fixed upwinding, tweaked loss notation
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diffphys.md
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diffphys.md
@ -238,8 +238,8 @@ The $\frac{ \partial L }{ \partial d}$ component is typically simple enough: we'
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$$
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\frac{ \partial L }{ \partial d}
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= \partial | \mathcal P ( d^{~0}, \mathbf{u}, t^e) - d^{\text{target}}|^2 / \partial d
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= 2 (d(t^e)-d^{\text{target}}).
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= \frac{ \partial | \mathcal P ( d^{~0}, \mathbf{u}, t^e) - d^{\text{target}}|^2 }{ \partial d }
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= 2 \big( d(t^e)-d^{\text{target}} \big).
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$$
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If $d$ is represented as a vector, e.g., for one entry per cell of a mesh,
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@ -284,8 +284,8 @@ gives the following:
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$$ \begin{aligned}
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& d_i(t+\Delta t) = d_i - u_i^+ (d_{i+1} - d_{i}) + u_i^- (d_{i} - d_{i-1}) \text{ with } \\
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& u_i^+ = \text{max}(u_i \Delta t / \Delta x,0) \\
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& u_i^- = \text{min}(u_i \Delta t / \Delta x,0)
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& u_i^+ = \text{min}(u_i \Delta t / \Delta x,0) \\
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& u_i^- = \text{max}(u_i \Delta t / \Delta x,0)
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\end{aligned} $$
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```{figure} resources/diffphys-advect1d.jpg
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@ -296,7 +296,8 @@ name: advection-upwind
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1st-order upwinding uses a simple one-sided finite-difference stencil that takes into account the direction of the motion
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```
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Thus, for a positive $u_i$ we have
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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 $u_i^-$.
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To pick the former case, for a negative $u_i$ we get
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$$
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\mathcal P ( d_i(t), \mathbf{u}(t), t + \Delta t) = (1 + \frac{u_i \Delta t }{ \Delta x}) d_i - \frac{u_i \Delta t }{ \Delta x} d_{i+1}
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@ -357,16 +358,16 @@ will give us a contribution to $\Delta \mathbf{u}$ which we can accumulate for a
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$$ \begin{aligned}
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\Delta \mathbf{u} =&
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\frac{ \partial d(t^e) }{ \partial \mathbf{u} }
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\frac{ \partial L }{ \partial d(t^e) }
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+
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\frac{ \partial L }{ \partial d(t^e) } \\
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& + \
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\frac{ \partial d(t^e - \Delta t) }{ \partial \mathbf{u}}
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\frac{ \partial d(t^e) }{ \partial d(t^e - \Delta t) }
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\frac{ \partial L }{ \partial d(t^e)}
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\\
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&
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+ \ \cdots \ + \\
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+ \ \cdots \ \\
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&
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\Big( \frac{ \partial d(t^0) }{ \partial \mathbf{u}} \cdots
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+ \ \Big( \frac{ \partial d(t^0) }{ \partial \mathbf{u}} \cdots
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\frac{ \partial d(t^e - \Delta t) }{ \partial d(t^e - 2 \Delta t) }
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\frac{ \partial d(t^e) }{ \partial d(t^e - \Delta t) }
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\frac{ \partial L }{ \partial d(t^e)} \Big)
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