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Deriving a Consistent Projection

Consider there are no product terms of $\hat{y}$ in f and $\hat{h}$,then one can write

\begin{eqnarray}
\html{eqn22}f'(\hat{y}) = f''(\dot{\hat{x}}, \dot{\bar{x}}, \ha...
 ...{x}, \tilde{y}) \\ \hat{h}'(\hat{y}) = \hat{h}''(\bar{x}, \hat{x})\end{eqnarray}


In case of linear relations, this can be written in matrix form

  \begin{eqnarray}
\html{eqn24}
F' \hat{y} = F'' z \\ 
\hat{H}' \hat{y} = \hat{H}''
\left[\begin{array}
{c}
\hat{x} \\ \bar{x}\end{array}\right]\end{eqnarray}


with, for reasons of conciseness, \begin{displaymath}
z =
\left[\begin{array}
{c}
\dot{\hat{x}} \\ \dot{\bar{x}} \\ \hat{x} \\ \bar{x} \\ \tilde{y}\end{array}\right].\end{displaymath}
In this system of equations the algebraic variables, $\hat{y}$, need to be eliminated to make the dependency in rate variables explicit. If F' spans $\hat{H}'$ [2] then there is a matrix G such that  \begin{displaymath}
G F' = \hat{H}'\end{displaymath}
and $\hat{y}$ can be solved by premultiplying Eq. (15) and combining the result with Eq. (16). This yields  \begin{displaymath}
G F'' z =
\hat{H}''
\left[\begin{array}
{c}
\hat{x} \\ \bar{x}\end{array}\right]\end{displaymath}
If the matrix F' has a pseudo-inverse F'+, then G can be found from Eq. (18) as  \begin{displaymath}
G = \hat{H}' F'^+\end{displaymath}
and, therefore, the dependency equations in Eq. (19) are given by  \begin{displaymath}
\hat{H}''
\left[\begin{array}
{c}
\hat{x} \\ \bar{x}\end{array}\right]
= \hat{H}' F'^+ F'' z\end{displaymath}
This equation establishes the possible dependency between rate variables. If there is dependency, the manifold constraints, $\bar{H} \bar{x}$can be differentiated to establish a system of independent equations.

To find the projection of the state variables, Eq. (21) can be integrated to remove all elements that are continuous in time (i.e., no algebraic constraints apply), $\{\dot{\hat{x}}, \hat{x}\}$,and those that are no time derivatives, $\{\bar{x}, \tilde{y}\}$,conform Eq. (8) through (10). This results in $\hat{H}' F'^+ F'' \int \dot{\bar{x}} dt = 0$which gives an equal number of constraints in $\bar{x}$ only,  \begin{displaymath}
\hat{H}' F'^+ F'' (\bar{x} - \bar{x}^0) = 0.\end{displaymath}
and describes the projection onto the manifold $\bar{H} \bar{x} = 0$.Note that the number of dependent state variables, $\bar{x}_i$,in Eq. (22) equals the number of rate dependencies, $\dot{\bar{x}_i}$,in Eq. (21). Because Eq. (21) can be solved by adding the differentiated form of $\bar{H} \bar{x}$,Eq. (22) can be solved by adding the original set of $\bar{H} \bar{x} = 0$.

Conjecture 1955

Eq. (21) can be solved for $\dot{\bar{x}}$ by adding $\dot{\bar{H}} \bar{x} + \bar{H} \dot{\bar{x}}$.

Lemma 1958 (Existence)

Eq. (22) can be solved for $\bar{x}$ by adding $\bar{H} \bar{x}$.


\begin{proof}
% latex2html id marker 476
The number of dependent variables in Eq...
 ...{H} \bar{x}$ 
is sufficient to solve the unknowns in Eq.~(\ref{xx0}).\end{proof}

Note that if F' = 0 then $\hat{y}$ is already solved from Eq. (15) and integrating $F'' \dot{\bar{x}}$ results in $F''(\bar{x} - \bar{x_0})$ which gives the conservation equations. If $\hat{H} = 0$, by definition, $\hat{y}$is empty and, again, integrating $F'' \dot{\bar{x}}$ gives the conservation equations. Furthermore, since F'+ is an inverse, $F'^+ \not = 0$.


next up previous
Next: The Algorithm Up: A Consistent Projection Previous: The DAE System
Pieter J. Mosterman ER
7/27/1998