USA-Chile Workshop on Nonlinear Analysis, Electron. J. Diff. Eqns., Conf. 06, 2001, pp. 257–266. http://ejde.math.swt.edu or http://ejde.math.unt.edu ftp ejde.math.swt.edu or ejde.math.unt.edu (login: ftp)

Heteroclinic connections for a class of non-autonomous systems ^{∗}

L. Sanchez

Abstract We prove the existence of heteroclinic connections for a system of ordinary differential equations, with time-dependent coefficients, which is reminiscent of the ODE arising in connection with traveling waves for the Fisher equation. The approach is elementary and it allows in particular the study of the existence of positive solutions for the same system that vanish on the boundary of an interval (t_{0}, +∞).

1 Introduction

When one looks for one-dimensional traveling waves u(x − ct) for the Fisher equation

\frac{∂u}{∂t} = \frac{∂^{2}u}{∂x^{2}} + f(u)

(that models a diffusion phenomenon in biomathematics), one finds the ordinary differential equation

u′′ + cu′ + f(u) = 0 .  (1)

Here c > 0 represents the admissible wave speed; the function f takes positive values between two zeros, say 0 and a (a > 0): see for example [7, 2].

In this paper we consider the following system, which is a non-autonomous multi-dimensional analogue of (1):

u′′_{i} + p_{i}(t)u′_{i} + f_{i}(u) = 0, i = 1, …, n ,  (2)

where u = (u_{1}, ⋯ u_{n}). The vector field f = (f_{1}, ⋯, f_{n}) is assumed to be defined in some n-dimensional box [0, a_{1}] × ⋯ × [0, a_{n}] (a_{i} > 0, ∀i = 1, …, n), the vertices (0, ⋯, 0) and (a_{1}, ⋯, a_{n}) being its only zeros. More precisely, we state the following basic assumptions:

(H1) For each i ∈ 1, ⋯, n, f_{i} : [0, a_{1}] × ⋯ × [0, a_{n}] → ℝ_{+} is a Lipschitz continuous function such that f_{i}(0, ⋯, 0) = 0 = f_{i}(a_{1}, ⋯ a_{n}) and f_{i}(u) > 0 if u_{i} > 0 and u ≠ a := (a_{1}, ⋯ a_{n}).

∗Mathematics Subject Classifications: 34B15, 34C37.

Key words: Heteroclinics, Fisher equation.

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

Published January 8, 2001.

Supported by Fundação para a Ciência e a Tecnologia and PRAXIS XXI.

(H2) The functions p_{i} : ℝ → ℝ_{+} will be assumed throughout to be continuous and

c_{i} := inf_{t∈ℝ} p_{i}(t) > 0.

We look for positive solutions u(t) = (u_{1}(t); ⋯, u_{n}(t)), i.e., solutions that have positive components. Accordingly, the word solution will be used to mean positive solution throughout.

The relevant problem is to find “monotonic” heteroclinics (in the sense that their components are decreasing functions) that connect the equilibria (a_{1}, ⋯ a_{n}) and 0. This problem has been very much studied for the autonomous scalar equation, various approaches being available in a vast literature: we refer the reader to [2, 3, 6] and the bibliography in those papers. The autonomous system has been dealt with in [1]; we owe a lot to the ideas there, and we would like to stress that our approach, which is also elementary, works in a slightly more general setting in the sense that it allows not only time dependence but also consideration of models where the vector field f may vanish to a higher order at u = 0. In addition we could equally consider a more general form of (2) where nonlinear terms b_{i}(t)f_{i}(u) replace f_{i}(u) and the functions b_{i} are bounded above and below by positive numbers (see [8]).

An important role is played by the functions g_{i}(u), defined (for those u such that 0 < u_{i} ≤ a_{i}) by

f_{i}(u) = g_{i}(u)u_{i}.

Study of the autonomous case (1) has shown that a sufficient condition for the

existence of the mentioned heteroclinic is a bound of the form sup_{0<u<a} \frac{f(u)}{u} :=

M ≤ \frac{c^{2}}{4} . In the case of (2) with p_{i}(t) ≡ c_{i} it has been shown in [1] that

sup g_{i}(u) ≤ \frac{c_{i}^{2}}{4} remains a sufficient condition for the existence of the heteroclinic. We shall show that this still holds for our system (2) as long as the meaning of c_{i} is that given in assumption (H2); in addition we remark that we do not need an assumption used in [1] according to which (the continuous extension of) g_{i} takes its maximum value at the origin.

When (2) reduces to a single equation, boundary value problems for (2) in a finite interval appear also in connection with nonlinear elliptic problems in an annulus (see [5], [4]). By analogy we consider a second problem: that of finding (nontrivial) positive solutions of (2) defined in an unbounded interval of the form [t_{0}, +∞) and satisfying u(t_{0}) = 0 = u(+∞). We shall apply the arguments used in the construction of heteroclinics to obtain as a by-product the existence of a continuum of solutions to this problem when f, defined in ℝ_{+}^{n}, is (componentwise) non-negative and has its only zero at the origin.

In section 2 we collect some auxiliary results. In section 3 we deal with the existence of heteroclinics. In section 4 we briefly consider the existence of positive solutions vanishing in the boundary of an infinite interval.

2 Some auxiliary results for scalar equations

We start with three very simple observations about the scalar equation

u′′ + p(t)u′ + f(u) = 0,  (2_{1})

where p is positive and f satisfies the one-dimensional analogue of (H1). Recall that solution means positive solution.

Remarks. 1) The solution for the initial value problem for (2_{1}) with u(t_{0}) = u_{0} ≥ 0 and u′(t_{0}) = u_{1} exists in the maximal interval [t_{0}, s) with s < +∞ only if lim_{t→s} u(t) = 0.

2) A nonconstant solution of (2_{1}) has at most a critical point, which must be an absolute maximum.

3) Let c > 0, μ > 0, c^{2} ≥ 4M and 0 ≤ ϵ ≤ μ(\frac{c}{2} + \frac{\sqrt{c^{2}−4M}}{2} ). Then the solution u(t) of the initial value problem

u′′ + cu′ + Mu = 0 (3) u(0) = μ, u′(0) = −ϵ (4)

is positive in [0, +∞) and tends to zero as t → +∞. (See [8].)

Lemma 2.1 Let continuous functions p, q, l, m be given such that p(t) ≥ q(t) > 0, 0 ≤ l(t) ≤ m(t) in the interval [t_{0}, t_{1}]. Let u and v be the respective solutions of

u′′ + p(t)u′ + l(t)u = 0, (5)
v′′ + q(t)v′ + m(t)v = 0 (6)

such that u(t_{0}) = v(t_{0}) ≥ 0 and u′(t_{0}) = v′(t_{0}). Assume in addition that p(t) ≡ q(t) in case u′(t_{0}) = v′(t_{0}) > 0. Then if v(t) ≥ 0 in [t_{0}, t_{1}] we have u(t) ≥ v(t) in [t_{0}, t_{1}].

Proof. If u(t_{0}) = v(t_{0}) = u′(t_{0}) = v′(t_{0}) = 0 or p ≡ q and l ≡ m there is nothing to prove. Otherwise, starting with u(t_{0}) = v(t_{0}) ≥ 0 and u′(t_{0}) = v′(t_{0}) + ϵ (ϵ > 0) it follows that u > v in some interval (t_{0}, t_{0} + δ). Suppose that there exists t̄ ≤ t_{1} such that u(t) > v(t) ∀t ∈ (t_{0}, t̄) and u(t̄) = v(t̄). Set P (t) := ∫_{t_{0}}^{t} p(s) ds. Multiplying (5) and (6) by e^{P (t)}, then (5) by v, (6) by u, integrating in [t_{0}, t̄] and subtracting we obtain, since v′(t) < 0 in case v′(t_{0}) ≤ 0 (according to Remark 2)

0 > [e^{P (t)}(u′(t)v(t) − u(t)v′(t)]t̄ ; t_{0} + ∫ t̄ ; t_{0}e^{P (t)}(p(t) − q(t))v′(t)u(t) dt

+ ∫ t̄ ; t_{0}e^{P (t)}(l(t) − m(t))u(t)v(t) dt = 0,

a contradiction. Hence u ≥ v in [t_{0}, t_{1}]. In the limit as ϵ → 0^{+} the statement follows.

Let us write, in accordance with previous notation for the n-dimensional case,

f(u) = ug(u), 0 ≤ u ≤ a.

Lemma 2.2 Assume p is continuous in ℝ, f is a Lipschitz continuous function in [0, μ] such that f(0) = 0 and f(u) > 0 if 0 < u < μ; and p(t) ≥ c ≥ 2\sqrt{M} where M := sup_{0<u<μ} g(u). Given t_{0} ∈ ℝ assume in addition that p(t) is bounded

for t > t_{0}. Then the solution of the Cauchy problem

u′′ + p(t)u′ + f(u) = 0, u(t_{0}) = μ, u′(t_{0}) = −ϵ

where we assume that μ and ϵ > 0 are as in Remark 3, is positive and strictly decreasing in [t_{0}, ∞) and vanishes at +∞.

Proof. Apply Lemma 2.1 with q(t) = c, l(t) = g(u(t)), m(t) = M. Take Remarks 2 and 3 into account. The fact that u(+∞) = 0 is an easy consequence of the boundedness of p, u and u′ since for t > t_{0} and some t^{∗} ∈ (t_{0}, t)

u′(t) + ϵ + p(t^{∗})(u(t) − u_{0}) + ∫_{t_{0}}^{t} f(u(s)) ds = 0, t > t_{0}

and we infer that ∫_{t_{0}}^{+∞} f(u(s)) ds converges.

Remark. It is immediately recognized that the above result still holds if ϵ = 0 provided f(u) > 0 ∀u ∈ (0, μ].

3 Heteroclinics

In this section we give a simple analytic argument to prove the existence of heteroclinics under hypotheses (H1)-(H2). For the sake of clarity we start with the case of the scalar equation

u′′ + p(t)u′ + f(u) = 0 ,  (2_{1})

where f : [0, a] → ℝ_{+} has the property (H1) for n = 1 and we write accordingly

c := inf_{t∈ℝ} p(t) > 0; f(u) = ug(u).

A basic assumption, which cannot be improved when p ≡ c is a constant and g(u) is decreasing, is c ≥ 2\sqrt{sup_{0<u<a} g(u)}. When one deals with other models,

namely when g(0) = 0, that lower bound can be improved. Condition (ii) in the following proposition is motivated by this setting.

Proposition 3.1 Assume p is continuous, f is Lipschitz continuous in some interval [0, μ], f(0) = 0, f(u) > 0 0 < u < μ, and that either

(i) c ≥ 2\sqrt{sup_{0<u<μ} g(u)} or

(ii) there exists ν ∈ (0, μ) such that N := sup_{0<u<μ} f(u) and M := sup_{0<u<ν} g(u) satisfy c^{2} ≥ 4M and

\frac{N}{c} ≤ ν(\frac{c}{2} + \frac{\sqrt{c^{2} − 4M}}{2} )  (7)

Suppose in addition that p(t) is bounded for t > t_{0}. Then for each sufficiently small ϵ > 0 the solution u(t, t_{0}, ϵ) of (2) such that u(t_{0}, t_{0}, ϵ) = μ and u′(t_{0}, t_{0}, ϵ) = −ϵ is positive in [t_{0}, +∞) and

lim_{t→+∞} u(t, t_{0}, ϵ) = 0.

Proof. In case (i) holds, this is only Lemma 2.2. Otherwise note that the solution u(t, t_{0}, ϵ) has no critical points and therefore is strictly decreasing. It cannot remain above a positive constant by the argument used at the end of the proof of Lemma 2.2. Let t_{1} be such that u(t_{1}, t_{0}, ϵ) = ν. The equation itself shows that u′(t_{1}, t_{0}, ϵ) ≥ −N/c (consider separately the cases where t_{1} lies in an interval of convexity or of concavity of the solution) and therefore Lemma 2.2 can be applied.

Remark. According to the remark after Lemma 2.1 it is obvious, via the same arguments, that the Proposition holds even if ϵ = 0 except in case f(μ) = 0.

Theorem 3.2 Assume (H1)-(H2) with n = 1 and, in addition to the hypotheses of proposition 3.1 with μ = a, that p(t) is bounded. Then (2_{1}) has a strictly decreasing heteroclinic solution connecting a and 0.

Proof. With respect to μ = a in Proposition 3.1 take a sequence t_{m} decreasing to −∞ and consider the solution u(., t_{1}, ϵ_{1}) where ϵ_{1} is a small positive number. According to proposition 3.1, 0 < u(t, t_{1}, ϵ_{1}) ≤ a for t ≥ t_{1} and there exists t̄ such that u(t̄, t_{1}, ϵ_{1}) = a/2. Claim: There exists m_{2} > 1 such that

u(t̄, t_{m_{2}}, ϵ_{1}) < a/2.

Proof of the Claim: Otherwise we would have u(t̄, t_{m}, ϵ_{1}) ≥ a/2 for all m > 1 and, by a t_{m} translation, this can be written

u_{m}(t̄− t_{m}) ≥ a/2  (8)

in terms of the solution of

u′′_{m} + p_{m}(t)u′_{m} + f(u_{m}) = 0, u_{m}(0) = a, u′_{m}(0) = −ϵ_{1} ,

where p_{m}(t) = p(t + t_{m}). The boundedness of p_{m}, u_{m} and u′_{m} and Ascoli’s theorem enable us, by extracting subsequences and a diagonal procedure, to suppose that (where we set d := sup_{t∈ℝ} p(t))

p_{m} → p_{∞} in L^{∞}weak-*, c ≤ p_{∞}(t) ≤ d
u_{m} → u in C^{1}(K), K any compact interval in [0, +∞).

Since

u′′ + p_{∞}(t)u′ + f(u) = 0, u(0) = 1, u′(0) = −ϵ_{1}

(and it is easy to see that proposition 3.1 still applies to solutions in the Carathéodory sense) there exists t̃ such that u(t̃) = a/4. Since u_{m} → u uniformly in [0, t̃] and t̄− t_{m} → +∞ this contradicts (8) and so the Claim holds.

To go on with the proof we observe that if δ > 0 is sufficiently small we have u(t̄, t_{m_{2}}, δ) > a/2, since u(., t_{m_{2}}, δ) → a as δ → 0^{+} in [t_{m_{2}}, t̄]. By the intermediate value theorem we can pick up 0 < ϵ_{2} < ϵ_{1} such that u(t̄, t_{m_{2}}, ϵ_{2}) = a/2. This argument can be iterated so as to construct decreasing sequences τ_{k} = t_{m_{k}} and ϵ_{k} with the property that u(t̄, τ_{k}, ϵ_{k}) = a/2.

Using again the boundedness of u(., τ_{k}, ϵ_{k}) and u′(., τ_{k}, ϵ_{k}) and the diagonal procedure we can pass to a subsequence (which for convenience is denoted by the same symbol) so that for any compact interval K ⊂ ℝ,

u(., τ_{k}, ϵ_{k}) → u in C^{1}(K).

The limit function u thus obtained is, of course, a decreasing solution to (1), such that u(t̄) = a/2 and 0 < u(t) < a ∀t ∈ ℝ (by the uniqueness theorem for the initial-value problem u cannot take the values 0 or a). Finally we can repeat the argument used in the proof of Lemma 2.2 to conclude that lim_{t→−∞} u(t) = a, lim_{t→+∞} u(t) = 0 and lim_{t→±∞} u′(t) = 0.

Remarks. 1) It can be shown, using an argument similar to the proof of the Claim, that ϵ_{k} → 0.

2) See [8] to see how in some instances (with g(0) = 0) assumption (ii) is an improvement over (i).

We now turn to the study of system (2).

Theorem 3.3 Assume (H1)-(H2), the functions p_{i} are bounded and

c_{i} ≥ 2\sqrt{M}_{i}, M_{i} := sup_{(0,a_{1})×···×(0,a_{n})} g_{i}(u); i = 1, …, n.

Then (2) has a heteroclinic solution, whose components are strictly decreasing, connecting a = (a_{1}, ⋯ a_{n}) and 0.

Proof. As in the previous proof, we start by considering solutions u(., t_{0}, ϵ) to (2) such that u(t_{0}, t_{0}, ϵ) = a and u′_{i}(t_{0}, t_{0}, ϵ) = −ϵ, i = 1, ⋯ n. Using Proposition 3.1, one easily sees that for ϵ > 0 sufficiently small such solutions are defined in [t_{0}, ∞), their components being strictly decreasing and vanishing at +∞. Now take a sequence t_{m} → −∞ and u(., t_{1}, ϵ_{1}) where ϵ_{1} is small. Selecting the first component, u_{1}, we easily establish, as in the proof of theorem 3.2, that there exists t̄ and subsequences τ_{k} = t_{m_{k}} → −∞, ϵ_{k} → 0^{+}) so that

u_{1}(t̄, τ_{k}, ϵ_{k}) = a_{1}/2  (9)

(it is sufficient to argue as in the proof of theorem 3.2 with respect to the equation for the first component).

Next consider the sequence u_{2}(., τ_{k}, ϵ_{k}) and let s_{k} be numbers such that

u_{2}(s_{k}, τ_{k}, ϵ_{k}) = a_{2}/2.

We claim that the sequence s_{k}−t̄is bounded: for suppose for instance that along a subsequence s_{k} − t̄ → +∞ (the case s_{k} − t̄ → −∞ is analogous); integrating the second equation of the system (2) in [t̄, s_{k}] we obtain

u′_{2}(s_{k}, τ_{k}, ϵ_{k}) − u′_{2}(t̄, τ_{k}, ϵ_{k}) + p_{2}(t_{k}^{∗})(u_{2}(s_{k}, τ_{k}, ϵ_{k}) − u_{2}(t̄, τ_{k}, ϵ_{k}))

+ ∫ s_{k} ; t̄ u_{2}(t, τ_{k}, ϵ_{k})g_{2}(u(t, τ_{k}, ϵ_{k})) dt = 0 ,

where t_{k}^{∗} ∈ [t̄, s_{k}]. Now the first factor in the integrand is greater than a_{2}/2; using (H1) we see that the second is bounded away from zero (because u_{1} takes values < a_{1}/2 while u_{2} takes values > a_{2}/2); therefore we have reached a contradiction.

Since this argument can be repeated with respect to the remaining components, along with (9) we construct sequences s_{k}^{(j)}, j = 2, ⋯, n such that

u_{j}(s_{k}^{(j)}, τ_{k}, ϵ_{k}) = a_{j}/2  (10)

and s_{k}^{(j)} − t̄ is bounded. Now, as in theorem 3.2 we go to the limit through a diagonal subsequence: u(., τ_{k}, ϵ_{k}) → v uniformly in compact intervals, and v is a solution of (2) with decreasing components. Moreover we may assume that s_{k}^{(j)} → t_{j}, j = 2, ⋯, n and therefore on account of (9)-(10) we obtain

v_{1}(t̄) = a_{1}/2, v_{j}(t_{j}) = a_{j}/2, j = 2, ⋯ n.  (11)

We assert that 0 < v_{i}(t) < a_{i} ∀t ∈ ℝ, and in particular v′_{i}(t) < 0 ∀t ∈ ℝ, i = 1, …, n. Indeed suppose for instance that v_{1}(t) = a_{1} for t ≤ t^{∗}. Then the first equation of (2) implies that g_{1}(a_{1}, v_{2}(t), ⋯, v_{n}(t)) = 0 if t ≤ t^{∗}, so that by (H1) we have v_{j}(t) = a_{j} for t ≤ t^{∗} and j = 2, ⋯, n. By uniqueness of solutions of the Cauchy problem, it follows that v ≡ a, contradicting (11). An easier argument shows that v_{j} cannot take the value 0. Then, arguing as in the proof of lemma 2.2 one concludes that v(−∞) = a and v(+∞) = 0. Finally we illustrate the proof of the fact that v′(±∞) = 0 by showing that v′_{1}(−∞) = 0. If this were not the case we could select sequences t_{k} < s_{k} → −∞ and a number δ > 0 with v′_{1}(t_{k}) → 0 and |v′_{1}(s_{k})| ≥ δ. Multiplying the first equation in (2) by v′_{1} and integrating yields

\frac{1}{2}[v_{1}^{′2}(s_{k}) − v_{1}^{′2}(t_{k})] + ∫_{t_{k}}^{s_{k}} p_{1}v_{1}^{′2} + ∫_{t_{k}}^{s_{k}} f_{1}(v)v′_{1} = 0,

where, by the mean value theorem and what has been already proved, the last summand tends to 0 as k → ∞; this is a contradiction and the proof is complete.

Remark. As in theorem 3.2, we could use a set of conditions like (7) to improve the lower bounds on c_{i}, i = 1, …, n in case the functions g_{i} approach 0 as u_{i} → 0.

4 Positive solutions vanishing at the endpoints of an unbounded interval

In this section we consider f satisfying

(H3) For each i ∈ 1, ⋯ n, f_{i} : R_{+}^{n} → ℝ_{+} is a locally Lipschitz continuous function such that f_{i}(0) = 0 and f_{i}(u) > 0 if u_{i} > 0.

We consider the problem of finding nontrivial positive solutions to

u′′_{i} + p_{i}(t)u′_{i} + f_{i}(u) = 0, i = 1, …, n (12) u(0) = 0 = u(+∞) (13)

where by nontrivial we mean that each component u_{i} of such solution is positive in (0, +∞). For definiteness the initial endpoint is taken to be t_{0} = 0, but our results can obviously be restated with an arbitrary left endpoint.

Before stating the result we note the following fact: denote by u(·, A) the solution of the Cauchy problem

u′′_{i} + p_{i}(t)u′_{i} + f_{i}(u) = 0, u_{i}(0) = 0, u′_{i}(0) = A;

then u_{i}(·, A) has, for every A > 0, a maximum μ_{i}(A) depending continuously on A and μ_{i}(0^{+}) = 0, μ_{i}(+∞) = +∞. To see this, take t^{∗} > 0 in a neighborhood of 0, A_{i}^{∗} = u′_{i}(t^{∗}, A) > 0, u_{i}^{∗} = u_{i}(t^{∗}, A) > 0. Let K_{i} be the least upper bound of the (scalar) solution of

z′′ + p_{i}(t)z′ = 0, z(t^{∗}) = u_{i}^{∗}, z′(t^{∗}) = A_{i}^{∗},

and define δ_{i} := inf{g_{i}(x) : u_{i}^{∗} ≤ x_{i} ≤ K_{i}; x_{j} ≤ K_{j} if j ≠ i} > 0. Comparing u_{i}(·, A) with the solution to

v′′ + p_{i}(t)v′ + δ_{i}v = 0, v(t^{∗}) = u_{i}^{∗}, v′(t^{∗}) = A_{i}^{∗},

it is easy to see, using Lemma 2.1, that (since v ≤ z) u_{i}(t, A) ≤ v(t) as long as u_{i}(t, A) > u_{i}^{∗}. But the behavior of v(t) implies that v(t) returns to the value u_{i}^{∗} and therefore u_{i}(t, A) attains a maximum. The assertion about μ_{i}(0^{+}) comes from the fact that K_{i} → 0 as A → 0^{+}. The other assertion is straightforward.

Proposition 4.1 Assume (H2)-(H3). Assume p is bounded in [0, ∞) and let c_{i} := inf_{t≥0} p_{i}(t). Given positive numbers α_{i}, i = 1, ⋯, n such that M_{i} :=

sup{g_{i}(u) : 0 < u_{i} < α_{i}} ≤ c_{i}^{2}/4, then there exists a number A_{0} > 0 such that whenever 0 < A ≤ A_{0} u(·, A) is a nontrivial solution of (12)-(13).

Proof. It suffices to define A_{0} := sup{A > 0 : (μ_{1}(A), ⋯, μ_{n}(A)) ∈ (0, α_{1}] × ⋯ × (0, α_{n}]}. Then if 0 < A ≤ A_{0} and u_{i}(t_{i}^{∗}, A) = μ_{i} it is easy to conclude, using lemma 2.2 and the remark after it, that each component u_{i}(·, A) remains positive in [t_{i}^{∗}, +∞) and vanishes at +∞.

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Luis Sanchez Universidade de Lisboa, Faculdade de Ciências Centro de Matemática e Aplicações Fundamentais Avenida Professor Gama Pinto, 2 1649-003 Lisboa, Portugal e-mail address: sanchez@lmc.fc.ul.pt