Quadratic model updating with gyroscopic structure from partial eigendata

Abstract

Quadratic eigenvalue model updating problem, which aims to match observed spectral information with some feasibility constraints, arises in many engineering areas. In this paper, we consider a damped gyroscopic model updating problem (GMUP) of constructing five n-by-n real matrices M,C,K,G and N, such that they are closest to the given matrices and the quadratic pencil Q(λ):=λ ² M+λ(C+G)+K+N possess the measured partial eigendata. In practice, M,C and K, represent the mass, damping and stiffness matrices, are symmetric (with M and K positive definite), G and N, represent the gyroscopic and circulatory matrices, are skew-symmetric. Under mild assumptions, we show that the Lagrangian dual problem of GMUP can be solved by a quadratically convergent inexact smoothing Newton method. Numerical examples are given to show the high efficiency of our method.

Appendices

Appendix A: Basic duality theory

Let $\mathcal{X}$ and $\mathcal{Y}$ be two real Hilbert spaces, each equipped with a scalar product 〈⋅,⋅〉 and its induced norm ∥⋅∥. Let $\mathcal{A}:\mathcal{X}\rightarrow\mathcal{Y}$ be a linear operator. Consider the following constrained best approximation problem,

$$ \begin{array}{@{}c@{\quad}c@{}}\textrm{min}& \displaystyle\frac{1}{2}\|x-c\|^2\\[7pt]\textrm{s.t.}& \mathcal{A}x=0,\\[5pt]&x\in\mathcal{Q},\end{array} $$

(29)

where $\mathcal{Q}$ is a closed convex cone in $\mathcal{X}$. The Lagrangian dual (see, e.g., Borwein and Lewis 1992 or Bonnans and Shaprio 2000) of the best approximation problem (29) is in the form of

$$ \begin{array} {@{}l@{\quad}l@{}} \textrm{max}& -\displaystyle\frac{1}{2}\bigl\|c+\mathcal{A}^*y\bigr\|^2 +\displaystyle \frac{1}{2}\bigl\|c+\mathcal{A}^*y-\varPi _{\mathcal{Q}}\bigl(c+\mathcal {A}^*y\bigr)\bigr\|^2+ \displaystyle\frac{1}{2}\|c\|^2\\[7pt]\textrm{s.t.}&y\in\mathcal{Y}, \end{array} $$

(30)

where $\mathcal{A}^{*}:\mathcal{Y}\rightarrow\mathcal{X}$ is the adjoint of $\mathcal{A}$, and for any $x\in\mathcal{X}$, $\varPi _{\mathcal{Q}}(x)$ is the metric projection of x onto $\mathcal{Q}$; i.e., $\varPi _{\mathcal{Q}}(x)$ is the unique optimal solution to

$$\begin{array}{@{}l@{\quad}l@{}}\textrm{min}& \displaystyle\frac{1}{2}\|u-x\|^2\\[7pt]\textrm{s.t.}&u\in\mathcal{Q}.\end{array} $$

Define $\theta:\mathcal{Y}\rightarrow\mathbb{R}$ by

$$\theta(y):=\frac{1}{2}\bigl\|c+\mathcal{A}^*y\bigr\|^2 -\frac{1}{2}\bigl\|c+\mathcal{A}^*y-\varPi _{\mathcal{Q}}\bigl(c+\mathcal {A}^*y\bigr)\bigl\|^2- \frac{1}{2}\|c\|^2.$$

Since $\mathcal{Q}$ is a closed convex cone, the function θ takes the following form,

$$ \theta(y):=\frac{1}{2}\bigl\|\varPi _{\mathcal{Q}}\bigl(c+\mathcal {A}^*y\bigr)\bigr\|^2- \frac{1}{2}\|c\|^2.$$

(31)

Note that θ(⋅) is a convex function (Borwein and Lewis 1992), and θ(⋅) is continuously differentiable (but not twice continuously differentiable) with

$$\nabla\theta(y)=\mathcal{A}\varPi _{\mathcal{Q}}\bigl(c+\mathcal{A}^*y\bigr),\quad y\in\mathcal{Y}.$$

Then the dual problem (30) becomes a smooth convex optimization problem with a simple constraint:

$$ \begin{array}{@{}l@{\quad}l@{}}\textrm{min}& \theta(y)\\[5pt]\textrm{s.t.}&y\in\mathcal{Y}.\end{array} $$

(32)

In order to apply a dual based optimization approach to solve problem (29), we need the following generalized Slater condition to hold:

$$ \left \{ \begin{array}{@{}l@{}}\mathcal{A}:\mathcal{X}\rightarrow\mathcal{Y}\textrm{ is onto},\\\exists\bar{x}\in\mathcal{X}\textrm{ such that }\mathcal{A}\bar {x}=0,\bar{x}\in\mathrm{int}(\mathcal{Q}),\end{array} \right .$$

(33)

where “int” denotes the topological interior of a given set. The following proposition is a well known result in the conventional duality theory for convex programming (Rockafellar 1974, Theorems 17 and 18).

Proposition 2

Under the generalized Slater condition (33), there exists at least one $y^{*}\in\mathcal{Y}$ that solves the dual problem (30), and the unique solution to original problem (29) is given by

$$x^*=\varPi _{\mathcal{Q}}\bigl(c+\mathcal{A}^*y^*\bigr).$$

Furthermore, for every real number τ, the level set $\{y\in\mathcal {Y}:\theta(y)\leq\tau\}$ is closed, bounded, and convex.

From Proposition 2, one may use any gradient based optimization method to find an optimal solution to the convex optimization problem (30), and then to solve problem (29), as long as the generalized Slater condition (33) holds. Since $\varPi _{\mathcal{Q}}(\cdot)$ is globally Lipschitz continuous, ∇θ(⋅) is globally Lipschitz continuous, which means one cannot use the classical Newton method to solve (30) directly. However, the semismooth Newton method or smoothing Newton method may be applicable if the function θ(⋅) is semismoothly differentiable.

Appendix B: Huber smoothing function

Suppose that the matrix $X\in\mathcal{S}^{n}$ has the spectral decomposition

$$ X=P\textrm{diag}(\lambda_1,\ldots, \lambda_n)P^T,$$

(34)

where λ ₁≥⋯≥λ _n are the eigenvalues of X and P is a corresponding orthogonal matrix of eigenvectors. Then,

$$\varPi _{\mathcal{S}_+^n}(X)=P\textrm{diag}\bigl(\textrm{max}(0,\lambda_1), \ldots ,\textrm{max}(0,\lambda_n)\bigr)P^T.$$

Define three index sets

$$\alpha:=\{i|\lambda_i>0\},\qquad\beta:=\{i|\lambda_i=0\},\qquad\gamma:=\{ i|\lambda_i<0\}.$$

Write P=[P _α   P _β   P _γ ] with P _α ,P _β and P _γ containing the columns in P indexed by α,β and γ, respectively. Let ϕ:ℝ×ℝ→ℝ be the following Huber smoothing function:

$$\phi(\varepsilon,t)=\left \{ \begin{array}{@{}l@{\quad}l@{}}t,&\textrm{if }t\geq\frac{|\varepsilon|}{2},\\[4pt]\frac{1}{2|\varepsilon|}(t+\frac{|\varepsilon|}{2})^2,&\textrm{if }|t|<\frac{|\varepsilon|}{2},\\[4pt]0,&\textrm{if }t\leq-\frac{|\varepsilon|}{2}.\end{array} \right .\quad(\varepsilon,t)\in\mathbb{R}\times\mathbb{R}.$$

For any ε∈ℝ, let

$$ \varPhi (\varepsilon,X):=P\textrm{diag}\bigl(\phi(\varepsilon,\lambda_1),\ldots ,\phi(\varepsilon,\lambda_n)\bigr)P^T.$$

(35)

Note that ϕ(0,t)=max(0,t), and similarly $\varPhi (0,X)=\varPi _{\mathcal{S}_{+}^{n}}(X)$. From Löwner (1934), we have that if β=∅ or ε≠0,

$$\varPhi '_{\varepsilon}(\varepsilon,X)=P\textrm{diag}\bigl(\phi_{\varepsilon }'(\varepsilon,\lambda_1),\ldots,\phi'_{\varepsilon}(\varepsilon,\lambda_n)\bigr)P^T,$$

and

$$\varPhi '_{X}(\varepsilon,X) (H)=P\bigl[\varOmega (\varepsilon,\lambda)\circ \bigl(P^THP\bigr)\bigr]P^T,\quad\forall H\in \mathcal{S}^n,$$

where “∘” denotes the Hadamard product, λ=(λ ₁,…,λ _n )^T, and the matrix Ω(ε,λ) is given by

$$\bigl[\varOmega (\varepsilon,\lambda)\bigr]_{ij}=\left \{ \begin{array}{@{}l@{\quad}l@{}}\frac{\phi(\varepsilon,\lambda_i)-\phi(\varepsilon,\lambda_j)}{\lambda_i-\lambda_j}\in[0,1],\quad &\textrm{if }\lambda_i\neq\lambda_j,\\\phi_{\lambda_i}'(\varepsilon,\lambda_i)\in[0,1],&\textrm{if}\lambda_i=\lambda_j,\end{array} \right .\quad i,j=1,\ldots,n.$$

Thus if β=∅ or ε≠0, Φ(⋅,⋅) is continuously differentiable around $(\varepsilon,X)\in\mathbb{R}\times\mathcal {S}^{n}$. Furthermore, Φ(⋅,⋅) is globally Lipschitz continuous and strongly semismooth at any $(0,X)\in\mathbb{R}\times\mathcal{S}^{n}$, which can be easily proved as in Sun et al. (2004).

Appendix C: Proof of Proposition 1

Proof

It is easy to see that, if (Δε,ΔY,ΔZ)=0, E′(ε,Y,Z)(Δε,ΔY,ΔZ)=0. Thus we only need to show, E′(ε,Y,Z)(Δε,ΔY,ΔZ)=0 implies (Δε,ΔY,ΔZ)=0.

From the definition of E, E′(ε,Y,Z)(Δε,ΔY,ΔZ)=0 is equivalent to

$$\left \{ \begin{array}{@{}l@{}}\varDelta \varepsilon=0,\\\varPsi '(\varepsilon,Y,Z)(\varDelta \varepsilon,\varDelta Y,\varDelta Z)=0,\end{array} \right .$$

which implies

$$\mathcal{A}\bigl(\varPhi _{X}'(\varepsilon,\hat{M})(M_{\varDelta }), C_{\varDelta },\varPhi _{X}'(\varepsilon,\hat{K}) (K_{\varDelta }),G_{\varDelta },N_{\varDelta }\bigr)=0,$$

where

$$\begin{cases}\hat{M}:=\varPhi (\varepsilon,M_0+\mathcal{A}_1^*(Y,Z)),\\[4pt]\hat{K}:=\varPhi (\varepsilon,K_0+\mathcal{A}_3^*(Y,Z)),\\[4pt]M_{\varDelta }:=\mathcal{A}_1^*(\varDelta Y, \varDelta Z),\\[4pt]C_{\varDelta }:=\mathcal{A}_2^*(\varDelta Y, \varDelta Z),\\[4pt]K_{\varDelta }:=\mathcal{A}_3^*(\varDelta Y, \varDelta Z),\\[4pt]G_{\varDelta }:=\mathcal{A}_4^*(\varDelta Y, \varDelta Z),\\[4pt]N_{\varDelta }:=\mathcal{A}_5^*(\varDelta Y, \varDelta Z).\end{cases} $$

Next, we consider

Since for any $H\in\mathcal{S}^{n}$, both 〈H,H〉 and $\langle H,\varPhi _{X}'(\varepsilon,X)(H)\rangle$ are nonnegative, we obtain

$$\left \{ \begin{array}{@{}l@{}}\langle M_{\varDelta },\varPhi _{X}'(\varepsilon,\hat{M})(M_{\varDelta })\rangle =0,\\[4pt]\langle K_{\varDelta },\varPhi _{X}'(\varepsilon,\hat{K})(K_{\varDelta })\rangle =0,\\[4pt]C_{\varDelta }=G_{\varDelta }=N_{\varDelta }=0.\end{array} \right .$$

From N _Δ =0, we get

$$\mathcal{A}_{5}^*(\varDelta Y,\varDelta Z)=\frac{1}{2} \left[\begin{array}{@{}c@{\quad}c@{}}\varDelta Y-\varDelta Y^T&-\varDelta Z\\\varDelta Z^T&0\end{array} \right]=0.$$

Thus ΔZ=0 and ΔY=ΔY ^T. From C _Δ =0, we get

$$\mathcal{A}_{2}^*(\varDelta Y,\varDelta Z)=\frac{1}{2} \left[\begin{array}{@{}c@{\quad}c@{}}\varDelta YS^T+S\varDelta Y^T&S\varDelta Z\\\varDelta Z^TS^T&0\end{array} \right]=0,$$

Similarly, from G _Δ =0, we have

$$\mathcal{A}_{4}^*(\varDelta Y,\varDelta Z)=\frac{1}{2} \left[\begin{array}{@{}c@{\quad}c@{}}\varDelta YS^T-S\varDelta Y^T&-S\varDelta Z\\\varDelta Z^TS^T&0\end{array} \right]=0.$$

Thus SΔY=0. Together with the fact that S=RΛR ⁻¹ is nonsingular, we obtain ΔY=0. The proof is complete. □