Hadamard variation of eigenvalues with respect to general domain perturbations

Takashi Suzuki Center of Mathematical Modeling and Data Science,
Osaka University,
Toyonaka, 560-8531, Japan suzuki@sigmath.es.osaka-u.ac.jp Takuya Tsuchiya Center of Mathematical Modeling and Data Science,
Osaka University,
Toyonaka, 560-8531, Japan tsuchiya.takuya.plateau@kyudai.jp

Abstract.

We study Hadamard variation of eigenvalues of Laplacian with respect to general domain perturbations. We show their existence up to the second order rigorously and characterize the derivatives, using associated eigenvalue problems in finite dimensional spaces. Then smooth rearrangement of multiple eigenvalues is explicitly given. This result follows from an abstract theory, applicable to general perturbations of symmetric bilinear forms.

Key words and phrases:

eigenvalue problem, perturbation theory of linear operators, domain deformation, Hadamard’s variational formulae, Garabedian-Schiffer’s formula
This work was promoted in RIMS program for joint research during 2019-2021. The authors thank Professors Hideyuki Azegami and Erika Ushikoshi for many detailed discussions at these occasions. The first author was supported by JSPS Grant-in-Aid for Scientific Research 19H01799. The second author was supported by JSPS Grant-in-Aid for Scientific Research 21K03372

2020 Mathematics Subject Classification:

Primary 35J25; Secondary 35R35

1. Introduction

Our purpose is to study Hadamard variation of eigenvalues of the Laplace operator with mixed boundary conditions. We characterize the first and the second derivatives, using associated finite dimensional eigenvalue problems, particularly, for multiple eigenvalues.

Hadamard’s variational formulae are used to provide effective numerical schemes for shape optimization and free boundary problems. In [12, 13] we have introduced a new variational formulation to filtration problem and applied the Hadamard variation. We have also derived Hadamard’s variational formulae for Green’s functions of the Poisson equation, extending Liouville’s volume and area formulae up to the second order [14, 15]. This paper is devoted to the eigenvalue problem concerning general perturbation of Lipschitz domains. Meanwhile we develop abstract theory applicable to other problems.

So far, $C^{1}$ and analytic rearrangements of multiple eigenvalues of self-adjoint operator have been discussed, for example, pp. 44–52 of [10] and p.487 of [1]. There are, however, several comments on the complexity of the proof on $C^{1}$ category, as in pp. 122–123 of [5] and p. 490 of [1]. In this paper we examine this process of rearrangement in details, using the above described characterization of the derivatives, to ensure their efficiencies in $C^{2}$ categories.

Let $\Omega$ be a bounded Lipschitz domain in $n$ -dimensional Euclidean space $\mathbb{R}^{n}$ for $n\geq 2$ . Suppose that its boundary $\partial\Omega$ is divided into two relatively open sets $\gamma_{0}$ and $\gamma_{1}$ , satisfying

\overline{\gamma_{0}}\cup\overline{\gamma_{1}}=\partial\Omega,\quad\overline{\gamma_{0}}\cap\overline{\gamma_{1}}=\emptyset.

(1)

As in [15], we thus avoid the technical difficulty described in p. 272 of [2], when $\gamma_{i}$ , $i=0,1$ have their boundaries on $\partial\Omega$ .

We study the eigenvalue problem of the Poisson equation with mixed boundary condition,

-\Delta u=\lambda u\ \mbox{in $\Omega$},\quad u=0\ \mbox{on $\gamma_{0}$},\quad\frac{\partial u}{\partial\nu}=0\ \mbox{on $\gamma_{1}$},

(2)

where

\Delta=\sum_{i=1}^{n}\frac{\partial^{2}}{\partial x_{i}^{2}}

is the Laplacian and $\nu$ denotes the outer unit normal vector on $\partial\Omega$ . This problem takes the weak form, finding $u$ satisfying

u\in V,\ B(u,u)=1,\quad A(u,v)=\lambda B(u,v),\ \forall v\in V

(3)

defined for

A(u,v)=\int_{\Omega}\nabla u\cdot\nabla v\ dx,\quad B(u,v)=\int_{\Omega}uv\ dx

and

V=\{v\in H^{1}(\Omega)\mid\left.v\right|_{\gamma_{0}}=0\}.

(4)

This $V$ is a closed subspace of $H^{1}(\Omega)$ under the norm

\|v\|_{V}=\sqrt{\|\nabla v\|_{2}^{2}+\|v\|_{2}^{2}}.

To justify the above weak formulation, we confirm several fundamental facts on the Lipschitz domain [6]. First, the trace operator

v\in C^{\infty}(\overline{\Omega})\mapsto\left.v\right|_{\partial\Omega}\in C(\partial\Omega)

defined for

C^{\infty}(\overline{\Omega})=\{v\in\overline{\Omega}\rightarrow\mathbb{R}\mid\exists\tilde{v}\in C_{0}^{\infty}(\mathbb{R}^{n}),\ \left.\tilde{v}\right|_{\overline{\Omega}}=v\}

is so extended as a bounded linear operator

v\in H^{1}(\Omega)\ \mapsto\ \left.v\right|_{\partial\Omega}\in H^{1/2}(\partial\Omega),

and there arises the isomorphism

v\in V/H^{1}_{0}(\Omega)\ \mapsto\left.v\right|_{\gamma_{1}}\in H^{1/2}(\gamma_{1}).

Second, if $\Delta v\in V^{\prime}$ is satisfied in the sense of distributions in $\Omega$ , the normal derivative of $v\in V$ on $\gamma_{1}$ is defined as in

\frac{\partial v}{\partial\nu}\in H^{-1/2}(\gamma_{1})=(H^{1/2}(\gamma_{1}))^{\prime},

and it holds that

\left\langle\varphi,\frac{\partial v}{\partial\nu}\right\rangle_{H^{1/2}(\gamma_{1}),H^{-1/2}(\gamma_{1})}=(\nabla v,\nabla\varphi)_{L^{2}(\Omega)}+\langle\varphi,\Delta v\rangle_{V,V^{\prime}},\quad\forall\varphi\in V,

where $\langle\cdot,\cdot\rangle_{Y,Y^{\prime}}$ denotes the paring between the Banach space $Y$ and its dual space $Y^{\prime}$ . See Theorem 2 of [15].

To confirm the well-posedness of (3), we note, first, that if $\gamma_{0}\neq\emptyset$ , there is coercivity of $A:V\times V\rightarrow\mathbb{R}$ , which means the existence of $\delta>0$ such that

A(v,v)\geq\delta\|v\|_{V}^{2},\quad\forall v\in V.

(5)

If $\gamma_{0}=\emptyset$ we replace $A$ by $A+B$ , denoted by $\tilde{A}$ . Then this $\tilde{A}:V\times V\rightarrow\mathbb{R}$ is coercive, and the eigenvalue problem

u\in V,\ B(u,u)=1,\quad\tilde{A}(u,v)=\tilde{\lambda}B(u,v),\ \forall v\in V,

is equivalent to (3) by $\tilde{\lambda}=\lambda+1$ . Henceforth, we assume (5), using this reduction if it is necessary.

Second, we note that $A:V\times V\rightarrow{\bf R}$ and $B:X\times X\rightarrow{\bf R}$ are bounded, coercive, and symmetric bilinear forms for $X=L^{2}(\Omega)$ . Since $V\hookrightarrow X$ is compact, there is a sequence of eigenvalues to (3), denoted by

0<\lambda_{1}\leq\lambda_{2}\leq\cdots\rightarrow+\infty.

The associated eigenfunctions, $u_{1},u_{2},\cdots$ , furthermore, form a complete ortho-normal system in $X$ , provided with the inner product induced by $B=B(\cdot,\cdot)$ :

B(u_{i},u_{j})=\delta_{ij},\quad A(u_{j},v)=\lambda_{j}B(u_{j},v),\ \forall v\in V,\quad i,j=1,2,\cdots.

The $j$ -th eigenvalue of (3) is given by the mini-max principle

\lambda_{j}=\min_{L_{j}}\max_{v\in L_{j}\setminus\{0\}}R[v]=\max_{W_{j}}\min_{v\in W_{j}\setminus\{0\}}R[v],

(6)

where

R[v]=\frac{A(v,v)}{B(v,v)}

is the Rayleigh quotient, and $\{L_{j}\}$ and $\{W_{j}\}$ denote the families of all subspaces of $V$ with dimension and codimension $j$ and $j-1$ , respectively. See [11], for example, for these fundamental facts.

Let

T_{t}:\Omega\rightarrow\Omega_{t}=T_{t}(\Omega),\quad|t|<\varepsilon_{0}

(7)

be a family of bi-Lipschitz homeomorphisms for $\varepsilon_{0}>0$ , satisfying $T_{0}=I$ , the identity mapping. We assume that $T_{t}x$ is continuous in $t$ uniformly in $x\in\Omega$ , and recall the following definition used in [15].

Definition 1.

The family $\{T_{t}\}$ of bi-Lipschitz homeomorphisms is said to be $p$ -differentiable in $t$ for $p\geq 1$ , if $T_{t}x$ is $p$ -times differentiable in $t$ for any $x\in\Omega$ and the mappings

\frac{\partial^{\ell}}{\partial t^{\ell}}DT_{t},\ \frac{\partial^{\ell}}{\partial t^{\ell}}(DT_{t})^{-1}:\Omega\rightarrow M_{n}(\mathbb{R}),\quad 0\leq\ell\leq p\\

are uniformly bounded in $(x,t)\in\Omega\times(-\varepsilon_{0},\varepsilon_{0})$ , where $DT_{t}$ denotes the Jacobi matrix of $T_{t}:\Omega\rightarrow\Omega_{t}$ and $M_{n}(\mathbb{R})$ stands for the set of real $n\times n$ matrices. This $\{T_{t}\}$ is said to be continuously $p$ -differentiable in $t$ if it is $p$ -differentiable and the mappings

t\in(-\varepsilon_{0},\varepsilon_{0})\mapsto\frac{\partial^{\ell}}{\partial t^{\ell}}DT_{t},\ \frac{\partial^{\ell}}{\partial t^{\ell}}(DT_{t})^{-1}\in L^{\infty}(\Omega\rightarrow M_{n}(\mathbb{R})),\quad 0\leq\ell\leq p

are continuous.

Putting

T_{t}(\gamma_{i})=\gamma_{it},\quad i=0,1,

(8)

we introduce the other eigenvalue problem

-\Delta u=\lambda u\ \mbox{in $\Omega_{t}$},\quad u=0\ \mbox{on $\gamma_{0t}$},\quad\frac{\partial u}{\partial\nu}=0\ \mbox{on $\gamma_{1t}$},

(9)

which is reduced to finding

u\in V_{t},\ \int_{\Omega_{t}}u^{2}\ dx=1,\quad\int_{\Omega_{t}}\nabla u\cdot\nabla v\ dx=\lambda\int_{\Omega_{t}}uv\ dx,\ \forall v\in V_{t}

(10)

for

V_{t}=\{v\in H^{1}(\Omega_{t})\mid\left.v\right|_{\gamma_{0t}}=0\}.

(11)

Let $\lambda_{j}(t)$ be the $j$ -th eigenvalue of the eigenvalue problem (9). In Section 4 we confirm that the eigenvalue problem (10)-(11) is reduced to

u\in V,\ B_{t}(u,u)=1,\quad A_{t}(u,v)=\lambda B_{t}(u,v),\ \forall v\in V

(12)

by the transformation of variables $y=T_{t}x$ for $V\subset H^{1}(\Omega)$ defined by (4), where

B_{t}(u,v)=\int_{\Omega}uva_{t}\ dx,\quad A_{t}(u,v)=\int_{\Omega}Q_{t}[\nabla u,\nabla v]a_{t}\ dx,

(13)

and

a_{t}=\det DT_{t},\quad Q_{t}=(DT_{t})^{-1}(DT_{t})^{-1T}.

(14)

Several representation formulae for

\lambda^{\prime}_{j}(t)=\lim_{h\rightarrow 0}\frac{1}{h}(\lambda_{j}(t+h)-\lambda_{j}(t))

and

\lambda_{j}^{\prime\prime}(t)=\lim_{h\rightarrow 0}\frac{1}{h}(\lambda^{\prime}_{j}(t+h)-\lambda^{\prime}_{j}(t))

have been derived in [3]. Here we characterize these derivatives, using associated finite dimensional eigenvalue problems (Theorems 1 and 4). We prove also the existence of derivatives, particularly, the second derivatives of multiple eigenvalues. These results follow from the differentiability of $A_{t}$ and $B_{t}$ in $t$ , defined by

\dot{A}_{t}(u,v)=\frac{d}{dt}A_{t}(u,v),\ \dot{B}_{t}(u,v)=\frac{d}{dt}B_{t}(u,v),\quad u,v\in V

(15)

and

\ddot{A}_{t}(u,v)=\frac{d^{2}}{dt^{2}}A_{t}(u,v),\ \ddot{B}_{t}(u,v)=\frac{d^{2}}{dt^{2}}B_{t}(u,v),\quad u,v\in V.

(16)

Then we show $C^{1}$ and $C^{2}$ rearrangements of eigenvalues, if these bilinear forms are continuous in $t$ (Theorems 3 and 6). We give the algorithm explicitly, as the transversal rearrangement in Definition 3. Consequently, no rearrangment is necessary to confirm $C^{1}$ or $C^{2}$ smoothness of eigenvalues, if their multiplicities are constant. Also elementary symmetric functions made by possible multiple eigenvalues are $C^{1}$ or $C^{2}$ . These properties are noticed by [7] and [8] in the real analytic category.

In [4], unilateral derivatives of the first order of the eigenvalue of the Stokes operator are calculated. It characterizes the derivatives using boundary integrals when the deformation of the domain is of normal direction. Our abstract theory reproduces and extends this result by the use of Piola transformation described in [9].

Our argument is executed in the $H^{1}$ category without requiring any further elliptic regularities. Hence the Lipschitz continuity of $\partial\Omega$ is sufficient to ensure all the results on (9) under the general perturbation of domains. We recall, in this context, that numerical computations on partial differential equations are mostly executed on Lipschitz domains. There, perturbation of the domain using Lipschitz continuous vector fields is often applied. This is the method of trial domains, efficient even to the case that normal perturbation of domains, called in [12], does not work because of the presence of corners on the boundary [12, 13].

2. Summary

As is noted in the previous section, $C^{1}$ and analytic categories for the smoothness of $\lambda_{j}(t)$ in $t$ have been discussed. Since each eigenvalue $\lambda$ of (12) is isolated, we can reduce this problem to a finite dimensional eigenvalue problem by the Lyapunov-Schmidt reduction as in p.486 of [1]. Rellich [10] in p.45 showed in this case that the continuous differentiablity in $t$ of $A_{t}$ and $B_{t}$ implies that of $\lambda_{j}(t)$ under a suitable change of its order. Kato [5] in p.123 then provided an alternative proof of this $C^{1}$ rearrangement.

The other category of analyticity is studied in Chapter II of [10] and p.370 of [5]. If $A_{t}$ and $B_{t}$ are analytic in $t$ , the eigenvlue problem is reduced to an algebraic equation with analytic coefficients in $t$ . The Puiseux expansion of $\lambda_{j}(t)$ at $t=0$ is described in p.31 of [10] and Chapter 2, Section 1 of [5]. Hence this $\lambda_{j}(t)$ is realized as an analytic function defined on a Riemann surface.

We study $C^{2}$ category. To begin with, we confirm Rellich’s theorem on $C^{1}$ category, the continuous differentiablity of rearranged eigenvalues. Here we show this rearrangement explicitly, to reach existence, characterization, and continuity of the second derivatives (Definition 3). In more details, first, the existence of $\dot{A}_{t}$ and $\dot{B}_{t}$ in (15) implies that of the first unilateral derivatives

\dot{\lambda}^{\pm}_{j}(t)=\lim_{h\rightarrow\pm 0}\frac{1}{h}(\lambda_{j}(t+h)-\lambda_{j}(t))

(17)

for each $j$ and $t$ . These derivatives, furthermore, are characterized by the other finite dimensional eigenvalue problems in accordance with the multiplicity of $\lambda_{j}(t)$ (Theorem 12). Second, if the above $\dot{A}_{t}$ and $\dot{B}_{t}$ are continuous in t, and if

\lambda_{k-1}(t)<\lambda_{k}(t)\leq\cdots\leq\lambda_{k+m-1}(t)<\lambda_{k+m}(t)

(18)

holds for $t\in I=(-\varepsilon_{0},\varepsilon_{0})$ , there are $C^{1}$ curves

\tilde{C}_{j},\ k\leq j\leq k+m-1,

made by at most countably many rearrangements of the $C^{0}$ curves

C_{j}=\{\lambda_{j}(t)\mid t\in I\},\quad k\leq j\leq k+m-1.

(Theorem 14).

In this paper, we notice two properties of the unilateral derivatives $\dot{\lambda}_{j}^{\pm}$ , for the proof of this Rellich’s theorem. First, there arises that

\dot{\lambda}_{j}^{+}(t)=\dot{\lambda}_{2\ell+n-1-j}^{-}(t),\quad\ell\leq j\leq\ell+n-1

\lambda_{\ell-1}(t)<\lambda_{\ell}(t)=\cdots=\lambda_{\ell+n-1}(t)<\lambda_{\ell+n}(t)

(19)

holds for $\ell\geq k$ , $\ell+n\leq m$ (Theorem 12). Second, the unilateral derivative $\dot{\lambda}_{j}^{\pm}$ are provided with the unilateral continuity as in

\lim_{h\rightarrow\pm 0}\dot{\lambda}_{j}^{\pm}(t+h)=\dot{\lambda}_{j}^{\pm}(t)

if both $\dot{A}_{t}$ and $\dot{B}_{t}$ are continuous in $t\in I$ (Theorem 13).

As for the second derivatives of eigenvalues, we assume the existence of $\ddot{A}_{t}$ and $\ddot{B}_{t}$ in (16) besides $\dot{A}_{t}$ and $\dot{B}_{t}$ . Then each $j=1,2,\cdots$ admits the existence of

\ddot{\lambda}_{j}^{\pm}(t)=\lim_{h\rightarrow\pm 0}\frac{2}{h^{2}}(\lambda_{j}(t+h)-\lambda_{j}(t)-h\dot{\lambda}_{j}^{\pm}(t)).

(20)

These limits are again unilateral and characterized by the other eigenvalue problem in a finite dimensional space (Theorem 15). The unilateral continuity of these $\ddot{\lambda}_{j}^{\pm}(t)$ is then assured under the continuity of $\ddot{A}_{t}$ and $\ddot{B}_{t}$ in $t$ , similarly. These properties induce $C^{2}$ smoothness of $\tilde{C}_{j}$ , $k\leq j\leq k+m-1$ , once their $C^{1}$ smoothness is achieved (Theorem 24).

This paper is composed of eight sections. The results on the Hadamard variation to (9) are described in the next section. In Section 4 we use the transformation of variables to introduce an abstract setting of the problem. Section 5 is concerned on the continuity of eigenvalues and eigenspaces. We study the first derivative of eigenvalues in Section 6, and then its $C^{1}$ rearrangement in Section 7. Finally, Section 8 is devoted to the second derivatives.

3. Hadamard variation

Here we state the results on the first and the second derivatives of $\lambda_{j}=\lambda_{j}(t)$ in (12). Fix $t\in I$ , and assume (19) for $\ell=k$ and $n=m$ for $k,m=1,2,\cdots$ with the convention $\lambda_{0}(t)=-\infty$ . Put

\lambda\equiv\lambda_{k}(t)=\cdots=\lambda_{k+m-1}(t),

(21)

and let $Y^{\lambda}_{t}$ , $\dim Y^{\lambda}_{t}=m$ , be the eigenspace corresponding to this eigenvalue $\lambda$ .

Theorem 1.

Assume the above situation, and let $\{T_{t^{\prime}}\}$ be $1$ -differentiable at $t^{\prime}=t$ . Then, there exist the unilateral limits

\dot{\lambda}_{j}^{\pm}=\lim_{h\rightarrow\pm 0}\frac{1}{h}(\lambda_{j}(t+h)-\lambda),\quad k\leq j\leq k+m-1,

which satisfies

\nu_{j}\equiv\dot{\lambda}_{j}^{+}=\dot{\lambda}_{2k+m-1-j}^{-},\quad k\leq j\leq k+m-1.

(22)

This $\nu_{j}$ is the $q$ -th eigenvalue of the matrix

G^{\lambda}_{t}=\left(E^{\lambda}_{t}(\tilde{\phi}_{i},\tilde{\phi}_{j})\right)_{1\leq i,j\leq m}

(23)

for $q=j-k+1$ , where $\{\tilde{\phi}_{j}\mid 1\leq j\leq m\}$ is a basis of $Y^{\lambda}_{t}$ , satisfying

B_{t}(\tilde{\phi}_{i},\tilde{\phi}_{j})=\delta_{ij},\quad 1\leq i,j\leq m

(24)

and

E^{\lambda}_{t}=\dot{A}_{t}-\lambda\dot{B}_{t}

for $\dot{A}_{t}$ and $\dot{B}_{t}$ defined by (13), (14), and (15).

Remark 1.

If $\langle\phi_{j}\mid 1\leq j\leq m\rangle$ is the other basis of $Y^{\lambda}_{t}$ satisfying

B_{t}(\phi_{i},\phi_{j})=\delta_{ij},\quad 1\leq i,j\leq m,

(25)

it holds that

\phi_{j}=\sum_{i=1}^{m}q^{i}_{j}\tilde{\phi}_{i},\ 1\leq j\leq m

with the orthogonal $m\times m$ matrix $Q=(q^{i}_{j})$ . Hence $\nu_{j}$ , $k\leq j\leq k+m-1$ , in (22) is determined, indpendent of the choice of $\langle\tilde{\phi}_{j}\mid 1\leq j\leq m\rangle$ .

Remark 2.

Under (21), it holds that

\dot{\lambda}_{k}^{+}(t)\leq\cdots\leq\dot{\lambda}_{k+m-1}^{+}(t),\quad\dot{\lambda}_{k}^{-}(t)\geq\cdots\geq\dot{\lambda}_{k+m-1}^{-}(t).

A direct consequnece of this theorem is the existence of the unilateral derivatives $\dot{\lambda}_{j}^{\pm}(t)$ in (17) for any $t\in I$ and $j=1,2,\cdots$ , if $\{T_{t}\}$ is $1$ -differentiable in $I$ . Then we obtain the following theorem.

Theorem 2.

If $\{T_{t}\}$ is continuously $1$ -differentiable in $t\in I$ , the functions $\dot{\lambda}^{\pm}_{j}=\dot{\lambda}^{\pm}_{j}(t)$ are unilaterally continuous, so that it holds that

\lim_{h\rightarrow\pm 0}\dot{\lambda}_{j}^{\pm}(t+h)=\dot{\lambda}_{j}^{\pm}(t)

for any $t$ and $j$ .

This fact ensures the following theorem of Rellich.

Theorem 3.

Let $\{T_{t}\}$ be continuously $1$ -differentiable in $t$ , and assume

\lambda_{k-1}(t)<\lambda_{k}(t)\leq\cdots\leq\lambda_{k+m-1}(t)<\lambda_{k+m}(t),\quad\forall t\in I

for some $k,m=1,2,\cdots$ . Let

C_{j}=\{\lambda_{j}(t)\mid t\in I\},\quad k\leq j\leq k+m-1

be $C^{0}$ curves. Then, there exist $C^{1}$ curves denoted by $\tilde{C}_{i}$ , $k\leq i\leq k+m-1$ , made by a rearrangement of

\{C_{j}\mid k\leq j\leq k+m-1\}

at most countably many times.

Turning to the second derivatives, we fix $t\in I$ again, and assume that $\{T_{t^{\prime}}\}$ is twice differentiable at $t^{\prime}=t$ . Suppose (19) for $\ell=k$ and $n=m$ , put $\lambda$ as in (21), and let $k\leq\ell<r\leq k+m$ be such that

\dot{\lambda}^{+}_{\ell-1}<\lambda^{\prime}\equiv\dot{\lambda}^{+}_{\ell}=\cdots=\dot{\lambda}^{+}_{r-1}<\dot{\lambda}^{+}_{r}

(26)

in Theorem 1. To state the finite dimensonal eigenvalue problem characterizing the second derivatives of $\lambda_{j}(t^{\prime})$ at $t^{\prime}=t$ for $\ell\leq j\leq r-1$ , we introduce the following definition.

Definition 2.

Let $R:X=L^{2}(\Omega)\rightarrow Y^{\lambda}_{t}$ be the orthogonal projection with respect to $B_{t}(\cdot,\cdot)$ and $P=I-R$ , where $I:X\rightarrow X$ is the identity operator. Let, furthermore, $A_{t}$ , $B_{t}$ , $\dot{A}_{t}$ , $\dot{B}_{t}$ , $\ddot{A}_{t}$ , and $\ddot{B}_{t}$ be as in (13), (14), (15), and (16). Then we define $w=\gamma(u)\in PV$ for $u\in V$ by

C_{t}(w,v)=-\dot{C}_{t}^{\lambda,\lambda^{\prime}}(u,v),\quad\forall v\in PV,

(27)

where

C_{t}=A_{t}-\lambda B_{t},\quad\dot{C}_{t}^{\lambda,\lambda^{\prime}}=\dot{A}_{t}-\lambda\dot{B}_{t}-\lambda^{\prime}B_{t}.

We put also

F_{t}^{\lambda,\lambda^{\prime}}(u,v)=(\ddot{A}_{t}-\lambda\ddot{B}_{t}-2\lambda^{\prime}\dot{B}_{t})(u,v)-2C_{t}(\gamma(u),\gamma(v)),\quad u,v\in V.

Remark 3.

To confirm the unique solvability of $w=\gamma(u)$ , let $Q:L^{2}(\Omega)\rightarrow Z^{k}_{t}$ be the orthogonal projection with respect to $B_{t}(\cdot,\cdot)$ , where $Z_{t}^{k}$ denotes the finite dimensional space of $L^{2}(\Omega)$ generated by the eigenfunctions corresponding to the eigenvalues $\lambda_{1}(t),\cdots,\lambda_{k-1}(t)$ . Let, furthermore, $V_{0}=QV$ and $V_{1}=(I-Q)RV$ . First, there is a unique $w_{0}\in V_{0}$ satisfying

C_{t}(w_{0},v)=-\dot{C}_{t}^{\lambda,\lambda^{\prime}}(u,v),\quad\forall v\in V_{0}

(28)

because $C_{t}$ is negative definite on $V_{0}\times V_{0}$ . Second, there is also a unique $w_{1}\in V_{1}$ satisfying

C_{t}(w_{1},v)=-\dot{C}_{t}^{\lambda,\lambda^{\prime}}(u,v),\quad\forall v\in V_{1}

(29)

because $C_{t}$ is positive definite on $V_{1}\times V_{1}$ . Then we obtain (27) for $w=w_{0}+w_{1}$ , because

C_{t}(w_{0},v)=C_{t}(w_{0},Qv),\ C_{t}(w_{1},v)=C_{t}(w_{1},(I-Q)v),\quad\forall v\in RV,

and hence

	$\displaystyle C_{t}(w,v)$	$\displaystyle=$	$\displaystyle C_{t}(w_{0},v)+C_{t}(w_{1},v)=C_{t}(w_{0},Qv)+C_{t}(w_{1},(I-Q)v)$
		$\displaystyle=$	$\displaystyle-\dot{C}_{t}^{\lambda,\lambda^{\prime}}(u,Qv)-\dot{C}_{t}^{\lambda,\lambda^{\prime}}(u,(I-Q)v)=-\dot{C}_{t}(u,v),\quad\forall v\in RV.$

We thus obtain $w\in RV$ satisfying (27). If $w\in RV$ is a solution to (27), conversely, then $w_{0}=Qw\in V_{0}$ and $w_{1}=(I-Q)w\in V_{1}$ solve (28) and (29), respectively. Then there arises the uniqueness of such $w=w_{0}+w_{1}$ because these $w_{0}\in V_{0}$ and $w_{1}\in V_{1}$ are unique.

Recall

Y^{\lambda}_{t}=\langle\tilde{\phi}_{j}\mid k\leq j\leq k+m-1\rangle

with (24).

Theorem 4.

Under the above situation of (19) and (26), there exist

\lambda^{\prime\prime}_{j}=\lim_{h\rightarrow+0}\frac{2}{h^{2}}(\lambda_{j}(t+h)-\lambda-h\lambda^{\prime}),\quad\ell\leq j\leq r-1.

(30)

This $\lambda^{\prime\prime}_{j}$ is the $q$ -th eigenvalue of the matrix

H^{\lambda,\lambda^{\prime}}_{t}=\left(F^{\lambda,\lambda^{\prime}}_{t}(\tilde{\phi}_{i},\tilde{\phi}_{j})\right)_{\ell\leq i,j\leq r-1}

(31)

for $q=j-\ell+1$ . If (26) is replaced by

\dot{\lambda}^{-}_{\ell-1}(t)>\lambda^{\prime}\equiv\dot{\lambda}^{-}_{\ell}(t)=\cdots=\dot{\lambda}_{r-1}^{-}(t)>\dot{\lambda}_{r}^{-}(t),

there arises that

\lambda_{j}^{\prime\prime}(t)=\lim_{h\rightarrow-0}\frac{2}{h^{2}}(\lambda_{j}(t+h)-\lambda-h\lambda^{\prime}),\quad\ell\leq j\leq r-1.

As in Remark 1 on the first derivative, the above $\lambda^{\prime\prime}_{j}$ , $\ell\leq j\leq r-1$ , are determined independent of the choice of

\langle\tilde{\phi}_{j}\mid\ell\leq j\leq r-1\rangle.

Theorems 1, 4 imply also the existence of the unilateral limits $\ddot{\lambda}_{j}^{\pm}(t)$ in (20) for any $t$ and $j$ if $\{T_{t}\}$ is $2$ -differentiable in $t\in I$ , that is,

\ddot{\lambda}_{j}^{\pm}(t)=\lim_{h\rightarrow\pm 0}\frac{2}{h^{2}}(\lambda_{j}(t+h)-\lambda_{j}(t)-h\dot{\lambda}_{j}^{\pm}(t)).

Remark 4.

By Liouville’s theorem on general perturbation of domains studied in [15], the matrix $G_{t}^{\lambda}$ in (23) is represented by the surface integrals of $\tilde{\phi}_{j}$ , $1\leq j\leq m$ . This property is confirmed by [4] for the Stokes operator with Dirichlet condition under a special perturbation of domains, called the normal perturbation in [14]. Similarly, the matrix $H_{t}^{\lambda,\lambda^{\prime}}$ in (31) is represented by the surface integrals of $\tilde{\phi}_{j}$ and $\gamma(\tilde{\phi}_{j})$ , $1\leq j\leq m$ .

Then we obtain the following theorems.

Theorem 5.

If $\{T_{t}\}$ is continuously $2$ -differentiable in $t$ , then $\ddot{\lambda}_{j}^{\pm}=\ddot{\lambda}_{j}^{\pm}(t)$ are unilaterally continuous, so that it holds that

\lim_{h\rightarrow\pm 0}\ddot{\lambda}_{j}^{\pm}(t+h)=\ddot{\lambda}_{j}^{\pm}(t)

for any $t\in I$ and $j=1,2,\cdots$ .

Theorem 6.

If $\{T_{t}\}$ is continuously $2$ -differentiable in $t$ , the $C^{1}$ curves $\tilde{C}_{j}$ , $k\leq j\leq k+m-1$ in Theorem 3 are $C^{2}$ .

Although the above theorems are to be extended to $p\geq 3$ of Definition 1, we restrict ourselves to $p=1,2$ in this paper. Yet these results on $p=2$ are efficient to examine the harmonic concavity of $\lambda_{j}(t)$ in $t$ studied by [3] for the first eigenvalue to (2) with $\gamma_{1}=\emptyset$ . We emphasize also that the treatment of the second derivatives is rather different from that of the first ones. See Remark 10 in Section 8.

4. Reduction to the abstract theory

For the moment, we fix $t$ and treat the bi-Lipschitz homeomorphism $T=T_{t}:\Omega\rightarrow\Omega_{t}=T_{t}\Omega$ . Let

\tilde{\Omega}=\Omega_{t},\quad f(y)=g(x),\ y=Tx,

and confirm the chain rule for this transformation of variables, that is,

\nabla g=\nabla f\ DT,\quad dy=(\det DT)dx

for $\nabla g=(\frac{\partial g}{\partial x_{1}},\cdots,\frac{\partial g}{\partial x_{n}})$ , $\nabla f=(\frac{\partial f}{\partial y_{1}},\cdots,\frac{\partial f}{\partial y_{n}})$ , and

DT=\left(\begin{array}[]{ccc}\frac{\partial y_{1}}{\partial x_{1}}&\cdots&\frac{\partial y_{1}}{\partial x_{n}}\\ \cdot&&\cdot\\ \frac{\partial y_{n}}{\partial x_{1}}&\cdots&\frac{\partial y_{n}}{\partial x_{n}}\end{array}\right).

Putting $\tilde{\gamma}_{i}=\gamma_{it}=T\gamma_{i}$ for $i=0,1$ , we take the eigenvalue problem

-\Delta u=\lambda u\ \mbox{in $\tilde{\Omega}$},\quad u=0\ \mbox{on $\tilde{\gamma}_{0}$},\quad\frac{\partial u}{\partial\nu}=0\ \mbox{on $\tilde{\gamma}_{1}$},

(32)

that is, (9) for $T=T_{t}$ . Let

\tilde{V}=\{v\in H^{1}(\tilde{\Omega})\mid\left.v\right|_{\tilde{\gamma}_{0}}=0\},

and introduce the weak form of (32),

u\in\tilde{V},\ \tilde{A}(u,v)=\lambda\tilde{B}(u,v),\quad\forall v\in\tilde{V},

(33)

where

\tilde{A}(u,v)=\int_{\tilde{\Omega}}\nabla_{y}u\cdot\nabla_{y}v\ dy,\quad\tilde{B}(u,v)=\int_{\tilde{\Omega}}uv\ dy.

(34)

Given $\phi\in V$ , put

\psi(y)=\phi(x),\quad y=Tx.

Then it holds that

\phi\in V\ \Leftrightarrow\ \psi\in\tilde{V}

for $V\subset H^{1}(\Omega)$ defined by (4). Writing

U(x)=u(y),\ V(x)=v(y),\quad y=Tx,

(35)

we obtain

$\displaystyle\nabla_{y}u\cdot\nabla_{y}v$	$\displaystyle=$	$\displaystyle[\nabla_{x}U(DT)^{-1}]\cdot[\nabla_{x}V(DT)^{-1}]$
	$\displaystyle=$	$\displaystyle(\nabla_{x}U)(DT)^{-1}(DT)^{-1T}(\nabla_{x}V)^{T}$
	$\displaystyle=$	$\displaystyle(\nabla_{x}U)Q(\nabla_{x}V)^{T}=Q[\nabla_{x}U,\nabla_{x}V]$

for $Q=(DT)^{-1}(DT)^{-1T}$ , where $F^{T}$ denotes the transpose of the matrix $F$ . Then, (33) means

\int_{\Omega}Q[\nabla_{x}U,\nabla_{x}V]a\ dx=\lambda\int_{\Omega}UVa\ dx

for $a=\det DT$ . The condition of normalization

\int_{\tilde{\Omega}}u^{2}\ dy=1

is also transformed into

\int_{\Omega}U^{2}a\ dx=1.

Under the family $\{T_{t}\}$ of homeomorphisms, therefore, the weak form (10) of (9), is equivalent to (12) for $V\subset H^{1}(\Omega)$ defined by (4) and $B_{t}$ , $A_{t}$ defined by (13)-(14). Here we confirm the following lemma.

Lemma 7.

The $j$ -th eigenvalue of (9) is equal to that of (12).

Proof.

For the moment, let $\tilde{\lambda}_{j}(t)$ be the $j$ -th eigenvalues of (9) and let $\lambda_{j}(t)$ be that of (12) for (13) and (14). By the mini-max principle (6), it holds that

\tilde{\lambda}_{j}(t)=\min_{\tilde{L}_{j}}\max_{v\in\tilde{L}_{j}\setminus\{0\}}\tilde{R}_{t}[v]=\max_{\tilde{W}_{j}}\min_{v\in\tilde{W}_{j}\setminus\{0\}}\tilde{R}_{t}[v],

(36)

where $V_{t}\subset H^{1}(\Omega_{t})$ is defined by (11),

\tilde{R}_{t}[v]=\frac{\tilde{A}_{t}(v,v)}{\tilde{B}_{t}(v,v)},\quad\tilde{A}_{t}(u,v)=\int_{\Omega_{t}}\nabla u\cdot\nabla v\ dx,\quad\tilde{B}_{t}(u,v)=\int_{\Omega_{t}}uv\ dx,

and $\{\tilde{L}_{j}\}$ and $\{\tilde{W}_{j}\}$ are the families of all subspaces of $V_{t}$ with dimension and codimension $j$ and $j-1$ , respectively.

It holds also that

\lambda_{j}(t)=\min_{L_{j}}\max_{v\in L_{j}\setminus\{0\}}R_{t}[v]=\max_{W_{j}}\min_{v\in W_{j}\setminus\{0\}}R_{t}[v],

(37)

for

R_{t}[v]=\frac{A_{t}(v,v)}{B_{t}(v,v)},

(38)

and $\{L_{j}\}$ and $\{W_{j}\}$ denote the families of all subspaces of $V$ with dimension and codimension $j$ and $j-1$ , respectively.

If the set $L$ is a $j$ -dimensional subspace of $V$ there is $\phi_{\ell}\in L$ , $1\leq\ell\leq j$ , such that

\int_{\Omega}\phi_{\ell}\phi_{\ell^{\prime}}dx=\delta_{\ell\ell^{\prime}}

and

\phi=\sum_{\ell=1}^{j}c_{\ell}\phi_{\ell},\quad c_{\ell}=\int_{\Omega}\phi\phi_{\ell}\ dx

for any $\phi\in L$ , which implies

\psi=\sum_{\ell=1}^{j}c_{\ell}\psi_{\ell}

for $\psi=\phi\circ T_{t}^{-1}$ and $\psi_{\ell}=\phi\circ T_{t}^{-1}$ . Hence we obtain $\mbox{dim}\ \tilde{L}_{t}\leq\mbox{dim}\ L$ for

\tilde{L}_{t}=\{\phi\circ T_{t}^{-1}\mid\phi\in L\}.

The reverse inequality follows similarly, and hence it holds that

\mbox{dim}\ \tilde{L}_{t}=\mbox{dim}\ L=j.

Since $T_{t}:\Omega\rightarrow\Omega_{t}$ is a bi-Lipschitz homeomorphism, furthermore, $L\subset V$ if and only if $\tilde{L}_{t}\subset V_{t}$ . We thus obtain

\tilde{\lambda}_{j}(t)=\lambda_{j}(t)

(39)

by (36)-(37). ∎

We are ready to develop an abstract theory, writing $L^{2}$ norm in $X=L^{2}(\Omega)$ as $|\ \cdot\ |_{X}$ . With $V$ in (4), we recall that $\|\ \cdot\ \|_{V}$ denotes the norm in $V$ and that the inclusion $V\hookrightarrow X$ is compact. It holds also that

|v|_{X}\leq K\|v\|_{V},\quad v\in V

(40)

for $K=1$ .

Henceforth, $C$ denotes the generic positive constant. The above $A_{t}:V\times V\rightarrow\mathbb{R}$ and $B_{t}:X\times X\rightarrow\mathbb{R}$ for $t\in I$ are symmetric bilinear forms, satisfying

|A_{t}(u,v)|\leq C\|u\|_{V}\|v\|_{V},\ A_{t}(u,u)\geq\delta\|u\|_{V}^{2},\quad u,v\in V

(41)

and

|B_{t}(u,v)|\leq C|u|_{X}|v|_{X},\ B_{t}(u,u)\geq\delta|u|_{X}^{2},\quad u,v\in X

(42)

for some $\delta>0$ . Then the eigenvalues of (12) are denoted by

0<\lambda_{1}(t)\leq\lambda_{2}(t)\leq\cdots\rightarrow+\infty.

The weak and the strong convergences of $\{u_{j}\}\subset Y$ to $u\in Y$ for $Y=X$ or $Y=V$ are, furthermore, indicated by

\mbox{w}\mathchar 45\relax\lim_{j\rightarrow\infty}u_{j}=u\ \mbox{in $Y$}

and

\mbox{s}\mathchar 45\relax\lim_{j\rightarrow\infty}u_{j}=u\ \mbox{in $Y$},

respectively.

5. Continuity of eigenvalues and eigenspaces

Let $t\in I$ be fixed. We begin with the following theorem valid under the abstract setting in the previous section.

Theorem 8.

The conditions

	$\displaystyle\lim_{h\rightarrow 0}\sup_{\\|u\\|_{V},\\|v\\|_{V}\leq 1}\|A_{t+h}(u,v)-A_{t}(u,v)\|=0$
	$\displaystyle\lim_{h\rightarrow 0}\sup_{\|u\|_{X},\|v\|_{X}\leq 1}\|B_{t+h}(u,v)-B_{t}(u,v)\|=0,$		(43)

imply

\lim_{h\rightarrow 0}\lambda_{j}(t+h)=\lambda_{j}(t)

(44)

for any $j=1,2,\cdots$ .

Proof.

We note that the $j$ -th eigenvalue of (9) is given by the mini-max principle as in (37), for the Rayleigh quotient $R_{t}[v]$ defined by (38).

Given $t$ , let

	$\displaystyle\alpha(h)=\sup_{\\|u\\|_{V},\\|v\\|_{V}\leq 1}\|(A_{t+h}-A_{t})(u,v)\|$
	$\displaystyle\beta(h)=\sup_{\|u\|_{X},\|v\|_{X}\leq 1}\|(B_{t+h}-B_{t})(u,v)\|.$		(45)

We obtain

(A_{t+h}-A_{t})(v,v)\geq-\alpha(h)\|v\|_{V}^{2}\geq-\frac{\alpha(h)}{\delta}A_{t}(v,v)

and

(A_{t+h}-A_{t})(v,v)\leq\alpha(h)\|v\|_{V}^{2}\leq\frac{\alpha(h)}{\delta}A_{t}(v,v)

by (41) and (43), which implies

(1-\delta^{-1}\alpha(h))A_{t}(v,v)\leq A_{t+h}(v,v)\leq(1+\delta^{-1}\alpha(h))A_{t}(v,v).

Similarly, there arises that

(1-\delta^{-1}\beta(h))B_{t}(v,v)\leq B_{t+h}(v,v)\leq(1+\delta^{-1}\beta(h))B_{t}(v,v)

for any $v\in V$ .

Then it follows that

(1-o(1))R_{t}[v]\leq R_{t+h}[v]\leq(1+o(1))R_{t}[v]

uniformly in $v\in V\setminus\{0\}$ , and hence

(1-o(1))\lambda_{j}(t)\leq\lambda_{j}(t+h)\leq(1+o(1))\lambda_{j}(t)

by (43). Thus we obtain (44). ∎

Let $u_{j}(t)\in V$ be the eigenfunction of (12) corresponding to the eigenvalue $\lambda_{j}(t)$ :

B_{t}(u_{j}(t),u_{j^{\prime}}(t))=\delta_{jj^{\prime}},\quad A_{t}(u_{j}(t),v)=\lambda_{j}(t)B_{t}(u_{j}(t),v),\ \forall v\in V.

(46)

Fix $t$ , assume (19), and define $\lambda$ by (21). Although this multiplicity $m$ is not stable under the perturbation of $t$ , we obtain the following theorem concerning the continuity of eigenspaces with respect to $t$ .

Let

Y_{t}^{\lambda}=\langle u_{j}(t)\mid k\leq j\leq k+m-1\rangle

(47)

be the subspace of $X$ generated by the above eigenfunctions $u_{j}(t)$ for $k\leq j\leq k+m-1$ .

Lemma 9.

Under the above situation, any $h_{\ell}\rightarrow 0$ admits a subsequence, denoted by the same symbol, such that the limits

\mbox{s}\mathchar 45\relax\lim_{\ell\rightarrow\infty}u_{j}(t+h_{\ell})=\phi_{j}\in Y_{t}^{\lambda}\ \mbox{in $V$},\quad k\leq j\leq k+m-1

(48)

exist. In particular it holds that

B_{t}(\phi_{j},\phi_{j^{\prime}})=\delta_{jj^{\prime}},\quad k\leq j,j^{\prime}\leq k+m-1.

(49)

Proof.

We have

\lim_{h\rightarrow 0}\lambda_{j}(t+h)=\lambda_{j}(t),\quad k\leq j\leq k+m-1

(50)

by Theorem 8. It holds also that

\|u_{j}(t^{\prime})\|_{V}\leq C,\quad k\leq j\leq k+m-1,\ |t^{\prime}|<\varepsilon_{0}

(51)

by (40)-(42) and (46):

\delta\|u_{j}(t^{\prime})\|_{V}^{2}\leq A_{t^{\prime}}(u_{j}(t^{\prime}),u_{j}(t^{\prime}))=\lambda_{j}(t^{\prime})B_{t}(u_{j}(t^{\prime}),u_{j}(t^{\prime}))=\lambda_{j}(t^{\prime}).

Given $h_{\ell}\rightarrow 0$ , therefore, we have a subsequence, denoted by the same symbol, which admits the weak limits

\mbox{w}\mathchar 45\relax\lim_{\ell\rightarrow\infty}u_{j}(t+h_{\ell})=\phi_{j}\ \mbox{in $V$},\quad k\leq j\leq k+m-1,

(52)

for some $\phi_{j}\in V$ . From (46) it follows that

A_{t}(\phi_{j},v)=\lambda_{j}(t)B_{t}(\phi_{j},v),\quad\forall v\in V

and hence

\phi_{j}\in Y_{t}^{\lambda},\quad k\leq j\leq k+m-1.

Since $V\hookrightarrow X$ is compact, the weak convergence (52) implies the strong convergence

\mbox{s}\mathchar 45\relax\lim_{\ell\rightarrow\infty}u_{j}(t+h_{\ell})=\phi_{j}\ \mbox{in $X$},

(53)

and hence (49). Now we improve this weak convergence (52) to the strong convergence (48) in $V$ , using (43).

For this purpose, we put

v=u_{j}(t+h_{\ell})-\phi_{j}

and recall (21) and (50). Then there arises that

$\displaystyle\delta\\|v\\|_{V}^{2}$	$\displaystyle\leq$	$\displaystyle A_{t+h_{\ell}}(u_{j}(t+h_{\ell})-\phi_{j},v)$
	$\displaystyle=$	$\displaystyle A_{t+h_{\ell}}(u_{j}(t+h_{\ell}),v)-(A_{t+h_{\ell}}-A_{t})(\phi_{j},v)-A_{t}(\phi_{j},v)$
	$\displaystyle=$	$\displaystyle\lambda_{j}(t+h_{\ell})B_{t+h_{\ell}}(u_{j}(t+h_{\ell}),v)-(A_{t+h_{\ell}}-A_{t})(\phi_{j},v)-\lambda_{j}(t)B_{t}(\phi_{j},v)$
	$\displaystyle=$	$\displaystyle(\lambda_{j}(t+h_{\ell})-\lambda_{j}(t))B_{t+h_{\ell}}(u_{j}(t+h_{\ell}),v)$
		$\displaystyle+\lambda_{j}(t)(B_{t+h_{\ell}}-B_{t})(u_{j}(t+h_{\ell}),v)+\lambda_{j}(t)B_{t}(u_{j}(t+h_{\ell})-\phi_{j},v)$
		$\displaystyle-(A_{t+h_{\ell}}-A_{t})(\phi_{j},v)$
	$\displaystyle\leq$	$\displaystyle C\|\lambda_{j}(t+h_{\ell})-\lambda_{j}(t)\|\cdot\|u_{j}(t+h_{\ell})\|_{X}\cdot K\\|v\\|_{V}$
		$\displaystyle+\lambda_{j}(t)\cdot\beta(h_{\ell})\cdot\|u_{j}(t+h_{\ell})\|_{X}\cdot K\\|v\\|_{V}$
		$\displaystyle+\lambda_{j}(t)\cdot C\|u_{j}(t+h_{\ell})-\phi_{j}\|_{X}\cdot K\\|v\\|_{V}+\alpha(h_{\ell})\\|\phi_{j}\\|_{V}\cdot\\|v\\|_{V},$

and hence

	$\displaystyle\\|v\\|_{V}$	$\displaystyle\leq$	$\displaystyle C\{\|\lambda_{j}(t+h_{\ell})-\lambda_{j}(t)\|+\beta(h_{\ell})+\|u_{j}(t+h_{\ell})-\phi_{j}\|_{X}+\alpha(h_{\ell})\}$
		$\displaystyle=$	$\displaystyle o(1)$

by (43) and (53). ∎

Remark 5.

If $m=1$ , the eigenfunction $u_{j}(t)$ in (58) is uniquely determined by (46) up to the multiplication of $\pm 1$ , which implies $\phi_{j}=\pm u_{j}(t)$ . In the other case of $m\geq 2$ , the eigenfunction which attains (39) does not satisfy this property. Hence we have

\phi_{j}=\sum_{i=1}^{m}q^{i}_{j}u_{i}(t)

for $Q=(q^{i}_{j})$ satisfying $Q^{T}=Q^{-1}$ in Theorem 9. In other words, the eigenfunction $u_{j}(t)$ corresponding to $\lambda_{j}(t)$ has more varieties than $\pm 1$ multiplication, although the eigenspace $Y_{t}^{\lambda}$ is determined. By this ambiguity the limits $\phi_{j}$ in (48) depend on the sequence $h_{\ell}\rightarrow 0$ , which makes the argument below to be complicated.

6. First derivatives

If $\{T_{t}\}$ is $1$ -differentiable in the setting of Section 1, we can put

\dot{A}_{t}(u,v)=\int_{\Omega}\dot{Q}_{t}[\nabla u,\nabla v]a_{t}+Q_{t}[\nabla u,\nabla v]\dot{a}_{t}\ dx,\quad u,v\in V

and

\dot{B}_{t}(u,v)=\int_{\Omega}uv\dot{a}_{t}\ dx,\quad u,v\in X.

These $\dot{A}_{t}:V\times V\rightarrow\mathbb{R}$ and $\dot{B}_{t}:X\times X\rightarrow\mathbb{R}$ in (13) are bilinear forms satisfying

	$\displaystyle\|\dot{A}_{t}(u,v)\|\leq C\\|u\\|_{V}\\|v\\|_{V},\quad u,v\in V$
	$\displaystyle\|\dot{B}_{t}(u,v)\|\leq C\|u\|_{X}\|v\|_{X},\quad u,v\in X$		(54)

and

	$\displaystyle\lim_{h\rightarrow 0}\frac{1}{h}\sup_{\\|u\\|_{V},\\|v\\|_{V}\leq 1}\left\|\left(A_{t+h}-A_{t}-h\dot{A}_{t}\right)(u,v)\right\|=0$
	$\displaystyle\lim_{h\rightarrow 0}\frac{1}{h}\sup_{\|u\|_{X},\|v\|_{X}\leq 1}\left\|\left(B_{t+h}-B_{t}-h\dot{B}_{t}\right)(u,v)\right\|=0.$		(55)

Hence Theorem 1 in Section 4 is reduced to the following abstract theorem. In this theorem, the assumption made by Theorem 8 is valid, and therefore, there arises that (44).

Theorem 10.

Let $X,V$ be Hilbert spaces over $\mathbb{R}$ , with compact embedding $V\hookrightarrow X$ . Let $A_{t}:V\times V\rightarrow\mathbb{R}$ and $B_{t}:X\times X\rightarrow\mathbb{R}$ be symmetric bilinear forms satisfying (41)-(42) for any $t\in I$ . Given $t$ , assume the existence of the bilinear forms $\dot{A}_{t}:V\times V\rightarrow\mathbb{R}$ and $\dot{B}_{t}:X\times X\rightarrow\mathbb{R}$ such that (54)-(55). Assume, finally, (19) with $\ell=k$ and $k=m$ for $\lambda_{j}(t)$ , $k\leq k+m-1$ , defined by (37)-(38). Then the conclusion of Theorem 1 holds.

Before proceeding to the rigorous proof, we develop a formal argument, writing (12) as

u_{t}\in V,\ B_{t}(u_{t},u_{t})=1,\quad A_{t}(u_{t},v)=\lambda_{t}B_{t}(u_{t},v),\ v\in V.

(56)

First, taking a formal differentiation in $t$ in this equality, we obtain

\dot{A}_{t}(u_{t},v)+A_{t}(\dot{u}_{t},v)=\dot{\lambda}_{t}B_{t}(u_{t},v)+\lambda_{t}\dot{B}_{t}(u_{t},v)+\lambda_{t}B_{t}(\dot{u}_{t},v),\quad\forall v\in V.

(57)

Putting $v=u_{t}$ , we obtain

\dot{A}_{t}(u_{t},u_{t})+A_{t}(\dot{u}_{t},u_{t})=\dot{\lambda}_{t}+\lambda_{t}\dot{B}_{t}(u_{t},u_{t})+\lambda_{t}B_{t}(\dot{u}_{t},u_{t})

(58)

B_{t}(u_{t},u_{t})=1.

(59)

To eliminate $\dot{u}_{t}$ in (58), second, we use (59) to deduce

\dot{B}_{t}(u_{t},u_{t})+2B_{t}(\dot{u}_{t},u_{t})=0.

(60)

From

\lambda_{t}=\lambda_{t}B_{t}(u_{t},u_{t})=A_{t}(u_{t},u_{t})

it is derived also that

\dot{\lambda}_{t}=\dot{A}_{t}(u_{t},u_{t})+2A_{t}(\dot{u}_{t},u_{t})

(61)

and then (58) is replaced by

\dot{A}_{t}(u_{t},u_{t})+\frac{1}{2}\dot{\lambda}_{t}-\frac{1}{2}\dot{A}_{t}(u_{t},u_{t})=\dot{\lambda}_{t}+\lambda_{t}\dot{B}_{t}(u_{t},u_{t})-\frac{1}{2}\lambda_{t}\dot{B}_{t}(u_{t},u_{t}),

or,

\dot{\lambda}_{t}=\dot{A}_{t}(u_{t},u_{t})-\lambda_{t}\dot{B}_{t}(u_{t},u_{t}).

(62)

As is noticed in Remark 5 in Section 5, if the eigenspace

Y_{t}^{\lambda}=\langle u_{j}(t)\mid k\leq j\leq k+m-1\rangle

corresponding to the eigenvalue $\lambda=\lambda_{j}(t)$ , $k\leq j\leq k+m-1$ , to (12), is one-dimensional as in $m=1$ , the eigenfunction $u_{t}$ in (56) is unique up to the multiplication of $\pm 1$ , and this ambiguity is canceled in (62). This property is valid even if $m\geq 2$ as in Remark 1.

Turning to the rigorous proof, we use the following lemma, recalling $u_{j}(t)\in V$ , $Y^{\lambda}_{t}$ , and $\phi_{j}$ in (46), (47), and (48), respectively.

Lemma 11.

Under the assumption of Theorem 10, any $h_{\ell}\rightarrow 0$ admits a subsequence, denoted by the same symbol, such that

\lim_{\ell\rightarrow\infty}\frac{1}{h_{\ell}}\{\lambda_{j}(t+h_{\ell})-\lambda_{j}(t)\}=\dot{A}_{t}(\phi_{j},\phi_{j})-\lambda_{j}(t)\dot{B}_{t}(\phi_{j},\phi_{j})

(63)

for $k\leq j\leq k+m-1$ . It holds also that

\dot{A}_{t}(\phi_{j},\phi_{j^{\prime}})-\lambda_{j}(t)\dot{B}_{t}(\phi_{j},\phi_{j^{\prime}})=0,\quad k\leq j\neq j^{\prime}\leq k+m-1.

(64)

Proof.

Let $k\leq j,j^{\prime}\leq k+m-1$ . Since

	$\displaystyle A_{t+h}(u_{j}(t+h)-\phi_{j},u_{j^{\prime}}(t+h)-\phi_{j^{\prime}})=A_{t+h}(u_{j}(t+h),u_{j^{\prime}}(t+h))$
	$\displaystyle\quad-A_{t+h}(u_{j}(t+h),\phi_{j^{\prime}})-A_{t+h}(\phi_{j},u_{j^{\prime}}(t+h))+A_{t+h}(\phi_{j},\phi_{j^{\prime}})$

and

	$\displaystyle A_{t}(u_{j}(t+h)-\phi_{j},u_{j^{\prime}}(t+h)-\phi_{j^{\prime}})=A_{t}(u_{j}(t+h),u_{j^{\prime}}(t+h))$
	$\displaystyle\quad-A_{t}(u_{j}(t+h),\phi_{j^{\prime}})-A_{t}(\phi_{j},u_{j^{\prime}}(t+h))+A_{t}(\phi_{j},\phi_{j^{\prime}}),$

it holds that

	$\displaystyle h(A_{t+h}-A_{t})\left(\frac{u_{j}(t+h)-\phi_{j}}{h},\frac{u_{j^{\prime}}(t+h)-\phi_{j^{\prime}}}{h}\right)$
	$\displaystyle\quad=\frac{1}{h}(A_{t+h}-A_{t})(u_{j}(t+h)-\phi_{j},u_{j^{\prime}}(t+h)-\phi_{j^{\prime}})$
	$\displaystyle\quad=\frac{1}{h}(A_{t+h}-A_{t})(u_{j}(t+h),u_{j^{\prime}}(t+h))+\frac{1}{h}(A_{t+h}-A_{t})(\phi_{j},\phi_{j^{\prime}})$
	$\displaystyle\qquad-\frac{1}{h}(A_{t+h}-A_{t})(u_{j}(t+h),\phi_{j^{\prime}})-\frac{1}{h}(A_{t+h}-A_{t})(\phi_{j},u_{j^{\prime}}(t+h)).$		(65)

In (65), first, we obtain

\frac{1}{h}(A_{t+h}-A_{t})(u_{j}(t+h),u_{j^{\prime}}(t+h))=\dot{A}_{t}(u_{j}(t+h),u_{j^{\prime}}(t+h))+o(1)

as $h\rightarrow 0$ by (51) and (55). Then (48) implies

	$\displaystyle\lim_{\ell\rightarrow\infty}\frac{1}{h_{\ell}}(A_{t+h_{\ell}}-A_{t})(u_{j}(t+h_{\ell}),u_{j^{\prime}}(t+h_{\ell}))$
	$\displaystyle\quad=\lim_{\ell\rightarrow\infty}\dot{A}_{t}(u_{j}(t+h_{\ell}),u_{j^{\prime}}(t+h_{\ell}))=\dot{A}_{t}(\phi_{j},\phi_{j^{\prime}})$		(66)

and also

\lim_{\ell\rightarrow\infty}\frac{1}{h_{\ell}}(A_{t+h_{\ell}}-A_{t})(\phi_{j},\phi_{j^{\prime}})=\dot{A}_{t}(\phi_{j},\phi_{j^{\prime}}).

(67)

Second, it holds that

	$\displaystyle A_{t+h}(u_{j}(t+h),\phi_{j^{\prime}})=\lambda_{j}(t+h)B_{t+h}(u_{j}(t+h),\phi_{j^{\prime}})$
	$\displaystyle A_{t}(u_{j}(t+h),\phi_{j^{\prime}})=\lambda_{j}(t)B_{t}(u_{j}(t+h),\phi_{j^{\prime}})$

by $\phi_{j^{\prime}}\in{Y_{t}^{\lambda}}$ , which implies

	$\displaystyle\frac{1}{h_{\ell}}(A_{t+h_{\ell}}-A_{t})(u_{j}(t+h_{\ell}),\phi_{j^{\prime}})$
	$\displaystyle\quad=\frac{1}{h_{\ell}}(\lambda_{j}(t+h_{\ell})B_{t+h_{\ell}}(u_{j}(t+h_{\ell}),\phi_{j^{\prime}})-\lambda_{j}(t)B_{t}(u_{j}(t+h_{\ell}),\phi_{j^{\prime}}))$
	$\displaystyle\quad=\frac{1}{h_{\ell}}(\lambda_{j}(t+h_{\ell})-\lambda_{j}(t))B_{t+h_{\ell}}(u_{j}(t+h_{\ell}),\phi_{j^{\prime}})$
	$\displaystyle\qquad+\frac{1}{h_{\ell}}\lambda_{j}(t)(B_{t+h_{\ell}}-B_{t})(u_{j}(t+h_{\ell}),\phi_{j^{\prime}})$
	$\displaystyle\quad=\frac{1}{h_{\ell}}(\lambda_{j}(t+h_{\ell})-\lambda_{j}(t))(\delta_{jj^{\prime}}+o(1))+\lambda_{j}(t)\dot{B}_{t}(\phi_{j},\phi_{j^{\prime}})+o(1)$		(68)

	$\displaystyle B_{t+h_{\ell}}(u_{j}(t+h_{\ell}),\phi_{j^{\prime}})$	$\displaystyle=$	$\displaystyle B_{t}(u_{j}(t+h_{\ell}),\phi_{j^{\prime}})+o(1)$
		$\displaystyle=$	$\displaystyle B_{t}(\phi_{j},\phi_{j^{\prime}})+o(1)=\delta_{jj^{\prime}}+o(1)$

and (55). Similarly, it follows that

	$\displaystyle\frac{1}{h_{\ell}}(A_{t+h_{\ell}}-A_{t})(\phi_{j},u_{j^{\prime}}(t+h_{\ell}))$
	$\displaystyle\quad=\frac{1}{h_{\ell}}(\lambda_{j}(t+h_{\ell})-\lambda_{j}(t))(\delta_{jj^{\prime}}+o(1))+\lambda_{j}(t)\dot{B}_{t}(\phi_{j},\phi_{j^{\prime}})+o(1).$		(69)

Finally, we obtain

	$\displaystyle\left\|h_{\ell}(A_{t+h_{\ell}}-A_{t})\left(\frac{u_{j}(t+h_{\ell})-\phi_{j}}{h_{\ell}},\frac{u_{j^{\prime}}(t+h_{\ell})-\phi_{j^{\prime}}}{h_{\ell}}\right)\right\|$
	$\displaystyle\quad\leq Ch_{\ell}^{2}\left\\|\frac{u_{j}(t+h_{\ell})-\phi_{j}}{h_{\ell}}\right\\|_{V}\cdot\left\\|\frac{u_{j^{\prime}}(t+h_{\ell})-\phi_{j^{\prime}}}{h_{\ell}}\right\\|_{V}$
	$\displaystyle\quad=C\\|u_{j}(t+h_{\ell})-\phi_{j}\\|_{V}\cdot\\|u_{j^{\prime}}(t+h_{\ell})-\phi_{j^{\prime}}\\|_{V}=o(1)$		(70)

by (48) and (55). Then equalities (63)-(64) follow from (66)-(70) as

	$\displaystyle 0$	$\displaystyle=$	$\displaystyle 2\dot{A}_{t}(\phi_{j},\phi_{j^{\prime}})-\frac{2}{h_{\ell}}(\lambda_{j}(t+h_{\ell})-\lambda_{j}(t))(\delta_{jj^{\prime}}+o(1))$
			$\displaystyle-2\lambda_{j}(t)\dot{B}_{t}(\phi_{j},\phi_{j^{\prime}})+o(1),\quad\ell\rightarrow\infty.$

. ∎

Below we confirm that the process of taking subsequence in the previous lemma is not necessary, if $h_{\ell}\rightarrow 0$ is unilateral as in $h_{\ell}\rightarrow+0$ or $h_{\ell}\rightarrow-0$ . Theorem 1 is thus reduced to the following theorem.

Theorem 12.

Under the assumption of Theorem 10, the unilateral limits

\dot{\lambda}_{j}^{\pm}(t)=\lim_{h\rightarrow\pm 0}\frac{1}{h}\{\lambda_{j}(t+h)-\lambda_{j}(t)\}

(71)

exist, and it holds that

\dot{\lambda}^{+}_{j}(t)=\mu_{j-k+1},\ \dot{\lambda}_{j}^{-}(t)=\mu_{k+m-j},\quad k\leq j\leq k+m-1.

(72)

Here, $\mu_{q}$ , $1\leq q\leq m$ , is the $q$ -th eigenvalue of

u\in Y_{t}^{\lambda},\quad E^{\lambda}_{t}(u,v)=\mu B_{t}(u,v),\ \forall v\in Y_{t}^{\lambda},

(73)

where $Y_{t}^{\lambda}$ is the $m$ -dimensional eigenspace of (12) corresponding to the eigenvalue $\lambda$ of (21) defined by (47), and

E^{\lambda}_{t}=\dot{A}_{t}-\lambda\dot{B}_{t}.

(74)

In particular, it holds that

\dot{\lambda}_{j}^{+}(t)=\dot{\lambda}_{2k+m-1-j}^{-}(t),\quad k\leq j\leq k+m-1.

Proof.

Since $Y_{t}^{\lambda}$ is $m$ -dimensional, the eigenvalue problem (73) admits $m$ -eigenvalues denoted by

\mu_{1}\leq\cdots\leq\mu_{m}.

By Lemma 11, on the other hand, any $h_{\ell}\rightarrow 0$ takes a subsequence, denoted by the same symbol, satisfying (63)-(64) for some

\phi_{j}\in Y_{t}^{\lambda},\quad k\leq j\leq k+m-1,

with (49).

This lemma ensures also the existence of

\tilde{\mu}_{j}=\lim_{\ell\rightarrow\infty}\frac{1}{h_{\ell}}(\lambda_{j}(t+h_{\ell})-\lambda_{j}(t)),

(75)

and the equalities

E^{\lambda}_{t}(\phi_{j},\phi_{j^{\prime}})=\delta_{jj^{\prime}}\tilde{\mu}_{j},\quad k\leq j,j^{\prime}\leq k+m-1.

(76)

We thus obtain

\phi_{j}\in{Y_{t}^{\lambda}},\quad B_{t}(\phi_{j},\phi_{j})=1,\quad E^{\lambda}_{t}(\phi_{j},v)=\tilde{\mu}_{j}B_{t}(\phi_{j},v),\ \forall v\in Y_{t}^{\lambda},

and therefore, $\mu=\tilde{\mu}_{j}$ is an eigenvalue of (73).

If $h_{\ell}\rightarrow+0$ , there arises that

\tilde{\mu}_{k}\leq\cdots\leq\tilde{\mu}_{k+m-1}

\lambda_{k}(t+h)\leq\cdots\leq\lambda_{k+m-1}(t+h),

and hence

\tilde{\mu}_{j}=\mu_{j-k+1},\quad k\leq j\leq k+m-1.

Then we obtain the result because the value $\tilde{\mu}_{j}$ in (75) is independent of the sequence $h_{\ell}\rightarrow+0$ .

In the other case of $h_{\ell}\rightarrow-0$ , we obtain

\tilde{\mu}_{j}=\mu_{k+m-j},\quad k\leq j\leq k+m-1,

and the result follows similarly. ∎

Theorem 2 is reduced to the following abstract theorem.

Theorem 13.

Let the assumption of Theorem 10 hold in $I$ . Fix $t\in I$ , and assume

	$\displaystyle\lim_{h\rightarrow 0}\sup_{\\|u\\|_{V},\\|v\\|_{V}\leq 1}\left\|\dot{A}_{t+h}(u,v)-\dot{A}_{t}(u,v)\right\|=0$
	$\displaystyle\lim_{h\rightarrow 0}\sup_{\|u\|_{X},\|v\|_{X}\leq 1}\left\|\dot{B}_{t+h}(u,v)-\dot{B}_{t}(u,v)\right\|=0.$		(77)

Then, it follows that

\lim_{h\rightarrow\pm 0}\dot{\lambda}_{j}^{\pm}(t+h)=\dot{\lambda}_{j}^{\pm}(t).

Proof.

Assume (19)-(21) and take $k\leq j\leq k+m-1$ . Since the assumption of Theorem 10 holds in $I$ , any $t^{\prime}\in I$ admits $u_{j}(t^{\prime})\in V$ such that

B_{t^{\prime}}(u_{j}(t^{\prime}),u_{j}(t^{\prime}))=1,\ A_{t^{\prime}}(u_{j}(t^{\prime}),u_{j}(t^{\prime}))=\lambda_{j}(t^{\prime})B_{t^{\prime}}(u_{j}(t^{\prime}),u_{j}(t^{\prime}))

(78)

and

\dot{\lambda}_{j}^{+}(t^{\prime})=\dot{A}_{t^{\prime}}(u_{j}(t^{\prime}),u_{j}(t^{\prime}))-\lambda_{j}(t^{\prime})\dot{B}_{t^{\prime}}(u_{j}(t^{\prime}),u_{j}(t^{\prime})).

(79)

by Lemma 11 and Theorem 12.

Given $t$ in this theorem, and take $h_{\ell}\rightarrow+0$ and $u_{j}(t^{\prime})$ in (78) for $t^{\prime}=t+h_{\ell}$ . Hence there is a subsequence denoted by the same symbol such that (48) with $\phi_{j}\in V$ . Then it holds that

\dot{\lambda}_{j}^{+}(t)=\dot{A}_{t}(\phi_{j},\phi_{j})-\lambda_{j}(t)\dot{B}_{t}(\phi_{j},\phi_{j}).

We thus obtain

$\displaystyle\lim_{\ell\rightarrow\infty}\dot{\lambda}_{j}^{+}(t+h_{\ell})$	$\displaystyle=$	$\displaystyle\lim_{\ell\rightarrow\infty}\{\dot{A}_{t+h_{\ell}}(u_{j}(t+h_{\ell}),u_{j}(t+h_{\ell}))$
		$\displaystyle-\lambda_{j}(t+h_{\ell})\dot{B}_{t+h_{\ell}}(u_{j}(t+h_{\ell}),u_{j}(t+h_{\ell}))\}$
	$\displaystyle=$	$\displaystyle\dot{A}_{t}(\phi_{j},\phi_{j})-\lambda_{j}(t)\dot{B}_{t}(\phi_{j},\phi_{j})=\dot{\lambda}_{j}^{+}(t)$

by (79), and hence

\lim_{h\rightarrow+0}\dot{\lambda}_{j}^{+}(t+h)=\dot{\lambda}_{j}^{+}(t)

because $h_{\ell}\rightarrow+0$ is arbitrary.

The proof of

\lim_{h\rightarrow-0}\dot{\lambda}_{j}^{-}(t+h)=\dot{\lambda}_{j}^{-}(t)

is similar. ∎

Remark 6.

The limits (75) exist for any $j$ under the conditions (41), (42), (54), and (55). If these conditions are satisfied for any $t\in(-\varepsilon_{0},\varepsilon_{0})$ , the limits (71) are unilaterally locally uniform in $t\in I$ . In fact, if not, there are, for example, $t_{k}\downarrow t_{0}\in$ $(-\varepsilon_{0},\varepsilon_{0})$ , $\delta>0$ , and $h_{\ell}\rightarrow 0$ , such that

\left|\frac{1}{h_{\ell}}(\lambda_{j}(t_{k}+h_{\ell})-\lambda_{j}(t_{k}))-\dot{\lambda}_{j}^{+}(t_{k})\right|\geq\delta.

Then we obtain

\left|\frac{1}{h_{\ell}}(\lambda_{j}(t_{0}+h_{\ell})-\lambda_{j}(t_{0}))-\dot{\lambda}_{j}^{+}(t_{0})\right|\geq\delta,

a contradiction with $\ell\rightarrow\infty$ .

7. Rearrangement of eigenvalues

For simplicity we introduce the following notations to prove Theorem 3. Recall $I=(-\varepsilon_{0},\varepsilon_{0})$ , and let $f_{j}\in C^{0}(I)$ , $1\leq j\leq m$ , satisfy

f_{1}(t)\leq\cdots\leq f_{m}(t),\quad t\in I.

(80)

Assume the existence of the unilateral limits

\dot{f}_{j}^{\pm}(t)=\lim_{h\rightarrow\pm 0}\frac{1}{h}(f_{j}(t+h)-f_{j}(t))

(81)

and

\lim_{h\rightarrow\pm 0}\dot{f}_{j}(t+h)=\dot{f}_{j}^{\pm}(t)

(82)

for any $j$ and $t$ . Assume, finally,

\dot{f}_{j}^{+}(t)=\dot{f}_{2k+n-j-1}^{-}(t),\quad k\leq j\leq k+n-1,

(83)

provided that

f_{k-1}(t)<f_{k}(t)=\cdots=f_{k+n-1}(t)<f_{k+n}(t),

(84)

where $1\leq n\leq m$ , $1\leq k\leq m-n+1$ , and $t\in I$ . In (84) we understand

f_{0}(t)=-\infty,\quad f_{m+1}(t)=+\infty.

We call

K=K_{k,n}(t)=\{k,\cdots,k+n-1\}

the $n$ -cluster at $t$ with entry $k$ if (84) arises, and also

p(K)=\max\{j\mid k\leq j\leq k+[\frac{n}{2}]-1,\ \dot{f}_{j}^{+}(t)<\dot{f}_{2k+n-1-j}^{+}(t)\}-k+1

its $p$ -value. Here, we understand $p(K)=0$ if

\dot{f}_{k}^{+}(t)=\dot{f}_{k+n-1}^{+}(t),

noting

k\leq i\leq j\leq k+n-1\quad\Rightarrow\quad\dot{f}_{i}^{+}(t)\leq\dot{f}_{j}^{+}(t).

We construct a rearrangement of $C^{0}$ -curves

C_{j}=\{f_{j}(t)\mid t\in I\},\quad 1\leq j\leq m,

denoted by $\tilde{C}_{j}$ , $1\leq j\leq m$ , so that are $C^{1}$ in $t\in I$ . This rearrangement is done only on

I_{1}=\{t\in I\mid\mbox{there exists a cluster $K$ at $t$ such that $p(K)\geq 1$}.\}.

To introduce this rearrangement, we note the following facts in advance. First, given $2\leq n\leq m$ , let

I_{1}^{n}=\{t\in I\mid\mbox{there exists an $n$-cluster $K$ at $t$ such that $p(K)\geq 1$}\}.

If $t\in I_{1}^{n}$ and $K=K_{k,n}(t)$ satisfies $p(K)\geq 1$ , it holds that

f_{k+n-1}(t^{\prime})>f_{k}(t^{\prime}),\quad 0<|t^{\prime}-t|\ll 1.

(85)

Hence this $t$ is an isolated point of $I_{1}^{n}$ . In particular, each $I_{1}^{n}$ , $2\leq n\leq m$ , is at most countable, and hence so is $I_{1}$ by

I_{1}=\bigcup_{n=2}^{m}I_{1}^{n}.

Second, given $t\in I_{1}$ and $1\leq j\leq m$ , if $j\in K=K_{k,n}(t)$ holds for some $K$ in $p(K)\geq 1$ , this $K$ is unique.

Definition 3.

The curves $\tilde{C}_{j}$ , $1\leq j\leq m$ , are called the transversal rearrangement of $C_{j}$ , $1\leq j\leq m$ , if the following operations are done.

(1)

If $t\in I_{1}$ , $1\leq j\leq m$ , and $j\in K$ hold for $K=K_{k,n}(t)$ with $p(K)\geq 1$ , the curve $C_{j}$ for $k\leq j\leq k+p-1$ and $k-n-p\leq j\leq k-n-1$ on the right, is connected to $C_{2k+n-j-1}$ on the left at $t$ , where $p=p(K)$ .
(2)

No rearrangements to $C_{j}$ , $1\leq j\leq m$ , are done otherwise.

The curves $\tilde{C}_{j}$ , $1\leq j\leq m$ , are uniquely constructed from $C_{j}$ , $1\leq j\leq m$ , by this transversal rearrangement. From the results in the previous section, Theorem 3 is a consequence of the following theorem.

Theorem 14.

Under the above situation, the $C^{0}$ curves $\tilde{C}_{j}$ , $1\leq j\leq m$ , made by the transversal rearrangement of $C_{j}$ , $1\leq j\leq m$ , are $C^{1}$ .

Proof.

This theorem is obvious if $m=1$ . Now we show it by an induction on $m$ , assuming the assertion up to $m-1$ .

Take $t_{0}\in I\setminus\overline{I_{1}}$ , and make the transversal rearrangement of $C_{j}$ , $1\leq j\leq q$ , toward left and right diections. Let $t_{\ell}$ , $\ell=1,2,\cdots$ , be the successive points of $I_{1}$ in the left direction:

t_{\ell}\in I_{1},\ t_{\ell-1}>t_{\ell},\ (t_{\ell},t_{\ell-1})\cap I_{1}=\emptyset,\quad\ell=1,2,\cdots.

If $\{t_{\ell}\}$ is finite, these $C_{j}$ ’s are successfully rearranged to $C^{1}$ curves on $(-\varepsilon_{0},t_{0})$ . If not, there is

t_{\ast}=\lim_{\ell\rightarrow\infty}t_{\ell}\in[-\varepsilon_{0},t_{0}].

Then the case $t_{\ast}=-\varepsilon_{0}$ ensures the same conclusion.

Letting $t_{\ast}>-\varepsilon_{0}$ , we show that $\{\tilde{C}_{j}\mid 1\leq j\leq m\}$ are $C^{1}$ curves on $(t_{\ast}-\delta,t_{\ast}+\delta)$ for $0<\delta\ll 1$ . Once this fact is proven, we can repeat this process up to $t=-\varepsilon_{0}$ by a covering argument. Turning to the right direction, we conclude that $\tilde{C}_{j}$ , $1\leq j\leq m$ , are $C^{1}$ curves on $I=(-\varepsilon_{0},\varepsilon_{0})$ .

To this end we distinguish the cases $t_{\ast}\in I_{1}$ and $t_{\ast}\not\in I_{1}$ .

If $t_{\ast}\in I_{1}$ , first, the above assertion follows from the assumption of induction. In fact, each $K=K_{k,n}(t_{\ast})$ with $p(K)\geq 1$ admits (85), while

\tilde{f}_{j}(t)=\left\{\begin{array}[]{ll}f_{j}(t),&0<t-t_{\ast}\ll 1\\ f_{2k+n-j-1}(t),&0<t_{\ast}-t\ll 1\end{array}\right.,\quad j=k,\ j=k+n-1

are $C^{1}$ around $t=t_{\ast}$ . Then we apply the assumption of induction to $C_{j}$ , $k+1\leq j\leq k+n-2$ , to get $n$ - $C^{1}$ curves in $(t_{\ast}-\delta,t_{\ast}+\delta)$ made by $C_{j}$ , $k\leq j\leq k+n-1$ . Operating this process to any $K=K_{k,n}(t_{\ast})$ with $p(K)\geq 1$ at $t=t_{\ast}$ , we get $C^{1}$ curves $\tilde{C}_{j}$ , $1\leq j\leq m$ , in $(t_{\ast}-\delta,t_{\ast}+\delta)$ by this transversal rearrangement of $C_{j}$ , $1\leq j\leq m$ , at $t=t_{\ast}$ .

If $t_{\ast}\not\in I_{1}$ , second, we take the cluster decomposition of $\{1,\cdots,m\}$ at $t=t_{\ast}$ , that is,

1=k_{1}<k_{1}+n_{1}=k_{2}<\cdots<k_{s-1}+n_{s-1}=k_{s}<k_{s}+n_{s}=m

satisfying

\bigcup_{r=1}^{s}K_{k_{r},n_{r}}(t_{\ast})=\{1,\cdots,m\}.

Since

p(K_{k_{r},n_{r}}(t_{\ast}))=0,\quad 1\leq r\leq s

holds by the assumption, there are $a_{r}$ , $1\leq r\leq s$ , such that

\dot{f}_{j}^{+}(t_{\ast})=a_{r},\quad\forall j\in K_{k_{r},n_{r}}.

We obtain, on the other hand,

f_{k_{r-1}}(t)<f_{k_{r}}(t),\ |t-t_{\ast}|\ll 1,\ 1\leq r\leq s+1

under the agreement

f_{k_{0}}(t)=-\infty,\quad f_{k_{s+1}}(t)=+\infty,

and hence $\tilde{C}_{j}$ , $1\leq j\leq m$ , are $C^{1}$ on $[t_{\ast},t_{\ast}+\delta)$ for $0<\delta\ll 1$ . If $t_{\ast}$ is not a right accumulating point of $I_{1}$ , therefore, these $\tilde{C}_{j}$ , $1\leq j\leq m$ , made by the transversal rearrangement of $C_{j}$ , $1\leq j\leq m$ , are $C^{1}$ on $(t_{\ast}-\delta,t_{\ast}+\delta)$ .

In the other case that $t_{\ast}$ is a right accumulating point of $I_{1}$ , these $\tilde{C}_{j}$ , $1\leq j\leq m$ , are $C^{1}$ on $(t_{\ast}-\delta,t_{\ast}]$ , similarly. Then it holds that

\dot{f}_{j}^{-}(t_{\ast})=a_{r},\quad\forall j\in K_{k_{r},n_{r}}=K_{k_{r},n_{r}}(t_{\ast})

by $p(K_{k_{r},n_{r}})=0$ , and hence $\tilde{C}_{j}$ , $1\leq j\leq m$ , are $C^{1}$ on $(t_{\ast}-\delta,t_{\ast}+\delta)$ . ∎

8. Second derivatives

If $T_{t}:\Omega\rightarrow\Omega_{t}$ is $2$ -differentiable, we have the other bilinear forms $\ddot{A}_{t}:V\times V\rightarrow\mathbb{R}$ and $\ddot{B}_{t}:X\times X\rightarrow\mathbb{R}$ satisfying

	$\displaystyle\|\ddot{A}_{t}(u,v)\|\leq C\\|u\\|_{V}\\|v\\|_{V},\quad u,v\in V$
	$\displaystyle\|\ddot{B}_{t}(u,v)\|\leq C\|u\|_{X}\|v\|_{X},\quad\ u,v\in X$		(86)

uniformly in $t$ and

	$\displaystyle\lim_{h\rightarrow 0}\frac{1}{h^{2}}\sup_{\\|u\\|_{V},\\|v\\|_{V}\leq 1}\left\|\left(A_{t+h}-A_{t}-h\dot{A}_{t}-\frac{h^{2}}{2}\ddot{A}_{t}\right)(u,v)\right\|=0$
	$\displaystyle\lim_{h\rightarrow 0}\frac{1}{h^{2}}\sup_{\|u\|_{X},\|v\|_{X}\leq 1}\left\|\left(B_{t+h}-B_{t}-h\dot{B}_{t}-\frac{h^{2}}{2}\ddot{B}_{t}\right)(u,v)\right\|=0$		(87)

for each $t$ . Hence Theorem 4 is reduced to the following abstract theorem.

Theorem 15.

In Theorem 10, assume, futhermore, (86)-(87). Then the conclusion of Theorem 4 holds.

For the moment we develop a formal argument as in the first derivative. Assuming (62), first, we deduce

	$\displaystyle\ddot{\lambda}_{t}$	$\displaystyle=$	$\displaystyle\ddot{A}_{t}(u_{t},u_{t})+2\dot{A}_{t}(\dot{u}_{t},u_{t})-\dot{\lambda}_{t}\dot{B}_{t}(u_{t},u_{t})-\lambda_{t}\ddot{B}_{t}(u_{t},u_{t})-2\lambda_{t}\dot{B}_{t}(\dot{u}_{t},u_{t})$		(88)
		$\displaystyle=$	$\displaystyle 2(\dot{A}_{t}-\lambda_{t}\dot{B}_{t})(\dot{u}_{t},u_{t})+D_{t}(u_{t},u_{t})$		(88)

for

D_{t}(u,v)=\ddot{A}_{t}(u,v)-\dot{\lambda}_{t}\dot{B}_{t}(u,v)-\lambda_{t}\ddot{B}_{t}(u,v),\quad u,v\in V.

(89)

Putting $v=\dot{u}_{t}$ in (57), second, we reach

(\dot{A}_{t}-\lambda_{t}\dot{B}_{t})(u_{t},\dot{u}_{t})=-(A_{t}-\lambda_{t}B_{t})(\dot{u}_{t},\dot{u}_{t})+\dot{\lambda}_{t}B_{t}(u_{t},\dot{u}_{t}).

(90)

Then, (60), (88), and (90) imply

	$\displaystyle\ddot{\lambda}_{t}=-2(A_{t}-\lambda_{t}B_{t})(z_{\ast},z_{\ast})+2\dot{\lambda}_{t}B_{t}(u_{t},z_{\ast})+D_{t}(u_{t},u_{t})$
	$\displaystyle=-2(A_{t}-\lambda_{t}B_{t})(z_{\ast},z_{\ast})-\dot{\lambda}_{t}\dot{B}_{t}(u_{t},u_{t})+D_{t}(u_{t},u_{t})$
	$\displaystyle=-2(A_{t}-\lambda_{t}B_{t})(z_{\ast},z_{\ast})+\ddot{A}_{t}(u_{t},u_{t})-2\dot{\lambda}_{t}\dot{B}_{t}(u_{t},u_{t})-\lambda_{t}\ddot{B}_{t}(u_{t},u_{t})$		(91)

for $z_{\ast}=\dot{u}_{t}$ .

We observe that $\dot{u}_{t}\in V$ is not uniquely determined by (57), which is derive formally also. It has, more precisely, the ambiguity of addition of an element in $Y_{t}^{\lambda}$ . This ambiguity, however, cancels in (91) by equality (92) below.

We now develop a rigorous argument valid even to (19). Define $\lambda$ by (21) and recall $Y_{t}^{\lambda}=\langle u_{j}(t)\mid k\leq j\leq k+m-1\rangle$ for $u_{j}(t)$ satisfying (46). Let, also,

C_{t^{\prime}}^{j}=A_{t^{\prime}}-\lambda_{j}(t^{\prime})B_{t^{\prime}},\quad k\leq j\leq k+m-1,\ t^{\prime}\in I.

Then we obtain

C_{t}^{j}=A_{t}-\lambda B_{t}\equiv C_{t}

and

C_{t}(u,v)=0,\quad\forall(u,v)\in Y_{t}^{\lambda}\times V.

(92)

By Lemma 11, given $h_{\ell}\rightarrow 0$ , which may not be of definite sign, we have a subsequence, denoted by the same symbol, satisfying (48),

\mbox{s}\mathchar 45\relax\lim_{\ell\rightarrow\infty}u_{j}(t+h_{\ell})=\phi_{j}\in Y_{t}^{\lambda}\ \mbox{in $V$},\quad k\leq j\leq k+m-1.

(93)

There exists

\dot{\lambda}_{j}^{\ast}=\lim_{\ell\rightarrow\infty}\frac{1}{h_{\ell}}(\lambda_{j}(t+h_{\ell})-\lambda_{j}(t)),\quad k\leq j\leq k+m-1

(94)

passing to a subsequence, with the equality

\dot{\lambda}_{j}^{\ast}\delta_{jj^{\prime}}=\dot{A}_{t}(\phi_{j},\phi_{j^{\prime}})-\lambda\dot{B}_{t}(\phi_{j},\phi_{j^{\prime}}),\quad k\leq j,j^{\prime}\leq k+m-1.

Remark 7.

If we take the subsequence of $h_{\ell}\rightarrow 0$ to be unilateral, the limit $\dot{\lambda}_{j}^{\ast}$ in (94) exists and is either $\dot{\lambda}_{j}^{+}(t)$ or $\dot{\lambda}_{j}^{-}(t)$ .

Let

\dot{C}_{t}^{\ast j}=\dot{A}_{t}-\dot{\lambda}_{j}^{\ast}B_{t}-\lambda\dot{B}_{t}.

It holds that

\lim_{\ell\rightarrow\infty}\sup_{\|u\|_{V}\leq 1,\|v\|_{V}\leq 1}\left|\frac{1}{h_{\ell}}(C_{t+h_{\ell}}^{j}-C_{t})(u,v)-\dot{C}_{t}^{\ast j}(u,v)\right|=0,\quad k\leq j\leq k+m-1

by (55), and also

\dot{C}_{t}^{\ast j}(u,v)=0,\quad\forall u,v\in Y_{t}^{\lambda},\ k\leq j\leq k+m-1

(95)

by Lemma 11. Let

z_{\ell}^{j}=\frac{1}{h_{\ell}}(u_{j}(t+h_{\ell})-\phi_{j}).

Lemma 16.

It holds that

\lim_{\ell\rightarrow\infty}C_{t}(z_{\ell}^{j},v)=-\dot{C}_{t}^{\ast j}(\phi_{j},v),\quad\forall v\in V.

(96)

Proof.

Given $v\in V$ , we obtain

	$\displaystyle C_{t}(z_{\ell}^{j},v)$	$\displaystyle=$	$\displaystyle\frac{1}{h_{\ell}}C_{t}(u_{j}(t+h_{\ell})-\phi_{j},v)=\frac{1}{h_{\ell}}C_{t}(u_{j}(t+h_{\ell}),v)$
		$\displaystyle=$	$\displaystyle-\frac{1}{h_{\ell}}(C_{t+h_{\ell}}^{j}-C_{t})(u_{j}(t+h_{\ell}),v)=-\dot{C}_{t}^{\ast j}(u_{j}(t+h_{\ell}),v)+o(1)$

by (51) and (92). Then (96) follows from (48). ∎

Recall that

R:X\rightarrow Y_{t}^{\lambda}=\langle u_{j}(t)\mid k\leq j\leq k+m-1\rangle

is the orthogonal projection with respect to $B_{t}(\cdot,\cdot)$ , and $P=I-R$ . There is a unique $z_{\ast}^{j}\in PV$ satisfying

C_{t}(z_{\ast}^{j},v)=-\dot{C}_{t}^{\ast j}(\phi_{j},v),\quad\forall v\in PV.

(97)

Remark 8.

Equality (97) ensures

\gamma_{\dot{\lambda}_{j}^{\ast}}(\phi_{j})=z_{\ast}^{j},\quad k\leq j\leq k+m-1

(98)

under the notation of Definition 2.

Lemma 17.

It holds that

\mbox{w}\mathchar 45\relax\lim_{\ell\rightarrow\infty}Pz_{\ell}^{j}=z_{\ast}^{j}\quad\mbox{in $V$},\quad k\leq j\leq k+m-1.

(99)

Proof.

Lemma 16 ensures that $\{Pz_{\ell}^{j}\}$ converges weakly in $PV$ , and hence is bounded there:

\|Pz_{\ell}^{j}\|_{V}\leq C.

Then, passing to a subsequence denoted by the same symbol, there is $\tilde{z}_{j}\in PV$ such that

\mbox{w}\mathchar 45\relax\lim_{\ell\rightarrow\infty}Pz_{\ell}^{j}=\tilde{z}_{j},

which satisfies

C_{t}(\tilde{z}_{j},v)=-\dot{C}_{t}^{\ast j}(\phi_{j},v),\quad\forall v\in V

(100)

by Lemma 16. Since such $\tilde{z}_{j}\in PV$ is unique, we obtain the result with $\tilde{z}_{j}=z_{\ast}^{j}$ . ∎

Remark 9.

Since (100) holds with $\tilde{z}_{j}=z^{j}_{\ast}$ , this $z_{j}^{\ast}\in PV$ defined by (97) satisfies

C_{t}(z_{\ast}^{j},v)=-\dot{C}_{t}^{\ast j}(\phi_{j},v),\quad\forall v\in V.

Remark 10.

Generally, the inequality

\|z_{\ell}^{j}\|_{V}\leq C

is not expected to hold, which causes the other difficulty in later arguments. In fact, if $m=1$ and $\phi_{j}=u_{j}(t)$ , for example, this property means

|B_{t}(u_{j}(t),w_{h_{\ell}}^{j})|\leq C,\quad w_{h}^{j}=\frac{1}{h}(u_{j}(t+h)-u_{j}(t)).

In the formal argument, we have actually (60), which, however, does not assure the actual convergence

\lim_{h\rightarrow 0}B_{t}(u_{j}(t),w_{h}^{j})=-\frac{1}{2}\dot{B}_{t}(u_{j}(t),u_{j}(t)).

(101)

In fact, the equality

1=B_{t+h}(u_{j}(t+h),u_{j}(t+h))=B_{t}(u_{j}(t),u_{j}(t))

just implies

$\displaystyle 0$	$\displaystyle=$	$\displaystyle\frac{1}{h}\{B_{t+h}(u_{j}(t+h),u_{j}(t+h))-B_{t}(u_{j}(t),u_{j}(t))\}$
	$\displaystyle=$	$\displaystyle\frac{1}{h}(B_{t+h}-B_{t})(u_{j}(t+h),u_{j}(t+h))$
		$\displaystyle+\frac{1}{h}\{B_{t}(u_{j}(t+h),u_{j}(t+h))-B_{t}(u_{j}(t),u_{j}(t))\}$
	$\displaystyle=$	$\displaystyle\dot{B}_{t}(u_{j}(t),u_{j}(t))+o(1)+\frac{1}{h}B_{t}(u_{j}(t+h)+u_{j}(t),u_{j}(t+h)-u_{j}(t))$
	$\displaystyle=$	$\displaystyle\dot{B}_{t}(u_{j}(t),u_{j}(t))+2B_{t}(\frac{u_{j}(t+h)+u_{j}(t)}{2},w_{h}^{j})+o(1)$

and hence

\lim_{h\rightarrow 0}B_{t}(\frac{u_{j}(t+h)+u_{j}(t)}{2},w_{h}^{j})=-\frac{1}{2}\dot{B}_{t}(u_{j}(t),u_{j}(t))

(102)

differently from (101). Here, the condition $\|w_{h}^{j}\|_{X}=O(1)$ is necessary to conclude

B_{t}(u_{j}(t+h),w_{h}^{j})=B_{t}(u_{j}(t),w_{h}^{j})+o(1)

in the left-hand side of (102) from $\phi_{j}=u_{j}(t)$ in (48). Our purpose, however, was to assure $\|w_{h}^{j}\|_{V}=O(1)$ , which is reduced to $\|w_{h}^{j}\|_{X}=O(1)$ by $\|P_{h}w_{h}^{j}\|_{V}=O(1)$ . This is a circular reasoning.

Lemma 18.

It holds that

\mbox{s}\mathchar 45\relax\lim_{\ell\rightarrow\infty}Pz_{\ell}^{j}=z_{\ast}^{j}\quad\mbox{in $V$},\qquad k\leq j\leq k+m-1.

(103)

Proof.

Since $V\hookrightarrow X$ is compact, we have

\mbox{s}\mathchar 45\relax\lim_{\ell\rightarrow\infty}Pz_{\ell}^{j}=z_{\ast}^{j}\quad\mbox{in $X$}

(104)

in the previous lemma. Then we obtain

	$\displaystyle\delta\\|Pz_{\ell}^{j}-z_{\ast}^{j}\\|_{V}^{2}$	$\displaystyle\leq$	$\displaystyle A_{t}(Pz_{\ell}^{j}-z_{\ast}^{j},Pz_{\ell}^{j}-z_{\ast}^{j})=C_{t}(Pz_{\ell}^{j}-z_{\ast}^{j},Pz_{\ell}^{j}-z_{\ast}^{j})+o(1)$
		$\displaystyle=$	$\displaystyle C_{t}(Pz_{\ell}^{j},Pz_{\ell}^{j}-z_{\ast}^{j})+o(1)=C_{t}(z_{\ell}^{j},Pz_{\ell}^{j}-z_{\ast}^{j})+o(1)$

by (99) and (104).

Since $\phi_{j}\in Y_{t}^{\lambda}$ it holds that

$\displaystyle C_{t}(z_{\ell}^{j},Pz_{\ell}^{j}-z_{\ast}^{j})$	$\displaystyle=$	$\displaystyle\frac{1}{h_{\ell}}C_{t}(u_{j}(t+h_{\ell}),Pz_{\ell}^{j}-z_{\ast}^{j})$
	$\displaystyle=$	$\displaystyle\frac{1}{h_{\ell}}(C_{t}-C_{t+h_{\ell}}^{j})(u_{j}(t+h_{\ell}),Pz_{\ell}^{j}-z_{\ast}^{j})$
	$\displaystyle=$	$\displaystyle-\dot{C}_{t}^{\ast j}(u_{j}(t+h_{\ell}),Pz_{\ell}^{j}-z_{\ast}^{j})+o(1)$
	$\displaystyle=$	$\displaystyle-\dot{C}_{t}^{\ast j}(\phi_{j},Pz_{\ell}^{j}-z_{\ast}^{j})+o(1)=o(1)$

by (48) and (99), because

v\in V\ \mapsto\ \dot{C}_{t}^{\ast j}(\phi_{j},v)\in\mathbb{R}

is a bounded linear mapping. Then the result follows as

\lim_{\ell\rightarrow\infty}\|Pz_{\ell}^{j}-z_{\ast}^{j}\|_{V}=0.

∎

Remark 11.

The limit $z_{\ast}^{j}$ in (103) depends on the sequence $h_{\ell}\rightarrow 0$ because it is prescribed by (93) and (98). The limit $\ddot{\lambda}_{j}^{\ast}$ in the following lemma depends also $h_{\ell}\rightarrow 0$ , but this ambiguity is canceled unilaterally. Hence these limits are uniquely determined as either $h\downarrow 0$ or $h\uparrow 0$ , because they are characterized by an eigenvalue problem on a finite dimsnsional space, similarly to the first derivative of $\lambda_{j}(t)$ . See Theorem 22 below.

Lemma 19.

There exists

\ddot{\lambda}_{j}^{\ast}\equiv\lim_{\ell\rightarrow\infty}\frac{2}{h_{\ell}^{2}}(\lambda_{j}(t+h_{\ell})-\lambda_{j}(t)-h_{\ell}\dot{\lambda}_{j}^{\ast}),\quad k\leq j\leq k+1-m

(105)

with

\ddot{\lambda}_{j}^{\ast}=(\ddot{A}_{t}-\lambda\ddot{B}_{t}-2\dot{\lambda}_{j}^{\ast}\dot{B}_{t})(\phi_{j},\phi_{j})-2C_{t}(z_{\ast}^{j},z_{\ast}^{j}).

Proof.

By (95) and Lemma 18, we have

	$\displaystyle C_{t}(z_{\ell}^{j},z_{\ell}^{j})$	$\displaystyle=$	$\displaystyle C_{t}(Pz_{\ell}^{j},Pz_{\ell}^{j})=C_{t}(z_{\ast}^{j},z_{\ast}^{j})+o(1)$		(106)
		$\displaystyle=$	$\displaystyle C_{t}(Pz_{\ell}^{j},z_{\ast}^{j})+o(1)=C_{t}(z_{\ell}^{j},z_{\ast}^{j})+o(1).$		(106)

It holds that

$\displaystyle C_{t}(z_{\ell}^{j},z_{\ast}^{j})$	$\displaystyle=$	$\displaystyle\frac{1}{h_{\ell}}C_{t}(u_{j}(t+h_{\ell})-\phi_{j},z_{\ast}^{j})=\frac{1}{h_{\ell}}C_{t}(u_{j}(t+h_{\ell}),z_{\ast}^{j})$
	$\displaystyle=$	$\displaystyle\frac{1}{h_{\ell}}(C_{t}-C_{t+h_{\ell}}^{j})(u_{j}(t+h_{\ell}),z_{\ast}^{j})=-\dot{C}_{t}^{\ast j}(u_{j}(t+h_{\ell}),z_{\ast}^{j})+o(1)$
	$\displaystyle=$	$\displaystyle-\dot{C}_{t}^{\ast j}(\phi_{j},z_{\ast}^{j})+o(1)$

by (92) and $\phi_{j}\in Y_{t}^{\lambda}$ , which implies

C_{t}(z_{\ast}^{j},z_{\ast}^{j})=-\dot{C}_{t}^{\ast j}(\phi_{j},z_{\ast}^{j})+o(1)

(107)

by (106). It holds also that

$\displaystyle\dot{C}_{t}^{\ast j}(\phi_{j},z_{\ast}^{j})$	$\displaystyle=$	$\displaystyle\frac{1}{h_{\ell}}\dot{C}_{t}^{\ast j}(\phi_{j},P(u_{j}(t+h_{\ell})-\phi_{j}))+o(1)$	(108)
	$\displaystyle=$	$\displaystyle\frac{1}{h_{\ell}}\dot{C}_{t}^{\ast j}(\phi_{j},u_{j}(t+h_{\ell})-\phi_{j})+o(1)$
	$\displaystyle=$	$\displaystyle\frac{1}{h_{\ell}}\dot{C}_{t}^{\ast j}(\phi_{j},u_{j}(t+h_{\ell}))+o(1)$

by (95) and $\phi_{j}\in Y_{t}^{\lambda}$ .

Here, we use the asymptotics

	$\displaystyle C_{t+h_{\ell}}^{j}(\phi_{j},u_{j}(t+h_{\ell}))=C_{t}^{j}(\phi_{j},u_{j}(t+h_{\ell}))+h_{\ell}\dot{C}_{t}^{\ast j}(\phi_{j},u_{j}(t+h_{\ell}))$
	$\displaystyle+\frac{1}{2}h_{\ell}^{2}\ddot{C}_{t,\ell}^{\ast j}(\phi_{j},u_{j}(t+h_{\ell}))+o(h_{\ell}^{2})$

for

\ddot{C}_{t,\ell}^{\ast j}=\ddot{A}_{t}-\lambda\ddot{B}_{t}-2\dot{\lambda}_{j}^{\ast}\dot{B}_{t}-\frac{2}{h_{\ell}^{2}}(\lambda_{j}(t+h_{\ell})-\lambda_{j}(t)-h\dot{\lambda}_{j}^{\ast})B_{t},

(109)

derived from (55) and (87). Since

C_{t+h_{\ell}}^{j}(\phi_{j},u_{j}(t+h_{\ell}))=C_{t}^{j}(\phi_{j},u_{j}(t+h_{\ell}))=0

holds by (92), we obtain

$\displaystyle C_{t}^{j}(z_{\ast}^{j},z_{\ast}^{j})$	$\displaystyle=$	$\displaystyle-\dot{C}_{t}^{\ast j}(\phi_{j},z_{\ast}^{j})+o(1)=-\frac{1}{h_{\ell}}\dot{C}_{t}^{\ast j}(\phi_{j},u_{j}(t+h_{\ell}))+o(1)$
	$\displaystyle=$	$\displaystyle\frac{1}{2}\ddot{C}_{t,\ell}^{\ast j}(\phi_{j},u_{j}(t+h_{\ell}))+o(1)=\frac{1}{2}\ddot{C}_{t,\ell}^{\ast j}(\phi_{j},\phi_{j})+o(1)$
	$\displaystyle=$	$\displaystyle\frac{1}{2}(\ddot{A}_{t}-\lambda\ddot{B}_{t}-2\dot{\lambda}_{j}^{\ast}\dot{B}_{t})(\phi_{j},\phi_{j})-\frac{1}{h_{\ell}^{2}}(\lambda_{j}(t+h_{\ell})-\lambda_{j}(t)-h_{\ell}\dot{\lambda}_{j}^{\ast})+o(1)$

by (107)-(108) and $B_{t}(\phi_{j},\phi_{j})=1$ . Then it follows that

\lim_{\ell\rightarrow\infty}\frac{2}{h_{\ell}^{2}}(\lambda_{j}(t+h_{\ell})-\lambda_{j}(t)-h_{\ell}\dot{\lambda}_{j}^{\ast})\\ =(\ddot{A}_{t}-\lambda\ddot{B}_{t}-2\dot{\lambda}_{j}^{\ast}\dot{B}_{t})(\phi_{j},\phi_{j})-2C_{t}(z_{\ast}^{j},z_{\ast}^{j}),

and the proof is complete. ∎

Lemma 20.

If $\dot{\lambda}_{\ast}\equiv\dot{\lambda}_{j}^{\ast}=\dot{\lambda}_{j^{\prime}}^{\ast}$ arises for some $k\leq j\neq j^{\prime}\leq k+m-1$ , then it holds that

(\ddot{A}_{t}-\lambda\ddot{B}_{t}-2\dot{\lambda}_{\ast}\dot{B}_{t})(\phi_{j},\phi_{j^{\prime}})=2C_{t}(z_{\ast}^{j},z_{\ast}^{j^{\prime}}).

Proof.

As in the previous lemma we obtain

$\displaystyle C_{t}(z_{\ast}^{j},z_{\ast}^{j^{\prime}})$	$\displaystyle=$	$\displaystyle C_{t}(Pz_{\ell}^{j},z_{\ast}^{j^{\prime}})+o(1)=C_{t}(z_{\ell}^{j},z_{\ast}^{j^{\prime}})+o(1)$
	$\displaystyle=$	$\displaystyle\frac{1}{h_{\ell}}C_{t}(u_{j}(t+h_{\ell})-\phi_{j},z_{\ast}^{j^{\prime}})+o(1)=\frac{1}{h_{\ell}}C_{t}(u_{j}(t+h_{\ell}),z_{\ast}^{j^{\prime}})+o(1)$
	$\displaystyle=$	$\displaystyle\frac{1}{h_{\ell}}(C_{t}-C_{t+h_{\ell}})(u_{j}(t+h),z_{\ast}^{j^{\prime}})+o(1)$
	$\displaystyle=$	$\displaystyle-\dot{C}_{t}^{\ast j}(u_{j}(t+h_{\ell}),z_{\ast}^{j^{\prime}})+o(1)=-\dot{C}_{t\ast}^{j}(\phi_{j},z_{\ast}^{j^{\prime}})+o(1)$
	$\displaystyle=$	$\displaystyle-\frac{1}{h_{\ell}}\dot{C}_{t}^{\ast j}(\phi_{j},P(u_{j^{\prime}}(t+h_{\ell})-\phi_{j^{\prime}}))+o(1)$
	$\displaystyle=$	$\displaystyle-\frac{1}{h_{\ell}}\dot{C}_{t}^{\ast j}(\phi_{j},u_{j^{\prime}}(t+h_{\ell})-\phi_{j^{\prime}})+o(1)$
	$\displaystyle=$	$\displaystyle-\frac{1}{h_{\ell}}\dot{C}_{t}^{\ast j}(\phi_{j},u_{j^{\prime}}(t+h_{\ell}))+o(1)$

by $\phi_{j},\phi_{j^{\prime}}\in Y_{t}^{\lambda}$ . Then it holds that

	$\displaystyle C_{t+h_{\ell}}^{j}(\phi_{j},u_{j^{\prime}}(t+h_{\ell}))=C_{t}(\phi_{j},u_{j^{\prime}}(t+h_{\ell}))+h_{\ell}\dot{C}_{t}^{\ast j}(\phi_{j},u_{j^{\prime}}(t+h_{\ell}))$
	$\displaystyle\quad+\frac{1}{2}h_{\ell}^{2}\ddot{C}_{t,\ell}^{\ast j}(\phi_{j},u_{j^{\prime}}(t+h_{\ell}))+o(h_{\ell}^{2})$

with

C_{t}(\phi_{j},u_{j^{\prime}}(t+h_{\ell}))=0

and

	$\displaystyle C^{j}_{t+h_{\ell}}(\phi_{j},u_{j^{\prime}}(t+h_{\ell}))$	$\displaystyle=$	$\displaystyle(C_{t+h_{\ell}}^{j}-C_{t+h_{\ell}}^{j^{\prime}})(\phi_{j},u_{j^{\prime}}(t+h_{\ell}))$
		$\displaystyle=$	$\displaystyle(\lambda_{j^{\prime}}(t+h_{\ell})-\lambda_{j}(t+h_{\ell}))B_{t+h_{\ell}}(\phi_{j},u_{j^{\prime}}(t+h_{\ell}))$

by (92), to conclude

	$\displaystyle C_{t}(z_{\ast}^{j},z_{\ast}^{j^{\prime}})$	$\displaystyle=$	$\displaystyle\frac{1}{h_{\ell}^{2}}(\lambda_{j}(t+h_{\ell})-\lambda_{j^{\prime}}(t+h_{\ell}))B_{t+h_{\ell}}(\phi_{j},u_{j^{\prime}}(t+h_{\ell}))$		(110)
			$\displaystyle+\frac{1}{2}\ddot{C}_{t,\ell}^{\ast j}(\phi_{j},u_{j^{\prime}}(t+h_{\ell}))+o(1).$		(110)

Then we use

	$\displaystyle\lambda_{j}(t+h_{\ell})=\lambda+h_{\ell}\dot{\lambda}_{j}^{\ast}+\frac{h_{\ell}^{2}}{2}\ddot{\lambda}_{j}^{\ast}+o(h_{\ell}^{2})$
	$\displaystyle\lambda_{j^{\prime}}(t+h_{\ell})=\lambda+h_{\ell}\dot{\lambda}_{j^{\prime}}^{\ast}+\frac{h_{\ell}^{2}}{2}\ddot{\lambda}_{j^{\prime}}^{\ast}+o(h_{\ell}^{2})$

with $\dot{\lambda}_{j}^{\ast}=\dot{\lambda}_{j^{\prime}}^{\ast}$ , to deduce

	$\displaystyle\lim_{\ell\rightarrow\infty}\frac{1}{h_{\ell}^{2}}(\lambda_{j}(t+h_{\ell})-\lambda_{j^{\prime}}(t+h_{\ell}))B_{t+h_{\ell}}(\phi_{j},u_{j^{\prime}}(t+h_{\ell}))$
	$\displaystyle\quad=\frac{1}{2}(\ddot{\lambda}_{j}^{\ast}+\ddot{\lambda}_{j^{\prime}}^{\ast})B_{t}(\phi_{j},\phi_{j^{\prime}})=0.$

Then the result follows from (109), (110), and the previous lemma. ∎

Recall $F^{\lambda,\lambda^{\prime}}_{t}$ in Definition 2.

Lemma 21.

Define $\tilde{\mu}_{k}\leq\cdots\leq\tilde{\mu}_{k+m-1}$ by

\{\tilde{\mu}_{j}\mid k\leq j\leq k+m-1\}=\{\dot{\lambda}_{j}^{\ast}\mid k\leq j\leq k+m-1\},

and assume $k\leq\ell<r\leq k+m$ be such that

\tilde{\mu}_{\ell-1}<\tilde{\mu}\equiv\tilde{\mu}_{\ell}=\cdots=\tilde{\mu}_{r-1}<\tilde{\mu}_{r}.

under the agreement of

\tilde{\mu}_{k-1}=-\infty,\quad\tilde{\mu}_{k+m}=+\infty.

Then, $\sigma=\ddot{\lambda}_{j}^{\ast}$ , $\ell\leq j\leq r-1$ , is an eigenvalue of

u\in Y_{\lambda,t}^{\ell,r},\quad F^{\lambda,\tilde{\mu}}_{t}(u,v)=\sigma B_{t}(u,v),\ \forall v\in Y_{\lambda,t}^{\ell,r},

for

Y_{\lambda,t}^{\ell,r}=\langle u_{j}(t)\mid\ell\leq j\leq r-1\rangle\subset Y^{\lambda}_{t}.

Proof.

Lemmas 19 and 20 imply

F^{\lambda,\tilde{\mu}}_{t}(\phi_{j},\phi_{j^{\prime}})=\delta_{jj^{\prime}}\ddot{\lambda}_{j}^{\ast},\quad\ell\leq j,j^{\prime}\leq r-1,

and hence the result follows from

Y_{\lambda,t}^{\ell,r}=\langle\phi_{j}\mid\ell\leq j\leq r-1\rangle,\quad B_{t}(\phi_{j},\phi_{j^{\prime}})=\delta_{jj^{\prime}}.

∎

Theorem 15 is now reduced to the following theorem.

Theorem 22.

Fix $t\in I$ , and assume (19) for $\ell=k$ and $n=m$ . Put (21) and let $k\leq\ell<r\leq k+m$ be such that

\dot{\lambda}^{+}_{\ell-1}(t)<\lambda^{\prime}\equiv\dot{\lambda}^{+}_{\ell}(t)=\cdots=\dot{\lambda}_{r-1}^{+}(t)<\dot{\lambda}_{r}^{+}(t).

(111)

Then there exists

\lambda_{j}^{\prime\prime}(t)=\lim_{h\rightarrow+0}\frac{2}{h^{2}}(\lambda_{j}(t+h)-\lambda-h\lambda^{\prime}),\quad\ell\leq j\leq r-1.

(112)

It holds, furthermore, that

\lambda_{j}^{\prime\prime}(t)=\sigma_{j-\ell+1},\quad\ell\leq j\leq r-1,

(113)

where $\sigma_{q}$ , $1\leq q\leq r-\ell$ , denotes the $q$ -th eigenvalue of

u\in Y_{\lambda,t}^{\ell,r},\qquad F_{t}^{\lambda,\lambda^{\prime}}(u,v)=\sigma B_{t}(u,v),\ \forall v\in Y_{\lambda,t}^{\ell,r}.

\dot{\lambda}^{-}_{\ell-1}(t)>\lambda^{\prime}\equiv\dot{\lambda}^{-}_{\ell}(t)=\cdots=\dot{\lambda}_{r-1}^{-}(t)>\dot{\lambda}_{r}^{-}(t),

there arises that

\lambda_{j}^{\prime\prime}(t)=\lim_{h\rightarrow-0}\frac{2}{h^{2}}(\lambda_{j}(t+h)-\lambda-h\lambda^{\prime}),\quad\ell\leq j\leq r-1

with (113).

Proof.

In the previous lemma, we obtain

\ddot{\lambda}_{\ell}^{\ast}\leq\cdots\leq\ddot{\lambda}_{r-1}^{\ast}.

Hence the result follows similarly to Theorem 12. ∎

Remark 12.

By the above theorem and Remark 7, the limits (20),

\ddot{\lambda}_{j}^{\pm}(t)=\lim_{h\rightarrow\pm 0}\frac{2}{h^{2}}(\lambda_{j}(t+h)-\lambda_{j}(t)-h\dot{\lambda}_{j}^{\pm}(t))

exist for any $j$ , provided that the conditions (41), (42), (54), (55), (86), and (87) hold. If these conditions hold for any $t$ locally uniformly in $I$ , these limits are locally uniform in $t\in I$ .

Theorem 5 is reduced to the following abstract theorem. The proof is similar to that of Theorem 13.

Theorem 23.

Let the assumption of Theorem 15 hold for any $t$ . Fix $t\in I$ , and assume, furthermore,

	$\displaystyle\lim_{h\rightarrow 0}\sup_{\\|u\\|_{V},\\|v\\|_{V}\leq 1}\left\|\ddot{A}_{t+h}(u,v)-\ddot{A}_{t}(u,v)\right\|=0$
	$\displaystyle\lim_{h\rightarrow 0}\sup_{\|u\|_{X},\|v\|_{X}\leq 1}\left\|\ddot{B}_{t+h}(u,v)-\ddot{B}_{t}(u,v)\right\|=0.$		(114)

Then it holds that

\lim_{h\rightarrow\pm 0}\ddot{\lambda}_{j}^{\pm}(t+h)=\ddot{\lambda}_{j}^{\pm}(t).

Finally, Theorem 6 is reduced to the following abstract theorem.

Theorem 24.

If (114) is valid to any $t$ in the previous theorem, $C^{1}$ curves $\tilde{C}_{j}$ , $1\leq j\leq m$ , in Theorem 3 are $C^{2}$ .

Proof.

Define $\tilde{\lambda}_{j}(t)$ by

\tilde{C}_{j}=\{\tilde{\lambda}_{j}(t)\mid t\in I\},\ 1\leq j\leq m.

From the proof of Theorem 14, Theorems 22 and 23 guarantee the existence of

\tilde{\lambda}_{j}^{\prime\prime}(t)=\lim_{h\rightarrow 0}\frac{1}{h^{2}}(\tilde{\lambda}_{j}(t+h)-\tilde{\lambda}_{j}(t)-h\tilde{\lambda}_{j}^{\prime}(t))

(115)

together with its continuity in $t$ ,

\lim_{h\rightarrow\pm 0}\tilde{\lambda}_{j}^{\prime\prime}(t+h)=\tilde{\lambda}_{j}^{\prime\prime}(t)

for any $t$ and $j$ . This convergence (115), furthermore, is locally uniform in $t\in I$ by Remark 12.

Then it follows that

	$\displaystyle\tilde{\lambda}_{j}(t+h)=\tilde{\lambda}_{j}(t)+h\tilde{\lambda}_{j}^{\prime}(t)+\frac{h^{2}}{2}\tilde{\lambda}_{j}^{\prime\prime}(t)+o(h^{2})$
	$\displaystyle\tilde{\lambda}_{j}(t)=\tilde{\lambda}_{j}(t+h)-h\tilde{\lambda}_{j}^{\prime}(t+h)+\frac{h^{2}}{2}\tilde{\lambda}_{j}^{\prime\prime}(t+h)+o(h^{2}),$

as $h\rightarrow 0$ , which implies

\lim_{h\rightarrow 0}\frac{1}{h}(\tilde{\lambda}^{\prime}_{j}(t+h)-\tilde{\lambda}^{\prime}_{j}(t))=\lim_{h\rightarrow 0}\frac{1}{2}(\tilde{\lambda}^{\prime\prime}_{j}(t+h)+\tilde{\lambda}^{\prime\prime}_{j}(t))=\tilde{\lambda}^{\prime\prime}_{j}(t)

for any $t\in I$ . Hence these $\tilde{C}_{j}$ ’s are $C^{2}$ . ∎

Acknowledgements. The authors thank the referees for valuable comments on the original manuscprit.

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	$\displaystyle\left\|h_{\ell}(A_{t+h_{\ell}}-A_{t})\left(\frac{u_{j}(t+h_{\ell})-\phi_{j}}{h_{\ell}},\frac{u_{j^{\prime}}(t+h_{\ell})-\phi_{j^{\prime}}}{h_{\ell}}\right)\right\|$
	$\displaystyle\quad\leq Ch_{\ell}^{2}\left\\|\frac{u_{j}(t+h_{\ell})-\phi_{j}}{h_{\ell}}\right\\|_{V}\cdot\left\\|\frac{u_{j^{\prime}}(t+h_{\ell})-\phi_{j^{\prime}}}{h_{\ell}}\right\\|_{V}$
	$\displaystyle\quad=C\\|u_{j}(t+h_{\ell})-\phi_{j}\\|_{V}\cdot\\|u_{j^{\prime}}(t+h_{\ell})-\phi_{j^{\prime}}\\|_{V}=o(1)$		(70)