User:Lothar.brendel/belt-notes - Revision history

Lothar.brendel: /* SEDNA: Ionization → Rate Equations */ +unstable y=1

2026-05-18T17:45:04Z

SEDNA: Ionization → Rate Equations: +unstable y=1

← Older revision		Revision as of 19:45, 18 May 2026
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	x^{n+1}=\frac{-B+\sqrt{\dots}}{2A}~,		x^{n+1}=\frac{-B+\sqrt{\dots}}{2A}~,
	\]		\]
	which is the solution with smaller absolute value. <code>QuadraticEquation()</code> yields \(x_2<0<x_1<\vert x_2\vert~\Rightarrow~x^{n+1}=x_1\).		which is the solution with smaller absolute value. <code>QuadraticEquation()</code> yields \(x_2<0\le x_1<\vert x_2\vert~\Rightarrow~x^{n+1}=x_1\).

	== \(y(t)\) ==		== \(y(t)\) ==
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	y^{n+1}=\frac{-B-\sqrt{\dots}}{2A}~,		y^{n+1}=\frac{-B-\sqrt{\dots}}{2A}~,
	\]		\]
	which again is the solution of smaller absolute value. <code>QuadraticEquation()</code> yields \(0<x_2<x_1~\Rightarrow~y^{n+1}=x_2\).		which again is the solution of smaller absolute value. <code>QuadraticEquation()</code> yields \(0\le x_2<x_1~\Rightarrow~y^{n+1}=x_2\).

			== Caveats ==

			Without radiation (\(a=0\)), \(1-x=y=1\) is a stationary solution, but it's unstable. Due to round off errors, \(y^n=1\Rightarrow y^{n+1}=1-\epsilon\), triggering the instability. On the other hand, \(x^n=1\Rightarrow x^{n+1}=1\) due to \(C=0\). Hence, in opaque regions, an \(x/y\)-inconsistency will we reported.

Lothar.brendel: /* SEDNA: Ionization → Rate Equations */ B/A

2026-05-18T06:05:27Z

SEDNA: Ionization → Rate Equations: B/A

← Older revision		Revision as of 08:05, 18 May 2026
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	The rate equations (44) and (45) in [https://ui.adsabs.harvard.edu/abs/2020ApJS..250...13K/abstract Kuiper etal 2020] are non-linear, namely quadratic in \(x\) and \(y\), respectively. Solving them with an implicit Euler scheme due to stability reasons (the rate equations seemingly are not considerd in the \(\Delta t\) control), yields quadratic equations, (46) and (53), for the values at \(t+\Delta t\). For each, the correct solution must be chosen, namely the one for which \(x^{n+1}-x^n=\mathcal O(\Delta t)\).		The rate equations (44) and (45) in [https://ui.adsabs.harvard.edu/abs/2020ApJS..250...13K/abstract Kuiper etal 2020] are non-linear, namely quadratic in \(x\) and \(y\), respectively. Solving them with an implicit Euler scheme due to stability reasons (the rate equations seemingly are not considerd in the \(\Delta t\) control), yields quadratic equations, (46) and (53), for the values at \(t+\Delta t\). For each, the correct solution must be chosen, namely the one for which \(x^{n+1}-x^n=\mathcal O(\Delta t)\).

	For \(B>0\), the function <code>Sedna1.1/RateEquations.c:QuadraticEquation(A,B,C,&x1,&x2)</code> assigns the solution with larger absolute value, which happens to be negative, to \(x_2\). For \(B<0\), the solution with larger absolute value is positive and is assigned to \(x_1\). The sign of the other solution, with smaller absolute value, depends on the sign of \(C\).		For \(B/A>0\), the function <code>Sedna1.1/RateEquations.c:QuadraticEquation(A,B,C,&x1,&x2)</code> assigns the solution with larger absolute value, which happens to be negative, to \(x_2\). For \(B/A<0\), the solution with larger absolute value is positive and is assigned to \(x_1\). The sign of the other solution, with smaller absolute value, depends on the sign of \(C\).

	== \(x(t)\) ==		== \(x(t)\) ==

Lothar.brendel: x->y

2026-05-17T13:16:15Z

x->y

← Older revision		Revision as of 15:16, 17 May 2026
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	* \(0<b+c=A\propto\Delta t\)		* \(0<b+c=A\propto\Delta t\)
	* \(B=1+a-b>0\) for not too large \(\Delta t\)		* \(B=1+a-b>0\) (for not too large \(\Delta t\)!)
	* \(C=-(x^n+a)<0\)		* \(C=-(x^n+a)<0\)

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	y^{n+1}=\frac{-B-\sqrt{\dots}}{2A}~,		y^{n+1}=\frac{-B-\sqrt{\dots}}{2A}~,
	\]		\]
	which again is the solution of smaller absolute value. <code>QuadraticEquation()</code> yields \(0<x_2<x_1~\Rightarrow~x^{n+1}=x_2\).		which again is the solution of smaller absolute value. <code>QuadraticEquation()</code> yields \(0<x_2<x_1~\Rightarrow~y^{n+1}=x_2\).

Lothar.brendel: /* SEDNA: Ionization → Rate Equations */ imp. Euler

2026-05-17T13:14:03Z

SEDNA: Ionization → Rate Equations: imp. Euler

← Older revision		Revision as of 15:14, 17 May 2026
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	= SEDNA: Ionization → Rate Equations =		= SEDNA: Ionization → Rate Equations =

	~~For~~ equations (46) and (53) in [https://ui.adsabs.harvard.edu/abs/2020ApJS..250...13K/abstract Kuiper etal 2020] , the correct solution must be chosen, namely the one for which \(x^{n+1}-x^n=\mathcal O(\Delta t)\).		The rate equations (44) and (45) in [https://ui.adsabs.harvard.edu/abs/2020ApJS..250...13K/abstract Kuiper etal 2020] are non-linear, namely quadratic in \(x\) and \(y\), respectively. Solving them with an implicit Euler scheme due to stability reasons (the rate equations seemingly are not considerd in the \(\Delta t\) control), yields quadratic equations, (46) and (53), for the values at \(t+\Delta t\). For each, the correct solution must be chosen, namely the one for which \(x^{n+1}-x^n=\mathcal O(\Delta t)\).

	For \(B>0\), the function <code>Sedna1.1/RateEquations.c:QuadraticEquation(A,B,C,&x1,&x2)</code> assigns the solution with larger absolute value, which happens to be negative, to \(x_2\). For \(B<0\), the solution with larger absolute value is positive and is assigned to \(x_1\). The sign of the other solution, with smaller absolute value, depends on the sign of \(C\).		For \(B>0\), the function <code>Sedna1.1/RateEquations.c:QuadraticEquation(A,B,C,&x1,&x2)</code> assigns the solution with larger absolute value, which happens to be negative, to \(x_2\). For \(B<0\), the solution with larger absolute value is positive and is assigned to \(x_1\). The sign of the other solution, with smaller absolute value, depends on the sign of \(C\).
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	y^{n+1}=\frac{-B-\sqrt{\dots}}{2A}~,		y^{n+1}=\frac{-B-\sqrt{\dots}}{2A}~,
	\]		\]
	which again is the solution ~~with~~ smaller absolute value. <code>QuadraticEquation()</code> yields \(0<x_2<x_1~\Rightarrow~x^{n+1}=x_2\).		which again is the solution of smaller absolute value. <code>QuadraticEquation()</code> yields \(0<x_2<x_1~\Rightarrow~x^{n+1}=x_2\).

Lothar.brendel: /* \(x(t)\) */ -redundancy

2026-05-17T11:38:43Z

\(x(t)\): -redundancy

← Older revision		Revision as of 13:38, 17 May 2026
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	\]		\]
	which is the solution with smaller absolute value. <code>QuadraticEquation()</code> yields \(x_2<0<x_1<\vert x_2\vert~\Rightarrow~x^{n+1}=x_1\).		which is the solution with smaller absolute value. <code>QuadraticEquation()</code> yields \(x_2<0<x_1<\vert x_2\vert~\Rightarrow~x^{n+1}=x_1\).

	~~Hence, \(x_1\) from <code>QuadraticEquation()</code> is used.~~

	== \(y(t)\) ==		== \(y(t)\) ==

Lothar.brendel: new

2026-05-17T11:38:18Z

new

New page

= SEDNA: Ionization → Rate Equations =

For equations (46) and (53) in [https://ui.adsabs.harvard.edu/abs/2020ApJS..250...13K/abstract Kuiper etal 2020] , the correct solution must be chosen, namely the one for which \(x^{n+1}-x^n=\mathcal O(\Delta t)\).

For \(B>0\), the function <code>Sedna1.1/RateEquations.c:QuadraticEquation(A,B,C,&x1,&x2)</code> assigns the solution with larger absolute value, which happens to be negative, to \(x_2\). For \(B<0\), the solution with larger absolute value is positive and is assigned to \(x_1\). The sign of the other solution, with smaller absolute value, depends on the sign of \(C\).

== \(x(t)\) ==

* \(0<b+c=A\propto\Delta t\)
* \(B=1+a-b>0\) for not too large \(\Delta t\)
* \(C=-(x^n+a)<0\)

\[
x^{n+1}=\frac{-B+\sqrt{\dots}}{2A}~,
\]
which is the solution with smaller absolute value. <code>QuadraticEquation()</code> yields \(x_2<0<x_1<\vert x_2\vert~\Rightarrow~x^{n+1}=x_1\).

Hence, \(x_1\) from <code>QuadraticEquation()</code> is used.

== \(y(t)\) ==

* \(0<b+c=A\propto\Delta t\)
* \(B=-1-a-b-2c<0\)
* \(C=y^n+a>0\)

\[
y^{n+1}=\frac{-B-\sqrt{\dots}}{2A}~,
\]
which again is the solution with smaller absolute value. <code>QuadraticEquation()</code> yields \(0<x_2<x_1~\Rightarrow~x^{n+1}=x_2\).