Fifth Mississippi State Conference on Differential Equations and Computational Simulations, Electronic Journal of Differential Equations, Conference 10, 2003, pp 71–78. http://ejde.math.swt.edu or http://ejde.math.unt.edu ftp ejde.math.swt.edu (login: ftp)

Nonlinear initial-value problems with positive global solutions ^{∗}

John V. Baxley & Cynthia G. Enloe

Abstract We give conditions on m(t), p(t), and f(t, y, z) so that the nonlinear

initial-value problem

\frac{1}{m(t)} (p(t)y′)′ + f(t, y, p(t)y′) = 0, for t > 0,

y(0) = 0, lim_{t→0^{+}} p(t)y′(t) = B,

has at least one positive solution for all t > 0, when B is a sufficiently small positive constant. We allow a singularity at t = 0 so the solution y′(t) may be unbounded near t = 0.

1 Introduction

We consider the initial-value problem

\frac{1}{m(t)}(p(t)y′)′ + f(t, y, p(t)y′) = 0, t > 0,  (1.1)
y(0) = 0, lim_{t→0^{+}} p(t)y′(t) = B, B > 0.  (1.2)

We allow a singularity at t = 0, and so y′(t) may not be bounded near t = 0. However, we require of a solution that it be continuous at t = 0, satisfy (1.1) a.e. on some interval (0, δ), and satisfy (1.2). The singularity may be caused by the behavior of m or p or f near t = 0 or by some combination of them.

This problem was considered earlier by Zhao [5] and by Maagli and Masmoudi [4]. In particular, [5] considered the case that m ≡ p ≡ 1 while [4] required that m ≡ p. In each of these papers, only one of the initial conditions (y(0) = 0) was imposed and conditions were specified which guaranteed that this “incomplete” initial-value problem has infinitely many positive solutions existing on the entire interval (0, ∞). Both papers viewed the problem as a boundary-value problem by imposing a condition at ∞, namely

lim_{t→∞} \frac{y(t)}{r(t)} = c > 0,  (1.3)

∗Mathematics Subject Classifications: 34A12, 34B15.

Key words: Nonlinear initial-value problems, positive global solutions, Carathéodory.

\begin{matrix}c \\ ◯ \end{matrix}2003 Southwest Texas State University.

Published February 28, 2003.

where r(t) = ∫_{0}^{t}(p(s))^{−1} ds. In [5], r(t) reduces to r(t) = t. In both [4] and [5], the Schauder fixed point theorem is the main tool and the hypotheses imposed allow the authors to prove existence of at least one solution of the boundaryvalue problem for c sufficiently small.

Here, we shall treat the problem in the initial-value form (1.1), (1.2). We shall impose conditions rather close to those of [4] and [5], and prove that our initial-value problem has at least one positive solution for B sufficiently small. Our methods use initial-value techniques, similar to those used already in [1, 2], and are completely different from the previous papers discussed above. To get started, we must have a local solution on some interval (0, δ) and for that purpose, we need a slight generalization of the classical theorem of Carathéodory [3], which we provide in Section 2.

In Section 3, we prove our main result, which we state below. Let r(t) = ∫_{0}^{t}(p(s))^{−1} ds and assume that

M1: p(t) and m(t) are positive and continuous on (0, ∞);

M2: \frac{1}{p(t)} ∈ L^{1}(0, 1);

M3: for some positive number, D < ∞,

f : (0, ∞) × (0, Dr(∞)) × (0, D) → ℝ

is a measurable function on (0, ∞) × (0, Dr(∞)) × (0, D) and f(t, ·, ·) is continuous on (0, Dr(∞)) × (0, D) for each fixed t ∈ (0, ∞);

M4:

|f(t, y, z)| ≤ h_{1}(t, y, z)y + h_{2}(t, y, z)z

where h_{1}(t, y, z) → 0 and h_{2}(t, y, z) → 0 as (y, z) → (0, 0), h_{1} and h_{2} are

nonnegative, and for α > 0, let h(t, y, z) = h_{1}(t, y, z)r(t) + h_{2}(t, y, z),
g_{α}(s) = sup{h(s, y, z) : 0 < y < αr(s), 0 < z < α}, s > 0,
and m(s)g_{α}(s) ∈ L^{1}(0, ∞) for sufficiently small α > 0.

Theorem 1.1 Under assumptions M1–M4, there exists γ > 0 so that B ∈ (0, γ) implies that the initial-value problem (1.1), (1.2) has at least one solution existing for 0 < t < ∞ and satisfying

\frac{B}{2} < p(t)y′(t) < \frac{3B}{2} ,

\frac{Br(t)}{2} < y(t) < \frac{3Br(t)}{2},

for 0 < t < ∞. Moreover, the two limits

lim_{t→∞} \frac{y(t)}{r(t)}, lim_{t→∞} p(t)y′(t)

exist, and if r(∞) = ∞, the two limits are equal.

Other than the fact that [4] requires that m ≡ p, the only substantive difference in our hypotheses is that we do not require that h_{1}, h_{2} be nondecreasing with respect to y and z, as they do. Of course, we prove existence for an initial-value problem, not a boundary-value problem as they do.

The key to our proof is that our local existence theorem in Section 2 is formulated carefully to provide a lower bound on the length of the interval of existence. In applying it in Section 3, we show that this lower bound gives us a uniform lower bound on the length of the interval of existence, regardless of where in the interval [0, ∞) we start the solution. Thus, we are able to step from 0 to ∞ inductively, without fear that the sum of the lengths of our intervals will converge, to complete the proof.

2 Local Solutions

In this section, we consider the initial-value problem

\frac{1}{m(t)}(p(t)y′)′ + f(t, y, p(t)y′) = 0, t > t_{0},  (2.1)
y(t_{0}) = A, lim_{t→t_{0}^{+}} p(t)y′(t) = B.  (2.2)

We use x_{1} = y, x_{2} = p(t)y′ to transform to the two-dimensional system

\begin{matrix}x′_{1} = \frac{x_{2}}{p(t)} \\ x′_{2} = −m(t)f(t, x_{1}, x_{2}) \end{matrix}  (2.3)

with initial conditions

lim_{t→t_{0}^{+}} x_{1}(t) = A, lim_{t→t_{0}^{+}} x_{2}(t) = B.  (2.4)

Let R(t) = ∫_{t_{0}}^{t}(p(s))^{−1} ds. We shall assume that

L1: There exists b > t_{0} such that p(t) and m(t) are positive and continuous on (t_{0}, b).

L2: \frac{1}{p(t)} ∈ L^{1}(t_{0}, b).

L3: f : S → ℝ, where S = {t_{0} < t ≤ b, A + cR(t) < y < A + dR(t), c < z < d}, and f is measurable in t for each fixed (y, z) and continuous in (y, z) for each fixed t.

L4: There exists h(t) ∈ L^{1}(t_{0}, b) such that m(t)|f(t, y, z)| ≤ h(t), almost everywhere on the set S.

We shall prove the following generalization of Carathéodory’s local existence theorem. The proof follows the same general lines as the well-known proof in Theorem 2.1 Suppose hypotheses L1-L4 are satisfied. Let 0 < d^{∗} < min{d − B, B − c} and suppose β ∈ (0, b) and satisfies ∫_{t_{0}}^{t_{0}+β} h(s)ds < d^{∗}. Then, the initial-value problem (2.1), (2.2) has a solution existing on the interval [t_{0}, t_{0}+β] and satisfies

A + cR(t) < x_{1}(t) < A + dR(t)
c < x_{2}(t) < d

for t_{0} < t ≤ t_{0} + β.

Proof: Choose a fixed integer n > 1. Let h_{n} = β/n and let t_{k} = t_{0} + kh_{n} for k = 1, 2, ⋯, n. Define

u_{2,n}(t) = B, for t_{0} ≤ t ≤ t_{1}.

Note that B − d^{∗} < u_{2,n}(t) < B + d^{∗} for t_{0} ≤ t ≤ t_{1}. Also define

u_{1,n}(t) = A + ∫_{t_{0}}^{t} \frac{u_{2,n}(s)}{p(s)} ds, for t_{0} ≤ t ≤ t_{1}.

It follows that

(B − d^{∗})R(t) < u_{1,n}(t) − A < (B + d^{∗})R(t)

and so

A + (B − d^{∗})R(t) < u_{1,n}(t) < A + (B + d^{∗})R(t).

Thus, (t, u_{1,n}(t), u_{2,n}(t)) ∈ S for t_{0} ≤ t ≤ t_{1}.

We extend the pair (u_{1,n}, u_{2,n}) to the entire interval [t_{0}, t_{0} +β] by recursively defining the pair on the subintervals [t_{j−1}, t_{j}]. Thus, for each j = 2, 3, ⋯, n, we define

u_{2,n}(t) = B − ∫_{t_{0}}^{t−h_{n}} m(s)f(s, u_{1,n}(s), u_{2,n}(s)) ds, for t_{j−1} ≤ t ≤ t_{j},

u_{1,n}(t) = A + ∫_{t_{0}}^{t} \frac{u_{2,n}(s)}{p(s)} ds, for t_{j−1} ≤ t ≤ t_{j}.

(The measurability of the integrand in the integral for u_{2,n} follows from L3 by approximating with simple functions.) Using L4, we have

|u_{2,n}(t) − B| ≤ ∫_{t_{0}}^{t−h_{n}} m(s)|f(s, u_{1,n}(s), u_{2,n}(s))|ds

≤ ∫_{t_{0}}^{t_{0}+β} h(s)ds < d^{∗},

and therefore, B − d^{∗} < u_{2,n}(t) < B + d^{∗}. Further,

(B − d^{∗})R(t) < u_{1,n}(t) − A < (B + d^{∗})R(t),

and so

A + (B − d^{∗})R(t) < u_{1,n}(t) < A + (B + d^{∗})R(t).

These inequalities show that (t, u_{1,n}(t), u_{2,n}(t)) remains in S on each subinterval and the recursive definition is allowed. Moreover, the two sequences {u_{1,n}}, {u_{2,n}} are uniformly bounded on t_{0} ≤ t ≤ t_{0} + β.

We shall show that these sequences are equicontinuous so that Ascoli’s theorem may be applied. Suppose t_{0} ≤ t ≤ t^{∗} ≤ t_{0} + β. Then

|u_{1,n}(t) − u_{1,n}(t^{∗})| = | ∫_{t}^{t^{∗}} \frac{u_{2,n}(s)}{p(s)} ds| ≤ ∫_{t}^{t^{∗}} \frac{Q}{p(s)} ds,

where Q = max{|B − d^{∗}|, |B + d^{∗}|}. Moreover,

|u_{2,n}(t) − u_{2,n}(t^{∗})| = | ∫_{t−h_{n}}^{t^{∗}−h_{n}} m(s)f(s, u_{1,n}(s), u_{2,n}(s)) ds|

≤ ∫_{t−h_{n}}^{t^{∗}−h_{n}} h(s) ds.

The desired equicontinuity follows from absolute continuity of the integral. Using Ascoli’s theorem, we may assume without loss of generality that both sequences converge uniformly on [t_{0}, t_{0} + β] to limit functions u_{1}(t), u_{2}(t). We may use the Lebesgue dominated convergence theorem (for u_{2,n} the dominating function is h(s); for u_{1,n}, the dominating function is (p(s))^{−1}) to take limits under each integral sign as n → ∞ to show that

u_{1}(t) = ∫_{t_{0}}^{t} \frac{u_{2}(s)}{p(s)} ds,

u_{2}(t) = B − ∫_{t_{0}}^{t} m(s)f(s, u_{1}(s), u_{2}(s)) ds,

for t_{0} ≤ t ≤ t_{0} + β, from which we obtain

u′_{2}(t) = −m(t)f(t, u_{1}(t), u_{2}(t)),

u′_{1}(t) = \frac{u_{2}(t)}{p(t)},

almost everywhere on [t_{0}, t_{0} + β].

The specific size of β provided by the hypotheses of this last theorem is crucial for our main proof in the next section.

3 Proof of Main Theorem

First note that the hypotheses M1-M4 imply that the earlier hypotheses L1-L4 hold on any interval (t_{0}, b) with 0 ≤ t_{0} < b < ∞, so we may apply Theorem 2.1 as needed.

From hypothesis M4, we have m(s)g_{α_{0}}(s) ∈ L^{1}(0, ∞) if α_{0} > 0 is sufficiently small. Further, M4 implies that m(s)g_{α}(s) ≤ m(s)g_{α_{0}}(s) whenever 0 < α < α_{0} and also that m(s)g_{α}(s) → 0 as α → 0, for all s > 0. Thus, by the Lebesgue Dominated Convergence Theorem,

∫_{0}^{∞} m(t)g_{α}(t)dt → 0 as α → 0.

Hence, there exists δ ∈ (0, α_{0}] such that 0 < α < δ implies

∫_{0}^{∞} m(t)g_{α}(t)dt < \frac{1}{4}.

We shall show that γ = \frac{1}{2} min\{D, δ\}, where D is the number from our hypothesis M3, satisfies the requirements of our theorem. To apply Theorem 2.1, we pick 0 < C ≤ γ, d^{∗} = C/2, c = 0, d = 2C, t_{0} = 0, b = 1, B = C, and A = 0. Note that d^{∗} = \frac{d}{4} < d − d/2 = d − C = d − B and d^{∗} < C = B − c. So d^{∗} < min{d − B, B − c}.

By absolute continuity of the integral, there exists β ∈ (0, b) = (0, 1) so that for k = 0, 1, ⋯,

∫_{kβ}^{(k+1)β} αm(s)g_{α}(s) ds < d^{∗}.  (3.1)

This last inequality, for k = 0 allows us to apply Theorem 2.1 to get a solution y_{1}(t) on [0, β] so that for 0 < t ≤ β,

0 < y_{1}(t) < 2Cr(t), 0 < p(t)y′_{1}(t) < 2C .  (3.2)

Integrating (1.1) from 0 to t and using M4, we obtain

|p(t)y′_{1}(t) − C| ≤ ∫_{0}^{t} m(s)|f(s, y_{1}(s), p(s)y′_{1}(s))| ds
= 2C ∫_{0}^{t} m(s)h(s, y_{1}(s), p(s)y′_{1}(s)) ds
< 2C ∫_{0}^{t} m(s)g_{2C}(s) ds < \frac{C}{2} ,

if t ∈ [0, β]. Hence,

\frac{C}{2} < p(t)y′_{1}(t) < \frac{3C}{2} .

Then

y_{1}(t) = ∫_{0}^{t} \frac{p(s)y′_{1}(s)}{p(s)}ds

and so

\frac{C}{2} r(t) < y_{1}(t) < \frac{3C}{2} r(t).

We claim that for k = 2, 3, …, there exists a solution y_{k}(t) of (1.1) on the interval (k − 1)β ≤ t ≤ kβ so that

y_{k+1}(kβ) = y_{k}(kβ),
y′_{k+1}(kβ) = y′_{k}(kβ),

for k ≥ 1, and

\begin{matrix}\frac{C}{2} r(t) < y_{k}(t) < \frac{3C}{2} r(t), \\ \frac{C}{2} < p(t)y′_{k}(t) < \frac{3C}{2} , \end{matrix}  (3.3)

for (k − 1)β ≤ t ≤ kβ.

Noting that y_{1}(t) has already been constructed, we continue by induction and assume that y_{1}(t), y_{2}(t), …, y_{n}(t) have been constructed. Next, we construct y_{n+1}(t). To use Theorem 2.1, we keep C, d^{∗}, c, d, b, and β as before and let t_{0} = nβ, A = A_{n} = y_{n}(nβ), and B = B_{n} = p(nβ)y′_{n}(nβ). Inequality (3.1) for k = n allows us to apply Theorem 2.1 to get a solution y_{n+1}(t) of (1.1) on [t_{0}, t_{0} + β] = [nβ, (n + 1)β] so that y_{n+1}(nβ) = y_{n}(nβ) and

A_{n} ≤ y_{n+1}(t) < A_{n} + 2C ∫_{t_{0}}^{t} \frac{1}{p(s)}ds,

0 < p(t)y′_{n+1}(t) < 2C.

To complete the induction, we must verify that y_{n+1}(t) satisfies (3.3). Define y(t) for 0 ≤ t ≤ (n + 1)β by y(t) = y_{k}(t) for (k − 1)β ≤ t ≤ kβ.

Since

\frac{C}{2} < p(s)y′(s) < \frac{3C}{2} , for 0 ≤ s ≤ nβ

and

0 < p(s)y′(s) < 2C, for nβ ≤ s ≤ (n + 1)β,

it follows that

0 < p(s)y′(s) < 2C

for the larger interval, 0 ≤ s ≤ (n + 1)β, and it follows by integrating that

0 < y(t) < 2Cr(t), for 0 ≤ t ≤ (n + 1)β.

The calculation appearing just after (3.2) may now be repeated to show that

\frac{C}{2} < p(t)y′(t) < \frac{3C}{2} , for 0 ≤ t ≤ (n + 1)β,

which implies, as before, that

\frac{C}{2} r(t) < y(t) < \frac{3C}{2} r(t), for 0 ≤ t ≤ (n + 1)β.

Finally, we define y(t) for 0 ≤ t < ∞ by y(t) = y_{k}(t) for (k − 1)β ≤ t < kβ, and each k = 1, 2, . . .. Clearly, y(t) is the desired solution.

To investigate the limit of p(t)y′(t) at infinity, we examine

p(t)y′(t) = p(t_{0})y′(t_{0}) + ∫_{t_{0}}^{t}(p(s)y′(s))′ ds
= p(t_{0})y′(t_{0}) − ∫_{t_{0}}^{∞}(m(s)f(s, y(s), p(s)y′(s))X_{[t_{0},t]}(s) ds.

Since (m(s)f(s, y(s), p(s)y′(s))X_{[t_{0},t]}(s) < m(s)g_{2C}(s), and m(s)g_{2C}(s) is in L^{1}(0, ∞), we can use the Lebesgue Dominated Convergence Theorem to take the limit of both sides as t approaches ∞ and conclude that lim_{t→∞} p(t)y′(t) exists. If r(∞) = ∞, then

lim_{t→∞} \frac{y(t)}{r(t)} = lim_{t→∞} \frac{y′(t)}{r′(t)} = lim_{t→∞} p(t)y′(t),

which we have already proven to exist.

If r(∞) < ∞, then y(t) is a monotone increasing function which is bounded above by \frac{3C}{2} r(∞). Hence, lim_{t→∞} y(t)/r(t) exists.

References

[1] Baxley, J. V. Existence and Uniqueness for Nonlinear Boundary Value Problems on Infinite Intervals, J. Math. Anal. Applic. 147 (1990), 122-133.

[2] Baxley, J. V., Boundary Value Problems on Infinite Intervals, in Boundary Value Problems for Functional Differential Equations (J. Henderson, ed.), World Scientific, Singapore, 49-62.

[3] Coddington, E. A. and N. Levinson, Theory of Ordinary Differential Equations, McGraw-Hill, New York, 1955.

[4] Maagli, H. and S. Masmoudi, Existence Theorem of Nonlinear Singular Boundary Value Problem, Nonlinear Analysis 46 (2001), 465-473.

[5] Zhao, Z., Positive Solutions of Nonlinear Second Order Ordinary Differential Equations, Proc. Amer. Math. Soc. 121 (1994), 465-469.

John V. Baxley (email: baxley@mthcsc.wfu.edu) Cynthia G. Enloe Department of Mathematics, Wake Forest University Winston-Salem, NC 27109, USA