2.1 Linearized PDE of the similarity condition

Obtaining the linearized PDE of the similarity condition for second order ode, which is used to solve for \(\xi ,\eta \) follows similar method as given earlier for the first order ode. The difference is that instead of \(y^{\prime }=\omega \left ( x,y\right ) \) the ode now \(y^{\prime \prime }=\omega \left ( x,y,y^{\prime }\right ) \).

\begin{align} \frac {d^{2}y}{dx^{2}} & =\omega \left ( x,y,\frac {dy}{dx}\right ) \nonumber \\ y^{\prime \prime } & =\omega \left ( x,y,y^{\prime }\right ) \tag {A}\end{align}

The linearized similarity condition for second order ode when \(\omega =0\) is

\[ \eta _{xx}+\left ( 2\eta _{xy}-\xi _{xx}\right ) y^{\prime }+\left ( \eta _{yy}-2\xi _{xy}\right ) \left ( y^{\prime }\right ) ^{2}-\xi _{yy}\left ( y^{\prime }\right ) ^{3}=0 \]

Which is polynomial in \(y^{\prime }\) hence all the coefficients must be zero giving

\begin{align*} 2\eta _{xy}-\xi _{xx} & =0\\ \eta _{yy}-2\xi _{xy} & =0\\ \xi _{yy} & =0\\ \eta _{xx} & =0 \end{align*}

And for general \(\omega \left ( x,y,y^{\prime }\right ) \), the linearized similarity condition is

\[ -\eta \omega _{y}+\left ( -3y^{\prime }\xi _{y}-2\xi _{x}+\eta _{y}\right ) \omega -\xi \omega _{x}+\left ( -y^{\prime }\eta _{y}+\left ( y^{\prime }\right ) ^{2}\xi _{y}+y^{\prime }\xi _{x}-\eta _{x}\right ) \omega _{y^{\prime }}+\eta _{xx}-\xi _{yy}\left ( y^{\prime }\right ) ^{3}+\left ( \eta _{yy}-2\xi _{yx}\right ) \left ( y^{\prime }\right ) ^{2}+\left ( 2\eta _{yx}-\xi _{xx}\right ) y^{\prime }=0 \]

To continue