Coupled Versus Uncoupled Blow-Up Rates in Cooperative n-Species Logistic Systems

Theorem 1.1.

Let z∈∂⁡Ω be such that (1.5) is satisfied, and consider the next partition of the subscripts set

Let k∈{1,…,n} be such that

Then, any positive solution of (1.1), u=(u1,…,un), satisfies

where

and

where A0,i stands for the unique positive solution of the equation

Remark 1.2.

The assignations ∂⁡Ω→[0,+∞), z↦αi⁢(z), are uniformly continuous for all 1≤i≤n.

Theorem 1.3.

Problem (1.1) admits a unique positive solution.

Theorem 2.1.

The following assertions are equivalent:

1. σ ⁢ [ 𝔏 , D ] > 0 .

2. There exists u ¯ ∈ [ 𝒞 2 ⁢ ( D ) ] n ∩ [ 𝒞 ⁢ ( D ¯ ) ] n such that u ¯ > 0 in D and

   𝔏 ⁢ u ¯ ≥ 0   in  ⁢ D ,

   and, for some 1 ≤ i 0 ≤ n , either u ¯ i 0 > 0 on ∂ ⁡ D , or else

   ( 𝔏 ⁢ u ¯ ) i 0 > 0   in  ⁢ D .

   Should this be the case, u ¯ is said to be a positive strict supersolution of 𝔏 in D.

3. The operator 𝔏 satisfies the strong maximum principle in D , in the sense that, for every h ∈ [ 𝒞 ν ⁢ ( D ¯ ) ] n , u∈[𝒞2+ν⁢(D¯)]n and w∈[𝒞2+ν⁢(∂⁡D)]n satisfying

   𝔏 ⁢ u = h ≥ 0   in  ⁢ D , u = w ≥ 0   on  ⁢ ∂ ⁡ D ,

   with some of these inequalities strict, one has that u ≫ 0 in D.

Theorem 2.2.

Suppose (2.2) admits a subsolution u¯∈[C2+ν⁢(Ω¯)]n and a supersolution u¯∈[C2+ν⁢(Ω¯)]n satisfying u¯≤u¯. Then (2.2) possesses a solution u∈[C2+ν⁢(Ω¯)]n such that u¯≤u≤u¯. Actually, (2.2) possesses a minimal and a maximal solution in the interval [u¯,u¯].

Theorem 2.3.

Problem (2.2) has a unique positive solution, θ[Ω,w]. Moreover, for every positive subsolution u¯ (resp. supersolution u¯) of (2.2),

Proof.

Adapting [2, Theorem 3.7], it suffices to construct a supersolution u¯ such that u¯i>0 for all 1≤i≤n. In the special case 𝔞i⁢(x)>0 for all x∈∂⁡Ω and 1≤i≤n, one can take u¯=(M,…,M) for some M>0 sufficiently large. In general, we may proceed as follows.

Since ∂⁡Ω is smooth, ∂⁡Ω possesses finitely many components, Γk, 1≤k≤m. For each ε>0 and 1≤k≤m, denote

Let

and let 𝔏¯ be the operator

Thanks to (2.1),

On the other hand, by the uniqueness of the principal eigenvalue,

Thus, since

where |Ωkε| stands for the Lebesgue measure of Ωkε, the Faber–Krahn inequality, going back to [11] and [18], yields

(see [21, Theorem 5.1]). Therefore, ε can be shortened, if necessary, so that min1≤k≤m⁡σ⁢[𝔏,Ωkε]>0. Fix ε>0 satisfying the last inequality and, for each 1≤k≤m, let

be a principal eigenfunction associated to σ⁢[𝔏,Ωkε]. As φk≫0, it is apparent that

Subsequently, we consider the auxiliary function Φ defined through

where g is any C2+ν-extension of the function φ1⊗⋯⊗φm to the open set Ωint with the special requirement that infΩint⁡g>0. Such a function exists because of (2.3). Then τ⁢Φ is a supersolution of (2.2) for sufficiently large τ>1. Indeed, by (2.3), there exists τ0≥1 such that

Moreover, for every 1≤k≤m and 1≤i≤n, we find that, in Ωkε/2∩Ω,

Lastly, in Ωint, we have that, for every 1≤i≤n,

for sufficiently large τ>1, because 𝔞i and gi are bounded away from zero in Ωint and pi>1 for all 1≤i≤n. This ends the proof. ∎

Theorem 2.4.

There exists a minimal and a maximal positive solution of (1.1), Lmin and Lmax, respectively, in the sense that any solution, L, of (1.1) satisfies

Moreover, the point-wise limit (2.4) provides us with the minimal solution

while the maximal solution is given by

where we have denoted

Lemma 3.1.

For each ε>0, there exists a constant C:=C⁢(ε,n) such that the function ψ¯ε:=(ψ¯ε,1,…,ψ¯ε,n), defined by

provides us with a supersolution of (3.1).

Proof.

By definition, ψ¯ε is smooth and satisfies the boundary conditions. Hence, ψ¯ε is a supersolution of (3.1) if, and only if,

for every 0<r<R and 0≤i≤n. Now, multiply (3.5) by (R-r)αi+2 if i∈I+∪I0, and by (R-r)-γi+αi⁢pi if i∈I-. Then, ψ¯ε is a supersolution if, and only if, for every 0<r<R,

if i∈I+∪I0, and

if i∈I-.

Let us show that

Indeed, if i∈I+ and j∈I+∪I0, j≠i, we have that

by the definition of I+ (see (3.2)). When j∈I-, we may proceed as follows. According to (3.2),

So, multiplying by pj-1 we deduce that γj<pj⁢μ1-2⁢pj-μ1, j∈I-, and dividing by pj yields

Equivalently,

and therefore

for every i∈I+ and j∈I-.

Analogously, the following estimates hold:

Indeed, by (3.3), αi=μi for all i∈I0. Similarly, since {1,…,k}⊂I+, we have αj=μj for all 1≤j≤k. Moreover, μj=μ1 for all 1≤j≤k. Thus,

for all i∈I0 and 1≤j≤k, which shows the validity of the identities of (3.10). In order to check the inequalities of (3.10), we can argue as follows. Pick i∈I0 and j∈{k+1,…,n}, j≠i. Suppose j∈I+∪I0. Then, by (3.3) and taking into account that, by construction, μj<μ1 for all k+1≤j≤n, we find that

because i∈I0. Now, suppose that j∈I-. Then, thanks to (3.3) and (3.9),

Therefore, (3.10) holds.

Next, we will see that

Indeed, by the definition of μi and since i∈I-,

Thus, adding γi at both sides of this inequality and taking common factor γi+2 yields

Hence, by (3.3),

whence (3.11).

Lastly, we will establish that

By (3.3), -γi+αi⁢pi=μ1 for all i∈I-. Thus,

for all i∈I- and 1≤j≤k, whence the identities of (3.12). Now, pick i∈I- and k+1≤j≤n, i≠j. Suppose j∈I+∪I0. Then, due to (3.3),

and hence (3.12) holds in this case. Now, suppose j∈I-. Then, by (3.9),

and therefore (3.12) is satisfied.

By (3.3) and the definition of μi,

Thus,

for all C≥0 and i∈I+∪I0. Hence, thanks to (3.8), (3.10) and (3.13), we can extend (3.6) to r=R by letting r↑R. Similarly, (3.7) can be extended to r=R by (3.11) and (3.12). More precisely, at r=R, (3.6) provides us with

while (3.7) at r=R becomes

Due to (3.4) and using that (1+ε)<(1+ε)pi (since pi>1 for all 1≤i≤n), it is easily seen that (3.14) and (3.15) are true. In the derivation one should note that αj=μj=μ1 for all 1≤j≤k, by construction. Actually, all inequalities in (3.14) and (3.15) are strict, because (1+ε)<(1+ε)pi. By continuity, this entails the existence of δ:=δ⁢(ε,n)>0 such that (3.6) and (3.7) are satisfied for all r∈[R-δ,R). Therefore, choosing a sufficiently large C>0, we can assume that (3.5) holds in (0,R), because, for each 1≤i≤n, the function bi⁢(R-r)γi is positive and bounded away from zero in [0,R-δ]. The proof is complete. ∎