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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.

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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.

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Thank you for your answer. –  Cristi Stoica May 28 '13 at 3:25
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