, t ) = sup (cid:26)(cid:90) t h ( γ ( s )) ds : γ ([0 , t ]) ⊂ (0 , ∞ ) , γ ( t ) = r, (cid:12)(cid:12)(cid:12)(cid:12) γ (cid:48) ( s ) + n − γ ( s ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:27) . (4.8)We can then prove that u ( x, t ) := φ ( | x | , t ) is the maximal viscosity solution to (C) in asimilar manner to that of Sections 4.1–4.2. We now provide a general version of Proposi-tion 4.6. Proposition 4.11.
Let u be the maximal viscosity solution to (C) . Assume that thereexists r > n − such that c := h ( r ) = max r ∈ [0 , ∞ ) h ( r ) . Then lim t →∞ u ( x, t ) t = c locally uniformly for x ∈ R n . roof. We only need to prove that φ ( · , t ) /t → c as t → ∞ locally uniformly in [0 , ∞ ).It is clear from the representation formula that φ ( r, t ) ≤ ct for ( r, t ) ∈ (0 , ∞ ) × (0 , ∞ ),which yields that lim sup t →∞ φ ( r, t ) t ≤ c uniformly for r ∈ (0 , ∞ ) . We now need to obtain the lower bound. Fix r ∈ (0 , ∞ ) and consider two cases. Case 1: r > r . Set T := r ( r − r ) r − ( n − . For t > T , consider the curve γ : [0 , t ] → (0 , ∞ ) as γ ( s ) := (cid:40) r for 0 < s < t − T ,r + ( s − t + T ) r − ( n − r for t − T < s < t. One can check that γ is admissible in formula (4.8) and hence φ ( r, t ) ≥ (cid:90) t h ( γ ( s )) ds ≥ (cid:90) t − T h ( γ ( s )) ds ≥ c ( t − T ) , as h is nonnegative, which is sufficient to get the conclusion. Case 2: < r ≤ r . We first consider the following ODE (cid:40) ξ (cid:48) ( s ) = − − n − ξ ( s ) for s > ,ξ (0) = r . Take T > ξ ( T ) = r . It is immediate that T ≤ r .For t > T , consider γ : [0 , t ] → (0 , ∞ ) as γ ( s ) := (cid:40) r for 0 ≤ s < t − T ,ξ ( s − t + T ) for t − T < s ≤ t. Clearly, γ is admissible in formula (4.8) and φ ( r, t ) ≥ (cid:90) t h ( γ ( s )) ds ≥ c ( t − T ) . (cid:3) Based on the above proposition and its proof, we have the following general result.
Theorem 4.12.
Set c := max r ∈ [ n − , ∞ ) h ( r ) and c = sup r ∈ ( n − , ∞ ) h ( r ) . Then, lim t →∞ u ( x, t ) t = c uniformly for x ∈ B (0 , n − , lim t →∞ u ( x, t ) t = c locally uniformly for x ∈ R n \ B (0 , n − . The proof of this theorem is similar to that of Proposition 4.11 hence omitted. It isimportant noticing that the large time average result of this theorem covers that of all thecases in Propositions 4.3, 4.6, and 4.8. It however does not give explicit/precise formulasof the maximal solution as in the mentioned propositions. emark 1. (i) One can easily generalize the surface evolution part to V = v ∞ ( ρ c κ + 1),where v ∞ , ρ c > x, t , we can reducethe problem to the case where V = κ + 1. (ii) One can also deal with a general initialdata which is bounded and uniformly continuous on R n instead of the constant initialdata. (iii) Theorem 4.12 is surprising as even though the source term is very thin, it veryessentially affects the growth of the crystal in a long time. More precisely, if we consider f ( x ) = c ∂B (0 ,R ) for R > n − t →∞ u ( x, t ) t = c locally uniformly for x ∈ R n . A framework to get estimates of growth rate
In a nonspherically symmetric case, it seems hard at this moment to obtain the preciselarge time average lim t →∞ u ( · , t ) /t , where u is the maximal viscosity solution of (C). Wewill point out why so in Section 7. Therefore, we here start to build up a framework to ob-tain rough estimates, namely the estimates on lim sup t →∞ u ( · , t ) /t and lim inf t →∞ u ( · , t ) /t first.We first try to understand the behavior of the top and bottom of a solution u to (C)as it should give an information of the behavior of the height of u .5.1. Motion of the top and the bottom of solutions.
Let v be a viscosity subsolutionof (C). By the comparison principle, we have v ∗ ( x, t ) ≤ ct in R n × [0 , ∞ ). For t ≥
0, set A max ( t ) := { x ∈ R n : v ∗ ( x, t ) = ct } , (5.1)which is a compact set for t > v ∗ is upper semicontinuous and compactly supported. Lemma 5.1.
Let A max ( t ) be the set defined by (5.1) . Then A max ( t ) is a set theoreticsubsolution of V = κ + 1 , i.e., h ( x, t ) := A max ( t ) ( x ) is a viscosity subsolution of (2.1) with f = 0 (see [5, Definition 5.1.1] for details) . Moreover, A max ( t ) ⊂ E for all t ∈ (0 , ∞ ) .Proof. We first notice that v c ( x, t ) := v ( x, t ) − ct is a viscosity subsolution of (2.1)with the right hand side f ≡
0, and v c ≤ R n × [0 , ∞ ). Moreover, A max ( t ) = { x ∈ R n : v ∗ c ( x, t ) = 0 } . Thus, it is clear to see that A max ( t ) is a set theoretic subsolutionof V = κ + 1 in view of [5, Theorem 5.1.6].We next prove that A max ( t ) ⊂ E for all t ∈ (0 , ∞ ). Suppose otherwise that there wouldexist x ∈ A max ( t ) ∩ E c for some t >
0. Then ϕ ( x, t ) := ct is a test function of v ∗ fromabove. This is a contradiction as c = ϕ t ( x , t ) + H ∗ ( Dϕ ( x , t ) , D ϕ ( x , t )) ≤ c E ( x ) = 0 , where H is defined by (2.3). (cid:3) Let w be a viscosity supersolution of (C). By the comparison principle again, we have w ∗ ( x, t ) ≥ R n × [0 , ∞ ). We set A min ( t ) := R n \ { x ∈ R n : w ∗ ( x, t ) = 0 } = { x ∈ R n : w ∗ ( x, t ) > } . (5.2) Lemma 5.2.
Let A min ( t ) be the set defined by (5.2) . Then A min ( t ) is a set theoreticsupersolution of V = κ + 1 , i.e., h ( x, t ) := A min ( t ) ( x ) is a viscosity supersolution of (2.1) with f = 0 . Moreover, int E ⊂ A min ( t ) for all t ∈ (0 , ∞ ) . roof. We only prove int E ⊂ A min ( t ) for all t ∈ (0 , ∞ ). Suppose otherwise that therewould exist x ∈ E ∩ A c min ( t ) for some t >
0. Then, w ( x , t ) = 0 which implies ϕ ( x, t ) ≡ w ∗ from below. This is a contradiction as0 = ϕ t ( x , t ) + H ∗ ( Dϕ ( x , t ) , D ϕ ( x , t )) ≥ c E ( x ) = c. (cid:3) In this manner, A max ( t ) and A min ( t ) are a set theoretic subsolution and supersolution,respectively, of obstacle problems of V = κ + 1 with A max ( t ) ⊂ E, and int E ⊂ A min ( t ) for all t ≥ . We give here a level set formulation for later use. See [5, Chapter 5] for more details. Thefunctions h ( x, t ) := A max ( t ) ( x ) , h ( x, t ) := A min ( t ) ( x ) are, respectively, a subsolution and asupersolution tomax (cid:26) v t − (cid:18) div (cid:18) Dv | Dv | (cid:19) + 1 (cid:19) | Dv | , v − E ( x ) (cid:27) = 0 in R n × (0 , ∞ ) , min (cid:26) v t − (cid:18) div (cid:18) Dv | Dv | (cid:19) + 1 (cid:19) | Dv | , v − int E ( x ) (cid:27) = 0 in R n × (0 , ∞ ) . We refer the readers to [11, 13] for some of related works concerning asymptotic behaviorof solutions of obstacle problems. Let us emphasize that even though [13] studies thelarge time behavior of obstacle problems for Hamilton-Jacobi equations with possiblydegenerate diffusion tr ( A ( x ) D u ), our problem here is not included since the degeneracyof the diffusion depends on the gradient of the solution.5.2. Upper and lower estimates.
From the heuristic observation by the Trotter-Katoapproximation, we realize that the motion of the top (5.1) and the bottom (5.2) can bedescribed by the obstacle problem of the surface evolution equation.For a closed set A ⊂ R n , we denote by F − [ A ]( t ) (resp., F + [ A ]( t )) the solution of thefront propagation of the obstacle problem V = κ + 1 with obstacle A, i.e., F − [ A ]( t ) ⊂ A (resp ., int A ⊂ F + [ A ]( t ))for any t ≥
0, and F ± [ A ](0) = A . We introduce two following geometric assumptions:(G1) there exist an open set D ⊃ E and t > F − [ D ]( t ) = ∅ ,(G2) F + [ E ]( t ) → R n as t → ∞ ,where E is the given set from the source.It is worthwhile emphasizing here that for each E ⊂ R n it is highly nontrivial to checkwhether (G1) and (G2) hold or not. This is a purely geometric problem which we have toinvestigate independently. In this subsection, we assume (G1) and (G2) first, and studyhow it gives an affect to the height of the solution to (1.1). In the next section, we willdiscuss more on (G1) and (G2).Recall that for any T > u ∈ USC ( R n × [0 , T ]) and there exists R T > u ( x, t ) = 0 for all ( x, t ) ∈ ( R n \ B (0 , R T )) × [0 , T ] . hus, x (cid:55)→ u ( x, T ) attains its maximum at some point x ∈ B (0 , R T ) and furthermore,in light of the subsolution built in (4.2), u ( x , T ) = max R n u ( · , T ) > . Lemma 5.3 (Upper estimate) . Assume (G1) holds, and let t be given by (G1) . Thereexists b ∈ (0 , c ) such that max x ∈ R n u ( x, t ) ≤ bt .Proof. Since A max ( t ) = ∅ , we have max x ∈ R n u ( x, t ) < ct . We set b := max x ∈ R n u ( x, t ) t to get the desired result. (cid:3) Theorem 5.4 (Global upper estimate) . Assume (G1) holds, and let t be given by (G1) .There exists b ∈ (0 , c ) such that u ( x, t ) ≤ bt + ( c − b ) t for all ( x, t ) ∈ R n × (0 , ∞ ) . In particular, lim sup t →∞ (cid:18) sup x ∈ R n u ( x, t ) t (cid:19) ≤ b. Proof.
Let w be the maximal solution to (2.1) with g = D . In light of Lemma 5.3, thereexists b ∈ (0 , c ) such that max x ∈ R n w ( x, t ) ≤ bt .By the comparison principle, u ≤ w on R n × [0 , ∞ ), which yields thatmax x ∈ R n u ( x, t ) ≤ max x ∈ R n w ( x, t ) ≤ bt . Using again the comparison principle and induction, we deduce that u ( x, mt + t ) ≤ w ( x, t ) + mbt on R n × [0 , ∞ ) for all m ∈ N . In particular, u ( x, mt ) ≤ mbt for any x ∈ R n and m ∈ N .For t ∈ ( mt , ( m + 1) t ), m ∈ N , we observe that u ( x, t ) ≤ u ( x, mt ) + c ( t − mt ) ≤ bmt + c ( t − mt )= bt + ( c − b )( t − mt ) ≤ bt + ( c − b ) t , which gives us the estimate on u ( x, t ) and also the estimate on lim sup. (cid:3) We get a lower bound in a similar manner. For R ≥ n and t >
0, set U R ( t ) := inf { u ∗ ( x, t ) : x ∈ B (0 , R ) ∪ E } . Lemma 5.5 (Lower estimate) . Assume (G2) holds. For R ≥ n , there exist a R > and t > such that U R ( t ) ≥ a R t . Theorem 5.6 (Global lower estimate) . Assume (G2) holds. For R ≥ n , there exist a R > and t > such that U R ( t ) ≥ a R ( t − t ) for all t ≥ . Thus, lim inf t →∞ (cid:18) inf x ∈ B (0 ,R ) u ∗ ( x, t ) t (cid:19) ≥ a R . emark 2. By the propagation property it is not difficult to see that a R can be takenindependent of R . Indeed, fix R > n and s > t . For t >
0, set A n ( t ) := { x ∈ R n : u ∗ ( x, t ) > a n ( s − t ) } . Then B (0 , n ) ⊂ A n ( s ). Note furthermore that A n ( t ) is a set theoretic supersolution of V = κ + 1. Hence, there exists t > s such that B (0 , R ) ⊂ A n ( s + t ) . In other words, t is the time it takes to transform B (0 , n ) into B (0 , R ) under the forcedmean curvature flow V = κ + 1. We conclude that u ∗ ( x, t ) ≥ a n ( t − t − t ) for all ( x, t ) ∈ B (0 , R ) × (0 , ∞ ) . Therefore, lim inf t →∞ (cid:18) inf x ∈ B (0 ,R ) u ∗ ( x, t ) t (cid:19) ≥ a n . The behavior of F ± [ E ]( t )In this section we investigate the precise behavior of the solution F ± [ E ]( · ) of the frontpropagation problems with obstacles. We in particular consider a family of squares in R ,i.e., E := E ( d ) = { ( x , x ) : | x i | ≤ d, i = 1 , } (6.1)for d > Proposition 6.1.
The followings hold :(i) If d < / √ , then u is bounded on R × [0 , ∞ ) . In particular, u ( · , t ) /t → uniformly in R as t → ∞ . (ii) If d > , then u ( · , t ) /t → c locally uniformly in R as t → ∞ . (iii) If d = 1 , then u ( · , t ) /t → c uniformly on B (0 , as t → ∞ . Note that in the case d = 1, Proposition 4.9 does not give the large time behavior of u ( x, t ) /t for x ∈ R n \ B (0 , / √ < d <
1, which is delicate. Our goal is to verifyassumptions (G1) and (G2), which in turn gives us some knowledge on the lim inf andlim sup behavior of u ( · , t ) /t as t → ∞ by using Theorems 5.4, 5.6.We first consider the behavior of F − [ E ]( t ). More precisely, we first study the behaviorof the solution to the obstacle problem for the graph:max (cid:26) y t − y xx y x ) − (1 + ( y x ) ) / , y − g ( x ) (cid:27) = 0 for ( x, t ) ∈ ( − D, D ) × (0 , ∞ ) , (6.2)where g ( x ) := −| x | , and D := √ d . We construct a viscosity supersolution w of (6.2) in[ − D, D ] × [0 , T ] for T > w ( x, ≥ g ( x ) − s in [ − D, D ] for some s > atisfying d + s <
1. Define W : [ − D, D ] → R by W ( x ) := (cid:40) −| x | for 1 / √ ≤ | x | ≤ D −√ − x ) / for | x | ≤ / √ , and set w ( x, t ) = λ ( t ) W (cid:18) xλ ( t ) (cid:19) , where λ : [0 , ∞ ) → R is the solution of the following ODE λ (cid:48) ( t ) = 1 λ ( t ) − t > ,λ (0) = s . (6.3)Pick T > λ ( T ) = d + s/ √ <
1. Clearly for t ∈ (0 , T ), λ (cid:48) ( t ) ≥ d + s/ √ − ε > , Thus T ≤ ε − < ∞ . Figure 4.
Graph of w ( x, t ) Lemma 6.2.
The function w defined above is a supersolution of (6.2) in [ − D, D ] × [0 , T ] and w ( x, ≥ g ( x ) − s in [ − D, D ] .Proof. It is straightforward from the definition that w ( x, ≥ g ( x ) − s .Let z = x/λ ( t ). For | z | < / √
2, we compute that w t = λ (cid:48) ( t )( W − zW (cid:48) ) = λ (cid:48) ( t ) (cid:18) −√ − z ) / − z − z (1 − z ) / (cid:19) = λ (cid:48) ( t ) (cid:18) −√ − z ) / (cid:19) ,w x = W (cid:48) = − z (1 − z ) / , w xx = W (cid:48)(cid:48) λ ( t ) = − − z ) / λ ( t ) . y using the above, for | z | < / √ w t − w xx w x ) − (1 + ( w x ) ) / = λ (cid:48) ( t ) (cid:18) −√ − z ) / (cid:19) + 1 λ ( t )(1 − z ) / − − z ) / = λ (cid:48) ( t ) (cid:18) −√ − z ) / (cid:19) + (cid:18) λ ( t ) − (cid:19) − z ) / = λ (cid:48) ( t ) (cid:18) −√ − z ) / (cid:19) ≥ , as λ (cid:48) ( t ) ≥ t ∈ [0 , T ].For 1 / √ ≤ | z | ≤ D , there is nothing to check as w already touches the obstacle. (cid:3) Theorem 6.3.
There exists t > such that F − [ E ]( t ) = ∅ for all t ≥ t .Proof. Recall that the original size of the obstacle (square) is 2 d ∈ ( √ , d + √ s <
2. By Lemma 6.2, F − [ E ]( T ) is containedin a ball of radius d + s/ √ <
1. We therefore deduce the existence of t > F − [ E ]( t ) = ∅ . (cid:3) Next, we consider the following obstacle problem:min (cid:26) y t − y xx y x ) − (1 + ( y x ) ) / , y − g ( x ) (cid:27) = 0 for ( x, t ) ∈ ( − D, D ) × (0 , ∞ ) . (6.4)Pick r ∈ (1 , D ). We construct a subsolution v of (6.4) in ( − D, D ) × [0 , T ] for T > v ( x,
0) = g ( x ) , ( v ( x, T ) + D ) + x ≥ r > x ∈ [ − D, D ] . Let φ : ( − r, r ) × [0 , ∞ ) → R such that φ ( x, t ) = − D + √ r − x + (cid:18) − r (cid:19) t. It is clear that φ is a separable subsolution of φ t − φ xx φ x ) − (1 + ( φ x ) ) / = 0 in ( − r, r ) × (0 , ∞ ) . Define the function v : R × [0 , ∞ ) → R by v ( x, t ) := (cid:26) max {−| x | , φ ( x, t ) } for | x | < r, −| x | for | x | ≥ r. igure 5. Picture of v ( x, t ) Lemma 6.4.
Let T = r (2 D − r ) r − . Then the function v defined as above is a subsolution of (6.4) in R × [0 , T ] with v ( x,
0) = −| x | in R and ( v ( x, T ) + D ) + x ≥ r > for all x ∈ [ − D, D ] . (6.5) Proof.
There is nothing to check in case | x | ≥ r as we always have v ( x, t ) = −| x | , whichis the obstacle part.We thus only need to check the case that | x | < r . In ( − r, r ) × (0 , ∞ ), v is defined as themaximum of two subsolutions of (6.4), which implies that v itself is also a subsolution.We can easily check that at T = r (2 D − r ) r − , v ( x, T ) := (cid:26) − r + √ r − x for | x | < r, −| x | for | x | ≥ r, which yields (6.5) immediately. (cid:3) Theorem 6.5.
Assume that < D < √ . Then F + [ E ]( t ) → R as t → ∞ .Proof. By Lemma 6.4, we have that F + [ E ]( T ) contains a ball of radius r >
1. As thisball of radius r expands to R under the forced mean curvature flow V = κ + 1 as timegoes to infinity, we get the conclusion. (cid:3) Remark 3.
If we consider the case where d < / √
2, then we can easily check that thefunction v defined by v ( x, t ) := √ − x for all ( x, t ) ∈ [ − D, D ] × [0 , ∞ ) is a supersolutionof (6.4) which is static. Therefore, the solution cannot grow up, which we have alreadygot in Proposition 6.1. On the other hand, if we consider the case where d > / √
2, thenwe cannot avoid to have shocks as in Figure 5, and therefore we cannot construct a staticsupersolution.In light of Theorems 6.3, 6.5, we conclude that
Proposition 6.6.
Assume that / √ < d < . Then there exists α, β such that < α <β < c and α ≤ lim inf t →∞ u ( x, t ) t ≤ lim sup t →∞ u ( x, t ) t ≤ β locally uniformly for x ∈ R . In a similiar manner to Theorem 6.5, we can get a slightly general result. roposition 6.7. Assume that, upon relabeling and reorienting the coordinates axes ifnecessary, there exist l > and functions g , g ∈ C ([ − l, l ] , R ) such that g < g on [ − l, l ] and { ( x, y ) : x ∈ [ − l, l ] and g ( x ) ≤ y ≤ g ( x ) } ⊂ E. Then F + [ E ]( t ) → R as t → ∞ . Remark 4.
One important problem in the crystal growth literature is to understandthe large time average of the crystal growth with two sources as well as the interactionbetween the sources. In the case that the two sources are the same and of circular shape,the equation becomes u t − (cid:18) div (cid:18) Du | Du | (cid:19) + 1 (cid:19) | Du | = c B (( a, ,R ) + c B (( − a, ,R ) in R × (0 , ∞ ) ,u ( · ,
0) = 0 on R , (6.6)where a, R > R > a , which means thatthe two sources overlap. As a corollary to Proposition 6.7, and Theorem 5.6, we getlim inf t →∞ u ( x, t ) t ≥ α locally uniformly for x ∈ R , for some α > a + R >
1. This is a tiny partial result toward this direction,which remains rather open so far. At least, we are able to give a first condition to have alocally uniform growth as t → ∞ . 7. Conclusion
We first established a well-posedness result for maximal viscosity solution of (2.1) inSection 2. Note that we are not able to prove uniqueness of viscosity solutions becauseof the discontinuity of the right hand side of (2.1) (see Remark 1). We believe that themaximal solution is the correct physical solution.In the spherically symmetric setting, we provide a complete analysis to understand thebehavior of the solution of (C) and its large time average in Propositions 4.3, 4.6, and4.8. We also study the case of inhomogeneous source f of (2.1) in Theorem 4.12.In the nonspherically symmetric case, it seems extremely hard to obtain the preciselarge time average of the maximal viscosity solution of (C). For instance, if we considerthe square case and use the Trotter-Kato product formula, then we could see clearly howcomplicated the behavior of the pyramid is as in Figure 6. This is of course cruciallydifferent from that of the spherically symmetric case which we observed in Section 3. igure 6 In the square case, we completely understood the case where d < / √ d > / √ < d < E is the union of the twoballs of same size in Remark 4. The precise growth speed in this case however is stillcompletely open. 8. Appendix
Lemma 8.1.
Let ψ : [0 , ∞ ) → R be a continuous function, which is C in (0 , R ) ∪ ( R, ∞ ) for some given R > . Assume further that ψ (cid:48) ( R − ) = a and ψ (cid:48) ( R +) = b. The followings hold (i) If a < b then for any φ ∈ C ( R n ) such that ψ ( | x | ) − φ ( x ) has a strict minimum at x ∈ ∂B (0 , R ) , then for some s ∈ [ a, b ] , Dφ ( x ) = s x R , and tr [ σ ( Dφ ( x )) D φ ( x )] ≤ ( n − sR . (ii) If a > b then for any φ ∈ C ( R n ) such that ψ ( | x | ) − φ ( x ) has a strict maximumat x ∈ ∂B (0 , R ) , then for some s ∈ [ b, a ] , Dφ ( x ) = s x R , and tr [ σ ( Dφ ( x )) D φ ( x )] ≥ ( n − sR . roof. We only prove (i). Without loss of generality, assume x = Re = ( R, , . . . , ψ ( R ) = 0. We only need to consider φ of the quadratic form φ ( x ) = p · ( x − x ) + A ( x − x ) · ( x − x ) , where A = ( a ij ) is a symmetric matrix. It is straightforward that Dφ ( x ) = p = s x R = se for some s ∈ [ a, b ] . We hence can rewrite φ ( x ) ≤ | x | = R as φ ( x ) = s ( x − R ) + a ij ( x i − δ i R )( x j − δ j R ) ≤ x + · · · x n = R , (8.1)where δ ij = 0 if i (cid:54) = j , and δ ii = 1. Let x = · · · = x n = 0 in the above to yield that, for x + x = R , s ( x − R ) + a ( x − R ) + 2 a ( x − R ) x + a ( R − x ) ≤ , which can be simplified further as a ( R − x ) − a x + a ( R + x ) ≤ s. Letting x → R (which also means that x →
0) to deduce that2 a ≤ sR . By using (8.1) in similar ways, we end up with2 n (cid:88) i =2 a ii ≤ ( n − sR . Note finally that σ ( p ) = I − p ⊗ p | p | = · · ·
00 1 · · · · · · , and thus tr [ σ ( Dφ ( x )) D φ ( x )] = n (cid:88) i =2 φ x i x i ( x ) = 2 n (cid:88) i =2 a ii ≤ ( n − sR . (cid:3) References
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Graduate School of Mathematical Sciences, University of Tokyo 3-8-1 Komaba,Meguro-ku, Tokyo, 153-8914, Japan
E-mail address : [email protected] (H. Mitake) Institute for Sustainable Sciences and Development, Hiroshima University1-4-1 Kagamiyama, Higashi-Hiroshima-shi 739-8527, Japan.
E-mail address : [email protected] (H. V. Tran) Department of Mathematics, University of Wisconsin, 480 Lincoln Dr.,Madison, WI 53706, USA.
E-mail address : [email protected]@math.wisc.edu