find the conservation law for the Euler-Lagrange equations from a given symmetry of the Lagrangian - Maple Programming Help

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JetCalculus[Noether] - find the conservation law for the Euler-Lagrange equations from a given symmetry of the Lagrangian

Calling Sequences

Noether(X, l)

Parameters

X         -  a vector field representing a symmetry of the Lagrangian $\mathrm{λ}$

λ         - a Lagrangian for a variational principle, defined as a top degree horizontal form of the jet space of a bundle $\mathrm{π}:E\to M$

Description

 • The celebrated theorem of E. Noether provides a formula for the calculation of a first integral or conservation law for any symmetry of the Lagrangian. This formula, which is very complicated for high order Lagrangians is easily implemented using the horizontal homotopy for the variational bicomplex.
 • Within the framework of the JetCalculus package conservation laws are represented by differential forms of degree $n-1$, where $n$ is the dimension of $M$, whose horizontal or total exterior derivative vanishes by virtue of the Euler Lagrange equations.
 • The vector field $X$ is a symmetry of the Lagrangian $\mathrm{λ}$ if the Lie derivative of l with respect to the prolongation of $X$ (to the order of λ) vanishes.
 • The command Noether is part of the DifferentialGeometry:-JetCalculus package. It can be used in the form Noether(...) only after executing the commands with(DifferentialGeometry) and with(JetCalculus), but can always be used by executing DifferentialGeometry:-JetCalculus:-Noether(...).

Examples

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

Example 1.

We set a simple single integral problem with 2 dependent variables and compute the Euler-Lagrange equations

 > $\mathrm{DGsetup}\left(\left[t\right],\left[x,y\right],E,2\right):$
 E > $L≔\frac{1\left({x}_{1}^{2}+{y}_{1}^{2}-\frac{1}{\sqrt{{x}_{[]}^{2}+{y}_{[]}^{2}}}\right)}{2}$
 ${L}{≔}\frac{{{x}}_{{1}}^{{2}}}{{2}}{+}\frac{{{y}}_{{1}}^{{2}}}{{2}}{-}\frac{{1}}{{2}{}\sqrt{{{x}}_{\left[\right]}^{{2}}{+}{{y}}_{\left[\right]}^{{2}}}}$ (2.1)
 E > $\mathrm{EL}≔\mathrm{EulerLagrange}\left(L\right)$
 ${\mathrm{EL}}{≔}\left[\frac{{{x}}_{\left[\right]}}{{2}{}{\left({{x}}_{\left[\right]}^{{2}}{+}{{y}}_{\left[\right]}^{{2}}\right)}^{{3}}{{2}}}}{-}{{x}}_{{1}{,}{1}}{,}\frac{{{y}}_{\left[\right]}}{{2}{}{\left({{x}}_{\left[\right]}^{{2}}{+}{{y}}_{\left[\right]}^{{2}}\right)}^{{3}}{{2}}}}{-}{{y}}_{{1}{,}{1}}\right]$ (2.2)
 E > $\mathrm{Eq}≔\mathrm{solve}\left(\mathrm{EL},\left\{{x}_{1,1},{y}_{1,1}\right\}\right)$
 ${\mathrm{Eq}}{≔}\left\{{{x}}_{{1}{,}{1}}{=}\frac{\sqrt{{{x}}_{\left[\right]}^{{2}}{+}{{y}}_{\left[\right]}^{{2}}}{}{{x}}_{\left[\right]}}{{2}{}\left({{x}}_{\left[\right]}^{{4}}{+}{2}{}{{x}}_{\left[\right]}^{{2}}{}{{y}}_{\left[\right]}^{{2}}{+}{{y}}_{\left[\right]}^{{4}}\right)}{,}{{y}}_{{1}{,}{1}}{=}\frac{\sqrt{{{x}}_{\left[\right]}^{{2}}{+}{{y}}_{\left[\right]}^{{2}}}{}{{y}}_{\left[\right]}}{{2}{}\left({{x}}_{\left[\right]}^{{4}}{+}{2}{}{{x}}_{\left[\right]}^{{2}}{}{{y}}_{\left[\right]}^{{2}}{+}{{y}}_{\left[\right]}^{{4}}\right)}\right\}$ (2.3)

The Lagrangian $L$ is invariant under rotations in the $x-y$ plane. Let us check this. To be technically correct we should work with the differential 1-form $\mathrm{λ}$ defined by $L$.

 E > $\mathrm{λ}≔L&mult\mathrm{Dt}$
 ${\mathrm{\lambda }}{≔}\frac{{{x}}_{{1}}^{{2}}{}\sqrt{{{x}}_{\left[\right]}^{{2}}{+}{{y}}_{\left[\right]}^{{2}}}{+}{{y}}_{{1}}^{{2}}{}\sqrt{{{x}}_{\left[\right]}^{{2}}{+}{{y}}_{\left[\right]}^{{2}}}{-}{1}}{{2}{}\sqrt{{{x}}_{\left[\right]}^{{2}}{+}{{y}}_{\left[\right]}^{{2}}}}{}{\mathrm{Dt}}$ (2.4)
 E > $X≔{x}_{[]}{\mathrm{D_y}}_{[]}-{y}_{[]}{\mathrm{D_x}}_{[]}$
 ${X}{≔}{-}{{\mathrm{D_x}}}_{\left[\right]}{}{{y}}_{\left[\right]}{+}{{\mathrm{D_y}}}_{\left[\right]}{}{{x}}_{\left[\right]}$ (2.5)
 E > $\mathrm{X1}≔\mathrm{Prolong}\left(X,1\right)$
 ${\mathrm{X1}}{≔}{-}{{y}}_{\left[\right]}{}{{\mathrm{D_x}}}_{\left[\right]}{+}{{x}}_{\left[\right]}{}{{\mathrm{D_y}}}_{\left[\right]}{-}{{y}}_{{1}}{}{{\mathrm{D_x}}}_{{1}}{+}{{x}}_{{1}}{}{{\mathrm{D_y}}}_{{1}}$ (2.6)
 E > $\mathrm{LieDerivative}\left(X,\mathrm{λ}\right)$
 ${0}{}{\mathrm{Dt}}$ (2.7)

Now we find the first integral associated to the symmetry $X$:

 E > $F≔\mathrm{Noether}\left(X,\mathrm{λ}\right)$
 ${F}{≔}{-}{{y}}_{{1}}{}{{x}}_{\left[\right]}{+}{{x}}_{{1}}{}{{y}}_{\left[\right]}$ (2.8)

To check that this is indeed a first integral, take the total derivative of $F$ with respect to $t$ and substitute from the Euler-Lagrange equations.

 E > $\mathrm{tdF}≔\mathrm{TotalDiff}\left(F,t\right)$
 ${\mathrm{tdF}}{≔}{-}{{x}}_{\left[\right]}{}{{y}}_{{1}{,}{1}}{+}{{y}}_{\left[\right]}{}{{x}}_{{1}{,}{1}}$ (2.9)
 E > $\mathrm{eval}\left(\mathrm{tdF},\mathrm{Eq}\right)$
 ${0}$ (2.10)

Example 2.

We use the command InfinitesimalSymmetriesOfGeometricObjectFields to find the symmetries of the Lagrangian for the wave equation in (2+1) dimensions.

We then use the command Noether to calculate the associated conservation laws.

 E > $\mathrm{with}\left(\mathrm{GroupActions}\right):$
 E > $\mathrm{DGsetup}\left(\left[x,y,t\right],\left[u\right],J,1\right)$
 ${\mathrm{frame name: J}}$ (2.11)
 J > $\mathrm{λ}≔\mathrm{evalDG}\left(\left({u}_{1}^{2}+{u}_{2}^{2}-{u}_{3}^{2}\right)\left(\mathrm{Dx}&w\mathrm{Dy}\right)&w\mathrm{Dt}\right)$
 ${\mathrm{\lambda }}{≔}\left({{u}}_{{1}}^{{2}}{+}{{u}}_{{2}}^{{2}}{-}{{u}}_{{3}}^{{2}}\right){}{\mathrm{Dx}}{}{\bigwedge }{}{\mathrm{Dy}}{}{\bigwedge }{}{\mathrm{Dt}}$ (2.12)
 J > $\mathrm{Γ}≔\mathrm{InfinitesimalSymmetriesOfGeometricObjectFields}\left(\left[\mathrm{λ}\right],\mathrm{output}="list"\right)$
 ${\mathrm{Γ}}{≔}\left[{-}{2}{}{x}{}{\mathrm{D_x}}{-}{2}{}{y}{}{\mathrm{D_y}}{-}{2}{}{t}{}{\mathrm{D_t}}{+}{{u}}_{\left[\right]}{}{{\mathrm{D_u}}}_{\left[\right]}{,}{{\mathrm{D_u}}}_{\left[\right]}{,}{t}{}{\mathrm{D_x}}{+}{x}{}{\mathrm{D_t}}{,}{t}{}{\mathrm{D_y}}{+}{y}{}{\mathrm{D_t}}{,}{\mathrm{D_t}}{,}{-}{y}{}{\mathrm{D_x}}{+}{x}{}{\mathrm{D_y}}{,}{\mathrm{D_y}}{,}{\mathrm{D_x}}\right]$ (2.13)

Let us find the conservation law $\mathrm{ω1}$ associated to the symmetry of infinitesimal translations in the dependent variable $u$. We check the horizontal exterior derivative of $\mathrm{ω1}$ vanishes on solutions to the 2+1 wave equation.

 J > $\mathrm{X1}≔{\mathrm{D_u}}_{[]}$
 ${\mathrm{X1}}{≔}{{\mathrm{D_u}}}_{\left[\right]}$ (2.14)
 J > $\mathrm{ω1}≔\mathrm{Noether}\left(\mathrm{X1},\mathrm{λ}\right)$
 ${\mathrm{ω1}}{≔}{2}{}{{u}}_{{3}}{}{\mathrm{Dx}}{}{\bigwedge }{}{\mathrm{Dy}}{+}{2}{}{{u}}_{{2}}{}{\mathrm{Dx}}{}{\bigwedge }{}{\mathrm{Dt}}{-}{2}{}{{u}}_{{1}}{}{\mathrm{Dy}}{}{\bigwedge }{}{\mathrm{Dt}}$ (2.15)
 J > $\mathrm{HorizontalExteriorDerivative}\left(\mathrm{ω1}\right)$
 ${-}\left({2}{}{{u}}_{{1}{,}{1}}{+}{2}{}{{u}}_{{2}{,}{2}}{-}{2}{}{{u}}_{{3}{,}{3}}\right){}{\mathrm{Dx}}{}{\bigwedge }{}{\mathrm{Dy}}{}{\bigwedge }{}{\mathrm{Dt}}$ (2.16)

Let us find the conservation law $\mathrm{ω2}$ associated to the symmetry of infinitesimal scaling of the independent and dependent variables. We check the the horizontal exterior derivative of $\mathrm{ω2}$ vanishes on solutions to the 2+1 wave equation.

 J > $\mathrm{X2}≔\mathrm{evalDG}\left(x\mathrm{D_x}+y\mathrm{D_y}+t\mathrm{D_t}-\frac{1{u}_{[]}{\mathrm{D_u}}_{[]}}{2}\right)$
 ${\mathrm{X2}}{≔}{x}{}{\mathrm{D_x}}{+}{y}{}{\mathrm{D_y}}{+}{t}{}{\mathrm{D_t}}{-}\frac{{{u}}_{\left[\right]}}{{2}}{}{{\mathrm{D_u}}}_{\left[\right]}$ (2.17)
 J > $\mathrm{ω2}≔\mathrm{Noether}\left(\mathrm{X2},\mathrm{λ}\right)$
 J > $\mathrm{HorizontalExteriorDerivative}\left(\mathrm{ω1}\right)$
 ${\mathrm{ω2}}{≔}{-}\left({t}{}{{u}}_{{1}}^{{2}}{+}{t}{}{{u}}_{{2}}^{{2}}{+}{t}{}{{u}}_{{3}}^{{2}}{+}{2}{}{x}{}{{u}}_{{1}}{}{{u}}_{{3}}{+}{2}{}{y}{}{{u}}_{{2}}{}{{u}}_{{3}}{+}{{u}}_{\left[\right]}{}{{u}}_{{3}}\right){}{\mathrm{Dx}}{}{\bigwedge }{}{\mathrm{Dy}}{-}\left({2}{}{t}{}{{u}}_{{2}}{}{{u}}_{{3}}{+}{2}{}{x}{}{{u}}_{{1}}{}{{u}}_{{2}}{-}{y}{}{{u}}_{{1}}^{{2}}{+}{y}{}{{u}}_{{2}}^{{2}}{+}{y}{}{{u}}_{{3}}^{{2}}{+}{{u}}_{\left[\right]}{}{{u}}_{{2}}\right){}{\mathrm{Dx}}{}{\bigwedge }{}{\mathrm{Dt}}{+}\left({2}{}{t}{}{{u}}_{{1}}{}{{u}}_{{3}}{+}{x}{}{{u}}_{{1}}^{{2}}{-}{x}{}{{u}}_{{2}}^{{2}}{+}{x}{}{{u}}_{{3}}^{{2}}{+}{2}{}{y}{}{{u}}_{{1}}{}{{u}}_{{2}}{+}{{u}}_{\left[\right]}{}{{u}}_{{1}}\right){}{\mathrm{Dy}}{}{\bigwedge }{}{\mathrm{Dt}}$
 ${-}\left({2}{}{{u}}_{{1}{,}{1}}{+}{2}{}{{u}}_{{2}{,}{2}}{-}{2}{}{{u}}_{{3}{,}{3}}\right){}{\mathrm{Dx}}{}{\bigwedge }{}{\mathrm{Dy}}{}{\bigwedge }{}{\mathrm{Dt}}$ (2.18)
 J > $\mathrm{factor}\left(\mathrm{HorizontalExteriorDerivative}\left(\mathrm{ω2}\right)\right)$
 $\left({{u}}_{{1}{,}{1}}{+}{{u}}_{{2}{,}{2}}{-}{{u}}_{{3}{,}{3}}\right){}\left({2}{}{{u}}_{{3}}{}{t}{+}{2}{}{{u}}_{{1}}{}{x}{+}{2}{}{{u}}_{{2}}{}{y}{+}{{u}}_{\left[\right]}\right){}{\mathrm{Dx}}{}{\bigwedge }{}{\mathrm{Dy}}{}{\bigwedge }{}{\mathrm{Dt}}$ (2.19)

Finally, let us find the conservation law $\mathrm{ω3}$ associated to the symmetry of infinitesimal boosts in the independent variables $x$ and $t$. We check that the horizontal exterior derivative of $\mathrm{ω3}$ vanishes on solutions to the 2+1 wave equation.

 J > $\mathrm{X3}≔\mathrm{evalDG}\left(x\mathrm{D_t}+t\mathrm{D_x}\right)$
 ${\mathrm{X3}}{≔}{t}{}{\mathrm{D_x}}{+}{x}{}{\mathrm{D_t}}$ (2.20)
 J > $\mathrm{ω3}≔\mathrm{Noether}\left(\mathrm{X3},\mathrm{λ}\right)$
 ${\mathrm{ω3}}{≔}{-}\left({2}{}{t}{}{{u}}_{{1}}{}{{u}}_{{3}}{+}{x}{}{{u}}_{{1}}^{{2}}{+}{x}{}{{u}}_{{2}}^{{2}}{+}{x}{}{{u}}_{{3}}^{{2}}\right){}{\mathrm{Dx}}{}{\bigwedge }{}{\mathrm{Dy}}{-}{2}{}{{u}}_{{2}}{}\left({t}{}{{u}}_{{1}}{+}{x}{}{{u}}_{{3}}\right){}{\mathrm{Dx}}{}{\bigwedge }{}{\mathrm{Dt}}{+}\left({t}{}{{u}}_{{1}}^{{2}}{-}{t}{}{{u}}_{{2}}^{{2}}{+}{t}{}{{u}}_{{3}}^{{2}}{+}{2}{}{x}{}{{u}}_{{1}}{}{{u}}_{{3}}\right){}{\mathrm{Dy}}{}{\bigwedge }{}{\mathrm{Dt}}$ (2.21)
 J > $\mathrm{factor}\left(\mathrm{HorizontalExteriorDerivative}\left(\mathrm{ω3}\right)\right)$
 ${2}{}\left({{u}}_{{1}{,}{1}}{+}{{u}}_{{2}{,}{2}}{-}{{u}}_{{3}{,}{3}}\right){}\left({t}{}{{u}}_{{1}}{+}{x}{}{{u}}_{{3}}\right){}{\mathrm{Dx}}{}{\bigwedge }{}{\mathrm{Dy}}{}{\bigwedge }{}{\mathrm{Dt}}$ (2.22)