Simulation-Based Classification; a Model-Order-Reduction Approach for Structural Health Monitoring

Abstract

We present a model-order-reduction approach to simulation-based classification, with particular application to structural health monitoring. The approach exploits (1) synthetic results obtained by repeated solution of a parametrized mathematical model for different values of the parameters, (2) machine-learning algorithms to generate a classifier that monitors the damage state of the system, and (3) a reduced basis method to reduce the computational burden associated with the model evaluations. Furthermore, we propose a mathematical formulation which integrates the partial differential equation model within the classification framework and clarifies the influence of model error on classification performance. We illustrate our approach and we demonstrate its effectiveness through the vehicle of a particular physical companion experiment, a harmonically excited microtruss.

Appendices

Appendix 1: Existence and Uniqueness of the Solutions to the Monitoring Problems

In this appendix, we discuss existence and uniqueness of solutions to the monitoring problems and indeed we develop an explicit—through not readily evaluated—expression for $g^\mathrm{opt,bk}$. For the sake of clarity, we recap the definition of $P_{w^{\mathrm{bk}}}$ in (28),

$$\begin{aligned} P_{w^{\mathrm{bk}}}( A) = \int _{{\mathcal {P}}^{\mathrm{bk}}} \mathbbm {1}_{A}(\mu ') \,w^{\mathrm{bk}}(\mu ') \, d\mu ', \quad A \subset {\mathcal {P}}^{\mathrm{bk}}, \end{aligned}$$

and of $P_{(\mu ,\xi )}^{\mathrm{exp}}$ in (44),

$$\begin{aligned} P_{(\mu ,\xi )}^{\mathrm{exp}}( A\times B ) = \int _{ {\mathcal {P}}^{\mathrm{bk}} \times {\mathcal {V}} } \; \mathbbm {1}_A(\mu ) \, \mathbbm {1}_B(\xi ) \,w^{\mathrm{bk}}(\mu ') \, p_{\xi } (\xi ') \, d\mu ', \, d\xi '. \quad A \subset {\mathcal {P}}^{\mathrm{bk}}, \; \; B \subset {\mathcal {V}}. \end{aligned}$$

We assume here that $P_{w^{\mathrm{bk}}}$ and $P_{(\mu ,\xi )}^{\mathrm{exp}}$ are Borel-measurable, that is they are defined on all open sets of ${\mathcal {P}}^{\mathrm{bk}}$ and ${\mathcal {P}}^{\mathrm{exp}}$, respectively. A sufficient condition for which $P_{w^{\mathrm{bk}}}$ and $P_{(\mu ,\xi )}^{\mathrm{exp}}$ are Borel-measurable is that $w^{\mathrm{bk}} \in L^1({\mathcal {P}}^{\mathrm{bk}})$ and $p_{\xi } \in L^1({\mathcal {V}})$. We further recall the random pair

$$\begin{aligned} (Z^{\mathrm{bk}},Y) := ({\mathbf {z}}^{\mathrm{bk}}(\mu ), f^{\mathrm{dam}}(\mu ) ), \qquad \mu \sim P_{w^{\mathrm{bk}}}, \end{aligned}$$

with probability distribution $P_{(Z^{\mathrm{bk}},Y)}$, and

$$\begin{aligned} (Z^{\mathrm{exp}},Y) := ({\mathbf {z}}^{\mathrm{exp}}(\mu ,\xi ), f^{\mathrm{dam}}(\mu ) ), \qquad (\mu ,\xi ) \sim P_{(\mu ,\xi )}^{\mathrm{exp}}, \end{aligned}$$

with probability distribution $P_{(Z^{\mathrm{exp}},Y)}$.

Next Lemma shows that if ${\mathbf {z}}^{\mathrm{bk}}$ and ${\mathbf {z}}^{\mathrm{exp}}$ are continuous, $f^{\mathrm{dam}}$ is Borel-measurable, and $P_{w^{\mathrm{bk}}}$ and $P_{(\mu ,\xi )}^{\mathrm{exp}}$ are Borel-measurable, then $P_{(Z^{\mathrm{bk}},Y)}$ and $P_{(Z^{\mathrm{exp}},Y)}$ are also Borel-measurable.

Lemma 1

Suppose that

• 1. the probability measures $P_{w^{\mathrm{bk}}}$ in (28) and $P_{(\mu ,\xi )}^{\mathrm{exp}}$ in (44) are Borel-measurable;

• 2. the bk features ${\mathbf {z}}^{\mathrm{bk}}: {\mathcal {P}}^{\mathrm{bk}} \rightarrow {\mathbb {R}}^Q$ and the experimental features ${\mathbf {z}}^{\mathrm{exp}}: {\mathcal {P}}^{\mathrm{exp}} \rightarrow {\mathbb {R}}^Q$ are continuous;

• 3. the discrete function $f^{\mathrm{dam}}: {\mathcal {P}}^{\mathrm{bk}} \rightarrow \{1,\ldots ,K\}$ is Borel-measurable.

Then, the probability measures $P_{(Z^{\mathrm{bk}},Y)}$ and $P_{(Z^{\mathrm{exp}},Y)}$ are Borel-measurable on ${\mathbb {R}}^Q \times \{1,\ldots ,K\}$.

Proof

We only prove that $P_{(Z^{\mathrm{bk}},Y)}$ is Borel-measurable. The proof of the measurability of $P_{(Z^{\mathrm{exp}},Y)}$ is analogous. Let $A \subset {\mathbb {R}}^Q$ be an open set and let $k \in \{1,\ldots ,K\}$. We must show that $P_{(Z^{\mathrm{bk}},Y)}( A \times \{ k \} )$ is well-defined. By construction, we have that

$$\begin{aligned} P_{(Z^{\mathrm{bk}},Y)}( A \times \{ k \} ) = \int _{(Z^{\mathrm{bk}},Y)^{-1} \left( A \times \{ k \} \right) } \, w^{\mathrm{bk}}(\mu ) \, d\mu . \end{aligned}$$

As a result, we must show that

$$\begin{aligned} \begin{array}{rl} (Z^{\mathrm{bk}},Y)^{-1} \left( A \times \{ k \} \right) \, & = \, \{ \mu \in {\mathcal {P}}^{\mathrm{bk}}: \; {\mathbf {z}}^{\mathrm{bk}}(\mu ) \in A, \; f^{\mathrm{dam}}(\mu )= k \}\\ & = \left( {\mathbf {z}}^{\mathrm{bk}} \right) ^{-1} (A) \, \cap \left( f^{\mathrm{dam}} \right) ^{-1} (k)\\ \end{array} \end{aligned}$$

is a Borel set. Since ${\mathbf {z}}^{\mathrm{bk}}$ is continuous and A is open, $\left( {\mathbf {z}}^{\mathrm{bk}} \right) ^{-1} (A)$ is also open. On the other hand, by assumption, $\left( f^{\mathrm{dam}} \right) ^{-1} (k)$ is Borel. Thesis follows by recalling that the intersection of Borel sets is also Borel. $\square$

Lemma 1 can be used to prove the following important result related to the existence and uniqueness of solutions to (20) and to (19). We state here only the result for (20). An analogous discussion applies also to the monitoring problem (19).

Proposition 2

The optimal solution $g^\mathrm{opt, bk}$ to (20) is given by

$$\begin{aligned} g^\mathrm{opt, bk}({\mathbf {z}}) = \hbox {arg} \max _{ k=1,\ldots ,K } \; P_{(Z^{\mathrm{bk}},Y)}(Y=k| Z^{\mathrm{bk}}={\mathbf {z}}). \end{aligned}$$

(55)

Here, $P_{(Z^{\mathrm{bk}},Y)} ( Y= k | \, Z^{\mathrm{bk}} = {\mathbf {z}} )$ denotes the conditional probability of the event $\{Y= k\}$ given $\{ Z^{\mathrm{bk}} = {\mathbf {z}} \}$.

Furthermore, if there exists $\epsilon >0$ such that

$$\begin{aligned} P_{Z^{\mathrm{bk}}} \left( P_{(Z^{\mathrm{bk}},Y)}(Y=g^\mathrm{opt, bk}({\mathbf {z}}) | Z^{\mathrm{bk}}={\mathbf {z}}) \ge \max _{ k \ne g^\mathrm{opt, bk}({\mathbf {z}}) } \, P_{(Z^{\mathrm{bk}},Y)}(Y=k | Z^{\mathrm{bk}}={\mathbf {z}}) + \epsilon \right) = 1, \end{aligned}$$

(56)

then any solution g to (20) satisfies $g^\mathrm{opt, bk}({\mathbf {z}}) = g({\mathbf {z}})$ for $P_{Z^{\mathrm{bk}}}$-almost every ${\mathbf {z}} \in {\mathbb {R}}^Q$.

Proof

We first show that (55) is measurable. Recalling [74, Theorem A.24 page 586], since $P_{(Z^{\mathrm{bk}},Y)}$ is Borel-measurable, if we denote by $P_{Z^{\mathrm{bk}}}$ the corresponding marginal distribution, there exists for $P_{Z^{\mathrm{bk}}}$—almost-every ${\mathbf {z}} \in {\mathbb {R}}^Q$ and for $k=1,\ldots ,K$ a measurable function $\xi _k^{\mathrm{bk}} : {\mathbb {R}}^Q \rightarrow {\mathbb {R}}$ such that $\xi _k^{\mathrm{bk}}({\mathbf {z}}) = P_{(Z^{\mathrm{bk}},Y)}(Y=k | Z^{\mathrm{bk}}={\mathbf {z}})$. Recalling that the pointwise maximum of measurable functions is also measurable, this implies that, under the hypotheses of Lemma 1, the function $g^\mathrm{opt, bk}$ in (55) is measurable.

We now observe that

$$\begin{aligned} \begin{array}{rl} R^{\mathrm{bk}}(g) & = \int _{{\mathbb {R}}^Q \times \{1,\ldots ,K\} } \, {\mathcal {L}}^{(0,1)} ( g({\mathbf {z}}), y) \, d P_{(Z^{\mathrm{bk}}, Y)}({\mathbf {z}}), y ) = \int _{{\mathbb {R}}^Q} \, \sum _{k \ne g({\mathbf {z}}) } \, P_{(Z^{\mathrm{bk}}, Y)}( Y=k | Z^{\mathrm{bk}}= {\mathbf {z}})) \, d P_{Z^{\mathrm{bk}}}({\mathbf {z}})\\ & = 1 - \int _{{\mathbb {R}}^Q} \, \, P_{(Z^{\mathrm{bk}}, Y)}( Y=g( {\mathbf {z}} ) | Z^{\mathrm{bk}}= {\mathbf {z}})) \, d P_{Z^{\mathrm{bk}}}({\mathbf {z}})\\& \ge 1 - \int _{{\mathbb {R}}^Q} \, \max _k \, P_{(Z^{\mathrm{bk}}, Y)}( Y=k | Z^{\mathrm{bk}}= {\mathbf {z}})) \, d P_{Z^{\mathrm{bk}}}({\mathbf {z}}) = R^{\mathrm{bk}}( g^\mathrm{opt,bk} ).\\ \end{array} \end{aligned}$$

Since $g^\mathrm{opt,bk}$ is measurable, this implies that $g^\mathrm{opt,bk}$ is a solution to (20).

Let g be a classifier such that $R^{\mathrm{bk}}( g^\mathrm{opt,bk} ) = R^{\mathrm{bk}}( g)$. Using the same reasoning as before, it is possible to verify that

$$\begin{aligned} R^{\mathrm{bk}}( g) = R^{\mathrm{bk}}( g^\mathrm{opt,bk} ) \, + \, \int _{{\mathbb {R}}^Q} \, \left( P_{(Z^{\mathrm{bk}},Y)}(Y=g^\mathrm{opt, bk}({\mathbf {z}}) | Z^{\mathrm{bk}}={\mathbf {z}}) - \, P_{(Z^{\mathrm{bk}},Y)}(Y=g({\mathbf {z}}) | Z^{\mathrm{bk}}={\mathbf {z}}) \right) \, d P_{Z^{\mathrm{bk}}}({\mathbf {z}}). \end{aligned}$$

Recalling (56), we find

$$\begin{aligned} R^{\mathrm{bk}}( g) \ge R^{\mathrm{bk}}( g^\mathrm{opt,bk} ) \, + \epsilon \int _{{\mathbb {R}}^Q} \, {\mathcal {L}}^{(0,1)} (g ({\mathbf {z}}), g^\mathrm{opt,bk} ({\mathbf {z}}) ) \, d \, P_{Z^{\mathrm{bk}}}({\mathbf {z}}). \end{aligned}$$

Then, we must have

$$\begin{aligned} \int _{{\mathbb {R}}^Q} \, {\mathcal {L}}^{(0,1)} (g ({\mathbf {z}}), g^\mathrm{opt,bk} ({\mathbf {z}}) ) \, d \, P_{Z^{\mathrm{bk}}}({\mathbf {z}})=0, \end{aligned}$$

which implies that $g({\mathbf {z}}) =g^\mathrm{opt,bk} ({\mathbf {z}})$ for $P_{Z^{\mathrm{bk}}}$-almost every ${\mathbf {z}} \in {\mathbb {R}}^Q$. $\square$

Appendix 2: Parametric-Affine Expansion for the Microtruss Problem

Below, we report the parameter-dependent coefficients $\{ {\varTheta }_q\}_{q=1}^{10}$ and the parameter-independent bilinear forms $\{ a^q\}_{q=1}^{10}$ associated with the microtruss problem considered in this work.

$$\begin{aligned} {\varTheta }_q(f,\mu =[\alpha ,\beta ,E, s_L,s_R]) = \left\{ \begin{array}{ll} (1 + i \omega _f \beta ) E & q =1, \\ (-\omega _f^2 + i \omega _f \alpha ) \rho L^2 & q =2, \\ (1 + i \omega _f \beta ) s_L \, E & q =3, \\ (1 + i \omega _f \beta ) E & q =4,\\ (1 + i \omega _f \beta )s_L^{-1} \, E & q =5, \\ \end{array} \right. \qquad \left\{ \begin{array}{ll} (-\omega _f^2 + i \omega _f \alpha ) \rho L^2 s_L & q =6, \\ (1 + i \omega _f \beta ) s_R \, E & q =7, \\ (1 + i \omega _f \beta ) \, E & q =8,\\ (1 + i \omega _f \beta )s_R^{-1} \, E & q =9, \\ (-\omega _f^2 + i \omega _f \alpha ) \rho L^2 s_R & q =10, \\ \end{array} \right. \end{aligned}$$

(57a)

and

$$\begin{aligned} a^q(u,v) = \left\{ \begin{array}{ll} b_{{\varOmega }_1^{ref}}(u,v) & q =1 \\ m_{{\varOmega }_1^{ref}}(u,v) & q =2, \\ { \int _{{\varOmega }_2^{ref}} \, \frac{1-\nu }{(1+\nu )(1-2\nu )} \, \frac{\partial u_1}{ \partial x_1 } \, \frac{\partial v_1}{ \partial x_1 } \, + \, \frac{1}{2(1+\nu )} \, \frac{\partial u_2}{ \partial x_1 } \, \frac{\partial v_2}{ \partial x_1 }\, dx } & q =3, \\ { \int _{{\varOmega }_2^{ref}} \, \frac{\nu }{(1+\nu )(1-2\nu )} \, \left( \frac{\partial u_1}{ \partial x_1 } \frac{\partial v_2}{ \partial x_2 } + \frac{\partial u_2}{ \partial x_2 } \frac{\partial v_1}{ \partial x_1 } \right) \, + \, \frac{1}{2(1+\nu )} \left( \frac{\partial u_1}{ \partial x_2 }\frac{\partial v_2}{ \partial x_1 } + \frac{\partial u_2}{ \partial x_1 } \frac{\partial v_1}{ \partial x_2 } \right) \, dx} & q =4,\\ { \int _{{\varOmega }_2^{ref}} \, \frac{1-\nu }{(1+\nu )(1-2\nu )} \, \frac{\partial u_2}{ \partial x_2 } \, \frac{\partial v_2}{ \partial x_2 } \, + \, \frac{1}{2(1+\nu )} \, \frac{\partial u_1}{ \partial x_2 } \, \frac{\partial v_1}{ \partial x_2 } \, dx} & q =5, \\ m_{{\varOmega }_2^{ref}}(u,v) & q =6, \\ { \int _{{\varOmega }_3^{ref}} \, \frac{1-\nu }{2(1+\nu )(1-2\nu )} \, \frac{\partial u_1}{ \partial x_1 } \, \frac{\partial v_1}{ \partial x_1 } \, + \, \frac{1}{2(1+\nu )} \, \frac{\partial u_2}{ \partial x_1 } \, \frac{\partial v_2}{ \partial x_1 }\, dx } & q =7, \\ { \int _{{\varOmega }_3^{ref}} \, \frac{\nu }{(1+\nu )(1-2\nu )} \, \left( \frac{\partial u_1}{ \partial x_1 } \frac{\partial v_2}{ \partial x_2 } + \frac{\partial u_2}{ \partial x_2 } \frac{\partial v_1}{ \partial x_1 } \right) \, + \, \frac{1}{2(1 + \nu )} \left( \frac{\partial u_1}{ \partial x_2 }\frac{\partial v_2}{ \partial x_1 } + \frac{\partial u_2}{ \partial x_1 } \frac{\partial v_1}{ \partial x_2 } \right) \, dx } & q =8,\\ { \int _{{\varOmega }_3^{ref}} \, \frac{1-\nu }{(1 + \nu )(1-2\nu )} \, \frac{\partial u_2}{ \partial x_2 } \, \frac{\partial v_2}{ \partial x_2 } \, + \, \frac{1}{2(1 + \nu )} \, \frac{\partial u_1}{ \partial x_2 } \, \frac{\partial v_1}{ \partial x_2 } \, dx} & q =9, \\ m_{{\varOmega }_3^{ref}}(u,v) & q =10, \\ \end{array} \right. \end{aligned}$$

(57b)