Skip to content
Permalink

Comparing changes

Choose two branches to see what’s changed or to start a new pull request. If you need to, you can also or learn more about diff comparisons.

Open a pull request

Create a new pull request by comparing changes across two branches. If you need to, you can also . Learn more about diff comparisons here.
base repository: mockingbirdnest/Principia
Failed to load repositories. Confirm that selected base ref is valid, then try again.
Loading
base: 74317597323e
Choose a base ref
...
head repository: mockingbirdnest/Principia
Failed to load repositories. Confirm that selected head ref is valid, then try again.
Loading
compare: 982274ef38cb
Choose a head ref
  • 12 commits
  • 2 files changed
  • 1 contributor

Commits on Jan 15, 2019

  1. Copy the full SHA
    8ad528e View commit details

Commits on Jan 16, 2019

  1. Copy the full SHA
    0f19491 View commit details
  2. Full proof of the lemma.

    pleroy committed Jan 16, 2019
    Copy the full SHA
    a777af8 View commit details

Commits on Jan 17, 2019

  1. Copy the full SHA
    c7f50c1 View commit details

Commits on Jan 19, 2019

  1. A bogus proof.

    pleroy committed Jan 19, 2019
    Copy the full SHA
    483e00d View commit details

Commits on Jan 20, 2019

  1. Proof with epsilons.

    pleroy committed Jan 20, 2019
    Copy the full SHA
    021f4c8 View commit details
  2. Full proof.

    pleroy committed Jan 20, 2019
    Copy the full SHA
    f6530c0 View commit details
  3. First round of review.

    pleroy committed Jan 20, 2019
    Copy the full SHA
    c44cb35 View commit details

Commits on Jan 22, 2019

  1. Second round of review.

    pleroy committed Jan 22, 2019
    Copy the full SHA
    75dc70b View commit details

Commits on Jan 26, 2019

  1. Another round of review.

    # Conflicts:
    #	documentation/Geopotential.pdf
    pleroy committed Jan 26, 2019
    Copy the full SHA
    34d5ecf View commit details
  2. Nitpicking.

    pleroy committed Jan 26, 2019
    Copy the full SHA
    be9c076 View commit details
  3. Merge pull request #2065 from pleroy/Lemma

    Proof of the damping condition
    pleroy authored Jan 26, 2019
    Copy the full SHA
    982274e View commit details
Showing with 161 additions and 5 deletions.
  1. BIN documentation/Geopotential.pdf
  2. +161 −5 documentation/Geopotential.tex
Binary file modified documentation/Geopotential.pdf
Binary file not shown.
166 changes: 161 additions & 5 deletions documentation/Geopotential.tex
Original file line number Diff line number Diff line change
@@ -18,7 +18,7 @@
Geopotential%
}%
}
\author{Pascal~Leroy (phl)}
\author{Pascal~Leroy (phl) and Robin~Leroy (egg)}
\begin{document}
\maketitle
\begin{sloppypar}
@@ -376,13 +376,169 @@ \subsection*{Sigmoid}
have constant sign for $r \in \intclos {s_0} {s_1}$.

Removing those factors that do not depend on $r$, this means that we must have
\begin{equation}
\forall r \in \intopen{s_0} {s_1}, \deriv[k+1] r {\gs\of r r^{-\pa{n+1}}} \neq 0\label{eqnderiv}\text.
\end{equation}

\begin{proposition}
The smallest value of $s_1>1.183s_0$ such that (\ref{eqnderiv}) holds for all $k>0$ and $n≥2$ is $s_1 = 3s_0$.
\begin{proof}
We start by writing:
\[
\forall r \in \intopen{s_0} {s_1}, \deriv[k+1] r {\gs\of r r^{-\pa{n+1}}} \neq 0\text.
\gs\of{r} = \sum{m = 0}[3] a_m r^m
\]

\begin{lemma}
The smallest value of $s_1$ such that the above property holds for all $k$ and $n$ is $s_1 = 3s_0$.
where:
\[
\begin{dcases}
a_0 &= \frac{\pa{s_1-3s_0}{s_1}^2}{\pa{s_1-s_0}^3} \\
a_1 &= \frac{6s_0 s_1}{\pa{s_1-s_0}^3} \\
a_2 &= \frac{-3\pa{s_0+s_1}}{\pa{s_1-s_0}^3} \\
a_3 &= \frac{2}{\pa{s_1-s_0}^3}
\end{dcases}
\]
The function being derived in (\ref{eqnderiv}) is:
\[
\gs\of{r} r^{-\pa{n+1}} = \sum{m = 0}[3] a_m r^{-\pa{n-m+1}}
\]
The p-th derivative of $r^{-q}$ for $q > 0$ is:
\[
\deriv[p] r {r^{-q}} = \pa{-1}^p \frac{\Factorial{\pa{q+p-1}}}{\Factorial{\pa{q-1}}} r^{-\pa{q+p}}
\]
Whence:
\begin{align*}
\deriv[k+1] r {\gs\of r r^{-\pa{n+1}}} &= \pa{-1}^{\pa{k+1}} \sum{m = 0}[3] a_m \frac{\Factorial{\pa{n+k-m+1}}}{\Factorial{\pa{n-m}}} r^{-\pa{n+k-m+2}} \\
&= \pa{-1}^{\pa{k+1}} r^{-\pa{n+k+2}} \frac{\Factorial{\pa{n+k-2}}}{\Factorial{n}}
\begin{aligned}[t]
&\;\Bigl[a_0 \pa{n+k+1}\pa{n+k}\pa{n+k-1} \\
&+a_1 \pa{n+k}\pa{n+k-1} n r \\
&+a_2 \pa{n+k-1} n \pa{n-1} r^2 \\
&+a_3 n \pa{n-1}\pa{n-2} r^3\Bigr]
\end{aligned}
\end{align*}
Substituting the values of the $\pa{a_i}$, we see that the zeroes of $\deriv[k+1] r {\gs\of r r^{-\pa{n+1}}}$ are those of the polynomial:
\begin{align*}
d_{n k}\of{s_0, s_1; r} &= \pa{s_1-3s_0}{s_1}^2 \pa{n+k+1}\pa{n+k}\pa{n+k-1} \\
&\quad + 6s_0 s_1 \pa{n+k}\pa{n+k-1} n r \\
&\quad -3\pa{s_0+s_1}\pa{n+k-1} n \pa{n-1} r^2 \\
&\quad + 2 n \pa{n-1}\pa{n-2} r^3
\end{align*}
The following two lemmata establish properties of the roots of $d_{n k}\of{s_0, s_1; r}$:
\begin{lemma}[necessary condition]
If $1.183s_0<s_1<3s_0$ then there exist $k$ and $n$ such that $d_{n k}\of{s_0, s_1; r}$ has a zero in $\intopen{s_0} {s_1}$.
\end{lemma}
\begin{lemma}[sufficient condition]
If $s_1 = 3s_0$ then the positive roots of $d_{n k}\of{s_0, s_1; r}$ are greater than $s_1$.
\end{lemma}
These lemmata trivially constitute necessary and sufficient conditions, respectively, for the proposition.
\end{proof}
\end{proposition}
\begin{proof}[of necessary condition]
In this proof we set $n=2$, which causes the term in $r^3$ of $d_{n k}\of{s_0, s_1; r}$ to vanish:
\begin{multline*}
d_{2,k}\of{s_0, s_1; r} = \pa{s_1-3s_0}s_1^2\pa{k+3}\pa{k+2}\pa{k+1} \\+ 12s_0 s_1\pa{k+2}\pa{k+1}r - 6\pa{s_0+s_1}\pa{k+1}r^2
\end{multline*}
The roots of $d_{2,k}\of{s_0, s_1; r}$ can be computed by solving:
\[
\pa{s_1-3s_0}s_1^2\pa{k+3}\pa{k+2} + 12s_0 s_1\pa{k+2}r - 6\pa{s_0+s_1}r^2 = 0
\]
The reduced discriminant of this equation is:
\begin{align*}
\gD' &= 36 s_0^2 s_1^2 \pa{k+2}^2 + 6\pa{s_0+s_1}\pa{s_1-3s_0}s_1^2\pa{k+2}\pa{k+3}\\
&= 6 s_1^2\pa{k+2}\pa{6 s_0^2\pa{k+2}+\pa{s_1^2-2 s_0 s_1-3s_0^2}\pa{k+3}}\\
&= 6 s_1^2\pa{k+2}\pa{k\pa{s_1^2-2 s_0 s_1 + 3s_0^2} + 3\pa{s_1-s_0}^2}
\end{align*}
and its solutions are:
\[
r_± = s_1\frac{6 s_0\pa{k+2} ± \sqrt{6\pa{k+2}\pa{k\pa{s_1^2-2 s_0 s_1 + 3s_0^2} + 3\pa{s_1-s_0}^2}}}{6\pa{s_0+s_1}}
\]
We will now define $k_1$ and $k_2$ such that the following claim holds:
\begin{claim}
For every $k \in \intopen{k_2}{k_1}\Intersection{\Naturals}$, $r_- \in \intopen{s_0}{s_1}$.
\end{claim}
First, consider the inequality $r_-<s_1$. It can be rewritten as follows:
\begin{align*}
&6 s_0\pa{k+2} - 6\pa{s_0+s_1} < \sqrt{6\pa{k+2}\pa{k\pa{s_1^2-2 s_0 s_1 + 3s_0^2} + 3\pa{s_1-s_0}^2}}
\end{align*}
If $k≥2$ the left-hand side is positive and we can further rewrite the inequality:
\begin{align*}
&6\pa{k s_0 + \pa{s_0-s_1}}^2 < \pa{k+2}\pa{k\pa{s_1^2-2 s_0 s_1 + 3s_0^2} + 3\pa{s_1-s_0}^2} \\
\iff &6\pa{k^2 s_0^2 + 2 k s_0 \pa{s_0 - s_1} + \pa{s_0 - s_1}^2} < \\
&\qquad k^2 \pa{s_1^2-2 s_0 s_1 + 3s_0^2} + k \pa{3\pa{s_1-s_0}^2 + 2\pa{s_1^2-2 s_0 s_1 + 3s_0^2}} + 6\pa{s_0 - s_1}^2 \\
\iff &k^2\pa{3s_0^2 + 2 s_0 s_1 - s_1^2} + k\pa{12 s_0\pa{s_0-s_1} - 3\pa{s_1-s_0}^2 - 2\pa{s_1^2-2 s_0 s_1 + 3s_0^2}} < 0 \\
\iff &k\pa{3s_0^2 + 2 s_0 s_1 - s_1^2} + \pa{3s_0^2 -2 s_0 s_1 - 5 s_1^2} < 0 \\
\iff &k\pa{3s_0-s_1}\pa{s_0+s_1} + \pa{3s_0 - 5s_1}\pa{s_0+s_1} < 0 \\
\iff &k < \frac{5s_1-3s_0}{3s_0-s_1}=:k_1
\end{align*}
Similarly, the inequality $s_0<r_-$ can be rewritten as follows:
\begin{align*}
&6 s_0 s_1\pa{k+2} - 6\pa{s_0+s_1}s_0 > s_1 \sqrt{6\pa{k+2}\pa{k\pa{s_1^2-2 s_0 s_1 + 3s_0^2} + 3\pa{s_1-s_0}^2}} \\
\iff &6 s_0^2\pa{k s_1 + \pa{s_1-s_0}}^2 > s_1^2\pa{k+2}\pa{k\pa{s_1^2-2 s_0 s_1 + 3s_0^2} + 3\pa{s_1-s_0}^2} \\
\iff &6 s_0^2\pa{k^2 s_1^2 + 2 k s_1\pa{s_1-s_0} + \pa{s_1-s_0}^2} > \\
&\qquad k^2 s_1^2\pa{s_1^2-2 s_0 s_1 + 3s_0^2} + k s_1^2\pa{3\pa{s_1-s_0}^2 + 2\pa{s_1^2-2 s_0 s_1 + 3s_0^2}} + 6 s_1^2\pa{s_0 - s_1}^2 \\
\iff &k^2 s_1^2\pa{3s_0^2 + 2 s_0 s_1 - s_1^2} + k s_1\pa{-5s_1^3 + 10 s_0 s_1^2 + 3 s_0^2 s_1 - 12 s_0^3} + 6\pa{s_0-s_1}^3\pa{s_0+s_1} > 0 \\
\iff &k^2 s_1^2\pa{3s_0-s_1}\pa{s_0+s_1} + k s_1\pa{-5s_1^2 + 15 s_0 s_1 - 12s_0^2}\pa{s_0+s_1} + 6\pa{s_0-s_1}^3\pa{s_0+s_1} > 0 \\
\iff &k^2 s_1^2\pa{3s_0-s_1} + k s_1\pa{-5s_1^2 + 15 s_0 s_1 - 12s_0^2} + 6\pa{s_0-s_1}^3 > 0
\end{align*}
When $s_1<3s_0$, the coefficient of $k^2$ in the polynomial above is positive. The constant term is negative, therefore the product of the roots is negative. For the inequality to be true, $k$ must be greater than the largest root the polynomial. The discriminant of the polynomial is:
\[
\gD = s_1^2\pa{\pa{-5s_1^2 + 15 s_0 s_1 - 12s_0^2}^2 - 24\pa{3s_0-s_1}\pa{s_0-s_1}^3}
\]
And the largest root is:
\[
k_2 = \frac{-s_1\pa{-5s_1^2 + 15 s_0 s_1 - 12s_0^2} + \sqrt{\gD}}{2s_1^2\pa{3s_0-s_1}}
\]
\begin{claim}
$\intopen{k_2}{k_1}\Intersection{\Naturals}$ is nonempty when $s_1>1.183s_0$.
\end{claim}
A necessary condition for the interval $\intopen{k_2}{k_1}$ to contain an integer is $k_1-k_2>1$. This inequality can be rewritten as follows:
\begin{align*}
&2s_1^2\pa{5s_1-3s_0} + s_1\pa{-5s_1^2 + 15 s_0 s_1 - 12s_0^2} - 2s_1^2\pa{3s_0-s_1} - \sqrt{\gD} > 0 \\
\iff &s_1\pa{7s_1^2 + 3 s_0 s_1 - 12s_0^2} > \sqrt{\gD}
\end{align*}
This inequality can only be true if the left-hand side is positive, that is, if $s_1\frac{\sqrt{345}-3}{14}s_0 \approx 1.112s_0$. Assuming that this condition is met we can further rewrite the inequality:
\begin{align*}
&\pa{7s_1^2 + 3 s_0 s_1 - 12s_0^2}^2 > \pa{-5s_1^2 + 15 s_0 s_1 - 12s_0^2}^2 - 24\pa{3s_0-s_1}\pa{s_0-s_1}^3 \\
\iff &49s_1^4 + 42 s_0 s_1^3 - 159 s_0^2 s_1^2 - 72 s_0^3 s_1 + 144 s_0^4 > \\
&\qquad s_1^4 - 6 s_0 s_1^3 + 57 s_0^2 s_1^2 - 120 s_0^3 s_1 + 72 s_0^4 \\
\iff &24\pa{s_1-s_0}\pa{2s_1^3 + 4 s_0 s_1^2 - 5 s_0^2 s_1 - 3s_0^3} > 0
\end{align*}
The sum of the roots of $2s_1^3 + 4 s_0 s_1^2 - 5 s_0^2 s_1 - 3s_0^3$ is $-2s_0$ and the product is $3s_0/2$. This means that this polynomial has a single positive root in $s_1$ and that when $s_1$ is greater than this root the inequality above holds. Solving the cubic using $\emph{Mathematica}$ yields this root:
\begin{align*}
\hat\gq &= \frac{1}{3} \arctan\of{\frac{3 \sqrt{2517}}{41}} \\
\hat{s}_1 &= \frac{s_0}{6}\pa{-4+\sqrt{138}\sin\hat\gq + \sqrt{46}\cos\hat\gq} \approx 1.183s_0
\end{align*}
Thus, if $s_1 \in \intopen{\hat{s}_1}{3s_0}$ it is possible to find an integer $k$ such that the smallest root of $d_{2,k}\of{s_0, s_1; r}$ is in $\intopen{s_0} {s_1}$.
\end{proof}
\begin{remark}
It is possible that $\intopen{k_2}{k_1}$ contains an integer value even if $k_1-k_2<1$, in which case more values $s_1 \in \intopen{s_0}{\hat{s}_1}$ could be excluded.
\end{remark}
\begin{proof}[of sufficient condition]
With the choice $s_1=3s_0$ the constant term of the polynomial $d_{n k}\of{s_0, s_1; r}$ vanishes and $r=0$ is a trivial root. We are left with the equation:
\[
6s_0 s_1 \pa{n+k}\pa{n+k-1} - 3\pa{s_0+s_1}\pa{n+k-1} \pa{n-1} r + 2 \pa{n-1}\pa{n-2} r^2 = 0
\]
or, substituting $s_1$:
\begin{equation}
9{s_0}^2 \pa{n+k}\pa{n+k-1} - 6 s_0 \pa{n+k-1} \pa{n-1} r + \pa{n-1}\pa{n-2} r^2 = 0 \label{eqn3s0}
\end{equation}
The reduced discriminant of equation (\ref{eqn3s0}) is:
\begin{align*}
\gD' &= 9 {s_0}^2 \pa{n+k-1}^2 \pa{n-1}^2 - 9 {s_0}^2 \pa{n+k}\pa{n+k-1}\pa{n-1}\pa{n-2}\\
&=9 {s_0}^2\pa{n+k-1}\pa{n-1}\pa{k+1}
\end{align*}
and its solutions are, when $n>2$:
\[
r_± = 3 s_0\frac{\pa{n+k-1}\pa{n-1} ± \sqrt{\pa{n+k-1}\pa{n-1}\pa{k+1}}}{\pa{n-1}\pa{n-2}}
\]
We want to prove that the smallest root, $r_-$ verifies $r_{-}>3s_0$. This inequality can be rewritten:
\begin{align*}
&\pa{n+k-1}\pa{n-1} - \pa{n-1}\pa{n-2} > \sqrt{\pa{n+k-1}\pa{n-1}\pa{k+1}} \\
\iff &\pa{n-1}\pa{k+1} > \sqrt{\pa{n+k-1}\pa{n-1}\pa{k+1}} \\
\iff &\pa{n-1}\pa{k+1} > n+k-1 \\
\iff &k\pa{n-2} > 0 \quad \text{which is trivally true.}
\end{align*}
When $n=2$ equation (\ref{eqn3s0}) further reduces to a first-degree equation with root $r = \frac{3}{2} s_0 \pa{k+2}$ which verifies $r≥3s_0$. Note however that the latter inequality is not strict, i.e., the value $3s_0$ may be reached if $k=0$ and $n=2$.
\end{proof}
Accordingly, we pick $s_1=3s_0$.
The expression for $\gs$ on $\intclos {s_0} {s_1}$ simplifies to\[
\gs\of r = \frac{r \pa{r-3s_0}^2} {4 s_0^3} = \frac{r^3}{4 s_0^3} - \frac{3 r^2}{2s_0^2} + \frac{9r}{4s_0}\text.