First order PDE, singular at a point - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T22:28:39Z http://mathoverflow.net/feeds/question/83948 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/83948/first-order-pde-singular-at-a-point First order PDE, singular at a point Kofi 2011-12-20T14:47:27Z 2011-12-21T15:18:04Z <p>I am pretty sure this should be text book material, but I couldn't find this anywhere; maybe I just don't know where to look.</p> <p><strong>Problem:</strong> Suppose we have a smooth vector field $X = a_i x^i \partial_i + O(|x|^2)$ with $a_i>0$, defined on some neighborhood of $0$ (or near some point on a manifold, that doesn't really matter since this is a local problem). Now I want to find all smooth functions $f \in C^{\infty}(U,\mathbb{R})$ and all $\alpha \in \mathbb{R}$ such that $$\partial_X f = \alpha \cdot f$$ on some neighborhood $U$ of $0$.</p> <p><strong>Conjecture:</strong> From looking at the linear example $X = a_ix^\partial_i$, I would think that either $f = 0$ or $\alpha = \sum k_i a_i$ for some $k \in \mathbb{N}_0^n$ and $f = C\cdot x^k + O(|x|^{k+1})$. (Because the exponent has to be integral for the solution to be smooth at $0$).</p> <p><strong>Remark:</strong> 1) This seems to be an eigenvalue problem of some sort, but I have no idea how to make this rigourous; like, in what space does the operator work, what is the domain etc.</p> <p>2) This is of course an ODE along the integral curves of the vector field $X$. The problem is though, that the curves don't start at the point $0$, just start there "asymptotically". One could transfer this into a singular ODE (with which substitution?) and try to go with integrating factors, but this gets tricky.</p> <p>/Edit: Renamed $\eta$ to $f$.</p> http://mathoverflow.net/questions/83948/first-order-pde-singular-at-a-point/84010#84010 Answer by Robert Bryant for First order PDE, singular at a point Robert Bryant 2011-12-21T15:18:04Z 2011-12-21T15:18:04Z <p>Your conjecture is true, and it can be proved by making a few observations.</p> <p>First, for each $\alpha\in\mathbb{R}$, let $V_\alpha$ be the vector space of germs of smooth functions $f$ at the origin that satisfy $X\ f = \alpha f$, and let $V_\alpha^k\subset V_\alpha$ denote the subspace that consists of those elements of $V_\alpha$ that vanish to order at least $k$ at $x=0$.</p> <p>Second, note that <code>$V_\alpha = V^0_\alpha \supseteq V^1_\alpha\supseteq \cdots \supseteq V^k_\alpha \supseteq V^{k+1}_\alpha\supseteq\cdots$</code>, that the operation of applying $X$ preserves each of the subspaces $V^k_\alpha$, and that the intersection of all of these subspaces is the zero subspace. (This last statement is the only one that isn't obvious, but it's not hard to prove, given that the $a_i$ are all positive.)</p> <p>Third, because $X$ preserves the nesting of the subspaces, it follows that $X$ induces a well-defined linear map $\bar X : W^k_\alpha \to W^k_\alpha$, where $W^k_\alpha = V^k_\alpha/V^{k+1}_\alpha$, which is simply multiplication by $\alpha$. However it's obvious from the action of $\bar X$ on the $x$-monomials of degree $k$ (which is obviously diagonal) that the eigenvalues of $\bar X$ on this space are all of the form $$\lambda = k_1\ a_1 + k_2\ a_2 + \cdots + k_n\ a_n$$ where $k = k_1 + k_2 + \cdots + k_n$ and each $k_i\ge 0$. Thus, $\alpha$ must be of this form if $W^k_\alpha$ is to be nonzero. </p> <p>Fourth, it's now clear that, for $k >>0$, we must have $W^k_\alpha=0$, so it follows that $V^k_\alpha = 0$ for all large $k$ as well. Since each $W^k_\alpha$ is finite dimensional (in fact, you can figure out an upper bound for its dimension quite easily), it follows that $V_\alpha$ is also finite dimensional. In fact, it is easy to see that the dimension of $V_\alpha$ is never more than the dimension of the corresponding vector space for the linearized vector field.</p>