From 5e5e7981221606d2d699e7552639196c8fd566a8 Mon Sep 17 00:00:00 2001 From: bangerth Date: Sun, 14 Feb 2010 00:01:49 +0000 Subject: [PATCH] Use @image instead of \image. git-svn-id: https://svn.dealii.org/trunk@20608 0785d39b-7218-0410-832d-ea1e28bc413d --- deal.II/examples/step-25/doc/results.dox | 34 ++++++++++++------------ 1 file changed, 17 insertions(+), 17 deletions(-) diff --git a/deal.II/examples/step-25/doc/results.dox b/deal.II/examples/step-25/doc/results.dox index aebb044ac6..4abd99cca9 100644 --- a/deal.II/examples/step-25/doc/results.dox +++ b/deal.II/examples/step-25/doc/results.dox @@ -9,22 +9,22 @@ each case, the respective grid is refined uniformly 6 times, i.e. $h\sim

An (1+1)-d Solution

The first example we discuss is the so-called 1D (stationary) breather solution of the sine-Gordon equation. The breather has the following -closed-form expression, as mentioned in the Introduction: +closed-form expression, as mentioned in the Introduction: \f[ -u_{\mathrm{breather}}(x,t) = -4\arctan \left(\frac{m}{\sqrt{1-m^2}} \frac{\sin\left(\sqrt{1-m^2}t +c_2\right)}{\cosh(mx+c_1)} \right), +u_{\mathrm{breather}}(x,t) = -4\arctan \left(\frac{m}{\sqrt{1-m^2}} \frac{\sin\left(\sqrt{1-m^2}t +c_2\right)}{\cosh(mx+c_1)} \right), \f] where $c_1$, $c_2$ and $m<1$ are constants. In the simulation below, we have chosen $c_1=0$, $c_2=0$, $m=0.5$. Moreover, it is know that the period of oscillation of the breather is $2\pi\sqrt{1-m^2}$, hence we have chosen $t_0=-5.4414$ and $t_f=2.7207$ so that we can observe three oscillations of the solution. Then, taking $u_0(x) = u_{\mathrm{breather}}(x,t_0)$, $\theta=0$ and $k=h/10$, the program computed the following solution. -\image html step-25.1d-breather.gif "Animation of the 1D stationary breather." width=5cm +@image html step-25.1d-breather.gif "Animation of the 1D stationary breather." width=5cm Though not shown how to do this in the program, another way to visualize the (1+1)-d solution is to use output generated by the DataOutStack class; it allows to "stack" the solutions of individual time steps, so that we get 2D space-time graphs from 1D time-dependent solutions. This produces the space-time plot below instead of the animation -above. +above. -\image html step-25.1d-breather_stp.png "A space-time plot of the 1D stationary breather." width=5cm +@image html step-25.1d-breather_stp.png "A space-time plot of the 1D stationary breather." width=5cm Furthermore, since the breather is an analytical solution of the sine-Gordon equation, we can use it to validate our code, although we have to assume that @@ -36,7 +36,7 @@ the numerical solution and the function described by the simulation shown in the two images above, the $L^2$ norm of the error in the finite element solution at each time step remained on the order of $10^{-2}$. Hence, we can conclude that the numerical method has been -implemented correctly in the program. +implemented correctly in the program.

A few (2+1)D Solutions

@@ -53,13 +53,13 @@ where $a_0$, $\vartheta$ and $\lambda$ are constants. In the simulation below we have chosen $a_0=\lambda=1$. Notice that if $\vartheta=\pi$ the kink is stationary, hence it would make a good solution against which we can validate the program in 2D because no reflections off the boundary of the -domain occur. +domain occur. -The simulation shown below was performed with $u_0(x) = u_{\mathrm{kink}}(x,t_0)$, $\theta=\frac{1}{2}$, $k=20h$, $t_0=1$ and $t_f=500$. The $L^2$ norm of the error of the finite element solution at each time step remained on the order of $10^{-2}$, showing that the program is working correctly in 2D, as well as 1D. Unfortunately, the solution is not very interesting, nonetheless we have included a snapshot of it below for completeness. +The simulation shown below was performed with $u_0(x) = u_{\mathrm{kink}}(x,t_0)$, $\theta=\frac{1}{2}$, $k=20h$, $t_0=1$ and $t_f=500$. The $L^2$ norm of the error of the finite element solution at each time step remained on the order of $10^{-2}$, showing that the program is working correctly in 2D, as well as 1D. Unfortunately, the solution is not very interesting, nonetheless we have included a snapshot of it below for completeness. -\image html step-25.2d-kink.png "Stationary 2D kink." width=5cm +@image html step-25.2d-kink.png "Stationary 2D kink." width=5cm -Now that we have validated the code in 1D and 2D, we move to a problem where the analytical solution is unknown. +Now that we have validated the code in 1D and 2D, we move to a problem where the analytical solution is unknown. To this end, we rotate the kink solution discussed above about the $z$ axis: we let $\vartheta=\frac{\pi}{4}$. The latter results in a @@ -71,22 +71,22 @@ to pick $\theta=\frac{2}{3}$ because for any $\theta\le\frac{1}{2}$ oscillations arose at the boundary, which are likely due to the scheme and not the equation, thus picking a value of $\theta$ a good bit into the "exponentially damped" spectrum of the time stepping schemes -assures these oscillations are not created. +assures these oscillations are not created. -\image html step-25.2d-angled_kink.gif "Animation of a moving 2D kink, at 45 degrees to the axes of the grid, showing boundary effects." width=5cm +@image html step-25.2d-angled_kink.gif "Animation of a moving 2D kink, at 45 degrees to the axes of the grid, showing boundary effects." width=5cm Another interesting solution to the sine-Gordon equation (which cannot be obtained analytically) can be produced by using two 1D breathers to construct -the following separable 2D initial condition: +the following separable 2D initial condition: \f[ - u_0(x) = - u_{\mathrm{pseudobreather}}(x,t_0) = + u_0(x) = + u_{\mathrm{pseudobreather}}(x,t_0) = 16\arctan \left( \frac{m}{\sqrt{1-m^2}} \frac{\sin\left(\sqrt{1-m^2}t_0\right)}{\cosh(mx_1)} \right) \arctan \left( \frac{m}{\sqrt{1-m^2}} - \frac{\sin\left(\sqrt{1-m^2}t_0\right)}{\cosh(mx_2)} \right), + \frac{\sin\left(\sqrt{1-m^2}t_0\right)}{\cosh(mx_2)} \right), \f] where $x=(x_1,x_2)\in{R}^2$, $m=0.5<1$ as in the 1D case we discussed above. For the simulation shown below, we have chosen $\theta=\frac{1}{2}$, @@ -94,7 +94,7 @@ $k=10h$, $t_0=-5.4414$ and $t_f=2.7207$. The solution is pretty interesting --- it acts like a breather (as far as the pictures are concerned); however, it appears to break up and reassemble, rather than just oscillate. -\image html step-25.2d-pseudobreather.gif "Animation of a 2D pseudobreather." width=5cm +@image html step-25.2d-pseudobreather.gif "Animation of a 2D pseudobreather." width=5cm -- 2.39.5