Projection based MOR: Difference between revisions

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Revision as of 15:15, 12 March 2013

The basic idea of almost all the model order reduction (MOR) methods is to find a subspace $S_{1}$ which approximates the manifold where the state vector $𝐱 (t)$ resides. Afterwards, $𝐱 (t)$ is approximated by a vector $\tilde{𝐱} (t)$ in $S_1$. The reduced model is produced by Petrov-Galerkin projection onto a subspace $S_2$, or by Galerkin projection onto the same subspace $S_{1}$ .

We use the system $E \frac{d x (t)}{d t} = A x (t) + B u (t),$ $y (t) = C x (t) + D u (t) .$ as an example to explain the basic idea. Assuming that an orthonormal basis $V = (v_{1}, v_{2}, \dots, v_{q})$ of the subspace $S_{1}$ has been found, then the approximation $\tilde{x} (t)$ in $S_{1}$ can be represented by the basis as $\tilde{x} (t) = V z (t)$ . Therefore $x (t)$ can be approximated by $x (t) \approx V z (t)$ . Here $z$ is a vector of length $q \ll n$.

Once $z (t)$ is computed, we can get an approximate solution $\tilde{x} (t) = V z (t)$ for $x (t)$ . The vector $z (t)$ can be computed from the reduced model which is derived by the following two steps.

@@ Line 6: / Line 6: @@
 We use the system
-<math>
-(E_0+s_1 E+s_2E_2+\ldots +s_pE_p)x=Bu(s_1,\ldots,s_p), \quad
-y=Cx,    \quad \quad \quad \quad (1)
-</math>
 <math>
 E \frac{dx(t)}{dt}=A x(t)+B u(t),</math>
@@ Line 16: / Line 11: @@
 y(t)=Cx(t)+Du(t).
 </math>
 as an example to explain the basic idea. Assuming that an orthonormal
 basis <math>V=(v_1,v_2, \ldots, v_q)</math> of the subspace <math>S_1</math> has been
-found, then the approximation <math>\tilde{\bf x}(t)</math> in <math>S_1</math> can be represented by
+found, then the approximation <math>\tilde x(t)</math> in <math>S_1</math> can be represented by
-the basis as <math>\tilde{\bf x}(t)=V{\bf z}(t)</math>. Therefore <math>{\bf x}(t)</math> can be approximated by <math>{\bf x}(t) \approx V{\bf z}(t)</math>. Here ${\bf z}$ is a vector
+the basis as <math>\tilde x(t)=V z(t)</math>. Therefore <math>x(t)</math> can be approximated by <math> x(t) \approx V z(t)</math>. Here <math>z</math> is a vector
 of length $q \ll n$.
-Once <math>{\bf z}(t)</math> is computed, we can get an
+Once <math>z(t)</math> is computed, we can get an
-approximate solution <math>\tilde{\bf x}(t)=V{\bf z}(t)</math> for <math>{\bf x}(t)</math>. The vector <math>{\bf z}(t)</math>
+approximate solution <math>\tilde x(t)=V z(t)</math> for <math>x(t)</math>. The vector <math>z(t)</math>
 can be computed from the reduced model which is  derived by the
 following two steps.