Asymptotic formulas of the eigenvalues for the linearization of a one-dimensional sinh-Poisson equation

We are concerned with a Neumann problem of a one-dimensional sinh-Poisson equation u′′+λsinhu=0for0<x<1,u′(0)=u′(1)=0,\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} {\left\{ \begin{array}{ll} u''+\lambda \sinh u=0 &{} \text {for}\ 0<x<1,\\ u'(0)=u'(1)=0, \end{array}\right. } \end{aligned}$$\end{document}where λ>0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\lambda >0$$\end{document} is a parameter. A complete bifurcation diagram of this problem is obtained. We also consider the linearized eigenvalue problem at every nontrivial solution u. We derive exact expressions of all the eigenvalues and eigenfunctions, using Jacobi elliptic functions and complete elliptic integrals. Then, we also derive asymptotic formulas of eigenvalues as λ→0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\lambda \rightarrow 0$$\end{document}. Exact eigenvalues and eigenfunctions for a Dirichlet problem are presented without proof. The main technical tool is an ODE technique.


Introduction and main results
We are concerned with a Neumann problem of a sinh-Poisson equation in an interval u + λ sinh u = 0 for 0 < x < 1, u (0) = u (1) = 0, (1.1) where λ > 0 is a parameter.The two-dimensional sinh-Poisson equation appears as a vorticity equation in hydrodynamics [8] and it is also related to a constant mean curvature surface in geometry [18].The Eq. (1.2) has received a considerable attention in recent years.See e.g.[2,4,5,9,10,17] for previous studies.The solutions of (1.1) give one-dimensional solutions of a Neumann problem in a rectangle and of a problem in a two-dimensional flat torus.Though they are special solutions of (1.2), a solution structure of (1.1) and spectral properties for linearization problems will be studied in details in this paper.A one-dimensional Dirichlet problem is also considered in Sect. 5.
In this paper we extensively use the Jacobi elliptic functions sn(x, k), cn(x, k) and dn(x, k) and the complete elliptic integrals of the first kind K (k), the second kind E(k) and the third kind (ν, k).We recall definitions and basic properties in Appendix.
The first main result is about the exact solutions of (1.1) and a complete bifurcation diagram of (1.1).Let R + := {x; x > 0} and X := C 2 (0, 1) ∩ C 1 [0, 1].Theorem 1.1 (Exact solutions) Let S ⊂ R + × X denote the solution set of (1.1).Then, where (1.6)  The curve C ± n is a graph of λ and it is defined on 0 < λ < λ n .In particular, the curve C ± n bifurcates from (λ n , 0) on the trivial branch C 0 and the bifurcation is a subcritical pitchfork type.If 0 < λ < λ n for n ≥ 1, then (1.1) has solutions u ± j for every j ≥ n.
We call u ± n n-mode solutions.Figure 1 shows a bifurcation diagram given by (1.3).Let (λ, u) be a nontrivial solution of (1.1).It follows from Theorem 1.1 that (λ, u) ∈ S\C 0 and u = u + n or u − n for some n ∈ {1, 2, . ..}.We consider the following linearized eigenvalue problem of (1.1) at u ϕ + λ(cosh u)ϕ = −μϕ for 0 < x < 1, ϕ (0) = ϕ (1) = 0. (1.9) Since cosh u + n (x, k) = cosh u − n (x, k), the eigenvalues for u + n are equal to those for u − n .Hence, {μ n j (k)} ∞ j=0 denotes the eigenvalues for both (λ n (k), u + n (x, k)) and (λ n (k), u − n (x, k)).Moreover, eigenfunctions for both are the same, and hence ϕ n j (x, k) denotes an eigenfunction for both In a general theory of functional analysis the spectrum for (1.9) consists only of real eigenvalues and the set of corresponding eigenfunctions is a complete orthogonal set in L 2 ([0, 1]).Though it is not easy to obtain exact expressions for general differential operators, in the case of (1.9) exact eigenvalues and eigenfunctions are obtained as follows: Theorem 1.2 (Exact eigenvalues and eigenfunctions) Let N 0 := {0, 1, 2, . ..}.The following hold: (i) The 0-th and n-th eigenvalues and eigenfunctions of (1.9) are given as follows: (1.10) and K (k) and (ν, k) are the complete elliptic integrals of the first and third kinds defined by Moreover, the Morse index of u, which is the number of negative eigenvalues, is n + 1 and u is nondegenerate, i.e., 0 is not an eigenvalue.(Fig. 2 shows a graph of A(μ, k).) (iii) [Eigenfunctions] For j ∈ N 0 \ {0, n}, a j-th eigenfunction ϕ n j (x, k) is given by (1.13) ) can be formally obtained by substituting j = 0 into (1.13).If we let the integral in (1.13) be 0 for j = n, then ϕ n n (x, k) can be also formally obtained from (1.13).
(ii) The eigenfunction (1.13) can be also written as follows: where In the studies of (1.2) a considerable attention has been attracted to the case λ → 0. The third main result is about asymptotic formulas of eigenvalues for (1.9) as k → 1.Note that λ → 0 as k → 1 because of (1.6) and Lemma A.3.

Theorem 1.4 (Asymptotic formulas)
The following asymptotic formulas hold: In particular, Remark 1.5 By Theorem 1.2 (i) we have the exact eigenvalues μ n 0 (k) and μ n n (k).We can check that, for j = 0, n, In Sect. 5 exact eigenvalues and eigenfunctions for a Dirichlet problem are presented without proof.
Let us mention technical details.A theory of exact expressions of eigenvalues and eigenfunctions associated to the linearization for one-dimensional elliptic problems was constructed in Wakasa-Yotsutani [11].This theory can be applied to a general nonlinear term f (u).We recall the theory in Sect. 3 of the present paper.In the theory there is a key ODE (3.6) which is a linear ODE of third order with variable coefficients.In general it is difficult to obtain a solution h(u) of the key ODE (3.6).However, in the case of our nonlinearity f (u) = sinh u we can find a solution h(u) which is given by (3.14).In this case all the eigenvalues and eigenfunctions can be written in terms of h(u).Specifically, the following function becomes an eigenfunction: for a certain constant ρ.
Since h(u ± n (x, k)) and ρ depend on μ, the LHS of the equation becomes a function in μ.A solution of the above equation gives the eigenvalue μ n j (k) for j ∈ N 0 \ {0, n}.Two cases j = 0, n are exceptional and eigenvalues and eigenfunctions are given in a different way.Since u ± n (x, k) can be also written explicitly, we have exact expressions of all the eigenvalues and eigenfunctions.This method was applied to (1) a Neumann problem of ε 2 u + sin u = 0 in [11,13], (2) a Dirichlet problem of ε 2 u + sin u = 0 in [12], (3) a Neumann problem of the Allen-Cahn equation ε 2 u + u − u 3 = 0 in [14,15], (4) a Neumann problem of the scalar field equation ε 2 u − u + u 3 = 0 in [6], (5) Dirichlet problems of u + λe ±u = 0 in [7], and it is applied to (6) a Neumann (and Dicrichlet) problem of u + λ sinh u = 0 in the present paper.
Another interesting technical aspect of this paper is the modified complete elliptic integral of the third kind M(ν, k) defined by (1.12).The function M(ν, k) was introduced in [16].In Theorem 1.2 and its proof we see that M(ν, k) plays a central role in the study of exact eigenvalues.It appears not only in our case (6) above but also in the cases (1), ( 2), (3) and (4) above, and hence M(ν, k) might be universal in some sense.Figure 3 in Appendix shows a graph of M(ν, k) in ν.We will recall some limit formulas for M(ν, k).
This paper consists of six sections.In Sect. 2 we obtain the exact solutions (λ, u) of (1.1).We show that dλ n (k)/dk < 0 for k ∈ (0, 1) and that 0 < λ n (k) < λ n for k ∈ (0, 1).Hence, C ± n is a graph of λ defined on 0 < λ < λ n .We prove Theorem 1.1.In Sect. 3 we recall the theory for exact eigenvalues and eigenfunctions which was developed in [11].In Sect. 4 we prove Theorems 1.2 and 1.4.Specifically, two exceptional eigenvalues and eigenfunctions are obtained in Lemma 4.1.We study a domain of A(μ, k), a monotonicity of A(μ, k) in μ and certain limits of A(μ, k).Using these properties of A(μ, k), we prove Theorem 1.2.Next, asymptotic formulas in Theorem 1.4 are proved.We will see that several limit formulas of complete elliptic integrals, which we recall in Appendix, are used in the proof.In Sect. 5 exact eigenvalues and eigenfunctions for a Dirichlet problem are presented without proof.Section A is an Appendix.We recall definitions and basic properties of sn(x, k), cn(x, k), dn(x, k), K (k), E(k) and (ν, k).

Proof of Theorem 1.1
Using a phase plane analysis, we obtain exact solutions of (1.1) and a complete bifurcation diagram of (1.1).

Lemma 2.1 All the nontrivial solutions of (1.1) can be expressed as
for some n ∈ {1, 2, . ..} and some k ∈ (0, 1), (2.1) where λ n (k) and u ± n (x, k) are defined by (1.6) and (1.7), respectively.Proof Let us consider the initial value problem with (u(0), v(0)) = (α, 0).We can easily see that (2.2) has only one equilibrium (u, v) = (0, 0) and that the other orbits are periodic.Since sinh(−u) = − sinh(u), each periodic orbit is symmetric with respect to both the u and v axes.Hence, we can define a quarter period of each periodic orbit.A solution of (1.1) corresponds to an orbit of (2.2) that starts from (α, 0) and reaches a point on the u-axis, which is (−α, 0) or (α, 0), at x = 1.Here, the orbit may rotate finitely many times around the origin.In summary, the orbit reaches the v-axis for the first time at x = 1/2n for some n ∈ {1, 2, . ..}, i.e., In other words the quarter period is 1/2n.If (2.3) holds, then u(x) is an n-mode solution.
Proof Because of Lemma A.2, we have

.11)
By Lemma A.1 (i) and (2.11) we have Proof of Theorem 1.1 Let λ n (k) and u ± n (x, k) be defined by (1.6) and (1.7), respectively.Let C 0 and C ± n , n = 1, 2, . .., be defined by (1.4) and (1.5), respectively.First, every element of C 0 is a trivial solution of (1.1).Next, we study nontrivial solutions.By Lemma 2.1, the solution set S can be written as (1.3).What we have to do is to prove a monotonicity of each nontrivial branch C ± n , n ≥ 1, which indicates that C ± n can be described as a graph of λ for a certain interval.It is clear that where λ n is defined by (1.8).By Lemma A.3 we have Because of Lemma 2.2, λ n (k) is a monotone function, and hence C ± n can be described as a graph of λ for 0 < λ < λ n .All the other statements in Theorem 1.1 follow from this result.The proof is complete.
3 Preliminaries for proofs of Theorems 1.2 and 1.4

Sturm-Liouville theory
Let f be an arbitrary C 1 -function.We consider the Neumann problem Let (λ, u) denote a solution of (3.1), where λ > 0. We define The linearized eigenvalue problem at (λ, u) becomes We recall basic properties about eigenvalues and eigenfunctions for (3.3).
Proposition 3.1 Let {μ j } ∞ j=0 denote the eigenvalues of (3.3), and let ϕ j (x), ϕ j (x) ≡ 0, be an eigenfunction corresponding to μ j .Then the following hold: (i) Each eigenvalue μ j is real and simple, and See [3, Theorem 2.1 in p.212] for details of a proof of Proposition 3.1.We omit the proof.

General nonlinearity
First, we look for an eigenfunction of the form ϕ We define Then, We consider the equation If φ(x) satisfies (3.4), then (x) = 0, and hence (x) is constant.In particular, 3) and it is a candidate of the eigenfunction.We show that almost all the eigenfunctions can be constructed even if (0) = 0. We look for a solution of (3.4) of the form φ(x) = h(u(x)), where h(u(x)) > 0 for 0 ≤ x ≤ 1.Here, h( • ) is an unknown function.In other words we construct an eigenfunction, using the shape of a solution u(x).Substituting h(u(x)) into (3.4),we see that h(u) satisfies the following key equation: where F(u) := u 0 f (s)ds and we use the following relations: It is worth noting that the independent variable of (3.6) becomes u and that the explicit x-dependence disappears.Thus, (3.6) does not depend on each solution u of (3.1).Substituting h(u(x)) into (3.5),we have and where we use u(0) = α and u (0) = 0. Hereafter, we construct an eigenfunction for (3.3).We assume that a solution h(u) of (3.6) that satisfies h(u(x)) > 0 for 0 ≤ x ≤ 1 is obtained.Then, we look for an eigenfunction of the form (3.9) The functions W (θ ) and θ(x) are defined later.Using (3.7), we have We assume that ρ > 0, where ρ is defined by (3.8).We consider the case where Since h(u(x)) is positive, θ(x) can be expressed as for some θ 0 .
Two cases j = 0, n are exceptional in our case f (u) = sinh u.

Proof of Theorems 1.2 and 1.4
For simplicity we write λ and u for λ n (k) and u ± n (x, k), respectively.In this section we study an exact expression of eigenvalues of (1.9) associated with an n-mode solution u.Let us recall α := u(0) defined by (3.2).A family of solutions (λ, u) can be parametrized by k, and it can be also parametrized by α.We use both k and α.Note that k and α are related by (2.7).

Lemma 4.1 The following hold:
(i) The 0-th eigenvalue μ n 0 (k) and a corresponding eigenfunction ϕ n 0 are given by (ii) The n-th eigenvalue μ n n (k) and a corresponding eigenfunction ϕ n n are given by we have where we use cosh u cosh u 2 − sinh u sinh u 2 = cosh u 2 .Since we have ϕ + λ(cosh u)ϕ + μϕ = 0. We easily see that ϕ (0) = ϕ (1) = 0. Since ϕ(x) > 0 for 0 ≤ x ≤ 1, by Proposition 3.1 (ii) we see that μ is the 0-th eigenvalue and that ϕ is a 0-th eigenfunction.We have The proof of (i) is complete.
Proof of Theorem 1.2 (i) The assertion follows from Lemma 4.1.

Asymptotic formulas
Proof of Theorem 1.4 (i) We consider the case j ∈ {1, 2, . . ., n − 1}.We define Then, μ n j (k) can be written as Because of (4.17), we see that a unique solution μ n j (k) of (4.12) satisfies μ n 0 (k) < μ n j (k) < μ n n (k).Since μ n 0 (k) < μ n j (k) < μ n n (k), we see that 0 < μ j < 4n 2 , and hence there is a sequence k i , which is denoted by k for simplicity, and μ * By Lemma A.6 we have The proof of (i) is complete.(ii) We consider the case j ∈ {n + 1, n + 2, . ..}.Because of (4.18), we see that a unique solution μ n j (k) satisfies μ n j (k) > 0. We have We define Since ν → ∞ as k → 1, by Lemma A.5 we have Thus, Here, the limit μ * j does not depend on a sequence k → 1.Since μ n j (k) = μ * j +o(1) as k → 1, we have The proof of (ii) is complete.