Routines in the liesymm package manipulate and produce results expressed in terms of ``inert derivatives'' by using Diff(). A key advantage of this inert representation is that the operands can be modified to reflect desired changes in dependencies prior to evaluation. The vfix() command changes the variable dependencies for functions appearing in these expressions.

One can use convert(...) and dvalue() to reformulate the resulting PDE.

The results of using diff() and Diff() often display the same way.  To determine which of these two is actually present, use lprint().  or has( ... , diff), etc..

This routine is ordinarily loaded via with(liesymm) but can be used in the ``package style'' as liesymm[vfix]()

$\mathrm{with}\left(\mathrm{liesymm}\right)\:$

$\mathrm{Diff}\left(h\left(t\,x\right)\,x\,x\right)=\mathrm{Diff}\left(h\left(t\,x\right)\,t\right)$

$\frac{{\partial }^2}{\partial x^2}h\left(t\,x\right)=\frac{\partial }{\partial t}h\left(t\,x\right)$

$\mathrm{eq}≔\mathrm{dvalue}\left(\right)$

$\mathrm{eq}≔\frac{{\partial }^2}{\partial x^2}h\left(t\,x\right)=\frac{\partial }{\partial t}h\left(t\,x\right)$

$\mathrm{map}\left(\mathrm{Diff}\,\mathrm{eq}\,t\right)$

$\frac{\partial }{\partial t}\left(\frac{{\partial }^2}{\partial x^2}h\left(t\,x\right)\right)=\frac{{\partial }^2}{\partial t^2}h\left(t\,x\right)$

$\mathrm{eq2}≔\mathrm{dvalue}\left(\right)$

$\mathrm{eq2}≔\frac{{\partial }^2}{\partial x^2}\left(\frac{\partial }{\partial t}h\left(t\,x\right)\right)=\frac{{\partial }^2}{\partial t^2}h\left(t\,x\right)$

$\mathrm{has}\left(\mathrm{eq2}\,\mathrm{diff}\right)$

$\mathrm{false}$

$\mathrm{has}\left(\mathrm{value}\left(\mathrm{eq2}\right)\,\mathrm{diff}\right)$

$\mathrm{true}$

$\mathrm{vfix}\left(\mathrm{eq}\,\left[\right]\,h\right)$

$\frac{{\ⅆ}^{2}h}{\ⅆx^2}=\frac{\ⅆh}{\ⅆt}$

$\mathrm{vfix}\left(\mathrm{eq}\,\left[x\right]\,h\right)$

$\frac{{\ⅆ}^2}{\ⅆx^2}h\left(x\right)=\frac{\partial }{\partial t}h\left(x\right)$

$\mathrm{convert}\left(\mathrm{eq}\,\mathrm{diff}\right)$

$\frac{{\partial }^2}{\partial x^2}h\left(t\,x\right)=\frac{\partial }{\partial t}h\left(t\,x\right)$

$\mathrm{has}\left(\,\mathrm{diff}\right)$

$\mathrm{true}$

$\mathrm{convert}\left(\mathrm{eq}\,\mathrm{D}\right)$

${\mathrm{D}}_{2,2}\left(h\right)\left(t\,x\right)={\mathrm{D}}_1\left(h\right)\left(t\,x\right)$

$\mathrm{convert}\left(\,\mathrm{Diff}\right)$

$\frac{{\partial }^2}{\partial x^2}h\left(t\,x\right)=\frac{\partial }{\partial t}h\left(t\,x\right)$