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Let $\gamma>0$ be a real number and $\Phi(r)=\frac{1}{r}-\frac{\pi}{4}\left(H_0(r/2)-Y_0(r/2)\right)$ defined on $[0,\infty)$, where $H_0$ is the Struve function of order zero and $Y_0$ is the Bessel function of second kind of order zero.

My goal is to find $\gamma_c$ such that for all $\gamma\geq\gamma_c$ the function $\Phi(r)-\frac{\gamma}{r^2}\leq0$ for all $r\in[0,\infty)$. Plots suggest that the critical $\gamma_c$ is about $0.5$. The root in this regime is about $r_0\approx 0.5$.

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  • $\begingroup$ Do you mean 1/r - (StruveH[0, r/2] - BesselK[0, r/2]) Pi/4, or something else? Y0 typically is used to represent a Bessel function, not a modified Bessel function. $\endgroup$
    – bbgodfrey
    May 31, 2016 at 14:31
  • $\begingroup$ My bad, I meant $Y_0$, the Bessel function of second kind of order zero. $\endgroup$
    – costalmu
    May 31, 2016 at 14:34
  • $\begingroup$ Welcome to Mathematica.SE! I hope you will become a regular contributor. To get started, 1) take the introductory tour now, 2) when you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge, 3) remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign, and 4) give help too, by answering questions in your areas of expertise. $\endgroup$
    – bbgodfrey
    May 31, 2016 at 14:42

1 Answer 1

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Defining 

f = 1/r - (StruveH[0, r/2] - BesselY[0, r/2]) Pi/4

the maximum value that f r^2 assumes is 

NMaximize[{f r^2, r > 0}, r]
(* {0.498508, {r -> 2.58341}} *)

Thus, the critical value of γ is 0.498508. The following illustrates this maximum.

Plot[r^2 f, {r, 0, 10}, AxesLabel -> {r, "f"}]

enter image description here

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