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Suppose you have this second order linear ordinary differential equation:

$$ \frac{\mathrm d}{\mathrm d x}\left[\left(1-\frac{2}{x}\right)\frac{\mathrm du(x)}{\mathrm d x}\right]-(l+1)(l+2) \frac{u(x)}{x^2}=S(x)$$

where $u(x)$ is the unknown, $S(x)$ is a source term and $l$ is an integer $l\geq0$. (It is a particular case of the Jacobi equation)

This equation has two regular singular points in $x=0,2,\infty$. I am looking for a solution defined in $x\in [2,\infty)$ so that $u(x)$ is regular in $x=2$ and:

$$\lim_{x\to\infty}u(x)=0$$

$S(x)$ is finite in $x=2$ and goes to $0$ for $x\to\infty$.

To satisfy the second boundary condition I found (using Mathematica) that you need to have for $S(x)$ the asymptotic behaviour$^1$:

$$S(x)\sim \frac{1}{x^3},\qquad(\text{for }x\to\infty) $$

Why is this the case?

More in general, given a non-homogeneous ODE and Dirichlet-like boundary conditions (like in my example), what should the behaviour of $S$ be on the boundary points so that the equation has a solution?


NOTES

$1)$ Here's how I found this:

  • Set $S(x) = \frac{1}{x}$ and $l$ to some integer number
  • Solved the equation with Mathematica getting two integration constants $c_1$ and $c_2$
  • Expanded in series the solution around $\infty$
  • The solution is a polynomial in $x$ (+ $\mathcal O(x^{-1})$ terms) where the coefficient of the $x$ term and the constant term are independent of those constants. So no matter what you can't get a convergent solution for $r\to\infty$
  • Repeat for $S(x) = \frac{1}{x^2}$ to get a solution that is constant at infinity and for $S(x) = \frac{1}{x^3}$ to get a solution that goes to $0$ as wanted.
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The general question for arbitrary ODE is quite difficult. But in this specific case, there is some estimate of $S(x)$ that you can obtain.

Note that $\lim_{x\rightarrow \infty} u(x) = 0$, thus for large $x$, your equation is approximated by $$x^2 \frac{d}{dx}\left[\left(1 - \frac 2x\right)\frac{du(x)}{dx}\right] = S(x)$$ Which is equivalent to $$\frac{d}{dx}\left[\left(1 - \frac 2x\right)\frac{du(x)}{dx}\right] = \frac{S(x)}{x^2}$$ Since $\lim_{x\rightarrow \infty} u(x) = 0$, the limit of its derivative must behave similarly, so $\lim_{x\rightarrow \infty} u'(x) = 0$, by barbalat's lemma. Consequently, the LHS is $0$. Thus, $$\lim_{x\rightarrow\infty} \frac{S(x)}{x^2} = 0$$ This is true only if $S(x)$ has order smaller than $x^2$, e.g. $S(x) = x$, etc. I can't see precisely how Mathematica comes up with $S(x) \sim \frac{1}{x^3}$ for large $x$, but that does satisfy the above condition.


A very informal analysis does gives a better asymptotic estimation though. Since $\lim_{x\rightarrow\infty} u(x) = 0$, suppose $u(x)$ is of order $\frac{1}{x}$. This means for large $x$, $u'(x)$ is of order $\frac{1}{x^2}$. "Patching" things together you have something like this $$x^2 \frac{d}{dx}\left[O\left(\frac{1}{x}\right) O\left(\frac{1}{x^2}\right)\right] - O\left(\frac{1}{x}\right) = S(x)$$ This is, loosely speaking, equivalent to $$O\left(x^2\right) O\left(\frac{1}{x^4}\right) - O\left(\frac{1}{x}\right) = S(x)$$ Or $$O\left(\frac{1}{x^2}\right) - O\left(\frac{1}{x}\right) = S(x)$$ Since $S(x)$ drives the equation, it has to goes to $0$ faster than any term on the left hand side, thus $S(x)$ is of order $\frac{1}{x^3}$ or lower. If it is of a lower order than $\frac{1}{x^3}$, then we can expect $u(x)$ to be of a lower order. But since $u(x) \sim \frac{1}{x}$ is perfectly fine assumption (without solving the equation and this is a very informal analysis), we conclude that $$S(x) \sim \frac{1}{x^3}$$


Edit: with the new information, then I think my very informal analysis makes more sense (since you were simply trying out different solution for $S(x)$).

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  • $\begingroup$ Yeah that's because I've mixed up two things... I have edited the post, now the equation is correct. Also added the explanation of what I did to find the $\frac{1}{x^3}$. $\endgroup$ – FrodCube May 2 '17 at 17:10
  • $\begingroup$ @FrodCube I edited it. $\endgroup$ – Paichu May 2 '17 at 17:27
  • $\begingroup$ Are there any theorems that can give this result like given some properties of the singular points of the homogeneous equation or is this the only way to get it? $\endgroup$ – FrodCube May 2 '17 at 17:45
  • $\begingroup$ I believe there are theorems for well-known equations, but I don't know of one specifically. $\endgroup$ – Paichu May 2 '17 at 17:51

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