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JetCalculus[HorizontalHomotopy] - apply the horizontal homotopy operator to a bi-form on a jet space

Calling Sequences

HorizontalHomotopy(${\mathrm{ω}}$, options)

Parameters

$\mathrm{\omega }$        - a differential bi-form on the jet space

options - any of  the optional arguments used in the commands DeRhamHomotopy

Description

 • Let be a fiber bundle, with base dimension $n$ and fiber dimension $m$ and let  be the infinite jet bundle of $E$. The space of $p$-forms ${\mathrm{Ω}}^{p}\left({J}^{\mathrm{∞}}\left(E\right)\right)$ decomposes into a direct sum , where  is the space of bi-forms of horizontal degree $r$ and vertical degree The horizontal exterior derivative  is a mapping  with the following properties. A form is called closed if and exact if there is a bi-form such that . Since every exact bi-form is closed.

[i] If $r, then every closed bi-form is exact,

[ii] If and and where $E$ is the Euler-Lagrange operator, then .

[iii] If and and where is the integration by parts operator, then .

There are a number of algorithms for finding the bi-form One approach is to use the horizontal homotopy operators . Similar to the DeRham homotopy operator, these homotopy operators satisfy the identities

[i]  if

[ii]  if  and and

[iii]  if  and and

 • If  is a bi-form of degree with  then HorizontalHomotopy(${\mathrm{ω}}$) returns a bi-form of degree (.
 • For  the operators ${h}_{{}^{}H}^{r,s}$ are total differential operators and therefore, unlike the usual homotopy operators for the de Rham complex or the vertical homotopy operators for bi-forms on jet spaces, do not involve any quadratures. For the horizontal homotopy does involve quadratures and the optional arguments used in the commands DeRhamHomotopy or VerticalHomotopy can be invoked.
 • The command HorizontalHomotopy is part of the DifferentialGeometry:-JetCalculus package. It can be used in the form HorizontalHomotopy(...) only after executing the commands with(DifferentialGeometry) and with(JetCalculus), but can always be used by executing DifferentialGeometry:-JetCalculus:-HorizontalHomotopy(...).
 Details Here are the explicit formulas for the horizontal homotopy operators. Let , ..., be a local system of jet coordinates and let . Let  be a $k$-th order bi-form with and let ${E}_{\mathrm{α}}^{I}\left(\mathrm{ω}\right)$be the higher (interior product) Euler operators. Let  be the multi-total derivative operator and let  . Then    .   For the horizontal homotopy operator is defined in terms of the vertical exterior derivative ${d}_{V}$  and the vertical homotopy operator  ${h}_{V}^{r,s}$ by  For further information, see: [i] Ian M. Anderson, Notes on the Variational Bicomplex. [ii] Niky. Kamran, Selected Topics in the Geometrical Study of Differential Equations, CBMS Lecture Series, 2002. [iii] Peter J. Olver, Applications of Lie Groups to Differential Equations, Chapter 5.

Examples

 > $\mathrm{with}\left(\mathrm{DifferentialGeometry}\right):$$\mathrm{with}\left(\mathrm{JetCalculus}\right):$

Example 1.

Create the jet space for the bundle with coordinates

 > $\mathrm{DGsetup}\left(\left[x\right],\left[u\right],E,3\right):$

Show that the EulerLagrange form for is 0 so that ${\mathrm{ω}}_{}{}_{1}$ is ${d}_{H}$ exact.

 E > $\mathrm{ω1}≔\mathrm{evalDG}\left(\left({u}_{1,1,1}{u}_{1}+x{u}_{1,1,1}{u}_{1,1}+2{u}_{1,1}{u}_{1,1,1}+x{u}_{1}{u}_{1,1,1,1}\right)\mathrm{Dx}\right)$
 ${\mathrm{ω1}}{≔}\left({x}{}{{u}}_{{1}}{}{{u}}_{{1}{,}{1}{,}{1}{,}{1}}{+}{x}{}{{u}}_{{1}{,}{1}{,}{1}}{}{{u}}_{{1}{,}{1}}{+}{{u}}_{{1}{,}{1}{,}{1}}{}{{u}}_{{1}}{+}{2}{}{{u}}_{{1}{,}{1}}{}{{u}}_{{1}{,}{1}{,}{1}}\right){}{\mathrm{Dx}}$ (3.1)
 E > $\mathrm{EulerLagrange}\left(\mathrm{ω1}\right)$
 ${0}{}{\mathrm{Dx}}{}{\bigwedge }{}{{\mathrm{Cu}}}_{\left[\right]}$ (3.2)

Apply the horizontal homotopy operator to ${\mathrm{ω}}_{1}$

 E > $\mathrm{η1}≔\mathrm{HorizontalHomotopy}\left(\mathrm{ω1}\right)$
 ${\mathrm{η1}}{≔}{x}{}{{u}}_{{1}{,}{1}{,}{1}}{}{{u}}_{{1}}{+}{{u}}_{{1}{,}{1}}^{{2}}$ (3.3)

Check that the horizontal exterior derivative of gives ${\mathrm{ω}}_{1}$.

 E > $\mathrm{ω1}&minus\left(\mathrm{HorizontalExteriorDerivative}\left(\mathrm{η1}\right)\right)$
 ${0}{}{\mathrm{Dx}}$ (3.4)

Example 2.

Show that the integration by parts operator for the type (1, 2)  bi-form is 0 so that is ${d}_{H}$ exact.

 E > $\mathrm{ω2}≔\mathrm{evalDG}\left(\left(\mathrm{Dx}&w\left({\mathrm{Cu}}_{1}\right)\right)&w\left({\mathrm{Cu}}_{1,1,1,1}\right)+\left(\mathrm{Dx}&w\left({\mathrm{Cu}}_{1,1}\right)\right)&w\left({\mathrm{Cu}}_{1,1,1}\right)\right)$
 ${\mathrm{ω2}}{≔}{\mathrm{Dx}}{}{\bigwedge }{}{{\mathrm{Cu}}}_{{1}}{}{\bigwedge }{}{{\mathrm{Cu}}}_{{1}{,}{1}{,}{1}{,}{1}}{+}{\mathrm{Dx}}{}{\bigwedge }{}{{\mathrm{Cu}}}_{{1}{,}{1}}{}{\bigwedge }{}{{\mathrm{Cu}}}_{{1}{,}{1}{,}{1}}$ (3.5)
 E > $\mathrm{IntegrationByParts}\left(\mathrm{ω2}\right)$
 ${0}{}{\mathrm{Dx}}{}{\bigwedge }{}{{\mathrm{Cu}}}_{\left[\right]}{}{\bigwedge }{}{{\mathrm{Cu}}}_{{1}}$ (3.6)

Apply the horizontal homotopy operator to ${\mathrm{ω}}_{2}.$

 E > $\mathrm{η2}≔\mathrm{HorizontalHomotopy}\left(\mathrm{ω2}\right)$
 ${\mathrm{η2}}{≔}{{\mathrm{Cu}}}_{{1}}{}{\bigwedge }{}{{\mathrm{Cu}}}_{{1}{,}{1}{,}{1}}$ (3.7)
 E > $\mathrm{ω2}&minus\left(\mathrm{HorizontalExteriorDerivative}\left(\mathrm{η2}\right)\right)$
 ${0}{}{\mathrm{Dx}}{}{\bigwedge }{}{{\mathrm{Cu}}}_{\left[\right]}{}{\bigwedge }{}{{\mathrm{Cu}}}_{{1}}$ (3.8)

Example 3.

Show that the Euler-Lagrange form for is 0 so that ${\mathrm{ω}}_{3}$ is exact.

 E > $\mathrm{HorizontalExteriorDerivative}\left(\frac{{u}_{1}{u}_{1,1,1}}{{u}_{1,1}^{4}}\right)$
 $\frac{{{u}}_{{1}}{}{{u}}_{{1}{,}{1}{,}{1}{,}{1}}{}{{u}}_{{1}{,}{1}}{-}{4}{}{{u}}_{{1}}{}{{u}}_{{1}{,}{1}{,}{1}}^{{2}}{+}{{u}}_{{1}{,}{1}{,}{1}}{}{{u}}_{{1}{,}{1}}^{{2}}}{{{u}}_{{1}{,}{1}}^{{5}}}{}{\mathrm{Dx}}$ (3.9)
 E > $\mathrm{ω3}≔\mathrm{map}\left(\mathrm{expand},\mathrm{evalDG}\left(\frac{\left({u}_{1,1}^{2}{u}_{1,1,1}-4{u}_{1,1,1}^{2}{u}_{1}+{u}_{1}{u}_{1,1,1,1}{u}_{1,1}\right)\mathrm{Dx}}{{u}_{1,1}^{5}}\right)\right)$
 ${\mathrm{ω3}}{≔}\left(\frac{{{u}}_{{1}{,}{1}{,}{1}}}{{{u}}_{{1}{,}{1}}^{{3}}}{-}\frac{{4}{}{{u}}_{{1}}{}{{u}}_{{1}{,}{1}{,}{1}}^{{2}}}{{{u}}_{{1}{,}{1}}^{{5}}}{+}\frac{{{u}}_{{1}}{}{{u}}_{{1}{,}{1}{,}{1}{,}{1}}}{{{u}}_{{1}{,}{1}}^{{4}}}\right){}{\mathrm{Dx}}$ (3.10)
 E > $\mathrm{EulerLagrange}\left(\mathrm{ω3}\right)$
 ${0}{}{\mathrm{Dx}}{}{\bigwedge }{}{{\mathrm{Cu}}}_{\left[\right]}$ (3.11)

Apply the horizontal homotopy operator to ${\mathrm{ω}}_{3}$. Because is singular at  we change the integration limits in the homotopy formula but still perform a radial integration.  See  DeRhamHomotopy for a detailed discussion.

 E > $\mathrm{eta3a}≔\mathrm{HorizontalHomotopy}\left(\mathrm{ω3},\mathrm{integrationlimits}=\left[\mathrm{∞},1\right]\right)$
 ${\mathrm{eta3a}}{≔}\frac{{{u}}_{{1}{,}{1}{,}{1}}{}{{u}}_{{1}}}{{{u}}_{{1}{,}{1}}^{{4}}}$ (3.12)

Check that .

 E > $\mathrm{ω3}&minus\left(\mathrm{HorizontalExteriorDerivative}\left(\mathrm{eta3a}\right)\right)$
 ${0}{}{\mathrm{Dx}}$ (3.13)

Instead of changing the limits of integration we can change the integration path to a sequence of coordinate lines.  See HorizontalExteriorDerivative for a detailed discussion.

 > $\mathrm{opt}≔\mathrm{intmethod}="ExteriorDerivativeHomotopy",\mathrm{path}="zigzag",\mathrm{variableorder}=\left[x,{u}_{[]},{u}_{1},{u}_{1,1},{u}_{1,1,1},{u}_{1,1,1,1},{u}_{1,1,1,1,1}\right],\mathrm{initialpoint}=\left[{u}_{1,1}=1\right]$
 ${\mathrm{opt}}{≔}{\mathrm{intmethod}}{=}{"ExteriorDerivativeHomotopy"}{,}{\mathrm{path}}{=}{"zigzag"}{,}{\mathrm{variableorder}}{=}\left[{x}{,}{{u}}_{\left[\right]}{,}{{u}}_{{1}}{,}{{u}}_{{1}{,}{1}}{,}{{u}}_{{1}{,}{1}{,}{1}}{,}{{u}}_{{1}{,}{1}{,}{1}{,}{1}}{,}{{u}}_{{1}{,}{1}{,}{1}{,}{1}{,}{1}}\right]{,}{\mathrm{initialpoint}}{=}\left[{{u}}_{{1}{,}{1}}{=}{1}\right]$ (3.14)
 E > $\mathrm{eta3b}≔\mathrm{HorizontalHomotopy}\left(\mathrm{ω3},\mathrm{opt}\right)$
 ${\mathrm{eta3b}}{≔}\frac{{{u}}_{{1}{,}{1}{,}{1}}{}{{u}}_{{1}}}{{{u}}_{{1}{,}{1}}^{{4}}}$ (3.15)

Example 4.

Create the jet space for the bundle with coordinates .

 E > $\mathrm{DGsetup}\left(\left[x,y\right],\left[u,v\right],\mathrm{E2},2\right):$

Define a type (1, 0) biform ${\mathrm{ω}}_{4}$ and check that it is closed.

 E2 > $\mathrm{ω4}≔\mathrm{evalDG}\left(\left({v}_{2,2}{u}_{1,1}+{u}_{1}{v}_{1,2,2}\right)\mathrm{Dx}+\left({v}_{2,2}{u}_{1,2}+{u}_{1}{v}_{2,2,2}\right)\mathrm{Dy}\right)$
 ${\mathrm{ω4}}{≔}\left({{u}}_{{1}}{}{{v}}_{{1}{,}{2}{,}{2}}{+}{{v}}_{{2}{,}{2}}{}{{u}}_{{1}{,}{1}}\right){}{\mathrm{Dx}}{+}\left({{u}}_{{1}}{}{{v}}_{{2}{,}{2}{,}{2}}{+}{{v}}_{{2}{,}{2}}{}{{u}}_{{1}{,}{2}}\right){}{\mathrm{Dy}}$ (3.16)
 E2 > $\mathrm{HorizontalExteriorDerivative}\left(\mathrm{ω4}\right)$
 ${0}{}{\mathrm{Dx}}{}{\bigwedge }{}{\mathrm{Dy}}$ (3.17)

Apply the horizontal homotopy operator to define ${\mathrm{η}}_{4}$.

 E2 > $\mathrm{η4}≔\mathrm{HorizontalHomotopy}\left(\mathrm{ω4}\right)$
 ${\mathrm{η4}}{≔}{{v}}_{{2}{,}{2}}{}{{u}}_{{1}}$ (3.18)

Check that .

 E2 > $\mathrm{ω4}&minus\left(\mathrm{HorizontalExteriorDerivative}\left(\mathrm{η4}\right)\right)$
 ${0}{}{\mathrm{Dx}}$ (3.19)

Example 5.

Define a type (2, 0) form  and check that its Euler-Lagrange form vanishes identically.

 E2 > $\mathrm{ω5}≔\mathrm{evalDG}\left(\left({v}_{2}{u}_{1,1}+{u}_{1}{v}_{1,2}-{v}_{1}{u}_{1,2,2}-{u}_{1,2}{v}_{1,2}\right)\mathrm{Dx}&w\mathrm{Dy}\right)$
 ${\mathrm{ω5}}{≔}\left({{u}}_{{1}}{}{{v}}_{{1}{,}{2}}{+}{{v}}_{{2}}{}{{u}}_{{1}{,}{1}}{-}{{u}}_{{1}{,}{2}}{}{{v}}_{{1}{,}{2}}{-}{{v}}_{{1}}{}{{u}}_{{1}{,}{2}{,}{2}}\right){}{\mathrm{Dx}}{}{\bigwedge }{}{\mathrm{Dy}}$ (3.20)
 E2 > $\mathrm{EulerLagrange}\left(\mathrm{ω5}\right)$
 ${0}{}{\mathrm{Dx}}{}{\bigwedge }{}{\mathrm{Dy}}{}{\bigwedge }{}{{\mathrm{Cu}}}_{\left[\right]}$ (3.21)
 E2 > $\mathrm{eta5a}≔\mathrm{HorizontalHomotopy}\left(\mathrm{ω5}\right)$
 ${\mathrm{eta5a}}{≔}\left(\frac{{7}}{{12}}{}{{v}}_{{1}}{}{{u}}_{{1}{,}{2}}{-}\frac{{1}}{{4}}{}{{v}}_{{1}}{}{{u}}_{{1}}{-}\frac{{1}}{{6}}{}{{v}}_{{1}{,}{1}}{}{{u}}_{{2}}{+}\frac{{1}}{{12}}{}{{u}}_{{1}}{}{{v}}_{{1}{,}{2}}{-}\frac{{1}}{{4}}{}{{v}}_{\left[\right]}{}{{u}}_{{1}{,}{1}}{-}\frac{{1}}{{4}}{}{{v}}_{\left[\right]}{}{{u}}_{{1}{,}{1}{,}{2}}{+}\frac{{1}}{{12}}{}{{v}}_{{1}{,}{1}{,}{2}}{}{{u}}_{\left[\right]}\right){}{\mathrm{Dx}}{+}\left({-}\frac{{1}}{{6}}{}{{v}}_{{1}}{}{{u}}_{{2}{,}{2}}{+}\frac{{3}}{{4}}{}{{v}}_{{2}}{}{{u}}_{{1}}{-}\frac{{1}}{{4}}{}{{v}}_{{2}}{}{{u}}_{{1}{,}{2}}{-}\frac{{1}}{{12}}{}{{v}}_{{1}{,}{2}}{}{{u}}_{{2}}{-}\frac{{1}}{{4}}{}{{v}}_{\left[\right]}{}{{u}}_{{1}{,}{2}}{-}\frac{{1}}{{4}}{}{{v}}_{\left[\right]}{}{{u}}_{{1}{,}{2}{,}{2}}{+}\frac{{1}}{{12}}{}{{v}}_{{1}{,}{2}{,}{2}}{}{{u}}_{\left[\right]}\right){}{\mathrm{Dy}}$ (3.22)
 E2 > $\mathrm{ω5}&minus\left(\mathrm{HorizontalExteriorDerivative}\left(\mathrm{eta5a}\right)\right)$
 ${0}{}{\mathrm{Dx}}{}{\bigwedge }{}{\mathrm{Dy}}$ (3.23)

So , but we can often find a much simpler answer.

 E2 > $\mathrm{opt}≔\mathrm{intmethod}="ExteriorDerivativeHomotopy",\mathrm{path}="zigzag",\mathrm{variableorder}=\mathrm{Tools}:-\mathrm{DGinfo}\left(\mathrm{E2},"FrameJetVariables"\right)$
 ${\mathrm{opt}}{≔}{\mathrm{intmethod}}{=}{"ExteriorDerivativeHomotopy"}{,}{\mathrm{path}}{=}{"zigzag"}{,}{\mathrm{variableorder}}{=}\left[{x}{,}{y}{,}{{u}}_{\left[\right]}{,}{{v}}_{\left[\right]}{,}{{u}}_{{1}}{,}{{u}}_{{2}}{,}{{v}}_{{1}}{,}{{v}}_{{2}}{,}{{u}}_{{1}{,}{1}}{,}{{u}}_{{1}{,}{2}}{,}{{u}}_{{2}{,}{2}}{,}{{v}}_{{1}{,}{1}}{,}{{v}}_{{1}{,}{2}}{,}{{v}}_{{2}{,}{2}}{,}{{u}}_{{1}{,}{1}{,}{1}}{,}{{u}}_{{1}{,}{1}{,}{2}}{,}{{u}}_{{1}{,}{2}{,}{2}}{,}{{u}}_{{2}{,}{2}{,}{2}}{,}{{v}}_{{1}{,}{1}{,}{1}}{,}{{v}}_{{1}{,}{1}{,}{2}}{,}{{v}}_{{1}{,}{2}{,}{2}}{,}{{v}}_{{2}{,}{2}{,}{2}}\right]$ (3.24)
 E2 > $\mathrm{eta5b}≔\mathrm{HorizontalHomotopy}\left(\mathrm{ω5},\mathrm{opt}\right)$
 ${\mathrm{eta5b}}{≔}{{v}}_{{1}}{}{{u}}_{{1}{,}{2}}{}{\mathrm{Dx}}{+}{{v}}_{{2}}{}{{u}}_{{1}}{}{\mathrm{Dy}}$ (3.25)
 E2 > $\mathrm{ω5}&minus\left(\mathrm{HorizontalExteriorDerivative}\left(\mathrm{eta5b}\right)\right)$
 ${0}{}{\mathrm{Dx}}{}{\bigwedge }{}{\mathrm{Dy}}$ (3.26)