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

Calling Sequences

     HorizontalHomotopy(ω, options)

Parameters

     ω        - a differential bi-form on the jet space

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

 

Description

Details

Examples

Description

• 

Let π:EM be a fiber bundle, with base dimension n and fiber dimension m and let π∞ :J∞E M  be the infinite jet bundle of E. The space of p-forms ΩpJ∞E decomposes into a direct sum  ΩpJ  = r+s =p Ωr,sJE, where  Ωr,s JE is the space of bi-forms of horizontal degree r and vertical degree s.  The horizontal exterior derivative  is a mapping dH :Ωr,sJE Ωr+1,sJE with the following properties. A form ω  Ωr,s JE is called dH  closed if dH ω = 0 and dH exact if there is a bi-form η Ωr1,s JE such that ω = dH η. Since dHdH =0, every dH  exact bi-form is dH closed.

[i] If r<n, then every dH closed bi-form &omega; is dH exact, ω &equals; dH &eta;&period;

[ii] If r &equals;n and s&equals;0 and E&omega; &equals;0&comma; where E is the Euler-Lagrange operator, then ω &equals; dH &eta;.

[iii] If r &equals;n and s&gt;0 and I&omega; &equals;0&comma; where I is the integration by parts operator, then ω &equals; dH &eta;.

There are a number of algorithms for finding the bi-form &eta;&period; One approach is to use the horizontal homotopy operators hHr&comma;s &colon;  Ωr&comma;sJE  Ωr1&comma;sJE. Similar to the DeRham homotopy operator, these homotopy operators satisfy the identities

[i]  hHr&plus;1&comma; s dH &omega; &plus; dHhHr&comma;s &omega; &equals; &omega;   if  r<n &semi;

[ii]   dHhHr&comma;s &omega; &equals; &omega;   if   r &equals;n  and s &equals;0 and E&omega; &equals;0. 

[iii]   dHhHr&comma;s &omega; &equals; &omega;   if   r &equals;n  and s &gt;0 and I&omega; &equals;0. 

• 

If &omega;  is a bi-form of degree r&comma;s with r&gt;0  then HorizontalHomotopy(&omega;) returns a bi-form of degree (r1&comma; s&rpar;.

• 

For s &gt;0  the operators hHr&comma;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 s&equals; 0 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 &lpar;xi&comma; u&alpha;&comma; ui&alpha;&comma; uij&alpha;, ..., uij  k&alpha;&comma; ....&rpar; be a local system of jet coordinates and let &Theta;&alpha; &equals; du&alpha;u&ell;&alpha;dx&ell;. Let ω  Ωr&comma;sJ E be a k-th order bi-form with s1 and let E&alpha;I&omega; &Omega;r&comma;s1J Ebe the higher (interior product) Euler operators. Let  DI &equals; Di1i2i&ell; be the multi-total derivative operator and let  &omega;j &equals; &iota;Dj&omega;. Then

 

hHr&comma;s&omega; &equals;  1sI &equals;0k1  &verbar;I&verbar;&plus;1nr&plus;I &plus;1DI &Theta;&alpha; E&alpha;Ij&omega;j.

 

For s&equals;0&comma; the horizontal homotopy operator is defined in terms of the vertical exterior derivative dV  and the vertical homotopy operator  hVr&comma;s by

hHr&comma;s&omega; &equals;   hVr1&comma; 1hHr&comma;1dV&omega;&period;

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

withDifferentialGeometry&colon;withJetCalculus&colon;

 

Example 1.

Create the jet space J3E for the bundle E with coordinates x&comma; u x&period;

DGsetupx&comma;u&comma;E&comma;3&colon;

 

Show that the EulerLagrange form for &omega;1 is 0 so that &omega;1 is dH exact.

E > 

&omega;1evalDGu1,1,1u1+xu1,1,1u1,1+2u1,1u1,1,1+xu1u1,1,1,1Dx

&omega;1_DGbiform&comma;E&comma;1&comma;0&comma;1&comma;xu1u1,1,1,1+xu1,1,1u1,1+u1,1,1u1+2u1,1u1,1,1&comma;_DGbiform&comma;E&comma;1&comma;0&comma;1&comma;xu1u1,1,1,1+xu1,1,1u1,1+u1,1,1u1+2u1,1u1,1,1&comma;_DGbiform&comma;E&comma;1&comma;0&comma;1&comma;xu1u1,1,1,1+xu1,1,1u1,1+u1,