Solving $x\partial_x f = 0$ over distributions - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T06:13:43Z http://mathoverflow.net/feeds/question/62364 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/62364/solving-x-partial-x-f-0-over-distributions Solving $x\partial_x f = 0$ over distributions Jacques Carette 2011-04-20T02:22:27Z 2012-04-21T19:18:09Z <p>Solving $x\partial_x f = 0$ over 'normal' functions is the same as solving $\partial_x f = 0$, i.e. one gets $f(x)=c_1$ as the complete answer. But over distributions (if my calculations are correct), $$f(x) = (c_2-c_1)H(x)+c_1$$ is the complete solution (with H being the Heaviside step function). For another comparison point, $x^2\partial_x f=0$ has solution $$f(x) = (c_2-c_1)H(x)+c_1+c_3\delta(x)$$ (with $\delta$ the Dirac $\delta$ function/distribution).</p> <p>My main question: are my computations correct? Are these in fact the most general solutions? [I have 3 different arguments showing that these are indeed solutions, although I am not sure that any of these constitute proper proofs, the last time I did anything with distributions was almost 20 years ago].</p> <p>Motivation: What I am actually trying to do is to get a differential equation for the density function for the <a href="http://en.wikipedia.org/wiki/Pareto_distribution" rel="nofollow">Pareto Distribution</a> (where 'distribution' here is used in a different sense). The only remaining problem is to properly take care of the 'jump' at $x_m$. The above should give me what I am missing to get there.</p> http://mathoverflow.net/questions/62364/solving-x-partial-x-f-0-over-distributions/67287#67287 Answer by Sylvain Bonnot for Solving $x\partial_x f = 0$ over distributions Sylvain Bonnot 2011-06-08T17:33:32Z 2011-06-08T17:33:32Z <p>Your answer seems correct to me. The equation $x.u=0$ (where $u$ is a distribution) has solution $u=c_{1}\delta_{0}$. Then the equation $\partial_{x}v=c_{1}\delta_{0}$ has solution $v=c_{1}H+c_{2}$... You can find the first result for example in the book by Friedlander and Joshi (thm 2.7.1).</p> http://mathoverflow.net/questions/62364/solving-x-partial-x-f-0-over-distributions/67815#67815 Answer by paul garrett for Solving $x\partial_x f = 0$ over distributions paul garrett 2011-06-14T23:37:33Z 2011-06-14T23:37:33Z <p>Of course, since the questioner does not divulge their arguments, their correctness cannot be confirmed or denied. In any case, there are some correct assertions which are fairly intuitive, and which follow from the definitions rather easily. A useful and simple example is that, for a smooth function f and distribution u, if f.u=0 then the support of u is contained in the zero-set of f.</p> <p>A second example is the iconic fact that distributions supported at a point are finite linear combinations of derivatives of Dirac's delta. This is proven many places: Rudin's functional analysis, my notes on functional analysis, ...</p> <p>This is subsumed by the idea is that distributions supported on a sub-manifold are <em>locally</em> values of distributions on the submanifold on transversal derivatives. (This is also in a small note on my functional analysis page. It is folkloric...)</p> <p>There are also useful and interesting things that can be said about <em>homogeneous</em> distributions, or equivariant ones... but this seems not in the questioner's interest.</p> http://mathoverflow.net/questions/62364/solving-x-partial-x-f-0-over-distributions/94766#94766 Answer by Bazin for Solving $x\partial_x f = 0$ over distributions Bazin 2012-04-21T19:18:09Z 2012-04-21T19:18:09Z <p>Let us work in $\mathbb R^n$. The distribution solutions of the equation $$(x\cdot \partial _x) u=0$$ are the distributions which are homogeneous of degree 0. Here $x\cdot \partial_x=\sum_{1\le j\le n}x_j\partial_{x_j}$ and the previous equation is the so-called Euler equation for homogeneous distributions with degree 0. The fact that $u$ satisfies the first equation is equivalent to $$\forall \lambda >0,\quad u(\lambda x) =u(x),$$ which means with brackets of duality $\langle u(x),\phi(x/\lambda)\rangle\lambda^{-n}=\langle u(x),\phi(x)\rangle.$ In one dimension, the homogeneous distributions with degree 0 are $$a H(x)+bH(-x),$$ where $H$ is the Heaviside function (characteristic function of $\mathbb R_+$). In fact the restriction to $\mathbb R_+$ (resp. $\mathbb R_-$) must be constant (respectively $a,b$) so that $u(x)-a H(x)-bH(-x)$ is homogeneous of degree 0 and supported at the origin. Since the distributions supported at 0 are finite linear combinations of derivatives of the Dirac mass and the Dirac mass in one dimension is homogeneous with degree $-1$, $u(x)-a H(x)-bH(-x)=0$. </p> <p>Note that the solutions of $$(x\cdot \partial _x) u=\mu u$$ are the distributions homogeneous with degree $\mu$, which means also for all $\lambda >0$ $$\langle u(x),\phi(x/\lambda)\rangle\lambda^{-n}=\lambda ^\mu\langle u(x),\phi(x)\rangle.$$ A simple example of a distribution homogeneous of degree $-n$ is the Dirac mass at 0 in $\mathbb R^n$.</p> <p>Bazin.</p>