Optimal second eigenvalue for Laplace operator with Dirichlet boudnary condition

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Optimal second eigenvalue for Laplace operator with Dirichlet boudnary condition

January 12, 2013 beni22sof Leave a comment Go to comments

It is known for some time that the problem of minimizing the ${k}$ -th eigenvalue of the Laplacian operator with Dirichlet boundary conditions

$\displaystyle \min_{|\Omega|=1} \lambda_k(\Omega)$

has a solution if the admissible sets ${\Omega}$ are all contained in a bounded open set ${D\subset \Bbb{R}^N}$ called design region (known as Buttazzo-Dal Maso Theorem).

The case ${D=\Bbb{R}^N}$ remained open for a long time until recently when D. Bucur and D. Mazzoleni, A. Pratelli gave two different proofs for the general case.

The case ${k=1}$ is known since 1923 with proofs given by G. Faber and E. Krahn. The usual proof relies on the Schwartz symmetrization of the domain and proves directly that the eigenvalue of the ball of the same volume is smaller than the eigenvalue of any other set with the same volume. In the case ${k=2}$ we have the following result:

Theorem. The minimum of ${\lambda_2(\Omega)}$ (the second eigenvalue of the Laplace-Dirichlet operator) among the quasi-open sets of volume ${m}$ given is attained for the union of two identical balls of volumes ${m/2}$ and this is the only minimizer.

Proof: Let ${\Omega}$ be an arbitrarily chosen quasi-open set and take ${\phi \in H_0^1(\Omega)}$ an eigenfunction associated to ${\lambda_2(\Omega)}$ which verifies

$\displaystyle \forall v \in H_0^1(\Omega),\ \int_\Omega \nabla \varphi \nabla v =\lambda_2(\Omega)\int_\Omega \varphi v,\ \int_\Omega \varphi^2=1,\ \int_\Omega \varphi\varphi_1=0,$

where ${\varphi_1}$ is an eigenfunction associated to the first eigenvalue of ${\Omega}$ . We can assume without loss of generality that ${\varphi_1 \geq 0}$ on ${\Omega}$ .

Denote ${\omega_1=[\varphi>0]}$ (the set where ${\varphi}$ is positive). We can assume that ${|\omega_1|>0}$ (if not, we change ${\varphi}$ by ${-\varphi}$ ). Therefore we have ${\psi=\varphi_{|_{\omega_1}}=\varphi^+ \in H_0^1(\omega_1)}$ . Apply the above relation to ${v \in H_0^1(\Omega)}$ and get

$\displaystyle \forall v \in H_0^1(\omega_1),\ \int_{\omega_1}\nabla \psi \nabla v=\lambda_2(\Omega)\int_{\omega_1}\psi v$

which tells us that ${\psi}$ is in fact an eigenfunction on ${\omega_1}$ of the corresponding eigenvalue ${\lambda_2(\Omega)}$ . Since this means that ${\lambda_2(\Omega)}$ is an eigenvalue for ${\omega_1}$ it follows that ${\lambda_2(\Omega) \geq \lambda_1(\omega_1)}$ .

Consider now ${\omega_2=\Omega \setminus \overline{\omega_1}}$ and ${\xi=\varphi_{|_{\omega_2}}=-\varphi^- \in H_0^1(\omega_2)}$ . If ${\xi\equiv 0}$ then because we have

$\displaystyle 0= \int_\Omega \varphi \varphi_1=\int_{\omega_1} \psi \varphi_1$

and ${\varphi,\varphi_1 \geq 0}$ we conclude that ${\varphi_1 \equiv 0}$ on ${\omega_1}$ . Since ${\varphi_1}$ is an eigenfunction it is not everywhere zero. Therefore ${\varphi_1 \in H_0^1(\omega_2), \varphi_1 \neq 0}$ . Writing now the variational characterization of the fact that ${\varphi_1}$ is an eigenfunction for ${\lambda_1(\Omega)}$ and using the fact that ${\varphi_1 \equiv 0}$ on ${\omega_1}$ we can see that ${\lambda_1(\Omega)}$ is also an eigenvalue for ${\omega_2}$ and therefore ${\lambda_2(\Omega) \geq \lambda_1(\Omega) \geq \lambda_1(\omega_2)}$ . If ${\xi}$ is not identically zero, we apply the initial variatioinal characterization for ${\lambda_2(\Omega)}$ and using a similar procedure as for ${\omega_1}$ we find that ${\lambda_2(\Omega)}$ is an eigenvalue of ${\omega_2}$ . This means that ${\lambda_2(\Omega) \geq \lambda_1(\omega_2)}$ .

We have found that in every case

$\displaystyle \lambda_2(\Omega) \geq \max \{ \lambda_1(\omega_1),\lambda_2(\omega_2)\}$

Consider now the domain ${\Omega^*}$ which is the union of two balls ${\omega_1^*,\omega_2^*}$ which have the same volumes as ${\omega_1,\omega_2}$ , respectively. Using similar arguments as above, we can see that every eigenvalue of ${\Omega^*}$ is an eigenvalue of ${\omega_1^*,\omega_2^*}$ . Conversely, if ${u}$ is an eigenfunction of ${\omega_1^*}$ then the function ${\bar u}$ equal to ${u}$ on ${\omega_1^*}$ and zero on ${\omega_2^*}$ is an eigenfunction of ${\Omega^*}$ for the same eigenvalue (the same for ${\omega_2^*}$ ). Using the monotonicity of eigenvalues for the inclusion we find that if, for example ${|\omega_1^*|\geq |\omega_2^*|}$ then ${\lambda_1(\Omega^*)=\lambda_1(\omega_1^*)}$ and ${\lambda_2(\Omega^*)\leq \lambda_1(\omega_2^*)}$ . Using the Faber Krahn inequality we have

$\displaystyle \lambda_2(\Omega^*) \leq \max \{\lambda_1(\omega_1^*),\lambda_1(\omega_2^*)\}\leq \max \{ \lambda_1(\omega_1),\lambda_2(\omega_2)\}\leq \lambda_2(\Omega)$

The above relation proves that the minimum must be searched among the unions of two balls. Denote by ${B_1}$ the unit ball, with the first two eigenvalues ${\lambda_1(B_1)=5.78,\lambda_2(B_1)=14.68}$ . Suppose the optimal set is ${\Omega^*=sB_1\cup (tB_1+v)}$ where ${\pi (s^2+t^2)=|\Omega|}$ . (where ${v}$ is chosen so that their intersection is void)

Then using the scaling formula ${\lambda_k(t\Omega)=\frac{1}{t^2}\lambda_k(\Omega)}$ , the first two eigenvalues of ${\Omega^*}$ are the smallest two elements in the set

$\displaystyle \left\{ \frac{5.78}{s^2},\frac{5.78}{t^2},\frac{14,68}{s^2},\frac{14,68}{t^2}\right\}$

Notice first that the smallest two elements from the above set, at the optimum, cannot be of the form ${5.78/s^2,14.68/s^2}$ where ${s\geq t}$ . Suppose it is the case. Then to minimize the second one ${14.68}$ we must make ${s}$ as big as possible and that happens if ${t=0}$ and the optimum is a ball. But then the union of two balls with equal radii has the first two eigenvalues equal to ${5.78\cdot \frac{2\pi}{|\Omega|}}$ whereas the second eigenvalue of the great ball is equal to ${14.68\cdot \frac{\pi}{\Omega}}$ . Contradiction.

If the smallest two elements from the above set, at the optimum, are of the form ${5.78/s^2,5.78/t^2}$ where ${s\geq t}$ , then it is obvious that ${5.78/t^2}$ is minimal when ${s=t}$ and only then.

We have seen that it is possible to find the exact optimal shapes for ${\lambda_1}$ and ${\lambda_2}$ by using direct methods. Unfortunately for ${k \geq 3}$ the optimal shapes are unknown. There are some conjectures and numerical computations, but no proof that those are indeed the optimal shapes exists, not even for ${k=3}$ where the optimal shape is conjectured to be a ball.

Bibliography

A. Henrot, M. Pierre, Variation et Optimization de Formes – Une analyse geometrique

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Categories: Calculus of Variations, Functional Analysis, Partial Differential Equations, shape optimization Tags: eigenvalues, laplace, optimization, shape optimization

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January 23, 2013 at 3:39 pm | #1

Numerical Results – Dirichlet Eigenvalues – Volume Constraint « Beni Bogoşel's blog

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Optimal second eigenvalue for Laplace operator with Dirichlet boudnary condition

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Optimal second eigenvalue for Laplace operator with Dirichlet boudnary condition

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