Analysis and Mathematical Physics (2023) 13:17 https://doi.org/10.1007/s13324-023-00780-3 Skeleton for the one-dimensional aggregation equation Juan Carlos Cantero1 · Joan Orobitg1 Received: 21 October 2022 / Revised: 25 October 2022 / Accepted: 26 November 2022 © The Author(s) 2023 Abstract For the aggregation equation in R, we consider the evolution of an initial density corresponding to the characteristic function of some set 0 . We study the limit measure at the blow up time 1 for 0 open or compact and we inspect the limit set (skeleton) where this measure is supported. Keywords Aggregation equation · Partial differential equations · Limit behaviour · Active scalars Mathematics Subject Classification Primary 35Q35 · 35F20 ; Secondary 35B30 1 Introduction Consider the aggregation equation ⎧ ⎪ ⎨ρt + div(ρv) = 0, v(·, t) = −∇ N ∗ ρ(·, t), (1) ⎪ ⎩ ρ(·, 0) = ρ0 , where N is the fundamental solution of the laplacian and ρ0 is a bounded compactly supported function. This problem has been a very active area of research in the literature (see [2, Sect. 1]). Juan Carlos Cantero and Joan Orobitg have contributed equally to this work. B Juan Carlos Cantero cantero@mat.uab.cat Joan Orobitg orobitg@mat.uab.cat 1 Departament de Matemàtiques, Universitat Autònoma de Barcelona, Bellaterra (Barcelona), Spain 0123456789().: V,-vol 17 Page 2 of 10 J. C. Cantero, J. Orobitg As in [2], we can compute an explicit solution to (1) along the flow map X (·, t) as  −1 1 ρ0 (α) ρ(X (α, t), t) = −t = , (2) ρ0 (α) 1 − tρ0 (α) where X (·, t) is defined, as usual, by the ODE  d dt X (α, t) = v(X (α, t), t), (3) X (α, 0) = α. We are interested in the case of the evolution of an aggregation patch, that is, when ρ0 = χ0 for some bounded domain 0 . In this case the explicit solution can be simply written, according to (2), as 1 ρ(·, t) = χ , (4) 1−t t where t = X (0 , t), 0 ≤ t < 1. That is, the set t is the evolution of 0 along X (·, t). With this expression, it is clear that the blow-up occurs at time t = 1. It is well known that equation (1) preserves the L 1 norm of the scalar ρ, and hence 1 ρ(·, t) L 1 = t  = ρ0  L 1 = 0  . 1−t Therefore t  = (1 − t) 0  → 0 as t → 1. In [1], they show that for an initial data of the form ρ0 = χ0 with 0 a domain of class C 1,γ then the domain t in the solution (4) is also of class C 1,γ when t < 1. A challenging question is to study in detail the structure of the skeleton, that is, the blow-up set 1 = lim X (0 , t). t→1− Explicit computations in [2, Sect. 4.1] shows that an elliptical patch collapses at time 1 to a measure supported in an interval contained in the straight line defined by the major axis. On the other hand, numerical simulations in [2, Sects. 4.2–4.4] suggest a much more cumbersome behaviour for other patches not that regular, even though they collapse into a complicated skeleton of codimension one. The dynamics of one-fold symmetric patches for the two-dimensional aggregation equation has been studied by Hmidi and Li [3]. They showed that for domains with a suitable symmetry structure the solution converges weakly towards a finite measure supported in the union of disjoint segments lying in the real axis. Since the arbitrary dimension problem is presently rather ambitious, we consider a toy model which corresponds to equation (1) but just for dimension 1. In this case, Skeleton for the one-dimensional aggregation equation Page 3 of 10 17 the fundamental solution of the laplacian is simply N (x) = 21 x , and so the kernel is  1 1/2 if x < 0, K (x) = −N x (x) = − sign(x) = 2 −1/2 if x ≥ 0. As we will see later, an advantage of considering this case is that the velocity v(X (α, t), t) is independent of t and hence it can be computed at the initial time as v0 (α) = (K ∗ χ D0 )(α). So the particle trajectories are “straight lines” as a function of t in R × [0, 1) in the sense that they can be written as X (α, t) = α + v0 (α)t. In dimension 2 or bigger we apparently lose this nice property which is fundamental to develop explicit computations concerning the behaviour of the skeleton. 1.1 Outline of the paper In Sect. 2 we prove the evolution of an open set towards a countable collection of Dirac deltas. In Sect. 3 we consider the evolution of a compact set. We prove that any compact set of vanishing Lebesgue measure is the skeleton of some well chosen compact set. In particular, Hausdorff dimensions can be distorted arbitrarily. Indeed, we prove in Theorem 3 a more general result involving measures. 2 Open set As we just said, we consider the one-dimensional aggregation equation. Explicitly, we have ⎧ ⎪ ⎨ρt + (ρv)x = 0, v(·, t) = −N x ∗ ρ(·, t), (5) ⎪ ⎩ ρ(·, 0) = ρ0 = χ0 , where ρ : R × R+ → R and v : R × R+ → R, and N is the fundamental solution of the Laplace operator, i.e. N (x) = 21 x, and therefore N x (x) = 21 sign(x). The following theorem, describing the evolution of a general bounded open set in R under equation (5), contains the choice of a representative in L ∞ (R) for the initial condition. ∞ Theorem 1 Let 0 = i=1 Ii ⊂ R, with Ii = (αi , βi ) pairwise disjoint, be a bounded open set. Let ρ and v be the solution of (5) with initial condition χ0 . Then, 1 = ∞ w ∞ i=1 {x i } and if dμt = ρ(·, t) dx, we have μt −→ μ1 = i=1 Ii  δxi , where δxi is the Dirac measure at xi . 17 Page 4 of 10 J. C. Cantero, J. Orobitg Proof of Theorem 1 In order to normalise a representative choice in L ∞ (R) for the initial condition, we assume without loss of generality that given x ∈ I j , y ∈ Ik with j = k, we have I (x, y) ∩ 0  < x − y , where I (x, y) indicates the minimum interval containing x and y. In particular, this implies that instead of (a1 , a2 ) ∪(a2 , a3 ) we will have (a1 , a3 ). It will also be seen from the proof, that if [a, b] is the smallest closed interval containing 0 and 0  = 2L > 0 then the skeleton 1 is contained in [a + L, b − L], because it is easily verified that v0 (a) = L and v0 (b) = −L. First of all, let us see that under these assumptions, there exists spatial derivative for the trajectory map at least for α ∈ 0 . Recall that, for any 0 ≤ t < 1 the trajectory map X (·, t) is the unique homeomorphism solution to the ODE  dX (α,t) dt = v(X (α, t), t), X (α, 0) = α. Hence, differentiating with respect to α the previous equation, we obtain   d dX (α, t) d = (v(X (α, t), t)) dα dt dα d 1 +∞ 1 X (α,t) = ρ(y, t) dy − ρ(y, t) dy (6) dα 2 X (α,t) 2 −∞ dX (α, t) 1 dX (α, t) = −ρ(X (α, t), t) =− ρ0 (α) . dα 1−t dα Interchanging derivatives we get     d dX (α, t) 1 dX (α, t) =− ρ0 (α) . dt dα 1−t dα Integrating and using that the homeomorphism X (·, 0) is the identity map, we obtain the spatial derivative (for any α ∈ 0 ). dX (α, t) = 1 − t. dα Secondly, we can prove that the velocity of a particle is constant along the trajectory, this is, v(X (α, t), t) = v(α, 0) =: v0 (α). (7) This just requires a simple computation, involving a change of variable y = X (α  , t). 1 +∞ v(X (α, t), t) = sign(X (α, t) − y)ρ(y, t) dy 2 −∞ 1 +∞ dX (α  , t)  = sign(X (α, t) − X (α  , t))ρ(X (α  , t), t) dα 2 −∞ dα Skeleton for the one-dimensional aggregation equation Page 5 of 10 17 1 +∞ 1 dX (α  , t)  = sign(X (α, t) − X (α  , t)) ρ0 (α  ) dα 2 −∞ 1−t dα 1 1 dX (α  , t)  = sign(X (α, t) − X (α  , t)) dα 2 α  ∈0 1−t dα 1 1 = sign(X (α, t) − X (α  , t)) (1 − t) dα  2 α  ∈0 1−t 1 = sign(X (α, t) − X (α  , t)) dα  2 α  ∈0 1 = sign(α − α  ) dα  = v0 (α); 2 α  ∈0 where we have used that X (α, t)− X (α  , t) and α −α  have the same sign since X (·, t) is a non-decreasing homeomorphism. Then, by (7) it is clear that all particle trajectory maps are straight lines. Indeed, for 0 ≤ t < 1, we have t t X (α, t) = α + v(X (α, s), s) ds = α + v0 (α) ds = α + v0 (α)t. 0 0 Now we can check that any x ∈ [αi , βi ] has the same limit point lim X (x, t). t→1− In fact, 1 ∞ 1 x v0 (x) = ρ0 (y) dy − ρ0 (y) dy 2 x 2 −∞  ∞ x αi x  1 = ρ0 (y) dy − ρ0 (y) dy − ρ0 (y) dy − ρ0 (y) dy 2 αi αi −∞ αi x = v0 (αi ) − ρ0 (y) dy = v0 (αi ) − (x − αi ), (8) αi where we have used the fact that ρ0 ≡ 1 in (αi , x). Hence, for any x ∈ [αi , βi ] , we have t→1− X (x, t) = x + (v0 (αi ) − (x − αi )) t −→ αi + v0 (αi ), (9) which does not depend on the choice of x. From now on, we denote the limit point for each interval I j = (α j , β j ) as x j := α j + v0 (α j ). Finally, we have to see the conver- ∞ gence of the measure μt defined as dμt = ρ(x, t) dx towards μ1 = i=1 Ii  δxi . In order to prove this, let f be a continuous function on R. Then, recall ∞  1 ρ(x, t) = χ(αi,t ,βi,t ) , 1−t i=1 17 Page 6 of 10 J. C. Cantero, J. Orobitg where αi,t = X (αi , t) and βi,t = X (βi , t). Summing up, we have ∞ 1  βi,t  f , μt  = f (x) dx. 1−t αi,t i=1 Let m i,t := inf f (x) and Mi,t := supx∈(αi,t ,βi,t ) f (x). Then, it is clear that x∈(αi,t ,βi,t ) for any i, 1 1 βi,t 1 m i,t (βi,t − αi,t ) ≤ f (x) dx ≤ Mi,t (βi,t − αi,t ). 1−t 1−t αi,t 1−t On the other hand, from (9) we have that βi,t − αi,t = (1 − t)(βi − αi ) and hence 1 βi,t m i,t (βi − αi ) ≤ f (x) dx ≤ Mi,t (βi − αi ). 1−t αi,t Both the left and the right hand sides of the previous inequality clearly tend to the same value f (xi )(βi − αi ), by definition of m i,t and Mi,t . Therefore we have 1 βi,t f (x) dx −→ f (xi )(βi − αi ), as t → 1− . 1−t αi,t Then  ∞    f , μt  −→ f, Ii  δxi , i=1 which proves the result.   We can also formulate the reciprocal result. Any bounded countable collection of points is the skeleton of some initial open set. Theorem 2 Let {x j }∞j=1 a bounded countable collection of points such that [c, d] is the smallest closed interval containing ∞ j=1 {x j }. Let μ1 = ∞ j=1 c j δx j , for c j > 0 and such that ∞ j=1 c j = 2L, where δ x j is the Dirac measure at x j . Then there exists a bounded open set 0 ⊆ [c − L, d + L] such that the solution of (5) with initial condition χ0 satisfies the following: ∞ i) 1 = i=1 {xi }. ii) for t and for the measure dμt = ρ(x, t) dx = 1−t χt (x) dx 1 we have that w μt −→ μ1 as t → 1− . Skeleton for the one-dimensional aggregation equation Page 7 of 10 17 Proof For any i ∈ N, let li := x j