ON A PERTURBED SYSTEM OF CHEMOTAXIS, II

We consider the blow up mechanism for a perturbed system of chemotaxis. First, using Moser’s iteration scheme the blow up point of the solution is characterized in terms of the local Zygmund norm. And then, by the analysis on the Gagliard-Nirenberg inequality together with the study of the Green’s function associated with elliptic part of system, we establish finiteness of blow up point.

tactic aggregation, that is, in Ω × (0, T ) 0 = ∆v − γv − f (v) + αu in Ω × (0, T ) ∂u ∂ν − χu ∂v ∂ν = 0 on ∂Ω × (0, T ) where Ω is a bounded domain in R n with smooth boundary ∂Ω, χ, α, γ are positive constants, f = f (s), g = g(s) are smooth functions, and ν is the unit outward normal vector on ∂Ω.The unknown functions u = u(x, t) and v = v(x, t) stand for the cell density and the concentration of the chemical substance at (x, t) ∈ Ω × [0, T ), respectively, and F = −∇u + χu∇v is the flux of u so that the effect of diffusion −∇ • ∇u and that of chemotaxis compete for u to vary.Another description is the movement from the gravitational equilibrium of polytropic fluid, see [4], [6].
For the problem (GCZ) with f (v) = g(v) = 0, Nagai [16] showed that the conjecture of Childress and Percus [9] concerning n = 2 is true for radially symmetric solutions, that is, the chemotaxis collapse can occur if the total cell number on Ω ⊂ R 2 is larger than 8π/αχ and can not occur in the other case, and then, Nagai, Senba and Yosida [17] and Senba and Suzuki [22] corrected this value to 4π/αχ in the general case, see also Biler [5] and Gajewski and Zacharias [11].Then, Senba and Suzuki [21] and Suzuki [23] showed the formation of collapses with the quantized mass for the blowup solution in finite time, refining the conjecture of Nanjundiah [19].
For the problem (GCZ) with f (v) = β|v| p−1 v and g(v) = 0, Chen and Zhong [8] studied the existence and the non-existence of the solution global in time, while Kurokiba and Suzuki [13] showed the finiteness of the blowup points of the blowup solution in finite time.
In this paper we impose the following conditions on (GCZ): for all s ∈ R, where C > 0, p 1 ∈ (1, p * ) and p 2 ∈ (1, p * * ) with (1.8) We assume the compatibility condition to the parabolic part: where v 0 denotes the solution to (2.1) for u = u 0 , see the following section.Then, we obtain the classical solution local in time: Theorem 1.If (1.1)-(1.9)hold, then there exists a unique non-negative classical solution (u, v) ∈ C 2+θ,1+θ/2 (Q T ) × C([0, T ]; C 2+θ (Ω)) to (GCZ) with θ ∈ (0, 1) provided that T is sufficiently small.Henceforth T max ∈ (0, +∞] and B stand for the maximal existence time of the solution and the blowup set of u, respectively:  Concerning the finiteness of B ∩ ∂Ω, see the final remark of this paper.This paper is composed of four sections.Theorems 1, 2, and 3 are proven in sections 2, 3, and 4, respectively.Henceforth, we put χ = α = γ = 1 without loss of generality.

Proof of Theorem 1
To begin with, we study the elliptic equation Proof.To confirm the uniqueness of v, let v 1 and v 2 be the classical solutions to (2.1) and set w := v 1 − v 2 .Then, it holds that and therefore, by (1.4).This implies To show the non-negativity v ≥ 0, we take v − = min(v, 0) ≤ 0. Since The existence of v is obtained by the variational method.We set and show that I attains the minimum in H 1 = H 1 (Ω), using (1.6)-(1.7).Since the assumption (1.6)-(1.7)guarantees the sub-criticalness of the nonlinearity, the minimizer will actually be in C 2+θ (Ω) by the bootstrap argument.
To confirm the weak lower semi-continuity of I, let {v k } ⊂ H 1 (Ω) be a minimizing sequence: Since {v k } is bounded from below by (2.6), we can extract a subsequence, denoted by the same symbol, such that for some v, and then it follows that (2.7) Meanwhile, we have the compact imbedding and the continuous imbedding W s,q (Ω) ֒→ W s− 1 q ,q (∂Ω) for 0 < s < 1 and 1 ≤ q < q * , 1 q * = 1 2 − 1−s n , see [1], which implies the compact imbeddings for p 1 , p 2 prescribed by (1.8).From this compactness, passing if necessary to yet another subsequence, we see that there exist h 1 ∈ L p 1 +1 (Ω) and h 2 ∈ L p 2 +1 (∂Ω) such that Since f and g are continuous and it follows from (1.6)-(1.7)that 2 (ξ)) for a.e.ξ ∈ Ω.
Then we have by the Lebesgue dominated convergence theorem together with the above facts, and consequently We obtain Finally, we show the regurality of v.We shall do it only in the case n ≥ 3, for it is done more easily in the case n = 2. Set We use (1.6)-(1.7) to get . From these inequalities and the L p estimates (see [2]), it follows that If 2n/(n + 2) < qk−1 ≤ n, then we define q k , r k , s k , qk recursively as follows: where For the above r k and s k , we similarly deduce Then we use the L p estimates, again, so that v ∈ W 2,q k (Ω).Repeating this finite times, we obtain and conclude the desired regularity v ∈ C 2+θ (Ω) by the Schauder estimate, see [2].The proof is complete.
Proof of Theorem 1.We define and v (0) , u (1) , v (1) , • • • inductively as follows: and where Q T := Ω × (0, T ) and Γ T := ∂Ω × (0, T ).We take and obtain the continuous imbedding where C l,l/2 (Q T ) (0 < l ∈ N ) is the standard Hölder space and its norm is denoted by [ • ] l,Q T , see [14].Set for M > 0 and 0 < T ≪ 1 to be decided later.From (2.11), (2.12) and Lemma 4, we can define v (k) as the solution to (2.9) with non-negativity if u denotes the positive constant which is monotone increasing on T and depends on M and given data, but not on u (k) and v (k) .Using the mean value theorem and standard elliptic regularity, we can find the inequality where is a solution to (2.9), then there exists a unique solution u by the W 2,1 q -theory, see [14].Note that C * 0 (T ) is independent of M and monotone increasing on T .We compute (2.16) It holds that by (2.14).Since f is smooth, we have (2.18) We remark the following fact (see [14]): If (2.11) holds and 0 < T ≪ 1 then where θ = 1 − (n + 2)/q and the constant C * 3 is independent of 0 < T ≪ 1.Using Hölder's inequality, we have and then Next, we estimate the boundary integral term as where C is independent of M and T. Since g is smooth and (2.14) holds, we have Following the notations of [14], we take the canonical set B ⊂ R n−1 , for instance the unit cube in R n−1 , and the finite family of the mappings Φ j : B → V j ⊂ ∂Ω (j ∈ J ) each of which is one-to-one, onto and smooth, and satisfies ∪ j∈J Φ j B = ∂Ω.Let {ρ j } j∈J , 0 ≤ ρ j ≤ 1, be a smooth partition of unity subject to {Φ j (B)} j∈J .Set for 0 < λ < 1, where and Φ * j is the pull back induced by Φ j .We set w (k) j = Φ * j (v (k) ρ j ) and may take the unit cube as the above B. We calculate according to the above definition and get From the elliptic regularity and (2.20) we have and then (2.28) One can see that if there exist constants A i (i = 1, 2) such that then there exists a constant C * 8 depending only on q, A 1 , A 2 , g and Ω such that by (2.26) and the trace embedding, and that by (2.25).Hence, we get (2.31) From (2.15), (2.21) and (2.31) we arrive at the following inequality:

.32)
We fix 0 < δ ≪ 1.Noting that C * 0 and C * 12 are monotone increasing on T, we can take M and T in (2.13) such that Then we can define a mapping To prove the existence of the solution in local-time, it suffices to prove that Ψ is a contraction mapping on X M,T .But we omit the proof since the calculations are similar to the above ones which assure that Ψ is a mapping on X M,T .We may need to replace T with T which is sufficiently smaller than the previous T .Fix M > 0 and 0 < T ≪ 1 such that Ψ is a contraction mapping on X M,T , and let (u, v) be a solution to (GCZ) satisfying u ∈ X M,T .The uniqueness of the solution is reduced to that of u by Lemma 4. Let (u 1 , v 1 ) be another solution to (GCZ).By the above argument, we see that there exist M 1 and T 1 such that Ψ is a contraction mapping on X M 1 ,T 1 .If we set then both u and u 1 are in X M ′ ,T ′ .Since Ψ is also a contraction mapping on X M ′ ,T ′ , the fixed point is unique in X M ′ ,T ′ , and therefore Otherwise, we take It holds that sup t∈[0,T 0 ) u 1 (t) W 2,1 q (Q T ) ≤ C for some C > 0. In fact, smoothness of the solution in Q T 0 follows from (2.11), (2.12), (2.14), Lemma 4 and the Schauder theory with (1.9), see [14].Therefore, there exists 0 < δ ≪ 1 such that u = u 1 in Q T 0 +δ , which is a contradiction by the definition of T 0 .Thus the uniqueness is shown.Smoothness of the solution is similar to the above.The proof of Theorem 1 is complete.

Proof of Theorem 2
We confirm several facts used later (see [21], [23]).The Gagliardo-Nirenberg inequality in two space dimentions is described by where K is a constant determined by Ω.We put To derive local in space estimates, we take the smooth cut-off function ϕ satisfying First, given x 0 ∈ Ω and 0 < R ′ < R with B 2R (x 0 ) ⊂ Ω, we take such ϕ by Given x 0 ∈ ∂Ω, next, we take a smooth conformal mapping X : B 2R ∩ Ω → R 2 satisfying x 0 → 0 and The above ϕ is sometimes written as where A and B are positive constants determined by R ′ and R. We use the following estimates derived from (3.1), see [21,23] for the proof.
Lemma 5.The following inequalities hold for any s > 1, where C > 0 is a constant: Lemma 5 is irrelevant to (GCZ) and the above u is not necessarily to be the solution.Here, we emphasize the mass conservation, i.e.
This property is derived from the first boundary condition of (GCZ) and the non-negativity of u.The second equation of (GCZ) is written as and hence It also holds that Then, we apply the L 1 estimate of [7] to obtain by Sobolev's and the trace embedding theorems, see [1].Henceforth, we assume T max < ∞.
Lemma 6.We obtain x 0 ∈ B if and only if for any 0 < R ≪ 1.
Proof.The 'if' part is obvious by the definition of B. To show the converse, we assume lim sup for some 0 < R ≪ 1 and show x 0 ∈ B. In fact, this implies Multiplying the fist equation of (GCZ) by uψ and integrating over Ω, we have 1 2 The first term of the right hand of (3.8) is equal to where Then, using the second equation of (GCZ) and (1.4)-(1.5),we obtain Using Young's inequality, we get the following inequalities: and Combining (3.12)-(3.15),we obtain 1 2 ) where C 1 > 0 is an absolute constant induced by (3.4).Now, we estimate I defined by (3.10).We see that there exists an absolute constant M 1 (only depending on g(0)) such that by (1.3)-(1.5).We take a constant L such that ) 2 and Combining (3.17)-(3.22),we get From (3.12) and (3.23), we have 1 2 Taking Now we use (3.7) and prescribe s ≫ 1 by Then it follows that by Lemma 5. We use Gronwall's inequality and our assumption We continue the process, multiplying the first equation of (GCZ) by u 2 ψ and integrating by parts.We have which is equivalent to where w := u 3/2 .Using the second equation of (GCZ), we have where Then, it follows that 2 where ε 1 and ε 2 are arbitrary positive constants.Inequalities (3.27)-(3.30)are summarized by where C 4 > 0 is an absolute constant induced by (3.4).If we take for R ′ ∈ (0, R), we can repeat the argument of the previous stage.We see replacing u, R and ψ = (ϕ x 0 ,R ′ ,R ) 6 by w, R ′ , and ψ = (ϕ x 0 ,R ′′ ,R ′ ) 6 respectively, where R ′′ ∈ (0, R ′ ) and r ∈ (0, R).From this stage, we use only the first equation of (GCZ).The rest of the proof is thus anagolous to Step 3 of Lemma 5 of [21] (see also [23]).Skipping details, we just mention that estimates (3.33)-(3.34)and the Moser's iteration scheme (see [3]) are used to obtain and therefore x 0 ∈ B.
Proof of Theorem 2. We shall prove that lim sup In fact, the proof of Lemma 6 is valid even to ϕ ≡ 1, and hence we can show that (3.35) implies lim sup If (3.37) is the case, then the standard theories (see [2], [12] and [14]) and (3.3)-(3.5)guarantee that the solution u to (GCZ) is continued after t = T .Thus, T max < ∞ implies (3.36) and hence we obtain the result lim t↑T Ω u log udx = ∞.Now, multiplying log u by the first equation of (GCZ), and we have Using Lemma 5 and (3.2), we see where s > 0 is arbitrary.Then, we take s = s(t) = exp(2K

Proof of Theorem 3
We take the Green's function G = G(x, x ′ ) defined by for x ∈ Ω, where δ denotes the Dirac' delta function, and shall use the following lemma (see [21], [23] for the proof): First, we show the following lemma: Proof.Multiplying ψ log u to the first equation of (GCZ), we have From the second equation of (GCZ) where We also have that Then we obtain the following inequality: Here, we recall the elementary inequality: Let α > 0 and 0 < β < 2. Then it holds that where k α,β is a positive constant determined by only α and β.
Finally, III, defined by (4.1), is estimated as follows: where M 1 and M 2 are positive constants determined by (3.17We are now in a position to prove the finiteness of blowup points. Proof of Theorem 3. We take x 0 ∈ B. From Lemmas 5 and 8, it follows that for g(ξ, t) = g(v(ξ, t)).An expected sharp estimate is concerned with the weak norm, see [10], v H 1 w (Ω) ≤ C in (2.1) for the prescribed u 1 = λ.Then, it will follow that and call each x 0 ∈ B the blowup point.The following theorem assures that T max < +∞ implies B = ∅: Theorem 2. If (1.1)-(1.9),n = 2 and T max < +∞ hold, then it holds that lim t↑Tmax Ω u log udx = ∞.Now, we state the main theorem.