Now we show how the main results of Section 1.3 can be applied when f (t, s) = a(t)g(s). To this aim, besides (1.4.2), we also suppose
a−(t) 6≡ 0. (1.4.5)
Consistently with assumption (a∗), we select, without loss of generality, the
endpoints of the intervals Ii+ in such a manner that
a(t) 6≡ 0 on each of the subintervals [σi, τi] and a(t) 6≡ 0 on all
In order to explain the rule that we have decided to follow so as to determine the endpoints of the intervals, let us consider the following weight function on the interval I = [0, 7π]
a(t) = (
sin t, if t ∈ [0, π] ∪ [3π, 4π] ∪ [6π, 7π];
0, if t ∈ [π, 3π] ∪ [4π, 6π]; (1.4.6)
(cf. Figure 1.5). Among the various possibilities that one could adopt to choose the endpoints of the intervals according to condition (a∗), the follow-
ing choice would fit with the above convention: σ1 = 0, τ1 = 3π, σ2 = 4π,
τ2 = 7π. t a(t) 0 −1 1 π 3π 4π 6π 7π
Figure 1.5: The figure shows the graph of the continuous function a(t) : [0, 7π] → R defined in (1.4.6).
To discuss another example, let us consider a function with a graph as that of Figure 1.3. It is clear that it satisfies (1.4.2) and (1.4.5), provided that we adopt a suitable choice of the points σi and τi. Typically, we shall
proceed in the following manner: if there is an interval where a(t) ≡ 0 between an interval where a(t) > 0 and an interval where a(t) < 0, we choose σi and τi in such a way that a(t) < 0 on ]τi, σi+1[ and we merge the
interval where a(t) ≡ 0 to an adjacent interval where a(t) ≥ 0.
We need also to introduce a further notation. For any weight function a(t) satisfying (a∗) (with the endpoints σi, τi chosen as described above), we
set
aµ(t) := a+(t) − µa−(t), t ∈ [0, T ],
where µ > 0 is a parameter. Notice that a(t) = aµ(t) for µ = 1 and,
moreover, for every µ > 0, it holds that aµ(t) satisfies (a∗) with the same σi
and τi chosen for a(t).
The introduction of the parameter µ is made only with the purpose to clarify the role of the negative humps of a(t) in order to produce multiplicity results. In other words, when we require that µ > 0 is sufficiently large, we
have a more precise manner to express the intuitive fact that the negative humps of a(t) are great enough.
Now we are in a position to present our main multiplicity result for the boundary value problem
(
u00+ aµ(t)g(u) = 0
u(0) = u(T ) = 0. (1.4.7)
Recall that we are assuming that a : I → R is a continuous weight function satisfying (1.4.2), (1.4.5) and (a∗) with the convention described above.
Theorem 1.4.3. Let g : R+→ R+ be a continuous function satisfying (g ∗)
and let a : I → R be a continuous function satisfying (a∗). Moreover, suppose
that
g0 < λ0 and a(t) 6≡ 0 on each Ii+ with g∞> max i=1,...,mλ
i 1.
Then there exists µ∗> 0 such that, for each µ > µ∗, problem (1.4.7) has at least 2m− 1 positive solutions.
Proof. From g0< λ0, we can choose δ > 0 such that g0 < λ0− δ. Let r0 > 0
be as in Lemma 1.1.2 and fix 0 < r ≤ r0 such that
g(s)
s < λ0− δ, ∀ 0 < s ≤ r. (1.4.8) Let R∗ > 0 as in Lemma 1.1.4 and fix R ≥ R∗.
Let I ⊆ {1, . . . , m}. Using the notation introduced in Section 1.3, con- sider the open and unbounded set ΓI. Moreover, consider an arbitrary L1-Carath´eodory function h : I × R+→ R such that
h(t, s) ≥ a(t)g(s), a.e. t ∈ [ i∈I Ii+, ∀ s ≥ 0; h(t, s) = a(t)g(s), a.e. t ∈ I \[ i∈I Ii+, ∀ s ≥ 0; (1.4.9)
and, as usual, define the completely continuous operator Ψh: C(I) → C(I),
(Ψhu)(t) := Z I G(t, ξ)˜h(ξ, u(ξ)) dξ, t ∈ I, where ˜ h(t, s) := ( h(t, s), if s ≥ 0; h(t, 0), if s ≤ 0.
We know that every fixed point of Ψh is a non-negative solution of
(
u00+ h(t, u) = 0
To prove the claim, we use Theorem 1.3.1. In particular we have to show that the triplet (I − Ψh, ΓI, 0) is admissible for each Carath´eodory function
h satisfying (1.4.9) and for each ∅ 6= I ⊆ {1, . . . , m}.
By the choice of R ≥ R∗ and by the convexity of the solution of (1.4.10) on each interval contained in I \Sm
i=1Ii+, we know that every fixed point u
of Ψh is contained in the open ball B(0, R), with R > 0 independent of the
particular choice of h(t, s) (see Lemma 1.1.3). Consequently it is sufficient to prove that Ψh has no fixed points in ∂(ΓI∩ B(0, R)), for µ sufficiently
large.
First, we note that if I = {1, . . . , m} there is nothing to prove, since all fixed points in ΓI = C(I) are contained in B(0, R). Then fix ∅ 6= I ( {1, . . . , m}. By contradiction, suppose that there is a fixed point u of Ψh in
∂(ΓI∩ B(0, R)). Due to what we have just remarked, this is equivalent to assuming the existence of a solution u of (1.4.10) with
max
t∈Ik+
u(t) = r, for some index k ∈ {1, . . . , m} \ I,
and such that maxt∈Iu(t) < R. Clearly u 6≡ 0 and, moreover, by the
concavity of u(t) in Ik+, we also have
u(t) ≥ r τk− σk
min{t − σk, τk− t}, ∀ t ∈ Ik+. (1.4.11)
In order to prove that our assumption is contradictory and hence that the topological degree is well-defined, we split our argument into three steps. Step 1. A priori bounds for |u0(t)| on Ik+. This part of the proof follows by adapting a similar estimate obtained in Theorem 1.2.2. Notice that h(t, u(t)) = a(t)g(u(t)) = a+(t)g(u(t)), for a.e. t ∈ Ik+. Hence
|u00(t)| ≤ γr(t) := a+(t) max 0≤s≤rg(s), a.e. t ∈ I + k, and, therefore |u0(y1) − u0(y2)| ≤ kγrkL1(I+ k), ∀ y1, y2∈ Ik+.
Let ˆt ∈ Ik+ = [σk, τk] be such that |u0(ˆt)| ≤ r/(τk−σk) (otherwise we have an
easy contradiction like in the proof of Theorem 1.2.2). Hence for all t ∈ Ik+
|u0(t)| ≤ |u0(ˆt)| + |u0(t) − u0(ˆt)| ≤ r τk− σk
+ kγrkL1(I+
k) =: Mk. (1.4.12)
Step 2. Lower bounds for u(t) on the boundary of Ik+. Let ϕkbe the positive
eigenfunction on Ik+ = [σk, τk] of
(
ϕ00+ λk1a+(t)ϕ = 0 ϕ|∂I+
with kϕkk∞ = 1, where λk1 is the first eigenvalue. Then ϕk(t) ≥ 0, for all
t ∈ Ik+, ϕk(t) > 0, for all t ∈ ]σk, τk[, and ϕ0k(σk) > 0 > ϕ0k(τk) (hence
kϕ0kk∞> 0).
By (1.4.8) and λ0 ≤ λk1, we know that
g(s) < (λk1 − δ)s, ∀ 0 < s ≤ r.
Then, by (1.4.11), we have
kϕ0kk∞(u(σk) + u(τk)) ≥
≥ u(σk)|ϕ0k(σk)| + u(τk)|ϕ0k(τk)| = u(σk)ϕ0k(σk) − u(τk)ϕ0k(τk)
=hu0(t)ϕk(t) − u(t)ϕ0k(t) it=τk t=σk = Z τk σk d dt h u0(t)ϕk(t) − u(t)ϕ0k(t) i dt = Z Ik+ h u00(t)ϕk(t) − u(t)ϕ00k(t) i dt = Z Ik+ h −h(t, u(t))ϕk(t) + u(t)λk1a+(t)ϕk(t) i dt = Z Ik+ h λk1u(t) − g(u(t))ia+(t)ϕk(t) dt > Z Ik+ δ r τk− σk min{t − σk, τk− t} a+(t)ϕk(t) dt = r δ τk− σk Z Ik+ min{t − σk, τk− t}a+(t)ϕk(t) dt .
Hence, from the above inequality, we conclude that there exists a con- stant ck> 0, depending on δ, Ik+ and a+(t), but independent of u(t) and r,
such that
u(σk) + u(τk) ≥ ckr > 0.
As a consequence of the above inequality, we have that at least one of the two inequalities 0 < ckr 2 ≤ u(τk) ≤ r, 0 < ckr 2 ≤ u(σk) ≤ r, (1.4.13) holds.
Step 3. Contradiction on an adjacent interval for µ large. Just to fix a case for the rest of the proof, suppose that the first inequality in (1.4.13) is true. In such a situation, we necessarily have τk < T (as u(T ) = 0). Now we
focus our attention on the right-adjacent interval [τk, σk+1], where a(t) ≤ 0.
Ii+, we have that a(t) is not identically zero on all right neighborhoods of τk.
Since g(s) > 0 for all s > 0, we can introduce the positive constant
νk:= ckrmin 4 ≤s≤R g(s) > 0 and define δk+:= min σk+1− τk, ckr 4Mk > 0,
where Mk> 0 is the bound for |u0| obtained in (1.4.12) of Step 1. Then, by
the convexity of u(t) on [τk, σk+1], we have that u(t) is bounded from below
by the tangent line at (τk, u(τk)), with slope u0(τk) ≥ −Mk. Therefore,
ckr
4 ≤ u(t) ≤ R, ∀ t ∈ [τk, τk+ δ
+ k].
We prove that for µ > 0 sufficiently large maxt∈[τk,σk+1]u(t) > R (which
is a contradiction to the upper bound for u(t)).
Consider the interval [τk, τk+ δk+] ⊆ [τk, σk+1]. For all t ∈ [τk, τk+ δk+]
we have u0(t) = u0(τk) + Z t τk µa−(ξ)g(u(ξ)) dξ ≥ −Mk+ µνk Z t τk a−(ξ) dξ, then u(t) = u(τk) + Z t τk u0(ξ) dξ ≥ ckr 2 − Mk(x − τk) + µνk Z t τk Z s τk a−(ξ) dξ ds. Hence, for t = τk+ δ+k, R ≥ u(τk+ δk+) ≥ ckr 2 − Mkδ + k + µνk Z τk+δ+k τk Z s τk a−(ξ) dξ ds.
This gives a contradiction if µ is sufficiently large, say
µ > µ+k := R + MkT νkA+k
,
where we have set
A+k := Z τk+δk+ τk Z s τk a−(ξ) dξ ds > 0, recalling that Rt τka −(ξ) dξ > 0 for each t ∈ ]τ k, σk+1].
A similar argument (with obvious modifications) applies if the second inequality in (1.4.13) is true (in such a case, we must have σk > 0, as
u(0) = 0). This time we focus our attention on the left-adjacent interval [τk−1, σk] where a(t) ≤ 0. Recall also that, by the convention we have
adopted in defining the intervals Ii+, we have that a(t) is not identically zero on all left neighborhoods of σk.
If we define δk−:= min σk− τk−1, ckr 4Mk > 0,
we obtain the same contradiction for
µ > µ−k := R + MkL νkA−k
,
where we have set
A−k := Z σk σk−δ−k Z σk s a−(ξ) dξ ds.
At the end, we define
µ∗:= max
k=1,...,mµ ± k
and we apply Theorem 1.3.1 with µ > µ∗. The proof is completed.
See Figure 1.6 for a numerical example concerning the multiplicity result stated in Theorem 1.4.3.
An immediate consequence of Theorem 1.4.3 is the following result which generalizes [94, Theorem 2.1].
Corollary 1.4.2. Let g : R+ → R+ be a continuous function such that
g(0) = 0 and g(s) > 0 for all s > 0. Suppose also that
lim s→0+ g(s) s = 0 and s→+∞lim g(s) s = +∞.
Let a±: I → R+ be continuous functions such that for some 0 = τ0 ≤ σ1 <
τ1 < σ2< τ2< . . . < σm< τm≤ σm+1 = T it holds that
a+(t) 6≡ 0, a−(t) ≡ 0, on [σi, τi], i = 1, . . . , m;
a−(t) 6≡ 0, a+(t) ≡ 0, on [τi, σi+1], i = 0, . . . , m.
Then there exists µ∗> 0 such that, for each µ > µ∗, problem (1.4.7) has at least 2m− 1 positive solutions.
Figure 1.6: The figure shows an example of multiple positive solutions for problem (1.4.7). For this numerical simulation we have chosen I = [0, 1], a(t) = sin(7πt), µ = 20 and g(s) = max{0, 500 s arctan |s|}. Notice that the weight function a(t) has 4 positive humps. We show the graphs of the 15 positive solutions of (1.4.7).
Corollary 1.4.3. Let g : R+→ R+ be a continuous function satisfying (g ∗)
and let a : I → R be a continuous function satisfying (a∗). Moreover, suppose
that
g0= 0 and g∞> 0.
Then, there exists ν∗ > 0 such that, for each ν > ν∗, there exists µ∗ = µ∗(ν) > 0 so that the boundary value problem
(
u00+ νa+(t) − µa−(t)g(u) = 0 u(0) = u(T ) = 0