Given a list with the infinitesimals of a symmetry generator or the corresponding infinitesimal generator operator, or a list of n of these lists or operators, possibly representing an n-dimensional symmetry group, InvariantTransformation command computes a transformation from the dependent and independent variables of the problem to variables that reduces the number of independent variables of PDESYS by one, or optionally by as many as indicated in the optional argument reduceinonego = n, where n is a positive integer; this option is valid when the first argument is a list of symmetries representing a multidimensional symmetry group.

When there is only one dependent variable, DepVars and NewVars can be functions; otherwise they must be a list of functions representing dependent variables. If NewVars are not given, InvariantTransformation will generate a list of globals to represent them.

By default, InvariantTransformation returns using function notation. It is sometimes convenient to have this output expressed in one of the jet notations available. For that purpose use ToJet or the optional argument jetnotation = ... where the right-hand-side is any of true, jetODE, jetvariables or withoutbrackets. For more information, see jet notations.

Optionally you can also pass a list of invariants or a list of lists of invariants instead of lists of infinitesimals, using the optional argument invariants = .... You can also specify a simplifier, or the dependency of the solutions expected to be obtained using these transformations; all these options are the same ones explained in the help page for InvariantSolutions.

To avoid having to remember the optional keywords, if you type the keyword misspelled, or just a portion of it, a matching against the correct keywords is performed, and when there is only one match, the input is automatically corrected.

$\mathrm{with}\left(\mathrm{PDEtools}\,\mathrm{InvariantTransformation}\,\mathrm{ChangeSymmetry}\,\mathrm{InfinitesimalGenerator}\right)$

$\left[\mathrm{InvariantTransformation}\,\mathrm{ChangeSymmetry}\,\mathrm{InfinitesimalGenerator}\right]$

$S≔\left[\mathrm{\_\ξ}\left[x\right]=x\,\mathrm{\_\ξ}\left[t\right]=1\,\mathrm{\_\η}\left[u\right]=u\right]$

$S≔\left[{\mathrm{\_\ξ}}_x=x\,{\mathrm{\_\ξ}}_t=1\,{\mathrm{\_\η}}_u=u\right]$

$G≔\mathrm{InfinitesimalGenerator}\left(S\,u\left(x\,t\right)\right)$

$G≔f\→x\left(\frac{\partial }{\partial x}f\right)\+\frac{\partial }{\partial t}f\+u\left(\frac{\partial }{\partial u}f\right)$

$\mathrm{InvariantTransformation}\left(S\,u\left(x\,t\right)\,v\left(r\,s\right)\right)$

$\left\{r=-\ln \left(x\right)+t\,s=t\,v\left(r\,s\right)=\frac{u\left(x\,t\right)}{x}\right\},\left\{t=s\,x={\ⅇ}^{-r+s}\,u\left(x\,t\right)=v\left(r\,s\right){\ⅇ}^{-r+s}\right\}$

$\mathrm{InvariantTransformation}\left(S\,u\left(x\,t\right)\,v\left(r\,s\right)\,\mathrm{jetnotation}\right)$

$\left\{r=-\ln \left(x\right)+t\,s=t\,v=\frac{u}{x}\right\},\left\{t=s\,u=v{\ⅇ}^{-r+s}\,x={\ⅇ}^{-r+s}\right\}$

$\mathrm{InvariantTransformation}\left(S\,u\left(x\,t\right)\,v\left(r\,s\right)\,\mathrm{jetnotation}=\mathrm{jetnumbers}\right)$

$\mathrm{jet\; notation:\; [1\; =\; x,\; 2\; =\; t,\; 3\; =\; r,\; 4\; =\; s]}$

$\mathrm{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}$

$\left\{r=-\ln \left(x\right)+t\,s=t\,v\left[\right]=\frac{u\left[\right]}{x}\right\},\left\{t=s\,x={\ⅇ}^{-r+s}\,u\left[\right]=v\left[\right]{\ⅇ}^{-r+s}\right\}$

$\mathrm{TR},\mathrm{NewVars}≔\left[2\right],\mathrm{map}\left(\mathrm{lhs}\,\left[1\right]\right)$

$\mathrm{TR},\mathrm{NewVars}≔\left\{t=s\,x={\ⅇ}^{-r+s}\,u\left(x\,t\right)=v\left(r\,s\right){\ⅇ}^{-r+s}\right\},\left\{r\,s\,v\left(r\,s\right)\right\}$

$\mathrm{ChangeSymmetry}\left(\mathrm{TR}\,S\,u\left(x\,t\right)\,\mathrm{NewVars}\right)$

$\left[{\mathrm{\_\ξ}}_r=0\,{\mathrm{\_\ξ}}_s=1\,{\mathrm{\_\η}}_v=0\right]$

$\mathrm{InfinitesimalGenerator}\left(\,v\left(r\,s\right)\right)$

$f\→\frac{\partial }{\partial s}f$

$\mathrm{InvariantTransformation}\left(S\,u\left(x\,t\right)\right)$

$\left\{\mathrm{\_t1}=-\ln \left(x\right)+t\,\mathrm{\_t2}=t\,\mathrm{\_u1}\left(\mathrm{\_t1}\right)=\frac{u\left(x\,t\right)}{x}\right\},\left\{t=\mathrm{\_t2}\,x={\ⅇ}^{-\mathrm{\_t1}+\mathrm{\_t2}}\,u\left(x\,t\right)=\mathrm{\_u1}\left(\mathrm{\_t1}\right){\ⅇ}^{-\mathrm{\_t1}+\mathrm{\_t2}}\right\}$