AN APPROXIMATION SOLVABILITY METHOD FOR NONLOCAL SEMILINEAR DIFFERENTIAL PROBLEMS IN BANACH SPACES

. A new approximation solvability method is developed for the s-tudy of semilinear diﬀerential equations with nonlocal conditions without the compactness of the semigroup and of the nonlinearity. The method is based on the Yosida approximations of the generator of C 0 − semigroup, the continuation principle, and the weak topology. It is shown how the abstract result can be applied to study the reaction-diﬀusion models.

1. Introduction.The paper deals with the nonlocal problem for semilinear differential equations of the form: in a reflexive Banach space E having a Schauder basis satisfying the property (π 1 ) and a strictly convex dual space E * , where A : D(A) ⊂ E → E is the generator of a C 0 −semigroup of contractions {S(t)} t≥0 ; f : [0, T ] × E → E and M : C([0, T ]; E) → E.
The nonlocal problem for a semilinear differential equation with a C 0 −semigroup generator was first studied by L. Byszewski [15].The technique used in [15] is based on the Banach fixed point theorem for contraction mappings.Then others fixed point theorems (for example, Leray-Schauder fixed point theorem for compact mappings, fixed point theorem for condensing mappings, Kakutani fixed point theorem for multivalued mappings...) are used for the study of semilinear differential equations and inclusions in Banach spaces with various boundary conditions (see, e.g.[5], [18] and [26] and the references therein).

IRENE BENEDETTI, NGUYEN VAN LOI AND VALENTINA TADDEI
In the present paper, we introduce a new method for the study of problem (1) which is the combination of the Yosida approximations of the generator of C 0 −semigroups and an approximation solvability method.This method is a generalization of the well known strong approximation used by F.E. Browder and W.V. Petryshyn in [14] and by W.V. Petryshyn in [23].It was developed in [20] for the study of periodic oscillations of differential inclusions and then it was combined with the bounding function method to investigate differential equations with nonlocal conditions in Hilbert spaces (see [6] and [7]).In this paper we extend it to Banach spaces.
The proofs of our main results are based on a continuation principle in Banach spaces due to Andres-Górniewicz [3].The employment of fixed point theorems or continuation principles requires strong compactness conditions, which are usually not satisfied in an infinite dimensional framework, if the evolution operator associated to A fails to be compact.In this paper making use of weak topologies we avoid any compactness assumptions on both the nonlinear term and on the C 0 -semigroup.Weak topology was first exploited to prove existence results of problems of type (1) in [9] and in [10].There, the proofs are based on a continuation principle in Fréchet spaces.It requires to prove the transversality condition, known as pushing condition (introduced in [17]), which is more difficult than proving the corresponding condition in a Banach space.Unlike these results, due to the approximation scheme which enables to reduce to a finite dimensional setting, in this paper we are able to consider the usual transversality condition (see (A2) in Section 3).Thus, we can handle nonlinearities with superlinear growth and we obtain, as a by-result, the existence of a solution in a prescribed bounded set.Furthermore, we prove the existence of a bounded solution of the equation in (1) on the whole half-line [0, +∞).
Finally, we show how the abstract results can be applied to study various reactiondiffusion models.More precisely, with our techniques we can handle diffusion problems of the form ∂u(t, ξ) ∂t = ∆u(t, ξ) + f t, ξ, u, in an open bounded domain with sufficiently regular boundary, with Neumann or Dirichlet conditions on the boundary and various boundary conditions.We consider three possible examples of map f : [0, T ] × Ω × R × R → R arising from various mathematical models in applied sciences.In particular, in the case of f t, ξ, u, Ω k(ξ, η) dη = h t, ξ, Ω k(ξ, η)u(t, η) dη we obtain a generalized version of the nonlocal FKPP equation.The same problem was studied also in [8] under different assumptions and techniques.The paper is organized as follows.In Section 2 we recall some notions and notation from the theory of functional analysis.The main results are presented in Section 3, in which first we prove the theorems of existence and uniqueness of mild solutions (see Theorem 3.1 and Theorem 3.2) to problem (1) on compact intervals, then the existence and uniqueness of bounded mild solutions to problem (1) on the half line for the case when f : [0, +∞) × E → E (see Theorem 3.3 and Theorem 3.4).In Section 4, applications to various reaction-diffusion models are considered.

2.
Preliminaries.Throughout this paper, I represents the real interval [0, T ].By E ω and •, • we denote respectively the space E endowed with the weak topology and the dual product between E and its dual.
We recall that a sequence {e n } ⊂ E is a Schauder basis for E if for every x ∈ E there exists a unique sequence of scalars n=1 a Schauder basis of E, for every n ∈ N let E n be the n−dimensional subspace of E generated by the basis {e k } n k=1 and P n : E → E n be the projection of E onto E n .We recall in particular that P n x = n k=1 α k (x)e k for every x ∈ E with the coefficients {α k (x)} linear and continuous, i.e. α k ∈ E * , ∀ k ∈ N (see [25, pp 18-20]), thus P n x j → P n x 0 for x j x 0 , i.e.P n : E ω → E n is continuous.It is well known that the sequence { P n } is bounded.The Schauder basis {e n } is said to satisfy property (π 1 ) if P n = 1 for every n ∈ N.
Remark 1.For every 1 < p < ∞, L p (Ω, R) is a reflexive Banach space with topological dual the strictly convex space L p (Ω, R), where p is the conjugate exponent of p, i.e. 1  p + 1 p = 1.Moreover for each bounded subset Ω ⊂ R n , L p (Ω, R) has a Schauder basis satisfying property (π 1 ) (see, e.g.[13]).Some of the main properties of the projection P n are contained in the following.They were proved in [7, Lemma 6] for Hilbert spaces, but are valid also in Banach spaces.
Lemma 2.1.The projection P n : E → E n satisfies the following properties: (a) Proof.Let Φ : L 1 (I, E) → R be a linear and bounded functional.Hence, there is We have Recall that f n L 1 (I,E) f and trivially P n g We can introduce the adjoint projections as , for every n ∈ N, see [25,Theorem 12.1]; moreover, being E a reflexive space it follows that P * n g → g, for every n ∈ N, see [25,Corollary 12.2].Furthermore, we denote with J : E → E * the duality map Since E * is a strictly convex Banach space, the duality map is single valued; moreover, since E has a basis satisfying property (π 1 ), it follows , for all n ∈ N and x ∈ E n (see [13]).
Given A ⊂ E, let A be the closure of A, while B E (0, R) denotes the closed ball By C(I, E) and L 1 (I, E) we denote respectively the Banach space of all continuous functions x : I → E with norm and the Banach space of summable functions with norm A ball of radius R centered at 0 in the space C(I, E) is denoted by B C (0, R).
Finally, with (L, • L ) we denote the Banach space of linear and bounded operators in E.
We recall the following characterization of weak convergence in the space of continuous functions.
Theorem 2.3.(see [12]) A sequence of continuous functions {x n } n x ∈ C(I; E) if and only if (i) there exists N > 0 such that, for every n ∈ N and t ∈ I, x n (t) ≤ N ; (ii) for every t ∈ I, x n (t) x(t).
It follows that {x n } n x ∈ C(I; E) implies that {x n } n x ∈ L 1 (I; E).Let S ⊆ R be a measurable subset.A subset A ⊂ L 1 (S, E) is called uniformly integrable if for every ε > 0 there is δ > 0 such that Ω ⊂ S and µ(Ω) < δ implies where µ is the Lebesgue measure on R.
We propose now the continuation principle that we use to prove the main result.
Theorem 2.4.(see, e.g.[3]) Let Q be a closed, convex subset of a Banach space F with nonempty interior and T : Q × [0, 1] → F be a compact map having a closed graph such that T (Q, 0) ⊂ intQ and T (•, λ) is fixed points free on the boundary of Q for all λ ∈ [0, 1).Then there exists y ∈ F such that y = T (y, 1).

3.1.
Existence results on compact intervals.We will study the existence of mild solutions to problem (1) under the following assumptions: (A1) M is a linear and bounded operator with M ≤ 1; (A2) there exist R 0 > r 0 > 0 such that J(z), f (t, z) ≤ 0, for a.e.∈ I and for z ∈ E : r 0 < z < R 0 ; (A3) for every z ∈ E the function f (•, z) : I → E is measurable; (A4) for every bounded subset Ω ⊂ E there exists a function v Ω ∈ L 1 (I, R) such that f (t, z) ≤ v Ω (t), for a.e.t ∈ I and all z ∈ Ω; (A5) for a.e.t ∈ I the function f (t, •) : for a.e.t ∈ I and all z, w ∈ B E (0, R), where R 0 , r 0 are the constants from Remark 2. Let us note that: (a) The transversality condition (A2) assures that every solution of the equation in (1) located in a suitable bounded region of E is strictly contained in such a region for t ∈ (0, T ], i.e. it isn't tangent to its boundary.This guarantees the absence of fixed points of the solution operator in the boundary of a given candidate set of solutions, which is a key ingredient when attaching solvability by means of fixed point theorems.Notice that, in the special case when E is a Hilbert space, condition (A2) reads as z, f (t, z) ≤ 0, for a.e.∈ I and for z ∈ E : r 0 < z < R 0 , which is an usual assumption in this setting; (b) It is clear that under conditions (A3) − (A5) [or, (A3) − (A4) and (A5) ] for every q ∈ Q the function f * (t) := f (t, q(t)) is in the space L 1 (I, E); (c) The class of boundary value problems with the operator M satisfying condition (A1) is sufficiently large.In particular, it includes the following well-known problems: (i) M x = 0 (the general Cauchy condition x(0) = x 0 can be replaced by condition z(0) = 0 by a transformation z = x − x 0 ); (ii) M x = ± x(T ) (periodic and anti-periodic problems); By mild solutions on Q of problem (1), we mean continuous functions x ∈ Q such that x = T (x).
for every t ∈ I and x ∈ E. Now, for every n ∈ N consider the problem To prove the existence of solutions u ∈ Q of the problem (3) we will use the bounding function technique and the approximation solvability method.In fact, consider the auxiliary problem in C(I, E m ) Step 1.Let us show that problem (4) has a (strong) solution u m ∈ Q (m) for each m ∈ N. Toward this goal, fix m and q ∈ Q (m) .Consider the linear Cauchy problem where λ ∈ [0, 1].It is clear that problem (5) has a unique solution, therefore we can define the solution operator Σ : Let us show that the operator Σ has closed graph and it is compact.Indeed, let {q k } ⊂ Q (m) and {λ k } ⊂ [0, 1] be two strongly convergent sequences to q 0 ∈ Q (m) and λ 0 ∈ [0, 1] respectively and assume that the sequence {Σ(q k , λ k )} strongly converges to x 0 ∈ C(I, E m ).We shall prove that x 0 = Σ(q 0 , λ 0 ).By the linearity and continuity of the operators M , P m and A n it follows that λ k P m M q k E → λ 0 P m M q 0 and P m A n q k (t) E → P m A n q 0 (t) for every t ∈ I.By virtue of the continuity of f and observing that E m is a finite dimensional space we have that P m f (t, q k (t)) → P m f (t, q 0 (t)) for a.e.t ∈ I.Moreover, notice that {q k } ⊂ Q (m) , then by the boundedness of the operators P m , M and A n and by (A4) we have for a.e.t ∈ I.By the Lebesgue Dominated Convergence Theorem we have that for every t ∈ I.By the uniqueness of the limit we obtain that x 0 (t) = λ 0 P m M q 0 + λ 0 t 0 P m A n q 0 (s) + P m f (s, q 0 (s)) ds, for every t ∈ I. Now, let {q k } ⊂ Q (m) and {λ k } ⊂ [0, 1] and denote x k = Σ(q k , λ k ).Recalling that {x k } is a sequence of strong solutions of (5), we have that for a.e.t ∈ I, and hence {x k } is uniformly integrable.It follows the equicontinuity of the sequence {x k }.Moreover, according to (A1), it also follows that i.e. the equiboundedness of {x k }.Then, applying the Ascoli-Arzelà Theorem we obtain the relative compactness of {x k } and the claimed result.
On the other hand, since R(n, A) L ≤ n −1 (see [26, Theorem 3.1.1]),we have J(q(t)), P m A n q(t) = J(q(t)), n 2 R(n, A)q(t) − n J(q(t)), q(t) However, giving a contradiction.Applying Theorem 2.4 we obtain that for each m ∈ N there exists u m ∈ Q (m) which is a solution to (4).
Step 2. In this step, we prove that for every n ∈ N problem (3) has a (strong) solution.Denote f m (t) = A n u m (t)+f (t, u m (t)).From the condition {u m } ⊂ Q and according to (A4) it follows that there exists ν * ∈ L 1 (I, E) such that f m (t) ≤ ν * (t) for a.e.t ∈ I and every m.Therefore, the sequence {f m } is bounded and uniformly integrable in L 1 (I, E).So, it is relatively weakly compact in L 1 (I, E).W.l.o.g.assume that From Lemma 2.2 we get that u m = P m f m L 1 (I,E) f 0 .Moreover, the set {u m (0) : m ∈ N} is bounded in the reflexive Banach space E. So, w.l.o.g.we can assume that Consider the absolute continuous function Clearly u 0 is absolutely continuous and u 0 (t) = f 0 (t) for a.e.t.Moreover, it is easy to see that for all t ∈ I.By (A5), Lemma 2.1 and the linearity and continuity of the operator A n we get that for a.e.t ∈ I and for every weak neighborhood Now, by the Mazur Lemma (see, e.g.[16], p. 16) there is a sequence of convex combinations {f (m) }, which converges to u 0 in L 1 (I, E).Applying e.g.[24, Theorem 38] we assume w.l.o.g that So, by the convexity of the set V we have Therefore, by the uniqueness of the weak limit we get u 0 (t) = A n u 0 (t) + f t, u 0 (t) for a.e.t ∈ I. Hence, Theorem 2.3 implies and thus, M u m E M u 0 .Consequently, again according to Lemma 2.1, u m (0) = P m M u m E M u 0 .From (6) we have M u 0 = γ 0 = u 0 (0), obtaining that u 0 is a solution to problem (3).
Step 3. We will show now that problem (1) has a mild solution in Q.For each n ∈ N let u n ∈ Q be a solution to the problem (3), i.e., From {u n } ⊂ Q, (A1) and the reflexivity of the space E we get that there exists u ∈ E such that, up to subsequence, M u n E u.Moreover, denoting f n (t) = f (t, u n (t)), from u n ∈ Q and (A4) it follows that there exists v * ∈ L 1 (I, R) such that Hence, we get that {f n } is bounded and uniformly integrable in L 1 (I, E).W.l.o.g.assume that For every t ∈ I, we have Consequently, the maps e (t−•)An f n and S(t − •)f belong to the space L 1 ([0, t]; E) for every t ∈ I. Now let us show that To this aim, let Φ : L 1 ([0, t]; E) → R be a linear and bounded functional.Hence, We have for t ∈ I and z * ∈ E * , where Notice that, from (7) we get that e (t−s)A * n ϕ(s) → S * (t − s)ϕ(s) for a.e.s ∈ [0, t].Moreover the convergence is dominated, thus and we get that for every t ∈ I. Thus, In fact, for every g ∈ E * we have g, e tAn z n − S(t)z = g, e tAn z − S(t)z + g, e tAn (z n − z) .
By virtue of (2) it follows that g, e tAn z − S(t)z → 0 as n → ∞.
Thus, according to (A1) and (A5), we get e tAn M u n E S(t)M u, Therefore, by the uniqueness of the weak limit we have obtaining the claimed result.
Step 4. Now we will show that the set of solutions to problem (1) is weakly compact in C(I, E).Let {u k } ⊂ Q be a sequence of solutions to the problem (1), i.e.,Moreover, from {u k } ⊂ Q, (A1) and the reflexivity of the space E we get that there exists u ∈ E such that, up to subsequence, M u k E u. Hence for every t ∈ I. Thus, u k C(I,E) u ∈ Q.
Thus, by the linearity and continuity of the semigroup S(t) according to (A1) and (A5), we get Therefore, by the uniqueness of the weak limit we have u(t) = S(t)M u + t 0 S(t − s)f (s, u(s))ds, for every t ∈ I.
Step 5. Finally, let the C 0 −semigroup {S(t)} t≥0 generated by A be compact and a ∈ (0, T ) be an arbitrary number.We will show that the set of solutions on Q to problem (1) is compact in C([a, T ], E).
Let {u k } ⊂ Q be a sequence of solutions to the problem (1), i.e., Reasoning as in Step 4, we get that the sequence h k,t defined as h k,t (s) = S(t − s)f (s, u k (s)) is bounded and uniformly integrable in L 1 ([0, t], E).Moreover, from the compactness of S(t − s) for every s < t, it follows that the set {h k,t (s)} is relatively compact for a.a.s < t.Since the operator P : ds is the Cauchy operator in the special case when the semigroup is identically equal to I, it satisfies conditions (i) and (ii) in [18, Theorem 5.1.1],thus according to the same theorem we obtain that the sequence r 0 h k,t (s) ds is relatively compact in C([0, t], E), hence, in particular, the sequence {v k } ⊂ C(I, E) defined as Now we can use Theorem 8.4.1 in [26] to get the conclusion.This ends our proof.
Remark 3. Let us note that (i) Theorem 3.1 can be proved also replacing (A1) with (A1 ) M is a weakly continuous map and M x ≤ x 0 for every x ∈ C(I, E). (ii) Whenever M is as in one of the specific cases (i), (ii) or (iii) with t 1 > 0 in item (c) of Remark 6 and the C 0 −semigroup generated by A is compact, then the set of mild solutions on Q of problem ( 1) is strongly compact in C([0, T ], E).In fact, given a sequence {u k } ⊂ Q of mild solutions, we already proved that it is equicontinuous in C([0, T ], E) as well as pointwise relatively compact for every t ∈ (0, T ].Since u k (0) = k0 i=1 α i u k (t i ) for some α i ∈ R and 0 < t 1 < • • • < t k0 ≤ T, the relative compactness of {u k (t i )} for every i = 1, ...k 0 yields also the relative compactness of {u k (0)}, thus the claimed result from Ascoli-Arzelá theorem.
We recall that given two Banach spaces E 1 and E 2 endowed with two measures of non compactness β 1 and β 2 , it is possible to define the (β 2 , β 1 )-norm of a bounded linear operator L : Notice that since L is a bounded and linear operator its (β 1 , β 2 )-norm is finite (see e.g.[2,Theorem 2.4.1]).If we consider the modulus of fiber of noncompactness as measure of non compactness in the space C([0, T ], E), i.e.
Proof.For fixed n ∈ N, reasoning as in Theorem 3.1 we can prove that for every m ∈ N problem (4) has a strong solution For any t ∈ [0, T ] we can compute the Hausdorff measure of noncompactness of {u m (t)}.Precisely, since {u m } is a bounded sequence and M is a bounded linear operator we have Since A n generates a semigroup of contraction, it is easy to prove that e tPmAn ≤ 1 for every t ∈ I, m ∈ N, hence (A5) implies that for a.e.s ∈ I, and hence, Thus, From M (χ,γ) + L 1 < 1 we obtain γ({u m }) = 0, and so χ({u m (t)}) = 0 for every t ∈ I. Hence, the set {u m (t)} is relatively compact for every t ∈ I. From u m ∈ Q (m) and (A5) we get that {u m } is relative compact in C(I, E m ) and {u m } is bounded and uniformly integrable in

Moreover, from (A5) and the continuity of
for every t ∈ I and the convergence is dominated, implying by the uniqueness of the weak limit.
So, we proved that for every n ∈ N problem (3) has a strong solution u n ∈ Q, i.e.
Reasoning as above we can prove that u n → u 0 ∈ Q.From (A1) and (A5) it follows that M u n → M u 0 and f (s, u n (s)) E → f (s, u 0 (s)) thus e (t−s)An f (s, u n (s)) → S(t − s)f (s, u 0 (s)) for every t ∈ I and a.e.s ≤ t and the convergence is dominated.Passing (10) to the limit we again obtain that Now, assume that there exist two mild solutions u and v of problem (1) in Q, i.e.
Therefore, for every t ∈ I: So, u ≡ v, i.e. problem (1) has a unique mild solution in Q.
Remark 4. We point out that in all classical conditions listed in Remark 2 it holds Indeed, consider the multipoint condition M x = k0 i=1 α i x(t i ), which includes the Cauchy condition (with α i = 0 for every i) and the periodic and antiperiodic conditions (with k 0 = 1, t 1 = T and α 1 = ±1) and let Ω ⊂ C(I, E) be bounded.From M (Ω) ⊂ k0 i=1 α i Ω(t i ) and the monotonicity, algebraically semiadditivity and semihomogeneity properties of the Hausdorff measure of noncompactness, it holds Take now the weighted condition M x = T 0 w(t)x(t)dt , including the mean value condition (with w(s) = 1 T for every s) and let Ω ⊂ C(I, E) be bounded.Then, for every s ∈ I, since In both cases we get M (χ,γ) ≤ M .

3.2.
Existence results on noncompact intervals.Given a function f : [0, +∞) × E → E, as consequence of our method, we discuss the existence of entirely bounded solution of the equation Then equation (11) admits at least one bounded mild solution on [0, +∞).
Proof.According to Theorem 3.1, for every k ∈ N + there exists a solution u k to the problem By construction it follows that By iterating this process we obtain, for every k ∈ N + , a mild solution ψ k : [0, k] → E for problem (12) such that for every integer k ≥ 2, we have Hence by the continuity of the maps ψ k for any k ∈ N + and by (13) we have that the map ψ : [0, +∞) → E defined as is a mild solution to equation ( 11) on [0, +∞).Moreover, ψ(t) ≤ R for every t ∈ [0, +∞), i.e. ψ is bounded on the half-line.
Easily we obtain an analogous result under the hypothesis (A5) .
Proof.According to Theorem 3.2 for every k ∈ N there exists a unique mild solution u k to the problem (12) such that u k (t) ≤ R for all k ∈ N and t ∈ [0, k].Hence the map defined as is the unique mild solution of ( 11) on [0, +∞).

Applications.
In this section we will apply our abstract result to study population diffusion models.Both our models develop from very simple and ancient population growth equations, respectively the confined exponential distribution u = a(N −u) and the logistic distribution u = au(1−u).These ordinary differential equations were then generalized to partial differential equations to take into account the fact that sometimes growth, transfer and diffusion all occur simultaneously in a phenomenon.For example, when an epidemic spreads through a geographical region, the number of infected people grows as the disease is transferred from those infected to susceptible.
Let Ω ⊂ R n be an open bounded domain with C 1 −boundary.Let us first consider a simple linear diffusion model of the form: This model is used to describe various mathematical models arising, for example, in technology transfer, social sciences, agriculture (see [4] and [21]).
Here, u(t, ξ), K(t, ξ) and a(t, ξ) represent the population size, the carrying capacity, which is the largest population that the resources in the environment can sustain, and the growth (or transfer) coefficient at time t and location ξ, respectively.The temporal change of the population size at location ξ is given by the diffusion term ∆u(t, ξ) and the growth component is a(t, ξ) K(t, ξ) − u(t, ξ) .
Let us mention that problem of reaction-diffusion with time and location dependent of carrying capacity and of growth coefficient and with initial condition u(0, ξ) = u 0 (ξ), for all ξ ∈ Ω is studied intensively by researchers (see, e.g.[19]).Here we will consider problem (14) with a multi-point condition.
Let E = L 2 (Ω, R).We denote with • the norm in E. For each t ∈ [0, T ] set Then problem ( 14) can be substituted by the following semilinear differential equation x (t) = Ax(t) + f (t, x(t)), for t ∈ (0, T ], where the operator A : D(A) ⊆ E → E is defined by It is well known (see, e.g.[26,Theorem 4.2.2]) that A is the generator of a C 0 −semigroup of contractions on E.
By a mild solution of problem (14) we mean a continuous function x ∈ C([0, T ]; E) that is a mild solution of (15).Proof of Theorem 4.1.It is easy to see that the map f satisfies conditions (A3) and (A5) with L(t) = a * for every t.Let us verify condition (A2).In fact, since E is a separable Hilbert space, so we have J(z) = z for every z ∈ E. For every (t, z) ∈ [0, T ] × E we have Therefore, from Theorem 3.2 and Remarks 2 and 4 we obtain the existence of a unique mild solution of problem (14).Now, let us consider again the reaction-diffusion equation in (14) for the case a(t, ξ) = a > 0 for all (t, ξ) ∈ [0, T ] × Ω, but with the growth component depending also on the total population size over the considered area.More precisely, the population dynamic is described by the following problem then problem ( 16) can be written as problem (15).
) the problem (16) has a unique mild solution.
In ( 17) u(t, ξ) represents the density of the population at time t and position ξ.The coefficient b > 0 is the death rate.The proliferation rate h depends on the time, the position and the total size of the population weighted by the kernel k corresponding to the probability of an individual to move from one position to an other.It contains also the forcing term l.The periodic condition aims to take under control the diffusion of the population.
Let E = L p (Ω, R), 1 < p < ∞.We denote with • p the norm in E. By means of a reformulation of this problem we will prove the existence of a mild solution to (17).
To this aim, for each t ∈ [0, 1], set x(t) = u(t, •); let W m,p (Ω, R) be the Sobolev space and W m,p 0 (Ω, R) the subspace containing all functions of W m,p (Ω, R) vanishing at the boundary ∂Ω.
Assume that: It is known that A generates a compact C 0 -semigroup of contractions {U (t)} t≥0 on E (see, [26,Theorem 4.1.3]).Moreover we remark that g is well-defined.Indeed, from Hölder inequality, (3) and (ii), for w ∈ E it follows h t, ξ, for a.e.t ∈ [0, 1] × Ω.Again, by a mild solution of problem (17) we mean a continuous function x ∈ C([0, T ]; E) that is a mild solution of (18).Proof.We show that all assumptions of Theorem 3.1 hold true.First of all, according to (19), for every w ∈ E, f (t, w) p ≤ [b + β(k w p )] w p + l.
The measurability of f (•, w) is trivially satisfied.So, conditions (A3) and (A4) hold true.We now prove that f (t, •) is weakly sequentially continuous for a.e.t.Indeed, it is sufficient to prove that g(t, •) is weakly sequentially continuous for a.e.i.e. the claimed result.Hence, condition (A5) holds true.Now, let us check the condition (A2).To this aim, we recall (see, e.g.[13]) that for w ∈ E, w p > 0, we have

Theorem 3 . 1 .
Let conditions (A1) − (A5) hold.Then the set of mild solutions on Q of problem (1) is nonempty and weakly compact in C(I, E).If, in addition, the C 0 −semigroup generated by A is compact, then the set of mild solutions on Q of problem (1) is also strongly compact in C([a, T ]; E) for each a ∈ (0, T ).Proof.For each n ∈ N, let A n := n 2 R(n, A) − nI be the Yosida approximation of A, where R(n, A) = (nI − A) −1 and I is the identity operator.It is well known (see, e.g.[26, Lemma 3.2.1 and 3.2.2])that A n ∈ L(E) and {e tAn } is a semigroup of contractions such that lim n→∞ e tAn x = S(t)x,

Theorem 4 . 1 .
If k i=1 |α i | + T ||a|| 0 < 1, then in B C (0, a * K * c ) the problem (14) has a unique mild solution.Remark 5. (a) The existence of mild solutions to problem (14) is a quite general result, since we do not need any other constrained condition on the functions a(t, ξ) and K(t, ξ); (b) the multi-point condition can be replaced by other conditions such that M ≤ 1 (see Remark 2).
Hence, we get that {f k } is bounded and uniformly integrable in L 1 (I, E).W.l.o.g.