From cddaf64b90493008f2fa82f2916a702f21aa00f2 Mon Sep 17 00:00:00 2001 From: Timo Heister Date: Wed, 19 Jun 2024 18:06:04 -0400 Subject: [PATCH] fix some doxygen formulas FYI @bangerth --- include/deal.II/sundials/ida.h | 30 +++++++++++++++--------------- 1 file changed, 15 insertions(+), 15 deletions(-) diff --git a/include/deal.II/sundials/ida.h b/include/deal.II/sundials/ida.h index 592f25646c..059458c231 100644 --- a/include/deal.II/sundials/ida.h +++ b/include/deal.II/sundials/ida.h @@ -235,11 +235,11 @@ namespace SUNDIALS * A more interesting example is a situation where the form $F(y', y, t) = 0$ * provides something genuinely more flexible than a typical ordinary * differential equation. Specifically, consider the equation - * @f[ + * @f{align*}{ * u'(t) &= av(t), * \\ * 0 &= v(t) - u(t). - * @f] + * @f} * One can combine the two variables into $y(t) = [u(t), v(t)]^T$. * Here, one of the two variables does not have a time derivative. In * applications, this is often the case when one variable evolves in @@ -249,19 +249,19 @@ namespace SUNDIALS * using the second equation, and would then just be left with the * equation * @f[ - * u'(t) &= au(t) + * u'(t) = au(t) * @f] * which has solution $u(t) = u(0)e^{at}$. But this is, in general, not * easily possible if the two variables are related by differential * operators. In fact, this happens quite frequently in application. Take, * for example, the time-dependent Stokes equations: - * @f[ + * @f{align*}{ * \frac{\partial \mathbf u(\mathbf x,t)}{\partial t} * - \nu \Delta \mathbf u(\mathbf x,t) + \nabla p(\mathbf x,t) * &= \mathbf f(\mathbf x,t), * \\ * \nabla \cdot \mathbf u(\mathbf x,t) &= 0. - * @f] + * @f} * Here, the fluid velocity $\mathbf u(\mathbf x,t)$ evolves over time, * and the pressure is always in equilibrium with the flow because the Stokes * equations are derived under the assumption that the speed of sound (at @@ -276,16 +276,16 @@ namespace SUNDIALS * Rather than show how to solve the trivial (linear) case above, let us * instead consider the situation where we introduce another variable $v$ that * is related to $u$ by the nonlinear relationship $v=u^p$, $p\ge 1$: - * @f[ + * @f{align*}{ * u'(t) &= a v(t)^{1/p}, * \\ * 0 &= v(t) - u(t)^p. - * @f] + * @f} * We will impose initial conditions as - * @f[ + * @f{align*}{ * u(0) &= 1 \\ * v(0) &= 1. - * @f] + * @f} * The problem continues to have the solution $u(t)=e^{at}$ with the * auxiliary variable satisfying $v(t)=[e^{at}]^p$. One would implement * all of this using the following little program where you have to recall @@ -367,13 +367,13 @@ namespace SUNDIALS * initial values in the form of a vector for $y(0)$, but also for * $y'(0)$, this is not a common situation. For example, for the Stokes * equations mentioned above, - * @f[ + * @f{align*}{ * \frac{\partial \mathbf u(\mathbf x,t)}{\partial t} * - \nu \Delta \mathbf u(\mathbf x,t) + \nabla p(\mathbf x,t) * &= \mathbf f(\mathbf x,t), * \\ * \nabla \cdot \mathbf u(\mathbf x,t) &= 0, - * @f] + * @f} * one generally might have an initial velocity field for * $\mathbf u(\mathbf x,0)$, but typically one does not have an initial * pressure field $p(\mathbf x,0)$ nor either of these variables' time @@ -382,17 +382,17 @@ namespace SUNDIALS * Fortunately, they can typically be computed via the relationship * $F(t,y,\dot y) = 0$. To illustrate how this can is done, let us * re-use the nonlinear example from the previous section: - * @f[ + * @f{align*}{ * u'(t) &= a v(t)^{1/p}, * \\ * 0 &= v(t) - u(t)^p. - * @f] + * @f} * If we now impose initial conditions for both variables, for * example - * @f[ + * @f{align*}{ * u(0) &= 1 \\ * v(0) &= 1, - * @f] + * @f} * then the only change necessary is to create the time stepper via * @code * SUNDIALS::IDA::AdditionalData data; -- 2.39.5