]> https://gitweb.dealii.org/ - dealii.git/commitdiff
Fix a mistake in the introduction of step-18.
authorWolfgang Bangerth <bangerth@colostate.edu>
Fri, 6 Dec 2019 23:17:36 +0000 (16:17 -0700)
committerWolfgang Bangerth <bangerth@colostate.edu>
Tue, 7 Jan 2020 04:48:49 +0000 (21:48 -0700)
The description of the problem is due to Ming Yang <ykevin09@gmail.com>.

examples/step-18/doc/intro.dox

index 9391cf8497e744c8591dff929e2afc268397ccff..79669864c2f8fa323b4de08ecb115e7220dd2e51 100644 (file)
@@ -60,7 +60,10 @@ to be specified for a unique solution:
 In above formulation, $\varepsilon(\mathbf{u})= \frac 12 (\nabla \mathbf{u} + \nabla
 \mathbf{u}^T)$ is the symmetric gradient of the displacement, also called the
 <em>strain</em>. $C$ is a tensor of rank 4, called the <em>stress-strain
-  tensor</em> that contains knowledge of the elastic strength of the material; its
+  tensor</em> (the inverse of the <a
+  href="https://en.wikipedia.org/wiki/Hooke%27s_law#Hooke's_law_for_continuous_media"><em>compliance
+  tensor</em></a>)
+that contains knowledge of the elastic strength of the material; its
 symmetry properties make sure that it maps symmetric tensors of rank 2
 (&ldquo;matrices&rdquo; of dimension $d$, where $d$ is the spatial dimensionality) onto
 symmetric tensors of the same rank. We will comment on the roles of the strain
@@ -193,7 +196,23 @@ such that
   (\mathbf{f}, \varphi)_{\Omega(t_{n-1})}
   -(\sigma^{n-1},\varepsilon(\varphi))_{\Omega(t_{n-1})}
   \\
-  &\qquad +(\mathbf{b}(\mathbf{x},t_n)-\mathbf{b}(\mathbf{x},t_{n-1}), \varphi)_{\Gamma_N}
+  &\qquad 
+  +(\mathbf{b}(\mathbf{x},t_n)-\mathbf{b}(\mathbf{x},t_{n-1}), \varphi)_{\Gamma_N}
+  +(\sigma^{n-1} \mathbf{n}, \varphi)_{\Gamma_N}
+  \\
+  &\qquad\qquad
+  \forall \varphi \in \{\mathbf{v}\in H^1(\Omega(t_{n-1}))^d: \mathbf{v}|_{\Gamma_D}=0\}.
+@f}
+Using that $\sigma^{n-1} \mathbf{n}
+            = [C \varepsilon(\mathbf{u}^{n-1})] \mathbf{n}
+            = \mathbf{b}(\mathbf x, t_{n-1})$,
+these equations can be simplified to
+@f{align*}
+  (C \varepsilon(\Delta\mathbf{u}^n), \varepsilon(\varphi) )_{\Omega(t_{n-1})}
+  &=
+  (\mathbf{f}, \varphi)_{\Omega(t_{n-1})}
+  -(\sigma^{n-1},\varepsilon(\varphi))_{\Omega(t_{n-1})}
+  +(\mathbf{b}(\mathbf{x},t_n),t_{n-1}), \varphi)_{\Gamma_N}
   \\
   &\qquad\qquad
   \forall \varphi \in \{\mathbf{v}\in H^1(\Omega(t_{n-1}))^d: \mathbf{v}|_{\Gamma_D}=0\}.
@@ -201,16 +220,17 @@ such that
   \qquad
   \textrm{[linear-system]}
 @f}
+
 We note that, for simplicity, in the program we will always assume that there
 are no boundary forces, i.e. $\mathbf{b} = 0$, and that the deformation of the
 body is driven by body forces $\mathbf{f}$ and prescribed boundary displacements
 $\mathbf{d}$ alone. It is also worth noting that when integrating by parts, we
 would get terms of the form $(C \varepsilon(\Delta\mathbf{u}^n), \nabla \varphi
-)_{\Omega(t_{n-1})}$, but that we replace it with the term involving the
+)_{\Omega(t_{n-1})}$, but that we replace them with the term involving the
 symmetric gradient $\varepsilon(\varphi)$ instead of $\nabla\varphi$. Due to
-the symmetry of $C$, the two terms are equivalent, but the symmetric version
-avoids a potential for round-off to render the resulting matrix slightly
-non-symmetric.
+the symmetry of $C$, the two terms are mathematically equivalent, but
+the symmetric version avoids the potential for round-off errors making
+the resulting matrix slightly non-symmetric.
 
 The system at time step $n$, to be solved on the old domain
 $\Omega(t_{n-1})$, has exactly the form of a stationary elastic

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