I am in the process of solving the equation $y''=0$ where $y=y(x)$ using symmetry methods. I have not studied this method and am trying to teach myself by following examples I find online.
UPDATE : I have been able to determine the following based on what I read in Bluman & Anco:
UPDATE: My Attempt: (based on what I found in Bluman & Anco)
Before we move further, we need to understand the concept of the Total Derivative Operator: \begin{equation}D=\frac{\partial}{\partial x}+\dot{y}\frac{\partial}{\partial y} \label{eq:5} \tag{5}\end{equation} NB:$DF(x,y)=F_x+y'F_y$
The $k^{th}$ prolongation is defined by:
\begin{equation}\zeta^{(k)}(x,y)=D\zeta^{(k-1)}(x,y)-y^kD\xi \label{eq:Prol} \tag{6}\end{equation} where $\zeta^{(0)}=\eta(x,y)$
Using the above definition we get the following:
First Prolongation \begin{align} \zeta^{(1)} & =D\eta(x,y) -\dot{y}D(\xi(x,y)) \nonumber \\ & = \eta_x +\dot{y}\eta_y-\dot{y} \left( \xi_x+\dot{y}\xi_y\right) \nonumber \\ & = \eta_x +\dot{y}\eta_y -\dot{y}\xi_x - \dot{y}^2\xi_y\\ \label{eq:ProlOne} \tag{7} \end{align} Second Prolongation \begin{align} \zeta^{(2)} & = D\zeta^{(1)}-\ddot{y}D\xi(x,y) \hspace{10mm} \text{$ \ddot{y}=0$ so the second term falls away} \nonumber \\ & = \eta_{xx}+\dot{y}\eta_{xy}-\dot{y}\xi_{xx}-\dot{y}^2\xi{xy}+\dot{y}\left(\eta_{xy}+\dot{y}\eta_{yy}-\dot{y}\xi_{xy}-\dot{y}^2\xi_{yy}\right) \nonumber\\ &=\eta_{xx} +\left(2\eta_{xy}-\xi_{xx}\right)\dot{y} +\left(\eta_{yy}-2\xi_{xy}\right)\dot{y}^2-\xi_{yy}\dot{y}^3\\ \label{ProlTwo} \tag{8} \end{align}
Is this correct?
I also need to find the Determining Equation
Am I correct in saying the Determining Equation is $\zeta^{(2)}=0$ ?
If so, I need to be able to show why this is the case. I am currently reading, and re-reading the sections in Bluman & Anco that apply.