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August 11th, 2018, 08:29 PM   #1
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Unhappy Computational nonlinear system equations understanding.

Here is a snippet of some mathematical article:

For a general, nonlinear dynamic system with parameters $\boldsymbol c$, running from time $0$ to time $T$

$\boldsymbol z\left(t+1\right)\boldsymbol=\boldsymbol s\left(\boldsymbol z\left(t\right),\boldsymbol u\left(t\right),\boldsymbol c\right)$ $\left(8\right)$

one may linearize a solution trajectory using the Jacobian of $\boldsymbol s$, $S$:

$\boldsymbol x\left(t+1\right)\boldsymbol=\boldsymbol S\left(t\right)\boldsymbol x\left(t\right)+\boldsymbol k\left(t\right)$ $(9)$

The derivative of $\boldsymbol z\left(T\right)$ with respect to $z\left(0\right)$ are implicit in:

$\boldsymbol x\left(T\right)=\boldsymbol S\left(T-1\right)\boldsymbol S\left(T-2\right)\cdots\boldsymbol S\left(0\right)\boldsymbol x\left(0\right)+\boldsymbol k'\left(t\right)$ $(10)$

and a similar formula may be derived (summing over $t$) for derivatives with respect to $\boldsymbol c$. From this, one can easily verify the validity of the following recursion formulas to determine the derivatives of a target variable $z_i\left(T\right)$ with respect to all of $z_j\left(0\right)$ (to appear in $x_j^{'}\left(0\right)$ ) and with respect to all the components of $\boldsymbol c$ (to appear in $\boldsymbol w\left(0\right)$):

$\boldsymbol x'\left(\boldsymbol T\right)=\boldsymbol e^{i^T}$ $(11a)$
$\boldsymbol x'\left(t\right)=\boldsymbol x'\left(t+1\right)\boldsymbol S\left(t\right)$ $(11b)$
$\boldsymbol w\left(T\right)=\mathbf0$ $(11c)$
$\boldsymbol w\left(t\right)=\boldsymbol w\left(t+1\right)+\boldsymbol x{'}\left(t\right)s_c^{'}\left(t\right)$, $(11d)$

where $s_c^{'}$ refers to the matrix of derivatives of $s_i$ with respect to $c_k$.

My questions are:
Is here $\boldsymbol x$ a solution of $\boldsymbol z$?
What does $\boldsymbol e^{i^T}$ mean?

Last edited by rubis; August 11th, 2018 at 08:41 PM.
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August 13th, 2018, 04:22 PM   #2
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Originally Posted by rubis View Post
My questions are:
Is here $\boldsymbol x$ a solution of $\boldsymbol z$?
What does $\boldsymbol e^{i^T}$ mean?
No $x$ is a solution to the linearized dynamical system. This the matrix difference equation induced by the Jacobian. It is the discrete time version of the first variational equation which you have problem seen before. For more details see:

Presumably the second part should read $e^{Tx}$ and refers to the matrix exponential of $x$. This is defined analagously to the scalar exponential. For a matrix, $x$, the definition is
\[ \exp(x) = \sum_{k = 0}^\infty \frac{x^k}{k!} \]
where $x^k$ in this case denotes powers of the matrix under the usual multiplication. For more details see:
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