We prove the uniform Hölder bounds of solutions to a singularly perturbed elliptic system arising in competing models in population dynamics. In this system, two species compete to some extent throughout the whole domain but compete strongly on a subdomain. The proof relies upon the blow up technique and the monotonicity formula by Alt, Caffarelli and Friedman.
A central problem in population ecology is the understanding of spatial behavior of interacting species, in particular in the case when the interactions are large and of competitive type. Spatial segregation may occur when two or more species interact in a highly competitive way. Such phenomenon has been studied using competition models (or its parabolic case) with positive parameter k → + ∞ :
− Δ u i = f i ( u i ) − k u i ∑ j ≠ i b i j u j in Ω . (1.1)
Here Ω is a smooth bounded domain in ℝ n , n ≥ 2 , u i denotes the density of the i-th population, whose internal dynamics is prescribed by f i ( u i ) , i = 1 , 2 , ⋯ , M , and M ≥ 2 is the number of the species. The positive constant k b i j is the interspecific competition rate between the population u i and u j , which is possibly symmetric.
In the study of system (1.1), we are mostly concerned with the asymptotic behavior of the solutions as the parameter k → + ∞ . It turns out that uniformly bounded solutions u k = ( u 1, k , u 2, k , ⋯ , u M , k ) of system (1.1) converge, as k → ∞ , to a limiting configuration in some weak sense, u = ( u 1 , u 2 , ⋯ , u M ) , the limit satisfies u i u j = 0 for i ≠ j , which is called the spatial segregation (cf. [
− Δ u i = f i ( u i ) − k u i ∑ j ≠ i b i j u j 2 in Ω . (1.2)
In recent years, people show a lot of interests in segregation phenomenon, and abroad literature is present: starting from [
Among the models proposed so far, the species compete strongly on the whole of Ω. However, in some heterogeneous environment, species may compete to some extent in the whole of a region Ω, but compete strongly on a subdomain A. To analysis the corresponding spatial segregation phenomenon governed by strong competition on A, Crooks and Dancer [
{ − Δ u = f ( u ) − s u v − k χ A u v , in Ω , − Δ v = g ( v ) − r u v − k χ A u v , in Ω , u = v = 0 , on ∂ Ω , (1.3)
where k is again a positive competition parameter, u and v denote the densities of two species, the self-interaction functions f and g are assumed to be continuously differentiable and such that f ( 0 ) = g ( 0 ) = 0 and f ( y ) < 0 , g ( y ) < 0 for large y. A is a nonempty open subset of Ω with smooth boundary such that A ¯ ⊂ Ω . The parameters r and s are assumed to nonnegative, and χ A is the characteristic function on A.
Due to the presence of the characteristic function χ A in (1.3), we cannot expect classical solutions in general. By a k-dependent solution of (1.3), we will mean a pair of functions ( u k , v k ) such that u k , v k ∈ W 2, p ( Ω ) , p > n , and satisfy (1.3) almost everywhere. The asymptotic behavior of solutions to system (1.3) has been investigated in [
α ∗ = { 2 / 3 , when n = 2 , 1 / 2 , when n ≥ 3. (1.4)
Due to the apparent of subdomain, we can not expect boundedness for every Hölder exponent. In fact we have the following.
Theorem 1.1. Let ( u k , v k ) be nonnegative solutions of (1.3), and α ∗ be defined in (4). Assume that for every k, there exists M > 0 , independent of k, such that
‖ ( u k , v k ) ‖ L ∞ ( Ω ) ≤ M .
Then for every α ∈ ( 0, α ∗ ) , there exists C > 0 , independent of k, such that
‖ ( u k , v k ) ‖ C 0, α ( A ¯ ) ≤ C .
Notations Throughout the paper, we denote by B R ( x 0 ) = { x ∈ ℝ n : | x − x 0 | < R } the open ball with center x 0 and radius R > 0 . If x 0 = 0 , we simply denote by B R : = B R ( 0 ) . We assume that any x ∈ ℝ n can be written as x = ( x ′ , x n ) , with x ′ ∈ ℝ n − 1 and x n ∈ ℝ . In this way, we denote by ℝ + n : = ℝ n ∩ { x n > 0 } . For any D ⊂ ℝ n we write D + : = D ∩ { x n > 0 } and ∂ + D : = ∂ D ∩ { x n > 0 } . We also denote by ∇ θ u and ∂ ν u the tangential derivative and the radial derivative of u, respectively.
The proof of Theorem 1.1 mainly follows the blow up method, developed by Terracini and her coauthors in [
The rest of this paper is organized as follows: in Section 2, we establish a monotonicity formula of ACF type, and by utilizing this monotonicity formula, we prove a Liouville type theorem for entire solutions to a semilinear system. In Section 3, we perform the blow up procedure and complete the proof of Theorem 1.1.
In this section, we prove some nonexistence result in ℝ n . The main tools will be the monotonicity formula by Alt, Caffarelli, Friedman originally stated in [
λ 1 ( E ) : = inf u ∈ H 0 1 ( E ) ∫ E | ∇ θ u | 2 ∫ E u 2 .
Here ∇ θ u stands for the tangential gradient of u on E.
Lemma 2.1. Let α ∗ be as in (1.4). We define the nondecreasing function γ ( x ) as
γ ( x ) = ( n − 2 2 ) 2 + x − n − 2 2 , ∀ x ∈ ℝ + ,
and the admissible set A by
A : = { ( E 1 , E 2 ) : E i ⊂ ∂ B 1 ( 0 ) , E 1 + ∩ E 2 + = ∅ } .
Then we have
inf ( E 1 , E 2 ) ∈ A ∑ i = 1 2 γ ( λ 1 ( E i ) ) ≥ 2 α ∗ .
Proof If n = 2 , it is obviously that
γ ( λ ( E 1 ) ) + γ ( λ ( E 2 ) ) = λ ( E 1 ) + λ ( E 2 ) .
A symmetrization argument gives that the optimal domain is a connected arc. Moreover, the longer the arc is, the smaller the first eigenvalue is. Thus the sum ∑ i = 1 2 γ ( λ 1 ( E i ) ) takes its minimum for two arcs E 1 , E 2 with
E 1 + ∪ E 2 + = ∂ + B 1 ( 0 ) and ∂ B 1 ( 0 ) \ ∂ + B 1 ( 0 ) ⊂ E i , i = 1 , 2.
If we assume that the length of E 1 is ( 1 + τ ) π , then the length of E 2 is ( 2 − τ ) π , and the corresponding eigenfunctions are
sin θ 1 + τ and sin θ 2 − τ , 0 < τ < 1.
Thus we have
γ ( λ ( E 1 ) ) + γ ( λ ( E 2 ) ) ≥ 1 1 + τ + 1 2 − τ ≥ 4 3 ,
and the equality holds if and only if τ = 1 / 2 .
If n > 2 , according to the argument in [
γ ( λ ( E ) ) ≥ ψ ( s ) ,
where E ⊂ ∂ B 1 ( 0 ) , s = m e a s ( E ) / m e a s ( ∂ B 1 ( 0 ) ) , and ψ ( s ) is convex and decreasing:
ψ ( s ) = { 1 2 log 1 4 s + 3 2 , s < 1 4 , 2 ( 1 − s ) , 1 4 < s < 1.
Setting s i = m e a s ( E i ) / m e a s ( ∂ B 1 ( 0 ) ) , i = 1 , 2 , we then have
γ ( λ ( E 1 ) ) + γ ( λ ( E 2 ) ) ≥ ψ ( s 1 ) + ψ ( s 2 ) ≥ 2 ψ ( ( s 1 + s 2 ) / 2 ) ≥ 2 ψ ( 3 / 4 ) = 1.
This completes the proof of Lemma 2.1. □
In the following, we shall prove an ACF type monotonicity formula associated with the following system
{ − Δ u = − χ T u v , in ℝ n , − Δ v = − χ T u v , in ℝ n . (2.1)
where T = ℝ + n , χ T is the characteristic function on T. As in [
f ( r ) = { 2 − n 2 r 2 + n 2 , r ≤ 1 , 1 r n − 2 , r > 1 ,
and denote m ( | x | ) = − Δ f ( | x | ) 2 . In this setting, we note that m ( | x | ) is bounded in ℝ n , vanishes in ℝ n \ B 1 ( 0 ) and m ( | x | ) ≥ 0 for a.e. x.
Under the previous notations, we can prove the following monotonicity formula.
Theorem 2.2. Let u , v ∈ H l o c 1 ( ℝ n ) ∩ C ( ℝ n ) be positive solutions of (2.1) and let ε > 0 be fixed. Then there exists r ¯ > 1 such that the function
J ( r ) ≜ 1 r 4 α ∗ − ε ∫ B r ( 0 ) ( f ( | x | ) ( | ∇ u | 2 + χ T u 2 v ) + m ( | x | ) u 2 ) × ∫ B r ( 0 ) ( f ( | x | ) ( | ∇ v | 2 + χ T u v 2 ) + m ( | x | ) v 2 ) .
is increasing for r ∈ ( r ¯ , + ∞ ) .
Proof The proof is inspired by [
J 1 ( r ) ≜ ∫ B r ( 0 ) ( f ( | x | ) ( | ∇ u | 2 + χ T u 2 v ) + m ( | x | ) u 2 ) ,
J 2 ( r ) ≜ ∫ B r ( 0 ) ( f ( | x | ) ( | ∇ v | 2 + χ T u v 2 ) + m ( | x | ) v 2 ) .
Then J ( r ) = r ε − 4 α * J 1 ( r ) J 2 ( r ) . Let us first evaluate the derivative of J ( r ) for r > 1 . A straightforward calculation leads to
J ′ ( r ) J ( r ) = − 4 α ∗ − ε r + ∫ ∂ B r ( 0 ) ( f ( | x | ) ( | ∇ u | 2 + χ T u 2 v ) + m ( | x | ) u 2 ) J 1 ( r ) + ∫ ∂ B r ( 0 ) ( f ( | x | ) ( | ∇ v | 2 + χ T u v 2 ) + m ( | x | ) v 2 ) J 2 ( r ) . (2.2)
By testing the equation for u in (2.1) with f ( | x | ) u on B r ( 0 ) , we obtain
∫ B r ( 0 ) f ( | x | ) ( | ∇ u | 2 + χ T u 2 v ) = − ∫ B r ( 0 ) ∇ ( u 2 2 ) ⋅ ∇ f ( | x | ) + ∫ ∂ B r ( 0 ) f ( | x | ) u ∂ ν u = − ∫ B r ( 0 ) m ( | x | ) u 2 + ∫ ∂ B r ( 0 ) ( f ( | x | ) u ∂ ν u − u 2 2 ∂ ν f ( | x | ) ) .
Thus we can rewrite the term J 1 ( r ) in a different way
J 1 ( r ) = ∫ ∂ B r ( 0 ) ( f ( | x | ) u ∂ ν u − u 2 2 ∂ ν f ( | x | ) ) = 1 r n − 2 ∫ ∂ B r ( 0 ) u ∂ ν u + n − 2 r n − 1 ∫ ∂ B r ( 0 ) u 2 2 . (2.3)
Now we define
Λ 1 ( r ) = r 2 ∫ ∂ B r ( 0 ) ( | ∇ θ u | 2 + χ T u 2 v ) ∫ ∂ B r ( 0 ) u 2 , Λ 2 ( r ) = r 2 ∫ ∂ B r ( 0 ) ( | ∇ θ v | 2 + χ T u v 2 ) ∫ ∂ B r ( 0 ) v 2 ,
where | ∇ θ u | 2 = | ∇ u | 2 − | ∂ ν u | 2 . Then for every δ ∈ ℝ , by Hölder inequality and Young’s inequality, there holds
| ∫ ∂ B r ( 0 ) u ∂ ν u | ≤ ( ∫ ∂ B r ( 0 ) u 2 ) 1 2 ( ∫ ∂ B r ( 0 ) ( ∂ ν u ) 2 ) 1 2 ≤ Λ 1 ( r ) 2 δ 2 r ∫ ∂ B r ( 0 ) u 2 + δ 2 r 2 Λ 1 ( r ) ∫ ∂ B r ( 0 ) ( ∂ ν u ) 2 ≤ 1 2 ( 1 δ 2 ∫ ∂ B r ( 0 ) ( | ∇ θ u | 2 + χ T u 2 v ) + δ 2 ∫ ∂ B r ( 0 ) ( ∂ ν u ) 2 r Λ 1 ( r ) ) .
Substituting in (2.3) we obtain
J 1 ( r ) ≤ 1 2 r n − 3 [ ( 1 δ 2 Λ 1 ( r ) + n − 2 Λ 1 ( r ) ) ∫ ∂ B r ( 0 ) ( | ∇ θ u | 2 + χ T u 2 v ) + δ 2 Λ 1 ( r ) ∫ ∂ B r ( 0 ) ( ∂ ν u ) 2 ] .
Now we choose δ in such a way that
1 δ 2 Λ 1 ( r ) + n − 2 Λ 1 ( r ) = δ 2 Λ 1 ( r ) .
After some calculation, we obtain
Λ 1 ( r ) δ 2 = γ ( Λ 1 ( r ) ) ,
where γ : ℝ + → ℝ is defined as
γ ( x ) = ( n − 2 2 ) 2 + x − n − 2 2 .
With this choice of δ we have
J 1 ( r ) ≤ 1 2 γ ( Λ 1 ( r ) ) ∫ ∂ B r ( 0 ) ( f ( | x | ) ( | ∇ u | 2 + χ T u 2 v ) ) .
Similarly, we also have
J 2 ( r ) ≤ 1 2 γ ( Λ 2 ( r ) ) ∫ ∂ B r ( 0 ) ( f ( | x | ) ( | ∇ v | 2 + χ T u v 2 ) ) .
Substituting in (2.2) we obtain
J ′ ( r ) J ( r ) ≥ − 4 α ∗ − ε r + 2 γ ( Λ 1 ( r ) ) r + 2 γ ( Λ 2 ( r ) ) r .
Therefore it only remains to prove that there exists a r ¯ > 1 such that for every r ≥ r ¯ there holds
γ ( Λ 1 ( r ) ) + γ ( Λ 2 ( r ) ) > 4 α ∗ − ε 2 . (2.4)
To this aim we define the functions u ( r ) ( θ ) , v ( r ) ( θ ) : ∂ B 1 ( 0 ) → ℝ as
u ( r ) ( θ ) ≜ u ( r θ ) , v ( r ) ( θ ) ≜ v ( r θ ) .
Then a change of variables gives
Λ 1 ( r ) = ∫ ∂ B 1 ( 0 ) ( | ∇ u ( r ) | 2 + r 2 χ T u ( r ) 2 v ( r ) ) ∫ ∂ B 1 ( 0 ) u ( r ) 2 , Λ 2 ( r ) = ∫ ∂ B 1 ( 0 ) ( | ∇ v ( r ) | 2 + r 2 χ T u ( r ) v ( r ) 2 ) ∫ ∂ B 1 ( 0 ) v ( r ) 2 .
Notice first of all that there exists a constant C > 0 such that ∫ ∂ B 1 ( 0 ) u ( r ) 2 ≥ C for r sufficiently large. Indeed assume by contradiction this is not true, then l i m r → ∞ 1 | ∂ B r ( 0 ) | ∫ ∂ B r ( 0 ) u = 0 , which implies u ( 0 ) = 0 since u is subharmonic, and this contradicts the assumption u > 0 . The same result clear holds also for v ( r ) .
Assume (2.4) does not hold, then there exists r n → ∞ such that
γ ( Λ 1 ( r n ) ) + γ ( Λ 2 ( r n ) ) ≤ 4 α * − ε 2 < 2 α * , (2.5)
in particular, Λ 1 ( r n ) , Λ 2 ( r n ) are bounded. We define
u ˜ ( r n ) ≜ u ( r ) ‖ u ( r ) ‖ L 2 ( ∂ B 1 ( 0 ) ) , v ˜ ( r n ) ≜ v ( r ) ‖ v ( r ) ‖ L 2 ( ∂ B 1 ( 0 ) ) .
Then there exists a constant C > 0 (independent of r n ) such that
C ≥ Λ 1 ( r n ) ≥ ∫ ∂ B 1 ( 0 ) | ∇ u ˜ ( r n ) | 2 , C ≥ Λ 2 ( r n ) ≥ ∫ ∂ B 1 ( 0 ) | ∇ v ˜ ( r n ) | 2 ,
which ensure the existence of u ¯ , v ¯ ≠ 0 such that up to a subsequence, we have u ˜ ( r n ) ⇀ u ¯ , v ˜ ( r n ) ⇀ v ¯ in H 1 ( ∂ B 1 ( 0 ) ) . Moreover, since
C ≥ Λ 1 ( r n ) = ∫ ∂ B 1 ( 0 ) ( | ∇ u ˜ ( r n ) | 2 + r n 2 χ T u ˜ ( r n ) 2 v ( r n ) ) ≥ r n 2 ∫ ∂ B 1 ( 0 ) χ T u ˜ ( r n ) 2 v ˜ ( r n ) .
We infer that u ¯ ⋅ v ¯ ≡ 0 on ∂ + B 1 ( 0 ) . Then Lemma 2.1 yields
lim inf n → ∞ [ γ ( Λ 1 ( r n ) ) + γ ( Λ 2 ( r n ) ) ] ≥ γ ( λ { supp ( u ¯ ) } ) + γ ( λ { supp ( v ¯ ) } ) ≥ 2 α ∗ ,
that is in contradiction with (2.5). □
As in [
Lemma 2.3. ( [
sup x , y ∈ ℝ n | u ( x ) − u ( y ) | | x − y | α < ∞ .
Then u is constant.
Theorem 2.4. Let u , v ∈ H l o c 1 ( ℝ n ) ∩ C ( ℝ n ) be nonnegative solutions of system (2.1). Assume that for some α ∈ ( 0, α ∗ ) there holds
sup x , y ∈ ℝ n | u ( x ) − u ( y ) | | x − y | α < ∞ , sup x , y ∈ ℝ n | v ( x ) − v ( y ) | | x − y | α < ∞ . (2.6)
Then one of the functions is identically zero and the other is a constant.
Proof We first note that, by (2.6) and Lemma 2.3, if one of the functions is identically zero or a positive constant, then the other must be a constant or 0 respectively. Hence we may assume by contradiction that neither u nor v is constant. Then by the maximum principle u and v are positive, and Theorem 2.2 ensures the existence of a constant C > 0 such that
∫ B r [ f ( | x | ) ( | ∇ u | 2 + χ T u 2 v ) + m u 2 ] ∫ B r [ f ( | x | ) ( | ∇ v | 2 + χ T u v 2 ) + m v 2 ] ≥ C r 4 α * − ε , (2.7)
for r sufficiently large. Let η = η r , 2 r ∈ C 0 ∞ ( ℝ n ) be any smooth, radial, cut-off function with the following properties: 0 ≤ η r ,2 r ≤ 1 , η r , 2 r = 1 in B r , η r , 2 r = 0 in ℝ n \ B 2 r and | ∇ η r ,2 r | ≤ C / r . Testing the equation − Δ u = − χ T u v with the function η 2 f u on B 2 r , we obtain
− ∫ B 2 r ( 2 η f u ∇ η ⋅ ∇ u + η 2 u ∇ f ⋅ ∇ u + η 2 f | ∇ u | 2 ) = ∫ B 2 r χ T η 2 f u 2 v .
Consequently,
∫ B 2 r η 2 f ( | ∇ u | 2 + χ T u 2 v ) = − ∫ B 2 r ( 2 η f u ∇ η ⋅ ∇ u + η 2 ∇ f ⋅ ∇ u 2 2 ) ≤ ∫ B 2 r ( 1 2 f η 2 | ∇ u | 2 + 2 f u 2 | ∇ η | 2 − η 2 ∇ u 2 2 ⋅ ∇ f ) = ∫ B 2 r ( 1 2 f η 2 | ∇ u | 2 + 2 f u 2 | ∇ η | 2 − ∇ ( η 2 u 2 2 ) ⋅ ∇ f + u 2 η ∇ η ⋅ ∇ f ) .
Testing the equality Δ f = − 2 m with the function η 2 u 2 / 2 on B 2 r , we obtain
− ∫ B 2 r ∇ ( η 2 u 2 2 ) ⋅ ∇ f = ∫ B 2 r − 2 m η 2 u 2 2 ,
which together with the previous inequality and the fact that m ≥ 0 , gives
∫ B 2 r η 2 [ f ( | ∇ u | 2 + χ T u 2 v ) + m u 2 ] ≤ 2 ∫ B 2 r ( 2 f u 2 | ∇ η | 2 + u 2 η ∇ η ⋅ ∇ f ) .
Now, recalling the definition of η and f and using assumption (2.6), we obtain
∫ B r η 2 [ f ( | ∇ u | 2 + χ T u 2 v ) + m u 2 ] ≤ C ∫ B 2 r \ B r u 2 | x | n ≤ C ∫ 0 2 r ρ 2 α ρ n ρ n − 1 d ρ ≤ C r 2 α .
Similarly,
∫ B r [ f ( | ∇ v | 2 + χ T u v 2 ) + m v 2 ] ≤ C r 2 α .
Thus we have
∫ B r [ f ( | ∇ u | 2 + χ T u 2 v ) + m u 2 ] ∫ B r [ f ( | ∇ v | 2 + χ T u v 2 ) + m v 2 ] ≤ C r 4 α ,
which contradicts with (2.7) for r large. □
Remark 2.5. If T = ℝ n , then u , v compete in the whole ℝ n . In this case we have α ∗ = 1 , see [
A similar nonexistence result is true when studying 2-tuple of subharmonic functions on ℝ n having disjoint supports on T
Corollary 2.6. Let u , v ∈ H 0 1 ( ℝ n ) ∩ C l o c ( ℝ n ) such that
{ − Δ u ≤ 0 , in ℝ n , − Δ v ≤ 0 , in ℝ n , u v = 0 , in T ,
and for some fixed α ∈ ( 0, α * ) , there exists a constant C > 0 such that
sup x , y ∈ Ω | u ( x ) − u ( y ) | | x − y | α ≤ C , sup x , y ∈ Ω | v ( x ) − v ( y ) | | x − y | α ≤ C ,
then one of the functions is identically zero and the other is a constant.
In this section, we shall establish the uniform Hölder bounds for solutions to system (1.3). Note that the strong competition effect of the system only occurs in subdomain A, while in the other regions, the equation does not contain the strong competition parameter k, so the solutions u k and v k are uniformly bounded independently of k in W 2, p ( Ω \ A ¯ ) , for each p ∈ [ 1, + ∞ ) , and in C 1, λ ( Ω \ A ¯ ) , for each λ ∈ ( 0,1 ) . Therefore, up to a subsequence, u k , v k converge strongly in C 1, λ ( Ω \ A ¯ ) . In order to improve the uniform convergence result obtained in [
Theorem 3.1. Let ( u k , v k ) be nonnegative solutions of system (1.3) uniformly bounded in L ∞ ( Ω ) . Then for every α ∈ ( 0, α * ) there exists C > 0 , independent of k, such that
‖ ( u k , v k ) ‖ C 0, α ( A ¯ ) ≤ C ,
for every k > 0 .
The proof of Theorem 3.1 is inspired from the work of [
L k = max { sup x , y ∈ A ¯ , x ≠ y | u k ( x ) − u k ( y ) | | x − y | α , sup x , y ∈ A ¯ , x ≠ y | v k ( x ) − v k ( y ) | | x − y | α } → ∞ .
We can assume that L k is achieved, say, by u k at the pair ( u k , v k ) . That is
L k = | u k ( x k ) − u k ( y k ) | | x k − y k | α → ∞ .
Let us define the rescaled functions
u ˜ k ≜ u k ( x k + r k x ) L k r k α , v ˜ k ≜ v k ( x k + r k x ) L k r k α , for x ∈ Ω k = Ω − x k r k ,
where r k → 0 will be chosen later. By direct calculation, ( u ˜ k and v ˜ k ) satisfy the following system
{ − Δ u ˜ k = r k 2 − α L k f ( L k r k α u ˜ k ) − s L k r k 2 + α u ˜ k v ˜ k − k L k r k 2 + α χ A k ( x ) u ˜ k v ˜ k , in Ω k , − Δ v ˜ k = r k 2 − α L k g ( L k r k α v ˜ k ) − r L k r k 2 + α u ˜ k v ˜ k − k L k r k 2 + α χ A k ( x ) u ˜ k v ˜ k , in Ω k ,
where A k = A − x k r k . We note that Ω k → ℝ n as k → + ∞ , and depending on the asymptotic behavior of the distance d ( x k , ∂ A ) , we have A k → T , where T is either ℝ n or an half-space (when d ( x k , ∂ A ) → ∞ or the limit is finite, respectively). We also observe that
max { sup x , y ∈ A ¯ k , x ≠ y | u ˜ k ( x ) − u ˜ k ( y ) | | x − y | α , sup x , y ∈ A ¯ k , x ≠ y | v ˜ k ( x ) − v ˜ k ( y ) | | x − y | α } = | u ˜ k ( 0 ) − u ˜ k ( y k − x k r k ) | | y k − x k r k | α = 1.
Since u k , v k are uniformly bounded in L ∞ ( Ω ) ∩ C 1, λ ( Ω \ A ¯ ) , r k → 0 and L k → + ∞ , by diect calculations it is easy to see that
r k 2 − α L k f ( L k r k α u ˜ k ) − s L k r k 2 + α u ˜ k v ˜ k , r k 2 − α L k g ( L k r k α v ˜ k ) − r L k r k 2 + α u ˜ k v ˜ k → 0 in L ∞ ( Ω k ) , (3.1)
| ∇ u ˜ k | , | ∇ v ˜ k | → 0 in L ∞ ( Ω k \ A ¯ k ) . (3.2)
In the following, we need to make different choices of the sequence r k . Once r k is chosen, we will use Ascoli-Arzelà’s Theorem to pass to the limit on compact sets. Now since the u ˜ k , v ˜ k ’s are uniform α-Hölder continuous, it is suffices to show that u ˜ k ( 0 ) and v ˜ k ( 0 ) are bounded in k. To begin with, we need the following technical lemma, which is proved in [
Lemma 3.2. [
{ − Δ u ≤ − H u , in B 2 R , u ≥ 0 , in B 2 R , u ≤ A , on ∂ B 2 R ,
where H is a positive constant, then for every δ ∈ ( 0,1 ) , it holds
‖ u ‖ L ∞ ( B R ) ≤ C A e − δ R H ,
where C > 0 is a constant, and only dependent on δ , R .
Lemma 3.3. Let r k → ∞ as k → ∞ be such that
(i) d ( x k , y k ) ≤ R ′ r k for some R ′ > 0 .
(ii) k L k r k 2 + α ↛ 0 .
Then u ˜ k ( 0 ) , v ˜ k ( 0 ) are uniformly bounded in k.
Proof We prove the estimate for u ˜ k ( 0 ) ; that for v ˜ k ( 0 ) follows similarly. Assume by contradiction that { u ˜ k ( 0 ) } is unbounded. Let R ≥ R ′ and choose k sufficiently large such that B R ( 0 ) ⊂ Ω k . Moreover since [ u ˜ k ] α , A ¯ k = 1 and k L k r k 2 + α ↛ 0 , we have
I k : = inf B R ( 0 ) ∩ A k k L k r k 2 + α u ˜ k → + ∞ .
Claim. ∀ R ≥ R ′ , ‖ k L k r k 2 + α u ˜ k v ˜ k ‖ L ∞ ( B R ( 0 ) ∩ A k ) → 0 .
Indeed, note that
− Δ v ˜ k = r k 2 − α L k g ( L k r k α v ˜ k ) − r L k r k 2 + α u ˜ k v ˜ k − k L k r k 2 + α χ A k ( x ) u ˜ k v ˜ k , in Ω k ,
and by (3.1), we have
− Δ v ˜ k ≤ − I k 2 v ˜ k in B R ( 0 ) ∩ A k . (3.3)
In order to simplify the notation, let K ≜ B R ( 0 ) ∩ A k and for each compact set K ′ ⋐ K , we choose a cut-off function η ∈ C 0 ∞ ( K ) such that η ≡ 1 on K ′ , η ≡ 0 on ℝ n \ K . Then by testing (3.3) with η 2 v ˜ k on K, we obtain
I k 2 ∫ K η 2 v ˜ k 2 + ∫ K η 2 | ∇ v ˜ k | 2 ≤ − 2 ∫ K η ∇ v ˜ k ⋅ ∇ η v ˜ k ≤ ∫ K η 2 | ∇ v ˜ k | 2 + ∫ K | ∇ η | 2 v ˜ k 2 ,
So, there exist two positive constants C 1 , C 2 such that
C 2 I k 2 inf x ∈ K v ˜ k 2 ≤ I k 2 ∫ K η 2 v ˜ k 2 ≤ ∫ K | ∇ η | 2 v ˜ k 2 ≤ C 1 sup x ∈ K v ˜ k 2 . (3.4)
Since [ v ˜ k ] α , K ≤ 1 , we have
inf x ∈ K v ˜ k ≥ sup x ∈ K v ˜ k − R α .
Then (3.4) implies that
C 2 I k 2 ( sup x ∈ K v ˜ k − R α ) 2 ≤ C 1 sup x ∈ K v ˜ k 2 .
If we choose k sufficient large such that C 1 I k ≤ C 2 8 , then
sup x ∈ K v ˜ k 2 ≤ 2 C 2 ( R ) C 1 ( R ) R 2 α ,
which implies the boundedness of u ˜ k in K. Thus we can apply Lemma 3.1, which gives
sup K ′ v ˜ k ≤ C e − C ′ I k ,
and the claim can be easily seen.
Define
u ^ ( x ) ≜ u ˜ k ( x ) − u ˜ k ( 0 ) .
We have u ^ k ( 0 ) = 0 , and it is Hölder continuous, and
| u ^ k ( y ¯ k ) − u ^ k ( 0 ) | | y ¯ k | α = 1, (3.5)
where y ¯ k = y k − x k r k . Moreover, by the claim
− Δ u ^ k ( x ) = ε k ,
where ε k → 0 , as k → ∞ , uniformly on any B R ( 0 ) .
From this equation, we can infer from L p theory and Sobolev embeddig theory that u ^ k ( x ) is uniformly Lipschitz continuous, that is, there exists L > 0 is a constant, independent with k such that
sup x , y ∈ B R ( 0 ) | u ^ k ( x ) − u ^ k ( y ) | | x − y | ≤ L .
We have that, up to a subsequence, y ¯ k → y ¯ , since y ¯ k = y k − x k r k ≤ R ′ . We claim that y ¯ ≠ 0 .
In fact, if y ¯ = 0 , then we obtain
1 = | u ^ k ( y ¯ k ) − u ^ k ( 0 ) | | y ¯ k | α = | u ^ k ( y ¯ k ) − u ^ k ( 0 ) | | y ¯ k | | y ¯ k | 1 − α ≤ L | y ¯ k | 1 − α → 0 ,
as k → ∞ , which is a contradiction. After passing to a subsequence, u ^ k converges to a continuous function u ^ ∞ on compacts and satisfying
− Δ u ^ ∞ ( x ) = 0 in ℝ n .
Moreover, (3.5) can be passed to the limit, which is
| u ^ ∞ ( y ¯ ) − u ^ ∞ ( 0 ) | | y ¯ | α = 1. (3.6)
Thus we have u ^ ∞ ≡ C by Lemma 2.3, this contradicts (3.6). So u ˜ k ( 0 ) is uniformly bounded. □
Lemma 3.4. Up to a subsequence, we have k L k | x k − y k | 2 + α → + ∞ .
Proof We assume by contradiction that there exists R ′ > 0 such that k L k | x k − y k | 2 + α ≤ R ′ . Let
r k = ( k L k ) − 1 2 + α , N k = L k r k 2 + α ,
then we obtain
k N k = 1 , lim k → ∞ r k = 0 , | x k − y k | α ≤ R ′ ( k L k ) − 1 = R ′ r k 2 + α ≤ R ′ r k ,
while k is sufficient large, so we can use Lemma 3.3 to conclude that u ˜ k ( 0 ) , v ˜ k ( 0 ) are uniformly bounded.
On the other hand, by the uniform Hölder continuity and the Ascoli-Arzelà theorem we have that, up to a subsequence, there exist u ˜ ∞ and v ˜ ∞ such that u ˜ k → u ˜ ∞ , v ˜ k → v ˜ ∞ uniformly on the compact set of ℝ n . Moreover, the choice of r k implies that the equations of u ˜ k and v ˜ k are
{ − Δ u ˜ k = r k 2 − α L k f ( L k r k α u ˜ k ) − s L k r k 2 + α u ˜ k v ˜ k − χ A k ( x ) u ˜ k v ˜ k , in Ω k , − Δ v ˜ k = r k 2 − α L k g ( L k r k α v ˜ k ) − r L k r k 2 + α u ˜ k v ˜ k − χ A k ( x ) u ˜ k v ˜ k , in Ω k . (3.7)
From the first equation, we can also obtain a uniform Lipschitz estimate of u ˜ k , and we also have
| u ˜ ∞ ( y ¯ ) − u ˜ ∞ ( 0 ) | | y ¯ | α = 1. (3.8)
Let k → ∞ in (3.7), we can obtain, up to a subsequence,
{ − Δ u ˜ ∞ = − χ T u ˜ ∞ v ˜ ∞ , in ℝ n , − Δ v ˜ ∞ = − χ T u ˜ ∞ v ˜ ∞ , in ℝ n ,
where T = ℝ + n , or T = ℝ n . So we can use Theorem 2.4 and Remark 2.5 to conclude that one of u ˜ ∞ and v ˜ ∞ is identically zero and the other is a constant, which contradicts (3.8). □
Now we come to the proof of Theorem 3.1.
The proof of Theorem 3.1. From Lemma 3.4, we must have k L k | x k − y k | 2 + α → + ∞ . Let r k = | x k − y k | . With this choice, we know that all the assumptions of Lemma 3.3 are satisfied and hence u ˜ k ( 0 ) and v ˜ k ( 0 ) are uniformly bounded. Again by the uniform Hölder continuity and the Ascoli-Arzelà theorem we have that, up to a subsequence, there exist u ˜ ∞ and v ˜ ∞ such that u ˜ k → u ˜ ∞ , v ˜ k → v ˜ ∞ uniformly on the compact set of ℝ n . Note that | y ¯ k | = | ( y k − x k ) / r k | = 1 , (3.5) implies that
| u ˜ ∞ ( y ¯ ) − u ˜ ∞ ( 0 ) | = 1 (3.9)
Moreover u ˜ k and v ˜ k satisfy the following inequalities
− Δ u ˜ k ≤ r k 2 − α L k f ( L k r k α u ˜ k ) − s L k r k 2 + α u ˜ k v ˜ k in Ω k , (3.10)
− Δ v ˜ k ≤ r k 2 − α L k g ( L k r k α v ˜ k ) − r L k r k 2 + α u ˜ k v ˜ k in Ω k . (3.11)
Let k → ∞ in (3.10) and (3.11), we obtain
{ − Δ u ˜ ∞ ≤ 0 in ℝ n , − Δ v ˜ ∞ ≤ 0 in ℝ n .
Now let K ⊂ ⊂ ℝ n be a compact set, we can choose k sufficient large such that K ⊂ ⊂ Ω k . Let us choose a cut-off function η ∈ C 0 ∞ ( ℝ n ) such that 0 < η < 1 and η ≡ 1 on K. Multiplying (3.10) by η and integrating by parts, we obtain
k L k r k 2 + α ∫ K χ A k ( x ) u ˜ k v ˜ k ≤ ∫ Ω k r k 2 − α L k f ( L k r k α u ˜ k ) η + ∫ Ω k u ˜ k Δ η ,
Since u ˜ k is uniformly Hölder continuous, it then form the boundedness of u ˜ k ( 0 ) that u ˜ k is uniformly bounded on compact set K. Therefore the right hand side of previous inequality is uniformly bounded. Because k L k | x k − y k | 2 + α → ∞ , we obtain
lim k → ∞ ∫ K χ A k ( x ) u ˜ k v ˜ k = 0 ,
which yields
χ T u ˜ ∞ v ˜ ∞ = 0 in ℝ n .
To sum up, we have
{ − Δ u ˜ ∞ ≤ 0 , in ℝ n , − Δ v ˜ ∞ ≤ 0 , in ℝ n , u ˜ ∞ v ˜ ∞ = 0 , in T .
Thus we can infer from Corollary 2.6 that one of the limiting functions is identically zero and the other is a constant, which contradicts (3.9). The proof of Theorem 3.1 is complete. □
The study of the asymptotic behavior of singular perturbed equations and system of elliptic or parabolic type is very broad and subject of research. In this paper, We study the large-interaction limit of solutions to a singularly perturbed elliptic system modeling the steady states of two species u and v which compete to some extent throughout a domain Ω but compete strongly on a subdomain A ⊂ Ω . We improve the uniform convergence result of [
Finally, we mention that there many interesting problems for further study. Note that we prove the uniform Hölder bounds to a singularly perturbed elliptic system, naturally to ask whether this result can be extended to the corresponding parabolic system? Up to our knowledge, the uniform Hölder bounds for parabolic setting is unknown, and both the asymptotics and the qualitative properties of the limit segregated profiles remain a challenge, this will be the object of a forthcoming paper.
The work is partially supported by PRC grant NSFC 11601224.
The authors declare no conflicts of interest regarding the publication of this paper.
Ni, J.H. and Zhang, S. (2023) Uniform Hölder Bounds for Competition Systems with Strong Interaction on a Subdomain. Journal of Applied Mathematics and Physics, 11, 1525-1540. https://doi.org/10.4236/jamp.2023.116100