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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:E→M 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,sJ∞E, where  Ωr,s J∞E is the space of bi-forms of horizontal degree r and vertical degree s.  The horizontal exterior derivative  is a mapping dH :Ωr,sJ∞E→ Ωr+1,sJ∞E with the following properties. A form ω ∈ Ωr,s J∞E is called dH  closed if dH ω = 0 and dH exact if there is a bi-form η∈ Ωr−1,s J∞E such that ω = dH η. Since dH∘dH =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;sJ∞E → Ωr−1&comma;sJ∞E. 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 (r−1&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 s≥1 and let E&alpha;I&omega;∈ &Omega;r&comma;s−1J∞ Ebe the higher (interior product) Euler operators. Let  DI &equals; Di1i2⋅⋅⋅i&ell; be the multi-total derivative operator and let  &omega;j &equals; &iota;Dj&omega;. Then

 

hHr&comma;s&omega; &equals;  1s∑I &equals;0k−1  &verbar;I&verbar;&plus;1n−r&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;   hVr−1&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

> 

with⁡DifferentialGeometry&colon;with⁡JetCalculus&colon;

 

Example 1.

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

> 

DGsetup⁡x&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;1≔evalDG⁡u1,1,1⁢u1+x⁢u1,1,1⁢u1,1+2⁢u1,1⁢u1,1,1+x⁢u1⁢u1,1,1,1⁢Dx

&omega;1≔x⁢u1⁢u1,1,1,1+x⁢u1,1,1⁢u1,1+u1,1,1⁢u1+2⁢u1,1⁢u1,1,1⁢Dx

(3.1)
E > 

EulerLagrange⁡&omega;1

0⁢Dx⁢⋀⁢Cu

(3.2)

 

Apply the horizontal homotopy operator to &omega;1

E > 

&eta;1≔HorizontalHomotopy⁡&omega;1

&eta;1≔x⁢u1,1,1⁢u1+u1,12

(3.3)

 

Check that the horizontal exterior derivative of &eta;1 gives &omega;1.

E > 

&omega;1&minusHorizontalExteriorDerivative⁡&eta;1

0⁢Dx

(3.4)

 

Example 2.

Show that the integration by parts operator for the type (1, 2)  bi-form &omega;1 is 0 so that &omega;2 is dH exact.

E > 

&omega;2≔evalDG⁡Dx&wCu1&wCu1,1,1,1+Dx&wCu1,1&wCu1,1,1

&omega;2≔Dx⁢⋀⁢Cu1⁢⋀⁢Cu1,1,1,1+Dx⁢⋀⁢Cu1,1⁢⋀⁢Cu1,1,1

(3.5)
E > 

IntegrationByParts⁡&omega;2

0⁢Dx⁢⋀⁢Cu⁢⋀⁢Cu1

(3.6)

 

Apply the horizontal homotopy operator to &omega;2&period;

E > 

&eta;2≔HorizontalHomotopy⁡&omega;2

&eta;2≔Cu1⁢⋀⁢Cu1,1,1

(3.7)
E > 

&omega;2&minusHorizontalExteriorDerivative⁡&eta;2

0⁢Dx⁢⋀⁢Cu⁢⋀⁢Cu1

(3.8)

 

Example 3.

Show that the Euler-Lagrange form for &omega;3 is 0 so that &omega;3 is dH exact.

E > 

HorizontalExteriorDerivative⁡u1⁢u1,1,1u1,14

(3.9)
E > 

&omega;3≔map⁡expand&comma;evalDG⁡u1,12⁢u1,1,1−4⁢u1,1,12⁢u1+u1⁢u1,1,1,1⁢u1,1u1,15⁢Dx

&omega;3≔u1,1,1u1,13−4⁢u1⁢u1,1,12u1,15+u1⁢u1,1,1,1u1,14⁢Dx

(3.10)
E > 

EulerLagrange⁡&omega;3

0⁢Dx⁢⋀⁢Cu

(3.11)

 

Apply the horizontal homotopy operator to &omega;3. Because &omega;3 is singular at  u1&comma;1 &equals; 0 we change the integration limits in the homotopy formula but still perform a radial integration.  See  DeRhamHomotopy for a detailed discussion.

E > 

eta3a≔HorizontalHomotopy⁡&omega;3&comma;integrationlimits=∞&comma;1

eta3a≔u1,1,1⁢u1u1,14

(3.12)

 

Check that dH &eta;3 &equals; &omega;3.

E > 

&omega;3&minusHorizontalExteriorDerivative⁡eta3a

0⁢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.

> 

opt≔intmethod=ExteriorDerivativeHomotopy,path=zigzag,variableorder=x&comma;u&comma;u1&comma;u1,1&comma;u1,1,1&comma;u1,1,1,1&comma;u1,1,1,1,1,initialpoint=u1,1=1

opt≔intmethod=ExteriorDerivativeHomotopy,path=zigzag,variableorder=x&comma;u&comma;u1&comma;u1,1&comma;u1,1,1&comma;u1,1,1,1&comma;u1,1,1,1,1,initialpoint=u1,1=1

(3.14)
E > 

eta3b≔HorizontalHomotopy⁡&omega;3&comma;opt

eta3b≔u1,1,1⁢u1u1,14

(3.15)

 

Example 4.

Create the jet space J2E for the bundle E with coordinates x&comma; u&comma; u&comma; v →x&comma;y.

E > 

DGsetup⁡x&comma;y&comma;u&comma;v&comma;E2&comma;2&colon;

 

Define a type (1, 0) biform &omega;4 and check that it is closed.

E2 > 

&omega;4≔evalDG⁡v2,2⁢u1,1+u1⁢v1,2,2⁢Dx+v2,2⁢u1,2+u1⁢v2,2,2⁢Dy

&omega;4≔u1⁢v1,2,2+v2,2⁢u1,1⁢Dx+u1⁢v2,2,2+v2,2⁢u1,2⁢Dy

(3.16)
E2 > 

HorizontalExteriorDerivative⁡&omega;4

0⁢Dx⁢⋀⁢Dy

(3.17)

 

Apply the horizontal homotopy operator to define &eta;4.

E2 > 

&eta;4≔HorizontalHomotopy⁡&omega;4

&eta;4≔v2,2⁢u1

(3.18)

 

Check that &omega;4  &equals;dH &eta;4.

E2 > 

&omega;4&minusHorizontalExteriorDerivative⁡&eta;4

0⁢Dx

(3.19)

 

Example 5.

Define a type (2, 0) form &omega;5  and check that its Euler-Lagrange form vanishes identically.

E2 > 

&omega;5≔evalDG⁡v2⁢u1,1+u1⁢v1,2−v1⁢u1,2,2−u1,2⁢v1,2⁢Dx&wDy

&omega;5≔u1⁢v1,2+v2⁢u1,1−u1,2⁢v1,2−v1⁢u1,2,2⁢Dx⁢⋀⁢Dy

(3.20)
E2 > 

EulerLagrange⁡&omega;5

0⁢Dx⁢⋀⁢Dy⁢⋀⁢Cu

(3.21)
E2 > 

eta5a≔HorizontalHomotopy⁡&omega;5

eta5a≔712⁢v1⁢u1,2−14⁢v1⁢u1−16⁢v1,1⁢u2+112⁢u1⁢v1,2−14⁢v⁢u1,1−14⁢v⁢u1,1,2+112⁢v1,1,2⁢u⁢Dx+−16⁢v1⁢u2,2+34⁢v2⁢u1−14⁢v2⁢u1,2−112⁢v1,2⁢u2−14⁢v⁢u1,2−14⁢v⁢u1,2,2+112⁢v1,2,2⁢u⁢Dy

(3.22)
E2 > 

&omega;5&minusHorizontalExteriorDerivative⁡eta5a

0⁢Dx⁢⋀⁢Dy

(3.23)

 

So &omega;5 &equals; dH&eta;5a , but we can often find a much simpler answer.

E2 > 

opt≔intmethod=ExteriorDerivativeHomotopy,path=zigzag,variableorder=Tools:-DGinfo⁡E2&comma;FrameJetVariables

opt≔intmethod=ExteriorDerivativeHomotopy,path=zigzag,variableorder=x&comma;y&comma;u&comma;v&comma;u1&comma;u2&comma;v1&comma;v2&comma;u1,1&comma;u1,2&comma;u2,2&comma;v1,1&comma;v1,2&comma;v2,2&comma;u1,1,1&comma;u1,1,2&comma;u1,2,2&comma;u2,2,2&comma;v1,1,1&comma;v1,1,2&comma;v1,2,2&comma;v2,2,2

(3.24)
E2 > 

eta5b≔HorizontalHomotopy⁡&omega;5&comma;opt

eta5b≔v1⁢u1,2⁢Dx+v2⁢u1⁢Dy

(3.25)
E2 > 

&omega;5&minusHorizontalExteriorDerivative⁡eta5b

0⁢Dx⁢⋀⁢Dy

(3.26)

See Also

DifferentialGeometry

JetCalculus

EulerLagrange

HorizontalExteriorDerivative

IntegrationByParts

VerticalExteriorDerivative

VerticalHomotopy

ZigZag