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agnumpde
dune-elasticity
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28031e65
Commit
28031e65
authored
4 years ago
by
oliver.sander_at_tu-dresden.de
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A bit of text on the dynamic model
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doc/relaxed-micromorphic-continuum.bib
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doc/relaxed-micromorphic-continuum.bib
doc/relaxed-micromorphic-continuum.tex
+135
-3
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doc/relaxed-micromorphic-continuum.tex
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and
3 deletions
doc/relaxed-micromorphic-continuum.bib
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28031e65
...
@@ -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}
,
}
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doc/relaxed-micromorphic-continuum.tex
+
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−
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28031e65
...
@@ -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
_{
i
3
}}{
\partial
x
_
2
}
-
\frac
{
\partial
P
_{
i3
}}{
\partial
x
_
2
}
,
\quad
\frac
{
\partial
P
_{
i
1
}}{
\partial
x
_
3
}
-
\frac
{
\partial
P
_{
i3
}}{
\partial
x
_
1
}
,
\quad
\frac
{
\partial
P
_{
i
3
}}{
\partial
x
_
2
}
-
\frac
{
\partial
P
_{
i
3
}}{
\partial
x
_
2
}
\Big
).
\frac
{
\partial
P
_{
i
2
}}{
\partial
x
_
1
}
-
\frac
{
\partial
P
_{
i
1
}}{
\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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