From: bangerth Date: Fri, 13 Apr 2012 12:49:54 +0000 (+0000) Subject: Minor edit. X-Git-Url: https://gitweb.dealii.org/cgi-bin/gitweb.cgi?a=commitdiff_plain;h=0c70ef312bfb2e9045f0889fbbe82d95ffd0e939;p=dealii-svn.git Minor edit. git-svn-id: https://svn.dealii.org/trunk@25413 0785d39b-7218-0410-832d-ea1e28bc413d --- diff --git a/deal.II/examples/step-15/doc/intro.dox b/deal.II/examples/step-15/doc/intro.dox index 526da2a527..5e1e4fccab 100644 --- a/deal.II/examples/step-15/doc/intro.dox +++ b/deal.II/examples/step-15/doc/intro.dox @@ -1,25 +1,22 @@ -/** - @page mse Minimal surface equation -

Introduction

Foreword

This programm deals with an example of a non-linear elliptic pde, the minimal -surface equation. You can imagine the solution as a soap bubble inside a -closed wire, where the wire isn't smooth, but curved. The soap bubble will +surface equation. You can imagine the solution as a soap bubble inside a +closed wire, where the wire isn't just a planar loop, but in in fact curved. The soap bubble will take a shape with minimal surface. The solution of the minimal surface equation -describes this shape with the wire as a boundary condition. +describes this shape with the wire as a boundary condition. -Because the equation is non-linear, we can't solve it directly, but have to use +Because the equation is non-linear, we can't solve it directly, but have to use the newton-method to compute the solution iterativly.

Classical formulation

-In a classical sense, the problem posseses following form: +In a classical sense, the problem posseses the following form: @par @f[ @@ -29,10 +26,10 @@ In a classical sense, the problem posseses following form: @f[ u=g \qquad\qquad on ~ \partial \Omega @f] - + In this example, we choose the unitball as our domain $\Omega$. -As described above, we have to formulate the Newton-method for this problem +As described above, we have to formulate the Newton-method for this problem with a damping parameter $\lambda$ to have a better global convergence behaviour: @par @@ -42,13 +39,13 @@ with a damping parameter $\lambda$ to have a better global convergence behaviour @f[ u^{n+1}=u^{n}+\delta u^{n} @f] - + with: @f[ F(u):= -\nabla \cdot \left( \frac{1}{\sqrt{1+|\nabla u|^{2}}}\nabla u \right) @f] - + and $F'(u,\delta u)$ the derivative of F in direction of $\delta u$: @f[ @@ -60,26 +57,26 @@ So we have to solve a linear elliptic pde in every Newton-step, with $\delta u$ the solution of: @f[ - - \nabla \cdot \left( \frac{1}{(1+|\nabla u^{n}|^{2})^{\frac{1}{2}}}\nabla - \delta u^{n} \right) + - \nabla \cdot \left( \frac{\nabla u^{n} \cdot - \nabla \delta u^{n}}{(1+|\nabla u^{n}|^{2})^{\frac{3}{2}}} \nabla u^{n} - \right) = + - \nabla \cdot \left( \frac{1}{(1+|\nabla u^{n}|^{2})^{\frac{1}{2}}}\nabla + \delta u^{n} \right) + + \nabla \cdot \left( \frac{\nabla u^{n} \cdot + \nabla \delta u^{n}}{(1+|\nabla u^{n}|^{2})^{\frac{3}{2}}} \nabla u^{n} + \right) = -\left( - \nabla \cdot \left( \frac{1}{(1+|\nabla u^{n}|^{2})^{\frac{1}{2}}} \nabla u^{n} \right) \right) @f] In order to solve the minimal surface equation, we have to solve this equation in every -Newton step. To solve this, we have to take a look at the boundary condition of this -problem. Assuming that $u^{n}$ already has the right boundary values, the Newton update +Newton step. To solve this, we have to take a look at the boundary condition of this +problem. Assuming that $u^{n}$ already has the right boundary values, the Newton update $\delta u^{n}$ should have zero boundary conditions, in order to have the right boundary condition after adding both. -In the first Newton step, we are starting with the solution $u^{0}\equiv 0$, the Newton -update still has to deliever the right boundary condition to the solution $u^{1}$. +In the first Newton step, we are starting with the solution $u^{0}\equiv 0$, the Newton +update still has to deliever the right boundary condition to the solution $u^{1}$. @par -Summing up, we have to solve the pde above with the boundary condition $\delta u^{0}=g$ +Summing up, we have to solve the pde above with the boundary condition $\delta u^{0}=g$ in the first step and with $\delta u^{n}=0$ in all the other steps. @@ -89,14 +86,14 @@ Starting with the strong formulation above, we get the weak formulation by multi both sides of the pde with a testfunction $\varphi$ and integrating by parts on both sides: @f[ - \left( \nabla \varphi , \frac{1}{(1+|\nabla u^{n}|^{2})^{\frac{1}{2}}}\nabla - \delta u^{n} \right)-\left(\nabla \varphi ,\frac{\nabla u^{n} \cdot \nabla - \delta u^{n}}{(1+|\nabla u^{n}|^{2})^{\frac{3}{2}}}\nabla u^{n} \right) + \left( \nabla \varphi , \frac{1}{(1+|\nabla u^{n}|^{2})^{\frac{1}{2}}}\nabla + \delta u^{n} \right)-\left(\nabla \varphi ,\frac{\nabla u^{n} \cdot \nabla + \delta u^{n}}{(1+|\nabla u^{n}|^{2})^{\frac{3}{2}}}\nabla u^{n} \right) = -\left(\nabla \varphi , \frac{1}{(1+|\nabla u^{n}|^{2})^{\frac{1}{2}}} \nabla u^{n} \right) @f] - -Where the solution $\delta u^{n}$ is a function in the infinte space $H^{1}(\Omega)$. + +Where the solution $\delta u^{n}$ is a function in the infinte space $H^{1}(\Omega)$. Reducing this space to a finite space with basis $\left\{ \varphi_{0},\dots , \varphi_{N-1}\right\}$, we can write the solution: @@ -108,10 +105,10 @@ Using the basis functions as testfunctions and defining $a_{n}:=\frac{1} {\sqrt{1+|\nabla u^{n}|^{2}}}$, we can rewrite the weak formualtion: @f[ - \sum_{j=0}^{N-1}\left[ \left( \nabla \varphi_{i} , a_{n} \nabla \varphi_{j} \right) - - \left(\nabla u^{n}\cdot \nabla \varphi_{i} , a_{n}^{3} \nabla u^{n} \cdot \nabla - \varphi_{j} \right) \right] \cdot U_{j}=\left( \nabla \varphi_{i} , a_{n} - \nabla u^{n}\right) \qquad \forall i=0,\dots ,N-1 + \sum_{j=0}^{N-1}\left[ \left( \nabla \varphi_{i} , a_{n} \nabla \varphi_{j} \right) - + \left(\nabla u^{n}\cdot \nabla \varphi_{i} , a_{n}^{3} \nabla u^{n} \cdot \nabla + \varphi_{j} \right) \right] \cdot U_{j}=\left( \nabla \varphi_{i} , a_{n} + \nabla u^{n}\right) \qquad \forall i=0,\dots ,N-1 @f] where the solution $\delta u^{n}$ is given by the coefficents $\delta U^{n}_{j}$. @@ -124,8 +121,8 @@ This linear equation system can be rewritten as: where the entries of the matrix $A^{n}$ are given by: @f[ - A^{n}_{ij}:= \left( \nabla \varphi_{i} , a_{n} \nabla \varphi_{j} \right) - - \left(\nabla u^{n}\cdot \nabla \varphi_{i} , a_{n}^{3} \nabla u^{n} \cdot \nabla + A^{n}_{ij}:= \left( \nabla \varphi_{i} , a_{n} \nabla \varphi_{j} \right) - + \left(\nabla u^{n}\cdot \nabla \varphi_{i} , a_{n}^{3} \nabla u^{n} \cdot \nabla \varphi_{j} \right) @f] @@ -135,20 +132,20 @@ and the right hand side $b^{n}$ is given by: b^{n}_{i}:=\left( \nabla \varphi_{i} , a_{n} \nabla u^{n}\right) @f] -The matrix A is symmetric, but it is indefinite. So we have to take a better look -at the solver we choose for this linear system. The CG-method needs +The matrix A is symmetric, but it is indefinite. So we have to take a better look +at the solver we choose for this linear system. The CG-method needs positive-definiteness of the matrix A, which is not given, so it can't be used. Using the symmetry of the matrix we can choose the minimal residual method as a solver, which needs symmetry but no definiteness.

Summary

-Starting with the function $u^{0}\equiv 0$, the first Newton update is computed by +Starting with the function $u^{0}\equiv 0$, the first Newton update is computed by solving the system $A^{0}U^{0}=b^{0}$ with boundary condition $\delta u^{0}=g$ on - $\partial \Omega$. The new approximation of the solution is given by - $u^{1}=u^{0}+\delta u^{0}$. The next updates are given as solution of - the linear system $A^{n}U^{n}=b^{n}$ with boundary condition $\delta u^{n}=0$ on + $\partial \Omega$. The new approximation of the solution is given by + $u^{1}=u^{0}+\delta u^{0}$. The next updates are given as solution of + the linear system $A^{n}U^{n}=b^{n}$ with boundary condition $\delta u^{n}=0$ on $\partial \Omega$ and the new approximation given by $u^{n+1}=u^{n}+\delta u^{n}$. -*/ \ No newline at end of file +*/