Complete description of the set of solutions to a

We give a complete description of the set of solutions to the boundary value problem ’ u 0 0 = f (u) in (0; 1) ; u (0) = u (1) = 0

Note that the differential operator u → (ϕ (u )) is linear if and only the function x → ϕ (x) is linear, hence the ODE in (1) is said strongly nonlinear.
In this work we will replace the growing conditions on ϕ and f at 0 and +∞ by global conditions on the convexity of ϕ and f.These new conditions which will play significant role in the proof of existence of solutions as well as in the proof of uniqueness of these solutions in some areas of C 1 ([0, 1]) , can appear very restrictive.However we think that this condition is usual, indeed this kind of assumption is often met in the literature when an exactitude result is aimed (see [3], [6] and [18]).
Our strategy is as follows: In a first stage, we locate the possible solutions of problem (1) in some subsets A ν k , (where for k ∈ N * and ν = +, − A ν k is defined in section 2) of C 1 ([0, 1]) and we give some properties of these solutions.An immediate consequence of these results is: u ∈ A + k is solution to problem (1) if and only if u is a positive solution to the problem Then we associate to problem (2) the auxiliary Sturm-Liouville problem ( such that u is positive solution to problem (2) if and only if v (x) = x 0 ϕ (u (t)) dt is a positive solution to the auxiliary Sturm-Liouville problem (3).Thus we are brought to investigate a nonlinear Sturm-Liouville problem for which after addition of a linear part containing a real parameter existence of a positive solution will be proved by the use of Rabinowitz global bifurcation theory (see [19], [20] and [21]).
At the end, we will use assumptions ( 5) and (7) to prove uniqueness of the solution in each subset A ν k .The paper is organized as follows: Section 2 is devoted to the statement of the main results and some necessary notations.In section 3 we expose some preliminary results we need in the proof of the principal results.In the last section we give the proofs of main results.

Notations and main results
In the following we denote by With this definition in mind, each function in S + k has exactly k humps such that the first one is positive, the second is negative, and so on with alterations.
Let A + k (k ≥ 1) the subset of S + k composed by the functions u satisfying: • Every hump of u is symmetrical about the center of the interval of its definition.
• Every positive (resp.negative) hump of u can be obtained by translating the first positive (resp.negative) hump.
• The derivative of each hump of u vanishes one and only one time. Let We recall that the boundary value problem: 2 .We will use in this work the so called Jensen inequality given by: Let S be the set of solutions to problem (1), then our main results are : Theorem 1 ( Superlinear case ) : Suppose the functions ϕ and f satisfy the following conditions: EJQTDE, 2003 No. 9, p. 3 the function s → f (s) s is increasing on (0, +∞) Then and for each integer k ≥ 1 there exists Theorem 2 ( Sublinear case ) : Suppose the functions ϕ and f satisfy the following conditions: f is increasing and concave in R + . Then and for each integer k ≥ 1 there exists

Remark 1
The above theorems give a complete description of the solution set of the problem (1), indeed the theorems state that there is no solution except the trivial solution and those belonging to ∪ k≥1 A k , and in each A ± k there is exactly one solution.
Remark 2 Hypothesis (7) is similar to (3-3) assumed in [4].To obtain the exact number of solutions to the boundary value problem according λ in a resonance interval, the author assumed the function s → f (s) s and Note that, hypothesis (7) implies that f is increasing, and if f is convex then hypothesis (7) is satisfied.
In the sublinear case, hypothesis (10) implies that the function s → f (s) s is decreasing on (0, +∞) .

EJQTDE, 2003
No. 9, p. 4 3 Some preliminary results: In this section we give some lemmas which will be crucial for the proof of our main results.Consider the boundary value problem where ϕ is an odd increasing homeomorphism of R and g is a function in We define a solution of problem ( 12) to be a function u If u is a solution to problem (12), then there exists a real constant C ≥ 0 such that where Note that Ψ the Legendre transform of the convex function Φ where Φ (s) = s 0 ϕ (t) dt, is even , Ψ (0) = 0 and Ψ (s) > 0 for all s = 0.
Then the first result in this section is: Lemma 3 Suppose that hypothesis (13) holds true.If u is a nontrivial solution to problem (12), then there exists an integer k ≥ 1 such that u ∈ A k .
Proof.Let u be a nontrivial solution to problem (12).We begin the proof by showing u (a) = 0. Let us suppose the contrary.Then, if we put x = a in equation ( 14), we get C = 0. Thus, for any Now, let us show that u has a finite number of zeros.Suppose the contrary and let (z n ) the infinite sequence of zeros of u and z * an accumulate point of (z n ) .Then we have Again, putting x = z * in equation ( 14) we get the same contradiction as above.
Let z 1 and z 2 two consecutive zeros of u, and suppose that u > 0 in (z 1 , z 2 ) and y * is a critical point of u in (z 1 , z 2 ) .It follows from equation ( 12) that (ϕ (u )) = −g (u) in (z 1 , z 2 ).Since ϕ is an increasing odd homeomorphism of R, u > 0 in (z 1 , y * ), u < 0 in (y * , z 2 ) and u (y * ) = 0. Thus y * is the unique critical point of u at which u reach its maximum value.EJQTDE, 2003 No. 9, p. 5 Let It follows from equation (14) that and where Ψ −1 + is the inverse of Ψ on R + .Then and Putting x = y * in equations ( 17) and ( 18) we get which yields For the symmetry of the (z 1 , z 2 ) −hump of u about , it suffices to show that for .This becomes very easy if we observe that x = (z 1 + z 2 ) − (z 1 + z 2 − x) and make use of equations ( 17) and ( 18) , then we get: in each To do this it suffices to prove that u (z Putting respectively x = z 1 + z 2 2 and x = z 3 + z 4 2 in equation ( 14) we deduce Since G is strictly increasing on (0, +∞) , u Making use of equations ( 17) and ( 18), we get : If we set v (x) = u ( z 3 + ( x − z 1 )) for all x ∈ [z 1 , z 2 ], then we have Observe that u and v are solutions of the problem , we have: EJQTDE, 2003 No. 9, p. 7 Using the symmetry of the function u we deduce that v (x) = u (x) for all x ∈ [z 1 , z 2 ].This completes the proof of the lemma.
Lemma 4 Suppose that hypothesis (13) holds true and g is odd.If u ∈ A + k ( resp.A − k ) is solution to problem (12) with k ≥ 2 then the first negative ( resp.positive ) hump of u is a translation of the first negative ( resp.positive ) of (−u) .
Proof.Let u ∈ A + k be a solution to problem (12) and (z i ) i=k i=0 the finite sequence of zeros of u such that 0 = z Since the positive ( resp.negative ) humps of u are translations of the first positive ( resp.negative ) hump one, it suffices to prove that u Let us prove that the two humps have the the same length.. Putting x = z 1 2 and Since G is even and increasing in R , as in the proof of Lemma 3 ) Setting v (x) = −u (z 1 + x) for all x ∈ [0, z 1 ] and arguing as in the proof of Lemma 3, we get u (x) = v (x) = −u (z 1 + x) for all x ∈ [0, z 1 ] .So the lemma is proved Proof.Let u 1 and u 2 be two solutions of the lemma.
We have Assume first that the first situation holds.We deduce from equation ( 14) : Since G is strictly increasing, we get , then (17) written for u 1 and u 2 gives: for any x ∈ a, a + b 2 .
Hence, u 1 (x) = u 2 (x) for all x ∈ a, a + b 2 .Since u 1 and u 2 are in A + 1 ; namely u 1 and u 2 are symmetrical about a+b 2 , u 1 (x) = u 2 (x) for all x ∈ [a, b], which contradicts the statement of the lemma.Now, suppose that u 1 (a) <u 2 (a).Since u 1 and u 2 are symmetrical about a + b 2 , we will prove that u 1 (x) < u 2 (x) for all x ∈ a, a + b 2 .
Then x 0 >a, indeed if x 0 = a and (x n ) is a sequence such that lim x n = x 0 n→+∞ , we get : which is impossible.Thus, let (y n ) be a sequence in (a, x 0 ) such that lim n→+∞ y n = x 0 .We get: Using again (14), we obtain: EJQTDE, 2003 No. 9, p. 9 so

Proof of the main results
Since the function f is odd and satisfies hypothesis (13) , it leads from lemma 3 any non trivial solution to problem (1) belongs to ∪ k≥1 A k .

Existence of solutions :
It arises from lemmas 3 and 4: to get a solution belonging to A + k (resp.A − k ) to problem (1) it suffices to prove that the problem admits a positive (resp.negative ) solution. 2et f + = max (f, 0) and consider the boundary value problem Observe that if v is a positive solution to problem (20) if and only if u (x) = x 0 ψ (u (t)) dt is a positive solution to the problem (19) 3 .
Hence, we are brought to look for positive solutions to the problem where a ∈ (0, 1) .Consider the boundary value problem where λ is a real parameter and u (x) = x 0 ψ (v (t)) dt.We mean by a solution of problem (22) and the boundary conditions v (0) = v (a) = 0.

Existence in the superlinear case:
Let ε > 0, we deduce from assumption (6) existence of δ > 0 such that Since ψ is an odd increasing function on R + , we have for v ∈ C 1 ([0, a]) and for all x ∈ [0, a] Therefore, Rabinowitz global bifurcation theory (see [19] and [20]) states: the pair Thus, to prove existence of a positive solution to problem (21) it suffices to show the following Before proving theorem 6, we need the following lemma: Proof.Let Φ be the first positive eigenfunction of Multiplying (22) by Φ and integrating on (0, a) we get: Then, two integrations by parts give

Proof of theorem 6
Suppose the contrary, and let (λ First Let us prove that v n is unbounded with the respect of the C 0 norm.Suppose the contrary; Since −v n = λ n v n + f (u n ) and v n is unbounded4 with the respect of the C 0 norm, u n is unbounded with the respect of the C 0 norm on [0, a] .
Let for any R > 0 We claim that there exist R 0 > 0 such that l .This is due to: Denote by θ n the real number belonging to [0, a] such that ϕ (u n (θ n )) = R and let Φ n and λ 1,n be respectively the first positive eigenfunction and the first eigenvalue of the problem Multiplying (22) by Φ n and integrating between θ n and a we get After two integrations by parts we obtain: We deduce from hypothesis (6) that lim Thus, we deduce from (23): Since ϕ is concave, Jensen inequality (4) leads to Thus, we deduce from (24): Now let us return to the equation satisfied by u n .We have Multiplying by u and integrating [x, a] , we get where ρ n = u n (a) and F + (x) = x 0 f + (t) dt.Then as in the proof of Lemma 3 we obtain Thus, on one hand, since is bounded in [0, R 0 ] and lim and θ n the real number belonging to [0, a] such that v n (θ n ) = R 0 .Thus, in one hand and on the other hand, which is impossible because from (27) we deduce that v n (θ n ) is bounded and (28) leads to v n (θ n ) is unbounded.This completes the proof of theorem 6.

Existence in the sublinear case:
Let ε > 0, we deduce from hypothesis (9) existence of χ > 0 such that Note that since ψ is concave and increasing, and f is increasing .
Let Φ be the first positive eigenfunction of and (λ, v) ∈ S + 1 .Arguing as in the proof of lemma 7 we get λ < λ 1 which means that S + 1 Ω don't meet (µ k ([0, a]) , +∞) with k ≥ 2. Now suppose that (λ n , v n ) is a sequence in S + 1 converging5 to (λ * , 0) with λ n > 0. Multiplying ( 22) by Φ and integrating on (0, a) we get This is impossible since which completes the proof of theorem 8.

Uniqueness in A ±
We will expose in this paragraph the proof of uniqueness in A ± k in the superlinear case.The other case will be treated similarly.
We deduce from Lemma 3 and Lemma 4 that: to show uniqueness of the solution to problem (1) First we deduce from Lemma 5 that u and v are ordered and from assumption (7) that f is increasing in R + .Then, if we suppose u < v in (0, 1) we get (ϕ (u ) − ϕ (v )) = In one hand, it follows from assumption (7) In the other hand, the concavity of ϕ involve that the function s → ϕ (s) s is decreasing on (0, +∞) , then Inequalities ( 31) and (32) contradict equation (32) , so uniqueness of the solution to problem (29) is proved.
2003 No. 9, p. 2 Let, for any integer k ≥ 1 and a < b S + k = u ∈ E : u admits exactly (k − 1) zeros in ]a, b[ all are simple, u (a) = u (b) = 0 and u (a) > 0 S − k = −S + k and S k = S + k ∪ S − k .Let u be a function belonging to C ([a, b]) which vanishes at x 1 and x 2 (x 1 < x 2 ).If u does not vanish at any point of the open interval I = ]x 1 , x 2 [ we call its restriction to this interval I-hump of u.When there is no confusion we say a hump of u.