MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Say $f:\mathbb R^{n+1}\to \mathbb R^p$ is a solution to an initial value problem, and $g:\mathbb R^{n+1}\to \mathbb R^q$, so that the components of $g$ can be expressed as polynomials in $f$, $f'$, and the partial derivatives of $f$.

Is $g$ a solution to an(other) initial value problem? Are there any references studying properties of $g$ in terms of those of $f$?

I am interested in the general case, and I don't think I can ask the question for a particular case, but I would appreciate even results under more specific conditions.

share|cite|improve this question

I guess this is not a complete answer, but there is a sufficient condition for the answer to be positive. Suppose that $E_1(j^{k_1} f(x)) = 0$ is the hyperbolic PDE of order $k_1$ defining your initial value problem. By $j^{k_1} f(x) = (x,f(x),\partial f(x),\ldots, \partial^{k_1} f(x))$ I mean the $k_1$-jet of $f$ at $x$. Similarly, let $g(x) = G_1(j^{l_1} f(x))$, be the variable locally defined in terms of $f$. Then $g(x)$ satisfies an initial value problem if there exist functions $G_2$ and $E_2$ such that $E_2(j^{k_2} G_1(j^{l_1} f(x))) = G_2(j^{l_2} E_1(j^{k_1} f(x)))$ and $E_2$ defines a hyperbolic PDE of degree $k_2$. Then, necessarily, if $E_1(j^{k_1} f(x)) = 0$ and $g(x) = G_1(j^{l_1} f(x))$, you have $E_2(j^{k_2} g(x)) = 0$ and so $g$ satisfies an initial value problem as well.

If you think of the $E_i$ and $G_i$ as maps between jet spaces (or simply as commutative operators) the condition in the above paragraph is just the "commutative square" condition $E_2 \circ G_1 = G_2 \circ E_1$ (the $E_i$ are the vertical maps, say, and the $G_i$ are the horizontal ones). I'm not sure what are the precise necessary conditions for the ability to complete such a commutative square given only $E_1$ and $G_1$.

I imagine that in general such a commutative square cannot always be completed. If it is completable, then the condition $g(x) = G_1(j^{l_1} f(x)) = 0$ is said to be a "constraint compatible with evolution" with respect to the equation defined by $E_1$, since, if its jet vanishes on an initial data surface, it vanishes everywhere in the evolution domain. And, AFAIK, such constraints are considered special.

share|cite|improve this answer
Thank you for your answer. – Cristi Stoica May 28 '13 at 3:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.