USA-Chile Workshop on Nonlinear Analysis, Electron. J. Diff. Eqns., Conf. 06, 2001, pp. 13–21 http://ejde.math.swt.edu or http://ejde.math.unt.edu ftp ejde.math.swt.edu or ejde.math.unt.edu (login: ftp)

Nonlinearities in a second order ODE ^{∗}

Pablo Amster

Abstract

In this paper we study the semilinear second order ordinary differential equation

u′′ + r(t)u′ + g(t, u) = f(t) .

Under a growth condition on g, we prove the existence and uniqueness for the Dirichlet problem and establish conditions for the existence of periodic solutions.

1 Introduction

The two-point boundary-value problem for a semilinear second order ODE

u′′ + ru′ + g(t, u) = 0, u(0) = u_{0}, u(T) = u_{T}

has been studied by many authors. In his pioneering work, Picard [7] proved the existence of a solution by an application of the well known method of successive approximations under a Lipschitz condition on g and a smallness condition on T. Sharper results were obtained by Hamel [2] in the special case of a forced pendulum equation (see also [4], [5]). The existence of periodic solutions for this equation was first considered by Duffing [1] in 1918. In the absence of friction (i.e. r = 0), variational methods have been applied by Lichtenstein [3], who considered the functional

I(u) = ∫_{0}^{T} \frac{(u′)^{2}}{2} − G(t, u(t))dt,

where G(t, x) = ∫_{0}^{x} g(t, s)ds. Finally, we want to mention the topological approach introduced in 1905 by Severini [8] who used a shooting method. He also presented and gave a survey of results obtained using Leray-Schauder techniques and degree theory. For further results, see [6].

In this work, we prove the existence and uniqueness of a solution to the Dirichlet problem under a growth condition on g. Then, we apply this result for finding periodic solutions.

∗Mathematics Subject Classifications: 34G20.

Key words: Second order ODE, Dirichlet and Periodic Problems.

\begin{matrix}c \\ ◯ \end{matrix}2001 Southwest Texas State University.

Published January 8, 2001.

Let S : H^{2}(0, T) → L^{2}(0, T) be the semilinear operator given by

Su = u′′ + ru′ + g(t, u) .

Assume that the function g satisfies the growth condition

\frac{g(t, u) − g(t, v)}{u − v} ≤ \frac{c_{p}}{p(t)} for t ∈ [0, T] and u, v ∈ ℝ (u ≠ v) ,  (1.1)

where p ∈ C^{1}([0, T]) is strictly positive, r_{0} := pr − p′ ∈ H^{1}(0, T) is nondecreasing, and c_{p} < λ_{p} with λ_{p} the first eigenvalue of the problem

−(pu′)′ = λ_{p}u, u(0) = u(T) = 0 .

To state a general existence and uniqueness result for the Dirichlet problem associated to our equation, we need the following apriori bounds.

Lemma 1.1 Assume that g satisfies (1.1) and let u, v ∈ H^{2}(0, T) with Tr(u) = Tr(v). Then

∥p(Su − Sv)∥_{2} ≥ (λ_{p} − c_{p})∥u − v∥_{2}

and

∥p(Su − Sv)∥_{2} ≥ \frac{λ_{p} − c_{p}}{\sqrt{λ_{p}}} ( ∫_{0}^{T} p(u′ − v′)^{2})^{1/2}

Proof. A simple computation shows that

∥p(Su − Sv)∥_{2}∥u − v∥_{2} ≥ ∫_{0}^{T} p(u′ − v′)^{2} − ∫_{0}^{T} r_{0}(u − v)(u′ − v′) − c_{p}∥u − v∥_{2}^{2}

and because − ∫_{0}^{T} r_{0}(u − v)(u′ − v′) = \frac{1}{2} ∫_{0}^{T} r′_{0}(u − v)^{2} ≥ 0, the result follows since ∥u − v∥_{2}^{2} ≤ \frac{1}{λ_{p}} ∫_{0}^{T} p(u′ − v′)^{2}. ♢

Remarks i) For simplicity and by the previous lemma, we may denote by k_{1} the best constant such that ∥u − v∥_{1,2} ≤ k_{1}∥p(Su − Sv)∥_{2} for u, v ∈ H^{2}(0, T) with Tr(u) = Tr(v). ii) In particular, if r ∈ H^{1}(0, T) is non-decreasing, the result holds for p ≡ 1 and c_{1} < λ_{1} = (\frac{π}{T} )^{2}.

Theorem 1.2 Let g satisfy (1.1). Then the Dirichlet problem

\begin{matrix}Su = f(t) in (0, T) \\ u(0) = u_{0}, u(T) = u_{T} \end{matrix}  (1.2)

is uniquely solvable in H^{2}(0, T) for any f ∈ L^{2}(0, T) and arbitrary boundary data.

Proof. Without loss of generality, we may suppose that p ≡ 1. For 0 ≤ σ ≤ 1 we consider the operator S_{σ} given by S_{σ}u := u′′ + ru′ + σg(t, u). We remark that if k_{σ} is the constant of lemma 1.1 for S_{σ}, then k_{σ} ≤ k_{1}.

From the theory of linear operators, for fixed \overline{u} ∈ H^{1}(0, T) we may define u = K\overline{u} as the unique solution of the problem

S_{0}u = f(t) − g(t, \overline{u}) in (0, T)
u(0) = u_{0}, u(T) = u_{T} .

Continuity of K : H^{1}(0, T) → H^{1}(0, T) follows immediately from the inequality

∥K\overline{u} − K\overline{v}∥_{1,2} ≤ k_{1}∥S_{0}(K\overline{u}) − S_{0}(K\overline{v})∥_{2} = k_{1}∥g(·, \overline{u}) − g(·, \overline{v})∥_{2}

and the fact that ∥g(·, \overline{u}) − g(·, \overline{v})∥_{2} → 0 for \overline{u} → \overline{v} in H^{1}(0, T) ↪ C([0, T]). Moreover, if φ(t) = \frac{u_{T} −u_{0}}{T} t + u_{0} we have that

∥K\overline{u} − φ∥_{1,2} ≤ k_{1}∥f − g(·, \overline{u}) − S_{0}φ∥_{2} ≤ C

for some constant C = C(R). Moreover, as

∥(K\overline{u})′′∥_{2} = ∥f − g(·, \overline{u}) − r(K\overline{u})′∥_{2}

it follows that K(B_{R}) is H^{2}-bounded. Thus, by the compactness of the imbedding H^{2}(0, T) ↪ H^{1}(0, T) we conclude that K is compact.

Let us assume that u = σKu for some σ ∈ (0, 1]. Then u′′ + ru′ + σg(t, u) = σf, and

∥u − σφ∥_{1,2} ≤ k_{1}∥S_{σ}u − S_{σ}(σφ)∥_{2} = k_{1}∥σf − S_{σ}(σφ)∥_{2}

This proves that the set {u : u = σKu} is uniformly bounded, and by Leray-Schauder theorem K has a fixed point. Uniqueness of the solution follows from lemma 1.1. ♢

As a simple consequence, we have an existence result for the general Dirichlet problem

\begin{matrix}Su = f(t, u, u′) in (0, T) \\ u(0) = u_{0}, u(T) = u_{T} \end{matrix}  (1.3)

Corollary 1.3 Let f be continuous and g satisfy (1.1). Assume that the growing condition

|f(t, u, x)| ≤ c|(u, x)| + d  (1.4)

holds for some constant c < \frac{1}{k_{1}∥p∥_{∞}}. Then (1.3) is solvable in H^{2}(0, T).

Proof. By (1.4) and the previous theorem, the operator K : H^{1}(0, T) → H^{1}(0, T) given by K\overline{u} = u, with u the unique solution of

Su = f(t, \overline{u}, \overline{u}′) in (0, T)
u(0) = u_{0}, u(T) = u_{T}

is well defined and compact. Moreover, as

∥K\overline{u} − φ∥_{1,2} ≤ k_{1}∥p(S(K\overline{u}) − Sφ)∥_{2} ≤ k_{1}∥p∥_{∞} (∥Sφ)∥_{2} + c∥\overline{u}∥_{1,2} + d)

then K(B_{R}) ⊂ B_{R} for R large and the result follows from Schauder Theorem.

2 Solutions to the periodic problem

In this section we’ll apply the previous results to the periodic problem

\begin{matrix}Su = f(t) in (0, T) \\ u(0) = u(T), u′(0) = u′(T) \end{matrix}  (2.1)

It is well known that the forced pendulum equation u′′ + b sin(u) = f admits periodic solutions for constant b if f is periodic and orthogonal to constants. We’ll show in the general case that in the presence of friction this orthogonality condition can be reinterpreted in terms of a certain p_{1} > 0. More precisely, we’ll show that in some cases -including the generalized pendulum equation- (2.1) is not solvable for any f such that ⟨p_{1}, f⟩ is large enough.

Lemma 2.1 For any c ∈ ℝ there exists a unique p_{c} such that p_{c}(0) = p_{c}(T) = c and p′_{c} − rp_{c} is constant. Furthermore, p_{c} = cp_{1}, and p_{1} is strictly positive.

Proof. From the equation p′_{c} − rp_{c} = k_{c} we obtain that

p_{c} = (c + k_{c} ∫_{0}^{t} e^{− ∫_{0}^{s} r}ds) e^{∫_{0}^{t} r}

and from the condition p_{c}(0) = p_{c}(T) = c we conclude that

k_{c} = c\frac{1 − e^{∫_{0}^{T} r}}{∫_{0}^{T} e^{∫_{s}^{T} r}ds} = ck_{1}

Thus, p_{c} = cp_{1}. Moreover, if k_{1} ≥ 0 it’s immediate that p_{1} > 0, and if k_{1} < 0, assuming that p_{1} vanishes there exists t_{0} ∈ (0, T) such that p_{1}(t_{0}) = 0 ≤ p′_{1}(t_{0}). Then k_{1} = p′_{1}(t_{0}) ≥ 0, a contradiction. ♢

Using the preceding lemma we’ll see that periodic solutions of Su = f satisfy an orthogonality condition. Indeed, from

u′′ + ru′ + g(t, u) = f

we obtain

(p_{1}u′)′ − k_{1}u′ + p_{1}g(t, u) = p_{1}f .

By the equality p_{1}u′|_{0}^{T} = u|_{0}^{T} = 0 we have

∫_{0}^{T} p_{1}g(t, u) = ∫_{0}^{T} p_{1}f .

Corollary 2.2 With the previous notation, let us assume that g(t, u) ≤ g_{max} for any t ∈ [0, T], u ∈ ℝ and some constant g_{max} ∈ ℝ (respectively, g(t, u) ≥ g_{min} for any t ∈ [0, T], u ∈ ℝ and some constant g_{min} ∈ ℝ). Then (2.1) is not solvable for any f ∈ L^{2}(0, T) such that ⟨p_{1}, f⟩ > g_{max}∥p_{1}∥_{1} (resp. ⟨p_{1}, f⟩ < g_{min}∥p_{1}∥_{1}).

Now we’ll give some existence results for (2.1), assuming that g satisfies (1.1). Our method is based in the existence and uniqueness result given by Theorem 1.2: indeed, for fixed s ∈ ℝ we may define u_{s} as the unique solution of the problem

Su = f(t) in (0, T)
u(0) = u(T) = s

Lemma 2.3 The mapping s → u_{s} is continuous for the H^{1}-norm.

Proof. For s → s_{0} and w_{s} = u_{s} − u_{s_{0}} we have

0 = ∫_{0}^{T} p(Su_{s} − Su_{s_{0}})w_{s}
≤ pw′_{s}w_{s}|_{0}^{T} − ∫_{0}^{T} p(w′_{s})^{2} + \frac{r_{0}w_{s}^{2}}{2} |_{0}^{T} − ∫_{0}^{T} r′_{0} \frac{w_{s}^{2}}{2} + c_{p} ∫_{0}^{T} w_{s}^{2}

Because ∫_{0}^{T} r′_{0} \frac{w_{s}^{2}}{2} ≥ 0, we conclude that

0 ≤ (1 − \frac{c_{p}}{λ_{p}}) ∫_{0}^{T} p(w′_{s})^{2} ≤ pw′_{s}w_{s}|_{0}^{T} + r_{0} \frac{w_{s}^{2}}{2} |_{0}^{T} .

Since w_{s}(0) = w_{s}(T) = s − s_{0} → 0 it suffices to prove that ∥w_{s}∥_{1,∞} is bounded. As ∥u_{s}−s∥_{1,2} ≤ k_{1}∥p(f−g(·, s))∥_{2}, we deduce that w_{s} is H^{1}-bounded. Moreover, from the equality u′′_{s} = f − ru′_{s} − g(t, u_{s}) we obtain that ∥w_{s}∥_{2,2} is bounded, and from the imbedding H^{2}(0, T) ↪ C^{1}([0, T]) the proof is complete. ♢

From the previous remarks, the solvability of (2.1) is equivalent to the solvability of the equation ψ(s) = ∫_{0}^{T} p_{1}f, where ψ : ℝ → ℝ is given by ψ(s) = ∫_{0}^{T} p_{1}g(t, u_{s}). Continuity of ψ follows immediately from the previous lemma, and hence (2.1) will admit a solution if and only if there exist s^{±} such that

ψ(s^{+}) ≥ ⟨p_{1}, f⟩ ≥ ψ(s^{−})

Remark. Writing u_{s}(t) − s = ∫_{0}^{t} p^{−1/2}p^{1/2}u′_{s} we obtain that

∥u_{s} − s∥_{∞} ≤ δ_{p}∥p(f − g(·, s))∥_{2}

for δ_{p} = (∫_{0}^{T} \frac{1}{p})^{1/2} ^{\frac{\sqrt{λ_{p}}}{λ_{p}−c_{p}}}.

Thus, if we consider the condition

∥pg(·, s)∥_{2} ≤ c|s| + d with cδ_{p} < 1  (2.2)

then u_{s}(t) ∈ J_{s}^{ε} for any t ∈ [0, T], where J_{s}^{ε} is the interval centered in s with radius δ_{p}(c|s| + d) + ε, ε = δ_{p}∥pf∥_{2}. As a simple consequence we have the following

Theorem 2.4 Let g satisfy (1.1)-(2.2), and assume that there exist s^{±} such that

g|_{[0,T]×J_{s^{+}}^{ε}} ≥ \frac{∫_{0}^{T} p_{1}f}{∥p_{1}∥_{1}} ≥ g|_{[0,T]×J_{s^{−}}^{ε}}

for ε = δ_{p}∥pf∥_{2}. Then (2.1) admits a solution u_{s} for some s between s^{−} and s^{+}.

In particular, if there exist s^{±} such that

g|_{[0,T]×J_{s^{+}}^{ε}} ≥ 0 ≥ g|_{[0,T]×J_{s^{−}}^{ε}}

then (2.1) admits a solution u_{s} for some s between s^{−} and s^{+} for any f ⊥ p_{1} such that δ_{p}∥pf∥_{2} ≤ ε.

Proof. As u_{s}^{±}([0, T]) ⊂ J_{s^{±}}^{ε}, we obtain:

∫_{0}^{T} p_{1}g(t, u_{s^{+}}) ≥ ∫_{0}^{T} p_{1}f ≥ ∫_{0}^{T} p_{1}g(t, u_{s^{−}})

and the result holds. ♢

Using the fact that |s|−δ_{p}(c|s|+d) → +∞ we deduce the following existence results:

Corollary 2.5 Let g satisfy (1.1)-(2.2), and assume, for some M > 0 that

g(t, x)sg(x) ≥ 0 for |x| ≥ M

or

g(t, x)sg(x) ≤ 0 for |x| ≥ M

Then (2.1) is solvable for any f ⊥ p_{1}.

Corollary 2.6 Let g satisfy (1.1)-(2.2), and assume that

lim_{|x|→+∞} g(t, x)sg(x) = +∞ or lim_{|x|→+∞} g(t, x)sg(x) = −∞

uniformly on t. Then (2.1) is solvable for any f.

Proof. Under the first assumption, there exists M such that

g(t, x)sg(x) ≥ \frac{| ∫_{0}^{T} p_{1}f|}{∥p_{1}∥_{1}} for |x| ≥ M

Hence, taking s > 0 such that s − δ_{p}(cs + d + ∥pf∥_{2}) ≥ M we have

∫_{0}^{T} p_{1}g(t, u_{s}) ≥ | ∫_{0}^{T} p_{1}f| ≥ ∫_{0}^{T} p_{1}f .

In the same way, for s < 0 with s + δ_{p}(−cs + d + ∥pf∥_{2}) ≤ −M we obtain ∫_{0}^{T} p_{1}g(t, u_{s}) ≤ ∫_{0}^{T} p_{1}f and the proof is complete. The case g(t, x)sg(x) → −∞ is analogous. ♢

Remark. In the previous corollaries (2.5)-(2.6), we also have that all the solutions belong to a compact arc of H^{1}(0, T), namely {u_{s} : −S ≤ s ≤ S} with

S = \frac{M + δ_{p}(d + ∥pf∥_{2}|)}{1 − δ_{p}c} .

We may also apply theorem (2.4) to the forced pendulum equation with friction

u′′ + ru′ + b sin(u) = f .  (2.3)

We first remark that in this case condition (1.1) reads

|b(t)| ≤ \frac{c_{p}}{p(t)} for any t ∈ [0, T]  (2.4)

for some p > 0 with pr − p′ nondecreasing and c_{p} < λ_{p}.

Theorem 2.7 With the previous notation, let us assume that

i) b satisfies (2.4) and does not vanish in (0, T). ii) ∥p(f ± b)∥_{2} ≤ \frac{c}{δ_{p}} for some c < \frac{π}{2} .

iii) | ∫_{0}^{T} p_{1}f| ≤ cos(c)∥p_{1}b∥_{1} Then there exist s_{1} ∈ [−\frac{π}{2} , \frac{π}{2} ], s_{2} ∈ [\frac{π}{2} , \frac{3}{2}π] such that u_{s_{i}} + 2kπ is a periodic solution of (2.3) for any integer k.

Proof. From the previous computations for s = \frac{π}{2} + kπ we obtain that

∥u_{s} − s∥_{∞} ≤ δ_{p}∥p(f − (−1)^{k}b∥_{2} ≤ c < \frac{π}{2}

As sin u_{s} = (−1)^{k} cos(u_{s} − s), taking k such that (−1)^{k}b > 0 we conclude that

∫_{0}^{T} p_{1}b sin u_{s} = ∫_{0}^{T} p_{1}|b| cos(u_{s} − s) ≥ cos(c)∥p_{1}b∥_{1} ≥ ∫_{0}^{T} p_{1}f

In the same way, for s = \frac{π}{2} + (k ± 1)π

∫_{0}^{T} p_{1}b sin u_{s} = − ∫_{0}^{T} p_{1}|b| cos(u_{s} − s) ≤ −cos(c)∥p_{1}b∥_{1} ≤ ∫_{0}^{T} p_{1}f

and the result holds. ♢

Remark. In particular, condition iii) is fulfilled if f is orthogonal to p_{1}.

If we assume that ∥pf∥_{2} ≤ \frac{π}{2δ_{p}} we also obtain existence under slightly different conditions.

Theorem 2.8 With the previous notation, let us assume that

i) b satisfies (2.4) and does not vanish in (0, T) ii) ∥pf∥_{2} ≤ \frac{π}{2δ_{p}}, ∥p(f − |b|)∥_{2} < \frac{c}{δ_{p}} for some c < \frac{π}{2} .

iii) sin(δ_{p}∥pf∥_{2}) ≤ ^{\frac{∫_{0}^{T} p_{1}f}{∥p_{1}b∥_{1}}} ≤ cos(δ_{p}∥p(f − |b|)∥_{2}). Then, if b > 0 (resp. b < 0) there exist s_{1} ∈ [0, \frac{π}{2} ], s_{2} ∈ [\frac{π}{2} , π] (resp. s_{1} ∈ [−\frac{π}{2} , 0], s_{2} ∈ [π, \frac{3}{2}π]) such that u_{s_{i}} + 2kπ is a periodic solution of (2.3) for any integer k. Moreover, if we replace ii) and iii) by ii’) ∥pf∥_{2} ≤ \frac{π}{2δ_{p}}, ∥p(f + |b|)∥_{2} < \frac{c}{δ_{p}} for some c < \frac{π}{2} .

iii’) sin(δ_{p}∥pf∥_{2}) ≤ ^{\frac{− ∫_{0}^{T} p_{1}f}{∥p_{1}b∥_{1}}} ≤ cos(δ_{p}∥p(f + |b|)∥_{2}) then if b < 0 (resp. b > 0) there exist s_{1} ∈ [0, \frac{π}{2} ], s_{2} ∈ [\frac{π}{2} , π] (resp. s_{1} ∈ [−\frac{π}{2} , 0], s_{2} ∈ [π, \frac{3}{2}π]) such that u_{s_{i}} + 2kπ is a periodic solution of (2.3) for any integer k.

Proof. It follows like in the previous theorem, using the fact that if s = kπ then ∥u_{s} − s∥_{∞} ≤ δ_{p}∥pf∥_{2}, and

| ∫_{0}^{T} p_{1}b sin u_{s}| ≤ ∥p_{1}b∥_{1} sin(δ_{p}∥pf∥_{2}) .

References

[1] Duffing, G., Erzwungene Schwingungen bei veränderlicher Eigenfrequenz, Vieweg. Braunschweig, 1918.

[2] Hamel, G., Über erzwungene Schwingungen bei endlichen Amplituden. Math. Ann., 86 (1922), 1-13.

[3] Lichtenstein, L., Über einige Existenzprobleme der Variationsrechnung. Methode der unendlichvielen Variabeln, J.Reine Angew. Math. 145 (1915), 24-85.

[4] Mawhin, J., Periodic oscillations of forced pendulum-like equations. Lecture Notes in Math., Springer, 964 (1982), 458-76.

[5] Mawhin, J., The forced pendulum: A paradigm for nonlinear analysis and dynamical systems. Expo. Math., 6 (1988), 271-87.

[6] Mawhin, J., Boudary value problems for nonlinear ordinary differential equations: from successive approximations to topology.

[7] Picard, E., Sur l’application des méthodes d’approximations succesives à l’étude de certaines équations différentielles ordinaires, J.Math. Pures Appl. 9 (1893), 217-271.

[8] Severini, C., Sopra gli integrali delle equazione differenziali del secondo ordine con valori prestabiliti in due punti dati, Atti R. Acc. Torino 40 (1904-5), 1035-40.

Pablo Amster Depto. de Matemática, CONICET Facultad de Cs. Exactas y Naturales Universidad de Buenos Aires Ciudad Universitaria, Pab. I - (1428) Buenos Aires, Argentina e-mail: pamster@dm.uba.ar