From: Lei Qiao Date: Thu, 5 Mar 2015 17:46:59 +0000 (-0600) Subject: fix mathbf and braces issues X-Git-Tag: v8.3.0-rc1~385^2~1 X-Git-Url: https://gitweb.dealii.org/cgi-bin/gitweb.cgi?a=commitdiff_plain;h=653bf1023e624eb294e197ff054eca9ec9ac422a;p=dealii.git fix mathbf and braces issues --- diff --git a/examples/step-33/doc/intro.dox b/examples/step-33/doc/intro.dox index d0abcef3ad..54c511e38f 100644 --- a/examples/step-33/doc/intro.dox +++ b/examples/step-33/doc/intro.dox @@ -131,16 +131,14 @@ We use a time stepping scheme to substitute the time derivative in the above equations. For simplicity, we define $ \mathbf{B}({\mathbf{w}_{n}})(\mathbf z) $ as the spatial residual at time step $n$ : @f{eqnarray*} - \mathbf{B}({\mathbf{w}_{n})(\mathbf z) &=& -- \int_{\Omega} \left(\mathbf{F}(\mathbf{w_n}), -\nabla\mathbf{z}\right) + h^{\eta}(\nabla \mathbf{w_n} , \nabla \mathbf{z}) -\\ -&& -+ -\int_{\partial \Omega} \left(\mathbf{H}(\mathbf{w_n}^+, -\mathbf{w}^-(\mathbf{w_n}^+), \mathbf{n}), \mathbf{z}\right) + \mathbf{B}(\mathbf{w}_{n})(\mathbf z) &=& +- \int_{\Omega} \left(\mathbf{F}(\mathbf{w}_n), +\nabla\mathbf{z}\right) + h^{\eta}(\nabla \mathbf{w}_n , \nabla \mathbf{z}) \\ +&& + +\int_{\partial \Omega} \left(\mathbf{H}(\mathbf{w}_n^+, +\mathbf{w}^-(\mathbf{w}_n^+), \mathbf{n}), \mathbf{z}\right) - -\int_{\partial \Omega} \left(\mathbf{G}(\mathbf{w_n}), +\int_{\partial \Omega} \left(\mathbf{G}(\mathbf{w}_n), \mathbf{z}\right) . @f} @@ -149,10 +147,10 @@ that the residual applied to any test function $\mathbf z$ equals zero: @f{eqnarray*} R(\mathbf{W}_{n+1})(\mathbf z) &=& -\int_{\Omega} \left(\frac{\mathbf{w}_{n+1} - \mathbf{w}_n}{\delta t}, +\int_{\Omega} \left(\frac{{\mathbf w}_{n+1} - \mathbf{w}_n}{\delta t}, \mathbf{z}\right)+ -\theta \mathbf{B}({\mathbf{w}_{n+1}) + (1-\theta) \mathbf{B}({\mathbf{w}_{n}) \\ -& = & 0 +\theta \mathbf{B}({\mathbf{w}}_{n+1}) + (1-\theta) \mathbf{B}({\mathbf w}_{n}) \\ +&=& 0 @f} where $ \theta \in [0,1] $ and $\mathbf{w}_i = \sum_k \mathbf{W}_i^k \mathbf{\phi}_k$. Choosing diff --git a/examples/step-33/step-33.cc b/examples/step-33/step-33.cc index bb6ed19cd8..3e466a14ec 100644 --- a/examples/step-33/step-33.cc +++ b/examples/step-33/step-33.cc @@ -1621,13 +1621,13 @@ namespace Step33 // residual read // $R_i = \left(\frac{\mathbf{w}^{k}_{n+1} - \mathbf{w}_n}{\delta t} , // \mathbf{z}_i \right)_K $ $ + - // \theta \mathbf{B}({\mathbf{w}^{k}_{n+1})(\mathbf{z}_i)_K $ $ + - // (1-\theta) \mathbf{B}({\mathbf{w}_{n}) (\mathbf{z}_i)_K $ where - // $\mathbf{B}({\mathbf{w})(\mathbf{z}_i)_K = + // \theta \mathbf{B}(\mathbf{w}^{k}_{n+1})(\mathbf{z}_i)_K $ $ + + // (1-\theta) \mathbf{B}(\mathbf{w}_{n}) (\mathbf{z}_i)_K $ where + // $\mathbf{B}(\mathbf{w})(\mathbf{z}_i)_K = // - \left(\mathbf{F}(\mathbf{w}),\nabla\mathbf{z}_i\right)_K $ $ // + h^{\eta}(\nabla \mathbf{w} , \nabla \mathbf{z}_i)_K $ $ // - (\mathbf{G}(\mathbf {w}), \mathbf{z}_i)_K $ for both - // ${\mathbf{w} = \mathbf{w}^k_{n+1}$ and ${\mathbf{w} = \mathbf{w}_{n}}$ , + // $\mathbf{w} = \mathbf{w}^k_{n+1}$ and $\mathbf{w} = \mathbf{w}_{n}$ , // $\mathbf{z}_i$ is the $i$th vector valued test function. // Furthermore, the scalar product // $\left(\mathbf{F}(\mathbf{w}), \nabla\mathbf{z}_i\right)_K$ is @@ -1763,8 +1763,8 @@ namespace Step33 // Next, in order to compute the cell contributions, we need to evaluate - // $F({\mathbf w}^k_{n+1})$, $G({\mathbf w}^k_{n+1})$ and - // $F({\mathbf w}_n)$, $G({\mathbf w}_n)$ at all quadrature + // $\mathbf{F}({\mathbf w}^k_{n+1})$, $\mathbf{G}({\mathbf w}^k_{n+1})$ and + // $\mathbf{F}({\mathbf w}_n)$, $\mathbf{G}({\mathbf w}_n)$ at all quadrature // points. To store these, we also need to allocate a bit of memory. Note // that we compute the flux matrices and right hand sides in terms of // autodifferentiation variables, so that the Jacobian contributions can @@ -1808,16 +1808,16 @@ namespace Step33 // \mathbf{w}_n)_{\text{component\_i}}}{\delta // t},(\mathbf{z}_i)_{\text{component\_i}}\right)_K // \\ &-& \sum_{d=1}^{\text{dim}} \left( \theta \mathbf{F} - // ({\mathbf{w^k_{n+1}}})_{\text{component\_i},d} + (1-\theta) - // \mathbf{F} ({\mathbf{w_{n}}})_{\text{component\_i},d} , + // ({\mathbf{w}^k_{n+1}})_{\text{component\_i},d} + (1-\theta) + // \mathbf{F} ({\mathbf{w}_{n}})_{\text{component\_i},d} , // \frac{\partial(\mathbf{z}_i)_{\text{component\_i}}} {\partial // x_d}\right)_K // \\ &+& \sum_{d=1}^{\text{dim}} h^{\eta} \left( \theta \frac{\partial - // \mathbf{w^k_{n+1}}_{\text{component\_i}}}{\partial x_d} + (1-\theta) - // \frac{\partial \mathbf{w_n}_{\text{component\_i}}}{\partial x_d} , + // (\mathbf{w}^k_{n+1})_{\text{component\_i}}}{\partial x_d} + (1-\theta) + // \frac{\partial (\mathbf{w}_n)_{\text{component\_i}}}{\partial x_d} , // \frac{\partial (\mathbf{z}_i)_{\text{component\_i}}}{\partial x_d} \right)_K // \\ &-& \left( \theta\mathbf{G}({\mathbf{w}^k_n+1} )_{\text{component\_i}} + - // (1-\theta)\mathbf{G}({\mathbf{w}_n} )_{\text{component\_i}} , + // (1-\theta)\mathbf{G}({\mathbf{w}_n})_{\text{component\_i}} , // (\mathbf{z}_i)_{\text{component\_i}} \right)_K , // @f} // where integrals are @@ -1971,7 +1971,7 @@ namespace Step33 } } // On the other hand, if this is an external boundary face, then the - // values of $W^-$ will be either functions of $W^+$, or they will be + // values of $\mathbf{W}^-$ will be either functions of $\mathbf{W}^+$, or they will be // prescribed, depending on the kind of boundary condition imposed here. // // To start the evaluation, let us ensure that the boundary id specified