Nonlocal interaction PDEs with repulsion and attraction:

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Nonlocal interaction PDEs with repulsion and attraction:

the quadratic diffusion case

Marco Di Francesco Departament de Matem`atiques Universitat Aut`onoma de Barcelona

Work in collaboration with M. Burger and M. Franek (University of M¨unster) Frontiers of Mathematics and Applications - Summer Course UIMP 2011

Santander (Spain), August 15-19, 2011

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The evolutionary PDE

We consider

∂_tρ=div(ρ∇(ερ−G ∗ρ)) (1)

posed on R^d with initial datum inL²(R^d)∩L¹(R^d) withρ≥0,ε >0.

Assumptions on the kernel G

G is radial, i. e. G(x) =g(|x|),

G is smooth, i. e. G ∈W^1,1∩L^∞∩C²(R^d), G is strictly attractive, i. e. g⁰(r)<0 for allr >0, G is bounded from below, i. e. (not restrictive)g ≥0, G is infinitesimal at infinity, i. e. limr→+∞g(r) = 0, g⁰⁰(0)<0.

Notice that the attractivity and smoothness assumptions imply g⁰(0) = 0.

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Basic properties

Conservation of non–negativity

Due to ρ₀ ≥0 we have ρ(x,t)≥0 almost everywhere for all t >0.

Mass Conservation

M :=

Z

ρ(x,t)dx = Z

ρ0(x)dx, for all t ≥0. We shall assume for simplicity that M = 1.

Conservation of the center of mass Let

CM[ρ(t)] :=

Z

xρ(x,t)dx,

then CM[ρ(t)] =CM[ρ₀] for allt ≥0. Crucial hypotheses needed:

G(−x) =G(x).

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Energy functional

We recall theenergy functional E[ρ] := ε

2 Z

R^d

ρ²(x)dx−1 2

Z

R^d

Z

R^d

G(x−y)ρ(y)ρ(x)dydx. (2) The following energy identitysatisfied by the solution to (1) easily follows by formal computation:

E[ρ(t)] + Z T

0

Z

R^d

ρ|∇(ερ−G∗ρ)|²dxdt =E[ρ0]. (3) The identity (3) can be proven rigorously in the context of the Wasserstein gradient flow theory developed in[Ambrosio, Gigli, Savar´e, Birkh¨auser 2003].

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Motivations

Interacting particles

Assume N particles in R^d, located on X₁(t), . . . ,X_N(t) respectively and having masses m1, . . . ,mN respectively, are subject to binary interactions of the form ^{1 2 3}

d

dtX_j(t) =−

N

X

j=1,j6=i

m_j∇I(X_i(t)−X_j(t)), j = 1, . . . ,N,

where the interaction force ∇I is decomposed into a repulsivepart and an attractive part, namely

∇I(x) =−∇G(x) +∇F(x).

In order to have attractive and repulsive effects, we require G(x) =g(|x|), g⁰(r)<0, as r >0, F(x) =f(|x|), f⁰(r)<0, as r >0.

1[Mogilner, Edelstein-Keshet - J. Math. Biol. 1999]

2[Morale, Capasso, Oelschlaeger - J. Math. Biol. 2005]

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Figure: N interacting particles

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Figure: Attractive and repulsive forces acting on two particles

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Repulsion with short range

Since the total mass M =PN

i=1mi is preserved along the flow, we shall assume once againM = 1.

We assume now (cf. [Burger, Capasso, Morale - Nonlinear Analysis RWA 2006]) F(x) :=λ^dV(λx), V(x) =v(|x|), v⁰(r)<0 asr >0, with V ∈L¹(R^d), V ≥0. By taking λ1, it is clear that the range or repulsion gets smaller and smaller (short repulsion range).

A typical situation occurs when λ=λ(N) is an increasing function of the number of particles. Here, the repulsion range gets smaller and smaller as the number of interacting individuals diverges.

The attraction range is taken independent onN. We refer to this as large attraction range.

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The empirical measure

Let us now consider the empirical measure µ(t) :=

N

X

i=1

m_iδ_X_i_(t).

It is an easy exercise to check that µsatisfies the following integro-PDE in the sense of distributions

∂µ

∂t =div(µ∇F ∗µ)−div(µ∇G∗µ).

By sending λ→+∞ one obtain (formally) that F * εδ0, ε:=kVk_L1,

and the limiting measure for µ (if it exists) satisfies the integro-PDE

∂µ

∂t =εdiv(µ∇µ)−div(µ∇G ∗µ).

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Motivations Mathematical motivation

A minimization problem

The minimization problem argmin_ρ∈L¹

+(R^d)

Z

R^d

Φ(ρ(x))dx−1 2

Z

R^d

Z

R^d

ρ(x)ρ(y)G(x−y)dxdy

was first studied in [Lions - Ann. Inst. H. Poincare 1984]. Existence of nontrivial minimizers was proven under the assumptions of

Total mass sufficiently large, Φ(tu)≤t^νΦ(u) with 1< ν <2,

G slow decaying at infinity, i. e. G(tx)≥t^−αG(x) withα∈(0,d), Φ(u) =o(u¹⁺^α^d) as u→0.

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A critical exponent

In [Bedrossian, preprint 2010]⁴, it is proven that nontrivial minimizers exist under the assumptions

G ∈L¹₊,

Φ(u) =cu²+o(u²) asu →0 with c >0, either c = 0 or 2c <R

G.

It turns out indeed that m= 2 is a criticalexponent in Φ(u) =u^m. Roughly speaking:

Ifm>2, then aggregation dominates and it produces formation of nontrivial stationary patterns,

Ifm<2, then diffusion dominates,

Ifm= 2, then the behavior depends on the ‘relative size’ of the diffusion and aggregation terms.

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A result in one space dimension

Clearly, the minimization problem is related to the existence of steady states of (1).

In the case d = 1, a partial result in the casem= 2 was recovered in [Burger, DF - NHM 2008], namely:

Ifε >kGk_L1, then no nontrivial steady states exist.

Ifε1 then there exists a nontrivial steady state, this is achieved by perturbing the case ε= 0 with an implicit function theorem approach in the pseudo-inverse variable (cf. one dimensional Wasserstein tools) Let us also mention [Primi, Stevens, Velazquez - Comm. PDE 2009]

(linear diffusion, small perturbation regime).

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Stationary solutions

A key question: large time behavior

How does the solution to (1) behave as t→+∞? There are (basically) three possibilities:

(i) Diffusion dominated case: ρ(t) decays to zero in someL^p norm with p >1. In this case, the repulsive effects dominates.

(ii) Aggregation dominated case: ρ(t) concentrates to a singular measure (delta) in finite or infinite time. Here, the aggregation effect dominates.

(iii) Balanced case: ρ(t) converges to some (stable) non trivialL¹ steady state for large times.

Unlike the Keller-Segel system, here no mass threshold phenomenon occurs, since the equation is quadratically homogeneous.

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Stationary states in multiple dimensions.

Threshold phenomenon

Parallel to [Bedrossian],we proved the following general property.

Let ε <kGk_L1. Then, there exists at least one non trivial L¹ steady state for (1), which is also a minimizer for the energyE[ρ].

Let ε≥ kGk_L1. Then, there exist no steady states for (1) except ρ≡0.

Finite time concentration is not possible under the present smoothness assumptions on G .

Stationary points of E[ρ] are steady states to (1)and vice-versa

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Nonexistence for ε > kG k

_L¹

Second derivative of E[ρ]

Let ρ∈L²∩ P. Then, the second order Gateaux derivative of E onρ satisfies

d²

dδ²E[ρ+δv]|_δ=0 =ε Z

R^d

v²(x)dx− Z

v(x)G∗v(x)dx, (4) for all v =div(ρV) andV ∈C_c¹(R^d).

Lemma

Let ε >kGk_L1. Then, there exists no stationary solutions to (1)in the space L²∩ P.

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Proof.

Assume ρ is a minimizer forE[ρ] under the constraintρ∈ P. Young inequality for convolutions implies

ε 2

Z

ρ²dx ≥E[ρ] = ε 2

Z

ρ²dx−1 2

Z

ρG∗ρdx ≥ ε− kGk_L1

2 Z

ρ²dx (5) with ε− kGk_L1 >0. Take a family ρλ(x)≥0 such thatR

ρλ(x)dx = 1 and R

ρ²_λ(x)dx →0 asλ→+∞. Clearly

E[ρλ]→0, as λ→ ∞.

Therefore, a minimizer ρ∞ for E[ρ] inP would imply thatE[ρ∞]>0 and we would necessarily have 0<E[ρ_λ]<E[ρ∞] for λlarge enough, which is a contradiction.

Now, assume that ρis a steady state. Then, due to (4) the functional E is convex, and therefore admits only one stationary point, which coincides with its global minimizer. But this contradicts the non existence of a

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The critical case ε = kG k

_L¹

Lemma

Let ε=kGk_L1. Then, there exists no stationary solutions to (1)in the space L²∩ P.

Proof.

Similarly to the previous case, a stationary solution must be a global minimizer because of the convexity ofE. Taking the same familyρλas before, with

0≤E[ρλ]≤^ε₂R

ρ²_λdx→0, a minimizerρ∞should then satisfyE[ρ∞] = 0. Now, Young inequality for convolutions says that this is possible only ifρ∞=cG∗ρ∞, for somec>0, and integration onR^d implies

ρ∞=H∗ρ∞, H:= 1 kGk_L1

G, Z

H(x)dx= 1.

Taking the Fourier transform on both side of the above identity, we get ρc∞(ξ) =H(ξ)b ρc∞(ξ), ξ∈R^d,

and sinceHis radial, nonnegative and with unit mass, it is easy to check that H(ξ)b <H(0) = 1 for allb ξ6= 0, which yieldsρc∞= 0 onR^d\ {0}, which is a

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Steady states for ε < kG k

_L¹

Theorem (Existence of minimizers)

Let ε <kGk_L1. Then, there exists a radially symmetric non-increasing minimizer ρ∞∈ P ∩L²(R^d) for the entropy functional E restricted to P with ρ6= 0.

For the proof in any d we refer to [Bedrossian]. We just prove Lemma

Let ε <kGk_L1. Then,inf_ρ∈P∩L2(R^d)E[ρ]<0.

Proof.

We consider the familyσλ(x) =_2λ¹χ[−λ,λ](x)∈ P ∩L². We have E[σλ] = ε

4λ− 1 8λ²

Z λ

−λ

Z λ

−λ

G(x−y)dydx= 1 4λ

ε−

Zλ

−λ

G(z)dz

, and there existsλsuch thatE[σ ]<0.

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The one dimensional case

Existence and uniqueness of steady states

With d = 1 we can characterize all the steady states as follows. From now one we shall assume kGk_L1 = 1 for simplicity andsupp(G) =R.

Theorem (Burger-DF-Franek - 2011 preprint)

Let ε <1. Then, there exists a unique ρ∈L²∩ P with zero center of mass which solves

ρ∂_x(ερ−G∗ρ) = 0.

Moreover,

ρ is symmetric and monotonically decreasing on x >0, ρ∈C²(supp[ρ]),

supp[ρ]is a bounded interval in R,

ρ has a global maximum at x = 0 and ρ⁰⁰(0)<0, ρ is the global minimizer of the energy

εR ₂ ₁R

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Comments

The above result is surprising for the following reasons:

The functionalE[ρ] is not convex when ε <kGk_L1.

Our model is aλ→+∞ limit of repulsive–attractive potentials W_λ(x) =λV(λx)−G(x), V,G ∈L¹₊, λ1 which generate very complex dynamics (stability vs. instability), cf.

[Fellner, Raoul - Carrillo et al.]

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The one dimensional case Necessary conditions

Continuity in one space dimension

Lemma (1d regularity)

Let ρ be an L²∩ P steady state to (1)in one space dimension. Thenρ is continuous onR.

Proof.

From the energy identity (3) we obtain (after some computations with the dissipation)

Z

R^d

ρρ²_xdx <+∞.

Then, the one dimensional Sobolev embedding theorem implies ρ continuous.

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Connected support

Lemma (Steady states have connected support) Let ρ solve

ρ∂_x(ερ−G ∗ρ) = 0 a.e. on R. (6)

Then, supp(ρ) is a connected set.

Proof.

LetAandB be to consecutive connected components ofsupp[ρ],a= supA,b= infB, a<b. Evaluate the evolution of the energyE[u(x,s)] along the solution to

us+ (uV)x = 0,u(x,0) =ρ(x), withV ∈C¹(R) such thatV =−1 on (−∞,a] and V = 1 on [b,+∞). _ds^dE[u(x,s)]|s=0= 0 and some computations imply

0 =ε Z

ρρxV = Z

ρVG⁰∗ρ=− Z a

−∞

Z +∞

b

ρ(x)G⁰(x−y)ρ(y)dydx +

Z +∞

b

dx Z a

−∞

dyρ(x)G⁰(x−y)ρ(y)dydx.

The assumptions onG imply thenρ(x)ρ(y) = 0 on (x,y) =A×B, which is a

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Symmetric rearrangement

Proposition

Let ρ∞ be a minimizer for the energy E[ρ] = ε

2 Z

ρ²(x)dx−1 2

Z Z

G(x−y)ρ(x)ρ(y)dydx under the constraint that the center of mass is zero. Then, ρ∞ is symmetric and monotonically decreasing on x >0.

Proof.

Assumeunot symmetric and decreasing onx>0, define

u^∗(x) = sup{t≥0 :meas({u>t})>2|x|}.We have to checkE[u^∗]<E[u]. It is easy to check thatR

(u^∗)²dx=R

u²dx. Then, the result follows from theRiesz’s rearrangement inequality, see e. g. [Lieb and Loss, AMS - 2007],

Z

R^d

Z

R^d

f(x)g(x−y)h(y)dydx≤ Z

R^d

Z

R^d

f^∗(x)g^∗(x−y)h^∗(y)dydx, (7) which holds for all nonnegative functionsf,g,hvanishing at infinity.

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An interpretation of the energy level

Lemma

Let ρ∈ P be a1d steady state, i. e.

ερ=G ∗ρ+C onsupp[ρ] (8) for some C ∈R. Then, C = 2E[ρ].

Proof.

Multiply (8) byρand integrate onsupp[ρ]:

ε Z

supp[ρ]

ρ²dx= Z

supp[ρ]

ρG∗ρ+C, and this proves the Lemma.

Lemma (Translation invariance)

Let ρ∈ P ∩L² and let x₀ ∈R. Letρ_x₀ be defined by ρ_x₀(x) :=ρ(x+x₀).

Then, E[ρ ] =E[ρ].

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Compact support

Lemma

Let ρ be a steady state for (6)with ε <1. Then, the support ofρ is compact.

Proof.

Suppose the (connected) support ofρis of the form (a,+∞). Then ερ=G∗ρ+ 2E[ρ], onsupp[ρ]

impliesE[ρ] = 0, otherwise the integration of the above formula onsupp[ρ] gives a contradiction. Therefore

0 =ερ(a) = Z

R

G(a−y)ρ(y)dy

which is a contradiction sincesupp[G] =R. A similar situation occurs if

supp[ρ] = (−∞,b). Therefore,supp[ρ] =Randερ=G∗ρonR. Integration overR gives a contradiction.

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Symmetrization

Lemma

Let ρ be a steady state. Then there exists a symmetric st. stateρes. t.

E[ρ] =e E[ρ].

Proof.

From previous lemmas we know thatsupp[ρ] = (a,b) for somea,b∈Rand

ερ(x) =G∗ρ(x) +C for x∈(a,b), (9)

withC=−Rb

a G(a−y)ρ(y)dy=−Rb

a G(b−y)ρ(y)dy. Letρ(x) =ρ(x+x0) with x0= (a+b)/2. Thenρis still a steady state and it satisfiesE[ρ] =E[ρ] thanks to a previous Lemma. supp[ρ] is symmetric. Leteρ(x) :=¹₂(ρ(x) +ρ(−x)). Clearly, supp[ρ] =e supp[ρ]. Using the symmetry ofG one can check that, for allx∈supp[ρ],e

ερ(xe ) =

Z (b−a)/2

(a−b)/2

G(x−z)ρ(z)dze +C.

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Minimizers have maximal support

Lemma (Support of a minimizer)

Let ρ∞ be a global minimizer to E . Letρ be a steady state such that meas(supp[ρ∞])≤meas(supp[ρ]).

Then ρ is also a minimizer.

Proof.

Due to the translation invariance, we can assumesupp[ρ∞]⊂supp[ρ]. The computation of the second variation ofE around the minimizerρ∞ along the directionρ∞−ρyields

0≤ d²

dδ²E[ρ∞+δ(ρ−ρ∞)]|δ=0=−2E[ρ] + 2E[ρ∞], and this completes the proof.

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The one dimensional case Existence and uniqueness of steady states

The Krein–Rutman Theorem

Theorem (Krein–Rutman Theorem, strong version)

Let X be a Banach space, K ⊂X a solid cone, i. e. such thatλK ⊂K for allλ≥0and such that K has a nonempty interior K0. Let T be a

compact linear operator which is strongly positive with respect to K , i. e.

such that T[u]∈K₀ if u∈K . Then,

(i) The spectral radius r(T) is strictly positive and r(T) is a simple eigenvalue with an eigenvector v ∈K0. There is no other eigenvalue with a corresponding eigenvector v ∈K .

(ii) |λ|<r(T)for all other eigenvalues λ6=r(T).

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The functional equation

Assume ρ is a symmetric steady states with unit mass, monotonically decreasing on the positive semi-axis,ε <1. This is equivalent to

ερ(x) = Z L

−L

G(x−y)ρ(y)dy +C, C = 2E[ρ]. (10) Taking the x-derivative on [−L,L] we obtain

ερ⁰(x) = d dx

Z L

−L

G(x−y)ρ(y)dy = d

dxG ∗ρ(x) = Z L

−L

G(x−y)ρ⁰(y)dy

=− Z _L

0

G(x+y)ρ⁰(y)dy+ Z _L

0

G(x−y)ρ⁰(y)dy

= Z L

0

[G(x−y)−G(x+y)]ρ⁰(y)dy.

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Application of KR Theorem

Assuming that ρ∈C¹([−L,L]), finding a steady state with the above assumptions is equivalent to find ρ on [0,L] such that

ρ(L) = 0, −ρ⁰(x) =u(x), x∈[0,L], u ≥0, andu solves εu =

Z L

0

H(x,y)u(y)dy, H(x,y) =G(x−y)−G(x+y).

Ingredients

Banach space X_L=

f ∈C¹([0,L]) : f(0) = 0 with the norm kfk_X_L =kfk_L^∞_([0,L])+kf⁰k_L^∞_([0,L]).

Compact operator H_L[u](x) :=RL

0 H(x,y)u(y)dy =RL

0(G(x−y)−G(x+y))u(y)dy. Solid cone K :={f ∈X : f ≥0}: any functionf ∈K with f⁰(0)>0 belongs to the interiorK0 of K.

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Strong positivity of the operator

Lemma

The compact operator H_L is strongly positive, i. e. it maps K into K₀. Proof.

SinceG is decreasing on the half-line [0,+∞) we get

H(x,y) =G(x−y)−G(x+y)≥0, on x,y ≥0.

Hence, for a givenu∈K, we haveHL[u](x) =RL

0 H(x,y)u(y)dy≥0 for allx∈[0,L]

and

HL[u](0) = Z L

0

H(0,y)u(y)dy= Z L

0

(G(−y)−G(y))u(y)dy= 0.

ThereforeHL[u]∈K. Moreover, (HL[u])⁰(0) =

Z L

0

(G⁰(−y)−G⁰(y))u(y)dy=−2 ZL

0

G⁰(y)u(y)dy>0, which provesHL[u]∈K0.

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Existence and uniqueness of eigenfunctions

By applying the KR Theorem, we have then proven what follows:

Proposition

For a fixed L>0there exists a unique symmetric function ρ∈C²([−L,L]) with unit mass and withρ⁰(x)≤0on x ≥0 such thatρ solves

ερ(x) = Z _L

−L

G(x−y)ρ(y)dy +C, C = 2E[ρ].

for someε=ε(L)>0. Such function ρ also satisfiesρ⁰⁰(0)<0.

Moreover, ε(L) is the largest eigenvalue of the compact operator G_L[ρ](x) :=RL

0 [G(x−y) +G(x+y)−G(L−y)−G(L+y)]ρ(y)dy on the space Banach YL:={ρ∈C([0,L]) : ρ(L) = 0}, and any other eigenfunction of G_L on Y_L with unit mass has the corresponding eigenvalue ε⁰ satisfying|ε⁰|< ε(L).

We observe that the uniqueness is achieved by the unit mass constraint (eigenvalues are

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Behavior of the function ε(L)

Proposition

The simple eigenvalue ε(L) found in the previous Proposition is uniquely determined as a function of L with the following properties

(i) ε(L) is strictly increasing with respect to L (ii) limL→+∞ε(L) = 1

(iii) ε(0) = 0.

The proof is omitted. (i) is obtained by the formula ε(L)uL(x) =HL[uL](x) =

Z L

0

H(x,y)uL(y)dx

forx ∈[0,L], multiplied byuL and integrated on [0,L]. In order to evaluate the monotonicity ofε(L) one considers the variationε(L+δ)−ε(L) and proves that it is positive whenδ >0.

(ii) is obtained by contradiction, assuming limL→+∞ε(L) =ε0<1 (using that for

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Main result

Theorem

Let ε <1. Then, there exists a unique ρ∈L² solution to ρ∂_x(ερ−G∗ρ) = 0,

with unit mass and zero center of mass. Moreover,

ρ is symmetric and monotonically decreasing on x >0, ρ∈C²(supp[ρ]),

supp[ρ]is a bounded interval in R,

ρ has a global maximum at x = 0 and ρ⁰⁰(0)<0, ρ is the global minimizer of the energy

E[ρ] = ₂^εR

ρ²dx −¹₂R

ρG∗ρdx .

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Proof of the main theorem I

We know that there exists a minimizer ρ∞ with unit mass and zero center of mass, which is symmetric and monotonically decreasing on x >0 and compactly supported on a certain [−L,L]

From the two above Propositions, we know that there exists a unique steady state with such properties, because the correspondence

ε=ε(L) is one-to-one.

So, the only possibility to violate uniqueness of steady states with unit mass and zero center of mass is to have a steady state which violates either the monotonicity property or the symmetry.

Suppose first that there exists a steady state with zero center of mass ρ which is not symmetric,supp[ρ] = [−L⁰,L⁰] (not restrictive).

Then, we know from one Lemma above that it is possible to

construct a symmetric steady state ρewith the same energy ofρ and with the same support ofρ. Now, there are two possibilities: eitherρe

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Proof of the main theorem II

Ifρeis a minimizer, then so isρ (they have the same energy!), and this is not possible becauseρ would be symmetric (every minimizer is symmetric by the rearrangement property).

Ifρeis not a minimizer, then the support ofρeis strictly contained in the support of ρ∞, andρeis not monotonically decreasing on x>0 because otherwise it would be the unique minimizer provided before.

Therefore,−ρe⁰ is an eigenfunction for H_L⁰ in the spaceX_L⁰ which is not belonging to the solid coneK

Therefore, KR and the fact thatε(L) is increasing imply that L⁰ >L, since ρ is an eigenfunction outside the solid coneK and it therefore should have an eigenvalue strictly less thanε(L⁰). This implies that ε(L)< ε(L⁰) and therefore L<L⁰.

Now this is a clearly a contradiction because we said before that the support or ρeis strictly contained in the support ofρ∞, so L>L⁰. The case in whichρ is symmetric but not monotonic onx >0 can be

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Concavity of ρ for small ε

Corollary

There exists a value ε₀∈(0,1)such that, for all ε∈(0, ε₀)the

corresponding stationary solution is concave on the whole interval [0,L].

Proof.

We can differentiate twice w.r.tx in ερ(x) =

Z L

−L

G(x−y)ρ(y)dy+C to obtain

ερ⁰⁰(x) = Z L

−L

G⁰⁰(x−y)ρ(y)dy

for allx ∈[−L,L]. Therefore,G⁰⁰is evaluated on the interval [−2L,2L] in the above integral. We know thatLis a monotonically increasing function ofεwith

limε&0L(ε) = 0. SinceG⁰⁰(0)<0, andG ∈C², then there existsL0>0 such that G⁰⁰<0 on [−2L,2L]. Letε be the eigenvalue inK corresponding toL=L. Then,

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Future perspectives

Future work

1 The casesupp[G] = [−l,l] bounded is of great interest since it allows for multiple steady states. We can prove that a necessary condition is that the distance between two connected components ofρ is at least 2l.

2 Large time stability of the nontrivial steady states (under preparation).

3 Large time decay in case of non-existence of steady states.

4 In the critical case ε=kGk_L1 we conjecture a large time dynamics of fourth order type (thin film).

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Future perspectives

End of the talk

Thank you for your attention!

Nonlocal interaction PDEs with repulsion and attraction: