Abstract
We are concerned with a Neumann problem of a one-dimensional sinh-Poisson equation
where \(\lambda >0\) 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 \(\lambda \rightarrow 0\). Exact eigenvalues and eigenfunctions for a Dirichlet problem are presented without proof. The main technical tool is an ODE technique.
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1 Introduction and main results
We are concerned with a Neumann problem of a sinh-Poisson equation in an interval
where \(\lambda >0\) 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 \(\textrm{sn}(x,k)\), \(\textrm{cn}(x,k)\) and \(\textrm{dn}(x,k)\) and the complete elliptic integrals of the first kind K(k), the second kind E(k) and the third kind \(\Pi (\nu ,k)\). 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 \(\mathbb {R}_+:=\{x;\ x>0\}\) and \(X:=C^2(0,1)\cap C^1[0,1]\).
Theorem 1.1
(Exact solutions) Let \(\mathcal {S}\subset \mathbb {R}_+\times X\) denote the solution set of (1.1). Then,
where
Let
The curve \(\mathcal {C}^{\pm }_n\) is a graph of \(\lambda \) and it is defined on \(0<\lambda <\lambda _n\). In particular, the curve \(\mathcal {C}^{\pm }_n\) bifurcates from \((\lambda _n,0)\) on the trivial branch \(\mathcal {C}_0\) and the bifurcation is a subcritical pitchfork type. If \(0<\lambda <\lambda _n\) for \(n\ge 1\), then (1.1) has solutions \(u^{\pm }_j\) for every \(j\ge n\).
We call \(u^{\pm }_n\) n-mode solutions. Figure 1 shows a bifurcation diagram given by (1.3).
Let \((\lambda ,u)\) be a nontrivial solution of (1.1). It follows from Theorem 1.1 that \((\lambda ,u)\in \mathcal {S}{\setminus }\mathcal {C}_0\) and \(u=u^+_n\) or \(u^-_n\) for some \(n\in \{1,2,\ldots \}\). We consider the following linearized eigenvalue problem of (1.1) at u
Since \(\cosh u^+_n(x,k)=\cosh u^-_n(x,k)\), the eigenvalues for \(u^+_n\) are equal to those for \(u^-_n\). Hence, \(\{\mu ^n_j(k)\}_{j=0}^{\infty }\) denotes the eigenvalues for both \((\lambda _n(k),u^+_n(x,k))\) and \((\lambda _n(k),u^-_n(x,k))\). Moreover, eigenfunctions for both \((\lambda _n(k),u^+_n(x,k))\) and \((\lambda _n(k),u^-_n(x,k))\) are the same, and hence \(\varphi ^n_j(x,k)\) denotes an eigenfunction for both \((\lambda _n(k),u^+_n(x,k))\) and \((\lambda _n(k),u^-_n(x,k))\).
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 \(\mathbb {N}_0:=\{0,1,2,\ldots \}\). The following hold:
-
(i)
The 0-th and n-th eigenvalues and eigenfunctions of (1.9) are given as follows:
$$\begin{aligned} \mu ^n_0(k)= & {} -4K(k)^2n^2\ \ \text {and}\ \ \varphi ^n_0(x,k)=\textrm{dn}(2nK(k)x,k),\\ \mu ^n_n(k)= & {} -4k^2K(k)^2n^2\ \ \text {and}\ \ \varphi ^n_n(x,k)=\textrm{cn}(2nK(k)x,k). \end{aligned}$$ -
(ii)
[Eigenvalues] For \(j\in \mathbb {N}_0\setminus \{0,n\}\), the j-th eigenvalue \(\mu ^n_j(k)\) is a unique solution of
$$\begin{aligned} \mathcal {A}(\mu ,k)=\frac{j\pi }{2n}. \end{aligned}$$(1.10)Here, \(\mathcal {A}(\mu ,k)\) is defined for \(\mu \in (\mu ^n_0(k),\mu ^n_n(k))\cup (0,\infty )\) and given by
$$\begin{aligned} \mathcal {A}(\mu ,k):= & {} \mathcal {M}(\nu ,k),\ \ \nu :=\frac{4k^2K(k)^2n^2}{\mu }, \end{aligned}$$(1.11)$$\begin{aligned} \mathcal {M}(\nu ,k):= & {} \sqrt{\frac{(\nu +1)(\nu +k^2)}{\nu }}\Pi (\nu ,k), \end{aligned}$$(1.12)and K(k) and \(\Pi (\nu ,k)\) are the complete elliptic integrals of the first and third kinds defined by
$$\begin{aligned} K(k):=\int _0^1\frac{ds}{\sqrt{(1-s^2)(1-k^2s^2)}},\ \ \Pi (\nu ,k):=\int _0^1\frac{ds}{(1+\nu s^2)\sqrt{(1-s^2)(1-k^2s^2)}}. \end{aligned}$$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\((\mu ,k)\).)
-
(iii)
[Eigenfunctions] For \(j\in \mathbb {N}_0\setminus \{0,n\}\), a j-th eigenfunction \(\varphi ^n_j(x,k)\) is given by
$$\begin{aligned} \varphi ^n_j(x,k)= & {} \sqrt{\left| \mu ^n_j(k)-\mu ^n_n(k)\textrm{sn}^2(2nK(k)x,k)\right| }\nonumber \\{} & {} \times \cos \left( \int _0^x\frac{\sqrt{\mu ^n_j(k)(\mu ^n_j(k)-\mu ^n_0(k))(\mu ^n_j(k)-\mu ^n_n(k))}}{\left| \mu ^n_j(k)-\mu ^n_n(k)\textrm{sn}^2(2nK(k)y,k)\right| }dy\right) . \nonumber \\ \end{aligned}$$(1.13)
Remark 1.3
-
(i)
\(\varphi ^n_0(x,k)\) 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 \(\varphi ^n_n(x,k)\) can be also formally obtained from (1.13).
-
(ii)
The eigenfunction (1.13) can be also written as follows:
$$\begin{aligned} \varphi ^n_j(x,k)= & {} C_0\sqrt{\left| \frac{1}{\nu _j}+\textrm{sn}^2(2nK(k)x,k)\right| }\\{} & {} \times \cos \left( \sqrt{\frac{(\nu _j+1)(\nu _j+k^2)}{\nu _j}}\right. \\{} & {} \left. \times \int _0^{\textrm{sn}(2nK(k)x,k)}\frac{ds}{(1+\nu _js^2)\sqrt{(1-s^2)(1-k^2s^2)}}\right) , \end{aligned}$$where \(\nu _j:=4k^2K(k)^2n^2/\mu ^n_j(k)\).
In the studies of (1.2) a considerable attention has been attracted to the case \(\lambda \rightarrow 0\). The third main result is about asymptotic formulas of eigenvalues for (1.9) as \(k\rightarrow 1\). Note that \(\lambda \rightarrow 0\) as \(k\rightarrow 1\) because of (1.6) and Lemma A.3.
Theorem 1.4
(Asymptotic formulas) The following asymptotic formulas hold:
-
(i)
If \(j\in \{1,2,\ldots ,n-1\}\), then
$$\begin{aligned} \mu ^n_j(k)=\left\{ -4n^2+4n^2(1-k^2)\sin ^2\left( \frac{j\pi }{2n}\right) +o((1-k^2))\right\} K(k)^2 \quad {\text {as}} \ \ k\rightarrow 1. \end{aligned}$$In particular,
$$\begin{aligned} \lim _{k\rightarrow 1}\frac{\mu ^n_j(k)}{K(k)^2}=-4n^2\ \ \text {for}\ \ j\in \{1,2,\ldots ,n-1\}. \end{aligned}$$ -
(ii)
If \(j\in \{n+1,n+2,\ldots \}\), then
$$\begin{aligned} \mu ^n_j(k)=\pi ^2(j-n)^2+o(1)\quad \text {as}\ \ k\rightarrow 1. \end{aligned}$$
Remark 1.5
By Theorem 1.2 (i) we have the exact eigenvalues \(\mu ^n_0(k)\) and \(\mu ^n_n(k)\). 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:
Since \(h(u^{\pm }_n(x,k))\) and \(\rho \) depend on \(\mu \), the LHS of the equation
becomes a function in \(\mu \). A solution of the above equation gives the eigenvalue \(\mu ^n_j(k)\) for \(j\in \mathbb {N}_0\setminus \{0,n\}\). Two cases \(j=0,n\) are exceptional and eigenvalues and eigenfunctions are given in a different way. Since \(u^{\pm }_n(x,k)\) 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 \(\varepsilon ^2u''+\sin u=0\) in [11, 13],
-
(2)
a Dirichlet problem of \(\varepsilon ^2 u''+\sin u=0\) in [12],
-
(3)
a Neumann problem of the Allen-Cahn equation \(\varepsilon ^2u''+u-u^3=0\) in [14, 15],
-
(4)
a Neumann problem of the scalar field equation \(\varepsilon ^2u''-u+u^3=0\) in [6],
-
(5)
Dirichlet problems of \(u''+\lambda e^{\pm u}=0\) in [7],
and it is applied to
-
(6)
a Neumann (and Dicrichlet) problem of \(u''+\lambda \sinh u=0\) in the present paper.
Another interesting technical aspect of this paper is the modified complete elliptic integral of the third kind \(\mathcal {M}(\nu ,k)\) defined by (1.12). The function \(\mathcal {M}(\nu ,k)\) was introduced in [16]. In Theorem 1.2 and its proof we see that \(\mathcal {M}(\nu ,k)\) 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 \(\mathcal {M}(\nu ,k)\) might be universal in some sense. Figure 3 in Appendix shows a graph of \(\mathcal {M}(\nu ,k)\) in \(\nu \). We will recall some limit formulas for \(\mathcal {M}(\nu ,k)\).
This paper consists of six sections. In Sect. 2 we obtain the exact solutions \((\lambda ,u)\) of (1.1). We show that \(d\lambda _n(k)/dk<0\) for \(k\in (0,1)\) and that \(0<\lambda _n(k)<\lambda _n\) for \(k\in (0,1)\). Hence, \(\mathcal {C}^{\pm }_n\) is a graph of \(\lambda \) defined on \(0<\lambda <\lambda _n\). 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 \(\mathcal {A}(\mu ,k)\), a monotonicity of \(\mathcal {A}(\mu ,k)\) in \(\mu \) and certain limits of \(\mathcal {A}(\mu ,k)\). Using these properties of \(\mathcal {A}(\mu ,k)\), 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 \(\textrm{sn}(x,k)\), \(\textrm{cn}(x,k)\), \(\textrm{dn}(x,k)\), K(k), E(k) and \(\Pi (\nu ,k)\).
2 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
where \(\lambda _n(k)\) and \(u^{\pm }_n(x,k)\) are defined by (1.6) and (1.7), respectively.
Proof
Let us consider the initial value problem
with \((u(0),v(0))=(\alpha , 0)\). 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 \((\alpha ,0)\) and reaches a point on the u-axis, which is \((-\alpha ,0)\) or \((\alpha ,0)\), 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 \(n\in \{1,2,\ldots \}\), i.e.,
In other words the quarter period is 1/2n. If (2.3) holds, then u(x) is an n-mode solution.
We consider (2.2) with \((u(0),v(0))=(\alpha , 0)\). Multiplying (1.1) by \(u'\) and integrating it in x over [0, x], we have
We consider the case \(\alpha >0\). Then we have
We use the change of variables
Since u(y) is monotone for \(0\le y\le 1/2n\), the mapping \(y\mapsto s\) defined by (2.6) for \(y\in [0,1/2n]\) is one-to-one. Therefore, this change of variables is valid if \(0\le x\le 1/2n\). Using (2.6), we see that (2.5) becomes
where \(\textrm{sn}^{-1}\) denotes the inverse function of \(\textrm{sn}\) and
We have
and hence
Substituting \(x=1/2n\) into (2.8), we see by (2.3) that
which leads to
Here we used (2.7). We denote \(\lambda \) by \(\lambda _n(k)\). Let
By (2.8) and (2.9) we see that \(\cosh u(x)=w(x)\). By a direct calculation we can check that
By (2.10) we have
which is denoted by \(u^+_n(x,k)\). In particular,
which is equivalent to (2.7). We can easily check that
which is denoted by \(u^-_n(x,k)\), is a solution of (1.1) with \((u(0),u'(0))=(-\alpha ,0)\). We have proved (2.1). Because of this construction of solutions, we see that \((\lambda ,u^{\pm }_n(x,k))\) consist of all the nontrivial solutions of (1.1). The proof is complete. \(\square \)
Lemma 2.2
Let \(\lambda _n(k)\) be defined by (1.6). Then, \(\frac{d\lambda _n(k)}{dk}<0\) for \(0<k<1\).
Proof
Because of Lemma A.2, we have
By Lemma A.1 (i) and (2.11) we have
\(\square \)
Proof of Theorem 1.1
Let \(\lambda _n(k)\) and \(u^{\pm }_n(x,k)\) be defined by (1.6) and (1.7), respectively. Let \(\mathcal {C}_0\) and \(\mathcal {C}^{\pm }_n\), \(n=1,2,\ldots \), be defined by (1.4) and (1.5), respectively.
First, every element of \(\mathcal {C}_0\) is a trivial solution of (1.1). Next, we study nontrivial solutions. By Lemma 2.1, the solution set \(\mathcal {S}\) can be written as (1.3). What we have to do is to prove a monotonicity of each nontrivial branch \(\mathcal {C}^{\pm }_n\), \(n\ge 1\), which indicates that \(\mathcal {C}^{\pm }_n\) can be described as a graph of \(\lambda \) for a certain interval. It is clear that
where \(\lambda _n\) is defined by (1.8). By Lemma A.3 we have
Because of Lemma 2.2, \(\lambda _n(k)\) is a monotone function, and hence \(\mathcal {C}^{\pm }_n\) can be described as a graph of \(\lambda \) for \(0<\lambda <\lambda _n\). All the other statements in Theorem 1.1 follow from this result. The proof is complete. \(\square \)
3 Preliminaries for proofs of Theorems 1.2 and 1.4
3.1 Sturm–Liouville theory
Let f be an arbitrary \(C^1\)-function. We consider the Neumann problem
Let \((\lambda ,u)\) denote a solution of (3.1), where \(\lambda >0\). We define
The linearized eigenvalue problem at \((\lambda ,u)\) becomes
We recall basic properties about eigenvalues and eigenfunctions for (3.3).
Proposition 3.1
Let \(\{\mu _j\}_{j=0}^{\infty }\) denote the eigenvalues of (3.3), and let \(\varphi _j(x)\), \(\varphi _j(x)\not \equiv 0\), be an eigenfunction corresponding to \(\mu _j\). Then the following hold:
-
(i)
Each eigenvalue \(\mu _j\) is real and simple, and \(\mu _0<\mu _1<\mu _2<\cdots \).
-
(ii)
For \(j\in \mathbb {N}_0\), the zero number of \(\varphi _j(x)\) on [0, 1] is j, i.e.,
$$\begin{aligned} \sharp \{x\in [0,1];\ \varphi _j(x)=0\}=j. \end{aligned}$$
See [3, Theorem 2.1 in p.212] for details of a proof of Proposition 3.1. We omit the proof.
3.2 General nonlinearity
First, we look for an eigenfunction of the form \(\varphi (x)=\sqrt{\phi (x)}\). Substituting \(\sqrt{\phi (x)}\) into the equation in (3.3), we see that \(\phi (x)\) satisfies
We define
Then,
We consider the equation
If \(\phi (x)\) satisfies (3.4), then \(\Phi '(x)=0\), and hence \(\Phi (x)\) is constant. In particular, \(\Phi (x)=\Phi (0)\), i.e.,
If \(\Phi (0)=0\), then \(\sqrt{\phi (x)}\) satisfies the equation in (3.3) and it is a candidate of the eigenfunction. We show that almost all the eigenfunctions can be constructed even if \(\Phi (0)\ne 0\).
We look for a solution of (3.4) of the form \(\phi (x)=h(u(x))\), where \(h(u(x))>0\) for \(0\le x\le 1\). Here, \(h(\,\cdot \,)\) 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 \(F(u):=\int _0^uf(s)ds\) 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 \(u(0)=\alpha \) 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 \(0\le x\le 1\) is obtained. Then, we look for an eigenfunction of the form
The functions \(W(\theta )\) and \(\theta (x)\) are defined later. Using (3.7), we have
Substituting (3.9) into the equation in (3.3), by (3.10) we have
We assume that \(\rho >0\), where \(\rho \) is defined by (3.8). We consider the case where
Since h(u(x)) is positive, \(\theta (x)\) can be expressed as
Moreover, it follows from (3.11) that W satisfies \(W''(\theta )+W(\theta )=0\). Therefore, \(W(\theta )=C_1\cos (\theta +\theta _1)\) for some \(C_1\) and \(\theta _1\). By (3.9) we have
Since \(\varphi \) satisfies the Neumann boundary condition \(\varphi '(0)=0\), we can take \(\theta _0+\theta _1=0\), and hence a candidate of an eigenfunction becomes
For each \(j\in \mathbb {N}_0\), let us consider the equation
Since \(\rho \) and h(u(y)) depend on \(\mu \), (3.13) is an equation for \(\mu \). If \(\mu \) satisfies (3.13) for some j, then \(\varphi '(1)=0\). Since \(\sqrt{h(u(x))}>0\) for \(0\le x\le 1\), \(\varphi (x)\) defined by (3.12) has exactly j zero(s) on \(0\le x\le 1\). By Proposition 3.1 (ii) we see that the corresponding \(\varphi (x)\) becomes a j-th eigenfunction, and hence \(\mu \) becomes the j-th eigenvalue.
3.3 Hyperbolic sine case
We find an exact solution h(u) of (3.6) such that \(h(u(x))>0\) for \(0\le x\le 1\). When \(f(u)=\sinh u\), (3.6) has an exact solution
Since \(|u(x)|\le \alpha \) for \(0\le x\le 1\), the following functions become positive for \(0\le x\le 1\):
where
Hence, (3.14) is a positive solution of (3.6). In the case \(f(u)=\sinh u\) the LHS of (3.13) can be defined provided that
In Sect. 4 we will see that when \(u(x)=u^{\pm }_n(x,k)\),
Two cases \(j=0,n\) are exceptional in our case \(f(u)=\sinh u\).
4 Proof of Theorems 1.2 and 1.4
For simplicity we write \(\lambda \) and u for \(\lambda _n(k)\) and \(u^{\pm }_n(x,k)\), respectively. In this section we study an exact expression of eigenvalues of (1.9) associated with an n-mode solution u. Let us recall \(\alpha :=u(0)\) defined by (3.2). A family of solutions \((\lambda ,u)\) can be parametrized by k, and it can be also parametrized by \(\alpha \). We use both k and \(\alpha \). Note that k and \(\alpha \) are related by (2.7).
4.1 Two exceptional eigenvalues
Lemma 4.1
The following hold:
-
(i)
The 0-th eigenvalue \(\mu ^n_0(k)\) and a corresponding eigenfunction \(\varphi ^n_0\) are given by
$$\begin{aligned} \mu ^n_0(k)=-4K(k)^2n^2\ \ \text {and}\ \ \varphi ^n_0(x,k)=\cosh \frac{u^{\pm }_n(x,k)}{2}=\frac{1}{\sqrt{1-k^2}}\textrm{dn}(2nK(k)x,k). \end{aligned}$$ -
(ii)
The n-th eigenvalue \(\mu ^n_n(k)\) and a corresponding eigenfunction \(\varphi ^n_n\) are given by
$$\begin{aligned} \mu ^n_n(k)=-4k^2K(k)^2n^2\ \ \text {and}\ \ \varphi ^n_n(x,k)=\sinh \frac{u^{\pm }_n(x,k)}{2}=\pm \frac{k}{\sqrt{1-k^2}}\textrm{cn}(2nK(k)x,k). \end{aligned}$$
Proof
(i) Let \(\mu =-4K(k)^2n^2\) and \(\varphi =\cosh \frac{u(x)}{2}\). Since
we have
where we use \(\cosh u\cosh \frac{u}{2}-\sinh u\sinh \frac{u}{2}=\cosh \frac{u}{2}\). Since
we have \(\varphi ''+\lambda (\cosh u)\varphi +\mu \varphi =0\). We easily see that \(\varphi '(0)=\varphi '(1)=0\). Since \(\varphi (x)>0\) for \(0\le x\le 1\), by Proposition 3.1 (ii) we see that \(\mu \) is the 0-th eigenvalue and that \(\varphi \) is a 0-th eigenfunction. We have
The proof of (i) is complete.
(ii) Let \(\mu =-4k^2K(k)^2n^2\) and \(\varphi =\sinh \frac{u(x)}{2}\). By a similar calculation we have
Since
we have \(\varphi ''+\lambda (\cosh u)\varphi +\mu \varphi =0\). We see that \(\varphi '(0)=\varphi '(1)=0\). Since u(x) has n zero(s) in \(0\le x\le 1\), \(\varphi (x)\) also has n zero(s) in \(0\le x\le 1\). It follows from Proposition 3.1 (ii) that \(\mu \) is the n-th eigenvalue and \(\varphi \) is an n-th eigenfunction. We have
Since an eigenfunction is smooth, we see that
The proof of (ii) is complete. \(\square \)
4.2 General eigenvalues
We define \(\mathcal {A}(\mu ,k)\) by
where \(\rho \) is defined by (3.8) and h is given by (3.14) depending on \(\mu \). We already saw in Sect. 3.3 that \(\mathcal {A}(\mu ,k)\) can be defined provided that
Lemma 4.2
Let \(\mathcal {A}(\mu ,k)\) be defined by (4.3). Then,
where
and \(\bar{\mu }^1\) is defined by (3.15). Moreover,
where \(\mathcal {M}(\nu ,k)\) is defined by (1.12) and
Proof
Since
by (4.4) we see that \(\mathcal {A}(\mu ,k)\) can be defined for \(\mu \in (\bar{\mu }^0,\bar{\mu }^1)\cup (0,\infty )\), and hence (4.5) is proved.
We consider the case \(\mu \in (0,\infty )\). Since \(\lambda >0\), we calculate \(\int _0^1dy/h_+(u(y))\). We use the change of variables (2.6). Note that this change of variables is valid for \(0\le x\le 1/2n\), since the n-mode solution u is monotone for \(0\le x\le 1/2n\). By (2.6), we have
Using this equality, by a direct calculation we have
where
Note that \(k^2\) given in (4.10) is equal to (2.7). We have
Using (4.11) and (4.9), we have
Then, (4.7) holds for \(\mu >0\) and (4.8) follows from (4.10).
Next, we consider the case \(\mu \in (\bar{\mu }^0,\bar{\mu }^1)\). We have
where \(k^2\) and \(\nu \) are defined by (4.10). Thus, \(\mathcal {A}(\mu ,k)=\mathcal {M}(\nu ,k)\).
In both cases \(\mu \in (0,\infty )\) and \(\mu \in (\bar{\mu }^0,\bar{\mu }^1)\) we obtain (4.7). \(\square \)
Thanks to Lemma 4.2 the equation (3.13) becomes
Therefore, the graph \(\mu \mapsto \mathcal {A}(\mu ,k)\), which is defined for \(\mu \in (\bar{\mu }^0,\bar{\mu }^1)\cup (0,\infty )\), becomes important.
Lemma 4.3
The following equalities hold:
-
(i)
\(\displaystyle \lim _{\mu \downarrow \bar{\mu }^0}\mathcal {A}(\mu ,k)=0\).
-
(ii)
\(\displaystyle \lim _{\mu \uparrow \bar{\mu }^1}\mathcal {A}(\mu ,k)=\frac{\pi }{2}\).
-
(iii)
\(\displaystyle \lim _{\mu \downarrow 0}\mathcal {A}(\mu ,k)=\frac{\pi }{2}\).
-
(iv)
\(\displaystyle \lim _{\mu \rightarrow \infty }\mathcal {A}(\mu ,k)=\infty \).
Proof
Because of (4.8), we have the following relations:
Since \(\mathcal {A}(\mu ,k)=\mathcal {M}(\nu ,k)\), the assertions (i), (ii), (iii) and (iv) follow from Lemma A.7 (i), (ii), (iii) and (iv), respectively. \(\square \)
Lemma 4.4
The function \(\mathcal {A}(\mu ,k)\) satisfies
Proof
We prove the following inequality instead of (4.13):
Since
it is enough to show that
Here, \(\bar{\mu }^0<\mu <\bar{\mu }^1\) corresponds to \(-1<\nu <-k^2\) and \(\mu >0\) corresponds to \(\nu >0\). By Lemma A.1 (ii) we have
Using Lemma A.2, we have
The inequality (4.14) follows from (4.16). The proof is complete. \(\square \)
Remark 4.5
Let \(\bar{\mu }^0\) and \(\bar{\mu }^1\) be defined by (4.6) and (3.15), respectively. By (1.6), (4.10) and Lemma 4.1 we see that
Proof of Theorem 1.2
-
(i)
The assertion follows from Lemma 4.1.
-
(ii)
By Remark 4.5 we see that \(\bar{\mu }^0=\mu ^n_0(k)\) and \(\bar{\mu }^1=\mu ^n_n(k)\). Let \(\mathcal {A}\) be defined by (4.3). Then, by (4.5) and (4.7) we see that \(\mathcal {A}(\mu ,k)\) is defined in \(\mu \in (\mu ^n_0(k),\mu ^n_n(k))\cup (0,\infty )\) and it can be written as (1.11). Since (3.13) is equivalent to (4.12), a solution of (1.10) gives a j-th eigenvalue. By Lemma 4.4 we see that \(\mathcal {A}(\mu ,k)\) is increasing for \(\mu \in (\mu ^n_0(k),\mu ^n_n(k))\cup (0,\infty )\). It follows from Lemma 4.3 that the range \(\{\mathcal {A}(\mu ,k);\ \mu ^n_0(k)<\mu <\mu ^n_n(k)\}\) is \((0,\pi /2)\) and the range \(\{\mathcal {A}(\mu ,k);\ \mu >0\}\) is \((\pi /2,\infty )\). This indicates that
$$\begin{aligned}&\text {for }j\in \{1,2,\ldots ,n-1\},~(4.12)~\text {has a unique solution}~\mu ^n_j(k)~\text {in}\nonumber \\&\quad ~(\mu ^n_0(k),\mu ^n_n(k)),~and \end{aligned}$$(4.17)$$\begin{aligned}&\text {for}~j\in \{n+1,n+2,\ldots \},~(4.12)~\text {has a unique solution}~\mu ^n_j(k)~\text {in}~(0,\infty ). \end{aligned}$$(4.18)Since \(\mu ^n_n(k)<0<\mu ^n_{n+1}(k)\), the number of the negative eigenvalues are \(n+1\) and 0 is not an eigenvalue. All the assertions in (ii) hold. The proof of (ii) is complete.
-
(iii)
We consider the case \(\mu _j>0\). Using (2.8), (3.14) and (4.11), we have
$$\begin{aligned} h_+(u(x))= & {} 2\frac{\mu _j}{\lambda }+\cosh \alpha -\cosh u(x)\\= & {} \frac{2k^2}{1-k^2}\left( -\frac{\mu _j}{\mu ^n_n(k)}+\textrm{sn}^2(2nK(k)x,k)\right) ,\\ \sqrt{\frac{\lambda \rho }{2}}= & {} \frac{2}{\lambda }\sqrt{\mu _j(\mu _j-\bar{\mu }^0)(\mu _j-\bar{\mu }^1)}\\= & {} \frac{2k^2}{1-k^2}\frac{-1}{\mu ^n_n(k)}\sqrt{\mu _j(\mu _j-\mu ^n_0(k))(\mu _j-\mu ^n_n(k))}. \end{aligned}$$Substituting these equalities into (3.12), we have
$$\begin{aligned} \varphi (x)= & {} C_1\sqrt{\frac{2k^2}{1-k^2}\left( -\frac{\mu _j}{\mu ^n_n}+\textrm{sn}^2(2nK(k)x,k)\right) }\\{} & {} \times \cos \left( \int _0^x\frac{\sqrt{\mu _j(\mu _j-\mu ^n_0(k))(\mu _j-\mu ^n_n(k))}}{\mu _j-\mu ^n_n(k)\textrm{sn}^2(2nK(k)y,k)}dy\right) . \end{aligned}$$
Let \(C_1=\sqrt{(1-k^2)(-\mu ^n_n(k))/2k^2}\). We obtain (1.13). The case \(\mu ^n_0(k)<\mu _j<\mu ^n_n(k)\) is almost the same. We omit the proof. \(\square \)
4.3 Asymptotic formulas
Proof of Theorem 1.4
-
(i)
We consider the case \(j\in \{1,2,\ldots ,n-1\}\). We define
$$\begin{aligned} \tilde{\mu }_j:=\frac{\mu ^n_j(k)}{K(k)^2}\quad \text {and}\quad \hat{\mu }_j:=\frac{\tilde{\mu }_j+4n^2}{1-k^2}. \end{aligned}$$(4.19)Then, \(\mu ^n_j(k)\) can be written as
$$\begin{aligned} \mu ^n_j(k)=\left\{ -4n^2+(1-k^2)\hat{\mu }_j\right\} K(k)^2. \end{aligned}$$(4.20)Because of (4.17), we see that a unique solution \(\mu ^n_j(k)\) of (4.12) satisfies \(\mu ^n_0(k)<\mu ^n_j(k)<\mu ^n_n(k)\). Since \(\mu ^n_0(k)<\mu ^n_j(k)<\mu ^n_n(k)\), we see that \(0<\hat{\mu }_j<4n^2\), and hence there is a sequence \(k_i\), which is denoted by k for simplicity, and \(\hat{\mu }^*_j\in [0,4n^2]\) such that \(k\rightarrow 1\) as \(i\rightarrow \infty \) and \(\hat{\mu }_j\rightarrow \hat{\mu }^*_j\) as \(i\rightarrow \infty \). Since
$$\begin{aligned} \nu (k):=\nu =\frac{\cosh \alpha -1}{2\mu ^n_j(k)}\lambda =\frac{4k^2K(k)^2n^2}{\mu ^n_j(k)}=\frac{-4k^2n^2}{4n^2-(1-k^2)\hat{\mu }^j}, \end{aligned}$$recall (4.8), we have
$$\begin{aligned} \nu ^*:=\lim _{i\rightarrow \infty }\frac{\nu +1}{1-k^2}=\lim _{i\rightarrow \infty }\frac{4n^2-\hat{\mu }_j}{4n^2-(1-k^2)\hat{\mu }_j}=1-\frac{\hat{\mu }^*_j}{4n^2}\in [0,1]. \end{aligned}$$By Lemma A.6 we have
$$\begin{aligned} \frac{j\pi }{2n}=\mathcal {A}(\mu ^n_j(k),k)=\mathcal {M}(\nu (k),k)\rightarrow \frac{\pi }{2} - \tan ^{-1}\sqrt{\frac{1-\frac{\hat{\mu }^*_j}{4n^2}}{1-\left( 1-\frac{\hat{\mu }^*_j}{4n^2}\right) }}\ \ \text {as}\ \ i\rightarrow \infty . \end{aligned}$$Then,
$$\begin{aligned} \tan \left( \frac{\pi }{2}-\frac{j\pi }{2n}\right) = \sqrt{\frac{1-\frac{\hat{\mu }^*_j}{4n^2}}{\frac{\hat{\mu }^*_j}{4n^2}}}. \end{aligned}$$(4.21)Solving (4.21) with respect to \(\hat{\mu }^*_j\), we have
$$\begin{aligned} \hat{\mu }^*_j=4n^2\sin ^2\left( \frac{j\pi }{2n}\right) . \end{aligned}$$(4.22)Note that the limit \(\hat{\mu }^*_j\) does not depend on a sequence \(k\rightarrow 1\). Substituting (4.22) into (4.20), we have
$$\begin{aligned} \mu ^n_j(k)=\left\{ -4n^2+4n^2(1-k^2)\sin ^2\left( \frac{j\pi }{2n}\right) +o((1-k^2))\right\} K(k)^2 \ \ \text {as}\ \ k\rightarrow 1. \end{aligned}$$The proof of (i) is complete.
-
(ii)
We consider the case \(j\in \{n+1,n+2,\ldots \}\). Because of (4.18), we see that a unique solution \(\mu ^n_j(k)\) satisfies \(\mu ^n_j(k)>0\). We have
$$\begin{aligned} \frac{j\pi }{2n}= & {} \mathcal {A}(\mu ^n_j(k),k)=\mathcal {M}(\nu ,k)\\\ge & {} \sqrt{\frac{(\nu +k^2)(\nu +k^2)}{\nu }}\int _0^1\frac{ds}{(1+\nu )\sqrt{(1-s^2)(1-k^2s^2)}}\\= & {} \frac{1}{\sqrt{\nu }}\frac{\nu +k^2}{\nu +1}K(k) \ge \frac{k^2K(k)}{\sqrt{\nu }}=\frac{\sqrt{\mu ^n_j(k)}k}{2n}, \end{aligned}$$recall (4.8). Since \(\mu ^n_j(k)\) is bounded as \(k\rightarrow 1\) by \(\sqrt{\mu ^n_j(k)}\le j\pi /k\), and since \(K(k)\rightarrow \infty \) as \(k\rightarrow 1\), we have
$$\begin{aligned} \nu =\frac{4k^2K(k)^2n^2}{\mu ^n_j(k)}\rightarrow \infty \ \ \text {as}\ \ k\rightarrow 1. \end{aligned}$$We define
$$\begin{aligned} J(\nu ,k):=\sqrt{\nu +1}\Pi (\nu ,k)-\frac{1}{\sqrt{\nu +1}}K(k). \end{aligned}$$(4.23)Multiplying (4.23) by \(\sqrt{(\nu +k^2)/\nu }\), we have
$$\begin{aligned} \sqrt{\frac{\nu +k^2}{\nu (\nu +1)}}K(k)= & {} \sqrt{\frac{(\nu +1)(\nu +k^2)}{\nu }}\Pi (\nu ,k) -\sqrt{\frac{\nu +k^2}{\nu }}J(\nu ,k)\nonumber \\= & {} \frac{j\pi }{2n}-\sqrt{\frac{\nu +k^2}{\nu }}J(\nu ,k). \end{aligned}$$(4.24)Since \(\nu \rightarrow \infty \) as \(k\rightarrow 1\), by Lemma A.5 we have
$$\begin{aligned} \sqrt{\frac{\nu +k^2}{\nu }}J(\nu ,k)\rightarrow \frac{\pi }{2}\ \ \text {as}\ \ k\rightarrow 1. \end{aligned}$$(4.25)Since \(\mu ^n_j(k)\) is bounded as \(k\rightarrow 1\), there exists \(\mu ^*_j\) such that \(\mu ^n_j(k)\rightarrow \mu ^*_j\) as \(k\rightarrow 1\). Then,
$$\begin{aligned} \sqrt{\frac{\nu +k^2}{\nu (\nu +1)}}K(k) =\sqrt{\frac{1+\frac{\mu ^n_j(k)}{4n^2K(k)^2}}{4k^2n^2+\frac{\mu ^n_j(k)}{K(k)^2}}{\mu _j(k) }}\rightarrow \frac{\sqrt{\mu ^*_j}}{2n} \ \ \text {as}\ \ k\rightarrow 1. \end{aligned}$$(4.26)It follows from (4.24)–(4.26) that as \(k\rightarrow 1\),
$$\begin{aligned} \frac{\sqrt{\mu ^*_j}}{2n}=\frac{j\pi }{2n}-\frac{\pi }{2}. \end{aligned}$$Thus,
$$\begin{aligned} \mu ^*_j=\pi ^2(j-n)^2. \end{aligned}$$Here, the limit \(\mu ^*_j\) does not depend on a sequence \(k\rightarrow 1\). Since \(\mu ^n_j(k)=\mu ^*_j+o(1)\) as \(k\rightarrow 1\), we have
$$\begin{aligned} \mu ^n_j(k)=\pi ^2(j-n)^2+o(1)\ \ \text {as}\ \ k\rightarrow 1. \end{aligned}$$
The proof of (ii) is complete. \(\square \)
5 Dirichlet problem
In this section we consider the Dirichlet problem
We present the results for the Dirichlet problem which correspond to Theorems 1.1, 1.2 and 1.4 in the Neumann problem. The proofs are almost the same, and hence we omit them.
Theorem 5.1
The solution set \(\tilde{\mathcal {S}}\) of (5.1) is
where \(\mathcal {C}_0\) is defined by (1.4),
and \(\lambda _n(k)\) is defined by (1.6). Here, \(\textrm{sd}(\xi ,k):=\textrm{sn}(\xi ,k)/\textrm{dn}(\xi ,k)\).
Let \((\lambda _n,\tilde{u}^{\pm }_n)\) be a nontrivial solution of (5.1). We consider the linearized eigenvalue problem
Let \(\{\tilde{\mu }^n_j(k)\}_{j=1}^{\infty }\) denote the eigenvalues and let \(\tilde{\varphi }^n_j(x,k)\) be an eigenfunction corresponding to \(\tilde{\mu }^n_j(k)\). Here, \(\tilde{\mu }^n_j(k)\) has no relation with \(\tilde{\mu }\) defined by (4.19). Note that \(\tilde{\mu }^n_1(k)\) is the principal eigenvalue in the Dirichlet problem, while \(\mu ^n_0(k)\) is that in the Neumann problem.
Theorem 5.2
Let \(\mathbb {N}:=\{1,2,\ldots \}\). The following hold:
-
(i)
The eigenvalues of (5.2) are given by
$$\begin{aligned} \tilde{\mu }^n_j(k)=\mu ^n_j(k)\ \ \text {for}\ \ j\in \mathbb {N}. \end{aligned}$$Therefore, the same asymptotic formulas given by Theorem 1.4 and Remark 1.5 hold for \(j\in \mathbb {N}\).
-
(ii)
An n-th eigenfunction of (5.2) is given by
$$\begin{aligned} \tilde{\varphi }^n_n(x,k)=\sinh \frac{{\tilde{u}^{\pm }_n}(x,k)}{2}=\pm k\textrm{sd}(2nK(k)x,k), \end{aligned}$$
and, for \(j\in \mathbb {N}\setminus \{n\}\), a j-th eigenfunction of (5.2) is given by
Here, \(\textrm{cd}(\xi ,k):=\textrm{cn}(\xi ,k)/\textrm{dn}(\xi ,k)\) and \(\bar{\mu }^0\) is defined by (4.6).
Note that we use \(\bar{\mu }^0\) in (5.3) since \(\tilde{\mu }^n_0(k)\) is not defined in the Dirichlet problem.
Data availability
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Acknowledgements
The authors are grateful to the anonymous referee for a very careful reading of the manuscript and following complicated calculations. His/her helpful suggestions make the presentation clear. YM was supported by JSPS KAKENHI Grant numbers 19H01797, 19H05599. TW was supported by JSPS KAKENHI Grant number 18K03374.
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Appendix A: Elliptic integrals and functions
Appendix A: Elliptic integrals and functions
1.1 A.1: Elliptic functions
Let \(k\in (0,1)\). The Jacobi elliptic function \(\textrm{sn}(x,k)\) is an odd, 4K(k)-periodic, 2K(k)-antiperiodic and analytic function for \(x\in \mathbb {R}\) and the inverse function \(\textrm{sn}^{-1}(y,k)\) is defined locally by
for \(0\le y\le 1\). The function \(\textrm{cn}(x,k)\) is an even, 4K(k)-periodic and 2K(k)-antiperiodic function defined locally by
for \(-K(k)\le x\le K(k)\) and \(\textrm{dn}(x,k)\) is an even and 2K(k)-periodic function defined by
for \(x\in \mathbb {R}\). In particular,
for \(x\in \mathbb {R}\) and \(k\in (0,1)\).
1.2 A.2: Complete elliptic integrals
Let \(k\in [0,1)\) and \(\nu \in \mathbb {C}{\setminus } (-\infty ,-1]\). The complete elliptic integral of the first kind K(k) is defined by
and it satisfies \(K(k)=\textrm{sn}^{-1}(1,k)\). The complete elliptic integrals of the second and third kinds are defined by
respectively. The function K(k) is monotonically increasing in k,
and E is monotonically decreasing in k,
We give standard formulas for K(k), E(k) and \(\Pi (\nu ,k)\) in Lemmas A.1–A.3 without proofs. See [1] for details.
Lemma A.1
Let \(k\in (0,1)\) and \(\nu \ne 0,-1,-k^2\). Then,
-
(i)
\(\displaystyle \frac{dK}{dk}(k)=\frac{E(k)-(1-k^2)K(k)}{k(1-k^2)}\).
-
(ii)
\(\displaystyle \frac{\partial \Pi }{\partial \nu }(\nu ,k)=-\frac{K(k)}{2\nu (\nu +1)}+\frac{E(k)}{2(\nu +1)(\nu +k^2)}+\frac{(k^2-\nu ^2)\Pi (\nu ,k)}{2\nu (\nu +1)(\nu +k^2)}\).
Lemma A.2
Let \(k\in (0,1)\). Then,
Lemma A.3
Let \(k\in (0,1)\). Then,
Lemmas A.4 and A.5 are formulas for \(\Pi (\nu ,k)\). Proofs can be found in [14].
Lemma A.4
Let \(k\in (0,1)\) and \(\nu >-1\). Then,
-
(i)
\(\displaystyle \lim _{\nu \downarrow -1}\sqrt{\nu +1}\Pi (\nu ,k)=\frac{\pi }{2\sqrt{1-k^2}}\).
-
(ii)
\(\displaystyle \lim _{\nu \rightarrow \infty }\sqrt{\nu +1}\Pi (\nu ,k)=\frac{\pi }{2}\).
Lemma A.5
Let \(J(\nu ,k):=\sqrt{\nu +1}\Pi (\nu ,k)-\frac{1}{\sqrt{\nu +1}}K(k)\). Then,
1.3 A.3: Modified complete elliptic integral of the third kind
The modified complete integral of the third kind
was introduced in [16]. This function is defined for \(k\in (0,1)\) and \(\nu \in \mathbb {C}{\setminus } ((-\infty ,-1]\cup [-k^2,0])\), and hence on the real axis it is defined for \(\nu \in (-1,-k^2)\cup (0,\infty )\). Figure 3 shows a graph of \(\mathcal {M}(\nu ,k)\) as a real function in \(\nu \) for \(k=1/\sqrt{2}\). This function appears in Theorem 1.2 and its proof in Sect. 4. The following Lemmas A.6 and A.7 are important and useful in the study of exact eigenvalues and eigenfunctions.
Lemma A.6
Let \(k\in (0,1)\). Suppose that \(\nu \) is a continuous function on (0, 1) with \(-1<\nu (k)<-k^2\) for \(k\in (0,1)\). Assume that there exists \(\nu ^*\in [0,1]\) such that
Then, for each \(\nu ^*\in [0,1]\),
See [14] for a proof of Lemma A.6.
Lemma A.7
Let \(k\in (0,1)\) and \(\nu \in (-1,-k^2)\cup (0,\infty )\). Then, the following hold:
-
(i)
\(\displaystyle \lim _{\nu \uparrow -k^2}\mathcal {M}(\nu ,k)=0\).
-
(ii)
\(\displaystyle \lim _{\nu \downarrow -1}\mathcal {M}(\nu ,k)=\frac{\pi }{2}\).
-
(iii)
\(\displaystyle \lim _{\nu \rightarrow \infty }\mathcal {M}(\nu ,k)=\frac{\pi }{2}\).
-
(iv)
\(\displaystyle \lim _{\nu \downarrow 0}\mathcal {M}(\nu ,k)=\infty \).
See Fig. 3. Lemma A.7 (ii) (resp. (iii)) follows from Lemma A.4 (i) (resp. (ii)). Lemma A.7 (i) and (iv) are trivial.
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Aizawa, S., Miyamoto, Y. & Wakasa, T. Asymptotic formulas of the eigenvalues for the linearization of a one-dimensional sinh-Poisson equation. J Elliptic Parabol Equ 9, 1043–1070 (2023). https://doi.org/10.1007/s41808-023-00233-9
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DOI: https://doi.org/10.1007/s41808-023-00233-9