Effective Shape Optimization of Laplace Eigenvalue Problems Using Domain Expressions of Eulerian Derivatives

Abstract

We consider to solve numerically the shape optimization models with Dirichlet Laplace eigenvalues. Both volume-constrained and volume unconstrained formulations of the model problems are presented. Different from the literature using boundary-type Eulerian derivatives in shape gradient descent methods, we advocate to use the more general volume expressions of Eulerian derivatives. We present two shape gradient descent algorithms based on the volume expressions. Numerical examples are presented to show the more effectiveness of the algorithms than those based on the boundary expressions.

Appendix A: Shape Sensitivity Analysis

Lemma A.1

(Theorem 2.5.3 [22]) Let $\varOmega$ be an open and bounded domain. Then, the Eulerian derivative reads:

$$\begin{aligned} |\varOmega |^\prime (\varOmega ;\mathcal {V})=\int _\varOmega \mathrm{div}\ \mathcal {V}\,\mathrm{d}x. \end{aligned}$$

If, moreover, $\varOmega$ is Lipschitz, then

$$\begin{aligned} |\varOmega |^\prime (\varOmega ;\mathcal {V})=\int _{\partial \varOmega } \mathcal {V}_n\,\mathrm{d}s. \end{aligned}$$

For a function $u(x):\varOmega \rightarrow \mathbb {R}$ and its perturbation $u_t(x_t):[0,\tau ]\times \mathbb {R}^d\rightarrow \mathbb {R}$, we have

Definition A.1

The material derivative of a state variable u in a direction $\mathcal {V}$ is denoted as

$$\begin{aligned} \dot{u}(\varOmega ;\mathcal {V}):=\lim _{t\searrow 0} \frac{u(\varOmega _t)\circ T_t(\mathcal {V})(\varOmega )-u(\varOmega )}{t} \end{aligned}$$

if the limit exists.

Definition A.2

The shape derivative of u in a direction $\mathcal {V}$ is defined by

$$\begin{aligned} u^\prime (\varOmega ;\mathcal {V}):=\dot{u}(\varOmega ;\mathcal {V})-\nabla u\cdot \mathcal {V}. \end{aligned}$$

In the following, we omit the subscript l of $\lambda _l$ and $u_l$ for simplicity. The following result holds (see Theorems 2.5.1 [22] and 4.2 [2]). We give a heuristic derivation for the sake of completeness.

Proposition A.1

Let $\varOmega$ be an open bounded domain. Assume that the first Dirichlet eigenvalue $\lambda =\lambda (\varOmega )$ is simple. Then, $\lambda (\varOmega )$ is shape differentiable and the Eulerian derivative

$$\begin{aligned} \lambda ^\prime (\varOmega ;\mathcal {V}) = \int _\varOmega \big ( - 2\nabla u\cdot \mathrm{D}\mathcal {V}\nabla u + \mathrm{div} \mathcal {V}(|\nabla u|^2-\lambda u^2)\big )\mathrm{d}x. \end{aligned}$$

(12)

If, furthermore, $\varOmega$ is convex or if it is of class $C^2$, then

$$\begin{aligned} \lambda ^\prime (\varOmega ;\mathcal {V})=-\int _{\partial \varOmega } \left( \frac{\partial u}{\partial n}\right) ^2 \mathcal {V}_n\mathrm{d}s. \end{aligned}$$

(13)

Proof

We only derive (13) formally below for simplicity. The derivation of volume formulation (12) can be performed similarly. For a test function $\phi \in H^1(\varOmega )$ with compact support $\omega \subset \varOmega$, we have

$$\begin{aligned} \int _\omega \nabla u\cdot \nabla \phi \, \mathrm{d}x = \lambda \int _\omega u\phi \,\mathrm{d}x. \end{aligned}$$

Differentiating with respect to $\varOmega$ implies that

$$\begin{aligned} \int _\omega \nabla u^\prime \cdot \nabla \phi \, \mathrm{d}x = \lambda ^\prime \int _\omega u\phi \mathrm{d}x+\lambda \int _\omega u^\prime \phi \mathrm{d}x, \end{aligned}$$

where $u^\prime :=u^\prime (\varOmega ;\mathcal {V})$ and $\lambda ^\prime :=\lambda ^\prime (\varOmega ;\mathcal {V})$. By integration by parts, we have

$$\begin{aligned} -\Delta u^\prime - \lambda u^\prime =\lambda ^\prime u\quad \mathrm{in}\ \varOmega . \end{aligned}$$

(14)

To derive the boundary condition, we have

$$\begin{aligned} \int _{\partial \varOmega } u\varphi \mathrm{d}s = 0\quad \forall \varphi \in C^\infty (\mathbb {R}^d). \end{aligned}$$

Taking the Eulerian derivative implies that

$$\begin{aligned} \int _{\partial \varOmega } \bigg [u^\prime \varphi +\bigg (\frac{\partial (u\varphi )}{\partial n}+\kappa u\varphi \bigg )\mathcal {V}_n\bigg ]\mathrm{d}s = 0\quad \forall \varphi \in C^\infty (\mathbb {R}^d), \end{aligned}$$

where $\kappa$ denotes the mean curvature. Thus,

$$\begin{aligned} u^\prime = -\frac{\partial u}{\partial n}\mathcal {V}_n \end{aligned}$$

(15)

since $u=0$ on $\partial \varOmega$.

Multiplying Eq. (14) by u and integrating by parts leads to

$$\begin{aligned} \int _{\varOmega } \nabla u^\prime \cdot \nabla u\mathrm{d}x =\lambda \int _{\varOmega } u^\prime u \mathrm{d}x+\lambda ^\prime \int _{\varOmega } u^2 \mathrm{d}x. \end{aligned}$$

Multiplying the equation for u by $u^\prime$ and integrating by parts leads to

$$\begin{aligned} \int _{\varOmega } \nabla u\cdot \nabla u^\prime \mathrm{d}x =\lambda \int _{\varOmega } u u^\prime \mathrm{d}x+\int _{\partial \varOmega }\frac{\partial u}{\partial n} u^\prime \mathrm{d}s. \end{aligned}$$

Thus, a combination of the two equations above yields that

$$\begin{aligned} \lambda ^\prime \int _{\varOmega } u^2 \mathrm{d}x = \int _{\partial \varOmega }\frac{\partial u}{\partial n} u^\prime \, \mathrm{d}s, \end{aligned}$$

from which we obtain the result after using (15) and the normalization $\int _\varOmega u^2\,\mathrm{d}x=1$. $\square$

Proposition A.2

Let $\varOmega$ be an open bounded domain. For $J(\varOmega )=|\varOmega |\lambda (\varOmega )$, we have

$$\begin{aligned} \begin{aligned} J^\prime (\varOmega ;\mathcal {V}) = \int _\varOmega \Big [ |\varOmega |\big (- 2\nabla u\cdot \mathrm{D}\mathcal {V}\nabla u + \mathrm{div} \mathcal {V}(|\nabla u|^2-\lambda u^2)\Big )+\lambda \mathrm{div} \mathcal {V} \big ]\,\mathrm{d}x. \end{aligned} \end{aligned}$$

If, moreover, $\varOmega$ is convex or $C^2$, then

$$\begin{aligned} \begin{aligned} J^\prime (\varOmega ;\mathcal {V})=\int _{\partial \varOmega }\bigg [\lambda -|\varOmega |\bigg (\frac{\partial u}{\partial n}\bigg )^2 \bigg ]\mathcal {V}_n\,\mathrm{d}s. \end{aligned} \end{aligned}$$

Proof

First, we perform the product rule for Eulerian derivatives

$$\begin{aligned} J^\prime (\varOmega ;\mathcal {V})= \lambda |\varOmega |^\prime + |\varOmega |\lambda ^\prime (\varOmega ;\mathcal {V}). \end{aligned}$$

Then, we obtain (6)–(7) and (9)–(10) by combining Proposition A.1 and Lemma A.1. $\square$