Diffusive logistic equations with harvesting and heterogeneity under strong growth rate

Theorem 1.1.

Assume b ⁢ ( x ) ≢ 0 in Ω.

1. If Ω 0 = ∅ , then for every a > λ 1 ⁢ ( Ω ) , equation ( 1.2 ) has a unique positive solution u a .

2. If Ω 0 ≠ ∅ , then for any a ∈ ( λ 1 ⁢ ( Ω ) , λ 1 ⁢ ( Ω 0 ) ) , equation ( 1.2 ) has a unique positive solution u a . In addition, if a ≥ λ 1 ⁢ ( Ω 0 ) , then ( 1.2 ) has no nonnegative solution except zero.

Furthermore, in either case the curve a → u a is continuous and increasing and the positive solution u a is stable, i.e., μ 1 ⁢ ( u a ) > 0 .

Theorem 1.2 ([8, Theorem 3.6]).

Let a 0 = λ 1 ⁢ ( Ω 0 ) . Then the following hold:

1. u a → + ∞ uniformly on Ω ¯ 0 as a ↗ a 0 .

2. u a → U ¯ a 0 uniformly as a ↗ a 0 on any compact subset of Ω ¯ ∖ Ω ¯ 0 , where U ¯ a 0 is the minimal positive solution of the boundary blow-up equation

   - Δ ⁢ u = a ⁢ u - b ⁢ ( x ) ⁢ u 2   in  ⁢ Ω ∖ Ω ¯ 0 , u = 0   on  ⁢ ∂ ⁡ Ω , u = + ∞   on  ⁢ ∂ ⁡ Ω 0 .

Theorem 1.3 ([18, Theorem 2.6]).

Suppose that λ 1 ⁢ ( Ω ) < a < λ 1 ⁢ ( Ω 0 ) , b ⁢ ( x ) ≢ 0 and h ⁢ ( x ) ≢ 0 . Then there exists c ^ a > 0 such that the following hold:

1. If 0 ≤ c < c ^ a , then equation ( 1.1 ) has a maximal positive solution u ¯ a , c . If c > c ^ a , then no solution of ( 1.1 ) stays positive in Ω.

2. The curve c → u ¯ a , c is decreasing with respect to the parameter c for c ∈ [ 0 , c ^ a ) and u ¯ a , c is stable, that is, μ 1 ⁢ ( u ¯ a , c ) > 0 . Furthermore, u ¯ a , c is the unique positive stable solution of ( 1.1 ).

Theorem 1.4 ([18, Theorem 2.8]).

Under the assumptions of Theorem 1.3, there exists ϵ > 0 such that for a ∈ ( λ 1 ⁢ ( Ω ) , λ 1 ⁢ ( Ω ) + ϵ ) , the following hold:

1. Equation ( 1.1 ) has exactly two positive solutions u ¯ a , c and u ¯ a , c for c ∈ ( 0 , c ^ a ) , exactly one positive solution u ^ a with c = c ^ a , and no positive solution for c > c ^ a .

2. The Morse index M ⁢ ( u ) is 1 for u = u ¯ a , c when c ∈ [ 0 , c ^ a ) , and u ^ a is degenerate with μ 1 ⁢ ( u ^ a ) = 0 .

3. All solutions lie on a smooth curve Σ that, on ( c , u ) space, starts from ( 0 , 0 ) , continues to the right, reaches the unique turning point at c = c ^ a where it turns back, then continues to the left without any turnings until it reaches ( 0 , u a ) , where u a is the unique positive solution of ( 1.1 ) with c = 0 .

Theorem 2.1 (Saddle-node bifurcation at a turning point [4]).

Let X and Y be Banach spaces and assume that ( λ 0 , u 0 ) ∈ R × X . Let F be a continuously differentiable mapping of an open neighborhood V of ( λ 0 , u 0 ) into Y. Suppose that the following hold:

1. dim ⁡ N ⁢ ( F u ⁢ ( λ 0 , u 0 ) ) = codim ⁡ R ⁢ ( F u ⁢ ( λ 0 , u 0 ) ) = 1 , N ⁢ ( F u ⁢ ( λ 0 , u 0 ) ) = span ⁡ { w 0 } ,

2. F λ ⁢ ( λ 0 , u 0 ) ∉ R ⁢ ( F u ⁢ ( λ 0 , u 0 ) ) .

If Z is a complement of span ⁡ { w 0 } in X, then the solutions of F ⁢ ( λ , u ) = F ⁢ ( λ 0 , u 0 ) near ( λ 0 , u 0 ) form a curve ( λ ⁢ ( s ) , u ⁢ ( s ) ) = ( λ 0 + τ ⁢ ( s ) , u 0 + s ⁢ w 0 + z ⁢ ( s ) ) , where s → ( τ ⁢ ( s ) , z ⁢ ( s ) ) ∈ R × Z is a continuously differentiable function near s = 0 and τ ⁢ ( 0 ) = τ ′ ⁢ ( 0 ) = 0 , z ⁢ ( 0 ) = z ′ ⁢ ( 0 ) = 0 . Moreover, if F is k times continuously differentiable, then so are τ ⁢ ( s ) and z ⁢ ( s ) .

Lemma 2.2 ([8]).

Suppose that u > 0 and v > 0 are, respectively, C 2 ⁢ ( Ω ) sub and super solutions of (2.1), that is,

Furthermore, assume that

Then u ≤ v in Ω.

Theorem 2.3 ([8, Theorem 2.4]).

For any a ∈ ( - ∞ , + ∞ ) , (2.2) has a minimal positive solution U ¯ a and a maximal positive solution U ¯ a , in the sense that any positive solution u of (2.2) satisfies U ¯ a ⁢ ( x ) ≤ u ⁢ ( x ) ≤ U ¯ a ⁢ ( x ) .

Theorem 2.4.

Suppose that λ 1 ⁢ ( Ω ) < a < λ 1 ⁢ ( Ω 0 ) and (A1) and (A2) hold. Then there exists c ¯ a > 0 such that the following hold:

1. If 0 ≤ c < c ¯ a , then equation ( 1.1 ) has a maximal solution u ¯ a , c , and if c > c ¯ a , then ( 1.1 ) has no solution.

2. The curve c → u ¯ a , c is decreasing with respect to the parameter c for c ∈ [ 0 , c ¯ a ) and u ¯ a , c is stable, that is, μ 1 ⁢ ( u ¯ a , c ) > 0 . Furthermore, u ¯ a , c is the unique stable solution of ( 1.1 ).

3. For c = c ¯ a , there exists a degenerate solution u ¯ a , i.e., μ 1 ⁢ ( u ¯ a ) = 0 , and equation ( 1.1 ) has another unstable solution for c near and to the left of c ¯ a .

Proof.

As arguments similar to the ones used in the proof of this result will be needed repeatedly later on, we will provide all the details here so as to be able to refer to them later on. First note that if equation (1.1) has a solution ( c 1 , u 1 ) , then either u 1 ≤ 0 or the comparison lemma above applied on the set { x : u 1 ⁢ ( x ) > 0 } implies that u 1 + ≤ u a , where u a is the unique positive solution of equation (1.2). Multiplying equation (1.1) with φ 1 , the first eigenfunction of - Δ in Ω with Dirichlet boundary condition, and then integrating over Ω, we obtain

implying that c ¯ a is well defined. Next for p > n let X = { u ∈ W 2 , p ⁢ ( Ω ) : u = 0 ⁢  on  ⁢ ∂ ⁡ Ω } and Y = L p ⁢ ( Ω ) . We define F : ℝ × X → Y by F ⁢ ( c , u ) = Δ ⁢ u + a ⁢ u - b ⁢ ( x ) ⁢ u 2 - c ⁢ h ⁢ ( x ) . Note that if F ⁢ ( c 1 , u 1 ) = 0 , that is, if u 1 is a solution of (1.1) with harvesting rate c = c 1 , and if μ 1 ⁢ ( u 1 ) > 0 , then by applying the implicit function theorem, we can continue the curve of solutions forward (i.e., for c ≥ c 1 ) with respect to c. Moreover, denoting the curve of solutions by ( c , u ) = ( c , u ⁢ ( c ) ) , it will be decreasing as v := ∂ ⁡ u ∂ ⁡ c ⁢ ( c 1 ) solves the equation

Hence, μ 1 ⁢ ( u ⁢ ( c ) ) = λ 1 ⁢ ( - a + 2 ⁢ b ⁢ ( x ) ⁢ u ⁢ ( c ) ) is decreasing with respect to c as well. Therefore, the curve of solutions starting at ( 0 , u a ) , can be continued to the right until a point ( c 0 , u 0 ) with μ 1 ⁢ ( u 0 ) = 0 . Next applying the saddle-node bifurcation (Theorem 2.1), one easily sees that the curve turns back at the degenerate solution u 0 , therefore generating a second unstable solution in a left neighborhood of c 0 . Furthermore, the curve of solutions obtained above is indeed the curve of maximal solutions. To see this, first note that if equation (1.1) has a solution at ( c 1 , u 1 ) , then (1.1) will have a maximal solution ( c 1 , u ¯ 1 ) with μ 1 ⁢ ( u ¯ 1 ) ≥ 0 . Indeed, u 1 , u a is a pair of sub-super solutions of (2.1), therefore the comparison lemma above applied on the set { x : u 1 ⁢ ( x ) > 0 } implies that u 1 + ≤ u a . Next, considering u 1 , u a as an ordered pair of sub-super solutions of (1.1) for c = c 1 , the standard iteration process starting at the super solution u a will provide the maximal solution u ¯ 1 = u ¯ 1 ⁢ ( c 1 ) of (1.1) for c = c 1 , which, by construction, will have μ 1 ⁢ ( u ¯ 1 ) ≥ 0 (see [17]). Also, as u ¯ 1 ⁢ ( c 1 ) is a sub-solution of (1.1) with c = c 2 , if c 1 > c 2 , it is clear that u ¯ ⁢ ( c ) is decreasing in c and therefore left continuous. In addition, for 0 ≤ c < c 0 , since u ⁢ ( c ) ≤ u ¯ ⁢ ( c ) , we have μ 1 ⁢ ( u ¯ ⁢ ( c ) ) ≥ μ 1 ⁢ ( u ⁢ ( c ) ) > 0 . This implies that the implicit function theorem applies at every point ( c , u ¯ ⁢ ( c ) ) , from which one easily concludes that u ¯ ⁢ ( c ) is right continuous as well. Hence, the curve of maximal solutions c → u ¯ ⁢ ( c ) is continuous and decreasing for 0 ≤ c < c 0 . Finally, as u a is the unique positive (and therefore maximal) solution of equation (1.1) for c = 0 , we have u ⁢ ( 0 ) = u ¯ ⁢ ( 0 ) = u a , which together with the fact that both curves ( c , u ⁢ ( c ) ) and ( c , u ¯ ⁢ ( c ) ) are continuous, decreasing and constitute of stable solutions yield u ⁢ ( c ) = u ¯ ⁢ ( c ) for 0 ≤ c < c 0 .

Next we will show that equation (1.1) does not have a solution for c > c 0 . Otherwise, if u 1 is a solution of (1.1) for some c 1 > c 0 , then as was shown above, (1.1) has a maximal solution u ¯ 1 at c 1 with μ 1 ⁢ ( u ¯ 1 ) ≥ 0 . Now depending on whether μ 1 ⁢ ( u ¯ 1 ) > 0 or μ 1 ⁢ ( u ¯ 1 ) = 0 , we can continue the curve of solutions from ( c 1 , u ¯ 1 ) backwards (i.e., for c < c 1 ) in an increasing fashion, initially using either the implicit function theorem or the saddle-node bifurcation theorem, and then (as μ 1 ⁢ ( u ⁢ ( c ) ) becomes positive) through the implicit function theorem all the way to c = c 0 . Thus, equation (1.1) will have a solution u at c = c 0 with μ 1 ⁢ ( u ) > 0 . Since u ≤ u ¯ ⁢ ( c 0 ) = u ⁢ ( c 0 ) , we get μ 1 ⁢ ( u ⁢ ( c 0 ) ) > 0 , which is a contradiction. Finally, a similar argument yields uniqueness of stable solutions. The proof of the theorem is now complete. ∎

Remark 2.5.

It is obvious that c ^ a ≤ c ¯ a . In the remarks on page 3614 of [15], Oruganti, Shi and Shivaji claim, in passing, that c ¯ a = c ^ a . Although the validity of this claim for arbitrary λ 1 ⁢ ( Ω ) < a < λ 1 ⁢ ( Ω 0 ) is not clear to us, we are able to show that if h ≡ 0 in Ω ¯ ∖ Ω 0 , then c ^ a = c ¯ a for a sufficiently close to λ 1 ⁢ ( Ω 0 ) (see Lemma 3.5).

Theorem 2.6.

Assume a ≥ λ 1 ⁢ ( Ω 0 ) . Then any solution u (not necessary positive) of (1.1) is unstable, that is, μ 1 ⁢ ( u ) < 0 .

Proof.

Let u 0 be a solution of equation (1.1) with c = c 0 > 0 . Let φ > 0 denote the first eigenfunction of the linearization of (1.1) at u 0 , that is,

Multiplying the above equation by φ 1 * , the first eigenfunction of - Δ in Ω 0 with Dirichlet boundary condition, and integrating over Ω 0 , we obtain

which readily yields μ 1 ⁢ ( u 0 ) < 0 . ∎

Remark 2.7.

Recalling the classical result (see [17]) that a solution u obtained between an ordered pair of sub and super solutions through the classical iteration procedure (starting at the sub or the super solution) will automatically satisfy μ 1 ⁢ ( u ) ≥ 0 , the above result, in particular, indicates that in our search for positive solutions of (1.1) in the strong growth rate case, we cannot utilize the technique of sub and super solutions.

Proposition 2.8.

Assume (A1) and (A2), and suppose that a > λ 1 ⁢ ( Ω 0 ) and a ≠ λ i ⁢ ( Ω ) for all i > 1 . Then equation (1.1) has a positive solution for 0 < c < σ (with some σ > 0 ) if and only if the linear equation

has a positive solution.

Remark 2.9.

In regards to the existence of a positive solution to (2.3), we may write h ⁢ ( x ) = h 1 ⁢ φ 1 + h ~ ⁢ ( x ) , where φ 1 is the first eigenfunction of - Δ in Ω with Dirichlet boundary condition and ∫ Ω h ~ ⁢ ( x ) ⁢ φ 1 = 0 . Then clearly v, the unique solution of equation (2.3), is given by

Therefore, there exists h 1 * > 0 , depending on a and h ~ , such that if h 1 > h 1 * , then v > 0 .

Theorem 3.1.

Assume that a ≥ λ 1 ⁢ ( Ω ) and h ≢ 0 in Ω ∖ Ω ¯ 0 . Then there exists 0 < c ¯ a < + ∞ such that equation (1.1) does not have a nonnegative solution for c > c ¯ a .

Proof.

Let u be a nonnegative solution of (1.1). Using the comparison Lemma 2.2, we have u ≤ U ¯ a in Ω ∖ Ω ¯ 0 , where U ¯ a is the minimal boundary blow-up solution given in Theorem 2.3. Let Ω n = { x ∈ Ω : d ⁢ ( x , Ω 0 ) > 1 n } . Clearly, h ≢ 0 in Ω n for all n large. Fix such an n and let φ n , 1 > 0 be the first eigenfunction of - Δ in Ω n with Dirichlet boundary condition. Multiplying equation (1.1) with φ n , 1 and then integrating over Ω n , we obtain

which implies

where ν is the outer normal vector of ∂ ⁡ Ω n . Therefore, we have

if a > λ 1 ⁢ ( Ω n ) , and

if a ≤ λ 1 ⁢ ( Ω n ) . Therefore, c has a bound independent of u. ∎

Theorem 3.2.

Assume that h ≡ 0 in Ω ∖ Ω ¯ 0 . Then c ^ a → + ∞ as a ↗ λ 1 ⁢ ( Ω 0 ) .

Proof.

Fix c > 0 arbitrary large. We will show that (1.1) has a positive solution for a sufficiently close λ 1 ⁢ ( Ω 0 ) by constructing an ordered pair of sub and super solutions. Clearly, u a is a super solution of (1.1). To construct a sub-solution, we consider

where m and n are large positive constants (given below), φ 1 * ⁢ ( x ) denotes the first eigenfunction of - Δ in Ω 0 with Dirichlet boundary condition, and w n is the positive solution of the equation

whose existence is shown in [8, Lemma 2.3]. We claim that u ¯ is a weak sub-solution of (1.1). First note that u ¯ is continuous. Next let 0 ≤ ϕ ∈ C c ∞ ⁢ ( Ω ) . We have

Therefore,

Now we assume that a ∈ [ λ 1 ⁢ ( Ω 0 ) - ϵ 0 , λ 1 ⁢ ( Ω 0 ) ) for some ϵ 0 small. First we take n large enough such that ( λ 1 ⁢ ( Ω 0 ) - ϵ 0 ) ⁢ n - c ⁢ ∥ h ∥ ∞ ≥ 1 . Next, using the fact that max ∂ ⁡ Ω 0 ⁢ ∂ ν ⁡ φ 1 * < 0 , we choose m sufficiently large so that m ⁢ ∂ ν ⁡ φ 1 * - ∂ ν ⁡ w n < 0 on ∂ ⁡ Ω 0 . Finally, we choose a close enough to λ 1 ⁢ ( Ω 0 ) such that u a > n + m ⁢ φ 1 * in Ω ¯ 0 and m ⁢ ( λ 1 ⁢ ( Ω 0 ) - a ) ⁢ ∥ φ 1 * ∥ ∞ < 1 . Then u ¯ is a weak sub-solution and, in addition, u ¯ ≤ u a (note that h ≡ 0 on Ω ∖ Ω ¯ 0 ). So, equation (1.1) has a positive solution between u ¯ and u a . This completes the proof. ∎

Lemma 3.3.

There exists 0 < δ < δ h such that if u n is a sequence of nonnegative solutions of equation (1.1), with a = a n , c = c n such that λ 1 ⁢ ( Ω ) < a n < λ 1 ⁢ ( Ω 0 ) + δ , c n ≥ 0 and ∥ u n ∥ ∞ → + ∞ . Then u n → + ∞ uniformly on Ω ¯ 0 .

Proof.

Up to a subsequence, a n → a * with λ 1 ⁢ ( Ω 0 ) ≤ a * ≤ λ 1 ⁢ ( Ω 0 ) + δ . We divide the proof into several steps. Step 1: u n → + ∞ uniformly in every compact subsets of Ω 0 . Note that c n ≲ ∥ u n ∥ ∞ . Thus, up to a subsequence, c n / ∥ u n ∥ ∞ → α < + ∞ . Let v n = u n / ∥ u n ∥ ∞ . From equation (1.1) we have

Since - Δ ⁢ v n ≤ a n ⁢ v n and ∥ v n ∥ ∞ = 1 , there exists v 0 ∈ H 0 1 ⁢ ( Ω ) such that v n → v 0 weakly in H 0 1 ⁢ ( Ω ) and strongly in L p ⁢ ( Ω ) for p > 1 . Multiplying (3.2) by v n / ∥ u n ∥ ∞ and then integrating over Ω, we get

Thus, ∫ Ω b ⁢ ( x ) ⁢ v 0 3 = 0 , and so v 0 ≡ 0 in Ω ¯ ∖ Ω 0 .

Next, multiplying (3.2) by ϕ ∈ C c ∞ ⁢ ( Ω 0 ) and integrating over Ω 0 , we have

Passing n to infinity, we obtain

Therefore, v 0 is a nonnegative weak solution of the equation

If a * = λ 1 ⁢ ( Ω 0 ) , then, thanks to the Fredholm alternative, we have α = 0 , since h ⁢ ( x ) ≥ 0 and h ⁢ ( x ) ≢ 0 . Therefore, v 0 = 0 or v 0 = φ 1 * > 0 in Ω 0 (recall that φ 1 * is the first eigenfunction of - Δ with Dirichlet boundary condition in Ω 0 ).

We claim that v 0 ≠ 0 . In fact, if v 0 = 0 , then, by equation (3.2), - Δ ⁢ v n ≤ a n ⁢ v n , so that

uniformly in Ω, as v n → 0 in L p ⁢ ( Ω ) for p > 1 . This contradicts ∥ v n ∥ ∞ = 1 .

Next if λ 1 ⁢ ( Ω 0 ) < a * ≤ λ 1 ⁢ ( Ω 0 ) + δ , then assuming that α = 0 , we have v 0 = 0 , as a * ≠ λ i ⁢ ( Ω 0 ) . But this will again yield a contradiction as above. Thus, α > 0 and v 0 = α ⁢ ψ a * > 0 , the unique positive solution of equation (3.1).

Now standard elliptic estimates (see [11]) imply that u n → + ∞ uniformly in every compact subset of Ω 0 , completing the proof of step 1.

Next we let u n ⁢ ( z n ) = min x ∈ Ω ¯ 0 ⁡ u n ⁢ ( x ) and show that u n ⁢ ( z n ) → ∞ . This will be done in the next two steps. First note that since by assumption ∂ ⁡ Ω 0 is C 2 , γ , it satisfies a uniform interior ball assumption, i.e., there exists R > 0 such that for every x ∈ ∂ ⁡ Ω 0 , there exists a ball B x = B R ⁢ ( y ) with radius R and center y such that B x ⊂ Ω ¯ 0 and B x ∩ ∂ ⁡ Ω 0 = { x } . Let K be a compact subset of Ω 0 such that B R / 2 ⁢ ( y ) ⊂ K for all x ∈ ∂ ⁡ Ω 0 .

Step 2: Let u n ⁢ ( z n ) = min x ∈ Ω ¯ 0 ⁡ u n ⁢ ( x ) . If { u n ⁢ ( z n ) } is bounded, then z n ∈ ∂ ⁡ Ω 0 for all n large. Using some of the ideas of the proof of [8, Lemma 3.3], we argue by contradiction by assuming that for a subsequence, still denoted by ( z n ) , we have z n ∈ Ω 0 . Then step 1 implies z n → ∂ ⁡ Ω 0 . Hence, there exists x n ∈ ∂ ⁡ Ω 0 such that z n ∈ B x n ∖ B R / 2 ⁢ ( y n ) . Next we consider the auxiliary functions

and show that there exists a sequence β n → ∞ such that

To start with, we fix a suitably chosen large σ > 0 , so that η n satisfy the following properties on B x n :

and

Note that the choice of σ depends only on R and the function h ⁢ ( x ) . Let α n = min K ⁡ u n ⁢ ( x ) . The rest of the proof proceeds by considering the two cases a * = λ 1 ⁢ ( Ω 0 ) and a * > λ 1 ⁢ ( Ω 0 ) separately.

If a * = λ 1 ⁢ ( Ω 0 ) , since u n / ∥ u n ∥ ∞ → φ 1 * > 0 in Ω 0 and c n / ∥ u n ∥ ∞ → 0 , then α n / c n → + ∞ . Hence, we can choose β n → ∞ so that c n ⁢ e σ ⁢ R 2 ≤ β n ≤ 1 2 ⁢ α n ⁢ e σ ⁢ R 2 / 4 . Therefore, for x ∈ B x n ∖ B R / 2 ⁢ ( y n ) , we have

and, in addition,

Since a n → λ 1 ⁢ ( Ω 0 ) < λ 1 ⁢ ( B x n ) , the maximum principle finally yields (3.3). Now taking x = z n in (3.3), we have β n ⁢ η n ⁢ ( z n ) ≤ 0 , which, since z n ∈ B x n ∖ B R / 2 ⁢ ( y n ) , contradicts (3.4).

In the second case, that is, a * > λ 1 ⁢ ( Ω 0 ) , by step 1 we have 0 < α = lim n → + ∞ ⁡ c n / ∥ u n ∥ ∞ and v n → v 0 = α ⁢ ψ a * . Thus,

From (3.1) and Remark 2.9, we have

Hence, by further decreasing δ, we can guarantee that

Now we may proceed as in the previous case, obtaining (3.3) and a contradiction as before. The proof of step 2 is now complete.

Step 3: Let u n ⁢ ( z n ) = min x ∈ Ω ¯ 0 ⁡ u n ⁢ ( x ) . Then u n ⁢ ( z n ) → ∞ . To argue by contradiction, we assume that u n ⁢ ( z n ) is bounded, i.e., u n ⁢ ( z n ) ≤ M for some M > 0 independent of n. Then, by step 2, we have z n ∈ ∂ ⁡ Ω 0 . We follow the argument in step 2 where now z n = x n , and conclude

for some sequence β n → ∞ . Lemma 2.3 in [8] implies that the equation

has a unique positive solution and by the comparison lemma, u n ⁢ ( x ) ≥ w n ⁢ ( x ) in Ω ¯ ∖ Ω 0 . Similarly, w, the unique positive solution of

satisfies w n ⁢ ( x ) ≤ w ⁢ ( x ) in Ω ¯ ∖ Ω 0 . Thus, ∥ w n ∥ L ∞ ⁢ ( Ω ∖ Ω ¯ 0 ) is bounded, and therefore standard elliptic estimates imply that { w n } is bounded in C 1 ⁢ ( Ω ¯ ∖ Ω 0 ) , and so, in particular, | ∇ ⁡ w n ⁢ ( x n ) | remains uniformly bounded. Since u n ⁢ ( x n ) = w n ⁢ ( x n ) and u n ⁢ ( x ) ≥ w n ⁢ ( x ) in Ω ¯ ∖ Ω 0 , we have

for some M 0 > 0 independent of n. On the other hand, using (3.6) and taking into account (3.5), we obtain

contradicting (3.7). This completes the proof of step 3, and therefore the proof of the lemma. ∎