Suppressing Chemotactic Blow-Up Through a Fast Splitting Scenario on the Plane

Transcription

1 Arch. Rational Mech. Anal. Digital Object Ientifier (DOI) Suppressing Chemotactic Blow-Up Through a Fast Splitting Scenario on the Plane Siming He & Eitan Tamor Communicate by A. Figalli Abstract We revisit the question of global regularity for the Patlak Keller Segel (PKS) chemotaxis moel. The classical 2D parabolic-elliptic moel blows up for initial mass M > 8π. We consier a more realistic scenario which takes into account the flow of the ambient environment inuce by harmonic potentials, an thus retain the ientical elliptic structure as in the original PKS. Surprisingly, we fin that alreay the simplest case of linear stationary vector fiel, Ax, with large enough amplitue A, prevents chemotactic blow-up. Specifically, the presence of such an ambient flui transport creates what we call a fast splitting scenario, which competes with the focusing effect of aggregation so that enough mass is pushe away from concentration along the x 1 -axis, thus avoiing a finite time blow-up, at least for M < 16π. Thus, the enhance ambient flow oubles the amount of allowable mass which evolve to global smooth solutions. Contents 1. Introuction A Fast Splitting Scenario Local Existence Weak Formulation Regularize Equation an Local Existence Theorems Proof of the Main Results The Three-Step Battle-Plan Step 1 Control of the Cell Density Distribution Step 2 Proof of the Main Theorem Uner the Constraine Setup (3.3) Step 3 Proof of the Main Theorem... References...

2 Siming He & Eitan Tamor 1. Introuction The Patlak Keller Segel (PKS) moel escribes the time evolution of colony of bacteria with ensity n(x, t) subject to two competing mechanisms aggregation triggere by the concentration of chemo-attractant riven by velocity fiel u := ( ) 1 n(, t), an iffusion ue to run-an-tumble effects, n t (nu) = n, u := ( ) 1 n. We focus on the two-imensional case where ( ) 1 n takes the general form of a funamental solution together with an arbitrary harmonic function ( ) 1 n(x, t) = (K n)(x, t) H(x, t), K (x) := 1 log x, H(, t). 2π The resulting PKS equation then reas n t (n c) b n = n, c = K n, x = (x 1, x 2 ) R 2, (1.1) subject to prescribe initial conitions n(x, ) = n (x). Here the ivergence free vector fiel b( ) represents the environment of a backgroun flui transporte with velocity b(x, t) := H(x, t). When b, the system is the classical parabolicelliptic PKS equation moeling chemotaxis in a static environment [21,27]. We recall the large literature on the static case b =, referring the intereste reaer to the review [19] an the follow-up representative works [3 8,1 15,2,28,29]. It is well-known that the large-time behavior of the static case (1.1) b epens on whether the initial total mass M := n 1 crosses the critical threshol of 8π: the equation amits global smooth solution in the sub-critical case M < 8π an it experiences a finite time blow-up if M > 8π [7] (when M = 8π, aggregation an iffusion exactly balance each other an solutions with finite secon moments form Dirac mass as time approach infinity [8]). 1 In this paper we stuy a more realistic scenario of the PKS moel (1.1), where we take into account an ambient environment ue to the flui transport by vector fiel b(, t). Surprisingly, we fin that alreay the simplest case of linear stationary vector fiel, b = A( x 1, x 2 ), corresponing to H(x) = 2 1 A(x2 2 x2 1 ), prevents chemotactic blow-up for M < 16π. As we shall see, the presence of such an ambient flui transport creates what we call a fast splitting scenario which competes with the focusing effect of aggregation so that enough mass is able to escape a finite time blow-up, at least for M < 16π. This scenario is likely to be enhance even further when larger amount of mass can be transporte by a more pronounce ambient fiel b(x, t) = H(, t) x q at x 1. We mention two other scenarios of non-static PKS with strong enough transport preventing blow-up for M > 8π. In[23] the authors exploit the relaxation enhancing of a vector fiel b with a large enough amplitue in orer to enforce 1 We let x enote the l 2 -size of vector x an let f p enote the L p -norm of a vector function f ( ).

3 Suppressing Chemotactic Blow-Up global smooth solutions. Here regularity follows ue to a mixing property over T 2 an T 3.In[2] it was shown how to exploit the enhance issipation effect of non-egenerate shear flow with large enough amplitue, [1], in orer to suppress the blow up in (1.1) ont 2, T 3, T R, T R 2. It is worth mentioning that the moel (1.1) is one among many attempts to take into account the unerlying flui transport effect, see, e.g. [16,18,24 26] A Fast Splitting Scenario Here, we exploit yet another mechanism that suppress the possible chemotactic blow up of the equation (1.1), where the unerlying flui flow splits the population of bacteria with ensity n exponentially fast, resulting in several isolate subgroups with mass smaller than the critical 8π. In this manner, an initial total mass greater than 8π is able to escape the finite-time concentration of Dirac mass. This provies a first no blow-up scenario over R 2,atleastforM up to 16π. We now fix the vector fiel riving a hyperbolic flow as the strain flow in [22]): b(x) := A( x 1, x 2 ). (1.2) Our aim is to show that a large enough amplitue, A 1, guarantees the global existence of solution of PKS (1.1) subject to initial mass M < 16π. Intuitively, the large enough amplitue A 1 is require so that the ambient fiel A( x 1, x 2 ) pushes away highly concentrate mass near the x 1 axis, namely, n (x)x 1. x 2 ɛ With this we can state the main theorem of the paper. Theorem 1.1. Consier the PKS equation (1.1), (1.2) subject to initial ata, n H s (s 2) with total mass, M := n 1 < 16π, such that (1 x 2 )n L 1 (R 2 ) an n log n L 1 (R 2 ). Assume n is symmetric about the x 1 -axis, an that the y-component of its center of mass in the upper half plane y (t) := 1 n(x, t)x 2 x, M := n(x, t)x M M x 2 x 2 2, is not too close to the x 1 -axis in the sense that y 2 () > 2 V (), V (t) := n(x, t) x 2 y 2 x. (1.3) M x 2 Then there exists a large enough amplitue, A = A(M, y (), V ()), such that the weak solution of (1.1), (1.2) exists for all time an the free energy (n 1 E[n](t) := log n 2 cn H(x)n(x, t)) x, H(x) = A 2 (x2 2 x2 1 ), (1.4) satisfies the issipation relation t E[n](t) n log n c b 2 xs E[n ]. (1.5) R 2

4 Siming He & Eitan Tamor We conclue the introuction with a series of remarks. Remark 1.1. (Why large enough stationary fiel prevents secon-moment collapse) Our main theorem extens the amount of critical mass, so that global regularity of (1.1), (1.3) prevails for M < 16π, provie A is large enough. To gain further insight why large enough A will prevent blow-up, we recall that the blow up phenomena in the static case, b is euce from the time evolution of the secon moment, V (t) := n(x, t) x 2 x. Inee, a straightforwar computation ( R 2 yiels, V (t) = 4M 1 M ) <, which implies that the positive V (t) ecreases 8π to zero in a finite time an hence rules out existence of global classical solutions for M > 8π. In contrast, the secon moment of our non-static PKS equation (1.1), (1.2), oes not ecrease to zero if A is chosen large enough. This is the content of our next lemma. Lemma 1.1. Let n(x, t) be the solution of (1.1) with vector fiel b(x) = A( x 1, x 2 ), subject to initial ata n such that W := n (x)(x 2 R 2 2 x2 1 )x is strictly positive. Then, if A is chosen large enough, the secon moment of the (classical) solution V (t) = n(x, t) x 2 x increases in time. R 2 Proof. First, the time evolution of V (t) can be calculate as follows: ( t V = 4M 1 M ) 2 x b n(x, t)x 8π ( = 4M 1 M ) (1.6) 2AW, W (t) = ( x 2 8π R 2 1 x2 2 )n(x, t)x. Next, we compute the time evolution of W (t): t W = ( x1 2 x2 2 ) ( n cn bn)x = 1 x y n(x, t)( 2x 1, 2x 2 ) n(y, t)xy 2AV 2π x y 2 = 1 (x1 y 1 ) 2 (x 2 y 2 ) 2 2π (x 1 y 1 ) 2 n(x, t)n(y, t)xy 2AV, (x 2 y 2 ) 2 where the last step follows by symmetrization. Since the first term on the right is boune from below by 2π 1 M2,wehave t W 1 2π M2 2AV. (1.7) Finally, notice that since W (an hence V ) are assume strictly positive, we can choose A large such that AV 1 4π M2, ( AW 2M 1 M ). (1.8) 8π Combining (1.8), (1.6) an (1.7) yiels that W (t) >, V (t) >.

5 Suppressing Chemotactic Blow-Up This shows that collapse may be prevente if A is large enough. Inee, our theorem 1.1 shows that collapse is avoie for large enough A: specifically (consult (3.4) below), A = M δ 2 with small enough δ such that δ (R 1) 2V () M, R 2 := M y 2 () 2V (). Observe that by (1.3) R > 1, which enables the choice of δ 1 an hence A 1 for the global regularity of (1.1), (1.3). We conjecture that a similar fast splitting scenario hols with even higher-egree harmonics, b(x, t) = H(, t) x q at x 1: the higher the egree q is, the larger amount of critical mass is expecte to be pushe away from the origin in ifferent irections, with even higher threshols than 16π. A main ifficulty, however, is the lack of closure to control the secon-orer moments in these higherorer cases. Remark 1.2. (On the free energy) We note that when b =, E[n] becomes the classical issipative free energy F = n log nx 1 ncx. (1.9) R 2 2 R 2 Due to the importance of the property (1.5), a weak solution of (1.1) satisfying (1.5) will be calle a free energy solution. One of the important properties of the PKS equation (1.1) with backgroun flow velocity (1.2) is the issipation of its free energy E[n]. The formal computation, inicating the energy issipation in non-static smooth solutions, is the content of our last lemma in this section. Lemma 1.2. Consier the PKS equation (1.1) with backgroun flui velocity (1.2). If the solution is smooth enough, the free energy E[n](t) is ecreasing. Proof. The time evolution of the free energy (1.4) can be compute in terms of the potential H = 2 1 A(x2 2 x2 1 ), t E[n](t) = n t (log n c H)x = n( log n c b) ( log n c H)x = n log n c b 2 x. This completes the proof of the lemma. Remark 1.3. (Smoothness) Arguing along the lines [17], one can prove that the free energy solution amits higher-orer integrability an consequently retains Sobolev smoothness for all positive time, n Cc ((, T ]; C x ) for all T <, thus our global weak solution is in fact a global strong solution.

6 Siming He & Eitan Tamor Remark 1.4. (Stability) We note that by symmetry, our 16π-threshol state in theorem 1.1 is evenly ivie between the upper- an lower-halves of the plane, each shoul contain at most 8π mass. We expect that this 16π threshol remains vali even for small asymmetric perturbations, as long as the mass in each half is kept separately below the 8π threshol. Our paper is organize as follows. In Section 2, we introuce the regularize problem to (1.1) which leas to the local existence results. In Section 3, we prove the main theorem, an in the appenix, we give etaile proofs of the results state in Section Local Existence 2.1. Weak Formulation It is stanar to unerstan the Keller Segel equation (1.1) with backgroun flui velocity (1.2) in the following weak formulation: Definition 2.1. (weak formulation) The function n is the weak solution of (1.1) if for ϕ Cc (R2 ), the following hols: ϕnx = ϕnx t R 2 R 2 1 ( ϕ(x) ϕ(y)) (x y) 4π R 2 R 2 x y 2 n(x, t)n(y, t)xy ϕ bnx. (2.1) R 2 Taking avantage of the assume symmetry across the x 1 -axis, one can further simplify the notion of a weak solution by restricting attention to the upper half plane, ={(x 1, x 2 ) x 2 }. Theorem 2.1. The function n is a weak solution of (1.1) if n := n1 x2 satisfies for ϕ Cc (R2 ), ϕn x = ϕn x t 1 ( ϕ(x) ϕ(y)) (x y) 4π R2 x y 2 n (x, t)n (y, t)xy c ϕn x ϕ bn x. (2.2) Here c (x) := ( )c x y 2π x y 2 n (y)y x y = ( )c 2π x y 2 n ( y)y, n := n1 x2.

7 Suppressing Chemotactic Blow-Up Proof. Rewrite (2.1) asfollows: ϕn x = ϕn x t 1 4π R2 1 2π R2 ϕ bn x. The thir term can be rewritten as ( ϕ(x) ϕ(y)) (x y) x y 2 n (x, t)n (y, t)xy ϕ(x) (x y) x y 2 n (x, t)n (y, t)xy c ϕ n (x)x, an we get (2.2) Regularize Equation an Local Existence Theorems In this section we introuce the local existence theorem an the blow up criterion for the Keller Segel equation with avection. The theorems are stanar, so the proofs are postpone to the appenix. The intereste reaer are referre to the papers [7,8] for further etails. In orer to prove the local existence theorem an the blow up criterion for the Keller Segel system with avection (1.1), we regularize the system as follows: n ɛ t (n ɛ c ɛ ) b n ɛ = n ɛ, c ɛ := K ɛ n, x R 2, t >, (2.3) with the regularize kernel, K ɛ, given by K ɛ (z) := K 1 ( z ɛ ) 1 2π log ɛ, K 1 (z) := { 1 2π log z, if z 4,, if z 1. (2.4) Noting that K ɛ (z) C ɛ for all z R 2, it follows that the solutions to the equation (2.3) exist for all time. The proof is similar to the corresponing proof in the classical case. We refer the intereste reaer to the paper [7] for more etails. Before stating the local existence theorems, we introuce the entropy of the solution S[n] := n log nx. (2.5) R 2 Now the local existence theorems are expresse as follows: Proposition 2.1. (Local Existence Criterion) Assume that b (x) C x, x R 2. Suppose {n ɛ } ɛ are the solutions of the regularize equation (2.3) on [, T ).If {S[n ɛ ](t)} ɛ is boune from above uniformly in ɛ an in t [, T ), then the cluster points of {n ɛ } ɛ, in a suitable topology, are non-negative weak solutions of the PKS system with avection (1.1) on [, T ) an satisfies the relation (1.5).

8 Siming He & Eitan Tamor Proposition 2.2. (Maximal Free-energy Solutions) Assume the bouneness of the vector fiel b (x) C x an the integrability of initial ata (1 x 2 )n L 1 (R2 ), n log n L 1 (R 2 ). Then there exists a maximal existence time T > of a free energy solution to the PKS system with avection (1.1), (1.5). Moreover, if T < then lim n log nx =. t T R 2 We conclue that if the entropy S[n](t) = n log n is boune, then the free energy solution of (1.1) exists locally. Moreover, if S[n](t) < for all t <, the solution exists for all time. 3. Proof of the Main Results 3.1. The Three-Step Battle-Plan We procee in three steps. The first step carrie in Section 3.2 below, is to control cell ensity istribution. From the last section, we see that an entropy boun is essential for erivation of local existence theorems for the PKS equation (1.1), (1.2). To this en, information about the istribution of cell ensity is crucial. The following lemma is the key to the proof of the main results. It shows that mass cannot concentrate along the the x 1 -axis, since we can fin a thin enough strip along the x 1 -axis with controlle amount of mass. For the rest of this section we fixe a small parameter <η 1 which will quantify the sharp estimates in the sequel. Lemma 3.1. Suppose a sufficiently smooth n is symmetric about the x 1 axis an assume that R 2 y 2 := M () > 1. (3.1) 2V () Fix a small enough <η 1. Then there exists δ = δ(y (), V (), M,η)such that if we choose A > M, the smooth solutions to the regularize (2.3) δ 2 ɛ satisfy, uniformly for small enough ɛ, n ɛ (1 η)2 (x, t)x 2R 2 M. (3.2) x 2 2δ Conition (3.2) implies, at least for M < 16π, that the mass insie that δ-strip is less than 8π. On the other han, it inicates the reason for the limitation R > 1: for if R < 1, then the boun (3.2) woul allow a concentration of mass M 8π 2R insie the strip x 2 2 2δ, which in turn leas to a finite-time blow-up. The proof of lemma 3.1 is base on the following simple observation. Given f with -center of mass at (, y f ) an variation V f = x 2 y f 2 f (x)x, we

9 Suppressing Chemotactic Blow-Up fin that its total mass outsie the strip S[y f, r] :={(x 1, x 2 ) x 2 y f r} with raius r = R 2V f /M f, oes not excee x 2 y f >r f (x)x = x 2 y f >r M f 2R 2 V f f (x) x 2 y f 2 x 2 y f 2 x f (x) x 2 y f 2 x = M f 2R 2. If we can fin the δ such that our target strip S δ := { x 2 2δ} is lying below an outsie the strip S[y f, r], then the total mass in the strip S δ woul be smaller than 1 2R 2 M f. When n ɛ (x, t) takes the role of f (x) with (y f, V f ) (y (t), V (t)),the aim is to boun the strip S[y (t), r(t)] with raius r(t) = R 2V (t)/m away from a fixe strip S δ. To this en we collect the necessary estimates on y (t), V (t) an complete the proof of the lemma in Section 3.2. The secon step, carrie in Section 3.3, is to prove the main theorem uner a constraine setup: equippe with lemma 3.1 we can control the entropy an prove a weaker form of our main theorem for any M < 16π (which is still larger than the 8π barrier), uner the constraint that the mass M concentrates far enough from the x 1 -axis. This is quantifie in our next theorem. We recall that a small parameter <η 1 was alreay quantifie in lemma 3.1. Theorem 3.1. Consier the PKS equation (1.1) with backgroun flui velocity (1.2) subject to H s (s 2) initial ata, symmetric about the x 1 -axis, with mass M = n 1 < 16π, an boune secon moment (1 x 2 )n L 1 (R 2 ).IfR 2 = y 2 M () is large enough so that 2V () R 2 > (1 η)2 M 16π M, (3.3) then there exists a large enough A = A(M, y (), V (), η), such that the free energy solution to PKS (1.1), (1.2) exists for all time. Observe that R is a imensionless parameter an R being large inicates that most of the mass concentrates away from the critical strip, S δ along the x 1 -axis, namely either the center of mass y is far enough an/or the mass variation V /M is small enough. Either way, we fin that for any M < 16π, ifr is large enough so that (3.3) hols, then (3.1) yiels global existence. Although theorem 3.1 is not as sharp as the main theorem, its proof is more illuminating an can be extene easily to prove the main theorem for the limiting case of any M < 16π. We therefore inclue its proof in Section 3.3. Finally, the thir step carrie in Section 3.4 presents the proof of the main theorem 1.1. We turn to a etaile iscussion of the three steps.

10 Siming He & Eitan Tamor 3.2. Step 1 Control of the Cell Density Distribution As pointe out before, the proof involves the calculation of y (t) an V (t), summarize in the following two lemmas. Here an below, we let A B enote the relation A CB with a constant C which is inepenent of δ. Lemma 3.2. Consier the regularize PKS equation (2.3) with backgroun flui velocity (1.2). Assume that the initial center of mass y () is boune away from the x 1 -axes in the sense that (3.1) hols. Then for any sufficiently small δ>, there exists a large enough amplitue of the ambient vector fiel, A = δ 2 M, an a constant C (inepenent of δ), such that y () Cδ > an the time evolution of (2.3), (1.2) pushes the center of mass, y (t), away from the critical strip, namely y (t) [ y () Cδ ] e At, A = M δ 2. (3.4) Lemma 3.3. Consier the regularize PKS equation (2.3) with backgroun flui velocity (1.2). Assume that the initial variation aroun the center of mass V () is not too large in the sense that (3.1) hols. Then for any sufficiently small δ>, there exists a constant C = C(V ()) such that the variation V (t) remains boune from above, V (t) [ CM δ V () ] e 2At. (3.5) We note that all the calculations mae below shoul be carrie out at the level of the regularize equation (2.3), but for the sake of simplicity, we procee at the formal level using the weak formulation (2.2). We explicitly point when there is a technical subtlety in the erivation ue to ifference between the regularize an weak formations. We begin with the proof of Lemma 3.2. To calculate the ynamics of the center of mass, one nees to formally test the equation with respect to ϕ(x) = x 2.To stay away from the critical strip S δ, however, we introuce an approximate test function 2 ϕ x 2 with a cut-off away from S δ x 2 x 2 (2δ, ), ϕ := x 2 (,δ), smooth x 2 (δ, 2δ). Note that there exists a constant C ϕ such that ϕ 2δ, x 2 2δ an ϕ δ 2 ϕ C ϕ. Here an below, we use C ϕ to enote ϕ-epenent constants that otherwise are inepenent of δ. Moreover, replacing x 2 nx with ϕnx, we lose information on the stripe {(x 1, x 2 ) x 2 2δ}; however, the contribution of this part is small in the sense that ϕn x x 2 n x 4M δ. (3.6) x 2 2δ x 2 2δ 2 To be precise, one shoul use here the usual argument of a further smooth cut-off for large ϕ at x 1 x 2 1, an recover the result uniformly with respect to that cut-off parameter.

11 Suppressing Chemotactic Blow-Up Next, one can use ϕ an the weak formulation (2.2) to extract information about y : ϕn x = ϕn x t 1 ( ϕ(x) ϕ(y)) (x y) 4π R2 x y 2 n (x, t)n (y, t)xy n c ϕx ϕ bn x = I II III IV. (3.7) Now we estimate the right han sie of (3.7) term by term. The first an secon terms are relatively easy to control from above, I = ϕn x R 2 C ϕ M, (3.8a) δ II 1 4π 2 ϕ n (x)n (y)xy 1 C ϕ R2 4π δ M2. (3.8b) Next we upper-boun the thir term in (3.7) in the following way: III = x2 c n x2 ϕx C ϕ M n x C ϕ M 2 2π δ 2π δ. (3.8c) This follows from the pointwise boun x2 c (x) 1 M shown below, an 2π x 2 the fact that supp(ϕ) is δ away from the x 1 -axis, an hence 1 x 2 1 δ, x2 c (x) = 1 2π (x y) 2 ( )c x y 2 n (y)y 1 1 x 2 2π x 2 ( )c x 2 y 2 n (y)y 1 2π M x 2 = 1 M 2π x 2. (3.9) Finally, we nee to aress aitional transport term IV in (3.7) to compete with the focusing effect. Recall that b = A( x 1, x 2 ) with A = M. First we write δ2 IV own explicitly: IV = b n ϕx = M δ 2 x 2 x2 ϕn x. Next we replace the right han sie by ϕn x. Due to the fact that x 2 x2 ϕ = x 2 = ϕ for x 2 > 2δ, the error introuce in this process originates from the thin 2δ-strip (x 2 x2 ϕ ϕ)n x x 2 x2 ϕ ϕ n x C ϕ δm, δ<x 2 <2δ δ<x 2 <2δ

12 Siming He & Eitan Tamor an we conclue that ( ) IV M δ 2 ϕn x C ϕ δm M R 2 δ 2 ϕn x C ϕ M 2. (3.1) R 2 δ Combining equation (3.7) with the upper-bouns (3.8) an the lower-boun (3.1) while recalling the efinition of A (3.4), we arrive at M 2 ϕn x C ϕ A ϕn x, t R 2 δ which implies that ( ) ϕn x ϕn x C ϕ M δ e At. (3.11) Finally, we calculate the center of mass of the upper half plane using the lower boun (3.11) an the error control (3.6) ( y (t) = 1 ) x 2 n x ϕn x ϕn x M x 2 2δ x 2 2δ 1 M x 2 n x ϕn x 1 ϕn x x 2 2δ x 2 2δ M ( ) 4δ 1 ϕn x C ϕ M δ e At M ( y () C ϕ δ ) e At. This completes the proof of lemma 3.2. Remark 3.1. The only ifference in estimating the regularize solutions (2.3) vs. the formal calculation we have one above is in terms II an III. In the calculation for the (2.3), we will nee the estimate K ɛ (z) 1 2π z, z R2. Here we show how to get a similar estimate for term II in (3.7) for the regularize equation (2.3): II = ϕ(x) x [K ɛ ( x y )]n ɛ (y)n ɛ (x)xy R2 = π R2 R2 ( ϕ(x) ϕ(y)) (x y) x y ϕ(x) ϕ(y) x y 1 4π 2 ϕ M 2 1 C ϕ 4π δ M2. K ɛ ( x y )n ɛ (x)n ɛ (y)xy n ɛ (x)n ɛ (y)xy

13 Suppressing Chemotactic Blow-Up The treatment of term III is similar to the one we gave above. Next we aress the proof of Lemma 3.3. The main goal is to calculate time evolution of the variation V (t) := x 2 y (t) 2 n(x, t)x. We again use C to enote constants which may change from line to line but are inepenent of δ. The first obstacle is that we cannot choose x 2 y 2 as a test function ue to the fact that y (t) epens on the solution. However, by the efinition of y we can expan the V -integran, ening up with the usual V (t) = x 2 2 n (x, t)x M y 2 (t). (3.12) Since we alreay know y, it is enough to calculate the R 2 x 2 2 n(x, t)x. For simplicity, we plug x 2 2 insie the weak formulation (2.2) an (3.12) to get the time evolution of V. Of course, what one really oes is to use a test function to approximate the x 2 2. Furthermore, when we use the weak formulation, we formally integrate by part twice, but since the value an the first erivative of the function x 2 2 are zero on the bounary, we will not create extra angerous bounary term. First combining (2.2) an (3.12) yiels t V = x 2 2 n (x, t)x M t R 2 t y2 (t) = ϕn x 1 ( ϕ(x) ϕ(y)) (x y) 4π R2 x y 2 n (x, t)n (y, t)xy c n ϕx ϕ bn x ( M (y ) 2) R 2 t = I II III IV M t y2 (t). (3.13) Next we estimate every term on the right han sie of (3.13). The first two terms are estimate as follows: I = (x 2 2 )n x = 2M, (3.14) an II = 1 4π R2 ( (x2 2) (y2 2 )) (x y) x y 2 n (x, t)n (y, t)xy

14 Siming He & Eitan Tamor 1 2n (x)n (y)xy 1 4π R2 2π M2. (3.15) Now for the thir term in (3.13), we use the previous upper-boun (3.9), x2 c (x), which yiels 1 2π M x 2 III = x2 c n 2x 2 x 1 2π 2 x 2 M x 2 n x 1 π M2. (3.16) Note that for the term II an III, we only estimate them formally above, one can prove the estimates explicitly using the same techniques as the one in Remark 3.1. For the IV term in (3.13), we use the (3.12) again to obtain IV = x2 2 bn x = 2A x2 2 n x = 2A(V M y 2 ).(3.17) Collecting equation (3.13) an all the estimates (3.14), (3.15), (3.16) an (3.17) above, we have the following ifferential inequality: ( ) ( ) 1 1 V (t) y 2 t M (t) C(1 M ) 2A V (t) y 2 M (t), which yiels ( 1 V (t)y 2 (t) C 1 1 ) δ 2 e 2At 1 V (t)e 2At y 2 M M M ()e2at. (3.18) Combining Lemma 3.2 with (3.18) yiels 1 [ V (t) (y () Cδ) e At] 2 1 V (t) y 2 M M (t) ( C 1 1 ) δ 2 e 2At M 1 V ()e 2At y 2 M ()e2at. By collecting similar terms, we finally have [ ( 1 V (t) 2Cδy () C 1 1 ) δ 2 1 ] V () e 2At, (3.19) M M M which completes the proof of Lemma 3.3. Equippe with the estimate on V (t), we can now conclue the proof of Lemma 3.1.

15 Suppressing Chemotactic Blow-Up Proof. (Lemma 3.1) Once <η 1 was fixe, we can clearly choose a small enough δ such that by (3.5) δ, there hols V (t) (1 η)v ()e 2At. (3.2) Now recalling that R = y () M 2V () further choose δ small enough to get y (t) [ R 2V (t) y () Cδ 1 η M = [( 1 2δe At 2δ. > 1, then we can use (3.4), (3.2) an R 1 η 1 1 η )y () Cδ ] 2V () M ] e At e At Thus, the thin δ-strip along the x 1 -axis, S δ := {(x 1, x 2 ) x 2 2δ}, lies outsie the strip centere aroun y (t), uniformly in time, S δ {(x 1, x 2 ) x 2 y (t) > R η (t)}, R η (t) := R 2V (t). 1 η M It follows that thanks to our choice of δ, the mass insie the δ-strip S δ oes not excee n (x, t)x n (x, t)x S δ { x 2 y >R η} R2 { x2 y >Rη} n (x, t) x 2 y 2 x 2 y 2 x (1 η)2 R 2 2V /M (1 η)2 2R 2 M. n (x, t) x 2 y 2 x = (1 η)2 R 2 2V /M V By symmetry, the mass insie the symmetric δ-strip, {(x 1, x 2 ) x 2 2δ} is smaller (1 η)2 (1 η)2 than 2R 2 2M = 2R 2 M, uniformly in time, which completes the proof of Lemma 3.1. Remark 3.2. One can o a similar computation to get the evolution for the higher moment estimates n(x, t) x 2k x, an erive similar results to Lemma 3.1.

16 Siming He & Eitan Tamor 3.3. Step 2 Proof of the Main Theorem Uner the Constraine Setup (3.3) With the Lemma 3.1 at our isposal, we can now turn to the proof of theorem 3.1 along the lines of [8]. Note that the actual calculation are to be carrie out with the regularize solutions n ɛ of (2.3), though for the sake of simplicity, we only o the formal calculation on n(x) = n(, t). The key is to use the logarithmic Hary Littlewoo Sobolev inequality to get a boun on the entropy S[n]. Theorem 3.2. (Logarithmic Hary Littlewoo Sobolev Inequality) [9] Let f be a nonnegative function in L 1 (R 2 ) such that f log f an f log(1 x 2 ) belong to L 1 (R 2 ).If R 2 f x = M, then f log f x 2 f (x) f (y) log x y xy C(M) (3.21) R 2 M R 2 R 2 with C(M) := M(1 log π log M). Remark 3.3. It is pointe out in [8] that by multiplying f by inicator functions, one can prove that the inequality (3.21) remains true with R 2 replace by any boune omains D R 2. The iea of the proof goes as follows. By observing that the mass in the upper half plane an lower half plane are subcritical ( n ± 1 < 8π), we plan to use the logarithmic Hary Littlewoo Sobolev inequality on these sub-omains to get uniform boun on the entropy. However, without extra information concerning the cell ensity istribution, naive application of logarithmic Hary Littlewoo Sobolev inequality fails. For this approach to work, the ensity istribution constraint require is that the cells in the upper an lower half plane are well-separate by a cell clear strip in which the total number of cells is sufficiently small. The strip is constructe through applying Lemma 3.1. Combining the logarithmic Hary Littlewoo Sobolev inequality an the cell seperation constraint, we can use a total entropy reconstruction trick introuce in [8] to obtain entropy boun. Now let s start the whole proof. Proof of Theorem 3.1. Accoring to propositions 2.1, 2.2, the life-span of freeenergy solution is etermine by the finite boun on the entropy S[n](t). Tothis en, we ecompose the free energy into three parts, ( E[n] 1 K ) n log nx 8π R 2 1 ( K n log nx 8π R 2 2 n(x)n(y) log x y xy R 2 R ( 2 =: 1 K ) S[n]I 1 I 2. 8π ) Hnx R 2 (3.22)

17 Suppressing Chemotactic Blow-Up Recall that since the free energy E[n](t) in (1.4) is ecreasing, ( 1 K ) S[n] E[n] I 1 I 2 E[n ] I 1 I 2, 8π then the esire entropy boun an hence a finite entropy solution follow provie we show the existence of a constant K < 8π, forwhichi 1 = I 1 (K ) is boune from below an I 2 is boune from above. Our main task is estimating I 1 (K ) from below, which is further ecompose into three terms I 1 (K ) = K n log nx 2 n(x)n(y) log x y xy R 2 R 2 R 2 K n log nx R 2 =: K I 11 I 12 K I 13. To estimate the various terms, we first construct the cell clear strip, Ɣ shown in Fig. 1, Ɣ := {x 2 x 2 > 2δ}, Ɣ := {x 2 x 2 < 2δ}, Ɣ := {x 2 x 2 2δ}. (3.23) Thus, region Ɣ contains points in the upper half plane which are 2δ away from the x 1 axis, whereas region Ɣ contains points in the lower half plane with the same property. Region Ɣ is a close 2δ-neighborhoo of the x 1 -axis. The δ neighborhoo of the Ɣ,Ɣ region is enote as follows: Ɣ (δ) = { x 2 x 2 >δ}, Ɣ (δ) } ={x 2 x 2 < δ. (3.24) Fig. 1. Regions Ɣ,Ɣ,Ɣ

18 Siming He & Eitan Tamor In the sequel, we will further ecompose Ɣ into subomains S ={x 2 δ<x 2 2δ}, S ={x 2 δ>x 2 2δ}, S ={x 2 x 2 δ}. (3.25) Estimating K I 11 I 12 from below. WefixK as { } K := max nx nx, nx nx, (3.26) Ɣ (δ) Ɣ Ɣ (δ) Ɣ an claim that K is amissible. Inee, lemma 3.1 with our choice of R 2 assume large enough so that (3.3) hols implies that the total mass insie Ɣ oes not excee nx = Ɣ {x 2 x 2 2δ} nx (1 η)2 2R 2 M 8π M 2. (3.27) Therefore, the Ɣ strip is the cell clear strip ; moreover, since the upper-half plane mass equals M 2, the amissibility of K follows, K M (1 η)2 2 2R 2 M < 8π. The motivation for this choice of K comes from the straightforwar boun nx n log nx Ɣ (δ) Ɣ (δ) nx n log nx nx n log nx Ɣ (δ) Ɣ (δ) Ɣ Ɣ nx n log nx nx Ɣ (δ) Ɣ (δ) n log nx ( )c (3.28) nx n log nx Ɣ R { 2 } max nx nx, Ɣ nx nx Ɣ n log nx =: K I 11. R 2 Ɣ (δ) Ɣ (δ) Inee, we procee along the lines of [8], appealing to the Logarithmic Hary Littlewoo Sobolev inequality in the three regions Ɣ (δ),ɣ(δ),ɣ, obtaining n(x)x nlog nx 2 n(x)n(y) log x y xy C, Ɣ (δ) Ɣ (δ) Ɣ (δ) Ɣ(δ) n(x)x nlog nx 2 n(x)n(y) log x y xy C, Ɣ (δ) Ɣ (δ) Ɣ (δ) Ɣ(δ) n(x)x nlog nx 2 n(x)n(y) log x y xy C. Ɣ Ɣ Ɣ Ɣ We now sum these three inequalities. By (3.28), the sum of their first three terms oes not excee K I 11 ; bookkeeping the overlap of the three omains Ɣ (δ) Ɣ(δ),Ɣ(δ)

19 Suppressing Chemotactic Blow-Up Ɣ (δ) an Ɣ Ɣ, consult Fig. 3 we fin the sum of the secon three terms is I 12 moulo the correction I 121 I 122 below, K I 11 I 12 C 4 ((Ɣ (δ) )c Ɣ ) (Ɣ (S S )) n(x)n(y) log x y xy 2 n(x)n(y) log x y xy (S S ) (S S ) =: C I 121 I 122. (3.29) Next applying the fact that x y δ for all (x, y) in the integral omain of I 121, we estimate the I 122 an I 122, I 121 4M 2 log δ, I n(x)n(y) log x y xy (S S ) (S S ) C n(x)n(y)(1 x 2 y 2 )xy (S S ) (S S ) C(M 2 2M n(x) x 2 x). (3.3) Combining (3.3) with (3.29) yiels K I 11 I 12 4M 2 log δ C(1 M 2 M n x 2 x). Estimating I 13 from below. We recall the well-known upper boun on the negative part of the entropy, [7,8], stating that for f positive function, f log f x 1 x 2 f x log(2π) f x 1 R 2 2 R 2 R 2 e ( ) C 1 M n x 2 x. (3.31) Combining (3.31) with our previous lower-boun of K I 11 I 12 yiels I 1 (K ) = K I 11 I 12 K I 13 4M 2 log δ C(M K ) ( 1 M ) n x 2 x. (3.32) It remains to upper-boun the I 2 term in (3.22). Since Hnx A n x 2 x, it is therefore suffices to show that the secon moment of n is boune for any finite time. Inee, the time evolution of the secon moment can be estimate as follows: n x 2 x 4AM 4A Hn(x)x 4AM A2 n x 2 x, t 2 an Gronwall inequality yiels the finite boun I 2 n(, t) x 2 x C(A, t) <.

20 Siming He & Eitan Tamor Finally, equippe with the lower boun on I 1 (K ) with K < 8π an with the upper boun on I 2 we revisit (3.23) to conclue that there exists a constant C = C(M, A, T )< such that the entropy S[n(, t)] is uniformly boune (inepenent of ɛ) for any finite time interval t [, T ], ( 1 S[n](t) (1 8π K ) E[n ]C 1 ) 2π M2 log δ, t [, T ]. (3.33) Now by the Propositions 2.1, 2.2, we have that the free energy solution exists on any time interval [, T ], T < Step 3 Proof of the Main Theorem In the proof of Theorem 3.1, we see that the cell population is separate by a cell clear zone near the x 1 axis. Since total mass in the cell clear zone is small, we can heuristically treat the total cell population as a union of two subgroups with subcritical mass (< 8π). However, since we lack sufficiently goo control over the total number of cells near the x 1 axis, we cannot use this iea to prove the optimal result as state in Theorem 1.1. The iea of proving Theorem 1.1 is that instea of consiering the total cell population as the union of two subgroups separate by one fixe cell clear zone, we treat it as the union of three subgroups with subcritical mass, namely, the cells in the upper half plane, the lower half plane an the neighborhoo of the x 1 axis, respectively. These three subgroups of cells are separate by two cell clear zones varying in time. The main ifficulty in the proof is setting up the three new regions such that: 1. mass insie each region is smaller than 8π; 2. the total mass of cells near their bounaries is well-controlle. Once the construction is complete, the remaining steps will be similar to step 2. Proof of Theorem 1.1. We start by constructing the three regions. First we note that the Lemma 3.1 implies that there exists δ>such that the following estimate is satisfie for a fixe R > 1 an η chosen small enough: x 2 2δ nx (1 η)2 R 2 nx 1 M, t >. (3.34) R 2 2 Now the region L ={(x 1, x 2 ) x 2 2δ} have total mass less than 2 1 M = M < 8π for all time. Seconly, we subivie the region L into J pieces: L = J 1 L i, L i := { (x 1, x 2 ) 2δ J (i) > x 2 2δ J } (i 1). Here J = J(M) 1, to be etermine later, epens on M. By the pigeon hole principle, there is at least three strips L i such that L i n(x)x 2 J M.

21 Suppressing Chemotactic Blow-Up Fig. 2. Regions Ɣ,Ɣ,Ɣ an Ɣ ω in the proof of the main theorem Suppose there are only two strips with mass smaller than 2 J M, then total mass in L will be bigger than (J 2) 2 J M > M, which is a contraiction. Now we pick from these three strips the one which is neither L 1 nor L J. As a result, this strip L i oes not touch the x 1 axis nor the bounary of L. We enote this i by i.thel i is the cell clear zone. Notice that here i = i (n, t) epens on time. Finally, we use this i to efine the regions epicte in Fig. 2. First we efine the three regions, each of which has total mass smaller than 8π: Ɣ = Next we set { x 2 2δ J i }, Ɣ = { x 2 2δ J i ρ = 2δ 3J, }, Ɣ = { x 2 2δ J (i 1) an efine the ρ neighborhoo of the above three regions as follows: { Ɣ (ρ) = x 2 > 2δ ( i 1 )} { Ɣ (ρ) J 3 = x 2 < 2δ ( i 1 )}, J 3 }.

22 Siming He & Eitan Tamor { Ɣ (ρ) = x 2 < 2δ ( i 2 )}. (3.35) J 3 Now we efine the complement Ɣ ω of the above three regions Ɣ (ρ) i, i =,, : { ( 2δ Ɣ ω = i 2 ) x 2 2δ ( i 1 )} = Ɣ ω Ɣ ω, Ɣ ω± = Ɣ ω R 2 ± J 3 J 3. of i {±,} Ɣ i an ecompose it into subo- Now we efine the complement Ɣ ω (ρ) mains: Ɣ (ρ) ω = ( ) c ( (ρ)) ( (ρ) i {±,} Ɣ i = Ɣ ω Ɣ ω ), Ɣ (ρ) ω± = Ɣ(ρ) ω R2 ±, Ɣ ω (ρ) = Ɣ ω S S S, S = Ɣ (ρ) \Ɣ, S = Ɣ (ρ) \Ɣ, S = Ɣ (ρ) \Ɣ. Remark 3.4. It is important to notice that the regions we are constructing are changing with respect to the given time t. Therefore, by oing the argument below, we can only show that the entropy is boune at time t, but since t is an arbitrary finite time, we have the boun on entropy for t [, T ], T <. We start estimating the entropy. By the free energy issipation, we obtain ( E[n ] 1 K ) n log nx 8π R 2 1 ( ) K nlog nx 2 n(x)n(y) log x y xy 8π R 2 R 2 R ( ) 2 K n log nx Hnx 8π R 2 R ( 2 =: 1 K ) S[n(T )]I 1 (K ) I 2 (K ). (3.36) 8π To erive entropy boun, we nee to estimate I 1 from below for K < 8π an estimate I 2 from above. We start by estimating I 1. Combining the efinition of L i an (3.34) yiels nx M = 1 M < 8π, (3.37) Ɣ (ρ) 2 nx M = 1 M < 8π, (3.38) Ɣ (ρ) 2 nx M = 1 M < 8π, (3.39) Ɣ (ρ) 2 Ɣ (ρ) ω nx L i nx 2 J (M ). (3.4)

23 Suppressing Chemotactic Blow-Up Now by the log-hary Littlewoo Sobolev inequality (3.21), we have that n(x)x n(x) log n(x)x Ɣ (ρ) i Ɣ (ρ) i 2 n(x)n(y) log x y xy C, i =,,, Ɣ (ρ) i Ɣ (ρ) i n(x)x n(x) log n(x)x Ɣ (ρ) ω± Ɣ (ρ) ω± 2 n(x)n(y) log x y xy C. Ɣ (ρ) ω± Ɣ(ρ) ω± Same as in Section 3.3, we recall the efinition of I 1 (3.36), an use the estimates above to reconstruct the entropy an the potential on the whole R 2 as follows: ( ) C K n(x) log n(x)x 2 n(x)n(y) log x y xy R 2 R 2 R 2 2 n(x)n(y) log x y xy 2 n(x)n(y) log x y xy (S S ) (S S ) (S S ) (S S ) =: 8πI 1 I 3 I 4. (3.41) The region an the integral omain of I 4 are inicate in Fig The K in (3.41) can be estimate using (3.4) as follows: ( K := M nx 1 2 ) M. (3.42) J Ɣ (ρ) ω By the assumption M < 8π, we can make J big such that K < 8π. This is where we choose the J = J(M). Applying the fact that x y 3J 2δ, (x, y),thei 3 an I 4 terms in (3.41) can be estimate as follows: I 3 CM 2 log 2δ 3J, I 4 2 n(x)n(y) log x y xy (S S ) (S S ) (S S ) (S S ) C n(x)n(y)(1 x 2 y 2 )xy (S S ) (S S ) (S S ) (S S ) 3 Region is the union of the following nine regions, ((1) (9)): 1)Ɣ (Ɣ (ρ) )c, 2)S (Ɣ Ɣ ω (ρ) )c, 3)Ɣ ω (Ɣ ω (ρ) )c, 1)4)S (Ɣ(ρ) Ɣ ω (ρ) )c, 5)Ɣ (Ɣ (ρ) )c, 6)S (Ɣ(ρ) Ɣ ω (ρ) )c, 1)7)Ɣ ω (Ɣ ω (ρ) )c, 8)S (Ɣ (ρ) Ɣ(ρ) ω )c, 9)Ɣ (Ɣ (ρ) )c.

24 Siming He & Eitan Tamor Fig. 3. Region in R 2 R 2 C(M 2 M n x 2 x). (3.43) Combining (3.41), (3.42) an (3.43) yiels I 1 CM 2 log 2δ 3J C(1 M2 M n x 2 x). (3.44) We estimate the I 2 term in (3.36)using(3.31) an the secon-moment boun of n as in the proof of (3.33): I 2 (t) C(M, A, T )<, t T. (3.45) Combining (3.36), (3.44), (3.45) an the secon moment boun of n, we obtain that ( 1 S[n](T ) (1 8π K ) E[n ]C(M, A, T ) CM 2 log 2δ ) <, T <. 3J (3.46) Once the entropy is boune for any finite time, the existence is guarantee by Propositions 2.1 an 2.2.

25 Suppressing Chemotactic Blow-Up Appenix A. In the appenix, we prove the two local existence theorems state in section 2.2. The proof follows along the same line, as the analysis in [8]. Appenix A.1. Proof of Proposition 2.1 Proof. For any fixe positive ɛ, following the argument as in section 2.5 of [7], we obtain the global solution in L 2 ([, T ], H 1 ) C([, T ], L 2 ) for the regularize Keller Segel system with avection (2.3). The goal is to use the Aubin Lions Lemma to show that the solutions {n ɛ } ɛ is precompact in certain topology. As in the paper [8], we ivie the proof into steps. Step 1. A priori estimates on n ɛ an c ɛ. In this step, we erive several estimates which we will nee later. First we estimate the secon moment V := n ɛ x 2 x. R 2 The time evolution of V can be estimate using the regularise equation (2.3) an the fact that the graient of the regularize kernel K ɛ is boune K ɛ (z) 1 2π z : t V = 4M n ɛ (x, t)n ɛ (y, t)(x y) K ɛ (x y)xy R 2 R 2 2 x bn ɛ (x)x 4M 2C n ɛ x 2 x. By Gronwall, we have that V (t) 4M 2C e2ct V C V (T )<, t T, (A.1) from which we obtain that (1 x 2 )n ɛ L ([, T ], L 1 ) uniformly in ɛ. Next, we estimate n ɛ log n ɛ L ([,T ];L 1 ) an n ɛ c ɛ x L t. Combining the assumption of the proposition 2.1 an (3.31) yiels n ɛ log n ɛ x n ɛ (log n ɛ x 2 )x 2log(2π)M 2 e. (A.2) Therefore, we prove that n ɛ log n ɛ L ([, T ], L 1 ) uniformly in ɛ. Recalling the bouneness of the secon moment (A.1) an the representation of c ɛ as c ɛ = K ɛ n ɛ, we euce the following estimate using the Young s inequality ab e a 1 b ln b for a, b 1: c ɛ (x) = K ɛ (x y)n ɛ (y)y C(M, V, nɛ log n ɛ L ([,T ],L 1 ))

26 Siming He & Eitan Tamor C(M) log(1 x ), (A.3) uniformly in ɛ. Combining this with the mass conservation of n ɛ an the secon moment control (A.1), we euce that n ɛ c ɛ x C, (A.4) L ([,T ]) where C is inepenent of ɛ. Next we erive the main a priori estimate, namely, the L 2 ([, T ] R 2 ) estimate of n ɛ c ɛ. First we calculate the time evolution of n ɛ c ɛ x: 1 n ɛ c ɛ x = 2 t n ɛ t cɛ x = ( n ɛ ( c ɛ n ɛ ) b n ɛ )c ɛ x = n ɛ c ɛ x n ɛ c ɛ 2 x bn ɛ c ɛ x. Integrating this in time yiels T n ɛ c ɛ 2 xt = 1 n ɛ c ɛ x(t ) 1 n ɛ c ɛ x() 2 2 T T n ɛ c ɛ xt n ɛ b c ɛ xt. (A.5) Now we estimate the right han sie of (A.5). We see from (A.4) that the first two terms on the right han sie is boune. Next we estimate the thir term on the right han sie of (A.5), which requires information erive from the entropy boun. By the property that b =, we formally calculate the time evolution of S[n](t) as S[n](t) = 4 n 2 x n 2 (t)x. (A.6) t The intereste reaer is referre to [7] for more etails. We nee to estimate the secon term in (A.6). Before oing this, note that, for K > 1, n K nx 1 log(k ) n log nx C =: η(k ) log(k ) (A.7) can be mae arbitrarily small. Now we can use the Gagliaro-Nirenberg-Sobolev inequality together with (A.7) to estimate the secon term in (A.6) as follows: ( ) 1/2 n 2 x MK n 2 x MK nx n 3/2 3 n K n K MK η(k ) 1/2 CM 1/2 n 2 2.

27 Suppressing Chemotactic Blow-Up Combining this with (A.6) an (A.7), we have t S[n](t) = (4 η(k )1/2 CM 1/2 ) n 2 x MK. (A.8) The factor (4 η(k ) 1/2 CM 1/2 ) can be mae non-positive for K large enough an therefore we have that T n 2 S[n]() S[n](T ) MKT xt (4 2η(K ) 1/2 M 1/2. C) It follows that n is boune in L 2 ([, T ] R 2 ). The erivation for n ɛ L 2 ([,T ];L 2 ) C is similar but more technical, an the intereste reaers are referre to [7] for more etails. As a consequence of the L 2 ([, T ] R 2 ) estimate on n ɛ an of the computation t S[nɛ ](t) = 4 n ɛ 2 x n ɛ ( c ɛ )x, we have the estimate T n ɛ ( c ɛ )xt C. (A.9) This completes the treatment of the thir term on the right han sie of (A.5). Next we estimate the last term T bnɛ c ɛ xt in (A.5). First we calculate the time evolution of Gn ɛ x as follows: Gn ɛ x = G ( n ɛ c ɛ n ɛ bn ɛ )x t = Gn ɛ x G c ɛ n ɛ x b 2 n ɛ x = b c ɛ n ɛ x b 2 n ɛ x. Now, integrating in time, we obtain T b c ɛ n ɛ x Gn ɛ x() Gn ɛ x(t ) T b 2 n ɛ xt. (A.1) From the assumption b x, we have that the right han sie of (A.1) can be boune in terms of the secon moment V : Gn ɛ x b 2 n ɛ x C n ɛ x 2 x. (A.11) Since the secon moment V is boune (A.1), we have that T b c ɛ n ɛ x C. (A.12)

28 Siming He & Eitan Tamor Applying estimates (A.4), (A.9) an (A.12) to(a.5), we obtain T n ɛ c ɛ 2 xt C <. (A.13) This conclues our first step. Step 2- Passing to the limit.asin [8], the following Aubin Lions compactness lemma is applie: Lemma A.1. (Aubin Lions lemma)[8] TakeT > an 1 < p <. Assume that ( f n ) n N is a boune sequence of functions in L p ([, T ]; H) where H is a Banach space. If ( f n ) n is also boune in L p ([, T ]; V ) where V is compactly imbee in H an f n / t L p ([, T ]; W ) uniformly with respect to n N where H is imbee in W, then ( f n ) n N is relatively compact in L p ([, T ]; H). Our goal now is to fin the appropriate spaces V, H, W for n ɛ. We subivie the proof into steps, each step etermines one space in the lemma. Space H: Boun on n ɛ L 2 ([,T ],L 2 ): We can estimate the n ɛ 2 2 by applying the following ecomposition trick: n ɛ = (n ɛ K ) min { n ɛ, K }. (A.14) The secon part in (A.14) is boune in L p, 1 p. The first part is boune in L 1 by (n ɛ K ) 1 nɛ log n ɛ 1 log K = η(k ), (A.15) which can be mae arbitrary small. Now we estimate the time evolution of (n ɛ K ) p x, 1 < p < as follows: 1 (n ɛ K ) p p t x = (n ɛ K ) p 1 ( nɛ ( c ɛ n ɛ ) b n ɛ )x 4(p 1) = p 2 (n ɛ K ) p 1 4(p 1) p 2 K (n ɛ K ) p 1 4(p 1) =: p 2 (n ɛ K ) p/2 2 x b (nɛ K ) x (n ɛ K ) p/2 2 x p 1 p cɛ x (n ɛ K ) p 1 ( cɛ n ɛ )x (n ɛ K ) p cɛ x (n ɛ K ) p/2 2 x I 1 I 2. (A.16)

29 Suppressing Chemotactic Blow-Up Note that since b, the term involving b vanishes. For the T 1 term in (A.16), using the facts that c ɛ = K ɛ n ɛ an K ɛ 1 is uniformly boune, we can estimate it as follows: I 1 (n ɛ K ) p K ɛ (n ɛ K ) x K (n ɛ K ) p x K ɛ 1 (n ɛ K ) p p1 K ɛ (n ɛ K ) p1 CK (n ɛ K ) p x C (n ɛ K ) p1 p1 CK (nɛ K ) p p. (A.17) Similarly, we can estimate the T 2 term in (A.16) asfollows: I 2 CK (n ɛ K ) p p CK 2 (n ɛ K ) p 1 p 1. (A.18) Combining (A.16), (A.17) an (A.18), we obtain 1 (n ɛ K ) p 1) x 4(p p t p 2 (n ɛ K ) p/2 2 x C (n ɛ K ) p1 p1 CK (nɛ K ) p p CK 2 (n ɛ K ) p 1 p 1. (A.19) For the highest orer term C (n ɛ K ) p1 p1 in (A.19), we use the following Gagliaro-Nirenberg-Sobolev inequality: f p1 x C ( f p/2 ) 2 x f x, R 2 R 2 R 2 f, (A.2) together with (A.15), to estimate it as follows: ( ) (n ɛ K ) p1 p1 C (n ɛ K ) p/2 2 2 (nɛ K ) 1 ( ) Cη(K ) (n ɛ K ) p/ (A.21) We can take K big such that it is absorbe by the negative issipation term in (A.19). Now applying Höler s inequality, Young s inequality an the Gronwall inequality to (A.19), we have that (n ɛ K ) L p(t) C(T )<, t [, T ], p (1, ). Applying a stanar argument, see e.g. [7] proof of Proposition 3.3, we obtain the estimate n ɛ L ([,T ];L p ) C(T ), p (1, ). (A.22) In particular, we set p = 2 an obtain that We conclue this step by setting H := L 2 (R 2 ). n ɛ L 2 ([,T ];L 2 ) C(T ). (A.23)

30 Siming He & Eitan Tamor Space V : Boun on n ɛ L 2 ([,T ] R 2 ): First we calculate the time evolution of the quantity n ɛ 2 2 : n ɛ 2 x = 2 n ɛ 2 x 2 n ɛ c ɛ n ɛ x (b(n ɛ ) 2 )x t = 2 n ɛ 2 x 2 n ɛ c ɛ n ɛ x. Now, integrating in time, we obtain the estimate n ɛ 2 x(t ) n ɛ 2 x() T T = 2 n ɛ 2 xt 2 n ɛ ( n ɛ c ɛ )xt T 2 n ɛ 2 xt ( T ) 1/2 ( T 1/2 2 n ɛ c ɛ 2 xt n ɛ xt) 2. (A.24) The terms on the left han sie of (A.24) are boune ue to (A.22). For the last term on the right han sie, we can estimate it as follows. The Hary Littlewoo Sobolev inequality yiels c ɛ 4 C n ɛ 4/3, (A.25) which implies n ɛ c ɛ 2 n ɛ 4 c ɛ 4 C n ɛ 4 n ɛ 4/3. Combining this an the L p boun (A.22) yiels the bouneness of n ɛ c ɛ in L ([, T ], L 2 ). Applying this fact an (A.22) in(a.24) an set X = ( T n ɛ 2 xt) 1/2, we obtain X 2 2 n ɛ c ɛ L 2 ((,T ) R 2 ) X n ɛ 2 x(t ) n ɛ 2 x() C. As a result, n ɛ L 2 ([,T ] R 2 ) C. (A.26) Same as in [8], we set V := H 1 (R 2 ) {n x n 2 L 1 }, which is shown to be compactly imbee in H there. Thanks to the boun (A.23), (A.26) an the (A.1), we have that n ɛ L 2 ([, T ]; V ). Space W : Boun for the t n ɛ : In orer to estimate the L 2 ([, T ]; H 1 ) norm of the function t n ɛ, we first nee to get a boun on the fourth moment of the solution V 4 := n(x 4 1 x4 2 )x. The time evolution of V 4 can be formally estimate as follows:

31 Suppressing Chemotactic Blow-Up t V 4 = ( n ( cn) b n)(x1 4 x4 2 )x = 12 n x 2 x 1 4(x 3 1 y1 3)(x 1 y 1 ) 4(x2 3 y3 2 )(x 2 y 2 ) 4π x y 2 n(x)n(y)xy bn (4x1 3, 4x3 2 )x C(M) n x 2 x C n(x1 4 x4 2 )x. (A.27) Combining the secon moment estimate (A.1) an the Gronwall inequality, we have that n(x, t) x 4 x C, t [, T ]. (A.28) One can aapt this calculation to the regularize solution n ɛ without much ifficulty. We leave the etails to the intereste reaer. Now we can estimate the L 2 ([, T ]; H 1 ) norm of the t n ɛ. Combining the L p boun on n ɛ (A.22), the boun on c (A.25) an the fourth moment control (A.28) an testing the equation (2.3) with f L 2 ([, T ], H 1 (R 2 )),wehave t n ɛ, f L 2 ([,T ] R 2 ) n ɛ L 2 ([,T ];L 2 ) f L 2 ([,T ];H 1 ) c ɛ n ɛ L 2 ([,T ];L 2 ) f L 2 ([,T ];H 1 ) bn ɛ L 2 ([,T ];L 2 ) f L 2 ([,T ];H 1 ) n ɛ L 2 ([,T ];L 2 ) f L 2 ([,T ];H 1 ) T 1/2 c ɛ L ([,T ];L 4 ) n ɛ L ([,T ];L 4 ) f L 2 ([,T ];H 1 ) CT 1/2 sup V 1/4 4 n ɛ 3/4 L t ([,T ];L 3 ) f L 2 ([,T ];H 1 ) C f L 2 ([,T ];H 1 ). As a result, we have that t n ɛ is uniformly boune in L 2 ([, T ]; H 1 ). Combining the results from all the steps above an the Aubin Lions lemma, we have that (n ɛ ) ɛ is precompact in L 2 ([, T ]; L 2 ). We enote n as the limit of one converging subsequence (n ɛ k ) ɛk. Moreover, combining (A.25) an the L 4/3 ([, T ] R 2 ) boun on n ɛ which can be erive from (A.22), we have that n ɛ k c ɛ k converge to n c in istribution sense. Step 3- Free energy estimates. By the convexity of the functional n R 2 n 2 x, the fact that n ɛ k c ɛ k converge to n c in istribution sense an weak semi-continuity, we have n 2 xt lim inf n ɛ k 2 xt, [,T ] R 2 k [,T ] R 2 n c 2 xt lim inf n ɛ k c ɛ k 2 xt. [,T ] R 2 k [,T ] R 2

32 Siming He & Eitan Tamor Moreover, it can be checke that S[n ɛ ](t) S[n](t) for almost every t, whose proof is similar to the one use in [7] Lemma 4.6. Next we show the free energy estimate (1.5) using the strong convergence of {n ɛ k } in L 2 ([, T ] R 2 ). The key is to show the following entropy issipation term is lower semi-continuous for the sequence (n ɛ k ): T n ɛ k log n ɛ k c ɛ k b 2 xt = 4 n ɛ k 2 xt n ɛ k c ɛ k 2 xt [,T ] R 2 [,T ] R 2 n ɛ k b 2 xt [,T ] R 2 2 (n ɛ k ) 2 xt 2 n ɛ k b c ɛ k xt [,T ] R 2 [,T ] R 2 =: I 1 I 2 I 3 I 4 I 5. (A.29) For the sake of simplicity, later we use n ɛ to enote n ɛ k. First, we estimate the I 3 term in (A.29). By the Fatou Lemma, we have the inequality n b 2 xt lim inf [,T ] R 2 ɛ k n ɛ k b 2 xt. [,T ] R 2 (A.3) This finishes the treatment of I 3. Next, we show that the term I 5 in (A.29) actually converges as ɛ k. We ecompose the ifference between I 5 an its formal limit into two parts: T b cn b c ɛ n ɛ xt R 2 T c bn c ɛ bn nb c ɛ b c ɛ n ɛ xt R 2 T T c c ɛ bn xt R 2 bn bn ɛ c ɛ xt R 2 =: I 51 I 52. (A.31) For the first term T 51 in (A.31), applying Höler an Hary Littlewoo Sobolev inequality, we can estimate it as follows: ( T ) 1/4 ( T I 51 (c c ɛ ) 4 4 t b ) 1/2 n 2 xt R ( 2 T ) 1/4 4 n xt R ( 2 T ) 1/4 ( T ) 1/2 C n n ɛ 4 4/3 t x 2 nxt R 2

33 ( T Suppressing Chemotactic Blow-Up ) 1/4 n 2 xt R ( 2 T ) 1/4 ( T ) 1/2 C n n ɛ 2 2 n nɛ 2 1 t x 2 nxt R ( 2 T ) 1/4 n 2 xt R ( 2 T ) 1/4 ( T ) 1/2 C n n ɛ 2 2 t sup n n ɛ 1/2 1 n x 2 xt t R ( 2 T ) 1/4 n 2 xt. R 2 From the L 2 ([, T ] R 2 ) strong convergence of (n ɛ ) ɛ, we have that the first factor goes to zero. Other factors are boune ue to (A.1), Fatou s Lemma, L 1 boun on n ɛ, n an n L 2 ([, T ] R 2 ). As a result, T 51 converges to zero. Next we estimate the I 52 term in (A.31). Applying the fact that c a c a, the Höler an Hary Littlewoo Sobolev inequality, we have T I 52 = b( n 2 n ɛ 2 ) c ɛ xt R 2 T b n( n n ɛ ) c ɛ xt R 2 T b n ɛ ( n n ɛ ) c ɛ xt R 2 c ɛ L 4 ([,T ] R 2 ) b ( T ) 1/4 n L 2 ([,T ] R 2 ) n n ɛ 4 xt R 2 c ɛ L 4 ([,T ] R 2 ) b T n ɛ L 2 ([,T ] R 2 )( n n ɛ 4 xt) 1/4 R 2 C sup n ɛ L 1 (R 2 ) 1/2 n ɛ 1/2 L t 2 ([,T ] R 2 ) ( T ) 1/2 (n ɛ n) x 2 xt n n ɛ 1/2 L 2 ([,T ] R 2 ). By the same reasoning as in T 51, we have that this term goes to zero. This finishes the treatment for the term. Now for the I 1, I 2, I 4 terms in (A.29), we can hanle them in the same way as in [7]. Combining all the estimates above we have that T T n log n c b 2 xt lim inf n ɛ k log n ɛ k c ɛ k b 2 xt. ɛ k (A.32) The remaining part of the proof is the same as the one in [7].