J Stat Phys (2018) 170:1215–1241 https://doi.org/10.1007/s10955-018-1980-6
Tail Behaviour of Self-Similar Profiles with Infinite Mass for Smoluchowski’s Coagulation Equation Sebastian Throm1
Received: 27 October 2017 / Accepted: 30 January 2018 / Published online: 8 February 2018 © Springer Science+Business Media, LLC, part of Springer Nature 2018
Abstract In this article, we consider self-similar profiles to Smoluchowski’s coagulation equation for which we derive the precise asymptotic behaviour at infinity. More precisely, we look at so-called fat-tailed profiles which decay algebraically and as a consequence have infinite total mass. The results only require mild assumptions on the coagulation kernel and thus cover a large class of rate kernels. Keywords Smoluchowski coagulation equation · Self-similar profiles · Decay behaviour · Fat tails
1 Introduction Smoluchowski’s coagulation equation is a kinetic model which describes the irreversible aggregation of clusters in particle systems that may originate from many different applications in physics, chemistry or biology. This model was originally derived by Marian von Smoluchowski in 1916 in the context of coagulation of gold particles in a colloidal solution which move according to Brownian motion [21,22]. The basic assumptions for the derivation of the equation are that each particle is completely characterised by a scalar quantity x ∈ (0, ∞) which is usually denoted as the size or mass of the cluster. Moreover, one assumes that the state of the whole system is fully described by the density φ(x, t) of clusters of size x at time t. Finally, one assumes that the system is dilute in the sense that in each coagulation process only two clusters are involved which means that the effect of higher order collisions can be neglected as a rare event.
B 1
Sebastian Throm
[email protected] Faculty of Mathematics, Technical University of Munich, Boltzmannstraße 3, 85748 Garching bei München, Germany
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The time evolution of such systems is described by Smoluchowski’s coagulation equation which reads ∞ 1 x ∂t φ(x, t) = K (x − y, y)φ(x − y, t)φ(y, t)dy − φ(x, t) K (x, y)φ(y, t)dy. 2 0 0 (1) The first integral on the right-hand side is denoted as the gain term and accounts for the creation of clusters of size x from clusters of sizes x − y and y while the factor 1/2 is due the symmetry of the process. The second integral on the right-hand side however corresponds to the loss of particles of size x due to the aggregation process and may thus be denoted as the loss term. Besides the specific application which Smoluchowski had in mind when he derived (1), the same model is used to describe a broad range of other phenomena in many different fields. More details on the model and its applications can be found in [2,10].
1.1 Assumptions on the Kernel K As one can see from (1), the dynamics of the system is completely determined by the integral kernel K . One prominent example of such a kernel is the one that Smoluchowski derived himself for the gold particles and which reads K (x, y) = x 1/3 + y 1/3 x −1/3 + y −1/3 . (2) The quantity x 1/3 corresponds to the radius, while x −1/3 relates to the diffusion constant of a spherical cluster of size x. Throughout this work we assume that the integral kernel K is a continuous function on (0, ∞) × (0, ∞) which is symmetric and homogeneous of degree λ ∈ (−1, 1), i.e. K (x, y) = K (y, x) and K (r x, r y) = r λ K (x, y) for all r, x, y ∈ (0, ∞).
(3)
Moreover, we require some upper and lower bound on the kernel K and we precisely assume that there exist constants c∗ , C∗ > 0 and α, β ∈ (−1, 1) with α ≤ β and λ = α + β as well as b, B ∈ (0, ∞) with b < B such that inf K (x, y) ≥ c∗ and 0 ≤ K (x, y) ≤ C∗ x α y β + x β y α . (4) x,y∈[b,B]
Remark 1 Note that for c∗ = 2, C∗ = 3 and α = −1/3, β = 1/3 this covers in particular the case (2) for each choice of b, B ∈ (0, ∞) with b < B. Remark 2 Note that the well-posedness of (1) has been shown for example in [18] for kernels satisfying K (x, y) ≤ ϕ(x)ϕ(y) for a continuous sublinear function ϕ : (0, ∞) → (0, ∞). This covers in particular the important kernel (2). Further results on the existence and uniqueness of solutions to (1) for a broad range of kernels can be found for instance in [5,8] where, among others, the class of kernels K (x, y) = x α y β + x β y α is considered. Remark 3 The restriction to homogeneity λ < 1 is due to the fact that for kernels with homogeneity λ > 1 a phenomenon known as gelation occurs where infinitely large clusters are created due to the fast growth of the kernel at infinity (see [4,20]). The case λ = 1 represents the borderline and has a special behaviour which is not accessible with our method (see for instance [9]). On the other hand, we restrict to λ > −1 in order to avoid certain integrals becoming too singular. This problem could be overcome if we specified the asymptotic behaviour of K close to zero more precisely. However, to simplify the presentation we do not pursue this issue here.
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1.2 Self-Similar Profiles One particular point of interest concerns the behaviour of solutions to (1) for t → ∞. For integral kernels satisfying (3) one expects that the long-time behaviour of solutions to (1) is self-similar. Precisely this means that there exist solutions φ for (1) which are of the form φ(x, t) =
1 f (s(t))θ
x s(t)
with s(t) −→ ∞ as t −→ ∞.
(5)
The so-called scaling hypothesis then states that the long-time behaviour of solutions φ to (1) is given by the profile f in the sense that θ s(t) φ(s(t)x, t) −→ f (x) as t −→ ∞.
(6)
This conjecture is unproven for kernels that one typically finds in applications such as (2). However, there are two specific solvable kernels with homogeneity λ ≤ 1 (i.e. K ≡ 2 and K (x, y) = x + y) for which the scaling hypothesis is known to be true due to the fact that explicit solution formulas are available in terms of the Laplace transform. Thus, these kernels allow to verify (6) in terms of weak convergence of measures (see [11]). Moreover, in [11] it is also shown that there exists a profile f 1 with exponential decay at infinity which is uniquely ∞ determined if we prescribe the total mass, i.e. the first moment 0 x f 1 (x). On the other hand one has a whole family of self-similar profiles f ρ which in general decay algebraically. If one considers for example the constant kernel K ≡ 2 and fixes the total mass to be one, f 1 is given by f 1 (x) = e−x . Conversely, the profiles f ρ for ρ ∈ (0, 1) have the asymptotic behaviour f ρ ∼ x −1−ρ for x → ∞. The latter holds in general only up to a constant factor which, however, can be normalised to one by rescaling, analogously to the total mass which is infinite for f ρ with ρ ∈ (0, 1). The existence of self-similar profiles with fast decay has been established for many nonsolvable kernels in [5,6] while existence of fat-tailed profiles could also be established recently for a broad class of coagulation kernels in [12,15]. The general approach for the latter consisted in applying a fixed-point argument to the set of non-negative measures μ such R that 0 xμ(dx) ∼ R 1−ρ for R → ∞. This means that the fat-tailed profiles have already been constructed as measures with a specific (weak) tail behaviour at infinity. Since there is up to now no general uniqueness statement for self-similar profiles available, one might think of the possibility that there exist other profiles with a different decay at infinity. That this cannot be the case is exactly the goal of this work, i.e. we show rigorously that each fat-tailed self-similar profile (up to a certain rescaling) has the decay behaviour as described before while for non-singular kernels, we require an additional integrability condition as explained in Sect. 1.4 below. As already indicated, except for special cases the question of uniqueness of self-similar profiles is completely open for non-solvable kernels. However, one example where uniqueness of self-similar profiles could be established recently is a perturbed model of the constant kernel, i.e. K = 2 + εW (x, y) with sufficiently small ε > 0 and W being analytic and behaving in a certain sense like (x/y)α + (y/x)α with α ∈ [0, 1/2) (for the precise conditions we refer to [13]). For such kernels it is possible to show uniqueness of self-similar profiles both in the case of finite mass [13] and for fat-tailed profiles [19]. The proof is based in both cases on Laplace transform methods and it turns out to be convenient to normalise the profiles according to their tail behaviour which might also be seen as a motivation for this work.
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1.3 The Equation for Self-Similar Profiles If one plugs the ansatz (5) into (1) one obtains by typical scaling arguments that, in order to obtain s(t) → ∞ for t → ∞, it should hold θ > 1 + λ (see (8) and Remark 4) while f solves the equation θ f (y) + y f (y) +
1 2
y
∞
K (y − z, z) f (y − z) f (z)dz − f (y)
0
K (y, z) f (z)dz = 0.
0
(7) For the scaling function s(t) we obtain up to a time shift that 1/(θ −1−λ) s(t) = (θ − 1 − λ)t .
(8)
Remark 4 Note that the choice θ = 1 + λ, leading to an exponential mean size s(t) = et , would also meet the requirements above. However, similarly to homogeneity λ = 1, this represents a critical boundary case which is not accessible to our method and has also been excluded in previous considerations (e.g. [12,15]). For K (x, y) = x α y β + x β y α with α, β > 0 and θ = 2, formal arguments in [20] suggest the behaviour f (x) ∼ C x −1−λ for x → 0 of self-similar profiles. Although fat-tailed profiles, i.e. θ ∈ (1 + λ, 2), have not been considered explicitly in [20], the reasoning can be easily adapted and still yields the expected asymptotic behaviour f (x) ∼ C x −1−λ as x → 0 in this case. As a consequence the integrals in (7) are not well-defined and one has to exploit some cancellation between them. In fact, multiplying (7) by y and integrating over [0, x] we find x
y y θ y f (y) + y 2 f (y) + K (y − z, z) f (y − z) f (z)dz 2 0 0 ∞ −y f (y) K (y, z) f (z)dz dy = 0.
(9)
0
From the discussion above, we expect f (x) ∼ C x −1−λ as x → 0 and thus in particular lim x→0 x 2 f (x) = 0. Integration by parts then yields θ 0
x
y f (y)dy + 0
x
y 2 f (y)dy = (θ − 2)
x
y f (y)dy + x 2 f (x).
(10)
0
On the other hand, splitting the integral, changing variables z → y−z and using the symmetry of K we obtain the relation y 2
y
K (y − z, z) f (y − z) f (z)dz 1 y 1 y = (y − z)K (y − z, z) f (y − z) f (z)dz + z K (y − z, z) f (y − z) f (z)dz 2 2 0 y0 (y − z)K (y − z, z) f (y − z) f (z)dz. = 0
0
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Thus, applying first Fubini’s Theorem, changing variables y → y + z and applying Fubini’s Theorem again, we find x y y K (y − z, z) f (y − z) f (z)dz 2 0 0 x x = (y − z)K (y − z, z) f (y − z) f (z)dydz 0
=
x
0
z x−z
y K (y, z) f (y) f (z)dydz =
0
x
0
0
y K (y, z) f (y) f (z)dzdy.
0
Together with (9) and (10) we thus obtain x y f (y)dy x 2 f (x) + (θ − 2) 0 x x−y =− y K (y, z) f (y) f (z)dzdy + 0
x−y
0
x
∞
y K (y, z) f (y) f (z)dzdy.
0
Combining the two integrals on the right-hand side and introducing ρ = θ − 1 we finally get x x ∞ y f (y)dy + y K (y, z) f (y) f (z)dzdy. (11) x 2 f (x) = (1 − ρ) 0
0
x−y
Note that due to our restrictions on θ we have ρ > λ. One fundamental property of (1) is that the total mass, i.e. the first moment (0,∞) xφ(x, t)dx is at least formally conserved over time provided that this quantity is finite for t = 0. In this case one expects the scaling ansatz (5) to have constant mass, too which fixes the parameter θ = 2 or equivalently ρ = 1 and the first moment of f is finite. Thus, the natural range for the parameter ρ is (λ, 1]. Since we will restrict in this work exclusively to the case of profiles with infinite mass we will only consider ρ ∈ (λ, 1). Note that in the case of finite mass, i.e. ρ = 1 the first term on the right-hand side of (11) vanishes and the tail behaviour of the profile is completely determined by the non-linear integral operator on the right-hand side. Formal considerations in [20] suggest that up to rescaling each self-similar profile f with finite mass has the behaviour f (x) ∼ Ax −λ e−x as x → ∞, where the constant A depends on the rate kernel K . Except for a small class of kernels with homogeneity zero (see [16]) this precise behaviour could not yet be established for non-solvable kernels but [3,7] provide exponential upper and lower bounds which are in accordance with the conjectured asymptotic behaviour. On the other hand, for profiles with infinite mass, i.e. for ρ ∈ (λ, 1) the asymptotic behaviour for large cluster sizes is completely different. Precisely, for ρ ∈ (λ, 1) the second term on the right-hand side of (11) is of lower order as x → ∞. Thus, if we neglect this term, the large-mass behaviour of f to first order is formally given by a simple linear ordinary differential equation, i.e. one obtains f (x) ∼ (1 − ρ)x −1−ρ as x → ∞. The goal of this work is to prove that for kernels K satisfying (3) and (4) each self-similar profile as specified in Definition 1 below necessarily has this behaviour. This works well for kernels exhibiting a singular behaviour at the origin, i.e. for α < 0. Unfortunately, for kernels with α ≥ 0 we have to require an integrability condition which is slightly stronger than the one we would naturally expect. More details on this can be found in Sect. 1.4 below. On the other hand, due to the different properties of (11) for fat-tailed profiles our proof covers a much broader class of kernels compared to the case of finite mass considered in [16]. We also emphasise at this point that in the existence results [12,15] mentioned above, the asymptotic behaviour of the constructed profiles is also determined but the important
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difference to the present result is that there, this behaviour has already been encoded in weak form in the construction process (see also Sect. 1.2) whereas here, we start with a general notion of self-similar profiles. This point is in particular important due to the lack of a uniqueness statement for most coagulation kernels. Finally, we note that the asymptotic behaviour close to zero in the finite-mass-case could be established rigorously in [1,7] for specific power law kernels.
1.4 Main Results Before we give the precise notion of self-similar profiles that we will use here, let us give a motivation for the Definition 1 below. For this, we first recall that in the case of finite mass a self-similar profile f is typically assumed to be an element of L 1 (xdx) (e.g. [7]) where ∞ the total mass, i.e. 0 x f (x)dx is a fixed constant. However, for fat-tailed profiles f we expect from the formal considerations in the previous section that f (x) ∼ x −1−ρ as x → ∞ with ρ < 1. Thus, the first moment is infinite. Yet, we still require an integrability condition at infinity in order to obtain finite integrals in (11). In fact, due to the upper bound on K ∞ in (4) a self-similar profile f should satisfy at least 1 x β f (x)dx < ∞. From the expected asymptotic behaviour of f we thus conclude that ρ > β. Moreover, if α > 0 one can easily check that the explicit power law f (x) = Aρ,λ x −1−λ satisfies (11) for an appropriate choice of the constant Aρ,λ > 0. However this function obviously has not the expected decay behaviour at infinity unless ρ = λ, which is the critical value which we will not discuss here (see Remark 4). In order to rule out this special solution we have to assume a higher integrability condition at infinity. The minimal choice would of ∞ course be 1 x λ f (x)dx < ∞. Unfortunately, it turns out that for α > 0 this condition is not sufficient to get our iteration method started. Instead we need the slightly stronger condition ∞ x γ f (x)dx < C(γ , f ) for some γ ∈ (λ, 1) such that γ ≥ β. (12) 1
Remark 5 We emphasise that ∞ for singular kernels, i.e. if α < 0 the condition (12) is an immediate consequence of 1 x β f (x)dx < ∞. In fact, if the latter is satisfied we directly conclude (12) with γ = β since α < 0 and thus λ = α + β < β. Thus, (12) is a natural requirement for singular kernels. Although for the borderline case α = 0 there does not exist an explicit ∞ power-law solution we still have to require (12) instead of the more natural condition 1 x β f (x)dx < ∞. The only exception is the case α = β = 0 where a completely different method applies which relies on the Laplace transform and does not exploit (12). Definition 1 For an integral kernel K satisfying assumptions (3) and (4) and ρ ∈ (max{λ, β}, 1) we denote a function f ∈ L 1loc (0, ∞) a self-similar profile (to (1)) or equivalently a solution to (11) if f is non-negative, f ≡ 0 and almost everywhere. (11) f satisfies ∞ Additionally, we require that f satisfies x f (x) ∈ L 1loc [0, ∞) and 1 x β f (x)dx < ∞. Remark 6 Note that if f is a self-similar profile according to Definition 1 then also the rescaled function f a (x) = a 1+λ f (ax) for all a > 0 is a self-similar profile. Remark 7 Note that we consider here rather general kernels with homogeneity λ ∈ (−1, 1). Thus, if β < 0 our results are even true for certain negative values of ρ, i.e. profiles f which decay slower than 1/x at infinity.
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Remark 8 Note that the results presented here are true and can be shown similarly if the self-similar profiles are only assumed to be non-negative Radon measures instead of L 1 functions. However, in order to simplify the presentation and to avoid certain technicalities we do not focus on this. As a first step towards the precise asymptotic behaviour of self-similar solutions we establish a bound on the integral of x f (x) over (0, R) which has the expected scaling behaviour. Proposition 1 Assume that K satisfies (3) and (4) and let f be a self-similar profile according to Definition 1. If α ≥ 0 and (α, β) = (0, 0) assume in addition that f satisfies (12). Then there exists a constant C f > 0 such that R y f (y)dy ≤ C f R 1−ρ for all R > 0. 0
With this estimate it will be rather straightforward to derive the asymptotic behaviour of 0 x f (x)dx which is already a weak form of our main result, i.e. Theorem 1 below. Note that this gives in particular also a lower bound on the integral for large values of R.
R
Proposition 2 Assume the same conditions as in Proposition 1. Then for each self-similar profile f there exists a suitable rescaling f a (x) := a 1+λ f (ax) such that R R y f a (y)dy ≤ R 1−ρ and y f a (y)dy ∼ R 1−ρ for R −→ ∞. 0
0
Note that f a is also a self-similar profile according to Remark 6. The main result of this work is the following statement which gives the precise asymptotic behaviour of self-similar profiles at infinity. Theorem 1 Assume that K satisfies (3) and (4) and let f be a self-similar profile according to Definition 1 and subject to the normalisation given by Proposition 2 (in particular we assume that f satisfies (12) if α ≥ 0 and (α, β) = (0, 0)). Then f is continuous on (0, ∞) and has the behaviour f (x) ∼ (1 − ρ)x −1−ρ as x → ∞. Remark 9 We note that Theorem 1 in the special case of homogeneity λ = 0, i.e. α = −β with β ∈ [0, 1) has already been treated in [19] using the same kind of arguments. In this work we will however generalise the result to a much broader class of kernels which may have different homogeneity.
1.5 Outline of the Article The remainder of this work is concerned with the proofs of these main results and as a first step, we will derive in Sect. 2 an estimate on the average of self-similar profiles which gives enough integrability in the region close to zero for what follows. In Sect. 3 we will provide the proofs of Propositions 1 and 2. More precisely, in the case of Proposition 1 the general strategy is to test the equation by the function x ρ−2 and conclude by some iteration argument. The only exception is the case α = β = 0 where slightly weaker assumptions on the integrability of self-similar profiles are needed because a completely different argument applies. More precisely, we take the Laplace transform of (11) and use the boundedness of the kernel to derive a differential inequality for the Laplace transformed self-similar profile. Solving this
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inequality explicitly, gives an upper bound on the Laplace transform from which the desired estimate can be derived easily. The proof of Proposition 2 is then an easy consequence of Proposition 1. Section 4 is finally devoted to the proof of Theorem 1. The key for this will be Proposition 3 which states that for large values of x each self-similar profile f is at least bounded as f (x) ≤ C x −1−ρ . Once this is established, one easily deduces that the non-linear term in (11) is of lower order as x → ∞ and the expected decay behaviour follows directly. However, the proof of Proposition 3 requires again some iteration argument and we have to consider additionally three different cases depending on the signs of the exponents α and β although the general argument in all cases is similar.
2 Regularity Close to Zero In this section we will prove the following lemma which provides a uniform estimate for self-similar profiles in the region close to zero. Lemma 1 Assume that K satisfies (3) and (4). There exists a constant C > 0 which only depends on c∗ , b and B such that each self-similar profile satisfies 2R x f (x)dx ≤ C R 1−λ for all R > 0. R
Remark 10 This lemma states that each self-similar profile behaves at least in an averaged sense not worse than x −1−λ close to zero. More precisely this is also true for large values of x while it will turn out that we can obtain better decay there. The same kind of result has already been used in the case of finite mass in [14]. However, the same proof also applies in the present situation where we consider fat-tailed profiles since it only relies on the structure of the non-linear term in (11). Yet, since we use slightly different assumptions on the kernel K here and for completeness we briefly recall the argument of [14]. Proof of Lemma 1 We may assume without loss of generality that B ≤ 2b in the assumption (4) since for B > 2b the lower bound in (4) is still valid with B replaced by 2b. We then divide (11) by x, integrate over [b R, B R] for any R > 0 and use the non-negativity of f as well as (1 − ρ) ≥ 0 to get BR x f (x)dx bR
≥
1 BR
BR
1 x bR BR x
= (1 − ρ)
bR
0
x
0 ∞
y f (y)dydx +
BR
bR
1 x
x
0
∞
y K (y, z) f (y) f (z)dzdydx
x−y
y K (y, z) f (y) f (z)dzdydx.
x−y
With Fubini’s Theorem and the non-negativity of the integrand, we can rewrite and estimate the right-hand side further as
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BR
bR
0 bR
=
x
0
≥
∞
(· · · )dzdydx
x−y BR
∞
y
BR
(· · · )dzdxdy +
bR x−y BR BR ∞
bR
1223
bR
BR y
∞
(· · · )dzdxdy
x−y
(· · · )dzdxdy.
x−y
Using Fubini’s Theorem again to rewrite the right-hand side we can estimate once more to get
BR
bR
≥ ≥
x
∞
(· · · )dzdydx
0 x−y B R B R−y
bR BR
0
bR
y BR
∞ B R−y
y+z
(· · · )dxdzdy +
BR
bR
∞ B R−y
BR
(· · · )dxdzdy
y
(· · · )dxdzdy.
y
Reducing the domain of integration in the variables y and z and using the homogeneity of the kernel K we thus find
BR bR
1 x f (x)dx ≥ BR
BR
bR
∞ B R−y
BR
dx y K (y, z) f (y) f (z)dzdy
y
y z
1 f (y) z f (z)dzdy (B R − y)y K R , R R R z bR (B−b)R b+B R B R
2 y z B − b λ−1 y f (y)z f (z)dzdy. R K ≥ , 2 2B R R bR (B−b)R ≥
1 BR
b+B 2
R
BR
Since we assume that B ≤ 2b and the integrand is non-negative, we have BR b R (· · · )dz. Together with (4) this further yields
BR
bR
BR
(B−b)R (· · · )dz
≥
z
y f (y)z f (z)dzdy R R bR bR BR b+B R 2 c∗ (B − b) λ−1 ≥ R y f (y)dy z f (z)dz. 2B 2 bR bR B − b λ−1 R 2B 2
x f (x)dx ≥
b+B 2
R
BR
K
y
,
From the latter estimate we conclude that
b+B 2
R
y f (y)dy ≤
bR
2B 2 R 1−λ . c∗ (B − b)
If we replace R by R/b this reads
b+B 2b
R
R
y f (y)dy ≤
2B 2 R 1−λ . c∗ (B − b)b1−λ
(13)
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To conclude the proof, we note that (b + B)/(2b) > 1. Therefore, we can fix a constant N +1 ≥ 2. Splitting the integral and using (13) we then obtain N ∈ N such that b+B 2b
2R
x f (x)dx ≤
R
N
n=0
b+B 2b
b+B 2b
n+1
n
R
x f (x)dx
R
N b + B (1−λ)n 1−λ 2B 2 R , ≤ c∗ (B − b)b1−λ 2b n=0
which yields the claim.
From Lemma 1 we directly infer the following result which provides estimates on the integral of f in the region close to zero. Lemma 2 If χ > λ there exists a constant Cχ > 0 such that 1 x χ f (x)dx < Cχ 0
for each self-similar profile f . Proof The proof follows from Lemma 1 together with a dyadic decomposition. We will use this kind of argument frequently in this article and we note that it has already been applied several times in previous work (e.g. [12,15]). In fact we have 1 ∞ 21−n x χ f (x)dx = x χ −1 x f (x)dx 0
n=1 2 ∞
≤C
−n
max 2(n−1)(1−χ ) , 2n(1−χ ) 2n(λ−1)
n=1 ∞
= C max 2χ −1 , 1 2n(λ−χ ) . n=1
Since χ > λ the sum on the right-hand side converges to 2λ−χ /(1 − 2λ−χ ) which ends the proof.
3 Averaged Tail Estimates 3.1 Proof of Proposition 1 for α = β = 0 We start with the proof of Proposition 1 for α = β = 0 which relies on the Laplace transform and is much easier than the corresponding proof for the remaining cases. The same argument is already contained in [19] while similar ideas have also been used in [17]. Proof of Proposition 1 For α = β = 0. Since α = β = 0, we obtain from (4) that K (x, y) ≤ 2C∗ for all x, y > 0. Following the notation of [11], we denote for q ≥ 0 by ∞ Q(q) := (1 − e−q x ) f (x)dx 0
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the desingularised Laplace ∞ transform of the self-similar profile f . Note that Q is well-defined due to the condition 1 f (x)dx < ∞ in Definition 1 and Lemma 2. Moreover, Q as a function of q is infinitely often differentiable on (0, ∞) and we have in particular ∞ ∞ Q (q) = x f (x)e−q x dx and Q
(q) = − x 2 f (x)e−q x dx. (14) 0
0
e−q x
We next multiply equation (11) by and integrate over (0, ∞) to obtain x ∞ ∞ 2 −q x −q x x f (x)e dx = (1 − ρ) e y f (y)dydx 0 0 ∞ 0 x ∞ e−q x y K (y, z) f (y) f (z)dzdydx. + 0
0
x−y
Applying Fubini’s Theorem on the right-hand side and using the relations given in (14) we obtain after some elementary rearrangements that Q satisfies the equation ∞ ∞ 1
−qy −qz − ∂q q Q (q) =−ρ Q (q)+∂q K (y, z) f (y) f (z)(1 − e )(1 − e )dzdy . 2 0 0 (15) We first note that the second term on the right-hand side is well-defined due to Definition 1 and Lemma 2. Moreover, it follows by monotone convergence that 1 ∞ ∞ K (y, z) f (y) f (z)(1 − e−qy )(1 − e−qz )dzdy −→ 0 as q −→ 0. 2 0 0 Similarly, we have Q(q) −→ 0 and q Q (q) −→ 0
as q −→ 0.
As a consequence, we can integrate equation (15) over (0, q) to get 1 ∞ ∞ K (y, z) f (y) f (z)(1 − e−qy )(1 − e−qz )dzdy. − q Q (q) = −ρ Q(q) + 2 0 0 (16) From the non-negativity of f and the bound K (y, z) ≤ 2C∗ we can thus derive the inequality −q Q (q) ≤ −ρ Q(q) + C∗ Q 2 (q). Since f is non-negative and non-trivial we have Q(q) > 0 for all q > 0 and thus we can rearrange the inequality to get −1 C∗
∂q q ρ Q(q) ≤ 0. − ρ Due to the fact that Q(q) → 0 as q → 0 we can fix q > 0 such that (1/Q(q) − C∗ /ρ) > 0 −1 := for all q ≤ q . If we then fix the constant C q ) − C∗ /ρ) > 0 and integrate q ρ (1/Q( over [ q, q ] we obtain after some rearrangement that Q(q) ≤
qρ C∗ ρ ρ q
−1 +C
ρ ≤ Cq
for all q ≤ q.
If we use this estimate together with (16) and the non-negativity of f we find ρ ρ−1 for all q ≤ Q (q) ≤ Q(q) ≤ Cρq q. q
123
1226
S. Throm
Now, if we recall also (14) we find for all R ≥ q −1 that
2R
x f (x)dx =
R
2R
x
x
x f (x)e− R e R dx ≤ e2
R
∞
2 R 1−ρ . x f (x)e− R dx ≤ Cρe x
(17)
0
On the other hand, for R ∈ (0, q −1 ] we find together with Lemma 1 that
2R
x f (x)dx ≤ C R 1−λ = C R ρ−λ R 1−ρ ≤ C q λ−ρ R 1−ρ .
R
Thus, together with (17) we have
2R
2, C x f (x)dx ≤ max Cρe q λ−ρ R 1−ρ for all R > 0.
R
By means of a dyadic decomposition we thus obtain
R
x f (x)dx =
0
∞
2−n R
−(n+1) R n=0 2
x f (x)dx
∞
2, C q λ−ρ 2−(1−ρ)(n+1) R 1−ρ ≤ max Cρe n=0 ∞
2, C = 2ρ−1 max Cρe q λ−ρ R 1−ρ 2(ρ−1)n . n=0
Since ρ < 1, the sum on the right-hand side converges and thus the claim follows. The following elementary lemma will be used frequently in the proof of Proposition 1. Lemma 3 For each ρ ∈ (λ, 1) there exists a constant Cρ such that
y+z y
x ρ−2 dx ≤ Cρ y ρ−1−a z a
for all y, z > 0 and all a ∈ [0, 1]. Proof For y ≤ z we get the estimate
y+z
1 ρ−1 y − (y + z)ρ−1 1−ρ 1 1 1 ≤ y ρ−1 = y ρ−1−a y a ≤ y ρ−1−a z a . 1−ρ 1−ρ 1−ρ
x ρ−2 dx =
y
On the other hand, if z ≤ y, we find y+z x ρ−2 dx ≤ y ρ−2 z = y ρ−2 z 1−a z a ≤ y ρ−a−1 z a . y
Thus, the claim follows with Cρ := max{1, 1/(1 − ρ)}.
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Tail Behaviour of Self-Similar Profiles with Infinite Mass…
1227
3.2 Proof of Proposition 1 for (α, β) = (0, 0) We will now give the proof of Proposition 1 for kernels where (α, β) = (0, 0) which relies essentially on an iterated moment estimate. Proof of Proposition 1 For (α, β) = (0, 0). Note first that in view of Lemma 2 it suffices to show the claim for R ≥ 1. Moreover, due to the assumptions of Proposition 1 and Remark 5 we have that f satisfies the estimate (12) for all (α, β) = (0, 0). We then multiply equation (11) by x ρ−2 and integrate over (0, R) to get R R x x ρ f (x)dx + (ρ − 1) x ρ−2 y f (y)dydx 0
=
R
x ρ−2
0
x 0
0 ∞
0
y K (y, z) f (y) f (z)dzdydx.
x−y
We next apply Fubini’s Theorem which is possible as the following estimates will show. This then yields R R R ρ x f (x)dx + (ρ − 1) y f (y) x ρ−2 dxdy 0
R
= 0
R
y
0 ∞
y
x ρ−2 y K (y, z) f (y) f (z)dzdxdy.
x−y
Due to the non-negativity of f and K , we can enlarge the domain of integration on the rightR hand side which together with the evaluation of the integral y x ρ−2 dx gives the estimate R ∞ ∞ R R ρ−1 y f (y)dy ≤ x ρ−2 y K (y, z) f (y) f (z)dzdxdy. 0
0
y
x−y
After another application of Fubini’s Theorem we finally arrive at R ∞ y+z R ρ−1 R y f (y)dy ≤ y K (y, z) f (y) f (z) x ρ−2 dxdzdy. 0
0
0
(18)
y
In order to proceed we have to derive an estimate on the integral on the right-hand side which requires to consider the regions close and far from zero separately. We thus split the integral R ∞ 0 0 (· · · )dzdy as 1 1 R 1 R ∞ (· · · )dzdy = (· · · )dzdy + (· · · )dzdy 0
0
0
0 1 ∞
+ 0
1
1
0 R
(· · · )dzdy + 1
∞
(· · · )dzdy.
(19)
1
We will now estimate the four integrals on the right-hand side separately and we start with the first one. Since we want to apply Lemma 3, we fix two parameters a1 and a2 such that a1 ∈ (α, ρ − β) ∩ [0, 1] and a2 ∈ (β, ρ − α) ∩ [0, 1]. Note that this is possible because we are assuming α + β = λ < ρ as well as α, β < 1 and α ≤ β < ρ. Moreover, the choice of a1 and a2 together with the relation α + β = λ guarantees that ρ + α − a1 > λ, β + a1 > λ, ρ + β − a2 > λ and α + a2 > λ.
(20)
123
1228
S. Throm
To estimate now the first integral on the right-hand side of (19), we use the upper bound on K which is given by (4) together with Lemma 3 once with a1 and one time with a2 which yields 1 1 y+z y K (y, z) f (y) f (z) x ρ−2 dxdzdy 0
0
≤C
y
1
1
y 1+α f (y)
0
z β f (z)y ρ−a1 −1 z a1 dzdy
0
1
+C
1
y 1+β f (y)
0
z α f (z)y ρ−a2 −1 z a2 dzdy.
0
If we rearrange the right-hand side and recall the estimates in (20), it follows together with Lemma 2 that 1 1 y+z y K (y, z) f (y) f (z) x ρ−2 dxdzdy 0
0
y
1
≤C
y ρ+α−a1 f (y)dy
0
1
z β+a1 f (z)dz
0
1
+C
y ρ+β−a2 f (y)dy
0
1
z α+a2 f (z)dz ≤ C.
(21)
0
We next consider the second integral on the right-hand side of (19) for which we again use the upper bound on K given by (4) as well as Lemma 3 with a = 1. This then yields y+z R 1 y K (y, z) f (y) f (z) x ρ−2 dxdzdy 1
0
≤C
∞ 1
1
y
y α+ρ−1 z 1+β + y β+ρ−1 z 1+α f (y) f (z)dzdy.
0
We now use that y ρ−1 ≤ 1 if y ≥ 1. Together with Definition 1 and Lemma 2 as well as the assumption α ≤ β this gives y+z R 1 y K (y, z) f (y) f (z) x ρ−2 dxdzdy 1 0 y (22) ∞
≤C
1
y α f (y)dy
1
∞
z 1+β f (z)dz + C
0
1
1
y β f (y)dy
z 1+α f (z)dz ≤ C.
0
The third integral on the right-hand side of (19) can be estimated similarly. Precisely, if we use the upper bound (4) on K and Lemma 3 with a = 0 together with Definition 1 and Lemma 2 we get 1 ∞ y+z y K (y, z) f (y) f (z) x ρ−2 dxdzdy 0
1
≤C 0
1 ∞
y
y α+ρ z β + y β+ρ z α f (y) f (z)dzdy ≤ C.
(23)
1
Note that we also used here that α + ρ > λ and β + ρ > λ which holds because ρ > β ≥ α and λ = α + β. It thus remains to estimate the fourth integral on the right-hand side of (19) for which we proceed similarly. Precisely, we fix γ > λ according to (12), use (4) and apply
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Tail Behaviour of Self-Similar Profiles with Infinite Mass…
1229
Lemma 3 once with a3 = min{γ − β, 1} and once with a4 = min{γ − α, 1}. Note that γ ≥ β ≥ α and thus in particular a3 , a4 ∈ [0, 1]. With this choice we then obtain
R
1
∞
1
y+z
y K (y, z) f (y) f (z)
R
≤C
∞
y 1+α f (y)
1
x ρ−2 dxdzdy
y
z β f (z)y ρ−min{γ −β,1}−1 z min{γ −β,1} dzdy
1
R
+C
∞
y 1+β f (y)
1
z α f (z)y ρ−min{γ −α,1}−1 z min{γ −α,1} dzdy.
1
Using that − min{A, B} = max{−A, −B} together with α + β = λ, the right-hand side can be rearranged such that we obtain
R
1
1
∞
≤C
y+z
y K (y, z) f (y) f (z)
x ρ−2 dxdzdy
y R
y ρ+max{λ−γ ,α−1} f (y)dy
1
+C
∞
z min{γ ,1+β} f (z)dz
1 R
y ρ+max{λ−γ ,β−1} f (y)dy
1
∞
z min{γ ,1+α} f (z)dz.
1
∞ γ From (12) we immediately obtain that 1 z f (z)dz ≤ C for all γ ≤ γ . Moreover, we have ρ+max{λ−γ ,α−1} ρ+max{λ−γ ,β−1} y ≤y for all y ≥ 1 since α ≤ β. Thus, we can estimate the right-hand side further to get 1
R
∞
y+z
y K (y, z) f (y) f (z)
1
x ρ−2 dxdzdy ≤ C
y
R
y ρ+max{λ−γ ,β−1} f (y)dy. (24)
1
To simplify the notation we introduce κ := min{γ − λ, 1 − β} = − max{λ − γ , β − 1} and note that the conditions λ < γ and β < 1 ensure that κ > 0. Thus, summarising (18)–(24) we find R
ρ−1
R
y f (y)dy ≤ C + C
0
R
y ρ−κ f (y)dy.
(25)
1
Now, we have to distinguish whether ρ − κ ≤ γ or ρ − κ > γ . In the first case, we can argue directly since due to (12) the right-hand side of (25) is bounded and we thus find R
ρ−1
R
y f (y)dy ≤ C or equivalently
0
R
y f (y)dy ≤ C R 1−ρ ,
0
which shows the claim in this case. On the other hand, if ρ − κ > γ we have to iterate the previous estimate. Precisely, we can further estimate the right-hand side of (25) together with (12) to get R ρ−1
0
R
y f (y)dy ≤ C + C
R
y ρ−κ−γ y γ f (y)dy ≤ C + C R ρ−κ−γ .
1
123
1230
S. Throm
In the last step we used that ρ − κ > γ , i.e. ρ − κ − γ > 0. Since we also assume R ≥ 1, the first constant on the right-hand side can be estimated by the second term and we further conclude R y f (y)dy ≤ C R 1−κ−γ for R ≥ 1. 0
By means of a dyadic argument we deduce ∞ ∞ 2n+1 x ν f (x)dx = x ν−1 x f (x)dx n n=0 2
1
≤ C max{1, 2ν−1 }
∞
2n(ν−1) 2(n+1)(1−κ−γ )
n=0
≤ Cν
∞
2n(ν−κ−γ ) ≤ Cν (κ, γ )
for all ν < κ + γ .
(26)
n=0
If we recall (25) we again have to distinguish two cases, namely ρ − κ < κ + γ and ρ − κ ≥ κ + γ . If the first inequality holds, the claim follows since then the right-hand side of (25) is uniformly bounded due to (26). On the other hand, if the second case applies, we have to iterate further. Precisely, due to (26) and Lemma 2 we obtain from (25) for each γ1 < κ + γ that R R y f (y)dx ≤ C + y ρ−κ−γ1 y γ1 f (y)dy ≤ Cγ1 R ρ−κ−γ1 R ρ−1 0
1
for R ≥ 1. Similarly as in (26), a dyadic argument gives ∞ x ν f (x)dx ≤ Cν (γ , κ) for all ν < 2κ + γ . 1
Continuing this procedure iteratively, due to κ > 0, we obtain after a finite number of steps that ∞ x ν f (x)dx ≤ Cν for all ν < nκ + γ 1
for some n ∈ N with ρ − κ < nκ + γ . Then the claim follows as above, since the right-hand side of (25) can be bounded uniformly with respect to R.
3.3 Proof of Proposition 2 We are now prepared to give the proof of Proposition 2. Proof of Proposition 2 According to Proposition 1 there exists for each self-similar profile R f a constant C f > 0 such that 0 x f (x)dx ≤ C f R 1−ρ for all R > 0. R For a given self-similar profile f we define M f (R) := 0 x f (x)dx and note that due to the non-negativity of f it follows immediately from (11) that M f (R) ≥
1−ρ M(R) or equivalently ∂ R R ρ−1 M f (R) ≥ 0. R
Thus, the function R → R ρ−1 M f (R) is non-decreasing and according to Proposition 1 we also have R ρ−1 M f (R) ≤ C f uniformly with respect to R.
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Tail Behaviour of Self-Similar Profiles with Infinite Mass…
1231
If we consider now the rescaling f a (x) = a 1+λ f (ax) one directly verifies that f a satisfies λ−ρ R 1−ρ for all R > 0 and all a > 0, i.e. R ρ−1 M (R) ≤ a λ−ρ C . fa f 0 x f a (x)dx ≤ C f a Moreover, since f ≥ 0 and f ≡ 0 according to Definition 1 and since R → R ρ−1 M fa (R) is non-decreasing, there exists the limit L fa := lim R→∞ R ρ−1 M fa (R) and L fa > 0. Thus, we can fix the parameter a > 0 such that R
L fa = lim R ρ−1 M fa (R) = 1. R→∞
R
This then already shows that 0 y f a (y)dy ∼ R 1−ρ for R → ∞. On the other hand, the monotonicity of R → R ρ−1 M fa (R) together with the choice of a immediately implies R
R ρ−1 M fa (R) ≤ 1 or equivalently 0 y f a (y)dy ≤ R 1−ρ , which finishes the proof. Remark 11 From now on we normalise all self-similar profiles according to Proposition 2.
3.4 Moment Estimates As a consequence of Proposition 1 and Lemma 1 we provide now several moment estimates which we will use frequently in the following. Lemma 4 Let f be a self-similar profile according to the normalisation given by Proposition 2 and let ν > 0. Then we have the estimates ∞ y χ f (y)dy ≤ C x χ −ρ for all χ < ρ and for all x > 0, (27) x ∞ y χ f (y)dy ≤ C x χ −λ for all χ < λ and for all x > 0, (28) x x y χ f (y)dy ≤ C x χ −λ for all χ > λ and for all x > 0, (29) 0 x y χ f (y)dy ≤ C x max{χ −ρ,1−ρ} for all χ > λ and for all x ≥ 1, (30) 0 x y χ f (y)dy ≤ Cν x max{χ −ρ+ν,0} for all χ > λ and for all x ≥ 1, (31) 0 ∞ y χ f (y)dy < C for all χ ∈ (λ, ρ). (32) 0
These estimates remain true also without the rescaling of Proposition 2 while then the constants depend on f . Proof These estimates follow easily from Proposition 2 and Lemma ∞1 together with a dyadic decomposition. More precisely, to show (27), we split the integral x (· · · )dy and use Proposition 2 to get ∞ ∞ 2n+1 x y χ f (y)dy = y χ −1 y f (y)dy x
n n=0 2 x
≤ 2max{χ −1,0}
∞
2(χ −1)n x χ −1
2n+1 x
y f (y)dy
0
n=0
≤ 2max{χ −ρ,1−ρ} x χ −ρ
∞
2(χ −ρ)n = C x χ −ρ .
n=0
123
1232
S. Throm
The sum in the last step is finite due to the condition χ < ρ. Using Lemma 1 instead of Proposition 2 we similarly obtain (28) and (29). The estimate (30) follows from (29) together with Proposition 2. Precisely, for x ≥ 1 we have 1 x x y χ f (y)dy = y χ f (y)dy + y χ −1 y f (y)dy 0
0
1
≤ C + C x max{χ −1,0} x 1−ρ ≤ C x max{χ −ρ,1−ρ} . In the last step we additionally used that 1 ≤ x max{χ −ρ,1−ρ} for x ≥ 1 since the exponent is non-negative. To prove (31), we deduce from (27) and (29) that x 1 x y χ f (y)dy = y χ f (y)dy + y χ −ρ+ν y ρ−ν f (y)dy 0 0 1 ∞ max{χ −ρ+ν,0} y ρ−ν f (y)dy ≤ Cν x max{χ −ρ+ν,0} . ≤C+x To obtain (32) we just use the splitting with (27) and (29).
∞ 0
1
(· · · )dy =
1 0
(· · · )dy +
∞ 1
(· · · )dy together
4 Asymptotic Tail Behaviour In this section we will provide the proof of Theorem 1. The key for this will be the following result which states that each self-similar profile decays at least as fast as x −1−ρ for large values of x. Proposition 3 Let f be a self-similar profile. There exist constants C, R∗ > 0 such that f (x) ≤ C R −1−ρ
if x ∈ [R, 2R]
(33)
for all R ≥ R∗ . In particular this means that f (x) ≤ 21+ρ C x −1−ρ for x ≥ R∗ . Furthermore, there exists δ > 0 such that x ∞ I [ f ] := y K (y, z) f (y) f (z)dzdy ≤ C x 1−ρ−δ (34) 0
x−y
if x ≥ 2R∗ . The proof of this result has to be split into three parts depending on the values of the exponents α and β. The main ingredient for the proof of Proposition 3 will be the following variant of Young’s inequality which can be found for example as Eq. (72) in [15]. Lemma 5 Let d, D > 0 be constants with d < D and p, q, r ∈ [1, ∞] with 1/ p + 1/q = 1/r + 1. If h ∈ L q (d/2, D) and g ∈ L p (0, D) then x h(y)g(x − y)dy ≤ h L q (d/2,D) g L p (0,D) . x/2
L r (d,D)
In [15] this result has been used qualitatively to show that self-similar profiles are locally bounded. We will proceed here similarly but we apply this estimate in a more quantitative way, i.e. we choose d and D as multiples of R and estimate precisely the dependence on R of the appearing constants.
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Tail Behaviour of Self-Similar Profiles with Infinite Mass…
1233
4.1 The Case α ≤ β < 0 We first consider the situation where the parameters α and β satisfy α ≤ β < 0. This is the easiest case since the statement follows rather directly, i.e. an iteration argument is not necessary. Proof of Proposition 3 for α ≤ β < 0. We first split the integral I [ f ] as x/2 ∞ x ∞ I [ f ](x) = (· · · )dydz + (· · · )dzdy. 0
x−y
x/2
(35)
x−y
Together with the upper bound on K in (4) this yields the estimate x/2 ∞ 1+α β z + y 1+β z α f (y) f (z)dzdy y I [ f ](x) ≤ C 0
+C
x
x/2 ∞
y 1+α z β + y 1+β z α f (y) f (z)dzdy.
(36)
x/2 0
From Lemma 4 (i.e. (27) and (32)) we recall that ∞ α α−ρ z f (z)dz ≤ C x and x/2
as well as
∞
z β f (z)dz ≤ C x β−ρ
(37)
z β f (z)dz ≤ C.
(38)
x/2
∞
α
z f (z)dz ≤ C
and
0
∞
0
Note that for the last two estimates it is essential that we have α, β < 0 to get α, β > λ which allows to apply (32). Using (37) and (38) in (36) we find x x/2 β−ρ 1+α 1+α x y I [ f ](x) ≤ C y + x α−ρ y 1+β f (y)dy + C + y 1+β f (y)dy. 0
x/2
(39) We next note that due to λ = α + β and α ≤ β < 0 we get for x ≥ 2 from Lemma 4 (i.e. (30)) that x/2 x/2 y 1+α f (y)dy ≤ C x 1−ρ and y 1+β f (y)dy ≤ C x 1−ρ . 0
x
0
x
Moreover, we have x/2 f (y)dy = x/2 yy f (y)dy ≤ C x −1 x 1−ρ = C x −ρ . If we use these estimates in (39) it follows I [ f ](x) ≤ C x 1+β−2ρ + x 1+α−2ρ + x 1+α−ρ + x 1+β−ρ ≤ C x 1−ρ+β . (40) Note that in the last step we used that x ≥ 2 and α ≤ β. This then already shows (34) with δ = −β which is positive by the assumption on β. To conclude the proof of the Proposition for α ≤ β < 0 we use that (11) reads as x f (x) = (1 − ρ)x −2 y f (y)dy + x −2 I [ f ](x). 0
123
1234
S. Throm
Thus, (40) and Proposition 1 imply f (x) ≤ C x −1−ρ + C x −1−ρ+β ≤ C x −1−ρ
for x ≥ 2
since β < 0. The estimate (33) can then be derived immediately for R∗ = 2.
4.2 The Case 0 < α ≤ β We next consider the case where both exponents α and β are positive and the general strategy is similar to that one in Sect. 4.1, i.e. we split the integral I [ f ] and estimate the terms separately. However, now we cannot argue that directly but we have to use some iteration argument based on Lemma 5 to conclude. Proof of Proposition 3 For 0 < α ≤ β. We recall the splitting (35) which yields together with (4) that
x/2 ∞
I [ f ](x) ≤ C 0
+C
y 1+α z β + y 1+β z α f (y) f (z)dzdy
x/2 x
∞
y 1+α f (y)
x/2
z β f (z)dzdy + C
x−y
x
y 1+β f (y)
x/2
∞
z α f (z)dzdy.
x−y
(41) We recall from Lemma 4 (i.e. (27)) that we have for all x > 0 that ∞ ∞ z χ f (z)dz ≤ C x χ −ρ and z χ f (z)dz ≤ C(x − y)χ −ρ x/2
if χ < ρ.
x−y
If we use these two estimates with χ = α and χ = β we deduce from (41) that I [ f ](x) ≤ C 0
x/2
+C
x
y 1+α x β−ρ + y 1+β x α−ρ f (y)dy x y 1+α (x − y)β−ρ f (y)dy + C y 1+β (x − y)α−ρ f (y)dy.
x/2
x/2
Since 0 < α ≤ β we can further estimate this together with Proposition 1 to get x x 1+λ−2ρ 1+α β−ρ 1+β +C x f (y)(x − y) dy + C x f (y)(x − y)α−ρ dy. I [ f ](x) ≤ C x x/2
x/2
(42) We now fix two parameters p > 1 and n ∗ ∈ N such that p =1+
1 1 . with n ∗ large enough such that 1 < p < n∗ ρ−α
(43)
The latter condition can be achieved since 0 < α ≤ β < ρ < 1 and it implies in particular that p < 1/(ρ − β). In terms of I [ f ] equation (11) reads x y f (y)dy + x −2 I [ f ](x). f (x) = (1 − ρ)x −2 0
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Tail Behaviour of Self-Similar Profiles with Infinite Mass…
1235
Thus, for n ≤ n ∗ we can estimate the L r -norm of f as 2n∗ +1 R f L r (2n R,2n∗ +1 R) ≤ C x −2r 2n R
+
x
0
2n ∗ +1 R
x
−2r
r 1/r y f (y)dy dx 1/r
|I [ f ](x)| dx r
2n R
.
Together with (42) and Proposition 1 we further obtain f L r (2n R,2n∗ +1 R) 2n∗ +1 R 1/r −r (1+ρ) ≤C x dx +C 2n R
+C
2n R
+C
2n ∗ +1 R
2n ∗ +1 R
2n R
x r (α−1) r (β−1) x
2n ∗ +1 R
x
r (λ−2ρ−1)
1/r dx
2n R
x x/2 x
r 1/r f (y)(x − y)β−ρ dy dx α−ρ
f (y)(x − y)
x/2
r 1/r dy dx .
Computing the first two integrals explicitly and using Lemma 5 we continue to estimate 1
1
f L r (2n R,2n∗ +1 R) ≤ C R −1−ρ+ r + C R λ−2ρ−1+ r
+ C R α−1 f L q (2n−1 R,2n∗ +1 R) z β−ρ L p (0,2n∗ +1 R) + C R β−1 f L q (2n−1 R,2n∗ +1 R) z α−ρ L p (0,2n∗ +1 R) .
(44)
Here p, q and r have to satisfy 1/ p + 1/q = 1 + 1/r . Next, we recall that ρ > λ which yields 1
1
R λ−2ρ−1+ r ≤ R −1−ρ+ r
for all R ≥ 1.
(45)
Moreover, since p ∈ (1, 1/(ρ − α)) by (43), and thus in particular p < 1/(ρ − β), we note that β−ρ 1 1 z p n +1 ≤ C R β−ρ+ p and z α−ρ p n +1 ≤ C R α−ρ+ p . (46) L (0,2 ∗ R) L (0,2 ∗ R) Thus, combining (44)–(46) and recalling λ = α + β, it follows 1
f L r (2n R,2n∗ +1 R) ≤ C R −1−ρ+ r + C R λ−1−ρ+ p f L q (2n−1 R,2n∗ +1 R) . 1
Since λ < ρ by assumption, we can further estimate the second term on the right-hand side for R ≥ 1 to get 1
f L r (2n R,2n∗ +1 R) ≤ C R −1−ρ+ r + C R −1+ p f L q (2n−1 R,2n∗ +1 R) . 1
(47)
To conclude the proof for 0 < α ≤ β we will now iteratively use this estimate. For this, we first take n = 0, r0 = p and q0 = 1. Then as an immediate consequence of Proposition 1 it follows 2n∗ +1 R 1 f L 1 (R/2,2n∗ +1 R) = (48) x f (x)dx ≤ C R −ρ . x R/2 Plugging this estimate in (47) yields f L r0 (R,2n∗ +1 R) ≤ C R
−1−ρ+ r1
0
for all R ≥ 1.
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1236
S. Throm p 2− p .
In the next step we take n = 1 and q1 = r0 which requires to choose r1 = combining the previous estimate with (47) it follows f L r1 (2R,2n∗ +1 R) ≤ C R
−1−ρ+ r1
1
+ CR
−1+ 1p + r1 −1−ρ 0
= CR
−1−ρ+ r1
1
Thus,
.
Proceeding in this way, we obtain for general n ≤ n ∗ that 1
f L rn (2n R,2n∗ +1 R) ≤ C R −1−ρ+ rn
with rn =
p . n(1 − p) + 1
Thus, after a finite number of steps we get rn ∗ = ∞ and it follows f L ∞ (2n∗ R,2n∗ +1 R) ≤ C R −1−ρ
for all R ≥ 1.
This then proves the estimate (33) for 0 < α ≤ β if we choose R∗ = 2n ∗ . To conclude the proof of the Proposition, it remains to show (34) which is now an immediate consequence of (33) and (42). Precisely, we have for x ≥ 2R∗ that x x (x − y)β−ρ dy + C x β−ρ (x − y)α−ρ dy I [ f ](x) ≤ C x 1+λ−2ρ + C x α−ρ x/2
= Cx
1−ρ+λ−ρ
x/2
.
Thus, since x ≥ 1 the claim follows with δ = ρ − λ since λ < ρ by assumption.
4.3 The Case α ≤ 0 ≤ β It remains to prove Proposition 3 for α ≤ 0 ≤ β which is the most complicated case. The approach will be the same as in Sect. 4.2, i.e. we iteratively apply Lemma 5 to get an estimate for the L ∞ -norm of f . However, depending on the values of ρ, λ and β the first bound that we obtain might not be enough and we have to start a second iteration to improve the L ∞ -bound further to conclude. Proof of Proposition 3 For α ≤ 0 ≤ β. We first collect some estimates and fix some notation that we will need in the following, i.e. we take ν ∈ (0, min{ρ − λ, 1 − β}). We then have the estimates ∞ (x − y)−β−ν z α f (z)dz ≤ C (49) (1 + (x − y))ρ−α−β−ν x−y and
∞ x−y
z β f (z)dz ≤ C
(x − y)−ν . (1 + (x − y))ρ−β−ν
To see this, we note that Lemma 4 (i.e. (27) and (32)) yields ∞ ∞ z α f (z)dz ≤ (x − y)−β−ν z α+β+ν f (z)dz x−y
123
x−y
if (x − y) ≤ 1 −β−ν C ν ≤ (x − y) α+β+ν−ρ if (x − y) ≥ 1. Cν (x − y)
(50)
Tail Behaviour of Self-Similar Profiles with Infinite Mass…
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From this, (49) immediately follows, while for (50), we can argue similarly using the splitting z β = z −ν z β+ν . If we use now (49) and (50) together with the splitting in (41) and Lemma 4 (i.e. (27)), we can deduce that x/2 1+α β−ρ I [ f ](x) ≤ C x + y 1+β x α−ρ f (y)dy y 0
+ C x 1+β + C x 1+α
x x/2 x x/2
f (y)
(x − y)−β−ν dy (1 + (x − y)ρ−α−β−ν
f (y)
(x − y)−ν dy. (1 + (x − y))ρ−β−ν
(51)
Next, we use that from Lemma 4 (i.e. (30)) together with α ≤ 0 ≤ β < 1 we have for all x ≥ 2 that x/2 x/2 y 1+α f (y)dy ≤ C x 1−ρ and y 1+β f (y)dy ≤ C x 1+β−ρ . 0
0
With these estimates and the fact that α ≤ 0 ≤ β the first integral in (51) can be estimated for x ≥ 2 as x/2 1+α β−ρ y x + y 1+β x α−ρ f (y)dy ≤ C x 1+β−2ρ + x 1+α+β−2ρ ≤ C x 1+β−2ρ . 0
Thus, in summary we conclude from (51) for x ≥ 2 that x (x − y)−β−ν f (y) dy I [ f ](x) ≤ C x 1+β−2ρ + C x 1+β (1 + (x − y)ρ−α−β−ν x/2 x (x − y)−ν + C x 1+α f (y) dy. (1 + (x − y))ρ−β−ν x/2
(52)
We then note that (11) in terms of I [ f ] can be written as x −2 y f (y)dy + x −2 I [ f ](x). f (x) = (1 − ρ)x 0
Thus, for r ∈ [1, ∞) and for n ≤ n ∗ to be fixed later, we can compute f L r (2n+1 R,2n∗ +2 R) 2n∗ +2 R −2r ≤C x
r 1/r 2n∗ +2 R 1/r y f (y)dy dx +C x r (β−1−2ρ) dx 2n+1 R 0 2n+1 R r 1/r x 2n∗ +2 R (x − y)−β−ν dx +C x r (β−1) f (y) dy ρ−α−β−ν (1 + (x − y)) 2n+1 R x/2 x 2n∗ +2 R 1/r r (x − y)−ν r (α−1) +C x f (y) dy dx . (1 + (x − y))ρ−β−ν 2n+1 R x/2 x
Note that in contrast to Sect. 4.2, we have to consider here the intervals (2n+1 R, 2n ∗ +2 R) instead of (2n R, 2n ∗ +1 R) because we exploited x ≥ 2 to derive (52) for example and the x latter is satisfied if x ∈ (2n+1 R, 2n ∗ +2 R) with R ≥ 1 and n ≥ 0. Since 0 y f (y)dy ≤ C x 1−ρ according to Proposition 1, we obtain together with Lemma 5 for 1/ p + 1/q = 1/r + 1 that
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1238
S. Throm 1
1
f L r (2n+1 R,2n∗ +2 R) ≤ C R −1−ρ+ r + C R β−1−2ρ+ r
z −β−ν f L q (2n R,2n∗ +2 R) +CR (1 + z)ρ−α−β−ν L p (0,2n∗ +2 R) z −ν α−1 f L q (2n R,2n∗ +2 R) +CR . (1 + z)ρ−β−ν p n +2 β−1
L (0,2
∗
R)
Since ρ > β by assumption and we also assume R ≥ 1, we have R β−ρ ≤ 1 such that we finally obtain f L r (2n+1 R,2n∗ +2 R)
1 ≤ C R −1−ρ+ r + C R β−1 f L q (2n R,2n∗ +2 R)
z −β−ν p n +2 ρ−α−β−ν (1 + z) L (0,2 ∗ R) −ν z + C R α−1 f L q (2n R,2n∗ +2 R) (53) (1 + z)ρ−β−ν p n +2 . L (0,2 ∗ R)
We now fix the parameter p > 1 as p = 1 + 1/n ∗ with n ∗ ∈ N large enough such that p <
1 . β +ν
(54)
Note that from the choice of ν, we have 1/(β + ν) > 1. Moreover, since β ≥ 0 we have in particular that pν < 1. With this, we find for R ≥ 1 that z −β−ν (1 + z)ρ−α−β−ν
L p (0,2n ∗ +2 R)
≤C
1
z
− p(β+ν)
0
dz +
1/ p
2n ∗ +2 R
z
p(α−ρ)
dz
1
α−ρ+ 1p ≤ C R max{α−ρ+1/ p,0} . ≤C 1+ R
(55)
Similarly, we get z −ν (1 + z)ρ−β−ν
L p (0,2n ∗ +2 R)
≤ C R max{β−ρ+1/ p,0} .
(56)
Combining (53), (55) and (56) and R ≥ 1 we find 1
f L r (2n+1 R,2n∗ +2 R) ≤ C R −1−ρ+ r + C R max{λ−ρ−1+1/ p,β−1} + R max{λ−ρ−1+1/ p,α−1} f L q (2n R,2n∗ +2 R) . We now define χ := max{λ − ρ − 1 + 1/ p, β − 1} and note that χ < 0 since β < 1, p > 1 and ρ > λ. Moreover, we get R max{λ−ρ−1+1/ p,α−1} ≤ R χ because α ≤ β and R ≥ 1. Thus, it further follows 1
f L r (2n+1 R,2n∗ +2 R) ≤ C R −1−ρ+ r + C R χ f L q (2n R,2n∗ +2 R) .
(57)
We now use this estimate iteratively until we reach r = ∞. Thus, as a first step, we choose n = 0, q0 = 1 and r0 = p and note that analogously to (48), Proposition 1 yields f L 1 (R,2n∗ +2 R) ≤ C R −ρ . Thus, we obtain from (57) that f L r0 (2R,2n∗ +2 R) ≤ C R
123
−1−ρ+ r1
0
+ C R χ −ρ .
(58)
Tail Behaviour of Self-Similar Profiles with Infinite Mass…
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In the next step we take n = 1 and q1 = r0 which requires r1 = p/(2 − p). We then find together with (57) and (58) that f L r1 (22 R,2n∗ +2 R) ≤ C R
−1−ρ+ r1
+ CR
≤ CR
−1−ρ+ r1
+ C R 2χ −ρ .
1
1
χ −1−ρ+ r1
0
+ C R 2χ −ρ
In the last step we used that χ + 1 − 1/ p ≤ 0 which yields χ + 1/r0 ≤ 1/r1 . If we iterate this procedure we obtain for n ≤ n ∗ and qn = rn−1 that 1
f L rn (2n+1 R,2n∗ +2 R) ≤ C R −1−ρ+ rn + C R (n+1)χ −ρ . p with rn = 1+n(1− p) . From the choice of p in (54) we have n ∗ = 1/( p − 1) and thus rn ∗ = ∞ which yields
f L ∞ (2n∗ +1 R,2n∗ +2 R) ≤ C R −1−ρ + C R (n ∗ +1)χ −ρ .
(59)
If (n ∗ + 1)χ ≤ −1 this then already shows the estimate (33) for R∗ = 2n ∗ +1 . On the other hand, if (n ∗ + 1)χ > −1 we have to iterate further. Precisely, we fix some constant m ∈ N such that (n ∗ + 1)χ + m max{λ − ρ, β − 1} ≤ −1. Then, we repeat the previous procedure with n˜ ∗ = n ∗ + m instead of n ∗ which gives f L ∞ (2n∗ +1 R,2n˜ ∗ +2 R) ≤ C R −1−ρ + C R (n ∗ +1)χ −ρ .
(60)
Moreover, we find that 1 2n˜ ∗ +2 R z −β−ν −β−ν α−ρ ≤ z dz + z dz (1 + z)ρ−α−β−ν 1 n˜ +2 0 1 L (0,2 ∗ R) ≤ CR Similarly, we get
z −ν (1 + z)ρ−β−ν
max{1+α−ρ,0}
L 1 (0,2n˜ ∗ +2 R)
(61)
.
≤ C R 1+β−ρ .
(62)
Thus, if we return to (53) and take p = 1 and r = q = ∞ this yields together with (61) and (62) that f L ∞ (2n+1 R,2n˜ ∗ +2 R) ≤ C R −1−ρ + C R max{λ−ρ,β−1} + R λ−ρ f L ∞ (2n R,2n˜ ∗ +2 R) ≤ C R −1−ρ + C R max{λ−ρ,β−1} f L ∞ (2n R,2n˜ ∗ +2 R) . If we now iterate this estimate m times and use (60), it follows f L ∞ (2n∗ +m+1 R,2n∗ +m+2 R) ≤ C R −1−ρ + C R m max{λ−ρ,β−1}+(n ∗ +1)χ −ρ ≤ C R −1−ρ . The last step follows from the choice of m together with the assumption R ≥ 1. The estimate (33) then directly follows with R∗ = 2n ∗ +m+1 . The upper bound (34) on I [ f ] is then an immediate consequence. Precisely, since the estimate (33) directly implies f (x) ≤ C x −1−ρ for x ≥ R∗ we deduce from (52) in combination with (61) and (62) that I [ f ](x) ≤ C x 1+β−2ρ + C x 1+β+max{1+α−ρ,0} + x 1+α+1+β−ρ x −1−ρ β−ρ =C x + x max{λ−ρ,β−1} + x λ−ρ x 1−ρ .
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S. Throm
Thus, the estimate (34) holds with δ = ρ − β since λ = α + β ≤ β < ρ < 1.
4.4 Proof of Theorem 1 The proof of Theorem 1 follows now easily from Proposition 3. Proof of Theorem 1 To show the continuity of self-similar profiles, one can proceed in the same way as in Lemma 4.2 of [15], i.e. one first proves that f ∈ L ∞ loc ((0, ∞)) using similar estimates as in the proof of Proposition 3. Once this is established, it is straightforward to show that f is even continuous on (0, ∞). Since the corresponding estimates are quite similar, we do not give further details on this here but only refer to [15] (see also Lemma 3.3 in [12]). Thus, it only remains to show how the asymptotic behaviour of self-similar profiles can be obtained from Propositions 2 and 3 (see also Proposition 3.6 in [12] and Lemma 4.3 in [15]). We recall that in terms of I [ f ] the equation (11) reads x x 2 f (x) = (1 − ρ) y f (y)dy + I [ f ](x). 0
This can be rewritten as x 1+ρ f (x) = x ρ−1 1−ρ
x
y f (y)dy +
0
1 x ρ−1 I [ f ](x). 1−ρ
(63)
ρ−1 I [ f ](x) → 0 as x → ∞ while due to the rescaling Due to Proposition 3 we have that x x ρ−1 in Proposition 2 we have x 0 y f (y)dy → 1 for x → ∞. Thus, taking the limit x → ∞ in (63) it follows
x 1+ρ f (x) −→ 1 1−ρ This then finishes the proof.
for x −→ ∞.
Acknowledgements The author thanks the two anonymous reviewers for their valuable comments which helped to improve this article. Funding This work has been supported through the CRC 1060 The mathematics of emergent effects at the University of Bonn that is funded through the German Science Foundation (DFG). Moreover, partial support through a Lichtenberg Professorship Grant of the VolkswagenStiftung awarded to Christian Kühn is acknowledged.
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