Riemannian metric for an ODE

Index of related atomic notes:

Riemannian metric for a first-order ODE — Construction for $u^{'} = ϕ (x, u)$
Riemannian metric for an autonomous second-order ODE — Construction for $u^{″} = ϕ (u, u_{1})$
Gauss curvature of an ODE-surface — Curvature formula
flatness-integrability theorem — Flat ⇔ quadratures
surface deformation of ODEs — Deformation family
energy foliation — Minimal foliation for 2nd-order ODEs

Original ideas (obsolete)

(Está obsoleta, revisar para ver cómo encaja con lo nuevo y luego borrar)
Cuando hacemos el cambio de variable

(t, c 1, c 2, c 3) \mapsto (x, u, u 1, u 2)

estamos transmitiendo la métrica usual de $R^{4}$ a una nueva métrica. Y la forma de volumen asociada a una nueva forma de volumen. Creo que con esa forma de volumen "heredada" sí se cumple que las 1-formas al estilo Basarab 1991, proposition 2 son directamente cerradas, sin necesidad de dividir.

¿Podrían tener utilidad estas ideas?

Tendría algo que ver aquí el Ricci curvature tensor? (las soluciones de la ODE son geodesics con la métrica inducida...)

Más de lo mismo: en el caso de orden 1, por ejemplo, tendríamos un $R^{2}$ junto con una métrica. Eso significaría que nuestro campo $A$ no está "en realidad", en $R^{2}$ , sino en otra superficie, con otra Gaussian curvature. Es interesante ver qué tipo de superficies salen (se pueden clasificar?). Las métricas no son cualquiera, pues todas tienen al vector $A = \partial x + ϕ \partial u$ "como de longitud 1".
Se pueden caracterizar estas métricas? Las métricas son una sección de un fibrado, estas métricas son un tipo especial?...

Más ideas. El conocimiento de las integrales primeras de la ecuación es equivalente al conocimiento de la métrica, pues con ellas podría rectificar el campo y obtener el cambio de variable como el de arriba.
¿Los vectores normales de las hipersuperficies integrales primera son los que rellenarían la base ortonormal que comienza con A? En realidad no, lo serían los normales a ellas pero con la "nueva" métrica... Necesitaría vectores que sean symmetry of a distribution de esas integrales primera... y que además formen el ángulo adecuado.

On the other hand, there are lots of metrics with these requirements. I don't need $∥ A ∥ = 1$ , but any constant. And I can reorganize the constants $c_{i}$ in such a way that give the same general solution. So we will obtain another metric. What have they got in common? Do they constitute an equivalence class in the metric space?

Hacer experimento con la ecuación de un péndulo doble, cuyo espacio de configuración es un toro, o el péndulo simple,... para ver si hay alguna relación entre la métrica asociada (la conocemos, pues sabemos la solución general) y la métrica del espacio "en el que viven"

Ideas para paper

Ver xournal_123 page 6, and maple_022.

Nuevo enfoque

See Maple_035 and xournal_139.

If we want to find the integral curves of a vector field, for example the vector field $A$ associated to an ODE, we can describe this "lines field" by local linear equations for every point of the plane, i.e., a 1-form. This 1-form, interpreted has a set of parallel lines in every point, probably has to be corrected by a factor, in order to have this parallel lines separated by the "same distance". This way we could find the first integral, i.e., a function whose level curves are tangents to the given lines field.

A way to perform this "correction" is by measuring the separation using the area (the standard volume form) and dividing by the length of the vector (computed with the standard Riemannian metric). But this only will work for special cases in which the desired integral curves are parallel curves with respect to the standard metric.

But what if this is just the case for every ODE, with the only trick that we have to consider a different metric? The Riemannian metric associated to the equation could be conformal to the standard one (in isothermal coordinates):

Λ^{2} I d

with

Λ^{2} = μ^{2} (1 + ϕ^{2})

where the ODE is given by $u_{1} = ϕ$ and $μ$ should be an integrating factor.

I think that for general vector fields $A = F \partial_{x} + G \partial_{u}$ in $R^{2}$ the formula is

Λ^{2} = μ^{2} (F^{2} + G^{2})

where $μ$ is an integrating factor for $ω = - G d x + F d u = Ω (A, -)$ where $Ω$ is the standard volume form.

Observe that in the new metric $g = Λ^{2} I d$ , the Riemannian volume form is $\tilde{Ω} = Λ^{2} Ω$ , therefore to "measure the local separation from $A$ " we use

ω = \frac{\tilde{Ω} (A, -)}{{‖ A ‖}_{n e w}} = \frac{μ^{2} (F^{2} + G^{2}) Ω (A, -)}{Λ {‖ A ‖}_{s t d}} = \frac{μ^{2} (F^{2} + G^{2}) Ω (A, -)}{μ (F^{2} + G^{2})}

= μ Ω (A, -),

which of course is closed.

I think that the quotient $Δ^{2} / {‖ A ‖}_{n e w}$ is an Jacobi last multiplier! See maple_035.

Relationship with symmetries

Suppose that $V$ is a symmetry of a distribution of $S ({A})$ . It is known that $\frac{1}{Ω (A, V)}$ is an integrating factor of the 1-form $Ω (A, -)$ (see Basarab 1991, proposition 2 and the slides of "04 Charla ODEs geometric viewpoint").

How does it fit in this context?
Pasted image 20220317103245.png
The volume spanned by $A$ y $V$ with the new metric is proportional to the length of $A$ with the new metric.

What about the length of $V$ with the new metric? I don't expect it to be constant, unless it is orthogonal to $A$ ...

On the other hand, the unit normal to $A$ should be a symmetry.

Open questions

What is the meaning of the curvature of this metric? Let us to classify equations?
Is there a determining equation for the metric? (see Maple_035 eq 19) Yes

(2 f ϕ - 1 - ϕ^{2}) f_{x} - (ϕ^{3} + ϕ) f_{u} - 2 f ϕ_{u} = 0

being $f = Λ^{2}$ .
Is it easier to solve that the equation for integrating factors?

What is the role of symmetries in this surface? I think that the commuting symmetries correspond to vector of constant length...
Is there an isometric embedding of this manifold in $R^{3}$ such that the curves are given by intersection of parallel planes with the surface? I think that at least they must be parallel curves inside this surface. The definition and importance is in this paper called "Computing parallel curves on parametric surfaces" by Galvez, Iglesias, Puig-Rey. Could what they call characteristic curves be the geodesics of the surface in this case?
Given the curvature, we have a surface. Can we study what equations would yield this surface?
Does the geodesics of this metric have a meaning?
For general vector fields $A = F \partial_{x} + G \partial_{u}$ in $R^{2}$ we have

f_{x} F^{3} + f_{x} F G^{2} + 2 F_{x} f G^{2} - 2 G_{x} F f G + f_{u} F^{2} G + f_{u} G^{3} - 2 F_{u} F f G + 2 G_{u} F^{2} f = 0

The relationship with an integrating factor of the associated 1-form $ω = - G d x + F d u$ .

Nuevo chispazo: orden $m$

(muy poco claro)
Dada una ODE $A$ , el conocimiento de una cinf-structure en diagonal form nos da una o varias métricas (tengo que pensar aún cómo se construye), con cierta curvatura. Con esta métrica el flujo de la ODE son geodésicas (hipótesis no confirmada).

Además, hopefully las integrales primeras están dando totally-geodesic manifolds, pues estarían hechas de geodeśicas.

Estamos describiendo el espacio con una especie de geodesic polar coordinates generalizadas a dimensión $n$ ?

La métrica puede ser deformada hasta una métrica plana (usando los symmetrising factors), y tenemos un frame, que contiene a $A$ , que es ortogonal y los vectores que lo forman commutan. Así pues, tenemos carta local tal que los vectores coordenados son esa base: hemos resuelto la ecuación.

En cierto sentido, sería como si demostrásemos que la cinf-structure da direcciones (familia de conformal metrics) y queremos buscar, de entre todas, aquella que sea plana, buscando las longitudes adecuadas (symmetrising factors).