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fixed upwinding, tweaked loss notation

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2021-11-03 14:23:09 +01:00
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commit d4507a0847
1 changed files with 10 additions and 9 deletions
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diffphys.md
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@@ -238,8 +238,8 @@ The $\frac{ \partial L }{ \partial d}$ component is typically simple enough: we'
$$ $$
\frac{ \partial L }{ \partial d} \frac{ \partial L }{ \partial d}
= \partial | \mathcal P ( d^{~0}, \mathbf{u}, t^e) - d^{\text{target}}|^2 / \partial d = \frac{ \partial | \mathcal P ( d^{~0}, \mathbf{u}, t^e) - d^{\text{target}}|^2 }{ \partial d }
= 2 (d(t^e)-d^{\text{target}}). = 2 \big( d(t^e)-d^{\text{target}} \big).
$$ $$
If $d$ is represented as a vector, e.g., for one entry per cell of a mesh, If $d$ is represented as a vector, e.g., for one entry per cell of a mesh,
@@ -284,8 +284,8 @@ gives the following:
$$ \begin{aligned} $$ \begin{aligned}
& d_i(t+\Delta t) = d_i - u_i^+ (d_{i+1} - d_{i}) + u_i^- (d_{i} - d_{i-1}) \text{ with } \\ & d_i(t+\Delta t) = d_i - u_i^+ (d_{i+1} - d_{i}) + u_i^- (d_{i} - d_{i-1}) \text{ with } \\
& u_i^+ = \text{max}(u_i \Delta t / \Delta x,0) \\ & u_i^+ = \text{min}(u_i \Delta t / \Delta x,0) \\
& u_i^- = \text{min}(u_i \Delta t / \Delta x,0) & u_i^- = \text{max}(u_i \Delta t / \Delta x,0)
\end{aligned} $$ \end{aligned} $$
```{figure} resources/diffphys-advect1d.jpg ```{figure} resources/diffphys-advect1d.jpg
@@ -296,7 +296,8 @@ name: advection-upwind
1st-order upwinding uses a simple one-sided finite-difference stencil that takes into account the direction of the motion 1st-order upwinding uses a simple one-sided finite-difference stencil that takes into account the direction of the motion
``` ```
Thus, for a positive $u_i$ we have Thus, for a negative $u_i$, we're using $u_i^+$ to look in the opposite direction of the velocity, i.e., _backward_ in terms of the motion. $u_i^-$ will be zero in this case. For positive $u_i$ it's vice versa, and we'll get a zero'ed $u_i^+$, and a backward difference stencil via $u_i^-$.
To pick the former case, for a negative $u_i$ we get
$$ $$
\mathcal P ( d_i(t), \mathbf{u}(t), t + \Delta t) = (1 + \frac{u_i \Delta t }{ \Delta x}) d_i - \frac{u_i \Delta t }{ \Delta x} d_{i+1} \mathcal P ( d_i(t), \mathbf{u}(t), t + \Delta t) = (1 + \frac{u_i \Delta t }{ \Delta x}) d_i - \frac{u_i \Delta t }{ \Delta x} d_{i+1}
@@ -357,16 +358,16 @@ will give us a contribution to $\Delta \mathbf{u}$ which we can accumulate for a
$$ \begin{aligned} $$ \begin{aligned}
\Delta \mathbf{u} =& \Delta \mathbf{u} =&
\frac{ \partial d(t^e) }{ \partial \mathbf{u} } \frac{ \partial d(t^e) }{ \partial \mathbf{u} }
\frac{ \partial L }{ \partial d(t^e) } \frac{ \partial L }{ \partial d(t^e) } \\
+ & + \
\frac{ \partial d(t^e - \Delta t) }{ \partial \mathbf{u}} \frac{ \partial d(t^e - \Delta t) }{ \partial \mathbf{u}}
\frac{ \partial d(t^e) }{ \partial d(t^e - \Delta t) } \frac{ \partial d(t^e) }{ \partial d(t^e - \Delta t) }
\frac{ \partial L }{ \partial d(t^e)} \frac{ \partial L }{ \partial d(t^e)}
\\ \\
& &
+ \ \cdots \ + \\ + \ \cdots \ \\
& &
\Big( \frac{ \partial d(t^0) }{ \partial \mathbf{u}} \cdots + \ \Big( \frac{ \partial d(t^0) }{ \partial \mathbf{u}} \cdots
\frac{ \partial d(t^e - \Delta t) }{ \partial d(t^e - 2 \Delta t) } \frac{ \partial d(t^e - \Delta t) }{ \partial d(t^e - 2 \Delta t) }
\frac{ \partial d(t^e) }{ \partial d(t^e - \Delta t) } \frac{ \partial d(t^e) }{ \partial d(t^e - \Delta t) }
\frac{ \partial L }{ \partial d(t^e)} \Big) \frac{ \partial L }{ \partial d(t^e)} \Big)