User:Lothar.brendel/belt-notes: Difference between revisions
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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 \(x_2<0<x_1<\vert x_2\vert~\Rightarrow~x^{n+1}=x_1\). | ||
== \(y(t)\) == | == \(y(t)\) == | ||
Revision as of 13:38, 17 May 2026
SEDNA: Ionization → Rate Equations
For equations (46) and (53) in 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 Sedna1.1/RateEquations.c:QuadraticEquation(A,B,C,&x1,&x2) 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. QuadraticEquation() yields \(x_2<0<x_1<\vert x_2\vert~\Rightarrow~x^{n+1}=x_1\).
\(y(t)\)
- \(0<b+c=A\propto\Delta t\)
- \(B=-1-a-b-2c<0\)
- \(C=y^n+a>0\)
\[
y^{n+1}=\frac{-B-\sqrt{\dots}}{2A}~,
\]
which again is the solution with smaller absolute value. QuadraticEquation() yields \(0<x_2<x_1~\Rightarrow~x^{n+1}=x_2\).