Correct minor typos

Check factors of 3 or 1/3 in friction definition

dual-problems.tex

@@ -362,15 +362,15 @@ \subsection{Slab of ice flowing into the ocean}
 	\includegraphics[width = 0.8\linewidth]{demos/slab/figures/marine_ice_sheet.pdf}
 	\caption{Setup for modelling a slab of ice on an inclined bed flowing into the ocean. At $x = 0$ we enforce a thickness $h = 500\,\mathrm{m}$ in order to approach a parallel slab of ice far upstream of the grounding line. The dotted line represents the sea level.}
 	\label{fig:domain_parallel_slab}
-\end{figure} 
+\end{figure}
 
-As a more challenging test in a flowline setting, we consider a slab of constant thickness ice flowing down an inclined slope and into the ocean, where it goes afloat at the grounding line, as illustrated in Figure \ref{fig:domain_parallel_slab}. We compute steady states under this configuration using the primal and dual methods described above. We set the ice thickness at $x = 0$ to $h = 500\, \mathrm{m}$ and the bedrock angle to $alpha = 0.5^\circ$. The bed is given by the expression
+As a more challenging test in a flowline setting, we consider a slab of ice of constant thickness flowing down an inclined slope into the ocean, where it goes afloat at the grounding line, as illustrated in Figure \ref{fig:domain_parallel_slab}. We compute steady states under this configuration using the primal and dual methods described above. We set the ice thickness at $x = 0$ to $h = 500\, \mathrm{m}$ and the bedrock angle to $\alpha = 0.5^\circ$. The bed is given by the expression
 %
 \begin{align}
 	b(x) = 1500 - x\tan{\alpha},
-\end{align} 
+\end{align}
 %
-with units in meters. Regarding the material parameters of the ice, we set $n = 3$, $A = 10^{-24}\,\mathrm{Pa}^{-3}\,\mathrm{s}^{-1}$, $C = 10^{-5}\,\mathrm{Pa}\,\mathrm{m}^{-3}\mathrm{s}^3$. For this problem, we do not only solve for the velocity $u$ or the velocity-stress pair $(u,M)$, but also for the thickness and the grounding line position. We therefore complement the momentum balance equations with the mass balance equation and the flotation condition, effectively yielding a free boundary problem. Far upstream from the grounding line, the ice adopts a constant thickness regime in which gravitational forces are balanced by basal stresses. As a result, the strain rate scalar tends to zero as $x$ approaches 0 and regularisation of the primal formulation becomes necessary. 
+with units in meters. For the material parameters, we set $n = 3$, $A = 10^{-24}\,\mathrm{Pa}^{-3}\,\mathrm{s}^{-1}$, $C = 10^{-5}\,\mathrm{Pa}\,\mathrm{m}^{-1/3}\mathrm{s}^{1/3}$. For this problem, we do not only solve for the velocity $u$ or the velocity-stress pair $(u,M)$, but also for the thickness and the grounding line position. We therefore complement the momentum balance equations with the mass balance equation and the flotation condition, effectively yielding a free boundary problem. Far upstream from the grounding line, the ice adopts a constant thickness regime in which gravitational forces are balanced by basal stresses. As a result, the strain rate scalar tends to zero as $x$ approaches 0 and regularisation of the primal formulation becomes necessary.
 
 We solve the free boundary problem with a primal method that seeks the velocity and thickness in $CG(1)\times CG(1)$, and with a dual method that computes the velocity, membrane stress and thickness in the space $CG(1)\times DG(0)\times CG(1)$. For the primal method, we consider a sequence of regularisation parameters $\epsilon$ between $10^{-4}\, \mathrm{yr}^{-1}$ and $10^{-16}\,\mathrm{yr}^{-1}$. The results for the grounding line position are displayed in Table \ref{tab:slab}. Additionally, we plot the iterations of the Newton residual in Figure \ref{fig:newton-its}. We can see that not only is the dual method just as accurate as the primal method solved using the lowest value of regularisation, but the Newton solver experiences a rate of convergence which is quickly lost for low values of $\epsilon$ when used with the primal method.
 
@@ -381,25 +381,25 @@ \subsection{Slab of ice flowing into the ocean}
 \label{tab:slab}
 \begin{tabular}{ccccc}
 \toprule
-solver & $\epsilon\,\mathrm{yr}^{-1}$ & $x_g\,\mathrm{km}$ & $h(x_g)\,\mathrm{m}$ & iterations \\ 
+solver & $\epsilon\,\mathrm{yr}^{-1}$ & $x_g\,\mathrm{km}$ & $h(x_g)\,\mathrm{m}$ & iterations \\
 \midrule
-\multirow{7}{*}{primal} & $10^{-4}$ & 189.08 & 163.64 & 7  \\ 
- & $10^{-6}$ & 203.19 & 297.90 & 6  \\ 
- & $10^{-8}$ & 214.86 & 408.98 & 7  \\ 
- & $10^{-10}$ & 215.19 & 412.18 & 9  \\ 
- & $10^{-12}$ & 215.20 & 412.22 & 19  \\ 
- & $10^{-14}$ & 215.20 & 412.22 & 17  \\ 
- & $10^{-16}$ & 215.20 & 412.22 & 24  \\ 
+\multirow{7}{*}{primal} & $10^{-4}$ & 189.08 & 163.64 & 7  \\
+ & $10^{-6}$ & 203.19 & 297.90 & 6  \\
+ & $10^{-8}$ & 214.86 & 408.98 & 7  \\
+ & $10^{-10}$ & 215.19 & 412.18 & 9  \\
+ & $10^{-12}$ & 215.20 & 412.22 & 19  \\
+ & $10^{-14}$ & 215.20 & 412.22 & 17  \\
+ & $10^{-16}$ & 215.20 & 412.22 & 24  \\
 \midrule
 dual & - & 215.20 & 412.22 & 7 \\
 \bottomrule
 \end{tabular}
-\end{table} 
+\end{table}
 
 \begin{figure}[t]
 	\centering
 	\includegraphics[width=\linewidth]{demos/slab/figures/newton_its_alpha0.50.pdf}
-	\caption{Results for the slab of ice flowing into the ocean. Norm of the (absolute) Newton residual for computations with the primal formulation with varying regularisation parameters $\epsilon$ and with the dual formulation..}
+	\caption{Results for the slab of ice flowing into the ocean. Norm of the (absolute) Newton residual for computations with the primal formulation with varying regularisation parameters $\epsilon$ and with the dual formulation.}
 	\label{fig:newton-its}
 \end{figure}