Sufficient conditions yielding the Rayleigh Conjecture for the clamped plate

Abstract

The Rayleigh Conjecture for the bilaplacian consists in showing that the clamped plate with least principal eigenvalue is the ball. The conjecture has been shown to hold in 1995 by Nadirashvili in dimension 2 and by Ashbaugh and Benguria in dimension 3. Since then, the conjecture remains open in dimension $d\ge 4$. In this paper, we contribute to answer this question, and show that the conjecture is true in any dimension as long as some special condition holds on the principal eigenfunction of an optimal shape. This condition regards the mean value of the eigenfunction, asking it to be in some sense minimal. This main result is based on an order reduction principle allowing to convert the initial fourth order linear problem into a second order affine problem, for which the classic machinery of shape optimization and elliptic theory is available. The order reduction principle turns out to be a general tool. In particular, it is used to derive another sufficient condition for the conjecture to hold, which is a second main result. This condition requires the Laplacian of the optimal eigenfunction to have constant normal derivative on the boundary. Besides our main two results, we detail shape derivation tools allowing to prove simplicity for the principal eigenvalue of an optimal shape and to derive optimality conditions. Finally, because our first result involves the principal eigenfunction of a ball, we are led to compute it explicitly.

Appendix

Proof of Proposition 15

For readability, we ommit the subscript B in $u_B$. According to [3], in B, the first eigenfunction is radially symmetric and of the form $\forall r\in [0,R)$,

$$\begin{aligned} u(r)=\left( aJ_\nu (k r)+bI_\nu (k r)\right) r^{-\nu }, \end{aligned}$$

where $k:=\Gamma (B)^{\frac{1}{4}}$. Then, using the identities $J_\nu '(x)=\frac{\nu J_\nu (x)}{x}-J_{\nu +1}(x)$ and $I_\nu '(x)=\frac{\nu I_\nu (x)}{x}+I_{\nu +1}(x)$ we find

$$\begin{aligned} \partial _r u(r)=\left( -aJ_{\nu +1}(kr)+bI_{\nu +1}(kr)\right) kr^{-\nu }. \end{aligned}$$

Now, for u to fulfill the condition $u(R)=\partial _r u(R)=0$ although being non trivial, one observes that the matrix

$$\begin{aligned} M=\left( \begin{array}{cc} J_\nu (kR) &{} I_\nu (kR) \\ -J_{\nu +1}(kR) &{} I_{\nu +1}(kR) \end{array}\right) \end{aligned}$$

needs having a non trivial kernel. In other words, its determinant needs to vanish, hence

$$\begin{aligned} f_\nu (kR)=J_\nu (kR)I_{\nu +1}(kR)+J_{\nu +1}(kR)I_\nu (kR)=0. \end{aligned}$$

Conversely, as soon as k satisfies this equation, u will be solution of an eigenvalue problem in B with Dirichlet boundary conditions. Consequently, k is necessarily the lowest positive solution of this equation, meaning that $kR=\gamma _\nu$. Hence $\Gamma (B)=\gamma _\nu ^4/R^4$.

Furthermore, since $I_\nu >0$ over $(0,\infty )$, $M\ne 0$ and it has a one-dimensional kernel. By virtue of the identity $u(R)=0$, the kernel is generated by the vector $(I_\nu (\gamma _\nu ),-J_\nu (\gamma _\nu ))$ or equivalently by the vector $R^\nu (J_\nu (\gamma _\nu )^{-1},-I_\nu (\gamma _\nu )^{-1})$. In other words, there exists a real number $\beta$ such that

$$\begin{aligned} \left( \begin{array}{c} a \\ b \end{array}\right) =\beta R^\nu \left( \begin{array}{c} J_\nu (\gamma _\nu )^{-1} \\ -I_\nu (\gamma _\nu )^{-1} \end{array}\right) . \end{aligned}$$

Finding the values of a and b is thus equivalent to determining $\beta$. For that purpose, we use the normalisation of u, i.e.

$$\begin{aligned} \begin{aligned} 1=\int _B u^2=\beta ^2R^{2\nu }|\mathbb {S}^{d-1}|&\left[ J_\nu (\gamma _\nu )^{-2}\int _0^RJ_\nu (k r)^2r^{d-2\nu -1}\right. \\&\left. +I_\nu (\gamma _\nu )^{-2}\int _0^RI_\nu (k r)^2r^{d-2\nu -1}\right. \\&\left. -2J_\nu (\gamma _\nu )^{-1}I_\nu (\gamma _\nu )^{-1}\int _0^RI_\nu (kr)J_\nu (kr)r^{d-2\nu -1}\right] . \end{aligned} \end{aligned}$$

(21)

As $d-2\nu -1=1$, it turns out that we need to compute the integral of product of Bessel functions against r. That’s why we use the Gradshteyn and Ryzhik collection [15, section 6.521, formula 1], that is, for all $\alpha \ne \beta \in \mathbb {C}$ and $\nu >-1$,

$$\begin{aligned} \begin{aligned} \int _0^1xJ_\nu (\alpha x)J_\nu (\beta x){} & {} =\frac{\beta J_{\nu -1}(\beta )J_\nu (\alpha )-\alpha J_{\nu -1}(\alpha )J_\nu (\beta )}{\alpha ^2-\beta ^2}\\{} & {} =\frac{\alpha J_{\nu +1}(\alpha )J_\nu (\beta )-\beta J_{\nu +1}(\beta )J_\nu (\alpha )}{\alpha ^2-\beta ^2}. \end{aligned} \end{aligned}$$

(22)

We apply this formula with $\alpha =i\gamma _\nu$ and $\beta =\gamma _\nu$, and find

$$\begin{aligned} \int _0^RI_\nu (k r)J_\nu (k r)r^{d-2\nu -1}=\frac{R^2}{2\gamma _\nu }[I_{\nu +1}(\gamma _\nu )J_\nu (\gamma _\nu )+J_{\nu +1}(\gamma _\nu )I_\nu (\gamma _\nu )]=\frac{R^2}{2\gamma _\nu }f_\nu (\gamma _\nu )=0. \end{aligned}$$

For the other integrals, we first remark that when ($\alpha ,\beta \in \mathbb {R}$ and) $\alpha \rightarrow \beta$ in (22), one obtains

$$\begin{aligned} \begin{aligned} \int _0^1 xJ_\nu (\beta x)^2&=\frac{J_\nu (\beta )^2}{2\beta }\frac{d}{d\beta }\left[ \frac{\beta J_{\nu +1}(\beta )}{J_\nu (\beta )}\right] \\&=\frac{1}{2}\left[ J_{\nu +1}(\beta )^2+J_\nu (\beta )^2-\frac{\nu }{\beta } J_{\nu +1}(\beta )J_\nu (\beta )\right] . \end{aligned} \end{aligned}$$

(23)

Hence, with $\beta =\gamma _\nu$, we find

$$\begin{aligned} \int _0^RJ_\nu (k r)^2r^{d-2\nu -1}=\frac{R^2}{2}\left[ J_{\nu +1}(\gamma _\nu )^2+J_\nu (\gamma _\nu )^2-\frac{\nu }{\gamma _\nu } J_{\nu +1}(\gamma _\nu )J_\nu (\gamma _\nu )\right] . \end{aligned}$$

But because both extremal members in (23) depend holomorphicly on $\beta$, this formula remains true even when $\beta \in {\mathbb {C}}$ thanks to the isolation of zeros, hence, we can apply it to $\beta =i\gamma _\nu$:

$$\begin{aligned} \int _0^RI_\nu (k r)^2r^{d-2\nu -1}=\frac{R^2}{2}\left[ -I_{\nu +1}(\gamma _\nu )^2+I_\nu (\gamma _\nu )^2-\frac{\nu }{\gamma _\nu } I_{\nu +1}(\gamma _\nu )I_\nu (\gamma _\nu )\right] . \end{aligned}$$

Finally, since $f_\nu (\gamma _\nu )=0$, the term between brackets in (21) becomes

$$\begin{aligned}{} & {} \frac{R^2}{2} \left[ J_\nu (\gamma _\nu )^{-2}\left( J_{\nu +1}(\gamma _\nu )^2+J_\nu (\gamma _\nu )^2-\frac{\nu }{\gamma _\nu }J_{\nu +1}(\gamma _\nu )J_\nu (\gamma _\nu )\right) \right. \\{} & {} \qquad \left. +I_\nu (\gamma _\nu )^{-2}\left( -I_{\nu +1}(\gamma _\nu )^2+I_\nu (\gamma _\nu )^2-\frac{\nu }{\gamma _\nu }I_{\nu +1}(\gamma _\nu )I_\nu (\gamma _\nu )\right) \right] \\{} & {} \quad = \frac{R^2}{2} \left[ 2+\left( \frac{J_{\nu +1}}{J_\nu }(\gamma _\nu )-\frac{I_{\nu +1}}{I_\nu }(\gamma _\nu )-\frac{\nu }{\gamma _\nu }\right) \left( \frac{J_{\nu +1}}{J_\nu }(\gamma _\nu )+\frac{I_{\nu +1}}{I_\nu }(\gamma _\nu )\right) \right] \\{} & {} \quad = R^2. \end{aligned}$$

Using $|\mathbb {S}^{d-1}|R^d=d|B|$, we have that $\beta ^{-2}=|\mathbb {S}^{d-1}|R^{2\nu +2}=d|B|$, hence

$$\begin{aligned} a=\frac{R^{\nu }}{J_\nu (\gamma _\nu )\sqrt{d|B|}},\qquad b=-\frac{R^{\nu }}{I_\nu (\gamma _\nu )\sqrt{d|B|}}. \end{aligned}$$

(24)

In particular,

$$\begin{aligned} u(r)=\frac{1}{\sqrt{d|B|}}\left( \frac{J_\nu (k r)}{J_\nu (\gamma _\nu )}-\frac{I_\nu (k r)}{I_\nu (\gamma _\nu )}\right) \left( \frac{r}{R}\right) ^{-\nu }, \end{aligned}$$

which corresponds to (17). After having obtained the expression of u, we would like to compute its integral. Observing that

$$\begin{aligned} \int _B u=\frac{1}{\Gamma (B)}\int _B\Delta ^2 u=\frac{1}{\Gamma (B)}\int _{\partial B}\partial _n\Delta u=\frac{R^{d-1}|\mathbb {S}^{d-1}|}{\gamma _\nu ^4/R^4}\partial _r\Delta u(R), \end{aligned}$$

it remains only to compute

$$\begin{aligned} \partial _r\Delta u(r)=[aJ_{\nu +1}(k r)+bI_{\nu +1}(k r)]k^3r^{-\nu }, \end{aligned}$$

for which we used the identities $J_{\nu +1}'(x)=J_\nu (x)-\frac{\nu +1}{x}J_{\nu +1}(x)$ and $I_{\nu +1}'(x)=I_\nu (x)-\frac{\nu +1}{x}I_{\nu +1}(x)$. As a result, we get

$$\begin{aligned} \int _B u =&\frac{R^{d-1}|\mathbb {S}^{d-1}|}{\gamma _\nu ^4/R^4}\frac{\gamma _\nu ^3/R^3}{\sqrt{d|B|}}\left[ \frac{J_{\nu +1}}{J_\nu }(\gamma _\nu )-\frac{I_{\nu +1}}{I_\nu }(\gamma _\nu )\right] \\ =&\frac{\sqrt{d|B|}}{\gamma _\nu }\left[ \frac{J_{\nu +1}}{J_\nu }(\gamma _\nu )-\frac{I_{\nu +1}}{I_\nu }(\gamma _\nu )\right] . \end{aligned}$$

Note that the last equality in (18) comes from the fact that $f_\nu (\gamma _\nu )=0$. $\square$