Multiple global bifurcation branches for nonlinear Picard

In this paper we prove the global bifurcation theorem for the nonlinear Picard problem. The right-hand side function ' is a Caratheodory map, not differentiable at zero, but behaving in the neighbourhood of zero as specified in details below. We prove that in some interval [a,b] � R the Leray-Schauder degree changes, hence there exists the global bifurcation branch. Later, by means of some approximation techniques, we prove that there exist at least two such branches.

Let Λ(q k ) denote the set of all λ ∈ [0, +∞), such that there exists a solution (λ, u) of ( 2), such that u = 0.As we can see the set Λ(q k ) is not empty.Let us also observe that 0 ∈ Λ(q k ).
The existence of bifurcation points and noncompact components of the set of solutions for boundary value problems (1) have been studied by many authors.The main ideas come from Krasnoselskii (see [10]) and Rabinowitz (see [12]).They studied the general nonlinear spectral problems in Banach spaces.Additionally Rabinowitz has studied the Sturm-Liouville problems (1) with ϕ linearizable at the origin.The problems with ϕ not differentiable at (0, 0) have also been studied (see e.g.[1], [2], [7], [13], [14]).In the mentioned papers the authors were mainly concentrated on the asymptotics such that ϕ(t, u, u , λ) ≥ 0 for u ≥ 0 and |u| + |u | small, which is not the case considered here.
The problems of the form for l representing Sturm-Liouville boundary conditions, where a is not necessarily of constant sign, were studied e.g. in [7] and [9].In [9] authors proved the important result for the linear case, which we are going to refer later.By means of the topological degree methods we may prove the following theorem: Caratheodory map, such that for each compact K ⊂ (0, +∞) Then there exists the noncompact component We can tell more about the structure of the solution set of the problem (1) when we study the linear eigenvalue problems and where a + ∈ L 1 (0, π) is the function given by a + (t) = sgn(sin(kt)) and a − = −a + .Both of the above problems are left-definite and right-indefinite (see [9]).The Dirichlet boundary conditions are self-adjoint and separated, so we may apply theorem 3.1 of [9].That is why there exists exactly one positive eigenvalue λ + > 0 of the problem (4) having the corresponding eigenvector with the constant sign.This eigenvalue is simple.Similarly there exists exactly one eigenvalue λ − > 0 of the problem (5) having the corresponding eigenvector with the constant sign.This eigenvalue is simple as well.
Let us observe that for both problems (4) and ( 5) there exists also the negative eigenvalue with the above properties, but we are interested only in the eigenvalues belonging to the interval (0, +∞).
f or (λ, u) ∈ C k the f unction u has exactly k − 1 zeroes (10) in (0, π), all zeroes of u are simple, and f unction u is positive in a neighborhood (0, δ) of 0.
The example given below shows the application of Theorem 2 to the simple situation where ϕ = λq3 in the neighbourhood of 0.
where p : R → [0, 1] is given by We will investigate the set of nontrivial solutions of (1) with ϕ given as above.
As additional observation we can state that for (λ, u) ∈ C + 1 parameter λ must be bounded.This is because each positive solution of (1) satisfies For λ > 0 this means that there exist zero of u in the interval (0, π 3 ) which is not possible for positive u.So, in this case, we can describe two components as half-lines, while the third may be explicitely described in the neighbourhood of zero.

Auxiliary lemmas
In this section we are going to show some facts that will be used in the proofs of Theorems 1 and 2, but first we are going to specify the basic assumptions and notations. Let EJQTDE, 2009 No. 33, p. 4 Then we can see that u = T h iff u is the solution of the boundary value problem for h ∈ L 1 (0, π).
For the problem (1) we may define the map f : (0, +∞) where Φ : (0, +∞) and Once we have this association we may define the closure R f (in (0, +∞)× C 1 [0, π]) of the set of all nontrivial zeroes of the map f and the set of bifurcation points B f of the map f , and observe that R f = R (1) and Let α, β ∈ (0, +∞) and α < β be such that (α, 0), (β, 0) ∈ B f .Then let us define the bifurcation index of the map f on the interval (α, β) by for r > 0 small enough.In the above formula deg(•) stands for the Leray--Schauder degree.We may extend this definition to the case of (α, 0), (β, 0) satisfying for some δ > 0. This may be done by The classical sufficient condition for the existence of bifurcation points and the theorem describing the structure of the set R f is given in [12].There exist numerous extensions and modifications of this theorem (for more detailed comments and the list of references see e.g.[3], [8], [11]).We will refer here to the theorem given in [4] for multivalued maps.The theorem given below is the slight modification of this theorem to the case of single valued maps:

EJQTDE, 2009 No. 33, p. 5
Now let us make the general observation that all zeroes of solution u of (1) are simple.
Proof.Let us observe that if then u = 0.This is because by (3), in some neighborhood of t0 the following estimation holds and for t close to t0 there must be u(t) = 0. Hence we may conclude that each zero of u must be isolated.
Let us further denote u k (t) = sin kt and let us define the homotopy We will show that for τ ∈ (0, 1] there are no zeros of h(τ, •).Assume, contrary to our claim, that h(τ, u) = 0 for some Hence . This contradicts (16) and proves that h(τ, u) = 0 for all τ ∈ (0, 1] and Now let us make one more observation related to the problem (2): let us observe that the positive solution of ( 4) is also the solution of (2) and, similarly, the negative solution of ( 5) is the solution of (2).This is formulated in the next proposition.
Proof.First let us assume that in the one of the intervals [ lπ k , (l+1)π k ) where l ∈ {0, 1, ..., k − 1} there are two adjacent zeroes t1, t2 ∈ [ lπ k , (l+1)π k ) of the function u k .Then we have for u with constant sign on (t1, t2).Hence there must be what is the contradiction with the lemma 2.
Similarly we can show that in each of the intervals ( lπ k , (l+1)π k ] where l ∈ {0, 1, ..., k − 1} there is at most one zero of the function u k .
Assume now that in the interval (0, π) there are exactly k − 1 zeroes of u.Because there are k − 1 intervals we may state that in each interval ( lπ k , (l+1)π k ] and [ lπ k , (l+1)π k ) there is exactly one zero of function u.From the above facts and u(0) = 0 we may conclude that there are no zero in the open interval (0, π k ), so there must be u( π k ) = 0. Hence λ = k 2 , what completes the proof.
The lemma below is in fact a classical result (cf.example 3.2(a) in section XI of Hartman's book [6]) and will be given without a proof.
Lemma 5 Let the function p ∈ L 1 (0, π) satisfy 0 < K ≤ p(t) ≤ L, for positive constants K, L ∈ (0, +∞).Let u be the solution of the linear differential equation with two adjacent zeroes t1, t2, then the distance between the zeroes t1 and t2 may be estimated as follows: Within the proof of the theorem 2 we will refer to the sequence of functions q n k : [0, π] × R → R given by denote the Nemytskii operator associated with q n k (n = 1, 2, ...).Let us consider the family of boundary value problems for n ∈ N, and associated completely continuous vector fields fn : (0, +∞)× ]. Lemma 6 If (λn, un) ∈ R fn and the sequence {(λn, un)} is bounded and {un} is bounded away from zero, then it contains subsequence convergent Proof.This is because and the sequence of maps {Q n k } is uniformly bounded.Hence, we can select convergent subsequence of {un}.We can also select convergent subsequence of {λn}.

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Let us now observe that for un → u0 the relation holds This completes the proof.
Proof.Let us fix ε ∈ (0, αm 2 ) and let r0 > 0, be such that Let us now denote by tn ∈ (0, π) the first zero of the function un.Then un(t) > 0 for t ∈ (0, tn) and < αm+ε < 3αm 2 .Then by lemma 5 we may estimate the distance between two adjacent zeroes of un by Hence un has no zero in the interval (0, π k ).Similarly we may observe that for the interval [ lπ k , ( ] for l = 1, ..., k − 1 the relations hold (*) for l odd and u( lπ k ) ≥ 0 there is at most one zero in the interval ]; (**) for l even and u( lπ k ) ≤ 0 there is at most one zero in the interval ] there exist at most 2 zeroes of the function u.EJQTDE, 2009 No. 33, p. 9 Let us now observe that if u changes sign in each interval ] function u has two zeroes, then there must exist the interval ] ( l ∈ {1, ..., l − 1}) with no zeroes of u.Similarly, between two intervals containing two zeroes of u there must exist the interval with no zero of u.
This is why we may conclude that in the interval [0, lπ k ] there are at most l zeroes of u.So function u, satisfying u(0) = u(π) = 0, has at most k − 2 zeroes in the open interval (0, π) what contradicts our assumption.
Proof.Let us fix ε ∈ (0, βm 2 ) and let r0 > 0, be such that Let us now denote by tn ∈ (0, π) the first zero of the function un and assume that un(t) > 0 for t ∈ (0, tn).
Similarly as in the proof of the lemma 7 let us consider, in the interval (0, tn), the equation Then by lemma 5 we may estimate the distance between two adjacent zeroes of un by This means that in the interval (0, π 2k ) there exists the zero of un.Assume un has exactly k − 1 zeroes in the open interval (0, π).As we know from lemma 1 all zeroes of un are simple (so un changes sign exactly k − 1 times), so in some neighborhood (π − δ, π) of the point π the relation holds q n k (t, un(t)) = un(t).So, we may repeat the above arguments and state that in the interval (π − π 2k , π) there exists the zero of un. ) for l = 1, ..., l0 − 1 has exactly one zero and there are exactly two zeroes in the interval [0, π k ), then u( lπ k ) < 0 for l odd and u( lπ k ) > 0 for l even.This implies that also in the interval [ l 0 π k , (l 0 +1)π k ) there is at least one zero of u, a contradiction.So at least one interval [ lπ k , (l+1)π k ) for l = 1, ..., l0 must contain at least two zeroes of u.So, the interval (0, (l 0 +1)π k ) contains at least l0 zeroes.

Moreover, if in the interval
) there are no zeroes of u, then u( l 0 π k ) > 0 for l0 odd and u( l 0 π k ) < 0 for l0 even, and u does not change sign in the interval [ l 0 π k , (l 0 +1)π k ), so there exists at least one zero in the interval [ (l 0 +1)π k , (l 0 +2)π k ).Similarly as above, between two intervals [ lπ k , (l+1)π k ) with no zero of u there exists at least one interval with two zeroes.So, each interval (0, lπ k + π 2k ) contains at least l zeroes of u.Because, as we have shown above, there exists zero of u in the interval (π − π 2k , π) the function u has at least k zeroes in the interval (0, π).A contradiction.

Proofs of the theorems
Proof.[Proof of theorem 1] Let us take the map f0 : (0, +∞) We are going to refer to the theorem A given in the section 2. First let us observe that by lemmas 2 and 3 Hence all assumptions of theorem A are satisfied and there exists the noncompact component Step 1.