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

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

# Problem

Say, we have a linear PDE

$$\hat{D}~F(x,y)=0,$$

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

$$\hat{D'}~\psi(x)=0.$$

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

• Use the chain rule for partial derivatives, assuming $y$ is a function of $x$ to rewrite any derivatives with respect to $y$ in the $2D$ operator to derivatives with respect to $x$. Jun 27, 2015 at 0:25
• I thought so, but... What would you take for $\partial_{xy}$? If $f(x)$ is not linear, then $\partial_{xy}\ne \partial_{yx}$.
– hayk
Jun 27, 2015 at 0:59
• Even more, the ODE would not be correct in any case. Say, we have $(\partial_x + \partial_y)F(x,y)=0$. We substitute $y=f(x)$, $\psi(x)=F(x,f(x))$; $\partial_y\rightarrow (f')^{-1}\partial_x$. And so we get $((f')^{-1}+1)\psi'(x)=0$. And $\psi=const$, as $f(x)$ is arbitrary. But $F(x,f(x))$ is NOT $const$ for all $f(x)$.
– hayk
Jun 27, 2015 at 1:04

• Well, yes, symmetries and invariants in general indeed reduce the dimension of PDE. The problem is, that I need to do it for any $y=f(x)$, or in general parametric form $x=f(t)$, $y=g(t)$. But maybe I'd be able to work out from that way, so I'll take a glance at those books. Thanks = )