Self-similar solutions with fat tails for Smoluchowski's coagulation equation with singular kernels
aa r X i v : . [ m a t h . A P ] N ov Self-similar solutions with fat tails for Smoluchowski’s coagulationequation with singular kernels
B. Niethammer S. Throm J. J. L. Vel´azquez
Institute of Applied Mathematics, University of Bonn, Endenicher Allee 60, 53115 Bonn, GermanyE-mail: [email protected] ; [email protected] ; [email protected] Abstract
We show the existence of self-similar solutions with fat tails for Smoluchowski’s coagulation equation forhomogeneous kernels satisfying C (cid:0) x − a y b + x b y − a (cid:1) ≤ K ( x, y ) ≤ C (cid:0) x − a y b + x b y − a (cid:1) with a > b <
1. This covers especially the case of Smoluchowski’s classical kernel K ( x, y ) = ( x / + y / )( x − / + y − / ).For the proof of existence we first consider some regularized kernel K ε for which we construct a sequenceof solutions h ε . In a second step we pass to the limit ε → K .The main difficulty is to establish a uniform lower bound on h ε . The basic idea for this is to consider thetime-dependent problem and choosing a special test function that solves the dual problem. Smoluchowski’s coagulation equation [13] describes irreversible aggregation of clusters through binary collisionsby a mean-field model for the density f ( ξ, t ) of clusters of mass ξ . It is assumed that the rate of coagulationof clusters of size ξ and η is given by a rate kernel K = K ( ξ, η ), such that the evolution of f is determined by ∂ t f ( ξ, t ) = 12 Z ξ K ( ξ − η, η ) f ( ξ − η, t ) f ( η, t )d η − f ( ξ, t ) Z ∞ K ( ξ, η ) f ( η, t )d η . (1)Applications in which this model has been used are numerous and include, for example, aerosol physics,polymerization, astrophysics and mathematical biology (see e.g. [1, 3]).A topic of particular interest in the theory of coagulation is the scaling hypothesis on the long-timebehaviour of solutions to (1). Indeed, for homogeneous kernels one expects that solutions converge to auniquely determined self-similar profile. This issue is however only well-understood for the solvable kernels K ( x, y ) = 2, K ( x, y ) = x + y and K ( x, y ) = xy . In these cases it is known that (1) has one fast-decayingself-similar solution with finite mass and a family of so-called fat-tail self-similar solutions with power-lawdecay. Furthermore, their domains of attraction under the evolution (1) have been completely characterizedin [9]. For non-solvable kernels much less is known and it is exclusively for the case γ <
1. In [5, 6] existenceof self-similar solutions with finite mass has been established for a large range of kernels and some propertiesof those solutions have been investigated in [2, 4, 7]. More recently, the first existence results of self-similarsolutions with fat tails have been proved, first for the diagonal kernel [10], then for kernels that are boundedby C ( x γ + y γ ) for γ ∈ [0 ,
1) [11]. It is the goal of this paper to extend the results in [11] to singular kernels,such as Smoluchowski’s classical kernel K ( x, y ) = ( x / + y / )( x − / + y − / ). Uniqueness of solutions withboth, finite and infinite mass is still one of the main problems for non-solvable kernels and in most cases anopen question. Only recently uniqueness has been shown in the finite mass case for kernels that are in somesense close to the constant kernel [12]. 1n order to describe our results in more detail, we first derive the equation for self-similar solutions. Suchsolutions to (1) for kernels of homogeneity γ < f ( ξ, t ) = βt α g ( x ) , α = 1 + (1+ γ ) β , x = ξt β (2)where the self-similar profile g solves − αβ g − xg ′ ( x ) = 12 Z x K ( x − y, y ) g ( x − y ) g ( y )d y − g ( x ) Z ∞ K ( x, y ) g ( y )d y . (3)It is known that for some kernels the self-similar profiles are singular at the origin, so that the integrals onthe right-hand side are not finite and it is necessary to rewrite the equation in a weaker form. Multiplyingthe equation by x and rearranging we obtain that a weak self-similar solution g solves ∂ x ( x g ( x )) = ∂ x (cid:20)Z x Z ∞ x − y yK ( y, z ) g ( z ) g ( y ) d z d y (cid:21) + (cid:18) (1 − γ ) − β (cid:19) xg ( x ) (4)in a distributional sense. If one in addition requires that the solution has finite first moment, then this alsofixes β = 1 / (1 − γ ) and in this case the second term on the right hand side of (4) vanishes.For the following it is convenient to go over to the monomer density function h ( x, t ) = xg ( x, t ) and tointroduce the parameter ρ = γ + β . Then equation (4) becomes ∂ x (cid:20)Z x Z ∞ x − y K ( y, z ) z h ( z ) h ( y ) d z d y (cid:21) − [ ∂ x ( xh ) + ( ρ − h ] ( x ) = 0 . (5)Our approach to find a solution to (5) requires to work with the corresponding evolution equation. Using asnew time variable log ( t ) which will be denoted as t from now on, the time dependent version of equation (5)becomes ∂ t h ( x, t ) + ∂ x (cid:20)Z x Z ∞ x − y K ( y, z ) z h ( z, t ) h ( y, t ) d z d y (cid:21) − [ ∂ x ( xh ) + ( ρ − h ] ( x, t ) = 0 , (6)with initial data h ( x,
0) = h ( x ) . (7) We now formulate our assumptions on the kernel K . We assume that K ∈ C ((0 , ∞ ) × (0 , ∞ )) , K ( x, y ) = K ( y, x ) ≥ x, y ∈ (0 , ∞ ) , (8) K is homogeneous of degree γ ∈ ( −∞ , K ( λx, λy ) = λ γ K ( x, y ) for all x, y ∈ (0 , ∞ ) , (9)and satisfies the growth condition C (cid:0) x − a y b + x b y − a (cid:1) ≤ K ( x, y ) ≤ C (cid:0) x − a y b + x b y − a (cid:1) for all x, y ∈ (0 , ∞ ) , (10)where a > b < γ = b − a , and C , C are positive constants. Furthermore we assume the followinglocally uniform bound on the partial derivative: for each interval [ d, D ] ⊂ (0 , ∞ ) there exists a constant C = C ( d, D ) > | ∂ x K ( x, y ) | ≤ C (cid:0) y − a + y b (cid:1) for all x ∈ [ d, D ] and y ∈ (0 , ∞ ) . (11)2et us first discuss what we can expect on the possible decay behaviours of self-similar solutions. If h ( x ) ∼ Cx − ρ as x → ∞ , then in order for R ∞ K ( x,y ) y h ( y ) dy < ∞ we need ρ > b = γ + a and ρ + a > . (12)Note that since γ can be negative, − a can be larger than b . Furthermore we need to assume that b < b > b = 1 is a borderline case that can also not betreated with our methods. The same assumption has also been made in related work, where, for example in[2], regularity of self-similar solutions with finite mass have been investigated. In addition it will turn out laterthat we have to assume ρ > Theorem 1.1.
Let K be a kernel that satisfies assumptions (8) - (11) for some b ∈ ( −∞ , and a > . Thenfor any ρ ∈ (max( − a, b, ,
1) = (max( b, , there exists a non-negative measure h ∈ M ([0 , ∞ )) that solves (5) in the sense of distributions. This solution decays in the expected manner in an averaged sense, i.e. itsatisfies R [0 ,R ] h d x ≤ R − ρ for all R > and for each δ > there exists R δ > such that (1 − δ ) R − ρ ≤ Z [0 ,R ] h d x for all R ≥ R δ , which together implies lim R →∞ R − ρ R [0 ,R ] h d x = 1 .Remark . One can in fact show that under the assumptions (8)-(11) the measure h has a continuous densityand satisfies h ( r ) ∼ (1 − ρ ) r − ρ as r → ∞ . This has been proved in the case of locally bounded kernels in [11]and the proof in the present case proceeds similarly. Furthermore, if K is more regular, one can also establishhigher regularity of h in (0 , ∞ ). In order to keep the present paper within a reasonable length we will givethe corresponding proofs in a subsequent separate paper. The proof of Theorem 1.1 consists of two main parts. In the first one which is contained in Section 2 we shiftthe singularities of the kernel by some ε > K ε that is bounded at the origin. The idea isthen to prove Theorem 1.1 with this modified kernel to get a solution h ε . The proof follows the one in [11],i.e. the existence of a stationary solution to (6) is shown by using the following variant of Tikhonov’s fixedpoint theorem. Theorem 1.3 (Theorem 1.2 in [5, 8]) . Let X be a Banach space and ( S t ) t ≥ be a continuous semi-group on X . Assume that S t is weakly sequentially continuous for any t > and that there exists a subset Y of X that is nonempty, convex, weakly sequentially compact and invariant under the action of S t . Then there exists z ∈ Y which is stationary under the action of S t . As most of the estimates from [11] remain valid for the shifted kernel K ε we only state the main definitionsand results and refer to [11] for the proofs. The only exception is the invariance of some lower bound definingthe set Y from Theorem 1.3. As this step cannot just be transferred to the present situation we will give thefull proof of this. The main idea here is to construct a special test function that solves the dual problem, forwhich one can derive some lower bounds that are sufficient to obtain the invariance (Section 2.5.1).In the second part which is contained in Section 3 we have to remove the shift in K ε , i.e we have to takethe limit ε →
0. The strategy here is similar to what is done in the first part as one of the main difficultiesconsists in showing a suitable lower bound (uniform in ε ) for h ε (Section 3.2.3). This will again be done byconstructing a suitable test function by solving the dual problem for which we get adequate estimates frombelow (Section 3.2.1). One difficulty then is to show that the functions h ε obtained before decay sufficientlyrapidly at the origin as ε → ε → Stationary solutions for the kernel K ε In this section we let ε > K ε ( y, z ) := K ( y + ε, z + ε ) . We prove the following Proposition:
Proposition 2.1.
For any ρ ∈ (max( b, , there exists a continuous function h ε : (0 , ∞ ) → [0 , ∞ ) that is aweak solution to (5) with K replaced by K ε . This solution satisfies Z r h ε ( x ) dx ≤ r − ρ and lim r →∞ R r h ε ( x ) dxr − ρ = 1 . h ε The proof of Proposition 2.1 follows closely the proof of Theorem 1.1 in [11]. As the estimates remain inprinciple the same here we just recall the strategy of the proof and state the main definitions and results whilefor proofs we refer to [11]. The only modification we have to establish, compared to [11], is the proof of theinvariance of some lower bound that cannot just easily be adapted and we will show this in Section 2.5.The strategy to find a solution to (5) (with K replaced by K ε ) will be to show that the evolution givenby (6) satisfies the assumptions of Theorem 1.3 (notice that it suffices that the respective properties hold in apossibly small time interval [0 , T ]). One key point in the application of this theorem is obviously an appropriatechoice of X and Y . Here we use for X the set of measures on [0 , ∞ ) and as Y the set of non-negative measureswhich satisfy the expected decay behaviour in an averaged sense (cf. Definition 2.4). As well-posedness of (6)is not so easy to show for K ε directly we introduce a regularized problem where we cut the kernel K ε (in asmooth way) for small and large cluster sizes in the following way: for λ > K λε ( x, y ) = K ε ( x, y ) , if λ ≤ min { x, y } and max { x, y } ≤ λ ,K λε ( x, y ) = 0 , if min { x, y } ≤ λ { x, y } ≥ λ ,K λε ≤ K ε . (13) Remark . Note that in [11] a slightly different cutoff was used but this does not cause any problem.In the rest of this section we assume that in all equations the kernel K is replaced by K λε (or later by K ε when we take the limit λ → K λε with λ >
0. We will consider theset of non-negative Radon measures that we will denote with some abuse of notation by h ( x ) d x and such thatthe norm defined in (14) is finite. We notice that this implies that h d x does not contain a Dirac at the origin.Since, however, h d x might contain Dirac measures away from the origin, we use the convention that integralssuch as R ba h ( x )d x are always understood in the sense R [ a,b ] h ( x )d x . Definition 2.3.
Given ρ ∈ (max { , b } ,
1) with b as in Assumption (10), we will denote as X ρ the set ofmeasures h ∈ M + ([0 , ∞ )) such that k h k := sup R ≥ R [0 ,R ] h ( x ) d xR − ρ < ∞ . (14)We introduce a suitable topology in X ρ . We define the neighbourhoods of h ∗ ∈ X ρ by means of theintersections of sets of the form N φ,ǫ := ( h ∈ X ρ : (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z [0 , ∞ ) ( h − h ∗ ) φ d x (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) < ǫ ) , φ ∈ C c ([0 , ∞ )) , ǫ > . (15)We now define the subset Y , for which we will show that it remains invariant under the evolution definedby (6). 4 efinition 2.4. Given R > δ > Y the family of measures h ∈ X ρ satisfying thefollowing inequalities Z [0 ,r ] h d x ≤ r − ρ , for all r ≥ Z [0 ,r ] h d x ≥ r − ρ (cid:18) − R δ r δ (cid:19) + for all r > . (17) Remark . We will not make the dependence of Y on the variables R and δ explicit.We then easily see that Lemma 2.6.
The sets
Y ⊂ X ρ defined in Definition 2.4 are weakly sequentially compact. We first have to make sure that the evolution equation (6)-(7) with K replaced by K λε as in (13) is well-posed.We are going to construct first a mild solution of (6). For that purpose we introduce the rescaling X = x e t , h ( x, t ) = H ( X, t ) (18)and get ∂ t H ( X, t ) − ρH ( X, t ) + ∂ X "Z X Z ∞ X − Y K ελ ( Y e − t , Z e − t ) Z H ( Z, t ) H ( Y, t ) d Z d Y = 0 (19) H ( X,
0) = h ( X ) . (20)Expanding the derivative ∂ X we find that (19) is equivalent to ∂ t H ( X, t ) + ( A [ H ] ( X, t )) H ( X, t ) − Q [ H ] ( X, t ) = 0 (21)with A [ H ] ( X, t ) := Z ∞ K λε ( X e − t , Y e − t ) Y H ( Y, t ) d Y − ρ Q [ H ] ( X, t ) := Z X K λε ( Y e − t , ( X − Y ) e − t ) X − Y H ( X − Y, t ) H ( Y, t ) d
Y .
Definition 2.7.
We will say that a function H ∈ C ([0 , T ] , X ρ ) is a mild solution of equation (21) if thefollowing identity holds in the sense of measures H ( · , t ) = T [ H ] for 0 ≤ t ≤ T, (22)where T [ H ] ( X, t ) = exp (cid:18) − Z t A [ H ] ( X, s ) d s (cid:19) h ( X )+ Z t exp (cid:18) − Z ts A [ H ] ( X, τ ) d τ (cid:19) Q H ( X, s ) d s. (23)For this notion of solutions we have the following existence result that follows by the contraction mappingprinciple. Theorem 2.8.
Let K satisfy Assumptions (8) - (10) and let K λε be as in (13) for λ > . Then there exists T > such that there exists a unique mild solution of (21) in (0 , T ) in the sense of Definition 2.7.
5e furthermore introduce weak solutions in the following sense:
Definition 2.9.
We say that h ∈ C ([0 , T ] , X ρ ) is a weak solution of (6), (7) if for any t ∈ [0 , T ] and any testfunction ψ ∈ C c ([0 , ∞ ) × [0 , t ]) we have Z ∞ h ( x, t ) ψ ( x, t ) d x − Z ∞ h ( x ) ψ ( x,
0) d x − Z t (cid:20)Z ∞ ∂ s ψ ( x, s ) h ( x, s ) d x (cid:21) d s + Z t (cid:20)Z ∞ ψ ( x, s ) Z ∞ K λε ( x, z ) z h ( z, s ) d zh ( x, s ) d x (cid:21) d s − Z t (cid:20)Z ∞ ψ ( x, s ) Z x K λε ( y, x − y ) x − y h ( x − y, s ) h ( y, s ) d y d x (cid:21) d s + Z t Z ∞ xh ( x, s ) ∂ x ψ ( x, s ) d x d s − ( ρ − Z t Z ∞ h ( x, s ) ψ ( x, s ) d x d s = 0 . (24)By changing variables one obtains the following lemma. Lemma 2.10.
Suppose H ∈ C ([0 , T ] , X ρ ) is a mild solution of (6) , (7) in the sense of Definition 2.7. Then h is a weak solution in the sense of Definition 2.9. We denote the value of the mild solution h of (6)-(7) obtained in Theorem 2.8 with initial data h as h ( x, t ) = S λε ( t ) h ( x ) , t > . (25)Note that S λε ( t ) defines a mapping from X ρ to itself. We also define the transformation that brings h ( x )to H ( X, t ) which solves (19), (20) and whose existence is given by Theorem 2.8. We will write H ( X, t ) = T λε ( t ) h , t > . (26)With this notation we can state the following Proposition giving the weak continuity of the evolutionsemi-group: Proposition 2.11.
The transformation S λε ( t ) defined by means of (25) for any t ∈ [0 , T ] is a continuousmap from X ρ into itself if X ρ is endowed with the topology defined by means of the functionals (15) .Remark . The continuity that we obtain is not uniform in λ .The idea to prove this is to use a special test function in the definition of weak solutions. As the trans-formation (18) is continuous in the weak topology it suffices to show that T λε is continuous. Since it is notlinear it is not enough to check continuity at h = 0. More precisely fix t ∈ [0 , T ] and consider a test function¯Ψ ( X ) with ¯Ψ ∈ C c ([0 , ∞ )). Suppose that we have H , H such that T λε h ,i = H i ( · , t ) for i = 1 ,
2. Usingthe definition of weak solutions and taking the difference of the corresponding equations we obtain after somemanipulations: Z ∞ ( H ( X, t ) − H ( X, t )) Ψ (
X, t ) d X − Z ∞ ( h , ( X ) − h , ( X )) Ψ ( X,
0) d X = Z t Z ∞ ( H ( X, s ) − H ( X, s )) [ ∂ s Ψ − T [Ψ]] ( X, s ) d X d s Thus choosing the test function Ψ in some suitable function space such that ∂ s Ψ − T [Ψ] = 0 and Ψ ( · , t ) = ¯Ψthe claim follows (for more details see [11, Proposition 2.8]).6 .4 Recovering the upper estimate - Conservation of (16) Proposition 2.13.
Suppose that h ∈ X ρ satisfies (16) . Let h ( x, t ) be as in (25) . Then h ( · , t ) satisfies (16) as well. This follows in the same way as in [11, Proposition 3.1] by integrating (19) and changing variables. (17)
The invariance of the lower bound will be shown in several steps. First we choose a special test function in thedefinition of weak solutions, more precisely the solution of the corresponding dual problem. Next we derivesuitable lower bounds and integral estimates for this function. Finally we show the invariance of (17).
In Definition 2.9 of weak solutions we choose ψ such that − Z t Z [0 , ∞ ) ∂ s ψ ( x, s ) h ( x, s ) d x d s − Z t Z [0 , ∞ ) ψ ( x, s ) Z ∞ K λε ( x, z ) z h ( z, s ) d zh ( x, s ) d x d s − Z t Z [0 , ∞ ) ψ ( x, s ) Z x K λε ( y, x − y ) x − y h ( x − y, s ) h ( y, s ) d y d x d s + Z t Z [0 , ∞ ) xh ( x, s ) ∂ x ψ ( x, s ) d x d s − ( ρ − Z t Z [0 , ∞ ) h ( x, s ) ψ ( x, s ) d x d s ≤ , (27)and thus obtain Z [0 , ∞ ) h ( x, t ) ψ ( x, t ) d x d s − Z [0 , ∞ ) h ( x ) ψ ( x,
0) d x ≥ . (28)After some rearrangement we find that (27) is satisfied if ψ solves the dual problem ∂ s ψ ( x, s ) + Z ∞ K λε ( x, z ) z h ( z, s ) [ ψ ( x + z, s ) − ψ ( x, s )] d z − x∂ x ψ ( x, s ) − (1 − ρ ) ψ ( x, s ) ≥ ψ ( x, t ) = ψ ( x ).With regard to (28), the idea to estimate R R h ( x, t ) d x is to take ψ as a smoothed version of χ ( −∞ ,R ] and to estimate ψ ( · ,
0) from below. Using then that (17) holds for h this will be enough to show that thisestimate is conserved under the action of S λε . Rescaling X := x e s − t and ψ ( x, s ) = e − (1 − ρ )( t − s ) Φ ( x e s − t , s ) wefind after some elementary computations that equation (29) together with the initial condition is equivalentto ∂ s Φ (
X, s ) + Z ∞ K λε ( X e t − s , Z e t − s ) Z h (cid:0) Z e t − s (cid:1) [Φ ( X + Z, s ) − Φ (
X, s )] d Z ≥
0Φ (
X, t ) = ψ ( X ) . (30)This is (the rescaled version of) the dual problem. Remark . Note that for applying Theorem 1.3 we only need the assumptions on a small interval [0 , T ]. Sowe may assume in the following that
T < .5.2 Construction of a solution to the dual problem
In the following we always assume without loss of generality that ε ≤ R ≥ R ≥ < κ < ϕ κ a non-negative, symmetric standard mollifiersuch that supp ϕ κ ⊂ [ − κ, κ ].The idea to construct a solution Φ to the dual problem is to replace the solution h and the integral kernel K λε in (30) by corresponding power laws (using (10)) multiplied by a sufficiently large constant and estimatingthe powers of X using X ∈ [0 , R ]. We therefore choose Φ as a solution to ∂ s Φ (
X, s ) + C ε − a max (cid:8) ε b , (cid:9) Z ∞ ˜ v ( Z ) Z [Φ ( X + Z ) − Φ ( X )] d Z + C ε − a (cid:2) max (cid:8) ε b , (cid:9) + max (cid:8) ε b , R b (cid:9)(cid:3) Z ∞ ˜ v ( Z ) Z [Φ ( X + Z ) − Φ ( X )] d Z = 0 (31)with initial condition Φ ( X, t ) = χ ( −∞ ,R − κ ] ∗ ϕ κ/ ( X ) and C > v i ( Z ) := Z − ω i for i = 1 ,
2, with ω = min { ρ − b, ρ } and ω = ρ . Here ∗ n denotes the n -fold convolution. Lemma 2.15.
There exists a solution Φ of (31) in C ([0 , t ] , C ∞ ( R )) .Proof. This is shown in Proposition B.8.We furthermore define G ( X, s ) := − ∂ X Φ (
X, s ) and ˜ G by G ( X, s ) = R ˜ G (cid:0) XR − κR , s (cid:1) . Then ˜ G solves ∂ s ˜ G ( ξ, s ) + C ε − a max (cid:8) ε b , (cid:9) Z ∞ ˜ v ( Rη ) η h ˜ G ( ξ + η ) − ˜ G ( ξ ) i d η + C ε − a (cid:2) max (cid:8) ε b , (cid:9) + max (cid:8) ε b , R b (cid:9)(cid:3) Z ∞ ˜ v ( Rη ) η h ˜ G ( ξ + η ) − ˜ G ( ξ ) i d η = 0 (32)with initial datum ˜ G ( · , t ) = δ ( · ) ∗ ϕ κ R . We also summarize the following properties for Φ and ˜ G given byRemark B.9 and the choice of the initial condition: Remark . The function Φ ( · , s ) given by Lemma 2.15 is non-increasing for all s ∈ [0 , t ] and satisfies:0 ≤ Φ ( · , s ) ≤ , supp Φ ( · , s ) ⊂ ( −∞ , R ] for all s ∈ [0 , t ] and Φ ( X, t ) = 1 for all X ∈ ( −∞ , R − κ ] . Furthermore ˜ G is non-negative and satisfiessupp ˜ G ( · , s ) ⊂ ( −∞ , κ/R ] and Z R ˜ G ( ξ, s ) d ξ = 1 for all s ∈ [0 , t ] . The following Lemma states the two integral bounds that will be the key in proving the invariance of (17):
Lemma 2.17.
There exist ω, θ ∈ (0 , such that for every µ ∈ (0 , and D > we have1. R − D −∞ ˜ G ( ξ, s ) d ξ ≤ C (cid:0) κRD (cid:1) µ + CtD ω R − θ ,2. R − | ξ | ˜ G ( ξ, s ) d ξ ≤ C (cid:0) κR (cid:1) µ + CtR − θ .Proof of Lemma 2.17. From the proof of Proposition B.8 we have that ˜ G is given as ˜ G = ˜ G ∗ ˜ G where ˜ G and ˜ G solve ∂ s ˜ G ( ξ, s ) + C ε − a max (cid:8) ε b , (cid:9) Z ∞ ˜ v ( Rη ) η h ˜ G ( ξ + η ) − ˜ G ( ξ ) i d η = 0 ∂ s ˜ G ( ξ, s ) + C ε − a (cid:2) max (cid:8) ε b , (cid:9) + max (cid:8) ε b , R b (cid:9)(cid:3) Z ∞ ˜ v ( Rη ) η h ˜ G ( ξ + η ) − ˜ G ( ξ ) i d η = 08ith initial data ˜ G ( · , t ) = ˜ G ( · , t ) = ϕ κ R . Then one has from Lemma B.11: Z − D −∞ ˜ G ( ξ, s ) d ξ ≤ Z − D/ −∞ ˜ G ( ξ, s ) d ξ + Z − D/ −∞ ˜ G ( ξ, s ) d ξ and Z − | ξ | ˜ G ( ξ, s ) d ξ ≤ Z κ R − − κ R | ξ | ˜ G ( ξ, s ) d ξ + Z κ R − − κ R | ξ | ˜ G ( ξ, s ) d ξ ≤ κR + Z − | ξ | ˜ G ( ξ, s ) d ξ + Z − | ξ | ˜ G ( ξ, s ) d ξ. From Lemma B.10 we obtain with N ( η ) = C ( ε ) R − ω η − − ω , N ( η ) = C ( ε ) (cid:2) max (cid:8) ε b , (cid:9) + max (cid:8) ε b , R b (cid:9)(cid:3) R − ω η − − ω the appropriate estimates for ˜ G i , i = 1 , ω i for i = 1 , θ = ω , θ = min { ω , b − ω } . Thus theclaim follows by setting θ := min { ω , ω , b − ω } and ω := min { ω , ω } and using also that κR ≤ (cid:0) κR (cid:1) µ for κ ≤ R ≥ Lemma 2.18.
For sufficiently large C the function Φ satisfies (30) .Proof. We have to show that for X ∈ [0 , R ] we have ∂ s Φ (
X, s ) + Z ∞ K λε ( X e t − s , Z e t − s ) Z e sβ h (cid:0) Z e t − s , s (cid:1) [Φ ( X + Z, s ) − Φ (
X, s )] d Z ≥ . By construction of Φ this is equivalent to − C ε − a max (cid:8) ε b , (cid:9) Z ∞ ˜ v ( Z ) Z [Φ ( X + Z ) − Φ ( X )] d Z − C ε − a (cid:2) max (cid:8) ε b , (cid:9) + max (cid:8) ε b , R b (cid:9)(cid:3) Z ∞ ˜ v ( Z ) Z [Φ ( X + Z ) − Φ ( X )] d Z + Z ∞ K λε ( X e t − s , Z e t − s ) Z e sβ h (cid:0) Z e t − s (cid:1) [Φ ( X + Z ) − Φ ( X )] dZ ≥ K λε for X ∈ [0 , R ] one has K λε ( X e t − s , Z e t − s ) Z e sβ ≤ C ( X e t − s + ε ) − a ( Z e t − s + ε ) b + ( X e t − s + ε ) b ( Z e t − s + ε ) − a Z ≤ Cε − a ( Z e t − s + ε ) b Z + C max (cid:8) ε b , R b (cid:9) ( Z e t − s + ε ) − a Z ≤ Cε − a max (cid:8) , ε b (cid:9) (cid:0) Z − χ [0 , ( Z ) + Z b − χ [1 , ∞ ) ( Z ) (cid:1) + C max (cid:8) ε b , R b (cid:9) ε − a Z − ≤ Cε − a (cid:2) max (cid:8) ε b , (cid:9) + max (cid:8) ε b , R b (cid:9)(cid:3) Z − + Cε − a max (cid:8) , ε b (cid:9) Z max { ,b }− . Defining w ( Z ) := h ( Z e t − s ) Z − max { ,b } and w ( Z ) := h ( Z e t − s ) Z C ε − a max (cid:8) ε b , (cid:9) Z ∞ ˜ v ( Z ) Z [Φ ( X ) − Φ ( X + Z )] d Z + C ε − a (cid:2) max (cid:8) ε b , (cid:9) + max (cid:8) ε b , R b (cid:9)(cid:3) Z ∞ ˜ v ( Z ) Z [Φ ( X ) − Φ ( X + Z )] d Z − Cε − a max (cid:8) , ε b (cid:9) Z ∞ w ( Z ) Z [Φ ( X ) − Φ ( X + Z )] d Z − Cε − a (cid:2) max (cid:8) ε b , (cid:9) + max (cid:8) ε b , R b (cid:9)(cid:3) Z ∞ w ( Z ) Z [Φ ( X ) − Φ ( X + Z )] d Z ≥ . Defining V i ( Z ) := Z ∞ Z v i ( Y ) Y d Y and W i ( Z ) := Z ∞ Z w i ( Y ) Y d Y we can rewrite this as ε − a max (cid:8) ε b , (cid:9) Z ∞ − ∂ Z ( C V ( Z ) − CW ( Z )) [Φ ( X ) − Φ ( X + Z )] d Z + ε − a (cid:2) max (cid:8) ε b , (cid:9) + max (cid:8) ε b , R b (cid:9)(cid:3) Z ∞ − ∂ Z ( C V ( Z ) − CW ( Z )) [Φ ( X ) − Φ ( X + Z )] d Z ≥ . Integrating by parts this is equivalent to ε − a max (cid:8) ε b , (cid:9) Z ∞ − ∂ Z Φ ( X + Z ) ( C V ( Z ) − CW ( Z )) d Z + ε − a (cid:2) max (cid:8) ε b , (cid:9) + max (cid:8) ε b , R b (cid:9)(cid:3) Z ∞ − ∂ Z Φ ( X + Z ) ( C V ( Z ) − CW ( Z )) d Z ≥ . Using that Φ is non-increasing it thus suffices to show C V i ( Z ) − CW i ( Z ) ≥ i = 1 ,
2. To see this notefirst that V i is explicitly given by V i ( Z ) = ω i Z − ω i for i = 1 ,
2. Furthermore using Lemma A.1 one has W ( Z ) ≤ CZ max { ,b }− ρ = CZ − ω and W ( Z ) ≤ CZ − ω . Thus choosing C sufficiently large the claim follows.We finally prove a technical Lemma that will be needed in the following. Lemma 2.19.
Let δ ∈ (0 , and ρ ∈ (max { , b } , . Then (cid:16) − κR + ξ (cid:17) − ρ − (cid:18) R R (cid:19) δ e − δt (cid:0) − κR + ξ (cid:1) δ ! ≥ − (cid:18) R R (cid:19) δ e − δt ! − (cid:12)(cid:12)(cid:12) ξ − κR (cid:12)(cid:12)(cid:12) holds for every ξ ∈ (cid:2) R R e − t − κR , κR (cid:3) .Proof. Let R R e − t − κR ≤ ξ ≤ κR , then (cid:0) − κR + ξ (cid:1) ∈ (cid:2) R R e − t , (cid:3) . Thus for 0 ≤ ρ < (cid:0) − κR + ξ (cid:1) − ρ ≥ (cid:0) − (cid:12)(cid:12) ξ − κR (cid:12)(cid:12)(cid:1) and therefore we can estimate (noting that the second term in bracketsis non-negative) (cid:16) − κR + ξ (cid:17) − ρ − (cid:18) R R (cid:19) δ e − δt (cid:0) − κR + ξ (cid:1) δ ! ≥ (cid:16) − κR + ξ (cid:17) − (cid:18) R R (cid:19) δ e − δt (cid:0) − κR + ξ (cid:1) δ ! = 1 − κR + ξ − (cid:18) R R (cid:19) δ e − δt − κR + ξ (cid:0) − κR + ξ (cid:1) δ ≥ − (cid:18) R R (cid:19) δ e − δt − (cid:12)(cid:12)(cid:12) ξ − κR (cid:12)(cid:12)(cid:12) as − κR + ξ ( − κR + ξ ) δ ≤ ξ as above and δ <
1. 10 .5.3 Invariance of the lower bound
We are now prepared to finish the proof of the invariance of the lower bound (17).
Proposition 2.20.
For sufficiently small δ > and sufficiently large R (maybe also depending on δ ), wehave Z R h ( x, t ) d x ≥ R − ρ − (cid:18) R R (cid:19) δ ! + . Proof.
From the special choice of the test function ψ we have according to (28) that Z R h ( x, t ) d x ≥ Z ∞ h ( x, t ) ψ ( x, t ) d x ≥ Z ∞ h ( x ) ψ ( x,
0) d x = e − (1 − ρ ) t Z ∞ h (cid:0) X e t (cid:1) Φ ( X,
0) e t d X. Defining H ( X ) := R X h ( Y ) d Y we obtain Z R h ( x, t ) ≥ e − (1 − ρ ) t Z ∞ h (cid:0) X e t (cid:1) Φ ( X,
0) e t d X = e − (1 − ρ ) t Z ∞ H ′ (cid:0) X e t (cid:1) Φ ( X,
0) d X = e − (1 − ρ ) t H (cid:0) X e t (cid:1) Φ ( X, (cid:12)(cid:12)(cid:12) ∞ + e − (1 − ρ ) t Z ∞ H (cid:0) X e t (cid:1) ( − ∂ X Φ ( X, X = e − (1 − ρ ) t Z ∞ H (cid:0) X e t (cid:1) ( − ∂ X Φ ( X, X where we integrated by parts and used that the boundary terms are zero as H (0) = 0 and the support of Φis bounded from the right. Using now that by assumption we have H ( X e t ) ≥ ( X e t ) − ρ (cid:16) − (cid:0) R X e t (cid:1) δ (cid:17) + weobtain Z R h ( x, t ) d x ≥ e − (1 − ρ ) t Z ∞ (cid:0) X e t (cid:1) − ρ − (cid:18) R X e t (cid:19) δ ! + G ( X,
0) d X = Z ∞ X − ρ − (cid:18) R X e t (cid:19) δ ! + R ˜ G (cid:18) XR − κR , (cid:19) d X = R − ρ Z κ/R − D (cid:16) − κR + ξ (cid:17) − ρ − (cid:18) R R (cid:19) δ e − δt (cid:0) − κR + ξ (cid:1) δ ! ˜ G ( ξ,
0) d ξ where we used the change of variables ξ = XR − κR and defined D := 1 − R R e − t − κR . Note that for κ ≤ (1 − e − t ) we have 0 ≤ D ≤ R ≥ Z R h ( x, t ) d x ≥ R − ρ Z κ/R − D − (cid:18) R R (cid:19) δ e − δt ! ˜ G ( ξ,
0) d ξ − R − ρ Z κ/R − D (cid:12)(cid:12)(cid:12) ξ − κR (cid:12)(cid:12)(cid:12) ˜ G ( ξ,
0) d ξ ≥ R − ρ − (cid:18) R R (cid:19) δ e − δt ! Z κ/R −∞ ˜ G ( ξ,
0) d ξ − Z − D −∞ ˜ G ( ξ,
0) d ξ ! − R − ρ Z − (cid:12)(cid:12)(cid:12) ξ − κR (cid:12)(cid:12)(cid:12) ˜ G ( ξ,
0) d ξ + κR Z κ/R ˜ G ( ξ,
0) d ξ ! Applying furthermore Remark 2.16 and Lemma 2.17 we get11 R h ( x, t ) d x ≥ R − ρ − (cid:18) R R (cid:19) δ e − δt ! (cid:18) − C (cid:16) κRD (cid:17) µ − CD ω R − θ (cid:19) − R − ρ (cid:18)Z − | ξ | ˜ G ( ξ,
0) d ξ + κR Z − ˜ G ( ξ,
0) d ξ + κR (cid:19) ≥ R − ρ − (cid:18) R R (cid:19) δ e − δt ! (cid:18) − C (cid:16) κRD (cid:17) µ − CtD ω R − θ (cid:19) − R − ρ (cid:18) C (cid:16) κR (cid:17) µ + CtR − θ + 2 κR (cid:19) . Inserting the definition of D and rearranging we thus obtain Z R h ( x, t ) d x ≥ R − ρ − (cid:18) R R (cid:19) δ e − δt ! − CR − ρ (cid:16) − (cid:0) R R (cid:1) δ e − δt (cid:17)(cid:16) − (cid:0) R R (cid:1) δ e − δt (cid:17) µ (cid:16) κR (cid:17) µ − CtR θ R − ρ (cid:16) − (cid:0) R R (cid:1) δ e − δt (cid:17)(cid:0) − (cid:0) R R (cid:1) e − t (cid:1) ω + 1 − CR − ρ (cid:16) κR + (cid:16) κR (cid:17) µ (cid:17) . Using that for δ, ω ∈ (0 ,
1) and
R > R we have (cid:16) − (cid:0) R R (cid:1) δ e − δt (cid:17) ≤ − R R e − t ≤ (cid:0) − R R e − t (cid:1) ω and thus (cid:18) − ( R R ) δ e − δt ( − R R e − t ) ω + 1 (cid:19) ≤ (cid:16) − ( R R ) δ e − δt (cid:17)(cid:16) − ( R R ) δ e − δt (cid:17) µ ≤
1, we therefore get Z R h ( x, t ) d x ≥ R − ρ − (cid:18) R R (cid:19) δ e − δt ! − CtR θ R − ρ − CR − ρ (cid:16) κR + (cid:16) κR (cid:17) µ (cid:17) . As for δt ≤ δ, t ≤
1) we can estimate 1 − (cid:0) R R (cid:1) δ e − δt ≥ − (cid:0) R R (cid:1) δ + (cid:0) R R (cid:1) δ δt , weobtain Z R h ( x, t ) d x ≥ R − ρ − (cid:18) R R (cid:19) δ ! + R − ρ (cid:18) R R (cid:19) δ δt e − CtR θ R − ρ − CR − ρ (cid:16) κR + (cid:16) κR (cid:17) µ (cid:17) . We now choose µ = θ and κ < R ≥ R ≥ κR ≤ (cid:0) κR (cid:1) µ = (cid:0) κR (cid:1) θ .Using this we can further estimate Z R h ( x, t ) d x ≥ R − ρ − (cid:18) R R (cid:19) δ ! + R − ρ (cid:18) R R (cid:19) δ δt e − CtR θ R − ρ − Cκ θ R θ R − ρ ≥ R − ρ − (cid:18) R R (cid:19) δ ! + R − ρ (cid:18) R R (cid:19) δ δt e − CR θ (cid:0) t + κ θ (cid:1)! . Thus it suffices to show (cid:0) R R (cid:1) δ δt e − CR θ (cid:0) t + κ θ (cid:1) ≥ R θ − δ ≥ C e t + κ θ R δ δt , but this is trueat least if we choose κ sufficiently small such that also κ θ < t , 0 < δ < θ and then R sufficiently large (notethat we have only to prove this for R ≥ R ). Proposition 2.21.
Let K satisfy Assumptions (8) - (10) . Then for any ρ ∈ (max { , b } , there exists a weakstationary solution h ε to (6) (with K replaced by K ε ). h λε ∈ Y that isstationary under the action of S λε ( t ), i.e. H λε is a stationary mild solution of (21). Then taking a subsequenceof h λε converging weakly to some h ε and passing to the limit λ → ε → h ε completingthe proof of Proposition 2.1: Proposition 2.22.
The solution h ε ∈ Y from Proposition 2.21 is continuous on (0 , ∞ ) and satisfies h ε ( x ) ∼ (1 − ρ ) x − ρ as x → ∞ . This can be proved in a similar way as the corresponding result in [11, Lemma 4.2 & Lemma 4.3]. ε → Our starting point is that we have a continuous positive function h ε that is a weak solution to ∂ x I ε [ h ε ] = ∂ x ( xh ε ) + ( ρ − h ε , with I ε [ h ε ] = Z x Z ∞ x − y K ε ( y, z ) z h ε ( y ) h ε ( z ) dz dy . (33)Furthermore we have the estimates Z r h ε ( x ) dx ≤ r − ρ and lim r →∞ Z r h ε ( x ) dx/r − ρ = 1 . (34)We introduce the following quantities: µ ε = Z h ε ( x ) ( x + ε ) − a dx , λ ε = Z h ε ( x ) ( x + ε ) b dx , L ε := max (cid:18) λ a ε , µ − b ε (cid:19) . (35)Up to passing to a subsequence we can in the following assume that either L ε converges or L ε → ∞ for ε → L ε → L := L ε if L ε L := 1 if L ε → L >
0. For the following let X = xL , h ε ( x ) = H ε ( X ) L − ρ . The strategy of the proof is the following: First we derive a uniform lower integral bound for H ε . Thiswill be done by constructing a special test function that provides us with some estimate from below that issufficient to conclude on a lower bound by some iteration argument. In order to obtain the same uniformlower bound for h ε we need to exclude the case L ε → ∞ . For this reason we will show that in this case H ε converges to some limit H solving some differential equation that has no solution satisfying the growthcondition R R H d X ≤ R − ρ . Note that at this point it is crucial to assume ρ >
0. Using the lower bound on h ε we can show some exponential decay of h ε near zero which is then enough to pass to the limit ε → < κ < ϕ κ a non-negative, symmetric standard mollifier withsupp ϕ κ ⊂ [ − κ, κ ]. H ε In this subsection we will show a uniform lower bound on H ε , i.e. we will prove: Proposition 3.1.
For any δ > there exists R δ > such that Z R H ε ( X )d X ≥ (1 − δ ) R − ρ for all R ≥ R δ . (36)13 .2.1 Construction of a suitable test function We start by constructing a special test function and therefore notice that for ψ = ψ ( x, t ) with ψ ∈ C andcompact support in [0 , T ] × [0 , ∞ ) we obtain from the equation on h ε :0 = Z T Z ∞ ∂ x ψI ε [ h ε ] d x d t − Z T Z ∞ x∂ x ψh ε d x d t + ( ρ − Z T Z ∞ ψh ε d x d t = Z T Z ∞ ∂ t ψh ε d x d t + Z ∞ ψ ( · , h ε d x − Z ∞ ψ ( · , T ) h ε d x. Choosing ψ such that Z T Z ∞ ∂ x ψI ε [ h ε ] d x d t − Z T Z ∞ x∂ x ψh ε d x d t + ( ρ − Z T Z ∞ ψh ε d x d t − Z T Z ∞ ∂ t ψh ε d x d t ≥ Z ∞ ψ ( · , h ε d x ≥ Z ∞ ψ ( · , T ) h ε d x. Rewriting (37) we obtain Z T Z ∞ h ε ( x ) (cid:26)Z ∞ K ε ( x, y ) y h ε ( y ) [ ψ ( x + y ) − ψ ( x )] d y − x∂ x ψ ( x ) + ( ρ − ψ ( x ) − ∂ t ψ ( x ) (cid:27) d x d t ≥ . (38)Defining W by ψ ( x, t ) = e − (1 − ρ ) t W ( ξ, t ) with ξ = xL e t we can rewrite the term in brackets and obtain that itsuffices to construct W such that ∂ t W (cid:16) xL e t , t (cid:17) ≤ Z ∞ K ε ( x, y ) y h ε ( y ) (cid:20) W (cid:18) x + yL e t , t (cid:19) − W (cid:16) xL e t , t (cid:17)(cid:21) d y. (39)For further use we also note that we only need this in weak form, i.e. we need that Z T Z ∞ e − (1 − ρ ) t h ε ( x ) (cid:26) ∂ t W (cid:16) xL e t , t (cid:17) − Z ∞ K ε ( x, y ) y h ε ( y ) (cid:20) W (cid:18) x + yL e t , t (cid:19) − W (cid:16) xL e t , t (cid:17)(cid:21) d y (cid:27) d x d t ≤ , (40)provided that we can justify the change from ψ to W .We furthermore list here some parameters that are frequently used in the following. For given ν ∈ (0 , β := ( b b ≥ νb b < , ω := min { ρ − b, ρ } , ω := ρ, ˜ b := max { , b } . The idea to construct the test function W is similar to the approach in Section 2.5.2, i.e. we replace theintegral kernel K ε and h ε by corresponding power laws. Due to the singular behaviour of K ε (for ε → Lemma 3.2.
For any constant ˜ C > there exists a function ˜ W ∈ C ([0 , T ] , C ∞ ( R )) solving ∂ t ˜ W ( ξ, t ) − C Z h ε ( z ) z (cid:16) L b A β ( z + ε ) − a + L − a A − νa ( z + ε ) b (cid:17) h ˜ W (cid:16) ξ + zL (cid:17) − ˜ W ( ξ ) i d z − CL ρ + a − max { ,b } Z ∞ A − νa η ω h ˜ W ( ξ + η ) − ˜ W ( ξ ) i d η − CL ρ − b Z ∞ A β η ω h ˜ W ( ξ + η ) − ˜ W ( ξ ) i d η = 0 with ˜ W ( · ,
0) = χ ( −∞ ,A − κ ] ∗ ϕ κ/ ( · ) . roof. This is shown in Proposition B.8.
Remark . As shown in the appendix ˜ W is non-increasing, has support in ( −∞ , A ], is non-negative andbounded by 1.As K ε might get quite singular at the origin for ε →
0, we define now W as the function W ( ξ, t ) :=˜ W ( ξ, t ) χ [ A ν , ∞ ) ( ξ ), i.e. we cut ˜ W at ξ = A ν in order to avoid integrating near the origin. Obviously W isnot in C and thus the corresponding ψ is also not differentiable. But as already mentioned it is enough toshow that (40) holds, provided we can justify the change from ψ to W (and reverse). This will be done next,i.e. we will first show that (39) holds for all ξ = A ν . Then by convolution in ξ with ϕ δ it is possible to changefrom ψ to W (and reverse). Finally taking the limit δ → Lemma 3.4.
For sufficiently large ˜ C , inequality (39) holds pointwise for all ξ = A ν .Proof. From the non-negativity of W the claim follows immediately for ξ < A ν (where W is identicallyzero). Thus it suffices to consider ξ > A ν . Using furthermore that supp W ⊂ ( −∞ , A ] it suffices toconsider ξ ∈ ( A ν , A ]. As W is non-increasing on ( A ν , A ] we can estimate − (cid:2) W (cid:0) ξ + yL e t (cid:1) − W ( ξ ) (cid:3) ≤− (cid:2) W (cid:0) ξ + yL (cid:1) − W ( ξ ) (cid:3) . On the other hand using the estimates on the kernel K we obtain − Z ∞ K ε ( L e t ξ, y ) y h ε ( y ) h W (cid:16) ξ + yL e t (cid:17) − W ( ξ ) i d y ≤ − C Z ∞ ( L e t ξ + ε ) − a ( y + ε ) b + ( L e t ξ + ε ) b ( y + ε ) − a y h ε ( y ) h W (cid:16) ξ + yL (cid:17) − W ( ξ ) i d y ≤ − C Z ∞ L − a A − νa ( y + ε ) b + L b A β ( y + ε ) − a y h ε ( y ) h W (cid:16) ξ + yL (cid:17) − W ( ξ ) i d y ≤ − C Z h ε ( y ) y h L b A β ( y + ε ) − a + L − a A − νa ( y + ε ) b i h W (cid:16) ξ + yL (cid:17) − W ( ξ ) i d y − CL ρ Z ∞ /L H ε ( η ) η h L b A β + L − a A − νa ( Lη + ε ) b i [ W ( ξ + η ) − W ( ξ )] d η ≤ − C Z h ε ( y ) y h L b A β ( y + ε ) − a + L − a A − νa ( y + ε ) b i h W (cid:16) ξ + yL (cid:17) − W ( ξ ) i d y − CA β L ρ − b Z ∞ H ε ( η ) η [ W ( ξ + η ) − W ( ξ )] d η − CA − νa L ρ + a − max { ,b } Z ∞ H ε ( η ) η − max { ,b } [ W ( ξ + η ) − W ( ξ )] d η. (41)As ξ > A ν we have by construction ∂ t W ( ξ, t ) = ∂ t ˜ W ( ξ, t ) = ˜ C Z h ε ( z ) z h L b A β ( z + ε ) − a + L − a A − νa ( z + ε ) b i h ˜ W (cid:16) ξ + zL (cid:17) − ˜ W ( ξ ) i d z + ˜ C A β L ρ − b Z ∞ η ω h ˜ W ( ξ + η ) − ˜ W ( ξ ) i d η + ˜ C A − νa L ρ + a − max { ,b } Z ∞ η ω h ˜ W ( ξ + η ) − ˜ W ( ξ ) i d η. (42)Thus in order to show (39), i.e. ∂ t W ( ξ, t ) − Z ∞ K ε ( L e t ξ, y ) y h ε ( y ) h W (cid:16) ξ + yL e t , t (cid:17) − W ( ξ, t ) i d y ≤ H ε . Lemma 3.5.
The change of variables from ψ to W (and reverse) is justified and inequality (40) holds. roof. We define W δ := W ( · , t ) ∗ ϕ δ ( · ). Then W δ ( · , t ) is smooth for all t and the change of variables fromthe corresponding ψ δ to W δ (and reverse) is justified. We next show that we can pass to the limit δ → ∂ t W δ (cid:0) xL e t , t (cid:1) is compactly supported and uniformly bounded (in δ ) it suffices to consider Z ∞ h ε ( x ) Z ∞ K ε ( x, y ) y h ε ( y ) (cid:20) W δ (cid:18) x + yL e t , t (cid:19) − W δ (cid:16) xL e t , t (cid:17)(cid:21) d y d x. (43)Changing variables, interchanging the order of integration and splitting the integral we have to consider Z ∞ Z ∞ L e t K ε ( L e t ξ, y ) y h ε (cid:0) L e t ξ (cid:1) h ε ( y ) h W δ (cid:16) ξ + yL e t (cid:17) − W δ ( ξ ) i d ξ d y = Z ∞ Z A ν − r ( · · · ) d ξ d y + Z ∞ Z ∞ A ν + r ( · · · ) d ξ d y + Z ∞ Z A ν + rA ν − r ( · · · ) d ξ d y =: ( I ) + ( II ) + ( III ) , where 2 δ < r is a fixed and sufficiently small constant. As W and W δ are compactly supported and bounded(uniformly in δ ) it is straightforward to pass to the limit δ → ξ -integrals (for fixed y > δ → y -integral while this will be done byusing Lebesgue’s Theorem. We therefore estimate the three integrands separately. First we have using that W is non-negative, bounded and has support in [ A ν − δ, A + δ ] as well as the estimate for K and Lemma A.1that (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z A ν − r L e t K ε ( L e t ξ, y ) y h ε (cid:0) L e t ξ (cid:1) h ε ( y ) h W δ (cid:16) ξ + yL e t (cid:17) − W δ ( ξ ) i d ξ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ CL e t χ [ L e t ( r − δ ) , ∞ ) ( y ) Z A ν − r ( L e t ξ + ε ) − a ( y + ε ) b y h ε (cid:0) L e t ξ (cid:1) h ε ( y ) W δ (cid:16) ξ + yL e t , t (cid:17) d ξ + CL e t χ [ L e t ( r − δ ) , ∞ ) ( y ) Z A ν − r ( L e t ξ + ε ) b ( y + ε ) − a y h ε (cid:0) L e t ξ (cid:1) h ε ( y ) W δ (cid:16) ξ + yL e t , t (cid:17) d ξ ≤ C ( ε, L, A ) h ε ( y ) y χ [ L e t r , ∞ ) ( y ) Z A ν − r h ε (cid:0) L e t ξ (cid:1) d ξ ≤ C ( ε, L, A ) (cid:0) L e t ( A ν − r ) (cid:1) − ρ h ε ( y ) y χ [ L e t r , ∞ ) ( y ) . As the right hand side is independent of δ and integrable due to Lemma A.1 we can pass to the limit δ → I ).To estimate the integrand in ( II ) note that we can bound h ε ( L e t ξ ) for ξ ∈ [ A ν , A + δ ] uniformly in t for t ∈ [0 , T ] as h ε is continuous. Furthermore as W is differentiable in ξ on [ A ν + r, ∞ ) with bounded derivative(for r > κ ) we can bound the L ∞ -norm of W δ and ∂ ξ W δ by the correspondingexpression of W . Thus we obtain (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z A + δA ν + r L e t K ε ( L e t ξ, y ) y h ε (cid:0) L e t ξ (cid:1) h ε ( y ) h W δ (cid:16) ξ + yL e t (cid:17) − W δ ( ξ ) i d ξ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ CL e t Z A + δA ν + r ( L e t ξ + ε ) − a ( y + ε ) b + ( L e t ξ + ε ) b ( y + ε ) − a y h ε (cid:0) L e t ξ (cid:1) h ε ( y ) (cid:12)(cid:12)(cid:12) W δ (cid:16) ξ + yL e t (cid:17) − W δ ( ξ ) (cid:12)(cid:12)(cid:12) d ξ ≤ C ( L, A, r, ε, κ ) (cid:16) k W k L ∞ + k ∂ ξ W k L ∞ ([ A ν + r, ∞ )) (cid:17) ( y + ε ) b + ( y + ε ) − a y h ε ( y ) min n yL e t , A − A ν + 1 − r o . Again the right hand side is independent of δ and integrable due to Lemma A.1. Thus we also can pass to thelimit δ → y -integral in ( II ). Before estimating the integrand of ( III ) we first derive an estimate forthe expression R A ν + rA ν − r (cid:12)(cid:12) W δ (cid:0) ξ + yL e t (cid:1) − W δ ( ξ ) (cid:12)(cid:12) d ξ , i.e. for y ∈ [0 , L e t r ] we have Z A ν + rA ν − r (cid:12)(cid:12)(cid:12) W δ (cid:16) ξ + yL e t (cid:17) − W δ ( ξ ) (cid:12)(cid:12)(cid:12) d ξ ≤ Z δ − δ ϕ δ ( η ) Z A ν + rA ν − r (cid:12)(cid:12)(cid:12) W (cid:16) ξ − η + yL e t (cid:17) − W ( ξ − η ) (cid:12)(cid:12)(cid:12) d ξ d η.
16e thus consider Z A ν + rA ν − r (cid:12)(cid:12)(cid:12) W (cid:16) ξ − η + yL e t (cid:17) − W ( ξ − η ) (cid:12)(cid:12)(cid:12) d ξ = Z A ν + ηA ν − r W (cid:16) ξ − η + yL e t (cid:17) d ξ + Z A ν + rA ν + η W ( ξ − η ) − W (cid:16) ξ − η + yL e t (cid:17) d ξ = Z A ν + yL e t A ν − r + yL e t − η W ( ξ ) d ξ + Z A ν + r − ηA ν W ( ξ ) d ξ − Z A ν + r − ηA ν + yL e t W ( ξ ) d ξ − Z A ν + r − η + yL e t A ν + r − η W ( ξ ) d ξ ≤ Z A ν + yL e t A ν W ( ξ ) d ξ − Z A ν + r − η + yL e t A ν + r − η W ( ξ ) d ξ ≤ k W k L ∞ yL e t . This shows R A ν + rA ν − r (cid:12)(cid:12) W δ (cid:0) ξ + yL e t (cid:1) − W δ ( ξ ) (cid:12)(cid:12) d ξ ≤ k W k L ∞ yL e t , i.e. for y ∈ [0 , L e t r ].On the other hand for y > L e t r we have the trivial estimate Z A ν + rA ν − r (cid:12)(cid:12)(cid:12) W δ (cid:16) ξ + yL e t (cid:17) − W δ ( ξ ) (cid:12)(cid:12)(cid:12) d ξ ≤ k W δ k L ∞ r ≤ k W k L ∞ r and thus altogether Z A ν + rA ν − r (cid:12)(cid:12)(cid:12) W δ (cid:16) ξ + yL e t (cid:17) − W δ ( ξ ) (cid:12)(cid:12)(cid:12) d ξ ≤ C min n yL e t , r o . Using this and again that h ε is continuous we can also estimate the integrand in ( III ), i.e. Z A ν + rA ν − r L e t K ε ( L e t ξ, y ) y h ε (cid:0) L e t ξ (cid:1) h ε ( y ) h W δ (cid:16) ξ + yL e t (cid:17) − W δ ( ξ ) i d ξ ≤ C ( L ) Z A ν + rA ν − r ( L e t ξ + ε ) − a ( y + ε ) b + ( L e t ξ + ε ) b ( y + ε ) − a y h ε (cid:0) L e t ξ (cid:1) h ε ( y ) (cid:12)(cid:12)(cid:12) W δ (cid:16) ξ + yL e t (cid:17) − W δ ( ξ ) (cid:12)(cid:12)(cid:12) d ξ ≤ C ( L, A, ε ) sup ξ ∈ [ A ν − r,A ν + r ] (cid:12)(cid:12) h ε (cid:0) L e t ξ (cid:1)(cid:12)(cid:12) ( y + ε ) b + ( y + ε ) − a y h ε ( y ) min n yL e t , r o . According to Lemma A.1 the right hand side is integrable and thus we can also bound the integrand in (
III )independently of δ by some integrable function. Thus due to Lebesgue’s Theorem we can pass to the limit δ → W . W Lemma 3.6.
There exists σ ∈ (max { b, ν } , and θ > such that − W ( A − A σ ) ≤ CtA − θ for sufficiently large A .Proof. From the construction in Section B.1 we know that ˜ W can be written as ˜ W ( ξ, t ) = R ∞ ξ G ( η, t ) d η with17 = − ∂ ξ ˜ W = G , ∗ G , ∗ G , where G , , G , and G solve ∂ t G , = CA − νa L ρ + a − max { ,b } Z ∞ η ω [ G , ( ξ + η ) − G , ( ξ )] d ηG , ( · ,
0) = δ ( · − A + κ ) ∗ ϕ κ/ ∂ t G , = CA β L ρ − b Z ∞ η ω [ G , ( ξ + η ) − G , ( ξ )] d ηG , ( · ,
0) = δ ( · ) ∗ ϕ κ/ ∂ t G = C Z h ε ( z ) z (cid:20) A − νa L a ( z + ε ) b + A β L β ( z + ε ) − a (cid:21) · h G (cid:16) ξ + zL (cid:17) − G ( ξ ) i d zG ( · ,
0) = δ ( · ) ∗ ϕ κ/ . (44)Then one has from Lemma B.10 for any µ ∈ (0 , Z − D −∞ G , ( ξ, t ) d ξ ≤ C (cid:16) κD (cid:17) µ + CA β tL ρ − b D ω and Z − D + A −∞ G , ( ξ, t ) d ξ = Z ( A − κ ) − ( D − κ ) −∞ G , ( ξ, t ) d ξ ≤ C (cid:18) κD − κ (cid:19) µ + CA − νa tL ρ + a − max { ,b } ( D − κ ) ω ≤ C (cid:16) κD (cid:17) µ + CA − νa tL ρ + a − max { ,b } D ω . In the last step we used that for any δ ∈ (0 ,
1) and D ≥ κ ≤ / D − κ ) − δ ≤ δ D − δ . One thusneeds an estimate for G . This will be quite similar to the proof of Lemma B.10 but due to the differentbehaviour for L ε → L ε G ( p, t ) := R R G ( ξ, t ) e p ( ξ − κ/ d ξ andmultiplying the equation for G in (44) by e p ( ξ − κ/ and integrating one obtains ∂ t ˜ G ( p, t ) = C Z h ε ( z ) z h ( z + ε ) − a L b A β + ( z + ε ) b L − a A − νa i · h e − pzL − i d z ˜ G ( p, t )=: M ( p, L ) ˜ G ( p, t ) . Thus ˜ G ( p, t ) = R R ϕ κ/ ( ξ ) e p ( ξ − κ/ d ξ exp ( − t | M ( p, L ) | ) and one can estimate: | M ( p, L ) | ≤ C Z h ε ( z ) z h ( z + ε ) − a L b A β + ( z + ε ) − a L − a A − νa i · pzL d z = Cp (cid:0) L b − A β µ ε + L − a − A − νa λ ε (cid:1) ≤ Cp (cid:0) A β + A − νa (cid:1) . For the last step note that due to our notation either L = L ε (in the case L ε
0) and then the estimate isdue to the definition of L ε . If L = 1 (in the case L ε →
0) one can assume without loss of generality that ε issuch small that L ε ≤ λ ε , µ ε ≤ p := D we obtain inthe same way as in the proof of Lemma B.10: Z − D −∞ G ( ξ, t ) d ξ ≤ C (cid:18)(cid:16) κD (cid:17) µ + tD (cid:0) A β + A − νa (cid:1)(cid:19) . Using these estimates on G , , G , and G one obtains from Lemma B.11 (note also Remark B.12):1 − ˜ W ( A − D ) = Z A − D −∞ ( G , ∗ G , ∗ G ) d ξ ≤ Z A − D −∞ G , d ξ + Z − D −∞ G , d ξ + Z − D −∞ G d ξ ≤ C κ µ D µ + CA − νa tL ρ + a − max { ,b } D ω + CA β tL ρ − b D ω + Ct (cid:0) A β + A − νa (cid:1) D . D = A σ (with A ≥
1) one has1 − ˜ W ( A − A σ ) ≤ Cκ µ A − µσ + CtL a + ω A − νa − σω + CtL ρ − b A β − σω + Ct (cid:0) A β − σ + A − νa − σ (cid:1) . In the case L = 1 (i.e. L ε →
0) it suffices to consider the exponents of A : • − µσ <
0, as µ, σ > • − νa − σω <
0, as ω , a > • β − σω = β − σρ = ( b − σρ b ≥ νb − σρ b < <
0, independently of the sign of b if we choose σ sufficiently closeto 1 and σ > ν (as b < ρ ). • β − σ = ( b − σ b ≥ νb − σ b < <
0, independently of the sign of b if we choose σ > b as ν < b < σ does not collide with the choice made before) • − νa − σ <
0, as a, σ > − θ to be the maximum of the (negative) exponents proves the claim in this case.If L = L ε (i.e. L ε
0) one has to consider also the exponents of L : • a + ω >
0, as a, ω > • ρ − b >
0, as by assumption b < ρ .Thus either the two terms containing L = L ε are bounded (if L ε is bounded) or converge to zero (if L ε → ∞ )and so in both cases with the same θ > In this section we will show Proposition 3.1. We therefore define F ε ( X ) := Z X H ε ( Y ) d Y while for L ε → F ε ( x ) = Z x h ε ( y ) d y. We first show the following Lemma, that will be the key in the proof of Proposition 3.1.
Lemma 3.7.
There exists θ > such that F ε ( A ) ≥ − CA ν (1 − ρ ) + F ε (cid:0) ( A − A σ ) e T (cid:1) e − (1 − ρ ) T (cid:18) − CA θ (cid:19) for A sufficiently large.Proof. From the choice of ψ and W respectively (using also the non-negativity and monotonicity propertiesof W ) F ε ( A ) = Z A H ε ( X ) d X ≥ Z ∞ W ( X, H ε ( X ) d X ≥ e − (1 − ρ ) T Z ∞ H ε ( X ) W (cid:18) X e T , T (cid:19) d X ≥ e − (1 − ρ ) T Z ∞ A ν ∂ X F ε ( X ) W (cid:18) X e T , T (cid:19) d X = − e − (1 − ρ ) T F ε ( A ν ) W (cid:18) A ν e T , T (cid:19) − Z ∞ A ν e − (1 − ρ ) T e − T F ε ( X ) ∂ ξ W (cid:18) X e T , T (cid:19) d X ≥ − CA ν (1 − ρ ) + e − (1 − ρ ) T Z ∞ A ν e − T F ε (cid:0) X e T (cid:1) ( G , ∗ G , ∗ G ) ( X, T ) d X W is bounded, ∂ ξ W = − G , ∗ G , ∗ G on ( A ν , ∞ )as well as F ε ( A ν ) ≤ A ν (1 − ρ ) . Noting that for σ > ν we have A ν e − T ≤ A − A σ for sufficiently large A andusing also the monotonicity of F ε we can further estimate F ε ( A ) ≥ − CA ν (1 − ρ ) + e − (1 − ρ ) T Z ∞ A − A σ F ε (cid:0) X e T (cid:1) ( G , ∗ G , ∗ G ) ( X, T ) d X ≥ − CA ν (1 − ρ ) + e − (1 − ρ ) T F ε (cid:0) ( A − A σ ) e T (cid:1) Z ∞ A − A σ ( G , ∗ G , ∗ G ) ( X, T ) d X = − CA ν (1 − ρ ) + F ε (cid:0) ( A − A σ ) e T (cid:1) e − (1 − ρ ) T W ( A − A σ ) ≥ − CA ν (1 − ρ ) + F ε (cid:0) ( A − A σ ) e T (cid:1) e − (1 − ρ ) T (cid:18) − CA θ (cid:19) , while in the last step Lemma 3.6 was applied.We are now prepared to prove Proposition 3.1. This will be done by some iteration argument usingrecursively Lemma 3.7. Proof of Proposition 3.1.
Let α := e T >
1. For any δ > R ǫ,δ > F ε ( R ) ≥ R − ρ (1 − δ )for all R ≥ R ε,δ . For A > (cid:16) αα − (cid:17) − σ we define a sequence { A k } k ∈ N by A k +1 := α ( A k − A σk ). From thechoice of A one obtains that A k is strictly increasing and one has A k → ∞ as k → ∞ . Furthermore αA k = A k +1 + αA σk and thus A k = A k +1 α (cid:18) α A σk A k +1 (cid:19) . By iteration one obtains for any N ∈ N : A = A N α N N − Y k =0 (cid:18) α A σk A k +1 (cid:19) . (45)For any N ∈ N and 0 ≤ k < N applying Lemma 3.7 one gets by induction: F ε ( A k ) ≥ F ε ( A N ) α − ( N − k )(1 − ρ ) N − Y n = k (cid:18) − CA θn (cid:19) − C N − X m = k α − ( m − k )(1 − ρ ) m − Y n = k (cid:18) − CA θn (cid:19)! A ν (1 − ρ ) m , (46)where we use the convention P uk = l a k = 0 and Q uk = l a k = 1 if u < l . Thus for k = 0 one particularly obtains F ε ( A ) ≥ F ε ( A N ) α − N (1 − ρ ) N − Y n =0 (cid:18) − CA θn (cid:19) − C N − X m =0 α − m (1 − ρ ) m − Y n =0 (cid:18) − CA θn (cid:19)! A ν (1 − ρ ) m = F ε ( A N ) α − N (1 − ρ ) N − Y n =0 (cid:18) − CA θn (cid:19) − C N − X m =0 α ( ν − − ρ ) m m − Y n =0 (cid:18) − CA θn (cid:19)! (cid:0) α − m A m (cid:1) ν (1 − ρ ) =: ( I ) − ( II ) . (47)We now estimate the two terms separately.Let δ ∗ := δ/
2. Choosing N sufficiently large such that A N ≥ R ǫ,δ ∗ one has, using also (45)( I ) ≥ (1 − δ ∗ ) A − ρN α − N (1 − ρ ) N − Y n =0 (cid:18) − CA θn (cid:19) = (1 − δ ∗ ) (cid:18) A N α N (cid:19) − ρ N − Y n =0 (cid:18) − CA θn (cid:19) = (1 − δ ∗ ) A − ρ Q N − n =0 (cid:16) − CA θn (cid:17)(cid:16)Q N − k =0 (cid:16) α A σk A k +1 (cid:17)(cid:17) − ρ . (48)20et 0 < D < D be parameters to be fixed later and assume A > D . One has A k +1 = α ( A k − A σk ). Thususing the monotonicity of A k A k +1 A k = α (cid:0) − A σ − k (cid:1) > α (cid:0) − D σ − (cid:1) =: β > D sufficiently large as α >
1. Using this, one has A k +1 > β A k and thus by iteration A k +1 > β k +10 A .We continue to estimate ( I ) and thus consider first Q N − n =0 (cid:16) − CA θn (cid:17) while we assume that D is sufficientlylarge such that CD θ < CA θn < A n . Taking the logarithm of the productone has using the estimate log (1 − x ) ≥ − x − x : N − X n =0 log (cid:18) − CA θn (cid:19) ≥ N − X n =0 − CA θn · − CA θn = − C N − X n =0 A θn − C ≥ − C N − X n =0 β nθ A θ − C ≥ − C N − X n =0 β θn D θ − C ≥ − C β θ β θ − D θ − C =: − C β D θ − C .
Thus one obtains using exp ( − x ) ≥ − x : N − Y n =0 (cid:18) − CA θn (cid:19) ≥ exp (cid:18) − C β D θ − C (cid:19) ≥ − C β D θ − C . (49)Considering Q N − k =0 (cid:16) α A σk A k +1 (cid:17) and applying again first the logarithm on the product and then using log (1 + x ) ≤ x one obtains N − X k =0 log (cid:18) α A σk A k +1 (cid:19) ≤ N − X k =0 α A σk A k +1 ≤ α N − X k =0 A σ − k ≤ αA σ − N − X k =0 (cid:0) β σ − (cid:1) k ≤ αD σ − ∞ X k =0 (cid:0) β σ − (cid:1) k =: C γ D σ − . Thus one can estimate N − Y k =0 (cid:18) α A σk A k +1 (cid:19)! − ρ ≤ exp (cid:2)(cid:0) C γ D σ − (cid:1) (1 − ρ ) (cid:3) = 1 + ∞ X n =1 (1 − ρ ) n C nγ D n ( σ − n ! ≤ D σ − ∞ X n =1 C nγ (1 − ρ ) n n ! ≤ C γ (1 − ρ )) D − σ . (50)Putting the estimates of (49) and (50) together one obtains Q N − n =0 (cid:16) − CA θn (cid:17)(cid:16)Q N − k =0 (cid:16) α A σk A k +1 (cid:17)(cid:17) − ρ ≥ − C β D θ − C exp( C γ (1 − ρ )) D − σ = 1 − C β D θ − C + exp( C γ (1 − ρ )) D − σ exp( C γ (1 − ρ )) D − σ ≥ − C β D θ − C − exp ( C γ (1 − ρ )) D − σ . (51)Together with (48) this shows( I ) ≥ (1 − δ ∗ ) A − ρ (cid:18) − C β D θ − C − exp ( C γ (1 − ρ )) D − σ (cid:19) . (52)21o estimate ( II ) we note first that one has Q m − n =0 (cid:16) − CA θn (cid:17) ≤ Q m − k =0 (cid:16) α A σk A k +1 (cid:17) ≥ m ∈ N . Using this as well as (45) for N = m and m = 0 , . . . , N − II ) = C N − X m =0 α ( ν − − ρ ) m m − Y n =0 (cid:18) − CA θn (cid:19)! (cid:0) α − m A m (cid:1) ν (1 − ρ ) ≤ C N − X m =0 α ( ν − − ρ ) m (cid:0) α − m A m (cid:1) ν (1 − ρ ) = C N − X m =0 α ( ν − − ρ ) m A ν (1 − ρ )0 (cid:16)Q m − k =0 (cid:16) α A σk A k +1 (cid:17)(cid:17) ν (1 − ρ ) ≤ C N − X m =0 α ( ν − − ρ ) m A ν (1 − ρ )0 ≤ A ν (1 − ρ )0 C ∞ X m =0 α ( ν − − ρ ) m =: C ν A ν (1 − ρ )0 (53)Putting together the estimates (52), (53) and (47) yields F ε ( A ) ≥ (1 − δ ∗ ) (cid:18) − C β D θ − C − exp ( C γ (1 − ρ )) D − σ (cid:19) A − ρ − C ν A ν (1 − ρ )0 ≥ A − ρ (cid:18) (1 − δ ∗ ) (cid:18) − C β D θ − C − exp ( C γ (1 − ρ )) D − σ (cid:19) − C ν D (1 − ν )(1 − ρ ) (cid:19) . (54)We choose now D sufficiently large such that one has • D ≥ (cid:0) C β − δδ + C (cid:1) /θ which is equivalent to C β D θ − C ≤ δ − δ • D ≥ (cid:0) C γ (1 − ρ )) − δδ (cid:1) − σ which is equivalent to exp( C γ (1 − ρ )) D − σ ≤ δ − δ • D ≥ (cid:16) C ν − δ/ − δδ (cid:17) − ν )(1 − ρ ) which is equivalent to C ν D (1 − ν )(1 − ρ ) ≤ (1 − δ/ δ − δ .Inserting these estimates in (54) together with δ ∗ = δ/ F ε ( A ) ≥ A − ρ (cid:18)(cid:18) − δ (cid:19) (cid:18) − δ − δ (cid:19) − (cid:18) − δ (cid:19) δ − δ (cid:19) = A − ρ (cid:18) − δ (cid:19) (cid:18) − δ − δ (cid:19) = (1 − δ ) A − ρ . This proves the claim with R δ = D . L ε → ∞ Before we can pass to the limit ε → L ε → ∞ as ε → h ε (instead of H ε ). This will be done by some contradiction argument. We furthermoreremark here that throughout this section we frequently use that we can bound Z ( x + ε ) − a h ε ( x ) d x ≤ L − bε and Z ( x + ε ) b h ε ( x ) d x ≤ L aε (55)due to the definition of L ε in (35). H ε We first show the following lemma, stating the convergence of a certain integral occurring later.22 emma 3.8.
Assume L ε → ∞ as ε → . Let Q ε be given by Q ε ( X ) := Z h ε ( y ) L ε K ε ( y, L ε X ) d y. Then there exists a (continuous) function Q such that Q ε → Q locally uniformly up to a subsequence.Proof. It suffices to show that both Q ε as well as Q ′ ε are uniformly bounded on each fixed interval [ d, D ] ⊂ R + .One has using (55) | Q ε ( X ) | ≤ C Z h ε ( z ) L ε (cid:16) ( L ε X + ε ) − a ( z + ε ) b + ( L ε X + ε ) b ( z + ε ) − a (cid:17) d z ≤ C L ε ( L ε X + ε ) − a Z ( z + ε ) b h ε ( z ) d z + C L ε ( L ε X + ε ) b Z ( z + ε ) − a h ε ( z ) d z ≤ C L aε ( L ε X + ε ) − a + C L − bε ( L ε X + ε ) b = C (cid:18) X + εL ε (cid:19) − a + C (cid:18) X + εL ε (cid:19) b (56)while the right hand side is clearly locally uniformly bounded under the given assumptions. Rewriting Q ε oneobtains Q ε ( X ) = Z h ε ( z ) L ε K ε ( L ε X, z ) d z = L γε Z h ε ( z ) L ε K εLε (cid:18) X, zL ε (cid:19) d z. Furthermore from (11) one has | ∂ y K ε ( y, z ) | ≤ C (cid:16) ( z + ε ) − a + ( z + ε ) b (cid:17) for all y ∈ [ a, A ]and hence, similarly as before | Q ′ ε ( X ) | ≤ CL γε Z h ε ( z ) L ε "(cid:18) z + εL ε (cid:19) − a + (cid:18) z + εL ε (cid:19) b d z = CL γ − aε Z ( z + ε ) − a h ε ( z ) d z + CL γ − − bε Z ( z + ε ) b h ε ( z ) d z ≤ C where we used also γ = b − a . Lemma 3.9.
Let ρ ∈ (max { , b } , and assume L ε → ∞ as ε → . Then there exists a measure H suchthat (up to a subsequence) H ε ∗ ⇀ H and H satisfies ∂ X ( XH ) + ( ρ − H − ∂ X ( Q ( X ) H ( X )) + H ( X ) Q ( X ) X = 0 (57) in the sense of distributions with Q ( X ) = lim ε → R h ε ( y ) L ε K ε ( y, L ε X ) d y .Proof. Transforming the equation ∂ x (cid:18)Z x Z ∞ x − y K ε ( y, z ) z h ε ( y ) h ε ( z ) d z d y (cid:19) = ∂ x ( xh ε ( x )) + ( ρ − h ε ( x )to the rescaled variables X = xL ε one obtains1 L ε ∂ X Z L ε X Z ∞ L ε X − y K ε ( y, z ) z h ε ( y ) h ε ( z ) d z d y ! = 1 L ε ∂ X (cid:18) L ε X H ( X ) L ρε (cid:19) + ( ρ − H ε ( X ) L ρε . ψ ∈ C ∞ c ( R + ) (in the rescaled X -variable), splitting the integral and interchanging the order ofintegration we can rewrite this as Z ∞ ( X∂ X ψ ( X ) − ( ρ − ψ ( X )) H ε ( X ) d X = 1 L ρ − γε Z ∞ Lε Z ∞ Lε K εLε ( Y, Z ) Z H ε ( Y ) H ε ( Z ) [ ψ ( Y + Z ) − ψ ( Y )] d Z d Y + Z ∞ Lε Z K ε ( L ε Y, z ) z h ε ( z ) H ε ( Y ) (cid:20) ψ (cid:18) Y + zL ε (cid:19) − ψ ( Y ) (cid:21) d z d Y + Z Z ∞ Z yLε + Z yLε ∂ X ψ ( X ) d X ! K ε ( y, L ε Z ) L ε Z h ε ( y ) H ε ( Z ) d Z d y = ( I ) + ( II ) + ( III ) . In the following we assume that supp ψ ⊂ [ d, D ] with d > D >
1. Furthermore we can assume that L ε > ε < L ε → ∞ for ε → I ) → ε → | ( I ) |≤ C L ρ − γε Z D Lε Z ∞ Lε (cid:16) Y + εL ε (cid:17) − a (cid:16) Z + εL ε (cid:17) b Z H ε ( Y ) H ε ( Z ) | ψ ( Y + Z ) − ψ ( Y ) | d Z d Y + C L ρ − γε Z D Lε Z ∞ Lε (cid:16) Y + εL ε (cid:17) b (cid:16) Z + εL ε (cid:17) − a Z H ε ( Y ) H ε ( Z ) | ψ ( Y + Z ) − ψ ( Y ) | d Z d Y ≤ ˜ b CL ρ − γε Z D Lε Z ∞ Lε Y − a Z b + Y b Z − a Z H ε ( Y ) H ε ( Z ) | ψ ( Y + Z ) − ψ ( Y ) | d Z d Y ≤ C k ψ ′ k L ∞ L ρ − γ − aε Z D Lε Z Lε (cid:0) Z b + Y b (cid:1) H ε ( Y ) H ε ( Z ) d Z d Y + C k ψ k L ∞ L ρ − γε " L aε Z D Lε H ε ( Y ) d Y Z ∞ Z b − H ε ( Z ) d Z + max (cid:8) L − bε , D b (cid:9) Z D Lε H ε ( Y ) d Y Z ∞ Z − a − H ε ( Z ) d Z ≤ CL γ + a − ρε max (cid:8) , L − bε , D b (cid:9) Z D Lε H ε ( Y ) d Y Z Lε H ε ( Z ) d Z + CL γ − ρε (cid:2) L aε D − ρ + max (cid:8) L − bε , D b (cid:9) D − ρ (cid:3) = CD − ρ L b − ρε max (cid:8) D b , L − bε (cid:9) + CD − ρ L b − ρε + CD − ρ L b − a − ρε max (cid:8) L − bε , D b (cid:9) → L ε → ∞ (i.e. for ε → II ) → R ∞ ∂ X ψ ( X ) H ( X ) Q ( X ) d X . As H ε is a sequence of locally uniformly bounded(non-negative Radon) measures there exists a (non-negative Radon) measure H such that H ε ∗ ⇀ H in thesense of measures. Using now Taylor’s formula for ψ one obtains ψ (cid:18) Y + zL ε (cid:19) − ψ ( Y ) = ψ ′ ( Y ) · zL ε + z L ε Z z ( z − t ) ψ ′′ (cid:18) Y + tL ε (cid:19) d t. II ) one gets( II ) = Z ∞ Lε Z K ε ( L ε Y, z ) z h ε ( z ) H ε ( Y ) · zL ε ψ ′ ( Y ) d z d Y + Z ∞ Lε Z K ε ( L ε Y, z ) z h ε ( z ) H ε ( Y ) · z L ε Z z ( z − t ) ψ ′′ (cid:18) Y + tL ε (cid:19) d t d z d Y = ( II ) a + ( II ) b We consider terms separately beginning with ( II ) b (and assuming L ε to be sufficiently large, i.e. L ε < d ): | ( II ) b | ≤ L ε Z ∞ Lε Z z h ε ( z ) K ε ( L ε Y, z ) H ε ( Y ) Z z (cid:12)(cid:12)(cid:12)(cid:12) ψ ′′ (cid:18) Y + tL ε (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) d t d z d Y ≤ C k ψ ′′ k ∞ L ε Z Dd − Lε Z h ε ( z ) H ε ( Y ) h ( L ε Y + ε ) − a ( z + ε ) b + ( L ε Y + ε ) b ( z + ε ) − a i d z d Y ≤ CD − ρ L ε h ( L ε d − ε ) − a L aε + max n ( L ε D + ε ) b , ( L ε d − ε ) b o L − bε i = CD − ρ (cid:16) d − L ε + εL ε (cid:17) − a L ε + max (cid:26)(cid:16) D + εL ε (cid:17) b , (cid:16) d − L ε + εL ε (cid:17) b (cid:27) L ε → ε → L ε → ∞ ). On the other hand (using the symmetry of K ε )( II ) a = Z ∞ Lε H ε ( Y ) ψ ′ ( Y ) Z h ε ( z ) L ε K ε ( L ε Y, z ) d z d Y = Z ∞ Lε H ε ( Y ) ψ ′ ( Y ) Q ε ( Y ) d Y. Thus one obtains ( II ) a → R ∞ H ( Y ) Q ( Y ) ψ ′ ( Y ) d Y directly from Lemma 3.8. It remains to show that( III ) → R ∞ H ( Y ) Q ( Y ) Y ψ ( Y ) d Y . We first rewrite ( III ) as(
III ) = Z ∞ Z Z yLε + Z yLε K ε ( y, L ε Z ) L ε Z h ε ( y ) H ε ( Z ) ∂ X ψ ( X ) d X d y d Z = Z ∞ Z K ε ( y, L ε Z ) L ε Z h ε ( y ) H ε ( Z ) (cid:20) ψ ( Z ) + ψ (cid:18) Z + yL ε (cid:19) − ψ ( Z ) − ψ (cid:18) yL ε (cid:19)(cid:21) d y d Z. Due to supp ψ ⊂ [ d, D ] we obtain for L ε sufficiently large (i.e. L ε ≥ d ) that ψ (cid:16) yL ε (cid:17) = 0 for all y ∈ [0 , Q ε we can rewrite ( III ) as(
III ) = Z ∞ ψ ( Z ) H ε ( Z ) Q ε ( Z ) Z d Z + Z ∞ Z K ε ( y, L ε Z ) L ε Z h ε ( y ) H ε ( Z ) (cid:20) ψ (cid:18) Z + yL ε (cid:19) − ψ ( Z ) (cid:21) d y d Z =: ( III ) a + ( III ) b . The integral (
III ) a converges (up to a subsequence) to R ∞ ψ ( Z ) H ( Z ) Q ( Z ) Z d Z according to Lemma 3.8. Itthus remains to show that ( III ) b converges to zero. To see this note that as ψ is smooth and compactlysupported we have for y ∈ [0 , (cid:12)(cid:12)(cid:12)(cid:12) ψ (cid:18) Z + yL ε (cid:19) − ψ ( Z ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ C ( ψ ) yL ε χ [ d − Lε , ∞ ) ( Z ) ≤ C ( ψ ) L ε χ [ d − Lε , ∞ ) ( Z ) . III ) b ≤ C ( ψ ) L ε Z ∞ d − /L ε H ε ( Z ) Q ε ( Z ) Z d Z. From the estimates on Q ε in (56) we obtain that the integral on the right hand side is bounded uniformly in ε and thus for L ε → ∞ the right hand side converges to zero, concluding the proof. For ρ ∈ (0 , there exists no solution H to (57) satisfying the lower bound (36) and R R H ( X ) d X ≤ R − ρ for each R ≥ .Proof. Assuming such a solution exists and rewriting (57) one has ∂ X (( X − Q ( X )) H ( X )) = (cid:18) (1 − ρ ) − Q ( X ) X (cid:19) H ( X ) . Defining F ( X ) := ( X − Q ( X )) H ( X ) this is equivalent to ∂ X F ( X ) = (1 − ρ ) − Q ( X ) X X − Q ( X ) F ( X ) and thus F ( X ) = C · exp Z XA (1 − ρ ) − Q ( Y ) Y Y − Q ( Y ) d Y ! for some constant C . Considering Q one can assume (up to passing to a subsequence of ε again denoted by ε )that either λ ε ≥ µ ε or µ ε ≥ λ ε for all ε , while both cases will be denoted as λ ≥ µ or µ ≥ λ either. One canthen estimate (using the definition of λ ε and µ ε ):0 ≤ Q ( X ) ≤ lim ε → C Z h ε ( y ) L ε (cid:16) ( y + ε ) − a ( L ε X + ε ) b + ( y + ε ) b ( L ε X + ε ) − a (cid:17) d y ≤ lim ε → C L ε (cid:16) L − bε ( L ε X + ε ) b + L a +1 ε ( L ε X + ε ) − a (cid:17) = lim ε → C (cid:18) X + εL ε (cid:19) b + (cid:18) X + εL ε (cid:19) − a ! = C (cid:0) X b + X − a (cid:1) . On the other hand Q ( X ) ≥ lim ε → C Z h ε ( y ) L ε (cid:16) ( y + ε ) − a ( L ε X + ε ) b + ( y + ε ) b ( L ε X + ε ) − a (cid:17) d y ≥ lim ε → C L ε ( L − bε ( L ε X + ε ) b µ ≥ λL aε ( L ε X + ε ) − a λ ≥ µ = C ( X b µ ≥ λX − a λ ≥ µ. This shows in particular that for sufficiently large A one has Q ( X ) < X for all X ≥ A and F is well definedfor X ≥ A . We claim now C >
0. To see this assume C ≤
0. Then F ≤ X − Q ( X ) ≥ X ≥ A . As H ( X ) = F ( X ) X − Q ( X ) one has (using R R H d X ≥ R − ρ for sufficiently large R due to Proposition 3.1and R A H d X ≤ A − ρ ):0 ≥ Z RA H ( X ) d X = Z R H ( X ) d X − Z A H ( X ) d X ≥ R − ρ − A − ρ = R − ρ − (cid:18) AR (cid:19) − ρ ! > R and thus a contradiction. Therefore we have C > X < A such that Q ( X ) = X and Q ( X ) < X for all X > X which is possibledue to the lower and upper bound for Q . We get that X − Q ( X ) X − X is bounded on [ X , ∞ ) and thus we have X − Q ( X ) ≤ K ( X − X ) for some K >
0. Furthermore as Q ( Y ) ∼ Y for Y → X we obtain − (1 − ρ ) + Q ( Y ) Y = ρ − Q ( Y ) Y ≥ ρ − δ > (cid:2) X , X (cid:3) for some X > X close to X and δ >
0. We then get Z AX (1 − ρ ) − Q ( Y ) Y Y − Q ( Y ) d Y := C (cid:0) X, A (cid:1) < ∞ . Using the definition of F we can then write H as H = CX − Q ( X ) exp − Z AX (1 − ρ ) − Q ( Y ) Y Y − Q ( Y ) d Y ! ≥ CK ( X − X ) exp − Z XX (1 − ρ ) − Q ( Y ) Y Y − Q ( Y ) d Y − Z AX (1 − ρ ) − Q ( Y ) Y Y − Q ( Y ) d Y ! ≥ CK ( X − X ) exp (cid:0) − C (cid:0) X, A (cid:1)(cid:1) exp ( ρ − δ ) Z XX K ( Y − X ) d Y ! = CK exp (cid:0) − C (cid:0) X, A (cid:1)(cid:1) (cid:0) X − X (cid:1) ρ − δK X − X ) ρ − δK = C (cid:0) A, X, X , K (cid:1) X − X ) α with α = ρ − δK >
0, contradicting the local integrability of H .This shows that L ε has to be bounded and thus by scale invariance we obtain from Proposition 3.1 alsothe lower bound for h ε , i.e. we have Proposition 3.11.
For any δ > there exists R δ > such that Z R h ε ( x ) dx ≥ (1 − δ ) R − ρ for all R ≥ R δ . (58) We will show in this section that h ε decays exponentially near zero in an averaged sense, a property that willbe crucial when passing to the limit ε → Lemma 3.12.
There exist constants C and c independent of ε such that Z D h ε ( y ) d y ≤ CD − ρ exp (cid:16) − c ( D + ε ) − a (cid:17) for any D ∈ (0 , and all ε > .Proof. Let δ = , then due to Proposition 3.11 there exists R ∗ > R B R ∗ h ε ( z ) d z ≥ ( B R ∗ ) − ρ forany B ≥
1. On the other hand one has R B R ∗ h ε ( z ) d z ≤ ( B R ∗ ) − ρ for any B ≥
0. Thus one has Z B R ∗ B R ∗ h ε ( z ) d z = Z B R ∗ h ε ( z ) d z − Z B R ∗ h ε ( z ) d z ≥ ( B R ∗ ) − ρ − ( B R ∗ ) − ρ ≥ B (depending on B ). Thus one can estimate Z R Z ∞ R − y K ε ( y, z ) z h ε ( y ) h ε ( z ) d z d y ≥ C Z R Z B R ∗ B R ∗ ( y + ε ) − a ( z + ε ) b + ( y + ε ) b ( z + ε ) − a z h ε ( y ) h ε ( z ) d z d y ≥ C ( R + ε ) a Z R h ε ( y ) d y Z B R ∗ B R ∗ ( z + ε ) b z h ε ( z ) d z ≥ C ( R + ε ) a Z R h ε ( y ) d y ( B R ∗ ) b − Z B R ∗ B R ∗ h ε ( z ) d z ≥ C ( R + ε ) a ( B R ∗ ) b − Z R h ε ( y ) d y. Using this and taking χ ( −∞ ,R ] (restricted to [0 , ∞ )) by some approximation argument as test function in theequation (1 − ρ ) h ε ( x ) − ∂ x ( xh ε ( x )) + ∂ x I ε [ h ε ] ( x ) = 0 we obtain0 = (1 − ρ ) Z R h ε ( x ) d x − Rh ε ( R ) + Z R Z ∞ R − y K ε ( y, z ) z h ε ( y ) h ε ( z ) d z d y ≥ (1 − ρ ) Z R h ε ( x ) d x − Rh ε ( R ) + C ( R + ε ) a ( B R ∗ ) b − Z R h ε ( x ) d x. Thus one has (1 − ρ ) Z R h ε ( x ) d x + C ( B R ∗ ) b − ( R + ε ) a Z R h ε ( x ) d x ≤ Rh ε ( R ) = R∂ R Z R h ε ( x ) d x or equivalently ∂ R Z R h ε ( x ) d x ! ≥ − ρR + C ( B R ∗ ) b − R ( R + ε ) a ! Z R h ε ( x ) d x ≥ − ρR + C ( B R ∗ ) b − ( R + ε ) a ! Z R h ε ( x ) d x, where we used R ( R + ε ) a ≥ R + ε ) a for R ∈ [ D, D,
1] and using R h ε d x ≤ ε ) − a ≤ Z D h ε ( x ) d x ≤ exp C ( B R ∗ ) b − a ! D − ρ exp − C ( B R ∗ ) b − a ( D + ε ) − a ! . Lemma 3.13.
For D ≤ and any α ∈ R one has the following estimate Z D ( x + ε ) α h ε ( x ) d x ≤ CD − ρ ( D + ε ) α exp (cid:16) − c ( D + ε ) − a (cid:17) if α ≥ Z D ( x + ε ) α h ε ( x ) d x ≤ ˜ CD − ρ exp (cid:16) − c D + ε ) − a (cid:17) if α < Proof.
The case α ≥ α < { h ε } ε> is a locally bounded sequence of non-negative Radon measures one can extract a subsequence(again denoted by ε ) such that h ε ∗ ⇀ h in the sense of measures. As a direct consequence of Lemma 3.13 oneobtains: 28 emma 3.14. For D ≤ and α ∈ R one has Z D x α h ( x ) d x ≤ ˜ CD − ρ exp (cid:16) − c D − a (cid:17) if α < Z D x α h ( x ) d x ≤ CD α − ρ exp (cid:0) − cD − a (cid:1) if α ≥ . Proof.
This follows from Lemma 3.13.As a consequence of Lemma 3.14 together with Lemma A.1 we obtain
Corollary 3.15.
For any α ∈ R and D > each limit h satisfies1. R ∞ x α h ( x ) d x < ∞ if α < ρ − ,2. R D x α h ( x ) d x < C ( D ) .Remark . We obtain corresponding results for h ε and h with x α replaced by ( x + ε ) α . ε → In this section we will finally conclude the proof of Theorem 1.1 by passing to the limit ε → I [ h ] is locally integrable: Lemma 3.17.
For h as given above one has I [ h ] ∈ L ([0 , ∞ )) .Proof. Let
D >
0. Then one has Z D I [ h ] ( x ) d x = Z D Z x Z ∞ x − y K ( y, z ) z h ( y ) h ( z ) d z d y d x ≤ C Z D Z D Z ∞ (cid:0) y − a z b − + y b z − a − (cid:1) h ( y ) h ( z ) d z d y d x ≤ C Z D Z D (cid:0) y − a + y b (cid:1) h ( y ) d y d x ≤ C ( D )where Corollary 3.15 was used. One similarly gets R N I [ h ] d x = 0 for bounded null sets N ⊂ [0 , ∞ ).To show that h is a (weak) self-similar solution it only remains to pass to the limit in the weak form ofthe equation ∂ x I ε [ h ε ] = ∂ x ( xh ε ) + ( ρ − h ε . Thus let ϕ ∈ C ∞ c ([0 , ∞ )). Then the weak form reads as Z ∞ ∂ x ϕ ( x ) Z x Z ∞ x − y K ε ( y, z ) z h ε ( y ) h ε ( z ) d z d y d x = Z ∞ ∂ x ϕ ( x ) xh ε ( x ) d x + (1 − ρ ) Z ∞ ϕ ( x ) h ε ( x ) d x. One can easily pass to the limit in the right hand side. To prove Theorem 1.1 it thus remains to show that onecan also take the limit in the left hand side of this equation. This will be done in the following Proposition.
Proposition 3.18.
For any ϕ ∈ C ∞ c ([0 , ∞ )) one has Z ∞ ∂ x ϕ ( x ) Z x Z ∞ x − y K ε ( y, z ) z h ε ( z ) h ε ( y ) d z d y d x −→ Z ∞ ∂ x ϕ ( x ) Z x Z ∞ x − y K ( y, z ) z h ( z ) h ( y ) d z d y d x as ε → . roof. Taking the difference of the two integrals and rewriting one obtains (cid:12)(cid:12)(cid:12)(cid:12)Z ∞ ∂ x ϕ ( x ) (cid:18)Z x Z ∞ x − y K ( y, z ) z h ( y ) h ( z ) − K ε ( y, z ) z h ε ( y ) h ε ( z ) d y d z (cid:19) d x (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12)Z ∞ ∂ x ϕ ( x ) (cid:18)Z x Z ∞ x − y K ( y, z ) − K ε ( y, z ) z h ( y ) h ( z ) d z d y (cid:19) d x (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)Z ∞ ∂ x ϕ ( x ) (cid:18)Z x Z ∞ x − y K ε ( y, z ) z h ( y ) ( h ( z ) − h ε ( z )) d z d y (cid:19) d x (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)Z ∞ ∂ x ϕ ( x ) (cid:18)Z x Z ∞ x − y K ε ( y, z ) z h ε ( z ) ( h ( y ) − h ε ( y )) d z d y (cid:19) d x (cid:12)(cid:12)(cid:12)(cid:12) =: ( I ) + ( II ) + ( III )We estimate the three terms separately and take
D > ϕ ⊂ [0 , D ]. Then due to Lebesgue’sTheorem (using also Corollary 3.15 and Lemma A.1) we obtain( I ) ≤ Z ∞ | ∂ x ϕ ( x ) | (cid:18)Z x Z ∞ x − y | K ( y, z ) − K ε ( y, z ) | z h ( y ) h ( z ) d z d y (cid:19) d x → ε → . To estimate the other two terms we will need some cutoff functions. Let
M, N ∈ N and ζ N , ζ N , ξ M , ξ M ∈ C ∞ ([0 , ∞ )) such that ζ N = 0 on (cid:20) , N (cid:21) ∪ [ N + 1 , ∞ ) , ζ N = 1 on (cid:20) N , N (cid:21) , ≤ ζ N ≤ , ζ N := 1 − ζ N ,ξ M = 0 on (cid:20) , M (cid:21) , ξ M = 1 on (cid:20) M , ∞ (cid:19) , ≤ ξ M ≤ , ξ M := 1 − ξ M . Defining K i,Nε ( y, z ) := K ε ( y, z ) · ζ Ni ( z ) for i = 1 , II ) ≤ (cid:12)(cid:12)(cid:12)(cid:12)Z ∞ Z ∞ K ,Nε ( y, z ) z h ( y ) ( h ( z ) − h ε ( z )) Z y + zy ∂ x ϕ ( x ) d x d y d z (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)Z ∞ Z ∞ K ,Nε ( y, z ) z h ( y ) ( h ( z ) − h ε ( z )) Z y + zy ∂ x ϕ ( x ) d x d y d z (cid:12)(cid:12)(cid:12)(cid:12) =: ( II ) a + ( II ) b We consider again terms separately and without loss of generality we assume ε <
1. Then using Corollary 3.15and Lemma A.1 we obtain( II ) b ≤ C k ∂ x ϕ k ∞ Z N Z D h ( y + ε ) − a ( z + ε ) b + ( y + ε ) b ( z + ε ) − a i h ( y ) ( h ( z ) + h ε ( z )) d y d z + C k ϕ k ∞ Z ∞ N Z D ( y + ε ) − a ( z + ε ) b + ( y + ε ) b ( z + ε ) − a z h ( y ) ( h ( z ) + h ε ( z )) d y d z ≤ k ∂ x ϕ k ∞ C ( D ) Z N (cid:16) ( z + ε ) b + ( z + ε ) − a (cid:17) ( h ( z ) + h ε ( z )) d z + C ( D ) k ϕ k ∞ Z ∞ N (cid:16) ˜ b z b − + z − a − (cid:17) ( h ( z ) + h ε ( z )) d z ≤ k ∂ x ϕ k ∞ C ( D ) " N − ρ (cid:18) N + ε (cid:19) ˜ b + 1 N − ρ + C ( D ) k ϕ k ∞ (cid:2) N b − ρ + N − a − ρ (cid:3) −→ , (59)for N → ∞ . Furthermore one has ( II ) a = (cid:12)(cid:12)(cid:12)(cid:12)Z ∞ ( h ( z ) − h ε ( z )) ψ Nε ( z ) d z (cid:12)(cid:12)(cid:12)(cid:12) (60)30ith ψ Nε ( z ) := R D K ,Nε ( y,z ) z h ( y ) [ ϕ ( y + z ) − ϕ ( y )] d y . We claim that ψ Nε → ψ N strongly in C ([0 , ∞ ))with ψ N ( z ) := R D K ( y,z ) z h ( y ) ζ N ( z ) [ ϕ ( y + z ) − ϕ ( y )] d y . Note that by construction we have supp ψ Nε ⊂ (cid:2) N , N + 1 (cid:3) for all ε >
0. To show (uniform) convergence we have to use a cutoff also in y , i.e. one canestimate (cid:12)(cid:12) ψ N ( z ) − ψ Nε ( z ) (cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z D K ( y, z ) − K ε ( y, z ) z ζ N ( z ) ξ M ( y ) h ( y ) [ ϕ ( y + z ) − ϕ ( y )] d y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z D K ( y, z ) − K ε ( y, z ) z ζ N ( z ) ξ M ( y ) h ( y ) [ ϕ ( y + z ) − ϕ ( y )] d y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) =: ( II ) a, + ( II ) a, . Using similar arguments as in (59) we get( II ) a, ≤ C ( N, ϕ ) (cid:20) M b − ρ + 1 M − ρ (cid:21) −→ M → ∞ and N fixed. As K is continuous on (cid:2) M , D (cid:3) × (cid:2) N , N + 1 (cid:3) for M, N ∈ N fixed, one has K ε → K uniformly on (cid:2) M , D (cid:3) × (cid:2) N , N + 1 (cid:3) for ε →
0. Thus we get ( II ) a, → ε → M, N fixed). Togetherwith (61) this shows that ψ Nε → ψ N strongly. Thus one can pass to the limit in (60) to obtain together with(59): ( II ) → ε → III ) → ε → Acknowledgements
The authors acknowledge support through the CRC 1060
The mathematics of emergent effects at the Universityof Bonn that is funded through the German Science Foundation (DFG).
A Moment estimates
Lemma A.1.
Let h ∈ X ρ and α ∈ R . Then one has the following estimates1. R D x α h ( x ) d x ≤ C k h k D − ρ + α for all D > if ρ − < α ,2. R ∞ D x α h ( x ) d x ≤ C k h k D − ρ + α for all D > if α < ρ − ,where k h k is defined in (14) .Proof.
1. The case α ≥ X ρ . For α ∈ ( ρ − ,
0) one has, using a dyadic decomposition,that Z D x α h ( x ) d x = ∞ X n =0 Z − n D − ( n +1) D x α h ( x ) d x ≤ ∞ X n =0 − α ( n +1) D α Z − n D − ( n +1) D h ( x ) d x ≤ k h k ∞ X n =0 − α ( n +1) D α (cid:0) − n D (cid:1) − ρ = 2 − α k h k D α − ρ ∞ X n =0 (cid:0) α − ρ (cid:1) − n = C ( α, ρ ) k h k D α − ρ .
2. This follows similarly using again a dyadic decomposition.31
Dual problems
B.1 Existence results
In this section we show the existence of solutions to some dual problems arising in the proof of the lowerbounds. Throughout this section we will use the following notation: M fin will denote the space of finitemeasures, M fin+ is the space on non-negative finite measures. Furthermore C nb denotes the space of bounded n -times differentiable functions with bounded derivatives. Let ω ∈ (0 , A ∈ R and consider the equation ∂ t f ( x, t ) − P Z ∞ y ω [ f ( x + y ) − f ( x )] d y = 0 (62)together with initial value f ( x,
0) = δ ( · − A ). Proposition B.1.
There exists a (weak) solution f ∈ C (cid:0) [0 , T ] , M fin+ (cid:1) of (62) with initial value f = δ ( · − A ) .Furthermore this f satisfies supp f ( · , t ) ⊂ ( −∞ , A ] and R R f ( · , t ) d x = 1 for all t ∈ [0 , T ] .Proof (Sketch). First we consider the regularized equation ∂ t f ( x, t ) = P Z ∞ y ω + ν [ f ( x + y, t ) − f ( x, t )] d yf ( · ,
0) = δ ( · − A ) (63)with ν >
0. In the second step we will pass to the limit ν →
0. We can reformulate (63) as the followingfixed-point problem: f ν ( x, t ) = δ ( x − A ) e − P R ∞ y ω + ν d y + Z t e − ( t − s ) R ∞ y ω + ν d y Z ∞ y ω + ν f ( x + y ) d y d s. (64)It is straightforward, applying the contraction mapping theorem, to obtain a solution f ∈ C (cid:0) [0 , T ] , M fin+ (cid:1) forany T >
0. Furthermore, one obtains R R f ν ( x, t ) d x = 1 for all t > ν > f ν satisfies equation (63) in weak form, i.e. Z R f ν ( x, t ) ψ ( x ) d x = ψ ( A ) + Z t Z R Z ∞ y ω + ν f ν ( x, s ) [ ψ ( x − y ) − ψ ( x )] d y d x d s (65)for all ψ ∈ C b ( R ) and for 0 < ˜ ω < ω taking ψ ( x ) = | x | ˜ ω and using (cid:12)(cid:12)(cid:12) | x − y | ˜ ω − | x | ˜ ω (cid:12)(cid:12)(cid:12) ≤ | y | ˜ ω we obtain (byapproximation) Z R f ν ( x, t ) | x | ˜ ω d x ≤ | A | ˜ ω + Z t Z R Z ∞ (cid:12)(cid:12)(cid:12) | x − y | ˜ ω − | x | ˜ ω (cid:12)(cid:12)(cid:12) y ω + ν f ν ( x, s ) d y d x d s ≤ | A | ˜ ω + Z t Z R Z ∞ | y | ˜ ω y ω + ν f ν ( x, s ) d y d x d s ≤ C ( T, ω, ˜ ω, A ) . Thus R R | x | ˜ ω f ν ( x, t ) d x is uniformly bounded (i.e. independent of ν and t ).Using this and that { f ν } ν> is uniformly bounded by 1, we can extract a subsequence { f ν n } n ∈ N (denotedin the following as { f n } n ∈ N ) such that f n ( · , t k ) converges in the sense of measures to some f ( · , t k ) for all k ∈ N , where { t k } k ∈ N = [0 , T ] ∩ Q .We next show that f n is equicontinuous in t as a distribution, i.e. from (65) we obtain for any ψ ∈ C c ( R ): (cid:12)(cid:12)(cid:12)(cid:12)Z R ( f n ( x, t ) − f n ( x, s )) ψ ( x ) d x (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)Z ts Z R f n ( x, r ) Z ∞ y ω + ν [ ψ ( x − y ) − ψ ( x )] d y d x d r (cid:12)(cid:12)(cid:12)(cid:12) ≤ Z ts Z R f n ( x, r ) (cid:20)Z k ψ ′ k L ∞ yy ω + ν d y + Z ∞ k ψ k L ∞ y ω + ν d y (cid:21) d x d r ≤ C ( ψ ) | t − s | , C ( ψ ) is a constant independent of ν but depending on ψ and ψ ′ . Using the equicontinuity of f n (as adistribution) one can show that f n converges to some limit f (in the sense of distributions) for all t ∈ [0 , T ].Using furthermore the uniform boundedness of R R | x | ˜ ω f n ( x, t ) d x one can show that f n converges alreadyin the sense of measures by approximating and cutting the test function for large values of | x | .Using similar arguments we can also show that for the limit f n ⇀ f we have f ∈ C (cid:0) [0 , T ] , M fin+ (cid:1) andtaking the limit n → ∞ in (65), f satisfies Z R f ( x, t ) ψ ( x ) d x = ψ ( A ) + Z t Z R Z ∞ y ω f ( x, s ) [ ψ ( x − y ) − ψ ( x )] d y d x d s (66)for each ψ ∈ C b ( R ) and all t ∈ [0 , T ].From the construction of f using the contraction mapping principle we immediately get supp f ( · , t ) ⊂ ( −∞ , A ] for all t ∈ [0 , T ]. To see R R f ( · , t ) d x = 1 for all t ∈ [0 , T ] we integrate equation (62) over R and useFubini’s theorem to obtain ∂ t R R G ( · , t ) d x = 0. Thus together with the initial condition the claim follows. Remark
B.2 . The analogous result holds true if f = − δ ( · − A ).As a direct consequence of Proposition B.1 we also obtain smooth solutions for smoothed initial data.Therefore for κ > ϕ κ a non-negative, symmetric standard mollifier withsupp ϕ κ ⊂ [ − κ, κ ]. Proposition B.3.
Let f := δ ( · − A ) . Then there exists a solution f ∈ C ([0 , T ] , C ∞ ( R )) to (62) withinitial datum f ∗ ϕ κ = ϕ κ ( · − A ) .Proof. This follows directly by convolution in x from Proposition B.1. Proposition B.4.
There exists a strong solution f ∈ C ([0 , T ] , C ∞ ( R )) to (62) with initial datum f := χ ( −∞ ,A ] ∗ ϕ κ .Proof. Let G be the solution given by Proposition B.3 for G := δ ( · − A ) ∗ ϕ κ . Then f ( x, t ) := R ∞ x G ( y, t ) d y solves (62) with the desired initial condition.In the same way as in the proofs of Proposition B.4 and Proposition B.4 we obtain the following existenceresult: Proposition B.5.
Let ε > , L > and λ , λ > be two constants (depending on some parameters). Thenthere exists a weak solution G ∈ C (cid:0) [0 , T ] , M fin+ (cid:1) and a strong solution W ∈ C ([0 , T ] , C ∞ ) of the equation ∂ t W ( ξ, t ) − Z h ε ( z ) z h λ ( z + ε ) − a + λ ( z + ε ) b i h W (cid:16) ξ + zL , t (cid:17) − W ( ξ, t ) i d z = 0 (67) together with initial condition G ( · ,
0) = δ ( · − A ) and W ( · ,
0) = χ ( −∞ ,A ] ∗ ϕ κ .Remark B.6 . The measure G has the same properties as the measure f in Proposition B.1. Remark
B.7 . By convolution we also obtain a strong solution G ∈ C ([0 , T ] , C ∞ ) of (67) with initial condition G ( · ,
0) = δ ( · − A ) ∗ ϕ κ .For further use we denote the integral kernels occurring in Proposition B.1 and Proposition B.5 by N ω ( z ) := z − − ω and N ε ( z ) := h ε ( z ) z h λ ( z + ε ) − a + λ ( z + ε ) b i . (68) Proposition B.8.
Let n ∈ N , R ∈ R and N i : (0 , ∞ ) → R ≥ either of the form N ω i for some ω i ∈ (0 , or N ε given by (68) (and then continued by to (0 , ∞ ) ) for i = 1 , . . . n . Let N := P ni =1 N i . Then there exists asolution f ∈ C ([0 , T ] , C ∞ ( R )) to the equation ∂ t f ( x, t ) = Z ∞ N ( z ) [ f ( x + z ) − f ( x )] d z (69) either with initial datum f = χ ( −∞ ,R ] ∗ n ϕ κ or f = δ ( · − R ) ∗ n ϕ κ , where ∗ n denotes the n -fold convolutionwith ϕ κ . roof. It suffices to consider the case n = 2 (otherwise argue by induction). Then by Proposition B.3 andProposition B.4 there exist solutions f i to equation (69) with N replaced by N i and initial datum f = δ ( · ) ∗ ϕ κ and f = χ ( −∞ ,R ] ∗ ϕ κ (or f = δ ( · − R ) ∗ ϕ κ ). A straightforward computation shows that the convolution f := f ∗ f satisfies (69) together with the correct initial condition. Remark
B.9 . Let G κ and f κ be the solutions given by Proposition B.8 with initial condition G κ ( · ,
0) = δ ( · − A ) ∗ n ϕ κ and f ( · ,
0) = χ ( −∞ ,A ] ∗ n ϕ κ . Then from the construction in the proof of Proposition B.8 andProposition B.1 we obtain:1. G κ ≥ R (in the sense of measures) and 0 ≤ f κ ≤ t ∈ [0 , T ],2. supp G κ ( · , t ) , supp f κ ( · , t ) ⊂ ( −∞ , A + nκ ] for all t ∈ [0 , T ],3. R R G ( · , t ) d x = 1 for all t ∈ [0 , T ],4. f κ is non-increasing. B.2 Integral estimates for subsolutions
In this section we will always assume that the integral kernel N is given as the sum of kernels of the form N ω i or N ε and we will prove several properties and estimates that are frequently used. We now prove someintegral estimates. Lemma B.10.
Let ω ∈ (0 , and G the solution of ∂ t G ( x, t ) = P Z ∞ N ω ( z ) [ G ( x + z ) − G ( x )] d zG ( · ,
0) = δ ( · − A ) ∗ ϕ κ = ϕ κ ( x − A ) (70) given by Proposition B.3, where P is a constant. Then for any µ ∈ (0 , one has1. R A − D −∞ G ( x, t ) d x ≤ C (cid:0) κD (cid:1) µ + C P tD ω for all D > and2. R AA − | x − A | G ( x, t ) d x ≤ C µ κ µ + C ω P t .Proof.
By shifting with A we can assume A = 0. Let Z >
0. Then testing equation (70) with e Z ( x − κ ) (notethat this is possible as supp G ⊂ ( −∞ , κ ]) one obtains ∂ t Z R G ( x, t ) e Z ( x − κ ) d x = P Z R Z ∞ N ω ( y ) [ G ( x + y ) − G ( x )] e Z ( x − κ ) d y d x = P Z ∞ N ω ( y ) (cid:0) e − Zy − (cid:1) d y Z R G ( x, t ) e Z ( x − κ ) d y =: M ω ( Z ) Z R G ( x, t ) e Z ( x − κ ) d x. Furthermore Z R G κ ( x,
0) e Z ( x − κ ) d x = Z R ϕ κ ( x ) e Z ( x − κ ) d x. Thus we obtain R R G ( x, t ) e Z ( x − κ ) d x = R R ϕ κ ( x ) e Z ( x − κ ) d x exp ( − t | M ω ( Z ) | ). Estimating M ω ( Z ) we obtain | M ω ( Z ) | ≤ P Z ∞ − e − Zy y ω d y = − Pω Z ∞ (cid:0) − e − Zy (cid:1) ∂∂y (cid:0) y − ω (cid:1) d y = P Zω Z ∞ e − Zy y ω d y = P Z ω ω Z ∞ y − ω e − y d y = P Γ (1 − ω ) ω Z ω = CP Z ω . G = 0 on ( κ, ∞ ) we get Z κ −∞ G ( x, t ) (cid:16) − e Z ( x − κ ) (cid:17) d x = Z R G ( x, t ) d x − Z R G ( x, t ) e Z ( x − κ ) d x = 1 − Z R ϕ κ ( x ) e Z ( x − κ ) d x exp ( − t | M ω ( Z ) | ) ≤ (cid:20)(cid:18) − Z R ϕ κ ( x ) e Z ( x − κ ) d x (cid:19) + Z R ϕ κ ( x ) e Z ( x − κ ) d x | M ω ( Z ) | t (cid:21) . As supp ϕ ⊂ [ − κ, κ ] we can estimate e − Zκ ≤ R R ϕ κ ( x ) e Z ( x − κ ) d x ≤
1. Then choosing Z = D and using alsothe estimate for M ω we obtain Z − D −∞ G ( x, t ) d x ≤ Z − D −∞ G ( x, t ) 1 − e x − κD − e − − κD d x ≤ Z κ −∞ ( · · · ) d x ≤ − e − − κD (cid:20)(cid:16) − e − κD (cid:17) + CP t D ω (cid:21) ≤ C (cid:16) κD (cid:17) µ + C P tD ω . To prove the second part we use a dyadic decomposition and the estimate from the first part to obtain Z − | x | G κ ( x, t ) d x = ∞ X n = − Z − − ( n +1) − − n | x | G κ ( x, t ) d x ≤ C ∞ X n = − − n h(cid:16) κ n +1 (cid:17) µ + P t ω ( n +1) i = C ∞ X n = − µ κ µ (cid:0) µ − (cid:1) n + 2 ω P t (cid:0) ω − (cid:1) n ≤ C µ κ µ + C ω P t.
We now consider the situation of Proposition B.8 where the integral kernel is given as the sum of differentkernels
Lemma B.11.
In the situation of Proposition B.8 with n = 2 one has1. R A − D −∞ G ( x, t ) d x ≤ R A − D/ −∞ G ( x, t ) d x + R − D/ −∞ G ( x, t ) d x R AA − | x − A | G ( x, t ) d x ≤ R A + κA − − κ | x − A | G ( x ) d x + R κ − − κ | x | G ( x ) d x .Proof. We consider again only the case A = 0, while the general result follows by shifting.1. One has Z − D −∞ G ( x, t ) d x = Z R Z R χ ( −∞ , − D ] ( x + y ) G ( x, t ) G ( y, t ) d x d y = Z R Z − D − y −∞ G ( x, t ) G ( y, t ) d x d y = Z ∞− D Z − D − y −∞ G ( x, t ) G ( y, t ) d x d y + Z − D −∞ Z − D − y −∞ G ( x, t ) G ( y, t ) d x d y ≤ Z − D −∞ G ( x, t ) d x Z R G ( y, t ) d y + Z − D −∞ G ( y, t ) d y Z R G ( x, t ) d x ≤ Z − D/ −∞ G ( x, t ) d x + Z − D/ −∞ G ( x, t ) d x where in the last step we used that G i is normalized for i = 1 , Z − | x | G ( x, t ) d x = Z R Z R χ [ − , ( x + y ) | x + y | G ( x, t ) G ( y, t ) d x d y = Z R Z − y − − y | x + y | G ( x, t ) G ( y, t ) d x d y = Z κ −∞ Z − y − − y | x + y | G ( x, t ) G ( y, t ) d x d y where we used that G = 0 on { y > κ } . Using also G = 0 on { x > κ } we have furthermore Z − | x | G ( x, t ) d x ≤ Z κ −∞ Z κ − − y | x + y | G ( x, t ) G ( y, t ) d x d y. Noting that for y < − − κ we have − − y > κ and thus the x -integral equal zero as G = 0 on { x > κ } we obtain (also using − − κ ≤ − − y for y ∈ [ − − κ, κ ]) Z − | x | G ( x, t ) d x ≤ Z κ − − κ Z κ − − κ | x + y | G ( x, t ) G ( y, t ) d x d y ≤ Z κ − − κ Z κ − − κ ( | x | + | y | ) G ( x, t ) G ( y, t ) d x d y ≤ Z κ − − κ | x | G ( x, t ) d x + Z κ − − κ | y | G ( y, t ) d y, where in the last step we used R κ − − κ G i ( x, t ) d x ≤ R R G i ( x, t ) d x = 1. Remark
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