Getting a differential equation for a function from a functional equation of its Mellin transform - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T08:33:18Z http://mathoverflow.net/feeds/question/36064 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/36064/getting-a-differential-equation-for-a-function-from-a-functional-equation-of-its Getting a differential equation for a function from a functional equation of its Mellin transform Armin Straub 2010-08-19T08:15:28Z 2010-09-03T04:10:52Z <p>If $f$ is a locally integrable function then its <a href="http://en.wikipedia.org/wiki/Mellin_transform" rel="nofollow">Mellin transform</a> $\mathcal{M}[f]$ is defined by $$\mathcal{M}[f] (s) = \int_0^{\infty} x^{s - 1} f (x) dx .$$ This integral usually converges in a strip $\alpha &lt; Re \; s &lt; \beta$ and defines an analytic function. For our purposes we can assume that $\mathcal{M}[f]$ converges in the right half-plane.</p> <p>Let us denote $F (s) =\mathcal{M}[f] (s)$. Provided that the corresponding Mellin transforms exist, the basic general theory tells us that, for instance, $$\mathcal{M} \left[ \frac{d}{d x} f (x) \right] = - (s - 1) F (s - 1),\quad \mathcal{M} \left[ x^{\mu} f (x) \right] = F (s + \mu) .$$ This allows us to translate a differential equation for $f (x)$ into a functional equation for its Mellin transform $F (s)$.</p> <p><em>Example</em>: For instance, the function $f (x) = e^{- x}$ satisfies the differential equation $$f' (x) + f (x) = 0$$ which translates to the functional equation $$- (s - 1) F (s - 1) + F (s) = 0$$ for its Mellin transform. Of course, the Mellin transform of $e^{- x}$ is nothing but the gamma function $\Gamma (s)$ which is well-known for satisfying exactly this functional equation.</p> <p>Now, let us assume that we are given a function $f (x)$ and its Mellin transform $F (s)$. Further, suppose that we know that $F (s)$, just as the gamma function, can be analytically extended to the whole complex plane with poles at certain nonpositive integers. We also know that $F (s)$ satisfies a functional equation which we would like to translate back into a differential equation for $f (x)$. Formally, we obtain, say, a third order differential equation with polynomial coefficients. Can we conclude that $f (x)$ solves this DE?</p> <p>The issue is that in our case the derivatives of $f (x)$ develop singularities in the domain and are no longer integrable. So Mellin transforms can't be defined in the usual way for them (and so we can't just use Mellin inversion).</p> <p>What I am looking for is conditions under which we can still conclude that the functional equation for $F (s)$ translates into a differential equation for $f (x)$. Preferably, these should be conditions on $F (s)$ and not on $f (x)$. If it helps, we can assume $f (x)$ to be compactly supported.</p> <p>Any help or references are greatly appreciated!</p> http://mathoverflow.net/questions/36064/getting-a-differential-equation-for-a-function-from-a-functional-equation-of-its/36076#36076 Answer by Jacques Carette for Getting a differential equation for a function from a functional equation of its Mellin transform Jacques Carette 2010-08-19T12:13:12Z 2010-08-19T12:13:12Z <p>As long as the singularities are not "too bad", the answer will be <em>yes</em>, $f(x)$ will <em>represent a solution of the differential equation</em>. The very same way that $$\sum_{x=0}^{\infty} (-1)^{x}n!x^{n+1}$$ <em>represents</em> a solution of $x^2y'+y=x$. One might object that the series diverges, but resummation theory says this is irrelevant, that sum nevertheless <em>represents</em> a unique function (in a sector) which is a solution of the differential equation.</p> <p>Balser's book, "From divergent series to analytic differential equations", would be a good place to start. Since the theory fundamentally uses Mellin transforms, you should be able to find what you want (perhaps indirectly) there.</p> http://mathoverflow.net/questions/36064/getting-a-differential-equation-for-a-function-from-a-functional-equation-of-its/37017#37017 Answer by Armin Straub for Getting a differential equation for a function from a functional equation of its Mellin transform Armin Straub 2010-08-29T03:11:57Z 2010-09-03T04:10:52Z <p>By defining the Mellin transform for distributions as for instance done in <a href="http://books.google.com/books?id=ufOYJWfhyeMC&amp;lpg=PA297&amp;ots=DmasoVx7OZ&amp;dq=%2522Transform%2520Analysis%2520of%2520Generalized%2520Functions%2522%2520by%2520O.%2520Misra%252C%2520J.%2520Lavoine&amp;pg=PA1#v=onepage&amp;q&amp;f=false" rel="nofollow">Transform Analysis of Generalized Functions</a> by O. Misra, J. Lavoine it follows that the functional equation for $F(s)$ translates into a differential equation of which $f(x)$ is a <a href="http://en.wikipedia.org/wiki/Weak_solution" rel="nofollow">weak solution</a>.</p>