From feccafbe0e00e8df595d3a4e86ce67ffb9ac0ee9 Mon Sep 17 00:00:00 2001 From: Wolfgang Bangerth Date: Fri, 6 Dec 2019 16:17:36 -0700 Subject: [PATCH] Fix a mistake in the introduction of step-18. The description of the problem is due to Ming Yang . --- examples/step-18/doc/intro.dox | 32 ++++++++++++++++++++++++++------ 1 file changed, 26 insertions(+), 6 deletions(-) diff --git a/examples/step-18/doc/intro.dox b/examples/step-18/doc/intro.dox index 9391cf8497..79669864c2 100644 --- a/examples/step-18/doc/intro.dox +++ b/examples/step-18/doc/intro.dox @@ -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 strain. $C$ is a tensor of rank 4, called the stress-strain - tensor that contains knowledge of the elastic strength of the material; its + tensor (the inverse of the compliance + tensor) +that contains knowledge of the elastic strength of the material; its symmetry properties make sure that it maps symmetric tensors of rank 2 (“matrices” 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 -- 2.39.5