From 17e2eabbf1c73714e4461697c73537fd7053ca6c Mon Sep 17 00:00:00 2001 From: Wolfgang Bangerth Date: Thu, 2 Apr 2015 16:33:59 -0500 Subject: [PATCH] Minor updates to the documentation. --- examples/step-15/doc/intro.dox | 25 +++++++++++++++++-------- 1 file changed, 17 insertions(+), 8 deletions(-) diff --git a/examples/step-15/doc/intro.dox b/examples/step-15/doc/intro.dox index 099982af18..23be1d960e 100644 --- a/examples/step-15/doc/intro.dox +++ b/examples/step-15/doc/intro.dox @@ -40,7 +40,7 @@ In a classical sense, the problem is given in the following form: -\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 + u&=g \qquad\qquad &&\textrm{on} ~ \partial \Omega. @f} $\Omega$ is the domain we get by projecting the wire's positions into $x-y$ @@ -109,7 +109,7 @@ 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} U_{j} \varphi_{j} + \delta u^{n}=\sum_{j=0}^{N-1} U_{j} \varphi_{j}. @f] Using the basis functions as test functions and defining $a_{n}:=\frac{1} @@ -119,14 +119,14 @@ Using the basis functions as test functions and defining $a_{n}:=\frac{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 + \nabla u^{n}\right) \qquad \forall i=0,\dots ,N-1, @f] where the solution $\delta u^{n}$ is given by the coefficients $\delta U^{n}_{j}$. This linear system of equations can be rewritten as: @f[ - A^{n}\; \delta U^{n}=b^{n} + A^{n}\; \delta U^{n}=b^{n}, @f] where the entries of the matrix $A^{n}$ are given by: @@ -134,17 +134,17 @@ 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 - \varphi_{j} \right) + \varphi_{j} \right), @f] and the right hand side $b^{n}$ is given by: @f[ - b^{n}_{i}:=\left( \nabla \varphi_{i} , a_{n} \nabla u^{n}\right) + b^{n}_{i}:=\left( \nabla \varphi_{i} , a_{n} \nabla u^{n}\right). @f] -

Solver issues

+

Questions about the appropriate solver

The matrix that corresponds to the Newton step above can be reformulated to show its structure a bit better. Rewriting it slightly, we get that it has the @@ -156,7 +156,7 @@ form \nabla \varphi_i, B \nabla \varphi_j - \right) + \right), @f] where the matrix $B$ (of size $d \times d$ in $d$ space dimensions) is given by the following expression: @@ -185,6 +185,15 @@ $\lambda_1=a_n \left(1-\frac{|\nabla u^n|^2}{1+|\nabla u^n|^2}\right) > 0$ while that are perpendicular to $v_1$ and each other are eigenvectors with eigenvalue $a_n$. Since all eigenvalues are positive, $B$ is positive definite and so is $A$. We can thus use the CG method for solving the Newton steps. +(The fact that the matrix $A$ is symmetric and positive definite should not come +as a surprise. It results from taking the derivative of an operator that +results from taking the derivative of an energy functional: the minimal +surface equation simply minimizes some non-quadratic energy. Consequently, +the Newton matrix, as the matrix of second derivatives of a scalar energy, +must be symmetric since the derivative with regard to the $i$th and $j$th +degree of freedom should clearly commute. Likewise, if the energy functional +is convex, then the matrix of second derivatives must be positive definite, +and the direct calculation above simply reaffirms this.) It is worth noting, however, that the positive definiteness degenerates for problems where $\nabla u$ becomes large. In other words, if we simply multiply -- 2.39.5