Lp Theory for the Multidimensional Aggregation Equation
Lp Theory for the Multidimensional Aggregation Equation
Lp Theory for the Multidimensional Aggregation Equation
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(ii) If α ≥ 2 <strong>the</strong>n <strong>the</strong> aggregation equation is globally well posed in P 2 (R d ) ∩ L p (R d ) <strong>for</strong> every p > 1.As a consequence we have existence and uniqueness <strong>for</strong> all potentials which are less singular than <strong>the</strong>Newtonian potential K(x) = |x| 2−d at <strong>the</strong> origin. In two dimensions this includes potentials with cuspsuch as K(x) = |x| 1/2 . In three dimensions this includes potentials that blow up such as K(x) = |x| −1/2 .From [5, 16] we know that <strong>the</strong> support of compactly supported solutions shrinks to a point in finite time,proving <strong>the</strong> second assertion in point (i) above. The first part of (i) and statement (ii) are direct corollaryof Theorem 1, Definition 2 and <strong>the</strong> fact that α ≥ 2 implies ∆K bounded.In <strong>the</strong> case where α = 1, i.e. K(x) ∼ |x| as |x| → 0, <strong>the</strong> previous Theorem gives local well posedness inP 2 (R d ) ∩ L p (R d ) <strong>for</strong> all p > p s =dd−1. The next Theorem shows that it is not possible to obtain local wellposedness in P 2 (R d ) ∩ L p (R d ) <strong>for</strong> p < p s =dd−1 .Theorem 4 (Critical p-exponent to generate instantaneous mass concentration). Suppose K(x) = |x| in aneighborhood of <strong>the</strong> origin, and suppose ∇K is compactly supported (or decays exponentially fast at infinity).Then, <strong>for</strong> any p < p s =dd−1 , <strong>the</strong>re exists initial data in P 2(R d ) ∩ L p (R d ) <strong>for</strong> which a delta Dirac appearsinstantaneously in <strong>the</strong> measure solution.In order to make sense of <strong>the</strong> statement of <strong>the</strong> previous Theorem, we need a concept of measure solution.The potentials K(x) = |x| is semi-convex, i.e. <strong>the</strong>re exist λ ∈ R such that K(x) − λ 2 |x|2 is convex. In[16], Carrillo et al. prove global well-posedness in P 2 (R d ) of <strong>the</strong> aggregation equation with semi-convexpotentials. The solutions in [16] are weak measure solutions - <strong>the</strong>y are not necessarily absolutely continuouswith respect to <strong>the</strong> Lebesgue measure. Theorems 3 and 4 give a sharp condition on <strong>the</strong> initial data in order<strong>for</strong> <strong>the</strong> solution to stay absolutely continuous with respect to <strong>the</strong> Lebesgue measure <strong>for</strong> short time. Theorem4 is proven in Section 4.Finally, in section 5 we consider a class of potential that will be referred to as <strong>the</strong> class of naturalpotentials. A potential is said to be natural if it satisfies thata) it is a radially symmetric potential, i.e.: K(x) = k(|x|),b) it is smooth away from <strong>the</strong> origin and it’s singularity at <strong>the</strong> origin is not worse than Lipschitz,c) it doesn’t exhibit pathological oscillation at <strong>the</strong> origin,d) its derivatives decay fast enough at infinity.All <strong>the</strong>se conditions will be more rigorously stated later. It will be shown that <strong>the</strong> gradient of naturalpotentials automatically belongs to W 1,q <strong>for</strong> q < d, <strong>the</strong>re<strong>for</strong>e, using <strong>the</strong> results from <strong>the</strong> sections 2 and 3,we have local existence and uniqueness in P 2 (R d ) ∩ L p (R d ), p >dd−1 .A natural potential is said to be repulsive in <strong>the</strong> short range if it has a local maximum at <strong>the</strong> originand it is said to be attractive in <strong>the</strong> short range if it has a local minimum at <strong>the</strong> origin. If <strong>the</strong> maximum(respectively minimum) is strict, <strong>the</strong> natural potential is said to be strictly repulsive (respectively strictlyattractive) at <strong>the</strong> origin. The main <strong>the</strong>orem of section 5 is <strong>the</strong> following:Theorem 5 (Osgood condition <strong>for</strong> global well posedness). Suppose K is a natural potential.(i) If K is repulsive in <strong>the</strong> short range, <strong>the</strong>n <strong>the</strong> aggregation equation is globally well posed in P 2 (R d ) ∩L p (R d ), p > d/(d − 1).(ii) If K is strictly attractive in <strong>the</strong> short range, <strong>the</strong> aggregation equation is globally well posed in P 2 (R d ) ∩L p (R d ), p > d/(d − 1), if and only ifr ↦→ 1k ′ (r)is not integrable at 0. (1.9)3
By globally well posed in P 2 (R d ) ∩ L p (R d ), we mean that <strong>for</strong> any initial data in P 2 (R d ) ∩ L p (R d ) <strong>the</strong>unique solution of <strong>the</strong> aggregation equation will exist <strong>for</strong> all time and will stay in P 2 (R d ) ∩ L p (R d ) <strong>for</strong> alltime.Notice that <strong>the</strong> exponent d/(d − 1) is not sharp in this <strong>the</strong>orem.Condition (1.9) will be refered as <strong>the</strong> Osgood condition. It is easy to understand why <strong>the</strong> Osgoodcondition is relevant while studying blowup: <strong>the</strong> quantityT (d) =∫ d0drk ′ (r)can be thought as <strong>the</strong> amount of time it takes <strong>for</strong> a particle obeying <strong>the</strong> ODE Ẋ = −∇K(X) to reach <strong>the</strong>origin if it starts at a distance d from it. For a potential satisfying <strong>the</strong> Osgood condition, T (d) = +∞, whichmeans that <strong>the</strong> particle can not reach <strong>the</strong> origin in finite time. The Osgood condition was already shownin [5] to be necessary and sufficient <strong>for</strong> global well posedness of L ∞ -solutions. Extension to L p requires L pestimates ra<strong>the</strong>r than L ∞ estimates. See also [43] <strong>for</strong> an example of <strong>the</strong> use of <strong>the</strong> Osgood condition in <strong>the</strong>context of <strong>the</strong> Euler equations <strong>for</strong> incompressible fluid.The “only if” part of statement (ii) was proven in [5] and [16]. In <strong>the</strong>se two works it was shown thatif (1.9) is not satisfied, <strong>the</strong>n compactly supported solutions will collapse into a point mass – and <strong>the</strong>re<strong>for</strong>eleave L p – in finite time. In section 5 we prove statement (i) and <strong>the</strong> ’if’ part of statement (ii).2 Existence of L p -solutionsIn this section we show that if <strong>the</strong> interaction potential satisfies∇K ∈ W 1,q (R d ), 1 < q < +∞, (2.10)and if <strong>the</strong> initial data is nonnegative and belongs to L p (R d ) (p and q are Hölder conjugates) <strong>the</strong>n <strong>the</strong>re existsa solution to <strong>the</strong> aggregation equation. Moreover, ei<strong>the</strong>r this solution exists <strong>for</strong> all times, or its L p -normblows up in finite time. The duality between L p and L q guarantees enough smoothness in <strong>the</strong> velocity fieldv = −∇K ∗ u to define characteristics. We use <strong>the</strong> characteristics to construct a solution. The argumentis inspired by <strong>the</strong> existence of L ∞ solutions of <strong>the</strong> incompressible 2D Euler equations by Yudovich [44] andof L ∞ solutions of <strong>the</strong> aggregation equation [4]. Section 3 proves uniqueness provided u ∈ P 2 . We prove inTheorem 18 of <strong>the</strong> present section that if u 0 ∈ P 2 <strong>the</strong>n <strong>the</strong> solution stays in P 2 . Finally we prove that if inaddition to (2.10), we haveess sup ∆K < +∞, (2.11)<strong>the</strong>n <strong>the</strong> solution constructed exists <strong>for</strong> all time.Most of <strong>the</strong> section is devoted to <strong>the</strong> proof of <strong>the</strong> following <strong>the</strong>orem:Theorem 6 (Local existence). Consider 1 < q < ∞ and p its Hölder conjugate. Suppose ∇K ∈ W 1,q (R d )and suppose u 0 ∈ L p (R d ) is nonnegative. Then <strong>the</strong>re exists a time T ∗ > 0 and a nonnegative functionu ∈ C([0, T ∗ ], L p (R d )) ∩ C 1 ([0, T ∗ ], W −1,p (R d )) such thatu ′ (t) + div ( u(t) v(t) ) = 0 ∀t ∈ [0, T ∗ ], (2.12)Moreover <strong>the</strong> function t → ‖u(t)‖ p Lis differentiable and satisfiesp∫ddt {‖u(t)‖p L p} = −(p − 1)The choice of <strong>the</strong> spacev(t) = −u(t) ∗ ∇K ∀t ∈ [0, T ∗ ], (2.13)u(0) = u 0 . (2.14)R d u(t, x) p div v(t, x) dx ∀t ∈ [0, T ∗ ]. (2.15)Y p := C([0, T ∗ ], L p (R d )) ∩ C 1 ([0, T ∗ ], W −1,p (R d ))is motivated by <strong>the</strong> fact that, if u ∈ Y p and ∇K ∈ W 1,q , <strong>the</strong>n <strong>the</strong> velocity field is automatically C 1 in spaceand time:4
Lemma 7. Consider 1 < q < ∞ and p its Hölder conjugate. If ∇K ∈ W 1,q (R d ) and u ∈ Y p <strong>the</strong>nu ∗ ∇K ∈ C 1 ( [0, T ∗ ] × R d) and ‖u ∗ ∇K‖ C1 ([0,T ∗ ]×R d ) ≤ ‖∇K‖ W 1,q (R d ) ‖u‖ Y p(2.16)where <strong>the</strong> norm ‖·‖ C 1 ([0,T ∗ ]×R d ) and ‖·‖ Y pare defined by∣ ∣ ∣ ∣∣∣‖v‖ C1 ([0,T ∗ ]×R d ) = sup∂vd∑ ∣∣∣|v| + sup[0,T ∗ ]×R d [0,T ∗ ]×R ∂t ∣ + ∂v ∣∣∣sup , (2.17)d [0,T ∗ ]×R ∂x d i‖u‖ Yp= sup ‖u(t)‖ <strong>Lp</strong> (R d ) +t∈[0,T ∗ ]supt∈[0,T ∗ ]i=1‖u ′ (t)‖ W −1,p (R d ) . (2.18)Proof. Recall that <strong>the</strong> convolution between a L p -function and a L q -function is continuous and sup x∈R d |f ∗g(x)| ≤ ‖f‖ L p ‖g‖ L q There<strong>for</strong>e, since ∇K and ∇K xi are in L q , <strong>the</strong> mappingf ↦→ ∇K ∗ fis a bounded linear trans<strong>for</strong>mation from L p (R d ) to C 1 (R d ), where C 1 (R d ) is endowed with <strong>the</strong> norm‖f‖ C 1= supx∈R d |f(x)| +d∑sup | ∂f (x)|.x∈R ∂x d iSince u ∈ C([0, T ∗ ], L p ) it is <strong>the</strong>n clear that u ∗ ∇K ∈ C([0, T ∗ ], C 1 ). In particular w(t, x) = (u(t) ∗ ∇K) (x)and ∂w∂x i(t, x) are continuous on [0, T ∗ ] × R d . Let us now show that ∂w∂t(t, x) exists and is continuous on[0, T ∗ ] × R d . Since u ′ (t) ∈ C([0, T ∗ ], W −1,p ) and ∇K ∈ W 1,q , we havei=1∂w∂t (t, x) = − (u′ (t) ∗ ∇K) (x) = −〈u ′ (t), τ x ∇K〉where 〈 , 〉 denote <strong>the</strong> pairing between <strong>the</strong> two dual spaces W −1,p (R d ) and W 1,q (R d ), and τ x denote <strong>the</strong>translation by x. Since x ↦→ τ x ∇K is a continuous mapping from R d to W 1,q it is clear that ∂w∂t(t, x) iscontinuous with respect to space. The continuity with respect to time come from <strong>the</strong> continuity of u ′ (t) withrespect to time. Inequality (2.16) is easily obtained.Remark 8. Let us point out that (2.12) indeed makes sense, when understood as an equality in W −1,p .Since v ∈ C([0, T ∗ ], C 1 (R d )) one can easily check that uv ∈ C([0, T ∗ ], L p (R d )). Also recall that <strong>the</strong> injectioni : L p (R d ) → W −1,p (R d ) and <strong>the</strong> differentiation ∂ xi : L p (R d ) → W −1,p (R d ) are bounded linear operators.There<strong>for</strong>e it is clear that both u and div(uv) belong to C([0, T ∗ ], W −1,p (R d )). <strong>Equation</strong> (2.12) has to beunderstand as an equality in W −1,p .The rest of this section is organized as follows. First we give <strong>the</strong> basic a priori estimates in subsection 2.1.Then, in subsection 2.2, we consider a mollified and cutted-off version of <strong>the</strong> aggregation equation <strong>for</strong> whichwe have global existence of smooth and compactly supported solutions. In subsection 2.3 we show that <strong>the</strong>characteristics of this approximate problem are uni<strong>for</strong>mly Lipschitz continuous on [0, T ∗ ]×R d , where T ∗ > 0is some finite time depending on ‖u 0 ‖ L p. In subsection 2.4 we pass to <strong>the</strong> limit in C([0, T ∗ ], L p ). To do thiswe need <strong>the</strong> uni<strong>for</strong>m Lipschitz bound on <strong>the</strong> characteristics toge<strong>the</strong>r with <strong>the</strong> fact that <strong>the</strong> translation byx, x ↦→ τ x u 0 , is a continuous mapping from R d to L p (R d ). In subsection 2.5 we prove three <strong>the</strong>orems. Wefirst prove continuation of solutions. We <strong>the</strong>n prove that L p -solutions which start in P 2 stay in P 2 as longas <strong>the</strong>y exist. And finally we prove global existence in <strong>the</strong> case where ∆K is bounded from above.2.1 A priori estimatesSuppose u ∈ Cc 1 ((0, T ) × R d ) is a nonnegative function which satisfies (2.12)-(2.13) in <strong>the</strong> classical sense.Suppose also that K ∈ Cc∞ (R d ). Integrating by part, we obtain that <strong>for</strong> any p ∈ (1, +∞):∫∫du(t, x) p dx = −(p − 1) u(t, x) p div v(t, x)dx ∀t ∈ (0, T ). (2.19)dt R d R d5
As a consequence we have:ddt ‖u(t)‖p L ≤ (p − 1)‖div v(t)‖ p L ∞‖u(t)‖p <strong>Lp</strong> ∀t ∈ (0, T ), (2.20)and by Hölder’s inequality:ddt ‖u(t)‖p L ≤ (p − 1)‖∆K‖ pL q‖u(t)‖p+1 L . (2.21)pWe now derive L ∞ estimates <strong>for</strong> <strong>the</strong> velocity field v = −∇K ∗ u and its derivatives. Hölder’s inequalityeasily givesSince ∂v∂t∂u= −∇K ∗∂t ∣|v(t, x)| ≤ ‖u(t)‖ L p‖∇K‖ L q ∀(t, x) ∈ (0, T ) × R d , (2.22)∥ ∂v j∣ (t, x)∂x i∣ ≤ ‖u(t)‖ L p∂ 2 K ∥∥∥L∥∀(t, x) ∈ (0, T ) × R d . (2.23)∂x i dx j q= ∇K ∗ div(uv) = ∆K ∗ uv we have∣∂v ∣∣∣∂t (t, x) ≤ ‖u(t)v(t)‖ L p‖∆K‖ L q ≤ ‖u(t)‖ L p‖v(t)‖ L ∞‖∆K‖ L q,which in light of (2.22) gives∣ ∂v ∣∣∣∣ ∂t (t, x) ≤ ‖u(t)‖ 2 L p‖∇K‖ L q‖∆K‖ L q ∀(t, x) ∈ (0, T ) × Rd . (2.24)2.2 Approximate smooth compactly supported solutionsIn this section we deal with a smooth version of equation (2.12)-(2.14). Suppose u 0 ∈ L p (R d ), 1 < p < +∞,and ∇K ∈ W 1,q (R d ). Consider <strong>the</strong> approximate problemu t + div(uv) = 0 in (0, +∞) × R d , (2.25)v = −∇K ɛ ∗ u in (0, +∞) × R d , (2.26)u(0) = u ɛ 0, (2.27)where K ɛ = J ɛ K, u ɛ 0 = J ɛ u 0 and J ɛ is an operator which mollifies and cuts-off, J ɛ f = (fM Rɛ ) ∗ η ɛ whereη ɛ (x) is a standard mollifier:η ɛ (x) = 1 ( x)∫ɛ d η , η ∈ Cc ∞ (R d ), η ≥ 0, η(x)dx = 1,ɛR dand M Rɛ (x) is a standard cut-off function: M Rɛ (x) = M( x R ɛ), R ɛ → ∞ as ɛ → 0,M ∈ C ∞ c (R d ),Let τ x denote <strong>the</strong> translation by x, i.e.:⎧⎨⎩M(x) = 1 if |x| ≤ 1,0 < M(x) < 1 if 1 < |x| < 2,M = 0 if 2 ≤ |x|.τ x f(y) := f(y − x).It is well known that given a fixed f ∈ L r (R d ), 1 < r < +∞, <strong>the</strong> mapping x ↦→ τ x f from R d to L r (R d )is uni<strong>for</strong>mly continuous. In (iv) of <strong>the</strong> next lemma we show a slightly stronger result which will be neededlater.Lemma 9 (Properties of J ɛ ). Suppose f ∈ L r (R d ), 1 < r < +∞, <strong>the</strong>n6
(i) J ɛ f ∈ C ∞ c (R d ),(ii) ‖J ɛ f‖ L r ≤ ‖f‖ L r,(iii) lim ɛ→0 ‖J ɛ f − f‖ L r = 0,(iv) The family of mappings x ↦→ τ x J ɛ f from R d to L r (R d ) is equicontinuous, i.e.: <strong>for</strong> each δ > 0, <strong>the</strong>re isa η > 0 independent of ɛ such that ‖τ x J ɛ f − τ y J ɛ f‖ L r ≤ δ if |x − y| ≤ η.Proof. Statements (i) and (ii) are obvious. If f is compactly supported, one can easily prove (iii) by notingthat fM Rɛ = f <strong>for</strong> ɛ small enough. If f is not compactly supported, (iii) is obtained by approximating f bya compactly supported function and by using (ii). Let us now turn to <strong>the</strong> proof of (iv). Using (ii) we obtain‖τ x J ɛ f − J ɛ f‖ L r ≤ ‖(τ x M Rɛ )(τ x f) − M Rɛ f)‖ L r≤ ‖τ x M Rɛ − M Rɛ ‖ L ∞‖τ x f‖ L r + ‖M Rɛ ‖ L ∞‖τ x f − f‖ L r.Because x ↦→ τ x f is continuous, <strong>the</strong> second term can be made as small as we want by choosing |x| smallenough. Since ‖τ x M Rɛ − M Rɛ ‖ L ∞ ≤ ‖∇M Rɛ ‖ L ∞|x| ≤ 1 R ɛ‖∇M‖ L ∞|x|, <strong>the</strong> first term can be made as smallas we want by choosing |x| small enough and independently of ɛ.Proposition 10 (Global existence of smooth compactly-supported approximates). Given ɛ, T > 0, <strong>the</strong>reexists a nonnegative function u ∈ C 1 c ((0, T ) × R d ) which satisfy (2.27) in <strong>the</strong> classical sense.Proof. Since u ɛ 0 and K ɛ belong to Cc∞function u ɛ satisfying(R d ), we can use <strong>the</strong>orem 3 p. 1961 of [28] to get <strong>the</strong> existence of au ɛ ∈ L ∞ (0, T ; H k ), u ɛ t ∈ L ∞ , (0, T ; H k−1 ) <strong>for</strong> all k, (2.28)u ɛ t + div (u ɛ (−∇K ɛ ∗ u ɛ )) = 0 in (0, T ) × R d , (2.29)u ɛ (0) = u ɛ 0, (2.30)u ɛ (t, x) ≥ 0 <strong>for</strong> a.e. (t, x) ∈ (0, T ), ×R d . (2.31)Statement (2.28) implies that u ɛ ∈ C((0, T ); H k−1 ). Using <strong>the</strong> continuous embedding H k−1 (R d ) ⊂ C 1 (R d )<strong>for</strong> k large enough we find that u ɛ and u ɛ x i, 1 ≤ i ≤ d, are continuous on (0, T ) × R d . Finally, (2.29) showsthat u ɛ t is also continuous on (0, T ) × R d . We have proven that u ɛ ∈ C 1 ((0, T ) × R d ). It is <strong>the</strong>n obvious thatv ɛ = −∇K ɛ ∗ u ɛ ∈ C 1 ((0, T ) × R d ). Note moreover that|v ɛ (x, t)| ≤ ‖u ɛ ‖ L∞ (0,T ;L 2 )‖∇K ɛ ‖ L 2<strong>for</strong> all (t, x) ∈ (0, T ) × R d . This combined with <strong>the</strong> fact that v ɛ is in C 1 shows that <strong>the</strong> characteristics arewell defined and propagate with finite speed. This proves that u ɛ is compactly supported in (0, T ) × R d(because u ɛ 0 is compactly supported in R d ).2.3 Study of <strong>the</strong> velocity field and <strong>the</strong> induced flow mapNote that K ɛ and u ɛ are in <strong>the</strong> right function spaces so that we can apply to <strong>the</strong>m to <strong>the</strong> a priori estimatesderived in section 2.1. In particular we have:ddt ‖uɛ (t)‖ L p ≤ (p − 1)‖∆K‖ L q‖u ɛ (t)‖ p+1L p ,‖u ɛ (0)‖ L p ≤ ‖u 0 ‖ L p.Using Gronwall inequality and <strong>the</strong> estimate on <strong>the</strong> supremum norm of <strong>the</strong> derivatives derived in section 2.1we obtain:7
Lemma 11 (uni<strong>for</strong>m bound <strong>for</strong> <strong>the</strong> smooth approximates). There exists a time T ∗ > 0 and a constantC > 0, both independent of ɛ, such that‖u ɛ (t)‖ L p ≤ C ∀t ∈ [0, T ∗ ], (2.32)|v ɛ t, x)|, |v ɛ x i(t, x)|, |v ɛ t(t, x)| ≤ C ∀(t, x) ∈ [0, T ∗ ] × R d . (2.33)From (2.33) it is clear that <strong>the</strong> family {v ɛ } is uni<strong>for</strong>mly Lipschitz on [0, T ∗ ] × R d , with Lipschitz constantC. We can <strong>the</strong>re<strong>for</strong>e use <strong>the</strong> Arzela-Ascoli Theorem to obtain <strong>the</strong> existence of a continuous function v(t, x)such thatv ɛ → v uni<strong>for</strong>mly on compact subset of [0, T ∗ ] × R d . (2.34)It is easy to check that this function v is also Lipschitz continuous with Lipschitz constant C. The Lipschitzand bounded vector field v ɛ generates a flow map X ɛ (t, α), t ∈ [0, T ∗ ], α ∈ R d :∂X ɛ (t, α)= v ɛ (X ɛ (t, α), t),∂tX ɛ (0, α) = α,where we denote by X t ɛ : R d → R d <strong>the</strong> mapping α ↦→ X ɛ (t, α) and by X −tɛ <strong>the</strong> inverse of X t ɛ.The uni<strong>for</strong>m Lipschitz bound on <strong>the</strong> vector field implies uni<strong>for</strong>m Lipschitz bound on <strong>the</strong> flow map andits inverse (see <strong>for</strong> example [4] <strong>for</strong> a proof of this statement) we <strong>the</strong>re<strong>for</strong>e have:Lemma 12 (uni<strong>for</strong>m Lipschitz bound on Xɛ t and Xɛ−t ). There exists a constant C > 0 independent of ɛsuch that:(i) <strong>for</strong> all t ∈ [0, T ∗ ] and <strong>for</strong> all x 1 , x 2 ∈ R d|Xɛ(x t 1 ) − Xɛ(x t 2 )| ≤ C|x 1 − x 2 | and |Xɛ−t (x 1 ) − x −tɛ (x 2 )| ≤ C|x 1 − x 2 |,(ii) <strong>for</strong> all t 1 , t 2 ∈ [0, T ∗ ] and <strong>for</strong> all x ∈ R d|X t1ɛ (x) − X t2ɛ (x)| ≤ C|t 1 , t 2 | and |Xɛ−t1(x) − X −t2ɛ (x)| ≤ C|t 1 − t 2 |.The Arzela-Ascoli Theorem <strong>the</strong>n implies that <strong>the</strong>re exists mapping X t and X −t such thatX t ɛ k(x) → X t (x) uni<strong>for</strong>mly on compact subset of [0, T ∗ ] × R d ,X −tɛ k(x) → X −t (x) uni<strong>for</strong>mly on compact subset of [0, T ∗ ] × R d .Moreover it is easy to check that X t and X −t inherit <strong>the</strong> Lipschitz bounds of X t ɛ and X −tɛ .Since <strong>the</strong> mapping X t : R d → R d is Lipschitz continuous, by Rademecher’s Theorem it is differentiablealmost everywhere. There<strong>for</strong>e it makes sense to consider its Jacobian matrix DX t (α). Because of Lemma12-(i) we know that <strong>the</strong>re exists a constant C independent of t and ε such thatsupαɛR d |det DX t (α)| ≤ C and supα∈R d |det DX t ɛ(α)| ≤ C.By <strong>the</strong> change of variable we <strong>the</strong>n easily obtain <strong>the</strong> following Lemma:Lemma 13. The mappings f ↦→ f ◦ X −t and f ↦→ f ◦ Xɛ−t , t ∈ [0, T ∗ ], ɛ > 0, are bounded linear operatorsfrom L p (R d ) to L p (R d ). Moreover <strong>the</strong>re exists a constant C ∗ independent of t and ɛ such that‖f ◦ X −t ‖ L p ≤ C ∗ ‖f‖ L p and ‖f ◦ X −tɛ ‖ L p ≤ C ∗ ‖f‖ L p <strong>for</strong> all f ∈ L p (R d ).8
Note that Lemma 12-(ii) impliesand <strong>the</strong>re<strong>for</strong>eThis gives us <strong>the</strong> following lemma:Lemma 14. let Ω be a compact subset of R d , <strong>the</strong>nwhere <strong>the</strong> compact set Ω + Ct is defined by2.4 Convergence in C([0, T ∗ ), L p )|X t ɛ(α) − α| ≤ Ct <strong>for</strong> all (t, α) ∈ [0, T ∗ ) × R d ,|X t (α) − α| ≤ Ct <strong>for</strong> all (t, α) ∈ [0, T ∗ ) × R d .X t ɛ(Ω) ⊂ Ω + Ct and X t (Ω) ⊂ Ω + Ct,Ω + Ct := {x ∈ R d : dist (x, Ω) ≤ Ct}.Since u ɛ and v ɛ = u ɛ ∗ ∇K are C 1 functions which satisfyu ɛ t + v ɛ · ∇u ɛ = −(div v ɛ )u ɛ and u ɛ (0) = u ɛ 0 (2.35)we have <strong>the</strong> simple representation <strong>for</strong>mula <strong>for</strong> u ɛ (t, x), t ∈ [0, T ∗ ), x ∈ R d :u ɛ (t, x) = u ɛ 0(Xɛ−t (x))e − R tdiv 0 vɛ (s,X −(t−s)ɛ(x))ds = u ɛ 0(X −t (x)) a ɛ (t, x).Lemma 15. There exists a function a(t, x) ∈ C 1 ([0, T ∗ ) × R d ) and a sequence ɛ k → 0 such thata ɛ k(t, x) → a(t, x) uni<strong>for</strong>mly on compact subset of [0, T ∗ ] × R d . (2.36)Proof. By <strong>the</strong> Arzela-Ascoli Theorem, it is enough to show that <strong>the</strong> familyb ɛ (t, x) :=∫ t0div v ɛ (s, Xɛ−(t−s) (x))dxis equicontinuous and uni<strong>for</strong>mly bounded. The uni<strong>for</strong>m boundedness simply come from <strong>the</strong> fact that|div v ɛ | = |u ɛ ∗ ∆K ɛ | ≤ ‖u ɛ ‖ L p‖∆K‖ L q.Let us now prove equicontinuity in space, i.e., we want to prove that <strong>for</strong> each δ > 0, <strong>the</strong>re is η > 0 independentof ɛ and t such that|b ɛ (t, x 1 ) − b ɛ (t, x 2 )| ≤ δ if |x 1 − x 2 | ≤ η.First, note that by Hölder’s inequality we haveɛ|b ɛ (t, x 1 ) − b ɛ (t, x 2 )| ≤∫ t0‖u ɛ (s)‖ L p‖τ ξ J ɛ ∆K − τ ζ J ɛ ∆K‖ L qdswhere ξ stands <strong>for</strong> Xɛ−(t−s) (x 1 ) and ζ <strong>for</strong> Xɛ−(t−s) (x 2 ). Then equicontinuity in space is a consequence ofLemma 9 (iv) toge<strong>the</strong>r with <strong>the</strong> fact thatwhere C is independent of t, s and ɛ.|Xɛ−(t−s) (x 1 ) − Xɛ −(t−s) (x 2 )| ≤ C|x 1 − x 2 |9
Let us finally prove equicontinuity in time. First not that, assuming that t 1 < t 2 ,b ɛ (t 1 , x) − b ɛ (t 2 , x) =−∫ t10∫ t2t 1div v ɛ (s, X −(t1−s) (x)) − div v ɛ (s, X −(t2−s) (x))dsdiv v ɛ (s, Xɛ−(t2−s) (x))ds.Since div v ɛ is uni<strong>for</strong>mly bounded we clearly have∫ t2∣ div v ɛ (s, Xɛ−(t2−s) (x))ds∣ ≤ C|t 1 − t 2 |.t 1The o<strong>the</strong>r term can be treated exactly as be<strong>for</strong>e, when we proved equicontinuity in space.Recall that <strong>the</strong> functionu ɛ (t, x) = u ɛ 0(Xɛ −t (x)) a ɛ (t, x)satisfies <strong>the</strong> ɛ-problem (2.27). We also have <strong>the</strong> following convergences:Define <strong>the</strong> functionu ɛ 0 → u 0 in L p (R d ), (2.37)Xɛ −tk(x) → X −t (x) unif. on compact subset of [0, T ∗ ] × R d , (2.38)a ɛ k(t, x) → a(t, x) unif. on compact subset of [0, T ∗ ] × R d , (2.39)v ɛ k(t, x) → v(t, x) unif. on compact subset of [0, T ∗ ] × R d . (2.40)u(t, x) := u 0 (X −t (x)) a(t, x). (2.41)Convergence (2.37)-(2.40) toge<strong>the</strong>r with Lemma 13 and 14 allow us to prove <strong>the</strong> following proposition.Proposition 16. : u, u ɛ , uv and u ɛ v ɛ all belong to <strong>the</strong> space C([0, T ∗ ), L p (R d )). Moreover we have:Proof. Straight <strong>for</strong>ward, see appendix at <strong>the</strong> end of <strong>the</strong> paper.We now turn to <strong>the</strong> proof of <strong>the</strong> main <strong>the</strong>orem of this section.u ɛ k→ u in C([0, T ∗ ), L p ), (2.42)u ɛ kv ɛ k→ uv in C([0, T ∗ ), L p ). (2.43)Proof of Theorem 6. Let φ ∈ C ∞ c(0, T ∗ ) be a scalar test function. It is obvious that u ɛ and v ɛ satisfy:∫ T∗∫ T∗− u ɛ (t) φ ′ (t) dt + div ( u ɛ (t) v ɛ (t) ) φ(t) dt = 0, (2.44)00v ɛ (t, x) = (u ɛ (t) ∗ ∇K ɛ )(x) <strong>for</strong> all (t, x) ∈ [0, T ∗ ] × R d , (2.45)u ɛ (0) = u ɛ 0, (2.46)where <strong>the</strong> integrals in (2.44) are <strong>the</strong> integral of a continuous function from [0, T ∗ ] to <strong>the</strong> Banach spaceW −1,p (R d ). Recall that <strong>the</strong> injection i : L p (R d ) → W −1,p (R d ) and <strong>the</strong> differentiation ∂ xi : L p (R d ) →W −1,p (R d ) are bounded linear operator. There<strong>for</strong>e (2.42) and (2.43) implyu ɛ k→ u in C([0, T ∗ ], W −1,p (R d )),div [u ɛ kv ɛ k] → div [uv] in C([0, T ∗ ], W −1,p (R d )), (2.47)hich is more than enough to pass to <strong>the</strong> limit in relation (2.44). To pass to <strong>the</strong> limit in (2.45), it is enoughto note that <strong>for</strong> all (t, x) ∈ [0, T ∗ ] × R d we have|(u ɛ (t) ∗ ∇K ɛ ) − (u(t) ∗ ∇K)(x)| ≤ ‖u ɛ (t) − u(t)‖ L p‖∇K ɛ ‖ L q + ‖u(t)‖ L p‖∇K ɛ − ∇K‖ L q, (2.48)10
and finally it is trivial to pass to <strong>the</strong> limit in relation (2.46).<strong>Equation</strong> (2.44) means that <strong>the</strong> continuous function u(t) (continuous function with values in W −1,p (R d ))satisfies (2.12) in <strong>the</strong> distributional sense. But (2.12) implies that <strong>the</strong> distributional derivative u ′ (t) is itselfa continuous function with value in W −1,p (R d ). There<strong>for</strong>e u(t) is differentiable in <strong>the</strong> classical sense, i.e, itbelongs to C 1 ([0, T ∗ ], W −1,p (R d )), and (2.12) is satisfied in <strong>the</strong> classical sense.We now turn to <strong>the</strong> proof of (2.15). The u ɛ ’s satisfies (2.19). Integrating over [0, t], t < T ∗ , we get∫ t ∫‖u ɛ (t)‖ p L = p ‖uɛ 0‖ p L − (p − 1) u ɛ (s, x) p div v ɛ (s, x) dxdt. (2.49)p0 R dProposition 16 toge<strong>the</strong>r with <strong>the</strong> general inequality()‖|f| p − |g| p ‖ L 1 ≤ 2p ‖f‖ p−1L + p ‖g‖p−1 L‖f − g‖ p L p (2.50)implies that(u ɛ ) p , u p ∈ C([0, T ∗ ], L 1 (R d )), (2.51)(u ɛ k) p → u p ∈ C([0, T ∗ ], L 1 (R d )). (2.52)On <strong>the</strong> o<strong>the</strong>r hand, replacing ∇K by ∆K in (2.48) we see right away thatCombining (2.51)-(2.54) we obtainSo we can pass to <strong>the</strong> limit in (2.49) to obtaindiv v, div v ɛ ∈ C([0, T ∗ ], L ∞ (R d )), (2.53)div v ɛ k→ div v in C([0, T ∗ ], L ∞ (R d )). (2.54)u p div v, (u ɛ ) p div v ɛ ∈ C([0, T ∗ ], L 1 (R d )), (2.55)(u ɛ k) p div v ɛ k→ u p div v in C([0, T ∗ ], L 1 (R d )). (2.56)∫ t ∫‖u(t)‖ p L = ‖u 0‖ p p L − (p − 1) u(s, x) p div v(s, x) dxdt.p0 R dBut (2.55) implies that <strong>the</strong> function t → ∫ u(t, x) p div v(t, x) dx is continuous, <strong>the</strong>re<strong>for</strong>e <strong>the</strong> functionR dt → ‖u(t)‖ p <strong>Lp</strong> is differentiable and satisfies (2.15).2.5 Continuation and conserved propertiesTheorem 17 (Continuation of solutions). The solution provided by Theorem 6 can be continued up to atime T max ∈ (0, +∞]. If T max < +∞, <strong>the</strong>n lim t→Tmax sup τ∈[0,t] ‖u(τ)‖ L p = +∞Proof. The proof is standard. One just needs to use <strong>the</strong> continuity of <strong>the</strong> solution with respect to time.Theorem 18 (Conservation of mass/ second moment). (i) Under <strong>the</strong> assumption of Theorem 6, and if weassume moreover that u 0 ∈ L 1 (R d ), <strong>the</strong>n <strong>the</strong> solution u belongs to C([0, T ∗ ], L 1 (R d )) and satisfies ‖u(t)‖ L 1 =‖u 0 ‖ L 1 <strong>for</strong> all t ∈ [0, T ∗ ].(ii) Under <strong>the</strong> assumption of Theorem 6, and if we assume moreover that u 0 has bounded second moment,<strong>the</strong>n <strong>the</strong> second moment of u(t) stays bounded <strong>for</strong> all t ∈ [0, T ∗ ].Proof. We just need to revisit <strong>the</strong> proof of Proposition 16. Since u 0 ∈ L 1 ∩ L p it is clear thatu ɛ 0 = J ɛ u 0 → u 0 in L 1 (R d ) ∩ L p (R d ). (2.57)11
Using convergences (2.57), (2.38), (2.39) and (2.40) we prove thatu ɛ k→ u in C([0, T ∗ ], L 1 ∩ L p ). (2.58)The proof is exactly <strong>the</strong> same than <strong>the</strong> one of Proposition 16 . Since <strong>the</strong> aggregation equation is a conservationlaw, it is obvious that <strong>the</strong> smooth approximates satisfy ‖u ɛ (t)‖ L 1 = ‖u ɛ 0‖ L 1. Using (2.58) we obtain‖u(t)‖ L 1 = ‖u 0 ‖ L 1.We now turn to <strong>the</strong> proof of (ii). Since <strong>the</strong> smooth approximates u ɛ have compact support, <strong>the</strong>ir secondmoment is clearly finite, and <strong>the</strong> following manipulation are justified:∫∫(∫) 1/2 (∫) 1/2d|x| 2 u ɛ (t, x)dx = 2 ⃗x · v ɛ du ɛ (x) ≤ 2 |x| 2 u ɛ (t, x)dx |v ɛ | 2 u ɛ (t, x)dxdt R d R d R d R(∫) d 1/2≤ C |x| 2 u ɛ (t, x)dx . (2.59)R dAssume now that <strong>the</strong> second moment of u 0 is bounded. A simple computation shows that if η ɛ is radiallysymmetric, <strong>the</strong>n |x| 2 ∗ η ɛ = |x| 2 + second moment of η ɛ . There<strong>for</strong>e∫∫|x| 2 u ɛ 0(x)dx ≤ |x| 2 ∗ η ɛ (x) u 0 (x)dxR d R∫d ∫≤ |x| 2 u 0 (x)dx + |x| 2 η ɛ (x)dxR d R∫d≤ |x| 2 u 0 (x)dx + 1 <strong>for</strong> ɛ small enough. (2.60)R dInequality (2.60) come from <strong>the</strong> fact that <strong>the</strong> second moment of η ɛ goes to 0 as ɛ goes to 0. Estimate (2.59)toge<strong>the</strong>r with (2.60) provide us with a uni<strong>for</strong>m bound of <strong>the</strong> second moment of <strong>the</strong> u ɛ (t) which only depends<strong>the</strong> second moment of u 0 . Since u ɛ converges to u in L 1 , we obviously have, <strong>for</strong> a given R and t:∫∫|x|≤R|x| 2 u(t, x)dx = limɛ→0∫|x|≤R|x| 2 u ɛ (t, x)dx ≤ lim supɛ→0R d |x| 2 u ɛ (t, x)dx.Since R is arbitrary, this show that <strong>the</strong> second moment of u(t, ·) is bounded <strong>for</strong> all t <strong>for</strong> which <strong>the</strong> solutionexists.Combining Theorem 17 and 18 toge<strong>the</strong>r with equality (2.15) we get:Theorem 19 (Global existence when ∆K is bounded from above). Under <strong>the</strong> assumption of Theorem 6,and if we assume moreover that u 0 ∈ L 1 (R d ) and ess sup ∆K < +∞, <strong>the</strong>n <strong>the</strong> solution u exists <strong>for</strong> all times(i.e.: T max = +∞).Proof. Equality (2.15) can be written∫ddt {‖u(t)‖p L p} = (p − 1)R d u(t, x) p (u(t) ∗ ∆K)(x) dx. (2.61)Since ∆K is bounded from above we have(u(s) ∗ ∆K)(x) ≤ (ess sup ∆K)∫u(s, x)dx = (ess sup ∆K) ‖u 0 ‖ L 1 .R d (2.62)Combining (2.61), (2.62) and Gronwall inequality gives‖u(t)‖ p L p ≤ ‖u 0‖ p L p e(p−1)(ess sup ∆K)‖u0‖ L 1 t ,so <strong>the</strong> L p -norm can not blow-up in finite time which, because of Theorem 17, implies global existence.12
3 Uniqueness of solutions in P 2 (R d ) ∩ L p (R d )In this section we use an optimal transport argument to prove uniqueness of solutions in P 2 (R d ) ∩ L p (R d ),when ∇K ∈ W 1,q <strong>for</strong> 1 < q < ∞ and p its Hölder conjugate. To do that, we shall follow <strong>the</strong> steps in [19],where <strong>the</strong> authors extend <strong>the</strong> work of Loeper [34] to prove, among o<strong>the</strong>r, uniqueness of P 2 ∩ L ∞ -solutionsof <strong>the</strong> aggregation equation when <strong>the</strong> interaction potential K has a Lipschitz singularity at <strong>the</strong> origin.One can easily check that solutions of <strong>the</strong> aggregation equation constructed in previous sections aredistribution solutions, i.e. <strong>the</strong>y satisfy∫ T ∫ ( )∫∂ϕ0 R ∂t (t, x) + v(t, x) · ∇ϕ(t, x) u(t, x) dx dt = ϕ(0, x)u 0 (x) dx (3.63)N R N<strong>for</strong> all ϕ ∈ C0 ∞ ([0, T ∗ ) × R N )). A function u(t, x) satisfying (3.63) is said to be a distribution solution to <strong>the</strong>continuity equation (2.12) with <strong>the</strong> given velocity field v(t, x) and initial data u 0 (x). In fact, it is uniquelycharacterized by∫∫u(t, x) dx = u 0 (x) dxBX −t (B)<strong>for</strong> all measurable set B ⊂ R d , see [1]. Here X t : R d → R d is <strong>the</strong> flow map associated with <strong>the</strong> velocityfield v(t, x) and X −t is its inverse. In <strong>the</strong> optimal transport terminology this is equivalent to say that X ttransports <strong>the</strong> measure u 0 onto u(t) (u(t) = X t #u 0 ).We recall, <strong>for</strong> <strong>the</strong> sake of completeness, [19, Theorem 2.4], where several results of [2, 35, 22, 1] are puttoge<strong>the</strong>r.Theorem 20 ([19]). Let ρ 1 and ρ 2 be two probability measures on R N , such that <strong>the</strong>y are absolutely continuouswith respect to <strong>the</strong> Lebesgue measure and W 2 (ρ 1 , ρ 2 ) < ∞, and let ρ θ be an interpolation measurebetween ρ 1 and ρ 2 , defined as in [34] byρ θ = ((θ − 1)T + (2 − θ)I R N ) # ρ 1 (3.64)<strong>for</strong> θ ∈ [1, 2], where T is <strong>the</strong> optimal transport map between ρ 1 and ρ 2 due to Brenier’s <strong>the</strong>orem [12] and I R Nis <strong>the</strong> identity map. Then <strong>the</strong>re exists a vector field ν θ ∈ L 2 (R N , ρ θ dx) such thatdi.dθ ρ θ + div(ρ θ ν θ ) = 0 <strong>for</strong> all θ ∈ [1, 2].∫ii. ρ θ |ν θ | 2 dx = W2 2 (ρ 1 , ρ 2 ) <strong>for</strong> all θ ∈ [1, 2].R Niii. We have <strong>the</strong> L p -interpolation estimate‖ρ θ ‖ L p (R N ) ≤ max { }‖ρ 1 ‖ L p (R N ), ‖ρ 2 ‖ L p (R N )<strong>for</strong> all θ ∈ [1, 2].Here, W 2 (f, g) is <strong>the</strong> Euclidean Wasserstein distance between two probability measures f, g ∈ P(R n ),{∫∫} 1/2W 2 (f, g) = inf|v − x| 2 dΠ(v, x) , (3.65)Π∈Γ R n ×R nwhere Π runs over <strong>the</strong> set of joint probability measures on R n × R n with marginals f and g.Now we are ready to prove <strong>the</strong> uniqueness of solutions to <strong>the</strong> aggregation equation.Theorem 21 (Uniqueness). Let u 1 , u 2 be two bounded solutions of equation (2.12) in <strong>the</strong> interval [0, T ∗ ]with initial data u 0 ∈ P 2 (R d ) ∩ L p (R d ), 1 < p < ∞ and assume that v is given by v = −∇K ∗ u, with Ksuch that ∇K ∈ W 1,q (R N ), p and q conjugates. Then u 1 (t) = u 2 (t) <strong>for</strong> all 0 ≤ t ≤ T ∗ .13
Here ∂ 0 K is <strong>the</strong> unique element of minimal norm in <strong>the</strong> subdifferential of K. Simply speaking, sinceK(x) = |x| is smooth away from <strong>the</strong> origin and radially symmetric, we have ∂ 0 K(x) = x|x|<strong>for</strong> x ≠ 0 and∂ 0 K(0) = 0, and thus:∫(∂ 0 x − yK ∗ µ)(x) =dµ(y). (4.77)|x − y|Note that, µ t being a measure, it is important <strong>for</strong> ∂ 0 K to be defined <strong>for</strong> every x ∈ R d so that (4.75) makessense. <strong>Equation</strong> (4.74) means that∫ +∞0∫y≠xR d ( dψdt (x, t) + ∇ψ(x, t) · v t(x))dµ t (x) dt = 0, (4.78)<strong>for</strong> all ψ ∈ C ∞ 0 (R d × (0, +∞)). From (4.77) it is clear that |v t (x)| ≤ 1 <strong>for</strong> all x and t, <strong>the</strong>re<strong>for</strong>e <strong>the</strong> aboveintegral makes sense.The main Theorem of this section is <strong>the</strong> following:Theorem 24 (Instantaneous mass concentration). Consider <strong>the</strong> initial datau 0 (x) ={L|x| d−1+ɛ if |x| < 1,0 o<strong>the</strong>rwise,(4.79)where ɛ ∈ (0, 1) and L :=( ∫|x| 0 we haveµ t ({0}) > 0,i.e., mass is concentrated at <strong>the</strong> origin instantaneously and <strong>the</strong> solution is no longer continuous with respectto <strong>the</strong> Lebesgue measure.Theorem 24 is a consequence of <strong>the</strong> following estimate on <strong>the</strong> velocity field:Proposition 25. Let (µ t ) t∈(0,+∞) be <strong>the</strong> unique measure solution of <strong>the</strong> aggregation equation with interactionpotential K(x) = |x| and with initial data (4.79). Then, <strong>for</strong> all t ∈ [0, +∞) <strong>the</strong> velocity field v t = −∂ 0 K ∗ µ tis focussing and <strong>the</strong>re exists a constant C > 0 such that|v t (x)| ≥ C|x| 1−ɛ <strong>for</strong> all t ∈ [0, +∞) and x ∈ B(0, 1). (4.80)By focussing, we mean that <strong>the</strong> velocity field points inward, i.e.λ t : [0, +∞) → [0, +∞) such that v t (x) = −λ t (|x|) ⃗x x .<strong>the</strong>re exists a nonnegative function4.1 Representation <strong>for</strong>mula <strong>for</strong> radially symmetric measure solutionsIn this section, we show that <strong>for</strong> radially symmetric measure solutions, <strong>the</strong> characteristics are well defined.As a consequence, <strong>the</strong> solution to (2.12) can be expressed as <strong>the</strong> push <strong>for</strong>ward of <strong>the</strong> initial data by <strong>the</strong> flowmap associated with <strong>the</strong> ODE defining <strong>the</strong> characteristics.In <strong>the</strong> following <strong>the</strong> unit sphere {x ∈ R d , |x| = 1} is denoted by S d and its surface area by ω d .Definition 26. If µ ∈ P(R d ) is a radially symmetric probability measure, <strong>the</strong>n we define ˆµ ∈ P([0, +∞)) by<strong>for</strong> all I ∈ B([0, +∞)).ˆµ(I) = µ({x ∈ R d : |x| ∈ I})16
Remark 27. If a measure µ is radially symmetric, <strong>the</strong>n µ({x}) = 0 <strong>for</strong> all x ≠ 0, and <strong>the</strong>re<strong>for</strong>e∫∇K(x − y) dµ(y) =R d \{x}∫∇K(x − y) dµ(y)R d <strong>for</strong> all x ≠ 0.In o<strong>the</strong>r words, <strong>for</strong> x ≠ 0, (∇K ∗ µ)(x) is well defined despite <strong>the</strong> fact that ∇K is not defined at x = 0. Asa consequence (∂ 0 K ∗ µ)(x) = (∇K ∗ µ)(x) if x ≠ 0 and (∂ 0 K ∗ µ)(0) = 0.Remark 28. If <strong>the</strong> radially symmetric measure µ is continuous with respect to <strong>the</strong> Lebesgue measure andhas radially symmetric density u(x) = ũ(|x|), <strong>the</strong>n ˆµ is also continuous with respect to <strong>the</strong> Lebesgue measureand has density û, whereû(r) = ω d r d−1 ũ(r). (4.81)Lemma 29 (Polar coordinate <strong>for</strong>mula <strong>for</strong> <strong>the</strong> convolution). Suppose µ ∈ P(R d ) is radially symmetric. LetK(x) = |x|, <strong>the</strong>n <strong>for</strong> all x ≠ 0 we have:(∫ +∞( ) ) |x|x(µ ∗ ∇K) (x) = φ dˆµ(ρ)(4.82)ρ|x|0where <strong>the</strong> function φ : [0, +∞) → [−1, 1] is defined byφ(r) = 1 ∫re 1 − yω d |re 1 − y| · e 1dσ(y). (4.83)Proof. This come from simple algebraic manipulations. These manipulations are shown in [5].In <strong>the</strong> next Lemma we state properties of <strong>the</strong> function φ defined in (4.83).Lemma 30 (Properties of <strong>the</strong> function φ).S d(i) φ is continuous and non-decreasing on [0, +∞). Moreover φ(0) = 0, and lim r→∞ φ(r) = 1.(ii) φ(r) is O(r) as r → 0. To be more precise:Proof. Consider <strong>the</strong> function F : [0, +∞) × S d → [−1, 1] defined byφ(r)lim = 1 − 1 (y · e 1 )r→0 r ω dr>0∫S 2 dσ(y). (4.84)d(r, y) ↦→ re 1 − y|re 1 − y| · e 1. (4.85)Since F is bounded, we have thatφ(r) = 1 ∫F (r, y)dσ(y) = 1 F (r, y)dσ(y).ω d S dω d∫S d \{e 1}If y ∈ S d \{e 1 } <strong>the</strong>n <strong>the</strong> function r ↦→ F (r, y) is continuous on [0, +∞) and C ∞ on (0, +∞). An explicitcomputation shows <strong>the</strong>n that∂F 1 − F (r, y)2(r, y) = ≥ 0, (4.86)∂r |re 1 − y|thus φ is non-decreasing and, by <strong>the</strong> Lebesgue dominated convergence, it is easy to see that φ is continuous,φ(0) = 0 and lim r→∞ φ(r) = 1, which prove (i).To prove (ii), note that <strong>the</strong> function ∂F∂r(r, y) can be extended by continuity on [0, +∞). There<strong>for</strong>e <strong>the</strong>right derivative with respect to r of F (r, y) is well defined:F (r, y) − F (0, y) 1 − F (0, y)2lim= = 1 − (y · e 1 ) 2 .r→0 r|y|r>017
and since ∂F/∂r is bounded on (0, +∞) × S d \{e 1 }, we can now use <strong>the</strong> Lebesgue dominated convergence<strong>the</strong>orem to conclude:φ(r)limr→0 rr>0φ(r) − φ(0)= limr→0 rr>0= limr→0r>0∫1ω dF (r, y) − F (0, y)S dr(Remark 31. Note that <strong>for</strong> r > 0 <strong>the</strong> function ρ ↦→ φrρdσ(y) = 1 ω d∫S d1 − (y · e 1 ) 2 dσ(y).)is non increasing and continuous. Indeed, it isequal to 1 when ρ = 0 and it decreases to 0 as ρ → ∞. In particular, <strong>the</strong> integral in (4.82) is well defined<strong>for</strong> any probability measure ˆµ ∈ P([0, +∞)).Remark 32. In dimension two, it is easy to check that lim r→1 φ ′ (r) = +∞ which implies that <strong>the</strong> derivativeof <strong>the</strong> function φ has a singularity at r = 1 and thus, that <strong>the</strong> function φ is not C 1 .Proposition 33 (Characteristic ODE). Let K(x) = |x| and let (µ t ) t∈[0,+∞) be a weakly continuous familyof radially symmetric probability measures. Then <strong>the</strong> velocity field{−(∇K ∗ µ t )(x) if x ≠ 0v(x, t) =(4.87)0 if x = 0is continuous on R d \{0} × [0 + ∞). Moreover, <strong>for</strong> every x ∈ R d , <strong>the</strong>re exists an absolutely continuousfunction t → X t (x), t ∈ [0, +∞), which satisfiesddt X t(x) = v(X t (x), t) <strong>for</strong> a.e. t ∈ (0, +∞), (4.88)X 0 (x) = x. (4.89)Proof. From <strong>for</strong>mula (4.82), Remark 31, and <strong>the</strong> weak continuity of <strong>the</strong> family (µ t ) t∈[0,+∞) , we obtaincontinuity in time. The continuity in space simply comes from <strong>the</strong> continuity and boundedness of <strong>the</strong>function φ toge<strong>the</strong>r with <strong>the</strong> Lebesgue dominated convergence <strong>the</strong>orem.Since v is continuous on R d \{0} × [0 + ∞) we know from <strong>the</strong> Peano <strong>the</strong>orem that given x ∈ R d \{0}, <strong>the</strong>initial value problem (4.88)-(4.89) has a C 1 solution at least <strong>for</strong> short time. We want to see that it is defined<strong>for</strong> all time. For that, note that by a continuation argument, <strong>the</strong> interval given by Peano <strong>the</strong>orem can beextended as long as <strong>the</strong> solution stays in R d \{0}. Then, if we denote by T x <strong>the</strong> maximum time so that <strong>the</strong>solution exists in [0, T x ) we have that ei<strong>the</strong>r T x = ∞ and we are done, or T x < +∞, in which case clearlylim t→Tx X t (x) = 0, and we can extend <strong>the</strong> function X t (x) on [0, +∞) by setting X t (x) := 0 <strong>for</strong> t ≥ T x .The function t → X t (x) that we have just constructed is continuous on [0, +∞), C 1 on [0, +∞)\{T x }and satisfies (4.88) on [0, +∞)\{T x }. If x = 0, we obviously let X t (x) = 0 <strong>for</strong> all t ≥ 0.Finally, we present <strong>the</strong> representation <strong>for</strong>mula, by which we express <strong>the</strong> solution to (2.12) as a push<strong>for</strong>wardof <strong>the</strong> initial data. See [1] or [41] <strong>for</strong> a definition of <strong>the</strong> push-<strong>for</strong>ward of a measure by a map.Proposition 34 (Representation <strong>for</strong>mula). Let (µ t ) t∈[0,+∞) be a radially symmetric measure solution of <strong>the</strong>aggregation equation with interaction potential K(x) = |x|, and let X t : R d → R d be defined by (4.87), (4.88)and (4.89). Then <strong>for</strong> all t ≥ 0,µ t = X t #µ 0 .Proof. In this proof, we follow arguments from [1]. Since <strong>for</strong> a given x <strong>the</strong> function t ↦→ X t (x) is continuous,one can easily prove, using <strong>the</strong> Lebesgue dominated convergence <strong>the</strong>orem, that t ↦→ X t #µ 0 is weaklycontinuous. Let us now prove that µ t := X t #µ 0 satisfies (4.78) <strong>for</strong> all ψ ∈ C ∞ 0 (R d × (0, ∞)). Giventhat <strong>the</strong> test function ψ is compactly supported, <strong>the</strong>re exist T > 0 such that ψ(x, t) = 0 <strong>for</strong> all t ≥ T . We<strong>the</strong>re<strong>for</strong>e have:∫∫()0 = ψ(x, T )dµ T (x) − ψ(x, 0)dµ 0 (x) = ψ(X T (x), T ) − ψ(x, 0) dµ 0 (x). (4.90)R d R∫R d d18
If we now take into account that from Proposition 33 <strong>the</strong> mapping t → φ(X t (x), t) is absolutely continuous,we can rewrite (4.90) as0 ====∫R d ∫ T0∫ T ∫R d∫ T0∫ T00( ddt ψ(X t(x), t))dt dµ 0 (x) (4.91)(∇ψ(X t (x), t) · v(X t (x), t) + dψ∫R d (∇ψ(X t (x), t) · v(X t (x), t) + dψdt (X t(x), t)dt (X t(x), t)(∇ψ(x, t) · v(x, t) +∫R dψ )dt (x, t) dµ t (x) dt.d)dt dµ 0 (x) (4.92))dµ 0 (x) dt (4.93)The step from (4.92) to (4.93) holds because of <strong>the</strong> fact that |v(x, t)| ≤ 1, which justifies <strong>the</strong> use of <strong>the</strong>Fubini Theorem.Remark 35 (Representation <strong>for</strong>mula in polar coordinates). Let µ t and X t be as in <strong>the</strong> previous proposition.Let R t : [0, +∞) → [0, +∞) be <strong>the</strong> function such that |X t (x)| = R t (|x|). Thenˆµ t = R t #ˆµ 0 . (4.94)Remark 36. Since φ is nonnegative (Lemma 30), from (4.82), (4.87) and (4.88) we see that <strong>the</strong> functiont ↦→ |X t (x)| = R t (|x|) is non increasing.4.2 Proof of Proposition 25 and Theorem 24We are now ready to prove <strong>the</strong> estimate on <strong>the</strong> velocity field and <strong>the</strong> instantaneous concentration result. Westart by giving a frozen in time estimate of <strong>the</strong> velocity field.Lemma 37. Let K(x) = |x|, and let u 0 (x) be defined by (4.79) <strong>for</strong> some ɛ ∈ (0, 1). Then <strong>the</strong>re exist aconstant C > 0 such that|(u 0 ∗ ∇K) (x)| ≥ C|x| 1−ɛ (4.95)<strong>for</strong> all x ∈ B(0, 1)\{0}.Proof. Note that if we do <strong>the</strong> change of variable s = |x|ρ|(u 0 ∗ ∇K) (x)| = |x|in equation (4.82), we find that∫ +∞0φ(s) û 0 ( |x|s )ds s 2 . (4.96)On <strong>the</strong> o<strong>the</strong>r hand, using (4.81) we see that <strong>the</strong> û 0 (r) corresponding to <strong>the</strong> u 0 (x) defined by (4.79) isû 0 (r) ={ωdr ɛ if r < 1,0 o<strong>the</strong>rwise.(4.97)Then, plugging (4.97) in (4.96) we obtain that <strong>for</strong> all x ≠ 0|(u 0 ∗ ∇K) (x)| = ω d |x| 1−ɛ ∫ +∞|x|φ(ρ)dρ. (4.98)ρ2−ɛ In light of statement (ii) of Lemma 30, we see that <strong>the</strong> previous integral converges as |x| → 0.|(u 0 ∗ ∇K) (x)| is O(|x| 1−ɛ ) as |x| → 0 and (4.95) follows.HenceFinally, <strong>the</strong> last piece we need in order to prove Proposition 25 from <strong>the</strong> previous lemma, is <strong>the</strong> followingcomparison principle:19
Lemma 38 (Temporal monotonicity of <strong>the</strong> velocity). Let (µ t ) t∈(0,+∞) be a radially symmetric measuresolution of <strong>the</strong> aggregation equation with interaction potential K(x) = |x| . Then, <strong>for</strong> every x ∈ R d \{0} <strong>the</strong>functiont ↦→ |(∇K ∗ µ t )(x)|is non decreasing.Proof. Combining (4.82) and (4.94) we see that|(µ t ∗ ∇K)(x)| =(∫ +∞0)φ( |x|R t (ρ) ) dˆµ 0(ρ) . (4.99)Now, by Lemma 30, φ is non decreasing and due to Remark 35, t ↦→ R t (ρ) is non increasing. Hence<strong>for</strong>th itis clear that (4.99) is itself non decreasing.At this point, Proposition 25 follows as a simple consequence of <strong>the</strong> frozen in time estimate (4.95) toge<strong>the</strong>rwith Lemma 38, and we can give an easy proof <strong>for</strong> <strong>the</strong> main result we introduced at <strong>the</strong> beginning of <strong>the</strong>section.Proof of Theorem 24. Using <strong>the</strong> representation <strong>for</strong>mula (Proposition 34) and <strong>the</strong> definition of <strong>the</strong> push <strong>for</strong>wardwe getµ t ({0}) = (X t #µ 0 )({0}) = µ 0 (Xt −1 ({0})).Then, note that <strong>the</strong> solution of <strong>the</strong> ODE ṙ = −r 1−ɛ reaches zero in finite time. There<strong>for</strong>e, from Proposition25 and 33 we obtain that <strong>for</strong> all t > 0, <strong>the</strong>re exists δ > 0 such thatX t (x) = 0 <strong>for</strong> all |x| < δ.In o<strong>the</strong>r words, <strong>for</strong> all t > 0, <strong>the</strong>re exists δ > 0 such that B(0, δ) ⊂ Xt−1 ({0}). Clearly, given our choice ofinitial condition, we have that µ 0 (B(0, δ)) > 0 if δ > 0, and <strong>the</strong>re<strong>for</strong>e µ t ({0}) > 0 if t > 0.4.3 Remark about <strong>the</strong> initial data 1/ |x| d−1An interesting open problem is we<strong>the</strong>r or not <strong>the</strong>re is instantaneous mass concentration when <strong>the</strong> initial datais defined by (4.79) with ɛ = 0. We right now can not answer this question, but below are some interestingremarks about this case.Lemma 39. Let u 0 be defined by (4.79) with ɛ = 0. Then <strong>the</strong>re exists constants C > 0 and β > 1 such that|(∇K ∗ u 0 )(x)| ≤ C|x||x|∣log β ∣ <strong>for</strong> all x ∈ B(0, 1). (4.100)Proof. From Lemma 30 it is clear that <strong>the</strong>re exists a constant α > 0 such that φ(r)/r ≤ α <strong>for</strong> all r ∈ (0, 1).We can <strong>the</strong>n use (4.98) with ɛ = 0 to obtain that, <strong>for</strong> |x| < 1,( ∫ 1 ∫ )φ(ρ) dρ ∞|(u 0 ∗ ∇K)(x)| = ω d |x|ρ ρ + φ(ρ)1 ρ 2 dρ≤ ω d |x|(α|x|∫ 1|x|dρρ + cst )= ω d |x| (−α log |x| + cst) = −C|x| log |x|β ,<strong>for</strong> some constants C > 0 and β > 1.Consider <strong>the</strong> push<strong>for</strong>ward S t (r) defined by <strong>the</strong> ODEddt S t(r) = −CS t (r)∣ log S t(r)β ∣ ,S 0 (r) = r.20
This ODE has an explicit solution: <strong>for</strong> r < 1 we have( ) eCtrS t (r) = ββThe push <strong>for</strong>ward of u 0 by <strong>the</strong> map S t can also be explicitly computed:(S t #u 0 )(r) ={ L(t)r α(t)if r < β 1−exp(Ct)0 o<strong>the</strong>rwisewhere α(t) = (d − 1) + ∫ eCt − 1β1−exp(Ct)e Ct , and L(t) =r −α(t) dr.0In dimension 2 and greater, α(t) ≤ d − 1 <strong>for</strong> t > 0, so S t #u 0 is less singular than u 0 .5 Osgood condition <strong>for</strong> global well-posednessThis section considers <strong>the</strong> global well-posedness of <strong>the</strong> aggregation equation in P 2 (R d ) ∩ L p (R d ), dependingon <strong>the</strong> potential K. We start by giving a precise definitions of “natural potential”, “repulsive in <strong>the</strong> shortrange” and “strictly attractive in <strong>the</strong> short range”, and <strong>the</strong>n we prove Theorem 5.Definition 40. A natural potential is a radially symmetric potential K(x) = k(|x|), where k : (0, +∞) → Ris a smooth function which satisfies <strong>the</strong> following conditions:(C1)sup |k ′ (r)| < +∞,r∈(0,∞)(C2) ∃ α > d such that k ′ (r) and k ′′ (r) are O(1/r α ) as r → +∞,(MN1) ∃ δ 1 > 0 such that k ′′ (r) is monotonic (ei<strong>the</strong>r increasing or decreasing) in (0, δ 1 ),(MN2) ∃ δ 2 > 0 such that rk ′′ (r) is monotonic (ei<strong>the</strong>r increasing or decreasing) in (0, δ 2 ).Remark 41. Note that monotonicity condition (MN1) implies that k ′ (r) and k(r) are also monotonic insome (different) neighborhood of <strong>the</strong> origin (0, δ). Also, note that (C1) and (MN1) imply(C3)limr→0 k′ (r) exists and is finite.+Remark 42. The far field condition (C2) can be dropped when <strong>the</strong> data has compact support.Definition 43. A natural potential is said to be repulsive in <strong>the</strong> short range if <strong>the</strong>re exists an interval (0, δ)on which k(r) is decreasing. A natural potential is said to be strictly attractive in <strong>the</strong> short range if <strong>the</strong>reexists an interval (0, δ) on which k(r) is strictly increasing.We would like to remark that <strong>the</strong> two monotonicity conditions are not very restrictive as, in order toviolate <strong>the</strong>m, a potential would have to exhibit some pathological behavior around <strong>the</strong> origin, like oscillatingfaster and faster as r → 0.5.1 Properties of natural potentialsAs a last step be<strong>for</strong>e proving Theorem 5, let us point out some properties of natural potentials, which show<strong>the</strong> reason behind <strong>the</strong> choice of this kind of potentials to work with.Lemma 44. If K(x) = k(|x|) is a natural potential, <strong>the</strong>n k ′′ (r) = o(1/r) as r → 0.21
Proof. First, note that since k(r) is smooth away from 0 we havek ′ (1) − k ′ (ɛ) =∫ 1ɛk ′′ (r)dr.Now, because of (MN1) we know that <strong>the</strong>re exists a neighborhood of zero in which k ′′ doesn’t change sign.There<strong>for</strong>e, letting ɛ → 0 and using (C3) we conclude that k ′′ is integrable around <strong>the</strong> origin. A simpleintegration by part, toge<strong>the</strong>r with (C3) gives <strong>the</strong>n that∫ r0k ′′ (s)s ds = −Dividing both sides by r and letting r → 0 we obtain1limr→0 r∫ r0∫ rwhich, combined with (MN2) implies lim r→0 k ′′ (r)r = 0.0k ′ (s)dr + k ′ (r)r.k ′′ (s)s ds = 0.The next lemma shows that <strong>the</strong> existence <strong>the</strong>ory developed in <strong>the</strong> previous section applies to this classof potentials.Lemma 45. If K is a natural potential <strong>the</strong>n ∇K ∈ W 1,q (R d ) <strong>for</strong> all 1 ≤ q < d. As a consequence, <strong>the</strong>critical exponents p s and q s associated to a natural potential satisfyProof. Recall that∇K(x) = k ′ (|x|) x|x|andq s ≥ d and p s ≤ dd − 1 .∂ 2 ()K(x) = k ′′ (|x|) − k′ (|x|) xi x j∂x i ∂x j |x| |x| 2 + δ ijk ′ (|x|),|x|where δ ij is <strong>the</strong> Kronecker delta symbol. In order to prove <strong>the</strong> lemma, it is enough to show thatk ′ (|x|), k ′′ (|x|) and k′ (|x|)|x|belong to L q (R d ) <strong>for</strong> all 1 ≤ q < d. To do that, observe that <strong>the</strong> decay condition(C2) implies that <strong>the</strong>y belong to L q (B(0, 1) c ) <strong>for</strong> all q ≥ 1. Then, we take into account that (C3)implies that k′ (r)r= O(1/r) as r → 0 and that we have seen in <strong>the</strong> previous Lemma that k ′′ (r) = o(1/r) asr → 0. This is enough to conclude, since <strong>the</strong> function x ↦→ 1/|x| is in L q (B(0, 1)) <strong>for</strong> all 1 ≤ q < d.The following Lemma toge<strong>the</strong>r with Theorem 19 gives global existence of solutions <strong>for</strong> natural potentialswhich are repulsive in <strong>the</strong> short range, whence part (i) of Theorem 5 follows.Lemma 46. Suppose K is a natural potential which is repulsive in <strong>the</strong> short range. Then ∆K is boundedfrom above.Proof. We will prove that <strong>the</strong>re is a neighborhood of zero on which ∆K ≤ 0. This combined with <strong>the</strong> decaycondition (C2) give <strong>the</strong> desired result. First, recall that ∆K(x) = k ′′ (|x|) + (d − 1)k ′ (|x|)|x| −1 . Then, sincek is repulsive in <strong>the</strong> short range, <strong>the</strong>re exists a neighborhood of zero in which k ′ ≤ 0. Now, we have twopossibilities: on one hand, if lim r→0 + k ′ (r) = 0, <strong>the</strong>n given r ∈ (0, +∞) <strong>the</strong>re exists s ∈ (0, r) such thatk ′ (r)r= k ′′ (s). Toge<strong>the</strong>r with (MN1), this implies that k ′′ is also non-positive in some neighborhood of zero.On <strong>the</strong> o<strong>the</strong>r hand if lim r→0 + k ′ (r) < 0, <strong>the</strong>n <strong>the</strong> fact that k ′′ (r) = o(1/r) implies that rk′′ (r)+(d−1)k ′ (r)risnegative <strong>for</strong> r small enough.Finally, <strong>the</strong> next Lemma will be needed to prove global existence <strong>for</strong> natural potentials which are strictlyattractive in <strong>the</strong> short range and satisfy <strong>the</strong> Osgood criteria.22
Lemma 47. Suppose that K is a natural potential which is strictly attractive in <strong>the</strong> short range and satisfies<strong>the</strong> Osgood criteria (1.9). If moreover sup x≠0 ∆K(x) = +∞ <strong>the</strong>n <strong>the</strong> following holds(Z1) lim r→0 + k ′′ (r) = +∞ and lim r→0 +k′ (r)r= +∞,(Z2) ∃δ 1 > 0 such that k ′′ (r) and k′ (r)rare decreasing <strong>for</strong> r ∈ (0, δ 1 ),(Z3) ∃δ 2 > 0 such that k ′′ (r) ≤ k′ (r)r<strong>for</strong> r ∈ (0, δ 2 ).Proof. Let us start by proving by contradiction thatIf we suppose thatlimsupr→0 + p∈(0,r)k ′ (r)rk ′ (r)r= +∞. (5.101)< C ∀r ∈ (0, 1], (5.102)<strong>the</strong>n given a sequence r n → 0 + <strong>the</strong>re will exist ano<strong>the</strong>r sequence s n → 0 + , 0 < s n < r n , such thatk ′ (r n )r n= k ′′ (s n ) < C.Since k ′′ (r) is monotonic around zero, this implies that k ′′ is bounded from above, and combining thiswith (5.102) we see that ∆K must also be bounded from above, which contradicts our assumption. Now,statements (Z1), (Z2), (Z3) follow easily: First note that if lim r→0 + k ′ (r) > 0, <strong>the</strong>n clearly <strong>the</strong> Osgoodcondition (1.9) is not satisfied, whence lim r→0 + k ′ (r) = 0. This implies that <strong>for</strong> all r > 0 <strong>the</strong>re existss ∈ (0, r) such thatk ′ (r)r= k ′′ (s). (5.103)Combining (5.103), (5.101) and <strong>the</strong> monotonicity of k ′′ we get that lim r→0 + k ′′ (r) = +∞ and k ′′ is decreasingon some interval (0, δ), which corresponds with <strong>the</strong> first part of (Z1) and (Z2). Now, going back to (5.103)we see that if 0 < s < r < δ <strong>the</strong>n k′ (r)rddr= k ′′ (s) ≥ k ′′ (r) which proves (Z3). This implies{ k ′ }(r)= 1 r r()k ′′ (r) − k′ (r)≤ 0rand <strong>the</strong>re<strong>for</strong>e k ′ (r)/r decreases on (0, δ). Thus, (5.101) implies lim r→0 + k′ (r)rcomplete.= +∞ and <strong>the</strong> proof is5.2 Global bound of <strong>the</strong> L p -norm using Osgood criteriaWe have already proven global existence when <strong>the</strong> potential is bounded from above. In this section we prove<strong>the</strong> following proposition, which allows us to prove global existence <strong>for</strong> potentials which are attractive in <strong>the</strong>short range, satisfy <strong>the</strong> Osgood criteria, and whose Laplacian is not bounded from above. From it, secondpart of Theorem 5 follows readily. This extends prior work on L ∞ -solutions [5] to <strong>the</strong> L p case.Proposition 48. Suppose that K is a natural potential which is strictly attractive in <strong>the</strong> short range, satisfies<strong>the</strong> Osgood criteria (1.9) and whose Laplacian is not bounded from above (i.e. sup x≠0 ∆K(x) = +∞). Letu(t) be <strong>the</strong> unique solution of <strong>the</strong> aggregation equation starting with initial data u 0 ∈ P 2 (R d ) ∩ L p (R d ), wherep > d/(d − 1). Define <strong>the</strong> length scale( ) q/d ‖u(t)‖L 1R(t) =.‖u(t)‖ L p23
Then <strong>the</strong>re exists positive constants δ and C such that <strong>the</strong> inequalityholds <strong>for</strong> all time t <strong>for</strong> which R(t) ≤ δ.dRdt ≥ −C k′ (R) (5.104)Remark 49. R(t) is <strong>the</strong> natural length scale associated with blow up of L p norm. Given that mass isconserved, R(t) > 0 means that <strong>the</strong> L p norm is bounded. The differential inequality (5.104) tell us that R(t)decays slower than <strong>the</strong> solutions of <strong>the</strong> ODE ẏ = −Ck ′ (y). Since k ′ (r) satisfies <strong>the</strong> Osgood criteria (1.9),solutions of this ODE do not go to zero in finite time. R(t) <strong>the</strong>re<strong>for</strong>e stays away from zero <strong>for</strong> all time whichprovides us with a global upper bound <strong>for</strong> ‖u(t)‖ L p.Proof. Equality (2.15) can be writtenUsing <strong>the</strong> chain rule we getddt∫ddt {‖u(t)‖p L p} = (p − 1){ }‖u(t)‖ − q dL= − p(p − 1)qpd≥ −R d u(t, x) p (u(t) ∗ ∆K)(x) dx.‖u(t)‖ − q d −pL p ∫R d u(t, x) p (u(t) ∗ ∆K)(x) dx (5.105)(p − 1)q‖u(t)‖ − q d<strong>Lp</strong>dsup {(u(t) ∗ ∆K)(x)} . (5.106)px∈R dTo obtain (5.104), we now need to carefully estimate sup x∈R d {(u(t) ∗ ∆K)(x)}:Lemma 50 (potential <strong>the</strong>ory estimate). Suppose that K(x) = k(|x|) satisfies (C2), (Z1), (Z2) and (Z3).Suppose also that p > . Then <strong>the</strong>re exists positive constants δ and C such that inequalitydd−1k ′ (R)sup (u ∗ ∆K) (x) ≤ C‖u‖ L 1x∈R Rd( ) q/d ‖u‖L 1where R =(5.107)‖u‖ L pholds <strong>for</strong> all nonnegative u ∈ L 1 (R d ) ∩ L p (R d ) satisfying R ≤ δ.Remark: By Lemma 47, potentials satisfying <strong>the</strong> conditions of Proposition 48 automatically satisfy (C2),(Z1), (Z2) and (Z3).Proof of Lemma 50. Recall that ∆K(x) = k ′′ (|x|) + (d − 1) k′ (|x|)|x|so that (Z3) implies that ∆K(x)
To go from (5.108) to (5.109) we have used <strong>the</strong> fact that k ′ is nonnegative in a neighborhood of zero (becauselim r→0 + k ′ (r)/r = +∞ ) and that u is also nonnegative. To go from (5.111) to (5.112) we have used <strong>the</strong>fact that k ′ is increasing in a neighborhood of zero (because lim r→0 + k ′′ (r) = +∞). Finally, to go from(5.112) to (5.113) we have used <strong>the</strong> fact that, since q < d, ∫ ɛ∫0 rd−1−q dr = ɛ d−q /(d − q). Let us now estimate|y|≥ɛ u(x − y)∆K(y)dy. On one hand k′′ (r) and k ′ (r)/r go to 0 as r → +∞. On <strong>the</strong> o<strong>the</strong>r hand k ′′ (r) andk ′ (r)/r go monotonically to +∞ as r → 0 + . There<strong>for</strong>e <strong>for</strong> ɛ small enough we have{}sup ∆K(x) = sup k ′′ (r) + (d − 1) k′ (r)|x|≥ɛr≥ɛrwhich gives, since u is nonnegative,∫|y|≥ɛu(x − y)∆K(y)dy ≤ d ‖u‖ L 1≤ k ′′ (ɛ) + (d − 1) k′ (ɛ)ɛCombining (5.113) and (5.114) we see that <strong>for</strong> ɛ small enough we have≤ d k′ (ɛ)ɛk ′ (ɛ). (5.114)ɛsup (u ∗ ∆K) (x) ≤ c k′ (ɛ)(ɛ d/q ‖u‖ L p + ‖u‖ L 1) (5.115)x∈R ɛd(‖u‖L 1) q/d.where c is a positive constant depending on d and q. To conclude <strong>the</strong> proof, choose ɛ = R =‖u‖ L pCombining (5.106) and (5.107) allows us to get (5.104), which concludes <strong>the</strong> proof of Proposition 48.6 ConclusionsThis paper develops refined analysis <strong>for</strong> well-posedness of <strong>the</strong> multidimensional aggregation equation <strong>for</strong>initial data in L p . An additional assumption of bounded second moment is needed as a decay condition <strong>for</strong>uniqueness; <strong>for</strong>tunately this condition is preserved by <strong>the</strong> dynamics of <strong>the</strong> equation. The results connectrecent <strong>the</strong>ory developed <strong>for</strong> L ∞ initial data [4, 5, 6] to recent <strong>the</strong>ory <strong>for</strong> measure solutions [16]. It turnsout that L p spaces provide a good understanding of <strong>the</strong> transition from a regular (bounded) solution to ameasure solution, which was proved to occur in [5] whenever <strong>the</strong> potential violates <strong>the</strong> Osgood condition.In [5], it is shown that <strong>for</strong> <strong>the</strong> special case of K(x) = |x|, no ‘first kind’ similarity solutions exist to describe<strong>the</strong> blowup to a mass concentration <strong>for</strong> odd space dimensions larger than one. A subsequent numerical studyof blowup [26], <strong>for</strong> K = |x|, illustrates that <strong>for</strong> dimensions larger than two, <strong>the</strong>re is a ‘second kind’ self-similarblowup in which no mass concentration occurs at <strong>the</strong> initial blowup time. Ra<strong>the</strong>r, <strong>the</strong> solution is in L p <strong>for</strong>some p < p s = d/(d − 1). The solution has asymptotic structure like <strong>the</strong> example constructed in Section 4of this paper, thus we expect after <strong>the</strong> initial blowup time, that it concentrates mass in a delta. We remarkthat <strong>the</strong>se results provide an interesting connection to classical results <strong>for</strong> Burgers equation. Our equation inone space dimension, with K = |x|, reduces to a <strong>for</strong>m of Burgers equation by defining w(x) = ∫ xu(y)dy [10].0Thus, an initial blowup <strong>for</strong> <strong>the</strong> aggregation problem is <strong>the</strong> same as a a singularity in <strong>the</strong> slope <strong>for</strong> Burgersequation. Generically, Burgers singularities <strong>for</strong>m by creating a |x| 1/3 power singularity in w, which gives a−2/3 power blowup in u. This corresponds to a blowup in L p <strong>for</strong> p > 3/2, but does not result in an initialmass concentration. However, as we well know, <strong>the</strong> Burgers solution <strong>for</strong>ms a shock immediately afterwards,resulting in a delta concentration in u. Thus <strong>the</strong> scenario described above is a multidimensional analogueof <strong>the</strong> well-known behavior of how singularities initially <strong>for</strong>m in Burgers equation. The delta-concentrationsare analogues of shock <strong>for</strong>mation in scalar conservation laws.Section 4 constructs an example of a measure solution with initial data in L p , p < p s , that instantaneouslyconcentrates mass, <strong>for</strong> <strong>the</strong> special kernel K(x) = |x| (near <strong>the</strong> origin). We conjecture that such solutions exist<strong>for</strong> more general power-law kernels K(x) = |x| α , 2 − d < α < 2. The proof of instantaneous concentrationuses some monotonicity properties of <strong>the</strong> convolution operator ⃗x/|x| which would need to be proved <strong>for</strong> <strong>the</strong>more general case.25
Several interesting open problems remain, in addition to proving sharpness of <strong>the</strong> exponent p s <strong>for</strong> moregeneral kernels. For initial data in L ps local well-posedness is not known and we do not have intuition <strong>for</strong>this.7 Appendix – Proof of Proposition 16Since all <strong>the</strong> convergences (2.38)-(2.40) take place on compact sets, one of <strong>the</strong> key ideas of this proof will beto approximate u 0 ∈ L p by a function with compact support and to use <strong>the</strong> fact that X t maps compact setsto compact sets (Lemma 14).PART I: We will prove that u(t, x) = u 0 (X −t (x))a(t, x) belongs to C([0, T ∗ ), L p (R d )). Assume first thatu 0 ∈ C c (R d ). The function u 0 is <strong>the</strong>n uni<strong>for</strong>mly continuous and, sinceit is clear that <strong>the</strong> quantitysup |X −t (x) − X −s (x)| ≤ C|t − s|,x∈R d‖u 0 (X −t (x)) − u 0 (X s (x))‖ L pcan be made as small as we want by choosing |t−s| small enough. We have <strong>the</strong>re<strong>for</strong>e proven that u 0 (X −t (x)) ∈C([0, T ∗ ), L p (R d ). Assume now that u 0 ∈ L p (R d ). Approximate it by a function g ∈ C c (R d ) and write:‖u 0 (X −t (x)) − u 0 (X −s (x))‖ L p ≤ ‖u 0 (X −t (x)) − g(X −t (x))‖ L p + ‖g(X −t (x)) − g(X −s (x))‖ L p= I + II + III.+‖g(X −s (x)) − u 0 (X −s (x))‖ L pAs we have seen above, <strong>the</strong> second term can be made as small as we want by choosing |t − s| small enough.Using Lemma 13 we getI, III ≤ C ∗ ‖u 0 − g‖ L p,which can be made as small as we want since C c (R d ) is dense in L p (R d ). We have <strong>the</strong>re<strong>for</strong>e proven that, ifu 0 ∈ L p (R d ), <strong>the</strong>n(t, x) ↦→ u 0 (X −t (x)) ∈ C([0, T ∗ ), L p (R d )). (7.116)Let us now consider <strong>the</strong> function u 0 (X −t (x))a(t, x).[0, T ∗ ] × R d . Write‖u 0 (X −t (x))a(t, x) − u 0 (X −s (x))a(s, x)‖ L p ≤‖u 0 (X −t (x)){a(t, x) − a(s, x)}‖ L pRecall that a(t, x) is continuous and bounded on+‖{u 0 (X −t (x)) − u 0 (X −s (x))}a(s, x)‖ L p= I + II.Since a(s, x) is bounded, (7.116) implies that II can be made as small as we want by choosing |t − s| smallenough. Let us now take care of I. Approximate u 0 by a function g ∈ C c (R d ) and writeI≤ ‖{u 0 (X −t (x)) − g(X −t (x))}{a(t, x) − a(s, x)}‖ L p+‖g(X −t (x)){a(t, x) − a(s, x)}‖ L p= A + B.From Lemma 13 we haveA ≤ 2 C ∗ ‖u 0 − g‖ L psup |a(t, x)|,(t,x)∈[0,T ∗ ]×R d26
and since C c (R d ) is dense in L p , A can be made as small as we want. Let Ω denote <strong>the</strong> compact support ofg. Using Lemmas 13 and 14 we obtain:(∫B =(≤Ω+Ct∣∣g(X −t (x)){a(t, x) − a(s, x)} ∣ ) 1/pp dxsup |a(t, x) − a(s, x)|x∈Ω+ct)C ∗ ‖g‖ L p.Since a(t, x) is uni<strong>for</strong>mly continuous on compact subset of [0, T ∗ ] × R d , it is clear that by choosing |t − s|small enough we can make B as small as we want. We have proven thatu ∈ C([0, T ∗ ], L p (R d ).PART II: In Part I we have proven that u ∈ C([0, T ∗ ], L p ). The same proof work to show that u ɛ , uvand u ɛ v ɛ belong to C([0, T ∗ ], L p ) (v(t, x) is continuous and bounded, so we can handle it exactly like a(t, x).)PART III: Let us now prove that u ɛ (t, x) = u ɛ 0(Xɛ−t (x))a ɛ (t, x) converges to u(t, x) = u 0 (X −t (x))a(t, x).For convenience we write ɛ instead of ɛ k . To do this, we will successively prove:u 0 (Xɛ −t (x)) → u 0 (X −t (x)) in C([0, T ∗ ), L p ), (7.117)u ɛ 0(Xɛ −t (x)) → u 0 (X −t (x)) in C([0, T ∗ ), L p ), (7.118)u ɛ 0(Xɛ −t (x))a ɛ (t, x) → u 0 (X −t (x))a(t, x) in C([0, T ∗ ), L p ). (7.119)To prove (7.117), approximate u 0 ∈ L p (R d ) by a function g ∈ C c (R d ) and write:sup t∈[0,T ∗ ) ‖u 0 (Xɛ −t (x) − u 0 (X −t (x))‖ L p ≤ sup t∈[0,T ∗ ) ‖u 0 (Xɛ−t (x)) − g(Xɛ −t (x))‖ L p +sup t∈[0,T ∗ ) ‖g(Xɛ−t (x)) − g(X −t (x))‖ L p + sup t∈[0,T ∗ ) ‖g(X −t (x)) − u 0 (X −t (x))‖ L p= I + II + III.From Lemma 13 it is clear that I and III can be made as small as we want by choosing an appropriatefunction g. Using Lemma 14, we see that, if Ω is <strong>the</strong> support of gII =sup ∣ g(X−tɛ (x)) − g(Xt∈[0,T(∫Ω+Ct−t (x)) ∣ 1/p dx) p .∗Using <strong>the</strong> uni<strong>for</strong>m continuity of g toge<strong>the</strong>r with <strong>the</strong> fact that Xɛ−t (x) converges uni<strong>for</strong>mly to X −t (x) on[0, T ∗ ] × Ω + CT ∗ , we can make II as small as we want by choosing ɛ small enough.This concludes <strong>the</strong> proof of (7.117). To prove (7.118), writesup ‖u ɛ 0(Xɛ −t (x)) − u 0 (X −t (x))‖ L p ≤ sup ‖u ɛ 0(Xɛ−t (x)) − u 0 (Xɛ−t (x)‖ L pt∈[0,T ∗ )t∈[0,T ∗ )= I + II.+ sup ‖u 0 (Xɛ−t (x)) − u 0 (X −t (x))‖ L pt∈[0,T ∗ )From (7.117) we know that II can be made as small as we want by choosing ɛ small enough. Using Lemma13 we obtainI ≤ C ∗ ‖u ɛ 0 − u 0 ‖ L p → 0 as ɛ → 0.27
Let us now prove (7.119). Writesup ‖u ɛ 0(Xɛ−t (x))a ɛ (t, x) − u 0 (X −t (x))a(t, x)‖ L pt∈[0,T ∗ ]≤supt∈[0,T ∗ ]‖{u ɛ 0(Xɛ−t (x)) − u 0 (X −t (x))}a ɛ (t, x)‖ L p+ sup ‖u 0 (X −t (y)){a ɛ (t, x) − a(t, x)}‖ L pt∈[0,T ∗ ]= I + II.Since a ɛ (t, x) is uni<strong>for</strong>mly bounded on [0, T ∗ ] × R d , it is clear from (7.118) that I → 0 as ɛ → 0. If u 0 is inC c (R d ), <strong>the</strong>n it is easy to prove that II → 0 as ɛ → 0. If u 0 ∈ L p (R d ), <strong>the</strong>n approximate it by g ∈ C c (R d )and proceed as be<strong>for</strong>e.Part IV: To prove that u ɛ kv ɛ kconverges to uv in C([0, T ∗ ], L p ), proceed exactly as in <strong>the</strong> proof of(7.119).AcknowledgmentsWe thank José Antonio Carrillo and Dejan Slepčev <strong>for</strong> useful comments. The authors were supportedby NSF grant DMS-0907931. JR acknowledge support from <strong>the</strong> project MTM2008-06349-C03-03 DGI-MCI (Spain) and 2009-SGR-345 from AGAUR-Generalitat de Catalunya. We thank <strong>the</strong> Centre de RecercaMatemàtica (Barcelona) <strong>for</strong> providing an excellent atmosphere <strong>for</strong> research.References[1] L.A. Ambrosio, N. Gigli, G. Savaré, Gradient flows in metric spaces and in <strong>the</strong> space of probabilitymeasures, Lectures in Ma<strong>the</strong>matics, Birkhäuser, 2005.[2] J.D. Benamou, Y. Brenier, A computational Fluid Mechanics solution to <strong>the</strong> Monge-Kantorovichmass transfer problem, Numer. Math., 84 (2000), pp. 375–393.[3] D. Benedetto, E. Caglioti, M. Pulvirenti, A kinetic equation <strong>for</strong> granular media, RAIRO Modél.Math. Anal. Numér., 31, (1997), 615–641.[4] A. Bertozzi, J. Brandman, Finite-time blow-up of L ∞ -weak solutions of an aggregation equation, toappear in Comm. Math. Sci., special issue in honor of Andrew Majda’s 60th birthday.[5] A. Bertozzi, J.A. Carrillo, T. Laurent, Blowup in multidimensional aggregation equations withmildly singular interaction kernels, Nonlinearity, 22, (2009), pp. 683-710.[6] A. Bertozzi, T. Laurent, Finite-time blow-up of solutions of an aggregation equation in R n , Comm.Math. Phys., 274 (2007), pp. 717–735.[7] P. Biler and A. Woyczyński Global and Expoding Solutions <strong>for</strong> Nonlocal Quadratic Evolution Problems,SIAM J. Appl. Math., 59 (1998), pp. 845-869.[8] A. Blanchet, J. A. Carrillo, and N. Masmoudi, Infinite Time <strong>Aggregation</strong> <strong>for</strong> <strong>the</strong> Critical Patlak-Keller-Segel model in R 2 , to appear in Comm. Pure Appl. Math.[9] A. Blanchet, J. Dolbeault, and B. Perthame, Two-dimensional Keller-Segel model: optimalcritical mass and qualitative properties of <strong>the</strong> solutions, Electron. J. Differential <strong>Equation</strong>s, (2006), No.44, 32 pp. (electronic).[10] M. Bodnar and J.J.L. Velázquez, An integro-differential equation arising as a limit of individualcell-based models, J. Differential <strong>Equation</strong>s 222, (2006), pp. 341–380.28
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