Let $$\begin{eqnarray}\nonumber f(y, t) &=& \frac{C}{\sigma ^2 t} \left[\frac{(1-\alpha) (b-y)}{\alpha t^{\alpha}} \, _1F_1\left[\frac{\alpha+1}{2 \alpha};\frac{3}{2};-\frac{ (b-y)^2}{2 \sigma^2 t^{2 \alpha}}\right]- \sqrt{2} \sigma \frac{\Gamma \left(\frac{3}{2}-\frac{1}{2 \alpha}\right)}{\Gamma \left(1-\frac{1}{2 \alpha}\right)} \, _1F_1\left[\frac{1}{2 \alpha};\frac{1}{2};-\frac{ (b-y)^2}{2 \sigma^2 t^{2 \alpha}}\right]\right] \end{eqnarray}$$ be defined in $f(y, t) \in (-\infty, b]$, where $0 < \alpha < 1$, $\sigma > 0$ and $t$ represents time and $C$ is some scalar value. Also where $_1F_1(a,b,z)$ denotes the Kummer confluent hypergeometric function with series expansion $$_1F_1 (a,b,z)=\sum _{k=0}^\infty \frac{(a)_k}{(b)_k}\left(\frac{z^k}{k!}\right),$$ where $(c)_k$ denote Pochhammer’s Symbol \begin{eqnarray}\nonumber (c)_0 &=& 1\\\nonumber (c)_n &=&c(c+1)(c+2)\cdots(c + n -1),\\\nonumber (c)_n &=&\frac{\Gamma(c + n)}{\Gamma(c)}\quad c \neq 0, -1, -2, \cdots \end{eqnarray} and $\Gamma(z)$ denotes the Gamma function which satisfies $$\Gamma (z)=\int _0^{\infty } e^{-t} t^{z-1} dt.$$

I am interested in showing convergence in the distributional sense, that for some $y_0$ the following limit holds $$\lim_{t \rightarrow 0}f\left(y, t\right) = \delta\left(y - y_0\right), \quad y_0 \in \left(- \infty, b\right).$$

The presence of Kummer confluent hypergeometric function have made evaluating Fourier transform of $f(y,t)$ or applying dominated convergence theorem too hard. I am having trouble proving this analytically. Any help?

Some observations which may be helpful $$\int_{-\infty}^b f(y, t) \, \mathrm{d}y = C t^{\alpha-1}.$$ $$\int_{-\infty}^\infty f(y, t) \, \mathrm{d}y = 0.$$ Contour integral representation of Kummer confluent hypergeometric function $$_1F_1 (a,b,z)= \frac{\Gamma (b)}{2 \pi \iota \Gamma (a)} \int_{\gamma -\iota\infty}^{\gamma +\iota\infty} \frac{(-z)^{-s} \Gamma (s) \Gamma (a-s)}{\Gamma (b-s)} \, \mathrm{d}s$$ where $ 0 <\gamma < \operatorname{Re}(a) \wedge |\arg(-z)| < \frac{\pi}{2}.$

Denoting $$ I(a,b,z) = \int_{\gamma -\iota\infty}^{\gamma +\iota\infty} \frac{(-z)^{-s} \Gamma (s) \Gamma (a-s)}{\Gamma (b-s)} \, \mathrm{d}s$$ Then the function of interest can be rewritten as \begin{eqnarray} f(y, t) &=& \frac{C}{\sigma ^2 t} \left[\frac{\sqrt{\pi }(1-\alpha) (b-y)}{2 \alpha t^{\alpha} \Gamma \left(\frac{\alpha+1}{2 \alpha}\right)} \, I\left[\frac{\alpha+1}{2 \alpha};\frac{3}{2};-\frac{ (b-y)^2}{2 \sigma^2 t^{2 \alpha}}\right] \right. \\[6pt] & & \left. {} - \sqrt{\frac{2}{\pi}} \sigma \sin \left(\frac{\pi }{2 H}\right) \Gamma \left(\frac{3}{2}-\frac{1}{2 \alpha}\right) I\left[\frac{1}{2 \alpha};\frac{1}{2};-\frac{ (b-y)^2}{2 \sigma^2 t^{2 \alpha}}\right] \vphantom{\frac{(1)}{\Gamma\left(\frac\alpha\alpha\right)}} \right] \end{eqnarray}

Solution attempt:

Let $\varphi \in C_{c}^{\infty}(\mathbb{R})$. Using the substitution $x = \frac{b-y}{\sigma t^\alpha}$ we obtain \begin{eqnarray}\nonumber f_t(x) &=& \frac{C}{\sigma t} \left[\frac{(1-\alpha)x}{\alpha} \, _1F_1\left[\frac{\alpha+1}{2 \alpha};\frac{3}{2};-\frac{x^2}{2}\right]- \sqrt{2} \frac{\Gamma \left(\frac{3}{2}-\frac{1}{2 \alpha}\right)}{\Gamma \left(1-\frac{1}{2 \alpha}\right)} \, _1F_1\left[\frac{1}{2 \alpha};\frac{1}{2};-\frac{x^2}{2}\right]\right] \\ \langle f_t,\varphi \rangle &=& \int_{-\infty}^\infty I_{\left(-\infty, b\right]} f(y,t) \varphi(y) \, \mathrm{d}y \\ &=& -\sigma t^\alpha\int_{-\infty}^\infty I_{\left[0, \infty\right)} f_t(x) \varphi(b - x \sigma t^\alpha) \, \mathrm{d}x \end{eqnarray} As $\varphi$ is continuous and compactly supported, this integrand is dominated by $? \|\varphi\|_{\infty}$ which integrates to $?\|\varphi\|_\infty$.

  • $\begingroup$ No I am afraid I do not, nit at least as of yet. $\endgroup$ Commented Sep 17, 2017 at 0:54
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    $\begingroup$ if $\int fdy=0$, how can it be a delta function? $\endgroup$ Commented Sep 17, 2017 at 7:09
  • $\begingroup$ The function is defined only on an interval, not only the complete real line, the integral does not have an indicator function to take into consideration, which probably makes my second observation useless $\endgroup$ Commented Sep 17, 2017 at 12:22
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    $\begingroup$ Have you considered using Laplace transforms instead of Fourier transforms? Given the one-sided nature of $f_t$ this seems a more appropriate. Note that $\mathcal{L}({_1 F_1}(a,b;\cdot))(s) = \tfrac{1}{s}{_2 F_1}(1,a,b;\frac{1}{s})$ so that there is hope of a analytical solution. $\endgroup$ Commented Sep 17, 2017 at 20:03
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    $\begingroup$ The reason is that I have a nice solution for the other sign ;-) I'm currently writing everything up. $\endgroup$ Commented Sep 18, 2017 at 0:08

1 Answer 1


Here is an attempt at a solution.

Continuing with your idea: Substitute $\xi:=\frac{b-y}{\sqrt{2}\sigma t^\alpha}$ to make everything nicer: $$\begin{align} f_{t}(\xi) &= \frac{\sqrt{2}C}{\sigma t} \left[\frac{(1-\alpha)}{\alpha} \xi\, _1F_1\left(\frac{\alpha+1}{2 \alpha};\frac{3}{2};-\xi^2\right)-\frac{\Gamma \left(\frac{3}{2}-\frac{1}{2 \alpha}\right)}{\Gamma \left(1-\frac{1}{2 \alpha}\right)} {_1F_1}\left(\frac{1}{2 \alpha};\frac{1}{2};-\xi^2\right)\right] \tag{1}\\ &= \frac{\sqrt{2}C}{\sigma t} \left[(-\tfrac{1}{2}+\tfrac{1}{2\alpha}) 2\xi {_1F_1}\left(\tfrac{1}{2}+\tfrac{1}{2 \alpha};\tfrac{3}{2};-\xi^2\right)-\frac{(\tfrac{1}{2}-\tfrac{1}{2\alpha})\Gamma(\tfrac{1}{2}-\tfrac{1}{2 \alpha})}{\Gamma(1-\frac{1}{2 \alpha})} {_1F_1}\left(\tfrac{1}{2 \alpha};\tfrac{1}{2};-\xi^2\right)\right] \tag{2}\\ &= \frac{\sqrt{2}C}{\sigma t} (\tfrac{1}{2}-\tfrac{1}{2\alpha})\left[-2\xi {_1F_1}\left(\tfrac{1}{2}+\tfrac{1}{2 \alpha};\tfrac{3}{2};-\xi^2\right)-\frac{\Gamma(\tfrac{1}{2}-\tfrac{1}{2 \alpha})}{\Gamma(1-\frac{1}{2 \alpha})} {_1F_1}\left(\tfrac{1}{2 \alpha};\tfrac{1}{2};-\xi^2\right)\right] \tag{3} \end{align}$$ Now we use the magic formula $$ {_1 F_1}(a;b;-z) = e^{-z} \cdot {_1 F_1}(b-a;b;z) \tag{4}$$ and get $$f_t(\xi) = \frac{\sqrt{2}C}{\sigma t} (\tfrac{1}{2}-\tfrac{1}{2\alpha})e^{-\xi^2}\left[-2\xi {_1F_1}\left(1-\tfrac{1}{2 \alpha};\tfrac{3}{2};\xi^2\right)-\frac{\Gamma(\tfrac{1}{2}-\tfrac{1}{2\alpha})}{\Gamma(1-\frac{1}{2 \alpha})} {_1F_1}\left(\tfrac{1}{2}-\tfrac{1}{2\alpha};\tfrac{1}{2};\xi^2\right)\right] \tag{5}$$

So far not much has happened. But now comes the part where I like the other sign more. If the original function indeed contains a sign error and we are really talking about $$f_t(\xi) = \frac{\sqrt{2}C}{\sigma t} (\tfrac{1}{2}-\tfrac{1}{2\alpha})e^{-\xi^2}\left[-2\xi {_1F_1}\left(1-\tfrac{1}{2 \alpha};\tfrac{3}{2};\xi^2\right) \color{red}{+}\frac{\Gamma(\tfrac{1}{2}-\tfrac{1}{2\alpha})}{\Gamma(1-\frac{1}{2 \alpha})} {_1F_1}\left(\tfrac{1}{2}-\tfrac{1}{2\alpha};\tfrac{1}{2};\xi^2\right)\right] \tag{6}$$ then we can use generalised Hermite polynomials $$H_\nu(z) = 2^\nu \sqrt{\pi} \left(\frac{{_1 F_1}(-\frac{\nu}{2};\frac{1}{2};z^2)}{\Gamma(\frac{1-\nu}{2})} - \frac{2z \cdot {_1 F_1}(\frac{1-\nu}{2};\frac{3}{2};z^2)}{\Gamma(-\frac{\nu}{2})} \right) \tag{7}$$ with $\nu:=1-\tfrac{1}{\alpha}$ ($\iff \tfrac{1-\nu}{2}=1-\tfrac{1}{2\alpha} \iff -\tfrac{\nu}{2}=\tfrac{1}{2}-\tfrac{1}{2\alpha}$). This gives us: $$f_t(\xi) = \frac{\sqrt{2}C}{\sigma t} (-\tfrac{\nu}{2})e^{-\xi^2} \Gamma(-\tfrac{\nu}{2}) 2^{-\nu} \pi^{-1/2} H_\nu(\xi) \tag{8}$$

In other words: The whole function is a constant multiple of $\tfrac{1}{t}e^{-\xi^2} H_\nu(\xi)$. Note that this is a true polynomial iff $\alpha\in\{1,\tfrac{1}{2},\tfrac{1}{3},\tfrac{1}{4},\ldots\}$ which may or may not be the case anyway in your situation.

Anyhow: Using the $\xi$-substitution, we find $$\langle f_t,\phi\rangle = const \int_0^\infty \frac{1}{t} e^{-\xi^2} H_\nu(\xi) \phi(b-\sqrt{2}\sigma t^\alpha \xi) d\xi$$ For $\alpha=1$ this goes to $const\cdot\phi(b)$ so that $f_t \to const\cdot \delta_b$ as desired (and the constant will probably work out alright). However, if $\alpha<1$, the integral goes to $sgn(\phi(b))\cdot\infty$.

  • $\begingroup$ +1, very insightful answer, I am gonna check my own computations and work through your answer and ping back, in the near future. $\endgroup$ Commented Sep 18, 2017 at 1:18
  • $\begingroup$ You were spot on right about the - being a + sign, i recomputed steps leading to this problem, and it sure is a +. You have amazing skills $\endgroup$ Commented Sep 19, 2017 at 0:44
  • $\begingroup$ I was wondering if it not too much trouble, can you possibly cite me the source where you found equation 7 formula. Thank you. $\endgroup$ Commented Sep 19, 2017 at 2:47
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    $\begingroup$ I found it by browsing through mathworld resources on hypergeometric functions. In effect it is the definition of $H_\nu$ that mathworld/wolframalpha/mathematica uses, see functions.wolfram.com/HypergeometricFunctions/HermiteHGeneral but it should be a relatively straight forward proof by induction to show that $H_\nu$ is the usual Hermite polynomial if $\nu\in\mathbb{N}$. $\endgroup$ Commented Sep 19, 2017 at 12:57

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