From e70f00a5cd1627dd7ce4558ced5371fdb8689dc1 Mon Sep 17 00:00:00 2001 From: Lei Qiao Date: Wed, 4 Mar 2015 22:33:09 -0600 Subject: [PATCH] fix bracket mismatch in result part of step-33 --- examples/step-33/doc/results.dox | 8 ++++---- 1 file changed, 4 insertions(+), 4 deletions(-) diff --git a/examples/step-33/doc/results.dox b/examples/step-33/doc/results.dox index 74407af04e..6736cfd913 100644 --- a/examples/step-33/doc/results.dox +++ b/examples/step-33/doc/results.dox @@ -264,14 +264,14 @@ The residual calulated in ConservationLaw::assemble_cell_term function reads $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 $ + \theta \mathbf{B}({\mathbf{w}^{k}_{n+1}})(\mathbf{z}_i)_K + + (1-\theta) \mathbf{B}({\mathbf{w}_{n}}) (\mathbf{z}_i)_K $ This means that we calculate the spatial residual twice at one Newton -iteration step: once respect to the current solution ${\mathbf{w}^{k}_{n+1}$ +iteration step: once respect to the current solution $\mathbf{w}^{k}_{n+1}$ and once more respect to the last time step solution $\mathbf{w}_{n}$ which remains the same during all Newton iterations through one timestep. Cache up the explicit part of residual - $ \mathbf{B}({\mathbf{w}_{n}) (\mathbf{z}_i)_K}$ + $ \mathbf{B}({\mathbf{w}_{n}}) (\mathbf{z}_i)_K$ during Newton iteration will save lots of labor. -- 2.39.5