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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 $π &colon; E \to M$ be a fiber bundle, with base dimension $n$ and fiber dimension $m$ and let $π^{\infty} &colon; J^{\infty} (E) \to M$ be the infinite jet bundle of $E$ . The space of $p$ -forms $Ω^{p} (J^{\infty} (E))$ decomposes into a direct sum $Ω^{p} (J^{\infty}) = ⨁_{r + s = p}^{} Ω^{(r, s)} (J^{\infty} (E))$ , where $Ω^{(r, s)} (J^{\infty} (E))$ is the space of bi-forms of horizontal degree $r$ and vertical degree $s &period;$ The horizontal exterior derivative is a mapping $d_{H} &colon; Ω^{(r, s)} (J^{\infty} (E)) \to Ω^{(r + 1, s)} (J^{\infty} (E))$ with the following properties. A form $ω \in Ω^{(r, s)} (J^{\infty} (E))$ is called $d_{H}$ closed if $d_{H} ω = 0$ and $d_{H}$ exact if there is a bi-form $η \in Ω^{(r - 1, s)} (J^{\infty} (E))$ such that $ω = d_{H} η$ . Since $d_{H} \circ d_{H} = 0,$ every $d_{H}$ exact bi-form is $d_{H}$ closed.

[i] If $r < n$ , then every $d_{H}$ closed bi-form $ω$ is $d_{H}$ exact, $ω = d_{H} η &period;$

[ii] If $r = n$ and $s = 0$ and $E (ω) = 0,$ where $E$ is the Euler-Lagrange operator, then $ω = d_{H} η$ .

[iii] If $r = n$ and $s > 0$ and $I (ω) = 0,$ where $I$ is the integration by parts operator, then $ω = d_{H} η$ .

There are a number of algorithms for finding the bi-form $η &period;$ One approach is to use the horizontal homotopy operators $h_{^{} H}^{r, s} &colon; Ω^{(r, s)} (J^{\infty} (E)) \to Ω^{(r - 1, s)} (J^{\infty} (E))$ . Similar to the DeRham homotopy operator, these homotopy operators satisfy the identities

[i] $h_{^{} H}^{r + 1, s} (d_{H} ω) + d_{H} (h_{^{} H}^{r, s} (ω)) = ω$ if $r < n &semi;$

[ii] $d_{H} (h_{^{} H}^{r, s} (ω)) = ω$ if $r = n$ and $s = 0$ and $E (ω) = 0.$

[iii] $d_{H} (h_{^{} H}^{r, s} (ω)) = ω$ if $r = n$ and $s > 0$ and $I (ω) = 0.$

•	If $ω$ is a bi-form of degree $(r, s)$ with $r > 0$ then HorizontalHomotopy( $ω$ ) returns a bi-form of degree ( $r - 1, s)$ .

•

For $s > 0$ 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 $s = 0$ 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 $(x^{i}, u^{α}, u_{i_{}}^{α}, u_{i_{} j}^{α}$ , ..., $u_{ij \cdot \cdot \cdot k}^{α}, .. ..)$ be a local system of jet coordinates and let $Θ^{α} = {du}^{α} - u_{&ell;}^{α} {dx}^{&ell;}$ . Let $ω \in Ω^{(r, s)} (J^{\infty} (E))$ be a $k$ -th order bi-form with $s \geq 1$ and let $E_{α}^{I} (ω)$ $\in Ω^{(r, s - 1)} (J^{\infty} (E))$ be the higher (interior product) Euler operators. Let $D_{I} = D_{i_{1} i_{2} \cdot \cdot \cdot i_{&ell;}}$ be the multi-total derivative operator and let $ω_{j} = ι_{D_{j}} (ω)$ . Then

$h_{H}^{r, s} (ω) = \frac{1}{s} \sum_{|I| = 0}^{k - 1}$ $\frac{| I | + 1}{n - r + |I| + 1} D_{I} (Θ^{α} \land E_{α}^{Ij} (ω_{j}))$ .

For $s = 0,$ 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

$h_{H}^{r, s} (ω) = h_{V}^{r - 1, 1} (h_{H}^{r, 1} (d_{V} ω)) &period;$

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): with(JetCalculus):

Example 1.

Create the jet space $J^{3} (E)$ for the bundle $E$ with coordinates $(x, u) \to x &period;$

>	DGsetup([x], [u], E, 3):

Show that the EulerLagrange form for $ω_{1}$ is 0 so that $ω_{}_{1}$ is $d_{H}$ exact.

E >	omega1 := evalDG((u[1,1,1]u[1] + xu[1,1,1]u[1,1] + 2u[1,1]u[1,1,1] + xu[1]u[1,1,1,1])Dx);

$ω1 ≔ (x u_{1} u_{1, 1, 1, 1} + x u_{1, 1, 1} u_{1, 1} + u_{1, 1, 1} u_{1} + 2 u_{1, 1} u_{1, 1, 1}) Dx$

(3.1)

E >	EulerLagrange(omega1);

$0 Dx ⋀ {Cu}_{[]}$

(3.2)

Apply the horizontal homotopy operator to $ω_{1}$

E >	eta1 := HorizontalHomotopy(omega1);

$η1 ≔ x u_{1, 1, 1} u_{1} + u_{1, 1}^{2}$

(3.3)

Check that the horizontal exterior derivative of $η_{1}$ gives $ω_{1}$ .

E >	omega1 &minus HorizontalExteriorDerivative(eta1);

$0 Dx$

(3.4)

Example 2.

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

E >	omega2 := evalDG(Dx &w Cu[1] &w Cu[1,1,1,1] + Dx &w Cu[1,1] &w Cu[1,1,1]);

$ω2 ≔ Dx ⋀ {Cu}_{1} ⋀ {Cu}_{1, 1, 1, 1} + Dx ⋀ {Cu}_{1, 1} ⋀ {Cu}_{1, 1, 1}$

(3.5)

E >	IntegrationByParts(omega2);

$0 Dx ⋀ {Cu}_{[]} ⋀ {Cu}_{1}$

(3.6)

Apply the horizontal homotopy operator to $ω_{2} &period;$

E >	eta2 := HorizontalHomotopy(omega2);

$η2 ≔ {Cu}_{1} ⋀ {Cu}_{1, 1, 1}$

(3.7)

E >	omega2 &minus HorizontalExteriorDerivative(eta2);

$0 Dx ⋀ {Cu}_{[]} ⋀ {Cu}_{1}$

(3.8)

Example 3.

Show that the Euler-Lagrange form for $ω_{3}$ is 0 so that $ω_{3}$ is $d_{H}$ exact.

E >	HorizontalExteriorDerivative((u[1]*u[1,1,1])/u[1,1]^4);

(3.9)

E >	omega3 := map(expand, evalDG((u[1,1]^2u[1,1,1] - 4u[1,1,1]^2u[1] + u[1]u[1,1,1,1]u[1,1])/u[1,1]^5Dx));

$ω3 ≔ (\frac{u_{1, 1, 1}}{u_{1, 1}^{3}} - \frac{4 u_{1} u_{1, 1, 1}^{2}}{u_{1, 1}^{5}} + \frac{u_{1} u_{1, 1, 1, 1}}{u_{1, 1}^{4}}) Dx$

(3.10)

E >	EulerLagrange(omega3);

$0 Dx ⋀ {Cu}_{[]}$

(3.11)

Apply the horizontal homotopy operator to $ω_{3}$ . Because $ω_{3}$ is singular at $u [1, 1] = 0$ we change the integration limits in the homotopy formula but still perform a radial integration. See DeRhamHomotopy for a detailed discussion.

E >	eta3a := HorizontalHomotopy(omega3, integrationlimits = [infinity, 1]);

$eta3a ≔ \frac{u_{1, 1, 1} u_{1}}{u_{1, 1}^{4}}$

(3.12)

Check that $d_{H} η_{3} = ω_{3}$ .

E >	omega3 &minus HorizontalExteriorDerivative(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, u[], u[1], u[1,1], u[1,1,1], u[1,1,1,1], u[1,1,1,1,1]], initialpoint = [u[1,1] = 1];

$opt ≔ intmethod = ExteriorDerivativeHomotopy, path = zigzag, variableorder = [x, u_{[]}, u_{1}, u_{1, 1}, u_{1, 1, 1}, u_{1, 1, 1, 1}, u_{1, 1, 1, 1, 1}], initialpoint = [u_{1, 1} = 1]$

(3.14)

E >	eta3b := HorizontalHomotopy(omega3, opt);

$eta3b ≔ \frac{u_{1, 1, 1} u_{1}}{u_{1, 1}^{4}}$

(3.15)

Example 4.

Create the jet space $J^{2} (E)$ for the bundle $E$ with coordinates $(x, u, u, v) \to (x, y)$ .

E >	DGsetup([x, y], [u, v], E2, 2):

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

E2 >	omega4 := evalDG((v[2, 2]u[1, 1] + u[1]v[1, 2, 2])Dx + (v[2, 2]u[1, 2] + u[1]v[2, 2, 2])Dy);

$ω4 ≔ (u_{1} v_{1, 2, 2} + v_{2, 2} u_{1, 1}) Dx + (u_{1} v_{2, 2, 2} + v_{2, 2} u_{1, 2}) Dy$

(3.16)

E2 >	HorizontalExteriorDerivative(omega4);

$0 Dx ⋀ Dy$

(3.17)

Apply the horizontal homotopy operator to define $η_{4}$ .

E2 >	eta4 := HorizontalHomotopy(omega4);

$η4 ≔ v_{2, 2} u_{1}$

(3.18)

Check that $ω_{4} = d_{H} η_{4}$ .

E2 >	omega4 &minus HorizontalExteriorDerivative(eta4);

$0 Dx$

(3.19)

Example 5.

Define a type (2, 0) form $ω_{5}$ and check that its Euler-Lagrange form vanishes identically.

E2 >	omega5 := evalDG((v[2]u[1, 1] + u[1]v[1, 2] - v[1]u[1, 2, 2] - u[1, 2]v[1, 2])* Dx &w Dy);

$ω5 ≔ (u_{1} v_{1, 2} + v_{2} u_{1, 1} - u_{1, 2} v_{1, 2} - v_{1} u_{1, 2, 2}) Dx ⋀ Dy$

(3.20)

E2 >	EulerLagrange(omega5);

$0 Dx ⋀ Dy ⋀ {Cu}_{[]}$

(3.21)

E2 >	eta5a := HorizontalHomotopy(omega5);

$eta5a ≔ (\frac{7}{12} v_{1} u_{1, 2} - \frac{1}{4} v_{1} u_{1} - \frac{1}{6} v_{1, 1} u_{2} + \frac{1}{12} u_{1} v_{1, 2} - \frac{1}{4} v_{[]} u_{1, 1} - \frac{1}{4} v_{[]} u_{1, 1, 2} + \frac{1}{12} v_{1, 1, 2} u_{[]}) Dx + (- \frac{1}{6} v_{1} u_{2, 2} + \frac{3}{4} v_{2} u_{1} - \frac{1}{4} v_{2} u_{1, 2} - \frac{1}{12} v_{1, 2} u_{2} - \frac{1}{4} v_{[]} u_{1, 2} - \frac{1}{4} v_{[]} u_{1, 2, 2} + \frac{1}{12} v_{1, 2, 2} u_{[]}) Dy$

(3.22)

E2 >	omega5 &minus HorizontalExteriorDerivative(eta5a);

$0 Dx ⋀ Dy$

(3.23)

So $ω_{5} = d_{H} η_{5 a}$ , but we can often find a much simpler answer.

E2 >	opt := intmethod = "ExteriorDerivativeHomotopy", path = "zigzag", variableorder = Tools:-DGinfo(E2, "FrameJetVariables");

$opt ≔ intmethod = ExteriorDerivativeHomotopy, path = zigzag, variableorder = [x, y, u_{[]}, v_{[]}, u_{1}, u_{2}, v_{1}, v_{2}, u_{1, 1}, u_{1, 2}, u_{2, 2}, v_{1, 1}, v_{1, 2}, v_{2, 2}, u_{1, 1, 1}, u_{1, 1, 2}, u_{1, 2, 2}, u_{2, 2, 2}, v_{1, 1, 1}, v_{1, 1, 2}, v_{1, 2, 2}, v_{2, 2, 2}]$

(3.24)

E2 >	eta5b := HorizontalHomotopy(omega5, opt);

$eta5b ≔ v_{1} u_{1, 2} Dx + v_{2} u_{1} Dy$

(3.25)

E2 >	omega5 &minus HorizontalExteriorDerivative(eta5b);

$0 Dx ⋀ Dy$

(3.26)

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