2.2.22.9 Selection of ansatz to try

The following are selection of ansatz to try for solving the linearized PDE above generated from the symmetry condition in order to solve for \(\xi \left ( x,y\right ) ,\eta \left ( x,y\right ) \). These use the functional form. As a general rule, the simpler that ansatz that works, the better it is.  Functional form of ansatz is better than explicit polynomials but much harder to use and implement. Maple’s symgen has 16 different algorithms include HINT option to support functional forms. The following are possible cases to use.

1. \(\xi =0,\eta =f\left ( x\right ) \)
2. \(\xi =0,\eta =f\left ( y\right ) \)
3. \(\xi =f\left ( x\right ) ,\eta =0\)
4. \(\xi =f\left ( y\right ) ,\eta =0\)
5. \(\xi =f\left ( x\right ) ,\eta =xg\left ( y\right ) \). An example: applied to \(y^{\prime }=\frac {x+\cos \left ( e^{y}+\left ( 1+x\right ) e^{-x}\right ) }{e^{y+x}}\) should give \(\xi =e^{x},\eta =xe^{-y}\) which leads to solution \(y=\ln \left ( 2\arctan \left ( \frac {e^{-\left ( c_{1}+e^{-x}\right ) -1}}{e^{-\left ( c_{1}+e^{-x}\right ) +1}}\right ) -\left ( 1+x\right ) e^{-x}\right ) \).
6. \(\xi =f\left ( x\right ) ,\eta =g\left ( y\right ) \)
7. \(\xi =0,\eta =f\left ( x\right ) g\left ( y\right ) \). For example, applied to \(y^{\prime }=\frac {x\sqrt {1+y}+\sqrt {1+y}+1+y}{1+x}\) should give \(f\left ( x\right ) =\sqrt {1+x},g\left ( y\right ) =\sqrt {1+y}\).
8. \(\xi =f\left ( x\right ) g\left ( y\right ) ,\eta =0\)