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Commit 28031e65 authored by oliver.sander_at_tu-dresden.de's avatar oliver.sander_at_tu-dresden.de
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A bit of text on the dynamic model

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...@@ -8,3 +8,11 @@ ...@@ -8,3 +8,11 @@
pages = {53--84}, pages = {53--84},
doi = {10.1093/qjmam/hbu027} doi = {10.1093/qjmam/hbu027}
} }
@Book{hairer_lubich_wanner:2006,
title = {Geometric Numerical Integration---Structure-Preserving Algorithms for Ordinary Differential Equations},
publisher = {Springer},
year = {2016},
author = {Ernst Hairer and Christian Lubich and Gerhard Wanner},
edition = {Zweite Auflage},
}
...@@ -88,10 +88,9 @@ matrix curl as ...@@ -88,10 +88,9 @@ matrix curl as
(\Curl P)_i (\Curl P)_i
\colonequals \colonequals
\Big( \frac{\partial P_{i3}}{\partial x_2} - \frac{\partial P_{i2}}{\partial x_3}, \quad \Big( \frac{\partial P_{i3}}{\partial x_2} - \frac{\partial P_{i2}}{\partial x_3}, \quad
\frac{\partial P_{i3}}{\partial x_2} - \frac{\partial P_{i3}}{\partial x_2}, \quad \frac{\partial P_{i1}}{\partial x_3} - \frac{\partial P_{i3}}{\partial x_1}, \quad
\frac{\partial P_{i3}}{\partial x_2} - \frac{\partial P_{i3}}{\partial x_2} \Big). \frac{\partial P_{i2}}{\partial x_1} - \frac{\partial P_{i1}}{\partial x_2} \Big).
\end{equation*} \end{equation*}
\todosander{Wie würde das Modell in 2d aussehen?}
We define the quantities We define the quantities
\begin{alignat*}{2} \begin{alignat*}{2}
...@@ -224,5 +223,138 @@ norm $\tnorm{\cdot}_\mathcal{X}$. ...@@ -224,5 +223,138 @@ norm $\tnorm{\cdot}_\mathcal{X}$.
\section{The dynamic case} \section{The dynamic case}
We now investigate a time-dependent relaxed micromorphic continuum, using the
Hamilton formalism. As shown above, the model is described by a quadratic potential energy,
and we assume the kinetic energy to be quadratic as well.
\subsection{The implicit midpoint rule for quadratic Hamiltonians}
Consider a model with quadratic potential energy
\begin{equation*}
U(q) = \frac{1}{2} q^TA q + b^T q
\end{equation*}
and quadratic kinetic energy
\begin{equation*}
T(q, \dot{q}) = T(\dot{q}) = \frac{1}{2} \dot{q}^T M \dot{q},
\end{equation*}
where $A$ and $M$ are symmetric and positive definite matrices.
As the kinetic energy is quadratic, the corresponding Hamiltonian
is simply the sum of the two energies~\cite[Chapter~VI.1.2]{hairer_lubich_wanner:2006}
Using that $p = M \dot{q}$ we get
\begin{equation*}
H(p,q)
=
T(\dot{q}(p)) + U(q)
=
\frac{1}{2} p^TM^{-T} p + \frac{1}{2} q^T A q + b^T q.
\end{equation*}
We want to integrate this system using an implicit midpoint rule. As is
well known, such an integrator is symplectic, second-order accurate~\cite[Theorem~VI.3.5]{hairer_lubich_wanner:2006}
and B-stable~\cite{}. One step of this method is
\begin{equation*}
\begin{pmatrix} p^{k+1} \\ q^{k+1} \end{pmatrix}
=
\begin{pmatrix} p^k \\ q^k \end{pmatrix}
+
\tau J^{-1} \nabla H\Big(\frac{p^k + p^{k+1}}{2}, \frac{q^k + q^{k+1}}{2}\Big),
\end{equation*}
where $\tau$ is the time step size, and $J^{-1}$ is the matrix
\begin{equation*}
J^{-1}
\colonequals
\begin{pmatrix} 0 & -I \\ I & 0 \end{pmatrix}.
\end{equation*}
With
\begin{equation*}
\nabla H(p,q)
=
\begin{pmatrix}
M^{-T} p \\ Aq + b
\end{pmatrix}
\end{equation*}
the integrator becomes
\begin{equation*}
\begin{pmatrix} p^{k+1} \\ q^{k+1} \end{pmatrix}
- \frac{\tau}{2} \begin{pmatrix} 0 & -I \\ I & 0 \end{pmatrix}
\underbrace{
\begin{pmatrix}
M^{-T} p^{k+1} \\ Aq^{k+1}
\end{pmatrix}
}_{ = \begin{pmatrix} M^{-T} & 0 \\ 0 & A \end{pmatrix} \begin{pmatrix} p^{k+1} \\ q^{k+1} \end{pmatrix} }
=
\begin{pmatrix} p^k \\ q^k \end{pmatrix}
+ \frac{\tau}{2} \begin{pmatrix} 0 & -I \\ I & 0 \end{pmatrix}
\begin{pmatrix}
M^{-T} p^k \\ Aq^k + 2b
\end{pmatrix}
\end{equation*}
This can be transformed to
\begin{equation*}
\Bigg[
\begin{pmatrix} I & 0 \\ 0 & I \end{pmatrix}
- \frac{\tau}{2}
\begin{pmatrix}
0 & -A \\ M^{-T} & 0
\end{pmatrix}
\Bigg]
\begin{pmatrix} p^{k+1} \\ q^{k+1} \end{pmatrix}
=
\begin{pmatrix}
p^k - \frac{\tau}{2} A q^k - \tau b\\
q^k + \frac{\tau}{2} M^{-T} p^k
\end{pmatrix},
\end{equation*}
and this in turn to
\begin{equation*}
\begin{pmatrix}
I & \frac{\tau}{2} A \\
- \frac{\tau}{2} M^{-T} & I
\end{pmatrix}
\begin{pmatrix} p^{k+1} \\ q^{k+1} \end{pmatrix}
=
\begin{pmatrix}
p^k - \frac{\tau}{2} A q^k - \tau b\\
q^k + \frac{\tau}{2} M^{-T} p^k
\end{pmatrix},
\end{equation*}
Eliminate the lower left block:
\begin{equation*}
\begin{pmatrix}
I & \frac{\tau}{2} A \\
0 & I + \frac{\tau^2}{4} M^{-T}A
\end{pmatrix}
\begin{pmatrix} p^{k+1} \\ q^{k+1} \end{pmatrix}
=
\begin{pmatrix}
p^k - \frac{\tau}{2} A q^k - \tau b\\
q^k + \frac{\tau}{2} M^{-T} p^k + \frac{\tau}{2} M^{-T}(p^k - \frac{\tau}{2}Aq^k - \tau b)
\end{pmatrix}.
\end{equation*}
This can be solved by backward substitution. In algorithmic form this reads:
\begin{enumerate}
\item Solve
\begin{equation*}
\Big(M^T + \frac{\tau^2}{4} A \Big) q^{k+1}
=
M^Tq^k + \tau p^k - \frac{\tau^2}{4}Aq^k - \frac{\tau^2}{2}b
\end{equation*}
\item Set
\begin{equation*}
p^{k+1}
=
p^k - \frac{\tau}{2} Aq^k - 2b - \frac{\tau}{2}Aq^{k+1}.
\end{equation*}
\end{enumerate}
\printbibliography \printbibliography
\end{document} \end{document}
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