Routines in the liesymm package manipulate and produce results expressed in terms of ``inert derivatives''. This is accomplished by using ``Diff'' instead of ``diff''. The dvalue() command forces the evaluation to take place as if ``diff'' had been used, and then rewrites the final result in terms of the inert ``Diff''.

The action of dvalue differs from that of value only in that the final result is expressed in terms of ``Diff'' rather than ``diff''.

One can also use convert(...) to reformulate a given PDE See the examples below.

One key advantage of the inert representation is that the operands can be modified to reflect desired changes in dependencies prior to evaluation.  This can be accomplished by use of vfix().

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[dvalue]()

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

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

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

$\mathrm{eq1}≔\mathrm{map}\left(\mathrm{Diff}\,\mathrm{eq}\,t\right)$

$\mathrm{eq1}≔\frac{\partial }{\partial t}\left(\frac{{\partial }^2}{\partial x^2}h\left(t\,x\right)\right)=\frac{{\partial }^2}{\partial t^2}h\left(t\,x\right)$

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

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

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

$\mathrm{false}$

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

$\mathrm{true}$

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

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

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

$0=0$

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

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

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

$\frac{{\ⅆ}^2}{\ⅆx^2}h\left(x\right)=0$

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

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

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

$\mathrm{true}$

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

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

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

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