Positive Answer to Berestycki’s Open Problem on the Unit Ball

Let Ik=[λk-M,λk+M] for every k∈N. The component Ck+ of Sk+∪(Ik×{0}), containing Ik×{0}, is unbounded and lies in Φk+∪(Ik×{0}) and the component Ck- of Sk-∪(Ik×{0}), containing Ik×{0}, is unbounded and lies in Φk-∪(Ik×{0}).

There exist two sequences λ1+<λ2+<⋯<λk+<⋯ and λ1-<λ2-<⋯<λk-<⋯ of simple half-eigenvalues for problem (1.3). The corresponding half-linear solutions are in {λk+}×Sk+ and {λk-}×Sk-. Furthermore, aside from these solutions and the trivial one, there is no other solution of problem (1.3).

If β⁢(r)≢-α⁢(r) in (0,1)∖Zi, then λi+≠λi- for any i∈N.

For any i, j∈N, if i≠j, then λj-≠λi+.

If (λ,u) is a solution of problem (1.2) under the assumptions (C1) and (C2) and u has a double zero, then u≡0.

Proof.

Let u be a solution of (1.2) and r*∈[0,1] be a double zero. We note that

If r*=0, then we have that

where ψ⁢(r)=|λ⁢u⁢(r)+F⁢(r,u⁢(r),λ)|. In view of (C1) and (C2), for any ε>0, there exists a constant δ>0 such that

uniformly with respect to r∈[0,1] and fixed λ when |s|∈[0,δ]. Hence,

where

By the Gronwall–Bellman inequality [4], we get u′⁢(r)≡0 on [0,1]. It follows that u⁢(r)≡0 on [0,1].

Next, we assume that r*>0. First, we consider r∈[0,r*]. Note that

Then

From the Gronwall–Bellman inequality, it follows that rN-1⁢|u⁢(r)|≡0 on [0,r*]. Therefore, u≡0 on (0,r*]. Furthermore, the continuity of u shows that u≡0 on [0,r*]. Now, for the case of r∈[r*,1], we have that

Again by the Gronwall–Bellman inequality, we get u≡0 on [r*,1]. ∎

Let ϵn, 0≤ϵn≤1, be a sequence converging to 0. If there exists a sequence (λn,un)∈(R×Skν) such that (λn,un) is a nontrivial solution of problem (2.1) corresponding to ϵ=ϵn, and (λn,un) converges to (λ,0) in R×E, then λ∈Ik.

Proof.

Let wn=un/∥un∥, then wn should be a solution of the following problem:

Let

for all r∈[0,1] and λ on bounded sets. Then g~ is nondecreasing with respect to u and

Further, from (2.3), it follows that

uniformly in r∈[0,1] and λ on bounded sets. Clearly, (C1) implies that

for all r∈[0,1]. It is obvious that (2.2), (2.4) and (2.5) imply that vn:=rN-1⁢wn′ is bounded in C1. Therefore, by the Arzelà–Ascoli theorem, we may assume that vn→v in C0. It follows that wn′→v~ in C0. Obviously,

Hence, wn is strong convergent in C1. Without loss of generality, we may assume that wn→w in C1, ∥w∥=1. Clearly, we have that w∈Skν¯.

We claim that w∈Skν. On the contrary, we suppose that w∈∂⁡Skν. Then w has at least one double zero r*∈[0,1]. It follows that wn⁢(r*)→0 and wn′⁢(r*)→0 as n→+∞. We note that

If r*=0, we have that

It follows that

where

In view of (C2) and the definition of wn, we can show that

uniformly in r∈[0,1]. Similarly, (C1) implies that

for all r∈[0,1]. Hence, there exists a positive constant K such that

for all r∈[0,1] and n∈ℕ large enough. Clearly, one has that

It follows that

Thus, we have that

So we have that

The Gronwall–Bellman inequality shows that limn→+∞⁡wn′⁢(r)≡0. From (2.6), it follows that limn→+∞⁡wn⁢(r)≡0. Therefore, we have that w⁢(r)≡0 on [0,1], which contradicts ∥w∥=1.

Next, we assume that r*>0. First, we consider r∈(0,r*]. It is easy to see that

Therefore,

Then we have that

By the Gronwall–Bellman inequality, we get that

This means that wn′⁢(r)→0 on [0,r*] as n→+∞. Similarly, for r∈[r*,1], we obtain that

It follows that wn′⁢(r)→0 on [r*,1] as n→+∞. It is obvious that

where ξ∈[r*,r] (or [r,r*]). Therefore, wn→0 in C1 as n→+∞, which is a contradiction. ∎

Proof of Theorem 1.

We only prove the case of 𝒞k+ because the case of 𝒞k- can be proved similarly. Let 𝒞k+ be the component of 𝒮k+∪(Ik×{0}) containing Ik×{0}. For any (λ,u)∈𝒞k+, there are two possibilities, either u∈Sk+ or u∈∂⁡Sk+. Clearly, in the former case, (λ,u)∈Φk+. While, the latter case implies that u has at least one double zero in [0,1]. From Lemma 5, it follows that u≡0. Hence, there exists a sequence (λn,un)∈Φk+ such that (λn,un) is a solution of problem (1.2), and (λn,un) converges to (λ,0) in ℝ×E. By Lemma 6, we have λ∈Ik, i.e., (λ,u)∈Ik×{0}. Hence, 𝒞k+⊆(Φk+∪(Ik×{0})).

To complete the proof, it remains to show that 𝒞k+ is unbounded in ℝ×E. Suppose, on the contrary, that 𝒞k+ is bounded. First, we claim that 𝒞k+ is compact in ℝ×E. Consider, for a given h∈C⁢[0,1], the following auxiliary problem:

It is well known that problem (2.7) can be equivalently written as u=G⁢(h)⁢(r), where G:C⁢[0,1]→E is continuous (see [8]). Define the Nemitskii operator H:ℝ×C⁢[0,1]→C⁢[0,1] by

Then it is clear that H is a continuous operator and problem (1.2) can be equivalently written as

Then Ψ:ℝ×E→E is completely continuous. Hence, 𝒞k+ is compact in ℝ×E because it is bounded. Applying a similar method as in the proof of [3, Theorem 1], with obvious changes, we may find an open bounded neighborhood 𝒪 of 𝒞k+ such that ∂⁡𝒪∩𝒮k+=∅.

Now, we consider the approximate problem (2.1) again. For ϵ>0, it is easy to show that the nonlinear term f⁢(r,u⁢|u|ϵ,λ)+g⁢(r,u,λ):=Fϵ⁢(r,u,λ) satisfies the condition

uniformly in r∈[0,1] and on bounded λ intervals. Define the Nemitskii operator Hε:ℝ×C⁢[0,1]→C⁢[0,1] by

Then Hε is continuous and problem (2.1) can be equivalently written as

where L⁢(u)=G⁢(u), H~⁢(λ,u)=G∘Hε⁢(λ,u). It is not difficult to verify that L:E→E is linear completely continuous and H~:ℝ×E→E is completely continuous with Φ=o⁢(∥u∥) near u=0 uniformly on bounded λ sets. Let

By [5, Theorem 2], there exist two continua 𝒞k,ϵ+ and 𝒞k,ϵ- of 𝒮ϵ, bifurcating from (λk,0), such that either 𝒞k,ϵ+ and 𝒞k,ϵ- are both unbounded or 𝒞k,ϵ+∩𝒞k,ϵ-≠{(λk,0)}. Theorem 4.1 of [8] shows that 𝒞k,ϵ+∪𝒞k,ϵ- is unbounded, lies in Φk and (𝒞k,ϵ+∪𝒞k,ϵ-)∩(ℝ×{0})={(λk,0)}. So 𝒞k,ϵ+ and 𝒞k,ϵ- are both unbounded and 𝒞k,ϵν⊆(Φkν∪{(λk,0)}).

So there exists (λϵ,uϵ)∈𝒞k,ϵ+∩∂⁡𝒪 for all ϵ>0. By compactness, one can find a sequence ϵn→0 such that (λϵn,uϵn) converges to a solution (λ,u) of problem (1.2), and so u∈Sk+¯. If u∈∂⁡Sk+, then by Lemma 5 we have u≡0. By Lemma 6, λ∈Ik, which contradicts the definition of 𝒪. On the other hand, if u∈Sk+, then (λ,u)∈(𝒮k+∩∂⁡𝒪) which contradicts ∂⁡𝒪∩𝒮k+=∅. Therefore, 𝒞k+ is unbounded in ℝ×E. ∎

Proof of Theorem 2.

By Theorem 1, we know that there exists at least one solution of problem (1.3), (λkν,ukν)∈(ℝ×Skν), for every k=1,2,…, and ν=±. The positive homogeneity of problem (1.3) implies that {(λkν,c⁢ukν),c>0} are half-linear solutions in {λkν}×Skν. Lemma 5 implies that any nontrivial solution u of problem (1.3) lies in some Skν.

We claim that for any solution (λ,u) of problem (1.3) with u∈Skν, we have that λ=λkν and u=c⁢ukν for some positive constant c. We may assume, without loss of generality, that the first zero of u⁢ukν to occur in (0,1] is a zero of u. That is, there exists ζ∈(0,1] such that u⁢(ζ)=0, u and ukν do not vanish and have the same sign in (0,ζ). By [3, Lemma 3], one has that λkν≤λ. On the other hand, by [3, Lemma 2], there exists an interval (ξ,η)⊆(0,1) such that u and ukν do not vanish and have the same sign in (ξ,η), and ukν⁢(ξ)=ukν⁢(η)=0. Again, by [3, Lemma 3], λ≤λkν. Hence, λ=λkν.