On Blaschke–Santaló diagrams for the torsional rigidity and the first Dirichlet eigenvalue

Abstract

We study Blaschke–Santaló diagrams associated with the torsional rigidity and the first eigenvalue of the Laplacian with Dirichlet boundary conditions. We work under convexity and volume constraints, in both strong (volume exactly one) and weak (volume at most one) form. We discuss some topological (closedness, simply connectedness) and geometric (shape of the boundaries, slopes near the point corresponding to the ball) properties of these diagrams, also providing a list of conjectures.

Appendix

This section is devoted to the proof of Proposition 2.7, namely to the computation of the second-order shape derivatives of T and $\lambda _1$ at $\mathbb B$ in dimension 2, with respect to deformations which preserve convexity and keep the volume unchanged. For the formulas of shape derivatives see [17, Chapter 5] and [6, 7, 22]. Similar computations in terms of Fourier coefficients can be found in [1, 5].

The representation (2.6) in terms of support functions accounts for the convexity constraint. As for the volume constraint, since we perform a second-order analysis, it is enough to impose that the first and second-order shape derivatives of the area vanish. These imply a constraint on the Fourier coefficients.

Lemma 6.1

Let V and W be two admissible deformations in ${\mathcal {A}}$. Denote by $\alpha$ and $\beta$ be the first and second variation of the support function, defined according to (2.5)–(2.6). Then,

$$\begin{aligned} \int _0^{2\pi }\alpha (\theta )\mathrm {d}\theta =0,\quad \int _0^{2\pi } \beta (\theta )\mathrm {d}\theta = \frac{1}{R}\int _0^{2\pi }[\dot{\alpha }(\theta )^2 -\alpha (\theta )^2] \mathrm {d}\theta . \end{aligned}$$

(6.1)

Proof

By assumption, for every $\epsilon$ small, the volume, denoted here by $\mathrm {Vol}$, is constant, namely $\mathrm {Vol}(\Omega _\epsilon )=\mathrm {Vol}(\mathbb B)$. In particular, $\mathrm {Vol}'(\mathbb B;V)=\mathrm {Vol}''(\mathbb B;V,W)=0$. In view of the well-known formulas for $\mathrm {Vol}'$ and $\mathrm {Vol}''$ (see for instance [17, §5.9.3 and §5.9.6]), we have

$$\begin{aligned} \mathrm {Vol}'(\mathbb B;V)=\int _{\partial \mathbb B} V\cdot n \, \mathrm {d}{\mathcal {H}}^1 = 0, \quad \mathrm {Vol}''(\mathbb B;V,W)= \int _{\partial \mathbb B} [\kappa (V\cdot n)^2 + Z + W\cdot n]\, \mathrm {d}{\mathcal {H}}^1=0, \end{aligned}$$

(6.2)

where $\kappa$ denotes the mean curvature, here equal to 1/R, and Z is the following function, defined on $\partial \mathbb B$:

$$\begin{aligned} Z:=( D_\Gamma n\, V_\Gamma ) \cdot V_\Gamma - 2 [\nabla _\Gamma (V\cdot n)] \cdot V_\Gamma . \end{aligned}$$

(6.3)

The subscript $\Gamma$ denotes the tangential component of a vector/operator: for a vector field U and a function f defined in the whole ${\mathbb {R}}^2$, there hold

$$\begin{aligned} U_\Gamma :=U - (U\cdot n) n\,,\quad D_\Gamma U :=D U - (DU\, n)\otimes n\,,\quad \nabla _\Gamma f:=\nabla f - (\nabla f \cdot n) n, \end{aligned}$$

where DU denotes the Jacobian matrix of U. Let us rewrite the boundary integrals in (6.2) in polar coordinates: in view of (2.5), we have $V\cdot n = \alpha$, $V\cdot \tau =\dot{\alpha }$, and $W\cdot n = \beta$, so that (6.2) reads

$$\begin{aligned} \int _0^{2\pi } \alpha (\theta )\mathrm {d}\theta = \int _0^{2\pi }[\alpha (\theta )^2 + R Z + R\beta ] \mathrm {d}\theta =0. \end{aligned}$$

(6.4)

Choosing any extension of n, $\tau$, and V to ${\mathbb {R}}^2$, we find $( D_\Gamma n\, V_\Gamma ) \cdot V_\Gamma =[\nabla _\Gamma (V\cdot n)] \cdot V_\Gamma =\dot{\alpha }^2/R$, so that

$$\begin{aligned} Z(R \cos \theta , R \sin \theta )=- \frac{\dot{\alpha }^2}{R}. \end{aligned}$$

(6.5)

Inserting this expression in (6.4), we conclude the proof. $\square$

Proof of Proposition 2.7

Throughout the proof, for brevity, we will omit the subscript $\mathbb B$ in the first eigenfunction and in the torsional rigidity, which will be denoted by $\varphi$ and w, respectively. The second-order shape derivatives of $\lambda _1$ and T at $\mathbb B$ are

$$\begin{aligned} \lambda _1''(\mathbb B;V,W)&= \int _{\partial \mathbb B} \left( - W\cdot n- Z + \kappa (V\cdot n)^2 \right) |\partial _\nu \varphi |^2 \mathrm {d}\mathcal H^1 + 2 \int _{\partial \mathbb B}\psi \partial _\nu \psi \mathrm {d}{\mathcal {H}}^1, \end{aligned}$$

(6.6)

$$\begin{aligned} T''(\mathbb B;V,W)&= \int _{\partial \mathbb B} \left[ \left( W\cdot n + Z - \kappa (V\cdot n)^2\right) |\partial _\nu w|^2 + 2 (V\cdot n)^2 |\partial _\nu w| \right] \mathrm {d}{\mathcal {H}}^1 - 2 \int _{\partial \mathbb B} v \partial _\nu v \, \mathrm {d}{\mathcal {H}}^1, \end{aligned}$$

(6.7)

where $\kappa$ is the curvature, Z is the function introduced in (6.3), and $\psi$ and v solve

$$\begin{aligned} \left\{ \begin{array}{lll} -\Delta \psi = \lambda _1(\mathbb B) \psi - \varphi \int _{\partial \mathbb B} |\partial _\nu \varphi |^2 V\cdot n\, \mathrm {d}{\mathcal {H}}^1 \quad &{} \hbox {in }\mathbb B\\ \psi = -(V\cdot n) \partial _\nu \varphi \quad &{} \hbox {on }\partial \mathbb B\\ \int _{\mathbb B} \psi \varphi =0 \end{array} \right. \qquad \left\{ \begin{array}{lll} \Delta v = 0\quad &{} \hbox {in }\mathbb B\\ v = -(V\cdot n) \partial _\nu w \quad &{} \hbox {on }\partial \mathbb B. \end{array} \right. \end{aligned}$$

(6.8)

We recall that the torsion function of the disk $\mathbb B$ is $w=(R^2- |x|^2)/4$ so that, on the boundary, we have $|\partial _\nu w| =R/2$. Similarly, since the first eigenfunction of the Dirichlet Laplacian, normalized in $L^2$, is $\varphi = J_0(j_{0,1} |x|/R )/|J_0'(j_{0,1})|$, we have $|\partial _\nu \varphi |=j_{0,1}/R$ on the boundary. Let us perform the change of variables in polar coordinates in the integrals above. Using the fact that $\kappa =1/R$, writing Z as in (6.5), recalling the expression (2.5) of V on $\partial \mathbb B$ in terms of $\alpha$, and exploiting the conditions (6.1) on $\alpha$ and $\beta$, we obtain a first simplification:

$$\begin{aligned} \lambda _1''(\mathbb B;V,W)&= \frac{2 j_{0,1}^2}{R^2} \int _0^{2\pi } \alpha ^2 \mathrm {d}\theta + 2 R \int _0^{2\pi }\psi \partial _\nu \psi \, \mathrm {d}\theta , \end{aligned}$$

(6.9)

$$\begin{aligned} T''(\mathbb B;V,W)&= \frac{R^2}{2} \int _0^{2\pi } \alpha ^2\mathrm {d}\theta - 2 R \int _0^{2\pi } v \partial _\nu v \, \mathrm {d}\theta . \end{aligned}$$

(6.10)

Let us now determine $\psi$ in terms of $\alpha$ and of its Fourier coefficients $a_m$ and $b_m$. First, we notice that, in view of the condition $\int \alpha =0$ in (6.1), the PDE solved by $\psi$ is $-\Delta \psi = \lambda _1(\mathbb B)\psi$. Therefore, it is natural to look for $\psi$ as a linear combination (possibly a series) of the eigenfunctions $J_m(j_{0,1} \rho /R)\cos (m\theta )$ and $J_m(j_{0,1} \rho /R)\sin (m\theta )$ associated to the eigenvalue $\lambda _1(\mathbb B)=j_{0,1}^2/R^2$, namely $\psi (\rho ,\theta )=\sum _{m\ge 0} [A_m \cos (m\theta ) + B_m \sin (m \theta )] J_m(j_{0,1}\rho /R)$. The orthogonality condition between $\psi$ and the radial function $\varphi$ gives $A_0=0$. Imposing the boundary condition $\psi (R,\theta )=j_{0,1}\alpha (\theta )/R$, we get

$$\begin{aligned} A_m=\frac{j_{0,1} a_m}{ R J_m (j_{0,1})}\,,\quad B_m=\frac{ j_{0,1} b_m}{R J_m (j_{0,1})}\,,\quad \forall m\ge 1\,. \end{aligned}$$

A direct computation leads to

$$\begin{aligned} \int _{0}^{2\pi } \psi \partial _\nu \psi \mathrm {d}\theta = \frac{\pi j_{0,1}^3}{R^3} \sum _{m\ge 1 }\frac{J_m'(j_{0,1})}{J_m(j_{0,1})}(a_m^2 + b_m^2). \end{aligned}$$

(6.11)

By combining (6.9) and (6.11), recalling that $\int _0^{2\pi } \alpha ^2 = \pi \sum _{m\ge 1}(a_m^2 + b_m^2)$ and using $j_{0,1} J_1'(j_{0,1})=- J_1(j_{0,1})$, we get

$$\begin{aligned} \lambda _1''(\mathbb B;V,W) = \frac{2\pi j_{0,1}^2}{R^2} \sum _{m\ge 2}\left[ \left( 1 + j_{0,1}\frac{J_m'(j_{0,1})}{J_m(j_{0,1})} \right) (a_m^2+b_m^2)\right] . \end{aligned}$$

Following the same procedure, we may derive v as a function of $a_m$ and $b_m$. Formally, v can be searched as the infinite sum of harmonic functions, namely $v(\rho ,\theta )=\sum _{m\ge 0} [C_m \cos (m\theta ) + D_m \sin (m\theta )] \rho ^m$. Imposing the boundary condition, we obtain

$$\begin{aligned} C_0=D_0=0\,,\quad C_m = \frac{a_m}{2 R^{m-1}} \,,\quad D_m = \frac{ b_m}{2R^{m-1}}\,,\quad \forall m\ge 1\,. \end{aligned}$$

In particular,

$$\begin{aligned} \int _0^{2\pi } v \partial _\nu v \mathrm {d}\theta = \frac{\pi R}{4} \sum _{m\ge 1} m (a_m^2 + b_m^2), \end{aligned}$$

and (6.10) reads

$$\begin{aligned} T''(\mathbb B;V,W) = \frac{\pi R^2}{2} \sum _{m\ge 2} \left[ (1-m)(a_m^2 + b_m^2) \right] . \end{aligned}$$

This concludes the proof. $\square$

Remark 6.2

At first sight, the equalities (2.8)–(2.9) might seem surprising, since W apparently does not play any role. Actually, as it is clear from the formulas used in the previous proof, in the second-order shape derivatives only the normal component of W appears, averaged with $|\nabla w_\mathbb B|^2$ or $|\nabla \varphi _\mathbb B|^2$ on the boundary. Since both norms of the gradients are constant, the relevant quantity is the average of $W\cdot n$. The average is nothing but $\int \beta = c_0$, which in turn can be written in terms of $\alpha$ or $\{a_m, b_m\}$, as we have proved in Lemma 6.1 and rephrased in (2.7).