1
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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 $x$. $\endgroup$
    – OnceUponACrinoid
    Jun 27, 2015 at 0:25
  • $\begingroup$ I thought so, but... What would you take for $\partial_{xy}$? If $f(x)$ is not linear, then $\partial_{xy}\ne \partial_{yx}$. $\endgroup$
    – hayk
    Jun 27, 2015 at 0:59
  • $\begingroup$ 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 $f(x)$. $\endgroup$
    – hayk
    Jun 27, 2015 at 1:04

1 Answer 1

Reset to default
1
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ Answers on this site should not be vague. Post a good answer, or don't post at all. Also, since the majority of your post is vague promotion ("hey, I can't answer your question, but buy this book"), this post should be flagged as spam. $\endgroup$
    – Mike Pierce
    Jun 27, 2015 at 2:04
  • $\begingroup$ 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 = ) $\endgroup$
    – hayk
    Jun 27, 2015 at 2:08
  • $\begingroup$ Mike, don't worry, I'm not going to buy this book, I already found it free :) $\endgroup$
    – hayk
    Jun 27, 2015 at 2:13
  • $\begingroup$ Dear Mike, It's not my book. The book is free. And the answer may be there. It is better to have a look there and maybe find the answer. $\endgroup$
    – Vasiliy Triandafilidi
    Oct 24, 2015 at 0:33

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .