AreSame - Maple Help

AreSame

check if two LHPDE objects are the same

 Calling Sequence AreSame( obj1, obj2, criteria = crit) AreSame( obj1, obj2, criterion = crit)

Parameters

 obj1, obj2, ... - LHPDE objects crit - (optional) a string: "sameOperator", "sameSystem", or "sameSolutions"

Description

 • The AreSame method checks if the two LHPDE objects obj1 and obj2 are the same.
 • This methods returns true if obj1 and obj2 are the same, in the sense they match the following criteria:
 1 criterion = "sameSystem" -- they are identical meaning that they have same independent variables, same dependent variables (with same dependencies) and same system of DEs.
 2 criterion = "sameOperator" -- they have the same operator. That is, LHPDO(obj1) matches LHPDO(obj2) apart from the ordering of DEs. See examples below.
 3 criterion = "sameSolutions" -- they have the same solutions.
 • The default criterion is "sameOperator". That is, AreSame(obj1, obj2) is equivalent to AreSame(obj1, obj2, criteria = "sameOperator").
 • In the second calling sequence, the word criterion is provided as alias for criteria.
 • This method is associated with the LHPDE object. For more detail, see Overview of the LHPDE object.

Examples

 > $\mathrm{with}\left(\mathrm{LieAlgebrasOfVectorFields}\right):$
 > $\mathrm{E2}≔\mathrm{LHPDE}\left(\left[\mathrm{diff}\left(\mathrm{\xi }\left(x,y\right),y,y\right)=0,\mathrm{diff}\left(\mathrm{\eta }\left(x,y\right),x\right)=-\mathrm{diff}\left(\mathrm{\xi }\left(x,y\right),y\right),\mathrm{diff}\left(\mathrm{\eta }\left(x,y\right),y\right)=0,\mathrm{diff}\left(\mathrm{\xi }\left(x,y\right),x\right)=0\right],\mathrm{indep}=\left[x,y\right],\mathrm{dep}=\left[\mathrm{\xi },\mathrm{\eta }\right]\right)$
 ${\mathrm{E2}}{≔}\left[\frac{{{\partial }}^{{2}}}{{\partial }{{y}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{\xi }}{}\left({x}{,}{y}\right){=}{0}{,}\frac{{\partial }}{{\partial }{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{\eta }}{}\left({x}{,}{y}\right){=}{-}\frac{{\partial }}{{\partial }{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{\xi }}{}\left({x}{,}{y}\right){,}\frac{{\partial }}{{\partial }{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{\eta }}{}\left({x}{,}{y}\right){=}{0}{,}\frac{{\partial }}{{\partial }{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{\xi }}{}\left({x}{,}{y}\right){=}{0}\right]{,}{\mathrm{indep}}{=}\left[{x}{,}{y}\right]{,}{\mathrm{dep}}{=}\left[{\mathrm{\xi }}{}\left({x}{,}{y}\right){,}{\mathrm{\eta }}{}\left({x}{,}{y}\right)\right]$ (1)
 > $\mathrm{E2p}≔\mathrm{LHPDE}\left(\left[\mathrm{diff}\left(\mathrm{\alpha }\left(x,y\right),y,y\right)=0,\mathrm{diff}\left(\mathrm{\beta }\left(x\right),x\right)=-\mathrm{diff}\left(\mathrm{\alpha }\left(x,y\right),y\right),\mathrm{diff}\left(\mathrm{\alpha }\left(x,y\right),x\right)=0\right],\mathrm{indep}=\left[x,y\right],\mathrm{dep}=\left[\mathrm{\alpha },\mathrm{\beta }\right]\right)$
 ${\mathrm{E2p}}{≔}\left[\frac{{{\partial }}^{{2}}}{{\partial }{{y}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{\alpha }}{}\left({x}{,}{y}\right){=}{0}{,}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{\beta }}{}\left({x}\right){=}{-}\frac{{\partial }}{{\partial }{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{\alpha }}{}\left({x}{,}{y}\right){,}\frac{{\partial }}{{\partial }{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{\alpha }}{}\left({x}{,}{y}\right){=}{0}\right]{,}{\mathrm{indep}}{=}\left[{x}{,}{y}\right]{,}{\mathrm{dep}}{=}\left[{\mathrm{\alpha }}{}\left({x}{,}{y}\right){,}{\mathrm{\beta }}{}\left({x}\right)\right]$ (2)

The two LHPDE objects are the same as operators:

 > $\mathrm{AreSame}\left(\mathrm{E2},\mathrm{E2p}\right)$
 ${\mathrm{true}}$ (3)

The method returns true as the dependent variable names are different, their LHPDOs (Delta, Delta1) are the same.

 > $\mathrm{\Delta }≔\mathrm{convert}\left(\mathrm{E2},'\mathrm{LHPDO}'\right)$
 ${\mathrm{\Delta }}{≔}\left({\mathrm{ξ}}{,}{\mathrm{η}}\right){→}\left[\frac{{\partial }}{{\partial }{y}}{}\left(\frac{{\partial }}{{\partial }{y}}{}{\mathrm{ξ}}\right){,}\frac{{\partial }}{{\partial }{x}}{}{\mathrm{η}}{+}\frac{{\partial }}{{\partial }{y}}{}{\mathrm{ξ}}{,}\frac{{\partial }}{{\partial }{y}}{}{\mathrm{η}}{,}\frac{{\partial }}{{\partial }{x}}{}{\mathrm{ξ}}\right]$ (4)
 > $\mathrm{Δ1}≔\mathrm{convert}\left(\mathrm{E2p},'\mathrm{LHPDO}'\right)$
 ${\mathrm{Δ1}}{≔}\left({\mathrm{α}}{,}{\mathrm{β}}\right){→}\left[\frac{{\partial }}{{\partial }{y}}{}\left(\frac{{\partial }}{{\partial }{y}}{}{\mathrm{α}}\right){,}\frac{{\partial }}{{\partial }{x}}{}{\mathrm{β}}{+}\frac{{\partial }}{{\partial }{y}}{}{\mathrm{α}}{,}\frac{{\partial }}{{\partial }{x}}{}{\mathrm{α}}\right]$ (5)

Clearly they have different dependent variables, so the systems are not identical.

 > $\mathrm{AreSame}\left(\mathrm{E2},\mathrm{E2p},\mathrm{criterion}="sameSystem"\right)$
 ${\mathrm{false}}$ (6)

Since they are same as operator, they definitely have the same solutions.

 > $\mathrm{AreSame}\left(\mathrm{E2},\mathrm{E2p},\mathrm{criteria}="sameSolutions"\right)$
 ${\mathrm{true}}$ (7)

Compatibility

 • The AreSame command was introduced in Maple 2020.