From: bangerth Date: Fri, 15 Feb 2013 17:17:05 +0000 (+0000) Subject: Improve file markup and formulas. X-Git-Url: https://gitweb.dealii.org/cgi-bin/gitweb.cgi?a=commitdiff_plain;h=bab6203b3c545ccdb076551d752fd287390b40de;p=dealii-svn.git Improve file markup and formulas. git-svn-id: https://svn.dealii.org/trunk@28407 0785d39b-7218-0410-832d-ea1e28bc413d --- diff --git a/deal.II/examples/step-42/doc/intro.dox b/deal.II/examples/step-42/doc/intro.dox index 8add995629..a02ffc0a2d 100644 --- a/deal.II/examples/step-42/doc/intro.dox +++ b/deal.II/examples/step-42/doc/intro.dox @@ -54,9 +54,9 @@ row component-by-component and in a pointwise sense. Furthermore we have to distinguish two cases. The continuous and convex function $\mathcal{F}$ denotes the von Mises flow function -$$\mathcal{F}(\tau) = \vert\tau^D\vert - \sigma_0¸\quad\text{with}\quad \tau^D -= \tau - \dfrac{1}{3}tr(\tau)I,$$ -$\sigma_0$ as yield stress and $\vert .\vert$ as the frobenius norm. If there +@f{gather*}\mathcal{F}(\tau) = \vert\tau^D\vert - \sigma_0¸\quad\text{with}\quad \tau^D += \tau - \dfrac{1}{3}tr(\tau)I,@f} +$\sigma_0$ as yield stress and $\vert .\vert$ as the Frobenius norm. If there are no plastic deformations in a particular point - that is $\lambda=0$ - this yields $\vert\sigma^D\vert < \sigma_0$ and otherwise if $\lambda > 0$ it follows that $\vert\sigma^D\vert = \sigma_0$. That means if the stress is smaller than the yield stress there are only elastic @@ -79,7 +79,7 @@ density $f$ which we will neglect in our example. The boundary of $\Omega$ separates as follows $\Gamma=\Gamma_D\bigcup\Gamma_C$ and $\Gamma_D\bigcap\Gamma_C=\emptyset$. At the boundary $\Gamma_D$ we have zero Dirichlet conditions. $\Gamma_C$ denotes the potential contact boundary.\\ The last two lines decribe the so-called Signorini contact conditions. If there is no contact the normal stress -$$ \sigma_n = \sigma n\cdot n$$ +@f{gather*} \sigma_n = \sigma n\cdot n@f} is zero with the outward normal $n$. If there is contact ($u_n = g$) the tangential stress $\sigma_t = \sigma\cdot n - \sigma_n n$ vanishes, because we consider a frictionless situation and the normal stress is negative. The gap $g$ comes with the start configuration of the obstacle and the @@ -90,31 +90,31 @@ deformable body. As a starting point to derive the equations above, let us imagine that we want to minimise an energy functional: -$$E(\tau) := \dfrac{1}{2}\int\limits_{\Omega}\tau A \tau d\tau,\quad \tau\in \Pi W^{\textrm{div}}$$ +@f{gather*}E(\tau) := \dfrac{1}{2}\int\limits_{\Omega}\tau A \tau d\tau,\quad \tau\in \Pi W^{\textrm{div}}@f} with -$$W^{\textrm{div}}:=\lbrace \tau\in -L^2(\Omega,\mathbb{R}^{\textrm{dim}\times\textrm{dim}}_{\textrm{sym}}):\textrm{div}(\tau)\in L^2(\Omega,\mathbb{R}^{\textrm{dim}})\rbrace$$ and -$$\Pi \Sigma:=\lbrace \tau\in \Sigma, \mathcal{F}(\tau)\leq 0\rbrace$$ +@f{gather*}W^{\textrm{div}}:=\lbrace \tau\in +L^2(\Omega,\mathbb{R}^{\textrm{dim}\times\textrm{dim}}_{\textrm{sym}}):\textrm{div}(\tau)\in L^2(\Omega,\mathbb{R}^{\textrm{dim}})\rbrace@f} and +@f{gather*}\Pi \Sigma:=\lbrace \tau\in \Sigma, \mathcal{F}(\tau)\leq 0\rbrace@f} as the set of admissible stresses which is defined by a continious, convex flow function $\mathcal{F}$. With the goal of deriving the dual formulation of the minimisation problem, we define a lagrange function: -$$L(\tau,\varphi) := E(\tau) + (\varphi, \textrm{div}(\tau)),\quad \lbrace\tau,\varphi\rbrace\in\Pi W^{\textrm{div}}\times V^+$$ +@f{gather*}L(\tau,\varphi) := E(\tau) + (\varphi, \textrm{div}(\tau)),\quad \lbrace\tau,\varphi\rbrace\in\Pi W^{\textrm{div}}\times V^+@f} with -$$V^+ := \lbrace u\in V: u_n\leq g \text{ on } \Gamma_C \rbrace$$ -$$V:=\left[ H_0^1 \right]^{\textrm{dim}}:=\lbrace u\in \left[H^1(\Omega)\right]^{\textrm{dim}}: u -= 0 \text{ on } \Gamma_D\rbrace$$ +@f{gather*}V^+ := \lbrace u\in V: u_n\leq g \text{ on } \Gamma_C \rbrace@f} +@f{gather*}V:=\left[ H_0^1 \right]^{\textrm{dim}}:=\lbrace u\in \left[H^1(\Omega)\right]^{\textrm{dim}}: u += 0 \text{ on } \Gamma_D\rbrace@f} By building the Fréchet derivatives of $L$ for both components we obtain the -dual formulation for the stationary case which is known as \textbf{Hencky-Type-Model}:\\ +dual formulation for the stationary case which is known as Hencky-Type-Model:\\ Find a pair $\lbrace\sigma,u\rbrace\in \Pi W\times V^+$ with -$$\left(A\sigma,\tau - \sigma\right) + \left(u, \textrm{div}(\tau) - \textrm{div}(\sigma)\right) \geq 0,\quad \forall \tau\in \Pi W^{\textrm{div}}$$ -$$-\left(\textrm{div}(\sigma),\varphi - u\right) \geq 0,\quad \forall \varphi\in V^+.$$ +@f{gather*}\left(A\sigma,\tau - \sigma\right) + \left(u, \textrm{div}(\tau) - \textrm{div}(\sigma)\right) \geq 0,\quad \forall \tau\in \Pi W^{\textrm{div}}@f} +@f{gather*}-\left(\textrm{div}(\sigma),\varphi - u\right) \geq 0,\quad \forall \varphi\in V^+.@f} By integrating by parts and multiplying the first inequality by the elastic tensor $C=A^{-1}$ we achieve the primal-mixed version of our problem: Find a pair $\lbrace\sigma,u\rbrace\in \Pi W\times V^+$ with -$$\left(\sigma,\tau - \sigma\right) - \left(C\varepsilon(u), \tau - \sigma\right) \geq 0,\quad \forall \tau\in \Pi W$$ -$$\left(\sigma,\varepsilon(\varphi) - \varepsilon(u)\right) \geq 0,\quad \forall \varphi\in V^+.$$ +@f{gather*}\left(\sigma,\tau - \sigma\right) - \left(C\varepsilon(u), \tau - \sigma\right) \geq 0,\quad \forall \tau\in \Pi W@f} +@f{gather*}\left(\sigma,\varepsilon(\varphi) - \varepsilon(u)\right) \geq 0,\quad \forall \varphi\in V^+.@f} Therein $\varepsilon$ denotes the linearised deformation tensor with $\varepsilon(u) := \dfrac{1}{2}\left(\nabla u + \nabla u^T\right)$ for small deformations.\\ Most materials - especially metals - have the property that they show some hardening effects during the forming process. There are different constitutive laws to describe those material behaviors. The @@ -124,8 +124,8 @@ $\eta$ is the norm of the plastic strain $\eta = \vert \varepsilon - A\sigma\vert$. It can be considered by establishing an additional term in our primal-mixed formulation:\\ Find a pair $\lbrace(\sigma,\xi),u\rbrace\in \Pi (W\times L^2(\Omega,\mathbb{R}))\times V^+$ with -$$\left(\sigma,\tau - \sigma\right) - \left(C\varepsilon(u), \tau - \sigma\right) + \gamma\left( \xi, \eta - \xi\right) \geq 0,\quad \forall (\tau,\eta)\in \Pi (W,L^2(\Omega,\mathbb{R}))$$ -$$\left(\sigma,\varepsilon(\varphi) - \varepsilon(u)\right) \geq 0,\quad \forall \varphi\in V^+,$$ +@f{gather*}\left(\sigma,\tau - \sigma\right) - \left(C\varepsilon(u), \tau - \sigma\right) + \gamma\left( \xi, \eta - \xi\right) \geq 0,\quad \forall (\tau,\eta)\in \Pi (W,L^2(\Omega,\mathbb{R}))@f} +@f{gather*}\left(\sigma,\varepsilon(\varphi) - \varepsilon(u)\right) \geq 0,\quad \forall \varphi\in V^+,@f} with the hardening parameter $\gamma > 0$. Now we want to derive a primal problem which only depends on the displacement $u$. For that purpose we @@ -134,28 +134,28 @@ theorem (see Grossmann, Roos: Numerical Treatment of Partial Differential Equations, Springer-Verlag Berlin Heidelberg, 2007 and Frohne: FEM-Simulation der Umformtechnik metallischer Oberflächen im Mikrokosmos, Ph.D. thesis, University of Siegen, Germany, 2011) on -$$\left(\sigma - C\varepsilon(u), \tau - \sigma\right) \geq 0,\quad \forall \tau\in \Pi W,$$ +@f{gather*}\left(\sigma - C\varepsilon(u), \tau - \sigma\right) \geq 0,\quad \forall \tau\in \Pi W,@f} which yields with the second inequality:\\ Find the displacement $u\in V^+$ with -$$\left(P_{\Pi}(C\varepsilon(u)),\varepsilon(\varphi) - \varepsilon(u)\right) \geq 0,\quad \forall \varphi\in V^+,$$ +@f{gather*}\left(P_{\Pi}(C\varepsilon(u)),\varepsilon(\varphi) - \varepsilon(u)\right) \geq 0,\quad \forall \varphi\in V^+,@f} with the projection: -$$P_{\Pi}(\tau):=@f{cases} +@f{gather*}P_{\Pi}(\tau):=@f{cases} \tau, & \text{if }\vert\tau^D\vert \leq \sigma_0 + \gamma\xi,\\ \hat\alpha\dfrac{\tau^D}{\vert\tau^D\vert} + \dfrac{1}{3}tr(\tau), & \text{if }\vert\tau^D\vert > \sigma_0 + \gamma\xi, - @f}$$ + @f}@f} with the radius -$$\hat\alpha := \sigma_0 + \gamma\xi .$$ +@f{gather*}\hat\alpha := \sigma_0 + \gamma\xi .@f} With the relation $\xi = \vert\varepsilon(u) - A\sigma\vert$ it is possible to eliminate $\xi$ inside the projection $P_{\Pi}$:\\ -$$P_{\Pi}(\tau):=@f{cases} +@f{gather*}P_{\Pi}(\tau):=@f{cases} \tau, & \text{if }\vert\tau^D\vert \leq \sigma_0,\\ \alpha\dfrac{\tau^D}{\vert\tau^D\vert} + \dfrac{1}{3}tr(\tau), & \text{if }\vert\tau^D\vert > \sigma_0, - @f}$$ -$$\alpha := \sigma_0 + \dfrac{\gamma}{2\mu+\gamma}\left(\vert\tau^D\vert - \sigma_0\right) ,$$ + @f}@f} +@f{gather*}\alpha := \sigma_0 + \dfrac{\gamma}{2\mu+\gamma}\left(\vert\tau^D\vert - \sigma_0\right) ,@f} with a further material parameter $\mu>0$ called shear modulus. We refer that this only possible for isotropic plasticity. So what we do is to calculate the stresses by using Hooke's law for linear elastic, isotropic materials -$$\sigma = C \varepsilon(u) = 2\mu \varepsilon^D(u) + \kappa tr(\varepsilon(u))I = \left[2\mu\left(\mathbb{I} -\dfrac{1}{3} I\otimes I\right) + \kappa I\otimes I\right]\varepsilon(u)$$ +@f{gather*}\sigma = C \varepsilon(u) = 2\mu \varepsilon^D(u) + \kappa tr(\varepsilon(u))I = \left[2\mu\left(\mathbb{I} -\dfrac{1}{3} I\otimes I\right) + \kappa I\otimes I\right]\varepsilon(u)@f} with the material parameter $\kappa>0$ (bulk modulus). The variables $I$ and $\mathbb{I}$ denote the identity tensors of second and forth order. In that notation $2\mu \varepsilon^D(u)$ is the deviatoric part and $\kappa @@ -178,7 +178,7 @@ method - inexact since we use an iterative solver for the linearised problems in

Linearisation of the constitutive law for the Newton method

For the Newton method we have to linearise the following semi-linearform -$$a(\psi;\varphi) := \left(P_{\Pi}(C\varepsilon(\varphi)),\varepsilon(\varphi)\right).$$ +@f{gather*}a(\psi;\varphi) := \left(P_{\Pi}(C\varepsilon(\varphi)),\varepsilon(\varphi)\right).@f} Because we have to find the solution $u$ in the convex set $V^+$, we have to apply an SQP-method (SQP: sequential quadratic programming). That means we have to solve a minimisation problem for a known $u^i$ in every SQP-step of the form @@ -189,13 +189,13 @@ to solve a minimisation problem for a known $u^i$ in every SQP-step of the form &\rightarrow& \textrm{min},\quad u^{i+1}\in V^+. @f} Neglecting the constant terms $ a(u^i;u^i)$ and $ a'(u^i;u^i,u^i)$ we obtain the -following minimisation problem $$\dfrac{1}{2} a'(u^i;u^{i+1},u^{i+1}) - F(u^i)\rightarrow \textrm{min},\quad u^{i+1}\in V^+$$ with -$$F(\varphi) := \left(a'(\varphi;\varphi,u^{i+1}) - a(\varphi;u^{i+1}) \right).$$ +following minimisation problem @f{gather*}\dfrac{1}{2} a'(u^i;u^{i+1},u^{i+1}) - F(u^i)\rightarrow \textrm{min},\quad u^{i+1}\in V^+@f} with +@f{gather*}F(\varphi) := \left(a'(\varphi;\varphi,u^{i+1}) - a(\varphi;u^{i+1}) \right).@f} In the case of our constitutive law the Fréchet derivative of the semi-linearform $a(.;.)$ at the point $u^i$ is -$$a'(u^i;\psi,\varphi) = -(I(x)\varepsilon(\psi),\varepsilon(\varphi)),\quad x\in\Omega,$$ $$ +@f{gather*}a'(u^i;\psi,\varphi) = +(I(x)\varepsilon(\psi),\varepsilon(\varphi)),\quad x\in\Omega,@f} @f{gather*} I(x) := @f{cases} 2\mu\left(\mathbb{I} - \dfrac{1}{3} I\otimes I\right) + \kappa I\otimes I, & \quad \vert \tau^D \vert \leq \sigma_0\\ @@ -203,9 +203,9 @@ I(x) := @f{cases} - \dfrac{\tau^D\otimes\tau^D}{\vert\tau^D\vert}\right) + \kappa I\otimes I, &\quad \vert \tau^D \vert > \sigma_0 @f} -$$ +@f} with -$$\tau^D := C\varepsilon^D(u^i).$$ +@f{gather*}\tau^D := C\varepsilon^D(u^i).@f} Remark that $a(.;.)$ is not differentiable in the common sense but it is slantly differentiable like the function for the contact problem and again we refer to Hintermueller, Ito, Kunisch: The primal-dual active set strategy as a semismooth newton method, SIAM J. OPTIM., 2003, Vol. 13, No. 3, pp. 865-888. @@ -220,8 +220,8 @@ system of equations finally. We introduce a Lagrange multiplier $\lambda$ and the convex cone $K\subset W'$, $W'$ dual space of the trace space $W$ of $V$ restricted to $\Gamma_C$, -$$K:=\{\mu\in W':\mu_T = 0,\quad\langle\mu n,v\rangle_{\Gamma_C}\geq 0,\quad -\forall v\in W, v \ge 0\text{ on }\Gamma_C \}$$ +@f{gather*}K:=\{\mu\in W':\mu_T = 0,\quad\langle\mu n,v\rangle_{\Gamma_C}\geq 0,\quad +\forall v\in W, v \ge 0\text{ on }\Gamma_C \}@f} of Lagrange multipliers, where $\langle\cdot,\cdot\rangle$ denotes the duality pairing, i.e. a boundary integral, between $W'$ and $W$. Intuitively, $K$ is the cone of all "non-positive functions", except that $ K\subset @@ -269,39 +269,39 @@ Newton method Now we describe an algorithm that combines the Newton-method, which we use for the nonlinear constitutive law, with the semismooth Newton method for the contact. It sums up the results of the sections before and works as follows: -@f{itemize} - \item[(0)] Initialize $\mathcal{A}_k$ and $\mathcal{F}_k$, such that $\mathcal{S} = \mathcal{A}_k \cup \mathcal{F}_k$ and $\mathcal{A}_k \cap \mathcal{F}_k = \emptyset$ and set $k = 1$. - \item[(1)] Assemble the Newton matrix $a'(U^k;\varphi_i,\varphi_j)$ and the right-hand-side $F(U^k)$. - \item[(2)] Find the primal-dual pair $(U^k,\Lambda^k)$ that satisfies +
    +
  1. Initialize $\mathcal{A}_k$ and $\mathcal{F}_k$, such that $\mathcal{S} = \mathcal{A}_k \cup \mathcal{F}_k$ and $\mathcal{A}_k \cap \mathcal{F}_k = \emptyset$ and set $k = 1$. +
  2. Assemble the Newton matrix $a'(U^k;\varphi_i,\varphi_j)$ and the right-hand-side $F(U^k)$. +
  3. Find the primal-dual pair $(U^k,\Lambda^k)$ that satisfies @f{align*} AU^k + B\Lambda^k & = F, &\\ \left[B^TU^k\right]_i & = G_i & \forall i\in\mathcal{A}_k,\\ \Lambda^k_i & = 0 & \forall i\in\mathcal{F}_k. @f} - \item[(3)] Define the new active and inactive sets by - $$\mathcal{A}_{k+1}:=\lbrace i\in\mathcal{S}:\Lambda^k_i + - c\left(\left[B^TU^k\right]_i - G_i\right) > 0\rbrace,$$ - $$\mathcal{F}_{k+1}:=\lbrace i\in\mathcal{S}:\Lambda^k_i + - c\left(\left[B^TU^k\right]_i - G_i\right) \leq 0\rbrace.$$ - \item[(4)] If $\mathcal{A}_{k+1} = \mathcal{A}_k$ and $\vert +
  4. Define the new active and inactive sets by + @f{gather*}\mathcal{A}_{k+1}:=\lbrace i\in\mathcal{S}:\Lambda^k_i + + c\left(\left[B^TU^k\right]_i - G_i\right) > 0\rbrace,@f} + @f{gather*}\mathcal{F}_{k+1}:=\lbrace i\in\mathcal{S}:\Lambda^k_i + + c\left(\left[B^TU^k\right]_i - G_i\right) \leq 0\rbrace.@f} +
  5. If $\mathcal{A}_{k+1} = \mathcal{A}_k$ and $\vert F\left(U^{k+1}\right) \vert < \delta$ then stop, else set $k=k+1$ and go to - step (1). -@f} + step (2). +
The mass matrix $B\in\mathbb{R}^{n\times m}$, $n>m$, is quadratic in our situation since $\Lambda^k$ is only defined on $\Gamma_C$: -$$B_{ij} = @f{cases} +@f{gather*}B_{ij} = \begin{cases} \int\limits_{\Gamma_C}\varphi_i^2(x)dx, & \text{if}\quad i=j\\ 0, & \text{if}\quad i\neq j. -@f}$$ +\end{cases}@f} So $m$ denotes the size of $\Lambda^k$ and $i$ a degree of freedom. In our programm we use the structure of a quadratic sparse for $B\in\mathbb{R}^{n\times n}$ and the length of $\Lambda^k$ is $n$ with $\Lambda^k_i = 0$ for $i>m$. The vector $G$ is defined by a suitable approximation $g_h$ of the gap $g$ -$$G_i = @f{cases} +@f{gather*}G_i = \begin{cases} \int\limits_{\Gamma_C}g_h(x)\varphi_i(x)dx, & \text{if}\quad i\leq m\\ 0, & \text{if}\quad i>m. -@f}$$ +\end{cases}@f} Compared to step-41, step (1) is added but it should be clear from the sections above that we only linearize the problem. In step (2) we have to solve a linear @@ -326,7 +326,7 @@ results it yields a quite reasonable adaptive mesh for the contact zone.

Implementation

-This tutorial is essentailly a mixture of step-40 and step-41 but instead of +This tutorial is essentially a mixture of step-40 and step-41 but instead of PETSc we let the Trilinos library deal with parallelizing the linear algebra (like in step-32). Since we are trying to solve a similar problem like in step-41 we will use the same methods but now in parallel. @@ -345,4 +345,5 @@ internal or external that keeps you going. This is the safest way to express motivation in Chinese. If your audience is Japanese, please see the other entry for motivation. This is a word in Japanese and Korean, but it means "motive power" or "kinetic energy" (without the motivation meaning that you are -probably looking for)"). \ No newline at end of file +probably looking for)"). +