From: bangerth Date: Wed, 23 May 2012 20:57:36 +0000 (+0000) Subject: Go over it once. X-Git-Url: https://gitweb.dealii.org/cgi-bin/gitweb.cgi?a=commitdiff_plain;h=c693a888862e4612009664ecaf343e8098d583ee;p=dealii-svn.git Go over it once. git-svn-id: https://svn.dealii.org/trunk@25541 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 d731db4eb7..90103af62d 100644 --- a/deal.II/examples/step-15/doc/intro.dox +++ b/deal.II/examples/step-15/doc/intro.dox @@ -13,58 +13,60 @@ is by him.

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 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. +This program deals with an example of a non-linear elliptic partial +differential equation, the minimal +surface equation. You can imagine the solution of this equation to describe +the surface spanned by a soap film that is enclosed by a +closed wire loop. We imagine the wire to not just be a planar loop, but in +fact curved. The surface tension of the soap film will then reduce the surface +to have minimal surface. The solution of the minimal surface equation +describes this shape with the wire's vertical displacement as a boundary +condition. For simplicity, we will here assume that the surface can be written +as a graph $u=u(x,y)$ although it is clear that it is not very hard to +construct cases where the wire is bent in such a way that the surface can only +locally be constructed as a graph but not globally. -Because the equation is non-linear, we can't solve it directly, but have to use -the newton-method to compute the solution iterativly. +Because the equation is non-linear, we can't solve it directly. Rather, we +have to use Newton's method to compute the solution iteratively.

Classical formulation

-In a classical sense, the problem posseses the following form: +In a classical sense, the problem is given in the following form: -@par - @f[ - -\nabla \cdot \left( \frac{1}{\sqrt{1+|\nabla u|^{2}}}\nabla u \right) = 0 \qquad - \qquad in ~ \Omega - @f] - @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 -with a damping parameter $\lambda$ to have a better global convergence behaviour: - -@par - @f[ - F'(u^{n},\delta u^{n})=- \lambda F(u^{n}) - @f] - @f[ - u^{n+1}=u^{n}+\delta u^{n} - @f] + @f{align*} + -\nabla \cdot \left( \frac{1}{\sqrt{1+|\nabla u|^{2}}}\nabla u \right) &= 0 \qquad + \qquad &&\textrm{in} ~ \Omega + \\ + u&=g \qquad\qquad &&\textrm{on} ~ \partial \Omega + @f} -with: +$\Omega$ is the domain we get by projecting the wire's positions into $x-y$ +space. In this example, we choose $\Omega$ as the unit disk. +As described above, we solve this equation using Newton's method in which we +compute the $n$th approximate solution from the $n-1$st one, and use +a damping parameter $\lambda^n$ to get better global convergence behavior: + @f{align*} + F'(u^{n},\delta u^{n})&=- F(u^{n}) + \\ + u^{n+1}&=u^{n}+\lambda^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[ F'(u,\delta u)=\lim \limits_{\epsilon \rightarrow 0}{\frac{F(u+\epsilon \delta u)- F(u)}{\epsilon}}. @f] -So we have to solve a linear elliptic pde in every Newton-step, with $\delta u$ as -the solution of: +Going through the motions to find out what $F'(u,\delta u)$ is, we find that +we have to solve a linear elliptic PDE in every Newton step, with $\delta u^n$ +as the solution of: @f[ - \nabla \cdot \left( \frac{1}{(1+|\nabla u^{n}|^{2})^{\frac{1}{2}}}\nabla @@ -76,39 +78,38 @@ the solution of: \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 -$\delta u^{n}$ 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 order to solve the minimal surface equation, we have to solve this equation +repeatedly, once per 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 $u^{1}$. -@par -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. +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 following steps.

Weak formulation of the problem

Starting with the strong formulation above, we get the weak formulation by multiplying -both sides of the pde with a testfunction $\varphi$ and integrating by parts on both sides: - +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 u^{n} - \right) + \right). @f] - -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: +Here the solution $\delta u^{n}$ is a function in $H^{1}(\Omega)$, subject to +the boundary conditions discussed above. +Reducing this space to a finite dimensional space with basis $\left\{ +\varphi_{0},\dots , \varphi_{N-1}\right\}$, we can write the solution: @f[ - \delta u^{n}=\sum_{j=0}^{N-1} \varphi_{j} \cdot U_{j} + \delta u^{n}=\sum_{j=0}^{N-1} U_{j} \varphi_{j} @f] Using the basis functions as testfunctions and defining $a_{n}:=\frac{1} @@ -122,10 +123,10 @@ Using the basis functions as testfunctions and defining $a_{n}:=\frac{1} @f] where the solution $\delta u^{n}$ is given by the coefficents $\delta U^{n}_{j}$. -This linear equation system can be rewritten as: +This linear system of equations can be rewritten as: @f[ - A^{n}U^{n}=b^{n} + A^{n}\; \delta U^{n}=b^{n} @f] where the entries of the matrix $A^{n}$ are given by: @@ -199,8 +200,9 @@ method we use here.

Summary

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 +solving the system $A^{0}\;\delta U^{0}=b^{0}$ 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 - $\partial \Omega$ and the new approximation given by $u^{n+1}=u^{n}+\delta u^{n}$. + $u^{1}=u^{0}+\lambda^0 \delta u^{0}$. The next updates are given as solution of + the linear system $A^{n}\;\delta U^{n}=b^{n}$ with boundary condition $\delta u^{n}=0$ on + $\partial \Omega$ and the new approximation given by $u^{n+1}=u^{n}+\lambda^n + \delta u^{n}$.