Hadamard’s Variational Formula for Simple Eigenvalues

Abstract

Here, we study Hadamard’s variational formula for simple eigenvalues under dynamical and conformal deformations. Particularly, harmonic convexity of the first eigenvalue of the Laplacian under the mixed boundary condition is established for a two-dimensional domain, which implies several new inequalities.

1. Introduction

The purpose of the present paper is to derive Hadamard’s variational formula for a simple eigenvalue of the Laplacian and to derive several new inequalities concerning the first eigenvalue concerning the mixed boundary condition.

In previous work [1], we studied Hadamard’s variational formula for general domain deformation and extended the results [2] on two or three-space dimensions under normal perturbations of the domain. There, we developed an abstract theory of perturbation of self-adjoint operators, refining the argument in [3].

To be precise, let $(X,|\left|\right)$ and $(V,\parallel \parallel )$ be a pair of Hilbert spaces over $\mathbb{R}$ with compact embedding $V\hookrightarrow X$. Henceforth, $C>0$ denotes a generic constant. Let $A_t:V\times V\to \mathbb{R}$ and $B_t:X\times X\to \mathbb{R}$ for $t\in I=(-{\epsilon }_0,{\epsilon }_0)$, ${\epsilon }_0>0$, be symmetric bilinear forms, satisfying

$$|A_t(u,v)|\le C\parallel u\parallel \parallel v\parallel ,A_t(u,u)\ge \delta {\parallel u\parallel }^2,u,v\in V$$

(1)

and

$$|B_t(u,v)|\le C|u\left|\right|v|,B_t(u,u)\ge \delta {\left|u\right|}^2,u,v\in X$$

(2)

for some $\delta >0$. We take the abstract eigenvalue problem

$$u\in V,B_t(u,u)=1,A_t(u,v)=\lambda B_t(u,v),\forall v\in V,$$

(3)

which ensures a sequence of eigenvalues denoted by

$$0<{\lambda }_1\left(t\right)\le {\lambda }_2\left(t\right)\le \cdots \to +\infty .$$

The associated normalized eigenfunctions,

$$u_1\left(t\right),u_2\left(t\right),\cdots ,$$

furthermore, form a complete ortho-normal system in X, provided with the inner product induced by $B_t=B_t(·,·)$. Hence, it holds that

$$B_t(u_i\left(t\right),u_j\left(t\right))={\delta }_{ij},A_t(u_j\left(t\right),v)={\lambda }_j{B}_t(u_j\left(t\right),v)$$

for any $v\in V$ and $i,j=1,2,\cdots$.

The abstract theory developed in [1] is stated as follows. First, if

$$\begin{array}{ccc}& & \underset{h\to 0}{lim}\underset{\parallel u\parallel ,\parallel v\parallel \le 1}{sup}|A_{t+h}(u,v)-A_t(u,v)|=0\hfill \\ & & \underset{h\to 0}{lim}\underset{\left|u\right|,\left|v\right|\le 1}{sup}|B_{t+h}(u,v)-B_t(u,v)|=0\hfill \end{array}$$

holds for fixed $t\in I$, we obtain

$$\underset{h\to 0}{lim}{\lambda }_j(t+h)={\lambda }_j\left(t\right)$$

at this t, where $j=1,2,\cdots$ are arbitrary (Theorem 8 of [1]). Second, if there exist bilinear forms ${\dot{A}}_t:V\times V\to \mathbb{R}$ and ${\dot{B}}_t:X\times X\to \mathbb{R}$ for any $t\in I$, such that

$$\begin{array}{ccc}& & |{\dot{A}}_t(u,v)|\le C\parallel u\parallel \parallel v\parallel ,u,v\in V\hfill \\ & & |{\dot{B}}_t(u,v)|\le C|u\left|\right|v|,u,v\in X,\hfill \end{array}$$

(4)

and if it holds that

$$\begin{array}{ccc}& & \underset{h\to 0}{lim}{\displaystyle \frac{1}{h}}\underset{\parallel u\parallel ,\parallel v\parallel \le 1}{sup}\left|\left(A_{t+h}-A_t-h{\dot{A}}_t\right)(u,v)\right|=0\hfill \\ & & \underset{h\to 0}{lim}{\displaystyle \frac{1}{h}}\underset{\left|u\right|,\left|v\right|\le 1}{sup}\left|\left(B_{t+h}-B_t-h{\dot{B}}_t\right)(u,v)\right|=0,\hfill \end{array}$$

(5)

for fixed $t\in I$, we obtain the existence of the unilateral derivatives

$${\dot{\lambda }}_j^{\pm }\left(t\right)=\underset{h\to \pm 0}{lim}{\displaystyle \frac{1}{h}}\{{\lambda }_j(t+h)-{\lambda }_j\left(t\right)\}$$

(6)

at this t, where $j=1,2,\cdots$ are arbitrary (Theorem 12 of [1]). If inequalities (5) are valid to any $t\in I$ and it also holds that

$$\begin{array}{ccc}& & \underset{h\to 0}{lim}\underset{{\parallel u\parallel }_V,{\parallel v\parallel }_V\le 1}{sup}\left|{\dot{A}}_{t+h}(u,v)-{\dot{A}}_t(u,v)\right|=0\hfill \\ & & \underset{h\to 0}{lim}\underset{{\left|u\right|}_X,{\left|v\right|}_X\le 1}{sup}\left|{\dot{B}}_{t+h}(u,v)-{\dot{B}}_t(u,v)\right|=0\hfill \end{array}$$

(7)

for the specified t, furthermore, the above unilateral derivatives satisfy

$$\underset{h\to \pm 0}{lim}{\dot{\lambda }}_j^{\pm }(t+h)={\dot{\lambda }}_j^{\pm }\left(t\right)$$

(Theorem 13 of [1]).

We assume, furthermore,

$${\lambda }_{k-1}\left(t\right)<{\lambda }_k\left(t\right)\le \cdots \le {\lambda }_{k+m-1}\left(t\right)<{\lambda }_{k+m}\left(t\right),\forall t\in I$$

(8)

for some $k,m=1,2,\cdots$, under the agreement of ${\lambda }_0\left(t\right)=-\infty$. Assume also that (5) and (7) hold for any $t\in I$. Then, there exists a family of $C^1$ curves denoted by ${\tilde{C}}_i$, $k\le i\le k+m-1$, created by a rearrangement of

$$\{C_j\mid k\le j\le k+m-1\}$$

at most countably many times in I, where

$$C_j=\{{\lambda }_j\left(t\right)\mid t\in I\},k\le j\le k+m-1$$

(Theorem 3, Theorem 14 of [1]).

Third, if we have the other bilinear forms ${\ddot{A}}_t:V\times V\to \mathbb{R}$ and ${\ddot{B}}_t:X\times X\to \mathbb{R}$ satisfying

$$\begin{array}{ccc}& & |{\ddot{A}}_t(u,v)|\le C\parallel u\parallel \parallel v\parallel ,u,v\in V\hfill \\ & & |{\ddot{B}}_t(u,v)|\le C|u\left|\right|v|,u,v\in X\hfill \end{array}$$

(9)

for any $t\in I$, and

$$\begin{array}{ccc}& & \underset{h\to 0}{lim}{\displaystyle \frac{1}{h^2}}\underset{\parallel u\parallel ,\parallel v\parallel \le 1}{sup}\left|\left(A_{t+h}-A_t-h{\dot{A}}_t-{\displaystyle \frac{h^2}{2}}{\ddot{A}}_t\right)(u,v)\right|=0\hfill \\ & & \underset{h\to 0}{lim}{\displaystyle \frac{1}{h^2}}\underset{\left|u\right|,\left|v\right|\le 1}{sup}\left|\left(B_{t+h}-B_t-h{\dot{B}}_t-{\displaystyle \frac{h^2}{2}}{\ddot{B}}_t\right)(u,v)\right|=0\hfill \end{array}$$

(10)

for the fixed t, then there are

$${\ddot{\lambda }}_j^{\pm }\left(t\right)=\underset{h\to 0}{lim}{\displaystyle \frac{2}{h^2}}({\lambda }_j(t+h)-{\lambda }_j\left(t\right)-h{\dot{\lambda }}_j^{\pm }\left(t\right))$$

for this t, where $j=1,2,\cdots$ are arbitrary (Remark 12 of [1]). These ${\ddot{\lambda }}_j^{\pm }\left(t\right)$, furthermore, satisfy

$$\underset{h\to \pm 0}{lim}{\ddot{\lambda }}_j^{\pm }(t+h)={\ddot{\lambda }}_j^{\pm }\left(t\right),$$

if inequalities (10) are valid to any $t\in I$, and it holds that

$$\begin{array}{ccc}& & \underset{h\to 0}{lim}\underset{\parallel u\parallel ,\parallel v\parallel \le 1}{sup}\left|{\ddot{A}}_{t+h}(u,v)-{\ddot{A}}_t(u,v)\right|=0\hfill \\ & & \underset{h\to 0}{lim}\underset{\left|u\right|,\left|v\right|\le 1}{sup}\left|{\ddot{B}}_{t+h}(u,v)-{\ddot{B}}_t(u,v)\right|=0\hfill \end{array}$$

(11)

for the above specified t (Theorem 23 of [1]). Furthermore, if (10) and (11) hold for any $t\in I$ and if it holds that (8), then the above described family of curves ${\tilde{C}}_i$, $k\le i\le k+m-1$, are $C^2$ (Theorem 24 of [1]).

Finally, these derivatives ${\dot{\lambda }}_j^{\pm }\left(t\right)$ and ${\ddot{\lambda }}_j^{\pm }\left(t\right)$ for $k\le j\le k+m-1$ are characterized as the eigenvalues of the associated eigenvalue problems on the m-dimensional space, $\langle u_j\left(t\right)\mid k\le j\le k+m-1\rangle$ (Theorem 12 and Theorem 15 of [1]). Henceforth, we assume (1), (2), (4), (5), (7), (9), (10), and (11) for any $t\in I$.

If $m=1$ holds in (8), for example, we obtain

$$\dot{\lambda }\equiv {\dot{\lambda }}_k^{\pm }\left(t\right)=({\dot{A}}_t-\lambda {\dot{B}}_t)({\varphi }_t,{\varphi }_t)$$

(12)

and

$$\ddot{\lambda }\equiv {\ddot{\lambda }}_k^{\pm }\left(t\right)=({\ddot{A}}_t-\lambda {\ddot{B}}_t-2\dot{\lambda }{B}_t)({\varphi }_t,{\varphi }_t)-2C_t^{\lambda }(\gamma \left({\varphi }_t\right),\gamma \left({\varphi }_t\right)),$$

where $\lambda ={\lambda }_k\left(t\right)$, ${\varphi }_t=u_k\left(t\right)$, and $C_t^{\lambda }=A_t-\lambda B_t$. Here, $w=\gamma \left(u\right)\in V_1$ is defined for $u\in V$ by

$$C_t^{\lambda }(w,v)=-{\dot{C}}_t^{\lambda ,\dot{\lambda }}(u,v),\forall v\in V_1,$$

where $V_1=PV$, $P=I-R$, and $R:X\to V_0=\langle {\varphi }_t\rangle$ is the orthogonal projection, and

$${\dot{C}}_t^{\lambda ,\dot{\lambda }}={\dot{A}}_t-\lambda {\dot{B}}_t-\dot{\lambda }{B}_t.$$

Thus, we obtain the following theorem because the bilinear form $C_t^{\lambda }$ is a non-negative definite on $V\times V$ if $\lambda ={\lambda }_1$.

Theorem 1. 

If $k=m=1$ holds in (8), there arises that

$$\dot{\lambda }=\dot{A}-\lambda \dot{B},\ddot{\lambda }\le \ddot{A}-\lambda \ddot{B}-2\dot{\lambda }\dot{B},$$

(13)

where

$$\dot{A}={\dot{A}}_t({\varphi }_t,{\varphi }_t),\ddot{A}={\ddot{A}}_t({\varphi }_t,{\varphi }_t),\dot{B}={\dot{B}}_t({\varphi }_t,{\varphi }_t),\ddot{B}={\ddot{B}}_t({\varphi }_t,{\varphi }_t)$$

(14)

for ${\varphi }_t=u_k\left(t\right)$.

As

$${\displaystyle \frac{d^2}{d{t}^2}}{\displaystyle \frac{1}{\lambda }}=-{\displaystyle \frac{\ddot{\lambda }{\lambda }^2-2\lambda {\dot{\lambda }}^2}{{\lambda }^2}}$$

Theorem 1 implies the following result.

Theorem 2. 

If $k=m=1$ and

$$2{\dot{A}}^2+{\lambda }^2\ddot{B}\ge \lambda (\ddot{A}+2\dot{A}\dot{B}),$$

(15)

it holds that

$${\displaystyle \frac{d^2}{d{t}^2}}{\displaystyle \frac{1}{\lambda }}\ge 0.$$

(16)

Harmonic convexity of the first eigenvalue, inequality (16), was noticed by [2] for the Dirichlet problem of Laplacian under the conformal deformations of the domain in two-space dimension. Here, we calculate the values $\dot{A}$, $\ddot{A}$, $\dot{B}$, $\ddot{B}$ and the validity of (15) under general setting of the deformations of the domain, and then turn to the dynamical and the conformal deformations. Taking preliminaries in Section 2, thus, we show the results on dynamical and conformal deformations in Section 3 (Theorem 5 and Theorem 6 for arbitrary space dimension) and Section 4 (Theorem 8 and Theorem 9 for two-space dimension), respectively. As applications, we show several new inequalities on the first eigenvalue of the two-dimensional problem.

2. General Deformations around $t=0$

Let $\Omega$ be a bounded Lipschitz domain in n-dimensional Euclidean space ${\mathbb{R}}^n$ for $n\ge 2$. Suppose that its boundary $\partial \Omega$ is divided into two relatively open disconnected sets ${\gamma }_0$ and ${\gamma }_1$, satisfying

$$\overline{{\gamma }_0}\cup \overline{{\gamma }_1}=\partial \Omega ,\overline{{\gamma }_0}\cap \overline{{\gamma }_1}=\varnothing .$$

(17)

We study the eigenvalue problem of the Laplacian with a mixed boundary condition,

$$-\Delta u=\lambda uin\Omega ,u=0\mathrm{on}{\gamma }_0,{\displaystyle \frac{\partial u}{\partial \nu }}=0\mathrm{on}{\gamma }_1,$$

(18)

where

$$\Delta =\sum _{i=1}^n{\displaystyle \frac{{\partial }^2}{\partial x_i^2}}$$

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

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

(19)

defined for

$$A(u,v)={\int }_{\Omega }\nabla u·\nabla vdx,u,v\in V$$

(20)

and

$$B(u,v)={\int }_{\Omega }uvdx,u,v\in X,$$

(21)

where

$$X=L^2(\Omega ),V=\{v\in H^1(\Omega )\mid {\left.v\right|}_{{\gamma }_0}=0\}.$$

(22)

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

$$\parallel v\parallel ={\left({\int }_{\Omega }{|\nabla v|}^2+v^{2}dx\right)}^{1/2}.$$

The above reduction in (18) to (19)–(22) is justified via the trace operator to the boundary as $\Omega$ is a bounded Lipschitz domain (Theorem 2 of [1]). Here, the assumption (17) is made just for formulating (18) as in (19)–(22), and is able to be weakend, see Remark 3 in Section 4.

To confirm the well-posedness of (3), we note, first, that if ${\gamma }_0\ne \varnothing$, there is a coercivity of $A:V\times V\to \mathbb{R}$, which means the existence of $\delta >0$, such that

$$A(v,v)\ge {\delta \parallel v\parallel }^2,\forall v\in V.$$

(23)

If ${\gamma }_0=\varnothing$ we replace A by $A+B$, denoted by $\tilde{A}$. Then, this $\tilde{A}:V\times V\to \mathbb{R}$ is coercive, and the eigenvalue problem

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

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

If the bounded domain $\Omega \subset {\mathbf{R}}^n$ is provided with the cone property, the inclusion $H^1(\Omega )\hookrightarrow L^2(\Omega )$ is compact by Rellich–Kondrachov’s theorem [4]. Thus, there is a sequence of eigenvalues to (3), denoted by

$$0<{\lambda }_1\le {\lambda }_2\le \cdots \to +\infty .$$

The associated eigenfunctions, $u_1,u_2,\cdots$, furthermore, form a complete ortho-normal system in X, provided with the inner product induced by $B=B(·,·)$:

$$B(u_i,u_j)={\delta }_{ij},A(u_j,v)={\lambda }_{j}B(u_j,v),\forall v\in V,i,j=1,2,\cdots .$$

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

$${\lambda }_j=\underset{L_j}{min}\underset{v\in L_j\setminus \left\{0\right\}}{max}R\left[v\right]=\underset{W_j}{max}\underset{v\in W_j\setminus \left\{0\right\}}{min}R\left[v\right],$$

(24)

where

$$R\left[v\right]={\displaystyle \frac{A(v,v)}{B(v,v)}}$$

is the Rayleigh quotient, and $\left\{L_j\right\}$ and $\left\{W_j\right\}$ denote the families of all subspaces of V with dimension and codimension j and $j-1$, respectively.

The following well-known fact is valid without the smoothness of $\partial \Omega$. The proof given in Appendix A for completeness.

Theorem 3. 

If $\Omega \subset {\mathbf{R}}^n$ is a bounded Lipschitz domain, the first eigenvalue ${\lambda }_1$ to (3) formulated to $A(u,v)$ and $B(u,v)$ defined by (20)–(22) is simple.

Coming back to the Lipschitz bounded domain $\Omega$, we introduce its deformation as follows. Let

$$T_t:\Omega \to {\Omega }_t=T_t(\Omega ),t\in (-{\epsilon }_0,{\epsilon }_0)$$

(25)

be a family of bi-Lipschitz homeomorphisms. We assume that $T_{t}x$ is continuous in t uniformly in $x\in \Omega$, and continue to use the following definition as in [1].

Definition 1. 

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

$${\displaystyle \frac{{\partial }^{\ell }}{\partial t^{\ell }}}D{T}_t,{\displaystyle \frac{{\partial }^{\ell }}{\partial t^{\ell }}}{\left(D{T}_t\right)}^{-1}:\Omega \to M_n\left(\mathbb{R}\right),0\le \ell \le p$$

are uniformly bounded in $(x,t)\in \Omega \times (-{\epsilon }_0,{\epsilon }_0)$, where $D{T}_t$ denotes the Jacobi matrix of $T_t:\Omega \to {\Omega }_t$ and $M_n\left(\mathbb{R}\right)$ stands for the set of real $n\times n$ matrices. This $\left\{T_t\right\}$ is, furthermore, said to be continuously p-differentiable in t if it is p-differentiable and the mappings

$$t\in (-{\epsilon }_0,{\epsilon }_0)\mapsto {\displaystyle \frac{{\partial }^{\ell }}{\partial t^{\ell }}}D{T}_t,{\displaystyle \frac{{\partial }^{\ell }}{\partial t^{\ell }}}{\left(D{T}_t\right)}^{-1}\in L^{\infty }(\Omega \to M_n\left(\mathbb{R}\right)),0\le \ell \le p$$

are continuous.

Putting

$$T_t\left({\gamma }_i\right)={\gamma }_{it},i=0,1,$$

(26)

in (18), we introduce the other eigenvalue problem

$$-\Delta u=\lambda u\mathrm{in}{\Omega }_t,u=0\mathrm{on}{\gamma }_{0t},{\displaystyle \frac{\partial u}{\partial \nu }}=0\mathrm{on}{\gamma }_{1t},$$

(27)

which is reduced to finding

$$u\in V_t,{\int }_{{\Omega }_t}{u}^{2}dx=1,{\int }_{{\Omega }_t}\nabla u·\nabla vdx=\lambda {\int }_{{\Omega }_t}uvdx,\forall v\in V_t$$

(28)

for

$$V_t=\{v\in H^1\left({\Omega }_t\right)\mid {\left.v\right|}_{{\gamma }_{0t}}=0\}.$$

(29)

Let ${\lambda }_j\left(t\right)$ be the j-th eigenvalue of the eigenvalue problem (27). Then, Lemma 7 of [1] ensures that the eigenvalue problem (28) and (29) is reduced to

$$u\in V,B_t(u,u)=1,A_t(u,v)=\lambda B_t(u,v),\forall v\in V$$

(30)

by the transformation of variables $y=T_{t}x$, where

$$\begin{array}{ccc}& & B_t(u,v)={\int }_{\Omega }uv{a}_{t}dx,u,v\in X\hfill \\ & & A_t(u,v)={\int }_{\Omega }{Q}_t[\nabla u,\nabla v]a_{t}dx,u,v\in V\hfill \end{array}$$

(31)

for V and X defined by (22), and

$$a_t=detD{T}_t,Q_t={\left(D{T}_t\right)}^{-1}{\left(D{T}_t\right)}^{-1T}.$$

(32)

Recall that $\Omega \subset {\mathbb{R}}^n$ is a bounded Lipshitz domain, and let $T_t:\Omega \to {\Omega }_t=T_t\Omega$, $t\in I=(-{\epsilon }_0,{\epsilon }_0)$ be a family of twice continuously differentiable bi-Lispchitz transformations. Then, the abstract theory described in Section 1 is applicable with

$${\dot{A}}_t(u,v)={\int }_{\Omega }{\displaystyle \frac{\partial }{\partial t}}\left(Q_t[\nabla u,\nabla v]a_t\right)dx,{\dot{B}}_t(u,v)={\int }_{\Omega }uv{\displaystyle \frac{\partial a_t}{\partial t}}dx$$

and

$${\ddot{A}}_t(u,v)={\int }_{\Omega }{\displaystyle \frac{{\partial }^2}{\partial t^2}}\left(Q_t[\nabla u,\nabla v]a_t\right)dx,{\ddot{B}}_t(u,v)={\int }_{\Omega }uv{\displaystyle \frac{{\partial }^2{a}_t}{\partial t^2}}dx$$

Henceforth, we write

$$G[\xi ,\eta ]={\xi }^{T}G\eta ,\xi ,\eta \in {\mathbb{R}}^n$$

for the symmetric matrix $G\in M_n\left(\mathbb{R}\right)$, where ${·}^T$ denotes the transpose of the vectors or matrices. The unit matrix is denoted by $E\in M_n\left(\mathbb{R}\right)$, and

$$G:F=\sum _{i,j=1}^n{g}_{ij}{f}_{ij}$$

for $G=\left(g_{ij}\right)$ and $F=\left(f_{ij}\right)\in M_n\left(\mathbf{R}\right)$. Let, furthermore, $I:\Omega \to \Omega$ be the identity mapping.

Theorem 4. 

Assume

$$T_t=I+tS+{\displaystyle \frac{t^2}{2}}R+o\left(t^2\right),t\to 0$$

(33)

uniformly on Ω, where $S,R:\Omega \to {\mathbb{R}}^n$ are Lipschitz continuous vector fields. Then, if $m=1$ in (8) for $t=0$, it holds that

$$\begin{array}{ccc}& & \dot{A}={\int }_{\Omega }(-(D{S}^T+DS)+(\nabla ·S)E)[\nabla \phi ,\nabla \phi ]\hfill \\ & & \ddot{A}={\int }_{\Omega }({(2{\left(DS\right)}^2-DR)}^T+(2{\left(DS\right)}^2-DR)\hfill \\ & & +2\left(DS\right){\left(DS\right)}^T-(\nabla ·S)(D{S}^T+DS)\hfill \\ & & +(\nabla ·R+{(\nabla ·S)}^2-D{S}^T:DS)E)[\nabla \varphi ,\nabla \varphi ]dx\hfill \\ & & \dot{B}={\int }_{\Omega }(\nabla ·S){\varphi }^{2}dx\hfill \\ & & \ddot{B}={\int }_{\Omega }(\nabla ·R+{(\nabla ·S)}^2-D{S}^T:DS){\varphi }^{2}dx,\hfill \end{array}$$

where $\dot{A}$, $\ddot{A}$, $\dot{B}$, and $\ddot{B}$ are defined by (14) with $t=0$, and $\varphi ={\varphi }_0$.

Proof. 

First, as

$$D{T}_t=E+tDS+{\displaystyle \frac{t^2}{2}}DR+o\left(t^2\right)$$

uniformly on $\Omega$, it holds that

$${\left(D{T}_t\right)}^{-1}=E-tD{S}^T+{\displaystyle \frac{t^2}{2}}(2{\left(DS\right)}^2-DR)+o\left(t^2\right)$$

uniformly on $\Omega$, which implies

$$a_t=detD{T}_t=1+t\nabla ·S+{\displaystyle \frac{t^2}{2}}(\nabla ·R+{(\nabla ·S)}^2-D{S}^T:DS)+o\left(t^2\right)$$

uniformly on $\Omega$ (Lemma 5 and Lemma 6 of [1]).  □

Second, we have

$$Q_t[\nabla u,\nabla v]=(D{T}_t^{-1}\nabla u,D{T}_t^{-1T}\nabla v),u,v\in V,$$

using the standard $L^2$ inner product $(,)$, and, therefore, it follows that

$$\begin{array}{ccc}& & {\left.{\displaystyle \frac{\partial Q_t}{\partial t}}[\nabla u,\nabla v]\right|}_{t=0}\hfill \\ & & ={\left.({\displaystyle \frac{\partial }{\partial t}}D{T}_t^{-1T}\nabla u,D{T}_t^{-1T}\nabla v)+(D{T}_t^{-1T}\nabla u,{\displaystyle \frac{\partial }{\partial t}}D{T}_t^{-1T}Dv)\right|}_{t=0}\hfill \\ & & =(-D{S}^T\nabla u,\nabla v)+(\nabla u,-D{S}^T\nabla v)=-(D{S}^T+DS)[\nabla u,\nabla v]\hfill \end{array}$$

and

$$\begin{array}{ccc}& & {\left.{\displaystyle \frac{{\partial }^2{Q}_t}{\partial t^2}}[\nabla u,\nabla v]\right|}_{t=0}=({\displaystyle \frac{{\partial }^2}{\partial t^2}}D{T}_t^{-1T}\nabla u,D{T}_t^{-1T}\nabla v)\hfill \\ & & {\left.+2({\displaystyle \frac{\partial }{\partial t}}D{T}_t^{-1T}\nabla u,{\displaystyle \frac{\partial }{\partial t}}D{T}_t^{-1T}\nabla v)+(D{T}_t^{-1T}\nabla u,{\displaystyle \frac{{\partial }^2}{\partial t^2}}D{T}_t^{-1T}Dv)\right|}_{t=0}\hfill \\ & & ={(2{\left(DS\right)}^2-DR)}^T\nabla u,\nabla v)+2(D{S}^T\nabla u,D{S}^T\nabla v)\hfill \\ & & +(\nabla u,{(2{\left(DS\right)}^2-DR)}^T\nabla v)\hfill \\ & & =({(2{\left(DS\right)}^2-DR)}^T+(2{\left(DS\right)}^2-DR)+2DSD{S}^T)[\nabla u,\nabla v].\hfill \end{array}$$

Using these equalities, finally, we obtain

$$\begin{array}{ccc}& & {\left.{\displaystyle \frac{\partial }{\partial t}}\left(Q_t[\nabla u,\nabla v]a_t\right)\right|}_{t=0}={\left.{\displaystyle \frac{\partial Q_t}{\partial t}}[\nabla u,\nabla v]a_t+Q_t[\nabla u,\nabla v]{\displaystyle \frac{\partial a_t}{\partial t}}\right|}_{t=0}\hfill \\ & & =-(D{S}^T+DS)[\nabla u,\nabla v]+(\nabla u,\nabla v)\nabla ·S\hfill \end{array}$$

and

$$\begin{array}{ccc}& & {\left.{\displaystyle \frac{{\partial }^2}{\partial t^2}}\left(Q_t[\nabla u,\nabla v]a_t\right)\right|}_{t=0}={\displaystyle \frac{{\partial }^2{Q}_t}{\partial t^2}}[\nabla u,\nabla v]a_t\hfill \\ & & {\left.+2{\displaystyle \frac{\partial Q_t}{\partial t}}[\nabla u,\nabla v]{\displaystyle \frac{\partial a_t}{\partial t}}+Q_t[\nabla u,\nabla v]{\displaystyle \frac{{\partial }^2{a}_t}{\partial t^2}}\right|}_{t=0}\hfill \\ & & =-((2{\left(DS\right)}^2-DR)+(2{\left(DS\right)}^2-DR)+2DSD{S}^T)[\nabla u,\nabla v]\hfill \\ & & -2(D{S}^T+DS)[\nabla u,\nabla v](\nabla ·S)\hfill \\ & & +(\nabla u,\nabla v)(\nabla ·R+{(\nabla ·S)}^2-D{S}^T:DS),\hfill \end{array}$$

and, hence, the conclusion.

3. Dynamical Deformations

We continue to suppose that $\Omega \subset {\mathbf{R}}^n$ is a bounded Lipschitz domain. Here, we study the dynamical deformation of domains introduced by [1].

To this end, we take a Lipschitz continuous vector field defined on a neighbourhood of $\overline{\Omega }$, denoted by $v=v\left(x\right)$. Then, the transformation $T_t:\Omega \to {\Omega }_t=T_t(\Omega )$ is made by $T_{t}x=X\left(t\right)$, where $X=X\left(t\right)$, $\left|t\right|\ll 1$, is the solution to

$${\displaystyle \frac{dX}{dt}}=v\left(X\right),{\left.X\right|}_{t=0}=x.$$

Then, we have the group property,

$$T_t\circ T_s=T_{t+s},\left|t\right|,\left|s\right|\ll 1,$$

and, therefore, the formulae derived for $t=0$ are shifted to $t=s$.

If v is a $C^{1,1}$ vector field, furthermore, this $T_t:\Omega \to {\Omega }_t$ is twice continuously differentiable, and it holds that (33) with

$$S=v,R=(v·\nabla )v$$

(34)

because of

$${\displaystyle \frac{d^{2}X}{d{t}^2}}=\left[(v·\nabla )\right]v\left(X\right).$$

Then, we obtain the following lemma.

Lemma 1. 

Under the above assumption, it holds that

$$\nabla ·R=D{S}^T:DS+(v·\nabla )(\nabla ·v).$$

Proof. 

Writing $v=\left(v^j\right)$, we obtain

$$\begin{array}{ccc}\hfill \nabla ·R& =& \nabla ·\left((v·\nabla )v\right)=\sum _{i,j}{\displaystyle \frac{\partial }{\partial x_i}}\left(v^j{\displaystyle \frac{\partial v^i}{\partial x_j}}\right)\hfill \\ & =& \sum _{i,j}({\displaystyle \frac{\partial v^j}{\partial x_i}}{\displaystyle \frac{\partial v^i}{\partial x_j}}+v^j{\displaystyle \frac{\partial }{\partial x_j}}{\displaystyle \frac{\partial v^i}{\partial x_i}})=D{S}^T:DS+(v·\nabla )(\nabla ·v)\hfill \end{array}$$

and the proof is complete.  □

Here, we take two categories that the vector fields are solenoidal and gradient. In the first category, we assume $\nabla ·v=0$ everywhere. Then, it holds that

$$|{\Omega }_t|=|\Omega |,|t|\ll 1.$$

(35)

Theorem 5. 

If $\nabla ·v=0$, it holds that $\dot{B}=\ddot{B}=0$ and

$$\begin{array}{ccc}& & \dot{A}=-{\int }_{\Omega }(D{S}^T+DS)[\nabla \varphi ,\nabla \varphi ]dx\hfill \\ & & \ddot{A}={\int }_{\Omega }{(2{\left(DS\right)}^2-DR)}^T+(2{\left(DS\right)}^2-DR)\hfill \\ & & +2\left(DS\right){\left(DS\right)}^T)[\nabla \varphi ,\nabla \varphi ]dx.\hfill \end{array}$$

(36)

in Theorem 4.

Proof. 

We recall (34). By the assumption we obtain $\nabla ·S=0$ and

$$\nabla ·R=D{S}^T:DS$$

by Lemma 1, which implies $\dot{B}=\ddot{B}=0$, and (36) by Theorem 4.  □

Remark 1. 

For the moment, we write

$$a_i={\displaystyle \frac{\partial a}{\partial x_i}},a_{ij}={\displaystyle \frac{{\partial }^{2}a}{\partial x_i\partial x_j}}$$

in short. Then, by the proof of Lemma 1, we obtain

$$DR={\left({\left(v^i{v}_i^j\right)}_k\right)}_{jk}={\left(DS\right)}^2+{\left(v^i{v}_{ik}^j\right)}_{jk}$$

and

$$D{R}^T={\left(DS\right)}^{2T}+{\left(v^i{v}_{ij}^k\right)}_{jk}$$

similarly. Hence, it holds that

$$\ddot{A}={\int }_{\Omega }({\left(DS\right)}^{2T}+{\left(DS\right)}^2+2\left(DS\right){\left(DS\right)}^T-G)[\nabla \varphi ,\nabla \varphi ]dx$$

for

$$G={\left(v^i(v_{ik}^j+v_{ij}^k)\right)}_{jk}.$$

In the second category that the vector field is a gradient of a scalar field, we obtain the following theorem.

Theorem 6. 

If $v=\nabla \mu$ for the $C^{1,1}$ scalar field μ, it holds that

$$\begin{array}{ccc}& & \dot{A}={\int }_{\Omega }(-2H+\left(trH\right)E)[\nabla \varphi ,\nabla \varphi ]dx\hfill \\ & & \ddot{A}={\int }_{\Omega }(6H^2-K-4\left(trH\right)H\hfill \\ & & +({\left(trH\right)}^2+{\displaystyle \frac{1}{2}}trK-H:H)E)[\nabla \varphi ,\nabla \varphi ]dx\hfill \end{array}$$

(37)

and

$$\dot{B}={\int }_{\Omega }\left(trH\right){\varphi }^{2}dx,\ddot{B}={\int }_{\Omega }({\left(trH\right)}^2+{\displaystyle \frac{1}{2}}trK-H:H){\varphi }^{2}dx,$$

(38)

where

$$H={\nabla }^2\mu ,K={\nabla }^2{\left(\right|\nabla \mu |}^2)$$

and

$${\left|H\right|}^2=H:H=\sum _{ij}{h}_{ij}^{2}forH=\left(h_{ij}\right).$$

Proof. 

In this case, we obtain $S=\nabla \mu$ and

$$R=((\nabla \mu )·\nabla )\nabla \mu ={\left(\sum _k{\mu }_k{\mu }_{ik}\right)}_i={\left({\left({\displaystyle \frac{1}{2}}\sum _k{\mu }_k^2\right)}_i\right)}_i={\displaystyle \frac{1}{2}}\nabla {|\nabla \mu |}^2.$$

(39)

Then, it holds that

$$\dot{B}={\int }_{\Omega }{\varphi }^2\nabla ·Sdx={\int }_{\Omega }{\varphi }^2\Delta \mu dx={\int }_{\Omega }\left(trH\right){\varphi }^{2}dx.$$

It also holds that

$$\begin{array}{ccc}\hfill (v·\nabla )(\nabla ·v)& =& \left(\right(\nabla \mu )·\nabla )(\Delta \mu )=\nabla \mu ·\Delta \nabla \mu \hfill \\ & =& {\displaystyle \frac{1}{2}}{\Delta \left(\right|\nabla \mu |}^2)-|{\nabla }^2{\mu |}^2,\hfill \end{array}$$

(40)

because

$${\Delta \left|a\right|}^2=2\Delta a·a+\sum _i{\displaystyle \frac{\partial a}{\partial x_i}}·{\displaystyle \frac{\partial a}{\partial x_i}}$$

is valid to the vector field a.

As

$$\ddot{B}={\int }_{\Omega }({(\nabla ·S)}^2+(v·\nabla )(\nabla ·v)){\varphi }^{2}dx$$

is valid by Lemma 1, and hence

$$\begin{array}{ccc}\hfill \ddot{B}& =& {\int }_{\Omega }{((\Delta \mu )}^2+{\displaystyle \frac{1}{2}}{\Delta \left(\right|\nabla \mu |}^2)-|{\nabla }^2\mu {|}^2){\varphi }^{2}dx\hfill \\ & =& {\int }_{\Omega }{\left(\mathrm{tr}H\right)}^2+{\displaystyle \frac{1}{2}}\mathrm{tr}K-{\left|H\right|}^2){\varphi }^{2}dx.\hfill \end{array}$$

Next, we obtain

$$D{S}^T+DS=2H$$

and hence

$$\begin{array}{ccc}\hfill \dot{A}& =& {\int }_{\Omega }(-2H+(\Delta \mu )E)[\nabla \varphi ,\nabla \varphi ]dx\hfill \\ & =& {\int }_{\Omega }(-2H+\left(trH\right)E)[\nabla \varphi ,\nabla \varphi ]dx.\hfill \end{array}$$

Finally, we divide $\ddot{A}$ as in

$$\ddot{A}=I+II+III$$

for

$$\begin{array}{ccc}& & I={\int }_{\Omega }\left(2({\left(DS\right)}^{2T}+{\left(DS\right)}^2+D{S}^{T}DS)\right)-(D{R}^T+DR))[\nabla \varphi ,\nabla \varphi ]dx\hfill \\ & & II={\int }_{\Omega }-2(D{S}^T+DS)[\nabla \varphi ,\nabla \varphi ]\nabla ·Sdx\hfill \\ & & III={\int }_{\Omega }(\nabla ·R+{(\nabla ·S)}^2-D{S}^T:DS){|\nabla \varphi |}^{2}dx.\hfill \end{array}$$

Here, we have

$${\left(DS\right)}^{2T}+{\left(DS\right)}^2+D{S}^{T}DS=3H^2.$$

Equality (39) now implies

$$R_j^i+R_i^j={\left(\right|\nabla \mu |}^2{)}_{ij}$$

for $R=\left(R^i\right)$, and hence

$$\nabla R^T+\nabla R={\nabla }^2{\left(\right|\nabla \mu |}^2)=K.$$

Thus, we obtain

$$I={\int }_{\Omega }(6H^2-K)[\nabla \varphi ,\nabla \varphi ]dx.$$

It holds also that

$$II={\int }_{\Omega }-4H[\nabla \varphi ,\nabla \varphi ](\Delta \mu )dx={\int }_{\Omega }-4\left(trH\right)H[\nabla \varphi ,\nabla \varphi ]dx.$$

Finally, we have

$$\begin{array}{ccc}\hfill III& =& {\int }_{\Omega }({(\nabla ·S)}^2+(v·\nabla )v){|\nabla \varphi |}^{2}dx\hfill \\ & =& {\int }_{\Omega }{((\Delta \mu )}^2+{\displaystyle \frac{1}{2}}{\Delta \left(\right|\nabla \mu |}^2)-|{\nabla }^2{\mu |}^2{\left)\right|\nabla \varphi |}^{2}dx\hfill \\ & =& {\int }_{\Omega }{(\left(trH\right)}^2+{\displaystyle \frac{1}{2}}trK-{\left|H\right|}^2{\left)\right|\nabla \varphi |}^{2}dx.\hfill \end{array}$$

We thus end up with the result by combining these equalities.  □

Remark 2. 

Theorem 5 and Theorem 6 are consistent if $v=\nabla \mu$ with

$$\Delta \mu =0.$$

In this case, it holds that

$$trH=0$$

(41)

and

$${\displaystyle \frac{1}{2}}trK-{\left|H\right|}^2=0$$

(42)

by (40):

$$0=(v·\nabla )(\nabla ·v)={\displaystyle \frac{1}{2}}{\Delta \left(\right|\nabla \mu |}^2)-|{\nabla }^2{\mu |}^2.$$

Then, we can reproduce the result of Theorem 5 from (38), as in

$$\dot{B}=0,\ddot{B}=0.$$

(43)

We have, furthermore,

$$\dot{A}={\int }_{\Omega }-{2H|\nabla \varphi |}^{2}dx,\ddot{A}={\int }_{\Omega }(6H^2-K){|\nabla \varphi |}^{2}dx$$

(44)

by (37). By (13), (43) and (44), first,

$$\dot{\lambda }=\dot{A}-\lambda \dot{B}=\dot{A}$$

cannot have a definite sign because of (41). Second, if

$$6H^2-K\le 0$$

(45)

we obtain $\ddot{A}\le 0$ by (13), which implies

$$\ddot{\lambda }\le \ddot{A}-\lambda \ddot{B}-2\dot{\lambda }\dot{B}=\ddot{A}\le 0$$

regardless of the form of ϕ. For inequality (45) to hold, however, it is necessary that

$$tr(6H^2-K)\ge 0,$$

or

$$3tr\left(H^2\right)\le {\left|H\right|}^2$$

by (42). This inequality implies $H=0$ because

$$tr\left(H^2\right)=\sum _{ij}{h}_{ij}^2={\left|H\right|}^2$$

for $H=\left(h_{ij}\right)$ with $h_{ji}=h_{ij}$. In other words, several possibilities can arise to the monotonicity or convexity of the first eigenvalue under the dynamical perturbation, provided with the property (35).

4. Conformal Deformations

Here, we say that $T_t:\Omega \to {\Omega }_t$ is conformal if

$${\left(D{T}_t\right)}^{-1}{\left(D{T}_t\right)}^{-1T}={\displaystyle \frac{1}{{\alpha }_t}}E$$

holds with a scalar field ${\alpha }_t>0$. It follows that ${\alpha }_{t}E=\left(D{T}_t\right){\left(D{T}_t\right)}^T$, and, therefore, the matrix $F_t={\alpha }_t^{-1/2}D{T}_t$ is orthogonal: $F_t{F}_t^T=E$.

Then, we assume, furthermore, that

$$det{F}_t=1$$

for simplicity. It follows that

$$a_t=detD{T}_t=det\left({\alpha }_t^{1/2}{F}_t\right)={\alpha }_t^{n/2},$$

(46)

and, hence, ${\alpha }_t=a_t^{2/n}$, which results in

$${\left(D{T}_t\right)}^{-1}{\left(D{T}_t\right)}^{-1T}=a_t^{-2/n}E,$$

and, hence

$$A_t(u,v)={\int }_{\Omega }(\nabla u·\nabla v)a_t^{1-2/n}dx,u,v\in V$$

and

$$B_t(u,v)={\int }_{\Omega }uv{a}_{t}dx.$$

Then, we can show the following theorem.

Theorem 7. 

Let $n=2$, $k=m=1$, and $T_t:\Omega \to {\Omega }_t$ be conformal at t. Assume, furthermore, that

$${\displaystyle \frac{{\partial }^2{a}_t}{\partial t^2}}\ge 0in\Omega .$$

Then, it holds that

$${\displaystyle \frac{d^2}{d{t}^2}}{\displaystyle \frac{1}{{\lambda }_1\left(t\right)}}\ge 0.$$

(47)

Proof. 

As $n=2$, we obtain

$$A_t(u,v)={\int }_{\Omega }\nabla u·\nabla vdx,B_t(u,v)={\int }_{\Omega }uv{a}_{t}dx.$$

(48)

Then, it holds that $\dot{A}=\ddot{A}=0$ and

$$\ddot{B}={\int }_{\Omega }{\varphi }^2{\displaystyle \frac{{\partial }^2{a}_t}{\partial t^2}}dx$$

in Theorem 1 for $\varphi ={\varphi }_t$. Then, the result follows from Theorem 2.  □

A consequence of Theorem 7 is the following isometric inequality.

Theorem 8. 

Let $\Omega \subset {\mathbf{R}}^2$ be a simply connected bounded Lipschitz domain and $f=f\left(z\right):D\to \Omega$ be a univalent bi-Lipschitz homeomorphism, where $D=\{z\in \mathbb{C}\mid |z|<1\}$. Let, furthermore,

$$f\left(z\right)=\sum _{n=0}^{\infty }{a}_n{z}^n,a_1\in \mathbb{R}.$$

Then, it holds that

$${\lambda }_1\left(D\right)\ge {\lambda }_1(\Omega )\left(2{\int }_D{\varphi }_1^{2}Re{f}^{\prime }\left(z\right)dx-1\right),$$

(49)

where ${\lambda }_1(\Omega )$ and ${\lambda }_1\left(D\right)$ are the first eigenvalue of (18):

$$-\Delta u=\lambda uin\Omega ,u=0on{\gamma }_0,{\displaystyle \frac{\partial u}{\partial \nu }}=0on{\gamma }_1$$

(50)

and that for $\Omega =D$,

$$-\Delta u=\lambda uinD,u=0in{\gamma }_0^{\prime },{\displaystyle \frac{\partial u}{\partial \nu }}=0on{\gamma }_1^{\prime }$$

(51)

respectively, with ${\gamma }_i=f\left({\gamma }_i^{\prime }\right)$, $i=1,2$, and furthermore, ${\varphi }_1={\varphi }_1\left(x\right)$ is the first eigenfunction of (51), such that

$${\int }_D{\varphi }_1^{2}dx=1.$$

Remark 3. 

Either ${\gamma }_0=\varnothing$ or ${\gamma }_1=\varnothing$ occurs if $\Omega \subset {\mathbb{R}}^2$ is simply connected and (17), and in the former case, it follows that ${\lambda }_1(\Omega )=0$. and the above theorem is trivial. This assumption (17), however, is used in [1] just to formulate (18) as in (3) with (20)–(22). Hence, we can exclude this assumption if we formulate (50) as in (19) with (20) and (21) for V standing for the closure of $V_0$ in $H^1(\Omega )$, where $v\in V_0$, if and only if, there is $\tilde{v}$, a smooth extension of v in a neighborhood of $\overline{\Omega }$, such that

$${\left.\tilde{v}\right|}_{{\gamma }_0}=0.$$

Then, we re-formulate (51) as in (30) with (31) for $t=1$, using the bi-Lipschits homeomorphism $T_1=f^{-1}:\Omega \to D$. Under this agreement of (50) and (51), we exclude the assumption (17) in the above theorem. Isoperimetric inequality of eigenvalues is a fundamental property in engineering, for example, in shape optimization, see [5]. To the best of our knowledge, however, most of them are restricted to the cases of Dirichlet or Neuman boundary conditions, and the above result on the mixed boundary condition seems to be new.

Proof of Theorem 8. 

Let

$$g_t\left(z\right)=(1-t)z+tf\left(z\right),0\le t\le 1$$

(52)

be the conformal mapping on D, and define $A_t$ and $B_t$ by (48) for (46) with $n=2$. We obtain $a_t={\left|g_t^{\prime }\left(z\right)\right|}^2$ in (46) and, hence, it follows that

$$|g_t^{\prime }{\left(z\right)|}^2=|1+t(f^{\prime }\left(z\right)-1){|}^2=1+2tRe(f^{\prime }\left(z\right)-1)+t^2{|f^{\prime }\left(z\right)-1|}^2$$

and hence

$$\begin{array}{ccc}& & {\displaystyle \frac{\partial }{\partial t}}|g_t^{\prime }{\left(z\right)|}^2=2Re(f^{\prime }\left(z\right)-1)+2t{|f^{\prime }\left(z\right)-1|}^2\hfill \\ & & {\displaystyle \frac{{\partial }^2}{\partial t^2}}|g_t^{\prime }{\left(z\right)|}^2=2{|f^{\prime }\left(z\right)-1|}^2\ge 0.\hfill \end{array}$$

(53)

Thus, we obtain

$${\displaystyle \frac{d^2}{d{t}^2}}{\displaystyle \frac{1}{{\lambda }_1\left(t\right)}}\ge 0,0\le t\le 1$$

for ${\lambda }_1\left(t\right)$ defined by (24) with $j=1$ and $A=A_t$ and $B=B_t$, which implies

$${\displaystyle \frac{1}{{\lambda }_1(\Omega )}}\ge {\displaystyle \frac{1}{{\lambda }_1\left(D\right)}}+{\left.{\left({\displaystyle \frac{1}{{\lambda }_1\left(t\right)}}\right)}^{\prime }\right|}_{t=0},$$

(54)

or

$${\displaystyle \frac{1}{{\lambda }_1(\Omega )}}\ge {\displaystyle \frac{1}{{\lambda }_1\left(D\right)}}+{\displaystyle \frac{1}{{\lambda }_1\left(D\right)}}{\int }_D{\phi }_1^2{\left.{\displaystyle \frac{\partial }{\partial t}}{\left|g_t^{\prime }\left(z\right)\right|}^2\right|}_{t=0}dx$$

by

$${\left({\displaystyle \frac{1}{{\lambda }_1}}\right)}^{\prime }=-{\displaystyle \frac{{\lambda }_1^{\prime }}{{\lambda }_1^2}}$$

and (12) with (48). Then, we obtain the result by (53).  □

Remark 4. 

If ${\gamma }_1=\varnothing$, inequality (49) is reduced to

$${\lambda }_1\left(D\right)\ge {\lambda }_1(\Omega )(2Re{a}_1-1),$$

(55)

As $f:D\to \Omega$ is univalent, it holds that

$$|\Omega |={\int }_D|f^{\prime }{\left(z\right)|}^{2}dx=\pi \sum _{n=1}^{\infty }n{\left|a_n\right|}^2,$$

and therefore, $|\Omega |=\left|D\right|$ if and only if

$$\sum _{n=1}^{\infty }n{\left|a_n\right|}^2=1.$$

This equality implies $|a_1|\le 1$ and, in particular,

$$2Re{a}_1-1\le 1$$

in (55).

Inequality (55) for ${\gamma }_1=\varnothing$ is proven in Appendix B. We conclude this section with an analogous result to (49).

Theorem 9. 

Under the assumption of the previous theorem, it holds that

$${\lambda }_1(\Omega )\ge {\lambda }_1\left(D\right)\left(2{\int }_D{\tilde{\varphi }}_1^{2}Re{f}^{\prime }\left(z\right)dx-1\right),$$

where ${\tilde{\varphi }}_1={\widehat{\varphi }}_1\circ f$ for the first eigenfunction ${\widehat{\varphi }}_1={\widehat{\varphi }}_1\left(x\right)$ to (50) such that

$${\int }_{\Omega }{\widehat{\varphi }}_1^{2}dx=1.$$

Proof. 

Similarly to (54), we obtain

$$\begin{array}{ccc}\hfill {\displaystyle \frac{1}{{\lambda }_1\left(D\right)}}& \ge & {\displaystyle \frac{1}{{\lambda }_1(\Omega )}}-{\left.{\left({\displaystyle \frac{1}{{\lambda }_1\left(t\right)}}\right)}^{\prime }\right|}_{t=1}={\displaystyle \frac{1}{{\lambda }_1(\Omega )}}\left(1+{\displaystyle \frac{{\lambda }_1^{\prime }\left(1\right)}{{\lambda }_1{(\Omega )}^2}}\right)\hfill \\ & =& {\displaystyle \frac{1}{{\lambda }_1(\Omega )}}\left(1-{\int }_D{\tilde{\varphi }}_1^2{\left.{\displaystyle \frac{\partial }{\partial t}}{\left|g_1^{\prime }\left(z\right)\right|}^2\right|}_{t=1}dx\right),\hfill \end{array}$$

which implies the result by

$${\int }_D{\tilde{\varphi }}_1^2{\left|f^{\prime }\left(z\right)\right|}^{2}dx={\int }_{\Omega }{\widehat{\varphi }}_1^{2}dx=1$$

and

$${\left.{\displaystyle \frac{\partial }{\partial t}}{\left|g^{\prime }\left(z\right)\right|}^2\right|}_{t=1}=2{\left|f^{\prime }\left(z\right)\right|}^2-2Re{f}^{\prime }\left(z\right).$$

□

5. Conclusions

We have derived several Hadamard’s variational formula for simple eigenvalues in explicit forms under dynamical and conformal deformations of the domain. Particularly, harmonic convexity of the first eigenvalue of the two-dimensional Laplacian is established, which implies several isoperimetric inequalities for the mixed boundary condtions.