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Code Issues Projects Releases Packages Wiki Activity Actions 5

Fix equations in docstrings for a few depletion integrations

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This commit is contained in:
Paul Romano 2022-04-21 09:53:50 -05:00
parent 6959303317
commit 03ee3d6e77
1 changed files with 10 additions and 9 deletions
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19
openmc/deplete/integrators.py
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@ -124,13 +124,13 @@ class CF4Integrator(Integrator):
.. math::
\begin{aligned}
\mathbf{A}_1 &= h\mathbf{A}(\mathbf{n}_0) \\
\hat{\mathbf{n}}_1 &= \exp \left ( \frac{\mathbf{A}_1}{2} \right ) \\
\mathbf{A}_1 &= h\mathbf{A}(\mathbf{n}_i) \\
\hat{\mathbf{n}}_1 &= \exp \left ( \frac{\mathbf{A}_1}{2} \right ) \mathbf{n}_i \\
\mathbf{A}_2 &= h\mathbf{A}(\hat{\mathbf{n}}_1) \\
\hat{\mathbf{n}}_2 &= \exp \left ( \frac{\mathbf{A}_2}{2} \right ) \\
\hat{\mathbf{n}}_2 &= \exp \left ( \frac{\mathbf{A}_2}{2} \right ) \mathbf{n}_i \\
\mathbf{A}_3 &= h \mathbf{A}(\hat{\mathbf{n}}_2) \\
\hat{\mathbf{n}}_3 &= \exp \left ( -\frac{\mathbf{A}_1}{2} + \mathbf{A}_3
\right ) \\
\right ) \hat{\mathbf{n}}_1 \\
\mathbf{A}_4 &= h\mathbf{A}(\hat{\mathbf{n}}_3) \\
\mathbf{n}_{i+1} &= \exp \left ( \frac{\mathbf{A}_1}{4} + \frac{\mathbf{A}_2}{6}
+ \frac{\mathbf{A}_3}{6} - \frac{\mathbf{A}_4}{12} \right )
@ -207,7 +207,8 @@ class CELIIntegrator(Integrator):
.. math::
\begin{aligned}
\mathbf{n}_{i+1}^p &= \exp \left ( h \mathbf{A}(\mathbf{n}_i ) \right ) \\
\mathbf{n}_{i+1}^p &= \exp \left ( h \mathbf{A}(\mathbf{n}_i ) \right )
\mathbf{n}_i \\
\mathbf{n}_{i+1} &= \exp \left( \frac{h}{12} \mathbf{A}(\mathbf{n}_i) +
\frac{5h}{12} \mathbf{A}(\mathbf{n}_{i+1}^p) \right)
\exp \left( \frac{5h}{12} \mathbf{A}(\mathbf{n}_i) +
@ -268,12 +269,12 @@ class EPCRK4Integrator(Integrator):
.. math::
\begin{aligned}
\mathbf{A}_1 &= h\mathbf{A}(\mathbf{n}_0) \\
\hat{\mathbf{n}}_1 &= \exp \left ( \frac{\mathbf{A}_1}{2} \right ) \\
\mathbf{A}_1 &= h\mathbf{A}(\mathbf{n}_i) \\
\hat{\mathbf{n}}_1 &= \exp \left ( \frac{\mathbf{A}_1}{2} \right ) \mathbf{n}_i \\
\mathbf{A}_2 &= h\mathbf{A}(\hat{\mathbf{n}}_1) \\
\hat{\mathbf{n}}_2 &= \exp \left ( \frac{\mathbf{A}_2}{2} \right ) \\
\hat{\mathbf{n}}_2 &= \exp \left ( \frac{\mathbf{A}_2}{2} \right ) \mathbf{n}_i \\
\mathbf{A}_3 &= h \mathbf{A}(\hat{\mathbf{n}}_2) \\
\hat{\mathbf{n}}_3 &= \exp \left ( \mathbf{A}_3 \right ) \\
\hat{\mathbf{n}}_3 &= \exp \left ( \mathbf{A}_3 \right ) \mathbf{n}_i \\
\mathbf{A}_4 &= h\mathbf{A}(\hat{\mathbf{n}}_3) \\
\mathbf{n}_{i+1} &= \exp \left ( \frac{\mathbf{A}_1}{6} + \frac{\mathbf{A}_2}{3}
+ \frac{\mathbf{A}_3}{3} + \frac{\mathbf{A}_4}{6} \right ) \mathbf{n}_i.