Topology optimization using an eigenvector aggregate

Abstract

Topology optimization problems with natural frequency or structural stability criteria often utilize objective or constraint functions computed from the eigenvalues of a generalized eigenvalue problem. However, design formulations involving the eigenvectors are not common, due to both the difficulties that occur in the presence of repeated eigenvalues and the computational cost of computing eigenvector derivatives. To address the formulation problem, a smoothly differentiable function is proposed that is computed based on the eigenvalues and eigenvectors of a generalized eigenvalue problem. This eigenvector aggregate is constructed to approximate a homogeneous quadratic function of the eigenvector associated with the smallest eigenvalue. To address the computational cost, a technique is proposed to compute high accuracy approximations of the derivative of the eigenvector aggregate by solving a sequence of related linear systems with a constrained Krylov method that incorporates orthogonal projection. The proposed eigenvector aggregate can be used to impose displacement and stress constraints on the eigenvectors. Results are shown for a tube and 2D topology optimization problems, each with bimodal lowest eigenvalue.

Appendices

Appendix 1

Here, we show that

$$\begin{aligned} Q \eta Q^T = \dfrac{B^{-1} \exp (-\rho A B^{-1}) }{{\text{tr}}(\exp (-\rho A B^{-1}))}. \end{aligned}$$

(55)

Since $\eta = \text {diag}\{\eta _{1}, \eta _{2}, \ldots , \eta _{n}\}$, combined with the definition of $\eta _i$ in (11), we have

$$\begin{aligned} \eta = \frac{\exp (-\rho \Lambda )}{{\text{tr}}(\exp (-\rho \Lambda ))}, \end{aligned}$$

(56)

where $\exp (\cdot )$ denotes the matrix exponential. Therefore,

$$\begin{aligned} Q \eta Q^T = \dfrac{Q \exp (-\rho \Lambda ) Q^T}{{\text{tr}}(\exp (-\rho \Lambda ))}. \end{aligned}$$

(57)

Using the definition of the matrix exponential, the numerator for (56) can be expressed as

$$\begin{aligned} \begin{aligned} \exp (-\rho \Lambda )&= \exp (-\rho Q^T A Q) \\&= \sum _{k=0}^\infty \dfrac{1}{k!} (-\rho Q^T A Q)^k \\&= \sum _{k=0}^\infty \dfrac{1}{k!} Q^T (-\rho A Q Q^T)^k Q^{-T} \\&= Q^T \exp (-\rho A Q Q^T) Q^{-T} \\&= Q^T \exp (-\rho A B^{-1}) Q^{-T}. \end{aligned} \end{aligned}$$

(58)

Using this relationship and the property $Q Q^{T} = B^{-1}$, (57) can be rewritten as

$$\begin{aligned} \begin{aligned} Q \eta Q^T&= \dfrac{Q Q^{T} \exp (-\rho A B^{-1}) Q^{-T} Q^T}{{\text{tr}}(Q^{T} \exp (-\rho A B^{-1}) Q^{-T})} \\&= \dfrac{B^{-1} \exp (-\rho A B^{-1}) }{{\text{tr}}(\exp (-\rho A B^{-1}))}. \end{aligned} \end{aligned}$$

(59)

Appendix 2

In this appendix, we show that

$$\begin{aligned}{} & {} \dfrac{{\text{tr}}\left( (D B^{-1} - h I) \dot{\Phi } \right) }{{\text{tr}}(\Phi )} = \nonumber \\{} & {} \qquad {\text{tr}}( ( E \circ (Q^{T} D Q - h I ) ) (Q^{T} \dot{A} Q - \Lambda Q^{T} \dot{B} Q)). \end{aligned}$$

(60)

The derivative of the matrix exponential taking the generic argument X with respect to a design parameter is

$$\begin{aligned} \begin{aligned} \dot{\exp }(X)&= \sum _{k=1}^{\infty } \frac{1}{k!} \sum _{l=0}^{k-1} X^{l} \dot{X} X^{k-l-1} \\&= \sum _{k=0}^{\infty } \sum _{l=0}^{k} \frac{1}{(k+1)!} X^{l} \dot{X} X^{k-l}. \end{aligned} \end{aligned}$$

(61)

Therefore, the derivative of $\Phi$ defined in (18) can be written as

$$\begin{aligned} \dot{\Phi }= & {} \sum _{k=0}^{\infty }\sum _{l=0}^{k} \frac{(-\rho )^{k+1}}{(k+1)!} (AB^{-1})^{l} (\dot{A}\nonumber \\{} & {} - A B^{-1} \dot{B} ) B^{-1} (A B^{-1})^{k-l}. \end{aligned}$$

(62)

Powers of the product $AB^{-1}$ can be expanded using the eigenvalues and eigenvectors as

$$\begin{aligned} (A B^{-1})^n = (A Q Q^T)^n = (B Q \Lambda Q^T)^n = B Q \Lambda ^n Q^T. \end{aligned}$$

(63)

Combining (62) and (63), the matrix product in the series can be simplified as follows:

$$\begin{aligned} \begin{aligned}&(A B^{-1})^{l} (\dot{A} - A B^{-1} \dot{B}) B^{-1} (A B^{-1})^{k-l} \\&\quad = B Q \Lambda ^l Q^T (\dot{A} - A B^{-1} \dot{B}) B^{-1} B Q \Lambda ^{k-l} Q^T \\&\quad = B Q \Lambda ^l (Q^T \dot{A} Q - Q^T A Q Q^T \dot{B} Q ) \Lambda ^{k-l} Q^T \\&\quad = B Q \Lambda ^l (Q^T \dot{A} Q - \Lambda Q^T \dot{B} Q) \Lambda ^{k-l} Q^T. \\ \end{aligned} \end{aligned}$$

As a result, the derivative of $\Phi$ is

$$\begin{aligned}{} & {} \dot{\Phi } = \nonumber \\{} & {} \quad \sum _{k=0}^{\infty } \sum _{l=0}^{k} \frac{(-\rho )^{k+1}}{(k+1)!} B Q \Lambda ^l (Q^T \dot{A} Q - \Lambda Q^T \dot{B} Q) \Lambda ^{k-l} Q^T. \end{aligned}$$

(64)

Using this result (64), we can express the ratio appearing in (60) as

$$\begin{aligned}{} & {} \frac{{\text{tr}}(D B^{-1} \dot{\Phi })}{{\text{tr}}(\Phi )} = \nonumber \\{} & {} \quad \quad \frac{1}{{\text{tr}}(\Phi )} \sum _{k=0}^{\infty } \sum _{l=0}^{k} \frac{(-\rho )^{k+1}}{(k+1)!} \text {tr} \left( \Lambda ^{k-l} Q^{T} D Q \Lambda ^l (Q^T \dot{A} Q \right. \nonumber \\{} & {} \qquad \left. - \Lambda Q^T \dot{B} Q)\right) . \end{aligned}$$

(65)

To further simplify the above equation (65), we introduce the following Hadamard product with the matrix E, whose components are derived in Appendix 1, along with the substitution $X = Q^{T}D Q$, such that

$$\begin{aligned} E \circ X = \frac{1}{{\text{tr}}(\Phi )} \sum _{k=0}^{\infty } \sum _{l=0}^{k} \frac{(-\rho )^{k+1}}{(k+1)!}\Lambda ^{k-l} X \Lambda ^l. \end{aligned}$$

(66)

With this definition, equation (65) can be rewritten as

$$\begin{aligned}{} & {} \frac{{\text{tr}}(D B^{-1} \dot{\Phi })}{{\text{tr}}(\Phi )} = \nonumber \\{} & {} \quad \quad {\text{tr}}( (E \circ (Q^{T} D Q)) (Q^{T} \dot{A} Q - \Lambda Q^{T} \dot{B} Q)). \end{aligned}$$

(67)

The second part of the ratio appearing in (60) can be expressed using (67) by making the substitution $D = B = (Q Q^{T})^{-1} = Q^{-T} Q^{-1}$ resulting in

$$\begin{aligned} \dfrac{{\text{tr}}\left( \dot{\Phi } \right) }{{\text{tr}}(\Phi )} = {\text{tr}}( ( E \circ I ) (Q^{T} \dot{A} Q - \Lambda Q^{T} \dot{B} Q)). \end{aligned}$$

(68)

Combining (67) and (68) gives the desired result (60).

Appendix 3

In this appendix, we show that

$$\begin{aligned} E_{ij} = \dfrac{\eta _{j} - \eta _{i}}{\lambda _{j} - \lambda _{i}}. \end{aligned}$$

(69)

From (66), the components of the matrix E can be expressed as the double series

$$\begin{aligned} E_{ij} = \frac{1}{{\text{tr}}(\Phi )} \sum _{k=0}^{\infty } \sum _{l=0}^{k} \dfrac{(-\rho )^{k+1}}{(k+1)!} \lambda _{i}^{k - l} \lambda _{j}^{l}. \end{aligned}$$

(70)

We use the following property of the infinite double series

$$\begin{aligned} \sum _{k=0}^{\infty } \sum _{l=0}^{k} \dfrac{(-\rho )^{k+1}}{(k+1)!} \lambda _{i}^{k - l} \lambda _{j}^{l} = \dfrac{e^{-\rho \lambda _{j}} - e^{-\rho \lambda _{i}}}{\lambda _{j} - \lambda _{i}}. \end{aligned}$$

Furthermore, since the denominator for (70) ${\text{tr}}(\Phi ) = {\text{tr}}(\exp (-\rho \Lambda )) = \sum _{i=1}^{n} e^{-\rho \lambda _{i}}$, we can combine it with $\eta _i$ defined in (11) to obtain the result shown in (69).