Precise Asymptotics for Bifurcation Curve of Nonlinear Ordinary Differential Equation

where λ > 0 is a parameter. In this paper, we mainly consider the case f (u) = log(1 + u). We know from [1] that, if f (u) is continuous in u ≥ 0 and positive for u > 0, then for a given α > 0, there exists a unique classical solution pair (λ, uα) of (1)–(3) satisfying α = ‖uα‖∞. Since λ is a continuous for α > 0, we write as λ = λ(α) for α > 0. Nonlinear eigenvalue and bifurcation problems have been one of the main topics in the study of nonlinear equations, and many authors have investigated the global behavior of bifurcation diagrams intensively. We refer to [2–7] and the references therein. The purpose of this paper is to establish the asymptotic expansion formula for λ(α) when α 1 up to the third term by time-map method. Our study is motivated by the following inverse bifurcation problem.

The purpose of this paper is to establish the asymptotic expansion formula for λ(α) when α 1 up to the third term by time-map method. Our study is motivated by the following inverse bifurcation problem.
We introduce some known results about global behavior of bifurcation curve. One of the most famous results was shown for the one-dimensional Gelfand problem, namely, the Equations (1)- (3) with f (u) = e u . It was shown in [9] that it has the exact solution u α (t) = α + log sech 2 2λ(α) 2 where sech x = 1/ cosh x. Then by time-map method, we explicitly obtain that for α > 0 (cf. [10]), Unfortunately, however, such explicit expression of bifurcation curves as (5) cannot be expected in general. For example, we introduce the following result.
As far as the author knows, however, the exact solution u α (t) of the equation in Theorem 1 is not known, although it is quite simple, and it also seems impossible to obtain the explicit formula for λ(α) as (5). From this point of view, one of the standard approaches for the better understanding of the global structure of λ(α) is to establish precise asymptotic expansion formula for λ(α) as α → ∞. In some cases, the asymptotic expansion formulas for λ(α) up to the second term like (6) have been obtained. We refer to [11,[14][15][16] and the references therein. However, to obtain the third term of λ(α), we need a very long and complicated calculation in general. We overcome this difficulty and establish the following results.

Remark 1.
From a view point of asymptotic expansion formula for λ(α), it is natural to expect that the following asymptotic formula for λ(α) holds.
Here, {b n } (n = 1, 2, · · · ) are expected to be constants, which are determined by induction. However, if the readers look at Section 3 below, then they understand immediately that it seems quite difficult to prove (9), since the calculation are quite long to obtain even the third term of (7).
The proof of Theorem 2 depends on the time-map method and Taylor expansion formula.

Second Term of λ(α) in Theorem 2
In this section, let α 1. In what follows, we denote by C the various positive constants independent of α. It is known that if (u α , λ(α)) ∈ C 2 (Ī) × R + satisfies (1)-(3), then By (1), we have By this, (11) and putting t = 0, we obtain This along with (12) implies that for −1 ≤ t ≤ 0, where By this and putting u α (t) = αs 2 , we obtain For 0 ≤ s ≤ 1, we have By this, (15) and Taylor expansion, we obtain We see from (20) that the second term of λ(α) in Theorem 2 follows from Lemma 1 below.
The proof of Lemma 1 is a conclusion of Lemmas 2 and 3 below. By (17), we have where (24) Proof. We put s = sin θ in (23). Then by integration by parts, Then by l'Hôpital's rule, we obtain By this, we see that Q 1 = 0. We next calculate Q 2 . For 0 ≤ s ≤ 1, by Taylor expansion, we obtain Then by direct calculation, we obtain the following (32) and (33).

The Third Term of λ(α) in Theorem 2
By (20), to obtain the third term of λ(α), we calculate We have We put Then M = I 1 + I 2 + I 3 .

By integration by parts and l'Hôpital's rule, we obtain
By this and (65), we obtain (63). Thus the proof is complete.

By integration by parts, we obtain
:= J 320 + J 321 .