I just started research in physics, but along the way, I want to show something like

$$\partial_x \beta(x) \sim -\frac{A}{x^2}[\alpha(x)+\beta(x)]\,\, as\,\, x\rightarrow\infty$$

where $\alpha(x)$ and $\beta(x)$ are non-negative monotonically decreasing much faster than $\frac{1}{x^n}$ (something like exponential decay or Gaussian) with a constant $A$.

After doing method of integrating factor, I am putting a boundary condition that $\beta(x)$ at infinity is equal to 0. Then I tried to bound the integral or show some asymptotic behavior, but it seems like my math is not sufficient enough.

Anyway, my goal is to show $\alpha$ is subdominant to $\beta$ at infinity or vice versa, how should I approach these kind of problem without losing generality (not assuming they are decaying like exp or gaus).

Any suggestion for how to show asymptotic behavior of beta would be appreciated! thank you

  • $\begingroup$ If the equation above is what you want to show, then what differential equation are you starting with? $\endgroup$ – user254433 Jun 30 '17 at 3:13
  • $\begingroup$ The equation that I wrote is the DE that I start with. Both alpha and beta goes to 0 as a boundary condition at infinity. But they are asymptotic to something faster than 1/x^n as x->infty. Can I obtain asymptotic behavior of beta in general in this case? $\endgroup$ – Duke Smith Jun 30 '17 at 15:42

Let $f=\frac{A}{x^2}$, $$\beta_x=-f\beta-\alpha f$$ $$\frac{\partial}{\partial x}e^{\int f dx}\beta=-e^{\int f dx}\alpha f$$

We observe that $$e^{\int f dx} = k e^{-\frac{A}{x}}$$

So $$ke^{-\frac{A}{x}}\beta=C - \int{k \frac{A}{x^2}e^{-\frac{A}{x}}\alpha dx}$$

Let $I$ be the right hand integral. Since $\alpha < Dx^{-n}$, $$0<I<kD\int{\frac{A}{x^{2+n}}e^{-\frac{A}{x}}dx}=\frac{kD}{A^{n+1}}\Gamma(n+1,\frac{A}{x})$$

That means, $$\beta>\left(\frac{C}{k}-\frac{D}{A^{n+1}}\Gamma(n+1,\frac{A}{x})\right)e^{\frac{A}{x}}$$ where $\Gamma(n+1, u)$ is the incomplete Gamma function. Using the relation, $$\Gamma(n+1,\frac{A}{x})=n!e^{-\frac{A}{x}}\sum_{m=0}^{n}\left(\frac{1}{m!}\left(\frac{A}{x}\right)^m\right)$$

Thus, $$\beta > \frac{C}{k}e^{\frac{A}{x}} -\frac{n!D}{A^{n+1}} \sum_{m=0}^{n}\left(\frac{1}{m!}\left(\frac{A}{x}\right)^m\right)$$

Upon expanding the exponential term into a series, we note that the coefficients for the $x^{-j}$ terms when $0<j<n$ are nonzero. Strictly speaking then, $\beta$ cannot fall as $O(x^{-n})$, and decays more slowly than $\alpha$, but does tend towards $0$ as $x\rightarrow \infty$.

  • $\begingroup$ Thank you very much, and I really apologize about the late reply (new to the forum). I've noticed you used incomplete gamma function with a indefinite integral. Is it possible to do that? $\endgroup$ – Duke Smith Jul 8 '17 at 3:16
  • $\begingroup$ Well, any integral limits from a definite integral would be absorbed into the constant $C$. On the other hand, I'd like to hear your thoughts on the assumption that $\alpha < Dx^{-n}$... This was my interpretation of your remark that $\alpha$ decays as $x^{-n}$. $\endgroup$ – player100 Jul 9 '17 at 1:29
  • $\begingroup$ Ok, Thank you. I think your logic is perfectly fine for bounding the integral. Because what I am interested is an asymptotic behavior, It is fine to set $\alpha$ is subdominant to $Dx^{-n}$ as z goes to infinity. However, since the equation that I started with is already an asymptotic, so my integration bound is something like from some z to infinity. And after your integrating factor treatment, when I plug in the limits of integral of LHS, does the upperbound (infinity) contributes nothing? therefore C in your equation become 0. Then I can no longer apply your logic because negative... $\endgroup$ – Duke Smith Jul 9 '17 at 19:28
  • $\begingroup$ ...sign from the lower bound cancels out the negative sign of the RHS. Does this case allow $\beta$ decay faster than $O(x^{-n})$ for positive n as z goes to infinity? $\endgroup$ – Duke Smith Jul 9 '17 at 19:32
  • $\begingroup$ I think you are asking how $\beta$ is bounded from above. To get an intuitive sense of that scenario, set $\alpha$ to $0$. Then it is clear as an upper bound that $\beta$ decays by $e^{\frac{A}{x}}$. The second thrust of your question seems to be looking for some $\alpha$ that can cause $\beta$ to decay on order of $x^{-n}$... I don't know off hand, but it is worth a look. Finally for any indefinite integral, $F=\int{f}dx$, the definite integral is simply $F(b)-F(a)$. In your case, you can apply the limits of infinity to $\beta$. In particular $\Gamma(n+1, 0)=n!$. Hope that helps? $\endgroup$ – player100 Jul 10 '17 at 1:51

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