From 4ef2087b74b54e793c67ba1db51a236feae9de2a Mon Sep 17 00:00:00 2001 From: David Wells Date: Sat, 27 Oct 2018 12:00:56 -0400 Subject: [PATCH] step-57: Reformat in-code TeX. I also added a missing parenthesis. --- examples/step-57/doc/intro.dox | 103 ++++++++++++++++++++++----------- 1 file changed, 69 insertions(+), 34 deletions(-) diff --git a/examples/step-57/doc/intro.dox b/examples/step-57/doc/intro.dox index 612aff39c3..4b1d14803d 100644 --- a/examples/step-57/doc/intro.dox +++ b/examples/step-57/doc/intro.dox @@ -76,7 +76,7 @@ where $\textbf{x}^{k+1}=\textbf{x}^{k}+\delta \textbf{x}^{k}$. Then we can evaluate the update term by solving the system @f{eqnarray*} -\nabla F(\textbf{x}^{k}) \delta \textbf{x}^{k} = -F(\textbf{x}^{k}). + \nabla F(\textbf{x}^{k}) \delta \textbf{x}^{k} = -F(\textbf{x}^{k}). @f} Here, the left of the previous equation represents the directional gradient of $F(\textbf{x})$ along $\delta @@ -85,31 +85,49 @@ directional gradient of $F(\textbf{x})$ along $\delta @f{eqnarray*} & &\nabla F(\mathbf{u}^{k}, p^{k}) (\delta \mathbf{u}^{k}, \delta p^{k}) \\ \\ - &=& \lim_{\epsilon \to 0} \frac{1}{\epsilon} (F(\mathbf{u}^{k}+\epsilon \delta \mathbf{u}^{k}, p^{k}+\epsilon\nabla\delta p^{k}) - (F(\mathbf{u}^{k}, p^{k}))\\ + &=& \lim_{\epsilon \to 0} \frac{1}{\epsilon} + \left( + F(\mathbf{u}^{k} + \epsilon \delta \mathbf{u}^{k}, + p^{k} + \epsilon \nabla \delta p^{k}) + - F(\mathbf{u}^{k}, p^{k}) + \right)\\ + \\ + &=& \lim_{\epsilon \to 0} \frac{1}{\epsilon} + \left( + \begin{array}{c} + - \epsilon \nu \Delta \delta \mathbf{u}^{k} + + \epsilon \mathbf{u}^{k} \cdot \nabla \delta \mathbf{u}^{k} + + \epsilon \delta \mathbf{u}^{k} \cdot \nabla \mathbf{u}^{k} + + \epsilon^{2} \delta \mathbf{u}^{k} \cdot \nabla \delta \mathbf{u}^{k} + + \epsilon \nabla \delta p^{k}\\ + - \epsilon \nabla \cdot \delta \mathbf{u}^{k}\\ + \end{array} + \right)\\ \\ - &=& \lim_{\epsilon \to 0} \frac{1}{\epsilon} \left( - \begin{array}{c} - - \epsilon\nu\Delta\delta \mathbf{u}^{k} + \epsilon\mathbf{u}^{k}\cdot\nabla\delta\mathbf{u}^{k}+\epsilon\delta\mathbf{u}^{k}\cdot\nabla\mathbf{u}^{k}+\epsilon^{2}\delta\mathbf{u}^{k}\cdot\nabla\delta\mathbf{u}^{k}+\epsilon \nabla\delta p^{k}\\ - - \epsilon \nabla \cdot\delta \mathbf{u}^{k}\\ - \end{array} - \right)\\ - \\ &=& \left( - \begin{array}{c} - - \nu\Delta\delta \mathbf{u}^{k} + \mathbf{u}^{k}\cdot\nabla\delta\mathbf{u}^{k}+\delta\mathbf{u}^{k}\cdot\nabla\mathbf{u}^{k}+ \nabla\delta p^{k}\\ - - \nabla \cdot\delta \mathbf{u}^{k}\\ - \end{array} - \right). + \begin{array}{c} + - \nu \Delta \delta \mathbf{u}^{k} + + \mathbf{u}^{k} \cdot \nabla \delta \mathbf{u}^{k} + + \delta \mathbf{u}^{k} \cdot \nabla \mathbf{u}^{k} + + \nabla\delta p^{k}\\ + - \nabla \cdot \delta \mathbf{u}^{k}\\ + \end{array} + \right). @f} Therefore, we arrive at the linearized system: @f{eqnarray*} -- \nu\Delta\delta\mathbf{u}^{k} + \mathbf{u}^{k}\cdot\nabla\delta\mathbf{u}^{k}+\delta\mathbf{u}^{k}\cdot\nabla\mathbf{u}^{k}+ \nabla\delta p^{k} = \mathbf{g}, \\ -- \nabla \cdot\delta \mathbf{u}^{k} = \nabla\cdot\mathbf{u}^{k}, + -\nu \Delta \delta \mathbf{u}^{k} + + \mathbf{u}^{k} \cdot \nabla \delta \mathbf{u}^{k} + + \delta \mathbf{u}^{k} \cdot \nabla \mathbf{u}^{k} + + \nabla \delta p^{k} + = \mathbf{g}, \\ + -\nabla \cdot\delta \mathbf{u}^{k} + = \nabla \cdot \mathbf{u}^{k}, @f} -where $\textbf{g} =\textbf{f}+\nu \Delta\textbf{u}^k -(\textbf{u}^k +where $\textbf{g} =\textbf{f} + \nu \Delta \textbf{u}^k - (\textbf{u}^k \cdot \nabla)\textbf{u}^k -\nabla p^k$ and $\textbf{u}^k$ and $p^k$ are the solutions from the previous iteration. Additionally, the right hand side of the second equation is not zero since the discrete @@ -159,15 +177,15 @@ as a staircase from the Stokes equations to the NSE we want to solve. That is, we first solve a Stokes problem @f{eqnarray*} -- \nu_{1} \Delta\textbf{u} + \nabla p &=& \textbf{f}\\ -- \nabla \cdot \textbf{u} &=& 0 + -\nu_{1} \Delta \textbf{u} + \nabla p &=& \textbf{f}\\ + -\nabla \cdot \textbf{u} &=& 0 @f} to get the initial guess for @f{eqnarray*} -- \nu_{1} \Delta\textbf{u} + (\textbf{u} \cdot \nabla)\textbf{u} + \nabla p &=& \textbf{f},\\ -- \nabla \cdot \textbf{u} &=& 0, + -\nu_{1} \Delta \textbf{u} + (\textbf{u} \cdot \nabla)\textbf{u} + \nabla p &=& \textbf{f},\\ + -\nabla \cdot \textbf{u} &=& 0, @f} which also acts as the initial guess of the continuation method. Here $\nu_{1}$ is relatively large so that the solution to the Stokes problem with viscosity $\nu_{1}$ @@ -176,15 +194,15 @@ can be used as an initial guess for the NSE in Newton's iteration. Then the solution to @f{eqnarray*} -- \nu_{i} \Delta\textbf{u} + (\textbf{u} \cdot \nabla)\textbf{u} + \nabla p &=& \textbf{f},\\ -- \nabla \cdot \textbf{u} &=& 0. + -\nu_{i} \Delta \textbf{u} + (\textbf{u} \cdot \nabla)\textbf{u} + \nabla p &=& \textbf{f},\\ + -\nabla \cdot \textbf{u} &=& 0. @f} acts as the initial guess for @f{eqnarray*} -- \nu_{i+1} \Delta\textbf{u} + (\textbf{u} \cdot \nabla)\textbf{u} + \nabla p &=& \textbf{f},\\ -- \nabla \cdot \textbf{u} &=& 0. + -\nu_{i+1} \Delta \textbf{u} + (\textbf{u} \cdot \nabla)\textbf{u} + \nabla p &=& \textbf{f},\\ + -\nabla \cdot \textbf{u} &=& 0. @f} This process is repeated with a sequence of viscosities $\{\nu_i\}$ that is @@ -250,8 +268,12 @@ side by $b$, we solve it iteratively with right preconditioning $P^{-1}$ as $GP^{-1}y = b$, where @f{eqnarray*} -P^{-1} = \left(\begin{array}{cc} \tilde{A} & B^T \\ - 0 & \tilde{S} \end{array}\right)^{-1}, +P^{-1} = \left( + \begin{array}{cc} + \tilde{A} & B^T \\ + 0 & \tilde{S} + \end{array} + \right)^{-1}, @f} with $\tilde{A} = A + \gamma B^TW^{-1}B$ and $\tilde{S}$ is the @@ -268,9 +290,24 @@ We decompose $P^{-1}$ as @f{eqnarray*} P^{-1} = -\left(\begin{array}{cc} \tilde{A}^{-1} & 0 \\ 0 & I \end{array}\right) -\left(\begin{array}{cc} I & -B^T \\ 0 & I \end{array}\right) -\left(\begin{array}{cc} I & 0 \\ 0 & \tilde{S}^{-1} \end{array}\right). + \left( + \begin{array}{cc} + \tilde{A}^{-1} & 0 \\ + 0 & I + \end{array} + \right) + \left( + \begin{array}{cc} + I & -B^T \\ + 0 & I + \end{array} + \right) + \left( + \begin{array}{cc} + I & 0 \\ + 0 & \tilde{S}^{-1} + \end{array} + \right). @f} Here two inexact solvers will be needed for $\tilde{A}^{-1}$ and @@ -290,10 +327,8 @@ The computational domain is the unit square and the right-hand side $f=0$. The boundary condition is @f{eqnarray*} -(u(x, y), v(x,y)) &=& (1,0) - \qquad\qquad \textrm{if}\ y = 1 \\ - (u(x, y), v(x,y)) &=& (0,0) - \qquad\qquad \textrm{otherwise}. + (u(x, y), v(x,y)) &=& (1,0) \qquad\qquad \textrm{if}\ y = 1 \\ + (u(x, y), v(x,y)) &=& (0,0) \qquad\qquad \textrm{otherwise}. @f} When solving this problem, the error consists of the nonlinear error (from -- 2.39.5