EXISTENCE RESULTS FOR THE PROBLEM(φ(u′))′ = f(t, u, u′) WITH NONLINEAR BOUNDARY

CONDITIONS.∗

Alberto Cabada† and Rodrigo L. PousoDepartamento de Analise Matematica

Facultade de MatematicasUniversidade de Santiago de Compostela

Spain

Key words and phrases: Nonlinear boundary value problem. p – Laplacian.

1. INTRODUCTION

In this paper we study existence results for the following problem, which we willrefer by (P ).

[φ(u′(t))]′ = f(t, u(t), u′(t)) for a.e. t ∈ I = [a, b], (1.1)0 = g(u(a), u′(a), u′(b)), (1.2)

u(b) = h(u(a)). (1.3)

The three following conditions hold:

(H1) f is a Caratheodory function, that is: f(t, ·, ·) is a continuous function fora. e. t ∈ I; f(·, u, v) is measurable for all (u, v) ∈ R2; and for every M > 0there exists a real-valued function ψ ≡ ψM ∈ L1(I) such that

|f(t, u, v)| ≤ ψ(t)

for a. e. t ∈ I and for every (u, v) ∈ R2 with |u| ≤M and |v| ≤M .

(H2) φ : R −→ R is continuous, increasing and φ(R) = R.

(H3) g : R3 −→ R is a continuous function, nondecreasing in the second variableand nonincreasing in the third one; h : R −→ R is a continuous andnondecreasing function.

∗Research partially supported by DGICYT, project PB94-0610.†Research partially supported by the EEC project grant ERB CHRX-CT94-0555.

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Problem (1.1) with different boundary conditions has been studied by differ-ent authors (see [1, 2] and references therein). Recently, Gao, Wang and Lin,following some ideas developed in [2], proved in [3] the existence of solution forthe problem (1.1) with f a continuous function and considering Dirichlet andmixed boundary conditions, in the presence of lower and upper solutions. Theseresults have been generalized in [4] to the case in which f is a Caratheodoryfunction. On the other hand, existence results for linear boundary data werealso presented by O’Regan in [5].

In [6], existence results for the Periodic and Neumann problems were proved.There we used the results given in [4]. However, the method of lower and uppersolutions was indirectly applied.

In this paper we give results, in the spirit of [3], that generalize all of theseexistence theorems for linear boundary conditions by considering the conditionsstudied in [7] for the second order problem u′′ = f(t, u, u′), together with equa-tions of the form of (1.1). Here, the method of lower and upper solutions and theNagumo condition, to obtain a priori bounds for the derivatives of the solutions,are applied, improving the results of [4] and [6].

Now, letW 1,1(I) be the Banach space of absolutely continuous functions on I.We say that y ∈ C1(I) is a lower solution for the problem (P ) if φy′ ∈W 1,1(I)and satisfies (φ(y′(t)))′ ≥ f(t, y(t), y′(t)) for a. e. t ∈ I, g(y(a), y′(a), y′(b)) ≥ 0and h(y(a)) = y(b).y is an upper solution of (P ) if the reversed inequalities hold. If equalities

hold, we say that y is a solution of (P ).Next we define the Nagumo condition we are going to use. Note that the

condition does not depend on the boundary data of the problem.

Definition 1.1 We say that f : I × R2 → R, a Caratheodory function, satisfiesa Nagumo condition relative to the pair α and β, with α, β ∈ C(I), α ≤ β inI, if there exist functions k ∈ Lp(I), 1 ≤ p ≤ ∞, and θ : [0,∞) −→ (0,∞)continuous, such that

|f(t, u, v)| ≤ k(t)θ(|v|) for a.e. (t, u, v) ∈ Ω,

where Ω =

(t, u, v) ∈ I ×R2 : α(t) ≤ u ≤ β(t)

, and also that∫ ∞φ(ν)

|φ−1(u)|p−1

p

θ(|φ−1(u)|)du,

∫ φ(−ν)

−∞

|φ−1(u)|p−1

p

θ(|φ−1(u)|)du > µ

p−1p ‖k‖p,

beingµ = max

t∈Iβ(t)−min

t∈Iα(t),

ν =max |α(a)− β(b)|, |α(b)− β(a)|

b− aand

‖k‖p =

supt∈I |k(t)| if p =∞[∫ ba|k(t)|pdt

] 1p

if 1 ≤ p <∞,

where (p− 1)/p ≡ 1 for p =∞.

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2. MAIN RESULT

Following the ideas of [7], before introducing the main result of existence ofsolutions of the problem (P ), suppose that hypotheses (H1) – (H3) and theNagumo condition relative to a lower solution α and an upper solution β, α ≤ βare satisfied. Now we start with the construction of the modified problem.

Firstly, we define p(t, x) = max α(t),min x, β(t) for all x ∈ R.One can find the next result, with its proof, in [8].

Lemma 2.1 For each u ∈ C1(I) the next two properties hold:

(a)d

dtp(t, u(t)) exists for a.e. t ∈ I.

(b) If u, um ∈ C1(I) and um → u in C1(I) then

d

dtp(t, um(t))→ d

dtp(t, u(t)) for a.e. t ∈ I.

By definition 1.1, we can find two real numbers, M− < 0 < M+, such that

M− < −ν ≤ ν < M+, M− < α′(t), β′(t) < M+ for all t ∈ I and

∫ φ(M+)

φ(ν)

|φ−1(u)|p−1

p

θ(|φ−1(u)|)du,

∫ φ(−ν)

φ(M−)

|φ−1(u)|p−1

p

θ(|φ−1(u)|)du > µ

p−1p ‖k‖p.

And we consider the following modified problem, which we denote by (P ∗).

[φ(u′(t))]′ = F (t, u(t),d

dtp(t, u(t))) for a.e. t ∈ I, (2.1)

u(a) = G(u(a), u′(a), u′(b)), (2.2)u(b) = h(u(a)), (2.3)

with F (t, x, y) = f(t, p(t, x), q(y)) + tanh (x− p(t, x)). Where function q isdefined as q(y) = max M−,min y,M+ for all y ∈ R and G(x, y, z) =p(a, x+ g(x, y, z)) for all (x, y, z) ∈ R3.

For the problem (P ∗) the following three lemmas rule, and we can concludethat every solution of (P ∗) is a solution of (P ) in the sector

[α, β] =v ∈ C1(I) : α(t) ≤ v(t) ≤ β(t) for all t ∈ I

,

and the set of derivatives of such solutions is bounded in C(I).

Lemma 2.2 If u is a solution of (P ∗) then u ∈ [α, β].

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Proof We shall only see that α(t) ≤ u(t) for every t ∈ I. An analogousreasoning shows that u(t) ≤ β(t) for all t ∈ I.

By definition of G, using (2.2), we have that α(a) ≤ u(a) ≤ β(a).Moreover, because h is nondecreasing and by the definition of lower and upper

solution,α(b) = h(α(a)) ≤ h(u(a)) ≤ h(β(a)) = β(b),

and then, using (2.3), we obtain

α(b) ≤ u(b) ≤ β(b).

If there exists t0 ∈ (a, b) such that

u(t0)− α(t0) = mint∈I(u− α)(t) < 0,

then, since u − α ∈ C1(I), we have (u − α)′(t0) = 0. Furthermore, there exista ≤ t1 < t0 < t2 ≤ b such that u < α in (t1, t2) and (u−α)(t1) = (u−α)(t2) = 0.Thus,

[φ(u′(t))]′ − [φ(α′(t))]′ ≤ f(t, α(t), α′(t)) + tanh[u(t)− α(t)]− f(t, α(t), α′(t)) =

= tanh[u(t)− α(t)] < 0 for all t ∈ (t1, t2).

In consequence φ(u′(t)) − φ(α′(t)) < φ(u′(t0)) − φ(α′(t0)) = 0 for all t ∈(t0, t2) and, since φ is increasing and, in particular, one-to-one, we have

u′(t) < α′(t) for all t ∈ (t0, t2).

Thus, (u− α)(t2) < (u− α)(t0) < 0, which is a contradiction. ut

Lemma 2.3 If u is a solution of (P ∗) then M− < u′(t) < M+ for every t ∈ I,where M+ and M− are the Nagumo constants that were chosen for the construc-tion of (P ∗), and only depend on α, β, φ, θ and k.

Proof Let u ∈ C1(I) be a solution of (P ∗). As we know, by lemma 2.2,u ∈ [α, β], and then

[φ(u′(t))]′ = f(t, u(t), q(u′(t))) for a.e. t ∈ I.

By the mean value theorem, there exists t0 ∈ (a, b) such that

u′(t0) =u(b)− u(a)

b− a,

and then

M− < −ν ≤α(b)− β(a)

b− a≤ u′(t0) ≤ β(b)− α(a)

b− a≤ ν < M+.

Let’s call ν0 = |u′(t0)|.Suppose that there exists a point in the interval I for which u′ > M+ or

u′ < M−. By the continuity of u′ we can choose t1 ∈ I verifying one of thefollowing situations:

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(i) u′(t0) = ν0, u′(t1) = M+ and ν0 ≤ u′(t) ≤M+ for all t ∈ (t0, t1),

(ii) u′(t1) = M+, u′(t0) = ν0 and ν0 ≤ u′(t) ≤M+ for all t ∈ (t1, t0),

(iii) u′(t0) = −ν0, u′(t1) = M− and M− ≤ u′(t) ≤ −ν0 for all t ∈ (t0, t1),

(iv) u′(t1) = M−, u′(t0) = −ν0 and M− ≤ u′(t) ≤ −ν0 for all t ∈ (t1, t0).

Assume that the situation (i) holds (the other cases admit analogous proofs).Since M− ≤ ν0 ≤ u′(t) ≤M+ for all t ∈ (t0, t1), we have

[φ(u′(t))]′ = f(t, u(t), q(u′(t))) = f(t, u(t), u′(t)) for a.e. t ∈ (t0, t1),

so, by the Nagumo condition,

|[φ(u′(t))]′| = |f(t, u(t), u′(t))| ≤ k(t)θ(|u′(t)|) for a.e. t ∈ (t0, t1). (2.4)

Note that φ−1(s) ≥ 0 for s ∈ [φ(ν0), φ(M+)]. On the other hand, we haveν0 ≤ ν and thus φ(ν0) ≤ φ(ν), what leads us to∫ φ(M+)

φ(ν0)

(φ−1(s))p−1

p

θ(φ−1(s))ds ≥

∫ φ(M+)

φ(ν)

(φ−1(s))p−1

p

θ(φ−1(s))ds > µ

p−1p ‖k‖p. (2.5)

Consider now the following function

ϕ : [t0, t1] −→ [φ(ν0), φ(M+)]

defined byϕ(r) = φ(u′(r)) for r ∈ [t0, t1].

By definition of solution u, ϕ is an absolutely continuous function. Further-more, the function under the integral is continuous in the interval (φ(ν0), φ(M+)),so we have (see [9]),∫ φ(M+)

φ(ν0)

(φ−1(s))p−1

p

θ(φ−1(s))ds =

∫ t1

t0

[φ(u′(t))]′(u′(t))p−1

p

θ(u′(t))dt ≤

≤∫ t1

t0

|[φ(u′(t))]′||u′(t)|p−1

p

θ(u′(t))dt,

using (2.4) and that u′ > 0 on (t0, t1) we have that∫ φ(M+)

φ(ν0)

(φ−1(s))p−1

p

θ(φ−1(s))ds ≤

∫ t1

t0

k(t)(u′(t))p−1

p dt

and, by Holder’s inequality,∫ φ(M+)

φ(ν0)

(φ−1(s))p−1

p

θ(φ−1(s))ds ≤ ‖k‖pµ

p−1p ,

which is a contradiction with (2.5). ut

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Lemma 2.4 If u is a solution of (P ∗) then g(u(a), u′(a), u′(b)) = 0.

Proof If we could show that α(a) ≤ u(a) + g(u(a), u′(a), u′(b)) ≤ β(a), then,by the boundary conditions of (P ∗), we would have

u(a) = G(u(a), u′(a), u′(b)) = u(a) + g(u(a), u′(a), u′(b)),

and thus g(u(a), u′(a), u′(b)) = 0.Suppose that u(a) +g(u(a), u′(a), u′(b)) < α(a), by definition of G, we obtain

u(a) = G(u(a), u′(a), u′(b)) = α(a)

and thenu(b) = h(u(a)) = h(α(a)) = α(b).

Then, since u − α ∈ C1(I), (u − α)(a) = (u − α)(b) = 0, and u ≥ α in I (lemma 2.2), we have (u− α)′(a) ≥ 0 ≥ (u− α)′(b).

Now, using (H3), we have that

u(a) + g(u(a), u′(a), u′(b)) = α(a) + g(α(a), u′(a), u′(b)) ≥

≥ α(a) + g(α(a), α′(a), α′(b)) ≥ α(a),

reaching a contradiction.An analogous reasoning proves that u(a) + g(u(a), u′(a), u′(b)) ≤ β(a). utNow we can prove our main result.

Theorem 2.1 Let α and β be a lower and an upper solution respectively forthe problem (P ) and such that α ≤ β in I. Assume that hypotheses (H1) – (H3)are satisfied, and the Nagumo condition relative to α and β holds.

Then (P ) has at least one solution u ∈ [α, β] and M− < u′(t) < M+ for all t ∈I, where M+ and M− only depend on α, β, φ, θ and k.

Proof By lemmas 2.2, 2.3 and 2.4, the proof will be over once we had shownthat (P ∗) admits a solution.

Let v ∈ C1(I) fixed, define gv : R −→ R such that

gv(x) =∫ b

a

φ−1

(x+

∫ r

a

Fv(s)ds)dr for all x ∈ R,

where Fv(s) ≡ F (s, v(s), ddsp(s, v(s))) for a. e. s ∈ I.Clearly gv is a continuous and increasing function.Note that there exists Ψ ∈ L1(I) such that

| Fu(s) |≤ Ψ(s) for a. e. s ∈ I and for all u ∈ C1(I),

and then ∣∣∣∣∫ t

a

Fu(s)ds∣∣∣∣ ≤ ‖Ψ‖1 for all t ∈ I and every u ∈ C1(I). (2.6)

6

Since φ−1 is increasing, we have, for each x ∈ R and each u ∈ C1(I), that

f−(x) ≡ (b− a)φ−1(x− ‖Ψ‖1) ≤ gu(x) ≤ (b− a)φ−1(x+ ‖Ψ‖1) ≡ f+(x). (2.7)

Functions f± are continuous, increasing and f±(R) = R; by the previousinequalities we have gu(R) = R for all u ∈ C1(I), and then for each u ∈ C1(I)there exists a unique τu satisfying∫ b

a

φ−1

(τu +

∫ r

a

Fu(s)ds)dr = h(Gu)−Gu, (2.8)

with Gu ≡ G(u(a), u′(a), u′(b)).Now call c(u)± = (f±)−1(h(Gu)−Gu). From (2.7) we deduce that

c(u)+ ≤ τu ≤ c(u)− for all u ∈ C1(I).

And now, since h(Gu)−Gu is bounded in C1(I) and (f±)−1 are continuousin R, there exists L > 0 such that

| τu |≤ L for all u ∈ C1(I) (2.9)

Now, we define the following operator T : C1(I) −→ C1(I) where, for eachu ∈ C1(I) and each t ∈ I it is

(Tu)(t) = Gu +∫ t

a

φ−1

(τu +

∫ r

a

Fu(s)ds)dr. (2.10)

If u ∈ C1(I) is a fixed point of T , then differentiating in (2.10) and using theregularity of Fu, we obtain that φu′ ∈W 1,1(I) and u satisfies (2.1). Expression(2.2) is trivially satisfied. Equation (2.3) follows from (2.8). Thus, if u is afixed point of T then u is a solution of (P ∗).

Now, we will prove that T has a fixed point u ∈ C1(I).We start showing that the operator T is continuous in C1(I).Suppose un → u in C1(I). Let τn corresponding to un by (2.8) and τ asso-

ciated to u, let’s see that limn→∞

τn = τ .By construction of τn and τ we have

h(Gn)− h(G) +G−Gn =∫ b

a

φ−1

(τn +

∫ r

a

Fn(s)ds)dr−

∫ b

a

φ−1

(τ +

∫ r

a

Fu(s)ds)dr,

(where Fn(s) ≡ Fun(s)). By the mean value theorem there exist ηn ∈ (a, b) such

that

1b− a

(h(Gn)− h(G) +G−Gn) = φ−1

(τn +

∫ ηn

a

Fn(s)ds)−

7

φ−1

(τ +

∫ ηn

a

Fu(s)ds).

By the continuity of G and h we have

limn→∞

φ−1

(τn +

∫ ηn

a

Fn(s)ds)− φ−1

(τ +

∫ ηn

a

Fu(s)ds)

= 0. (2.11)

Using (2.6), (2.9) and (H2), we know that there exists a compact interval Jsuch that

φ−1

(τn +

∫ ηn

a

Fn(s)ds), φ−1

(τ +

∫ ηn

a

Fu(s)ds)∈ J for all n ∈ N.

By (H2) again, we can affirm that φ is uniformly continuous in J .Take ε > 0 and choose δ > 0, associated by the definition of uniformly

continuous function. By (2.11) there exists nδ ∈ N such that∣∣∣∣φ−1

(τn +

∫ ηn

a

Fn(s)ds)− φ−1

(τ +

∫ ηn

a

Fu(s)ds)∣∣∣∣ < δ for all n ≥ nδ,

and then ∣∣∣∣τn − τ +∫ ηn

a

[Fn(s)− Fu(s)]ds∣∣∣∣ < ε for all n ≥ nδ.

From the above discussion we have

limn→∞

τn − τ +

∫ ηn

a

[Fn(s)− Fu(s)]ds

= 0. (2.12)

On the other hand, since un → u in C1(I) and F is a Caratheodory function,we have, by lemma 2.1, that

Fn(t)→ Fu(t) in a.e. t ∈ I,

so, by (2.6) and the dominated convergence theorem, we conclude

Fn → Fu in L1(I).

Moreover

0 ≤∣∣∣∣∫ ηn

a

[Fn(t)− Fu(t)]dt∣∣∣∣ ≤ ‖Fn − Fu‖1 for all n ∈ N,

and then limn→∞

∫ ηn

a

[Fn(t)− Fu(t)]dt = 0. Now, using (2.12), we conclude that

limn→∞

τn = τ.

8

Furthermore,

τn +∫ t

a

Fn(s)ds→ τ +∫ t

a

Fu(s)ds for all t ∈ I.

Now, since∣∣∣∣τn +∫ t

a

Fn(s)ds− τ −∫ t

a

Fu(s)ds∣∣∣∣ ≤ |τn − τ |+ ‖Fn − Fu‖1 for all t ∈ I,

the convergence is uniform on I. And, by the uniform continuity of φ−1 oncompact intervals,

(Tun)′ → (Tu)′ uniformly on I

and henceTun → Tu uniformly on I.

Now we are going to prove that T (C1(I)) is a relatively compact set in C1(I).Using (2.6), (2.9) and (H2), we have that there exists Q > 0 such that

φ−1(−Q) ≤ (Tu)′(t) ≤ φ−1(Q) for all t ∈ I and all u ∈ C1(I),

i.e., the set of derivatives of T (C1(I)) is uniformly bounded on I, and then wehave that T (C1(I)) is a bounded subset of C1(I) and it is an equicontinuoussubset of C(I).

Finally, to see that(T (C1(I))

)′ =y′ : y ∈ T (C1(I))

is equicontinuous on

I, we use that φ−1 is uniformly continuous on [−Q,Q] and expression (2.6).By Schauder fixed point theorem (see [10]), the operator T has at least a fixed

point, which concludes the proof. ut

3. FINAL REMARKS

A more general result is obtained if we change the hypothesis (H2) by

(H∗2 ) There exist K− < 0 < K+ such that φ : [K−,K+] −→ R is a continuousand increasing function in [K−,K+].

Note that the solutions (and also the lower and upper solutions) must beelements of the set

C =v ∈ C1(I) : K− < v′ < K+ and φ v′ ∈W 1,1(I)

,

so, in their definitions, this condition cannot be disregarded.Instead of definition 1.1 we must assume that

K− < −ν ≤ ν < K+,

9

and there exist k ∈ Lp(I), 1 ≤ p ≤ ∞ and θ : [0,∞) −→ (0,∞) continuous,such that

|f(t, x, y)| ≤ k(t)θ(|y|) for a.e. (t, x, y) ∈ Ω,

and ∫ φ(−ν)

φ(K−)

|φ−1(s)|p−1

p

θ(|φ−1(s)|)ds,

∫ φ(K+)

φ(ν)

|φ−1(s)|p−1

p

θ(|φ−1(s)|)ds > µ

p−1p ‖k‖p,

being µ, ν and ‖k‖p as in definition 1.1.Although this condition seems to be a restriction over the first Nagumo con-

dition it is not true, since definition 1.1 is equivalent to request that there existsM− < 0 < M+ such that∫ φ(−ν)

M−

|φ−1(s)|p−1

p

θ(|φ−1(s)|)ds,

∫ M+

φ(ν)

|φ−1(s)|p−1

p

θ(|φ−1(s)|)ds > µ

p−1p ‖k‖p.

The idea of the proof consists simply on defining Φ satisfying (H2) and φ ≡ Φon [K−,K+], and showing that theorem 2.1 can be used to prove that there existssome solution u for the problem with Φ and such that K− < u′ < K+ on I.Hence u will be a solution for the former problem.

On the other hand, one can obtain an analogous existence result for the prob-lem

(Q)

[φ(u′(t))]′ = f(t, u(t), u′(t)) for a.e. t ∈ I,r(u(a), u′(a), u(b), u′(b)) = 0,s(u(a), u′(a), u(b), u′(b)) = 0.

With r : R4 −→ R continuous and nondecreasing in the second and thirdvariables; and s : R4 −→ R continuous, nondecreasing in the first variableand nonincreasing in the fourth one. Furthermore (H1), (H2) and the Nagumocondition hold for a pair of a lower solution α and an upper solution β, suchthat α ≤ β in I.

In this case we say that y is lower solution for the problem (Q) if φ y′ ∈W 1,1(I) and

[φ(y′(t))]′ ≥ f(t, y(t), y′(t)) for a. e. t ∈ I,

r(y(a), y′(a), y(b), v) ≥ 0 for all v ∈ R,

s(y(a), u, y(b), y′(b)) ≥ 0 for all u ∈ R.

If the reversed inequalities hold, we say that y is an upper solution.To prove that if there exist α ≤ β lower and upper solutions of (Q) then

problem (Q) admits a solution u ∈ [α, β], we construct the following modifiedproblem

(Q∗)

[φ(u′(t))]′ = F (t, u(t), ddtp(t, u(t))) for a. e. t ∈ I,R(u(a), u′(a), u(b), u′(b)) = u(a),S(u(a), u′(a), u(b), u′(b)) = u(b).

10

with F and p defined as in theorem 2.1, R(x, u, y, v) = p(a, x+ r(x, u, y, v)) andS(x, u, y, v) = p(b, y + s(x, u, y, v)).

The solutions of this problem are given as the fixed points of the operatorT : C1(I) −→ C1(I) which is defined, for every v ∈ C1(I) and t ∈ I, as follows:

(Tv)(t) = Rv +∫ t

a

φ−1

(τv +

∫ r

a

F (s, v(s),d

dsp(s, v(s)))ds

)dr.

Here τv ∈ R is the unique solution of the equation∫ b

a

φ−1

(τv +

∫ r

a

F (s, v(s),d

dsp(s, v(s)))ds

)dr = Sv −Rv,

with Rv ≡ R(v(a), v′(a), v(b), v′(b)) and Sv ≡ S(v(a), v′(a), v(b), v′(b)).It is possible to extend this result to the more general case in which the func-

tion φ is defined, continuous and increasing just on a compact interval [K−,K+]with K− < 0 < K+. The way to establish the existence of solution in this newsituation is to proceed as it was shown in the previous comment over problem(P ).

Finally, note that problem (P ) includes the Periodic problem, and problem(Q) generalizes Dirichlet, Neumann, and Mixed problems. Note that the periodicconditions are also included in (Q), but the definitions of α and β are not validin this case.

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REFERENCES

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