Proof of Theorem 1.1. We extend each to be a nonnegative continuous function, which is still denoted by , defined on ℝ in the following way: if , then .
Let us define
(3.1)
where . Then it follows from (A1) that is continuous, and is always a solution of (1.1). Moreover, (A2) implies that F is differentiable at , and
(3.2)
where . By (A2), all entries of J are positive. Therefore Lemma 2.3 yields the result that J has a positive principal eigenvalue , the corresponding eigenvector satisfying (). Moreover, it is not difficult to verify that
(3.3)
where . This implies that is a positive eigenvector of the operator . Similarly, has the same principal eigenvalue and the corresponding eigenvector is , where () is a positive constant. Obviously,
Hence when , is not invertible and is a potential bifurcation point. More precisely, the null space
is one dimensional. In addition, it is easy to see that and exist for .
We divide the rest of the proof into two steps.
Step 1. We show that is actually a bifurcation point.
Indeed, the proof of this is similar to the proof of Theorem A(ii), we state it here for the readers’ convenience.
Suppose . Then there exists such that
(3.4)
Let us consider the adjoint eigenvalue equation
(3.5)
where , . Multiplying the system (3.4) by , multiplying the system (3.5) by , integrating on Ω and subtracting, then we obtain
(3.6)
Thus if and only if (3.6) holds, which implies that is one dimensional.
Next, we verify that
(3.7)
Otherwise, we have
(3.8)
Since
(3.9)
multiplying the system (3.9) by and using (3.8), we can get a contradiction that
By using [[13], Theorem 1.7], we conclude that is a bifurcation point. Furthermore, by the Rabinowitz global bifurcation theorem [14], there exists a continuum of positive solutions of (1.1), which joins to infinity in . Clearly,
(3.10)
since (1.1) has only the trivial solution when .
Step 2: We claim that cannot blow up at some finite .
Otherwise, a sequence can be taken such that
(3.11)
where . Let be the Green operator of −Δ subject to Dirichlet boundary conditions, i.e., if and only if
By the elliptic regularity, satisfies
(3.12)
Here also denotes the Nemytski operator generated by itself. Clearly, (3.12) is equivalent to
(3.13)
where . It is well known that is continuous and compact, and so is continuous and compact on .
Let (). Then in Ω and . Dividing both sides of (3.13) with , we have
(3.14)
For each , from (A2) and (A3) it follows that is bounded in . Moreover, we have
(3.15)
Therefore,
is bounded in X. This together with the compactness of implies that has a subsequence, denoted by itself, satisfying, in X,
Obviously, () in Ω and . In addition, we have . Or else, let , then by (3.14) we get () in Ω, which contradicts .
We define
(3.16)
Then for each ,
by Lebesgue control convergence theorem, we get
which together with (3.15) yields
(3.17)
On the other hand, we know from (A2) and (3.15) that
(3.18)
Hence we conclude from (3.17) and (3.18) that
(3.19)
Now, let in (3.14), using (3.19) and the fact that we can obtain
which contradicts .
Finally, by (3.10), the connectness of and above arguments, we can find some such that (1.1) has no positive solution for , and (1.1) has at least one positive solution for . □
To prove Theorem 1.2, we need the following lemmas as required.
By Remark 1.1 and Lemma 2.3, the matrices and have the principal eigenvalues and , respectively, and the corresponding positive eigenvectors are and . Moreover, it is easy to obtain
(3.20)
and
(3.21)
where is given as in (2.6). By Lemma 2.2, the matrices and have principal eigenvalues and , respectively, the associated positive eigenvectors are and . We can easily verify that
(3.22)
and
(3.23)
Let be the closure of the set of positive solutions to (1.1). We extend each to be a function defined on ℝ by
(3.24)
then on ℝ. Let be a solution of
(3.25)
Then by (3.24), for each ,
where K is given as in the proof of Theorem 1.1. Hence is a nonnegative solution of (3.25). Moreover, from (3.24) it follows that is a solution of (1.1). On the other hand, (3.25) has no half-trivial solutions. Otherwise, U must have trivial and nontrivial components, and so there is a such that in Ω, and by the maximum principle of elliptic boundary value problems, we have
which is a contradiction. Therefore, the closure of the set of nontrivial solutions of (3.25) is exactly Σ.
In the following, we shall apply the Leray-Schauder degree theory, mainly to the mapping ,
where is given as in (3.13), and
is the associated Nemytski operator. For , let , let denote the degree of on with respect to 0.
Lemma 3.1 Let be a compact interval with . Then there exists a such that
Proof Suppose on the contrary that there exist sequences and so that
(3.27)
(3.28)
Apparently, (3.27) is equivalent to
(3.29)
and () in Ω, and therefore , . Furthermore, it follows from (A2)′ that, for sufficiently small, . This together with (3.28) implies that, for k large enough,
(3.30)
Multiplying (3.29) by , multiplying (3.22) by , integrating on Ω and adding, using (3.30) and the fact , , we know that, for k large enough,
(3.31)
and so for k sufficiently large. Similarly, by (3.23) and (3.30), we can deduce that for k large enough. Consequently, for k sufficiently large we get , which contradicts . □
Corollary 3.2 For and , .
Proof Lemma 3.1, applied to the interval , guarantees the existence of such that, for ,
Hence for any ,
□
Lemma 3.3 Suppose that . Then there exists such that
where .
Proof Suppose on the contrary that there exist and a sequence with and such that
which is
(3.32)
Clearly, () in Ω. Multiplying (3.22) by , multiplying (3.32) by , integrating over Ω and adding, then by (3.30) we know that, for k large enough,
Hence for k sufficiently large, which contradicts . □
Corollary 3.4 For and , .
Proof Let , where is the constant given as in Lemma 3.3. Since is bounded in , there exists a constant such that , . By Lemma 3.3, we get
Hence,
□
Proof of Theorem 1.2. For such that , let , and . For any , it is easy to see that the assumptions of Lemma 2.4 are all satisfied. Therefore there exists a continuum of solutions of (3.25) containing , and either
-
(i)
is unbounded in ; or
-
(ii)
.
By Lemma 3.1, the case (ii) cannot occur, and hence is unbounded bifurcated from . Note that (3.25) has only trivial solutions when , and therefore . Moreover, from Lemma 3.1 it follows that for a closed interval , if , then in X is impossible. Thus must be bifurcated from . Finally, applying similar methods to the proof of Step 2 of Theorem 1.1, we can show that
Consequently, (1.1) has at least one positive solution for . □