2 to 1 dimension in linear PDE with non-constant coefficients

Question

I have a question that can majorly help in my physics.

Problem

Say, we have a linear PDE

\begin{equation} \hat{D}~F(x,y)=0, \end{equation}

with $\hat{D}$ being a (second order) differential operator, containing, in general, non-constant coefficients, $\partial_x$, $\partial_y$, $\partial_x^2$, $\partial_y^2$ and $\partial_{xy}$.

Suppose, I know a solution $F(x,y)$ - surface, and I want to cut it with a curve $y=f(x)$, so that I get a function $\psi(x)=F(x,f(x))$.

Question

Is there any general way, to transform the $\hat{D}$ 2D operator to $\hat{D'}$ 1D operator, containing $x$, $\partial_x$ and $\partial_x^2$, so that

\begin{equation} \hat{D'}~\psi(x)=0. \end{equation}

Would much appreciate any close answer or even a probable reference where could I read about this.

Answer 1

My answer will be very vague. But you need to find invariants of this equation and after you find go to canonical coordinates and it will reduce the dimensionality of the equation(will make it ODE) Prof. Bluman is one of the most prominent scientist in the field of symmetries you can check out books on his website for more information. https://www.math.ubc.ca/~bluman/

I recommend this book: Bluman, G, Cheviakov, A & Anco, S, Applications of Symmetry Methods to Partial Differential Equations, 417pp. Springer, New York, Vo. 168, Appl. Math. Sci. 2010.