For $x^2y''+axy'+by=0$ where $a,b\in\mathbb{R}$, find conditions on $a,b$ such that all solutions are bounded as $x\to0$.

First write $y''+\frac{a}{x}y'+\frac{b}{x^2}y=0$ and guess $y=x^r$. Plug it in and simplify I got $r^2+(a-1)r+b=0$, thus $r=\frac{1-a}{2}\pm\sqrt{\left(\frac{1-a}{2}\right)^2-b}$. Since the solution to this equation looks like $y=c_1x^{r_1}+c_2x^{r_2}$, I think if $r_1,r_2\geq 0$, all solutions are bounded as $x\to0$. Similarly, if one of $r_1,r_2$ is positive, then the solution is partially bounded as $x\to0$.

My question is, how can I find the condition on $a,b$ such that $r_1,r_2\geq 0$? I figured if $a=1,b=0$, $r_1=r_2=0$ and all the solutions are bounded. But what's next?


I would assume that your solution is required to be real.

Does your solution need to be well defined at $x = 0$? Nothing was stated about the range of validity.

Your statement that if one is positive then it is partially bounded is suspect. If one of the roots is less than zero then one term will be complex for values of $x < 0$ so the limit from the left hand side may not be well defined in $\Bbb R$ (in the complex plane maybe).

So let's say the two conditions are positivity and real valued.

The discriminant needs to be positive or zero $$ \implies \frac{(1 - a)^2}{4} \ge b $$

if b is negative then the negative root will be less than zero, only one possible solution with the positive root (thus you'd need to search for a second solution by some other means).

If $b$ is positive then for both roots to be positive we need $a \le 1$

For $b = 0$, $a = 1$, as you pointed out.

For now let's look at the positive solutions. The conditions are:

$$a \ge 1$$ and $$\frac{(1 - a)^2}{4} \ge b > 0 $$

So there is a whole family of solutions.

I am not sure that there is a next step. We used the desire to have $\lim_\limits{x\to 0}y(x)$ be finite and to impose these conditions. You should check that the limits are the same form both sides (but we used that to constrain a and b).


There are 3 cases to consider here, based on the sign of the determinant

$$ \Delta = (a-1)^2 - 4b $$

Case 1: If $\Delta = 0$, there is a double root, $r = \frac{1-a}{2}$, and the general solution is

$$ y(x) = c_1x^r + c_2x^r\ln x $$

For this to be bounded at $x=0$, you want $\lim_\limits{x\to 0}x^r\ln x = 0 $, which satisfies for $r > 0$ (but not $r = 0$). Therefore the condition for this is $a < 1$ and $b = \frac{(a-1)^2}{4} > 0$

Case 2: If $\Delta < 0$, the roots are complex conjugates of the form $r = \alpha \pm i\beta$, where $\alpha = \frac{1-a}{2}$ and $\beta = \sqrt{|\Delta|}$. Then the general solution is

$$ y(x) = x^{\alpha}\big[c_1\cos(\beta\ln x) + c_2\sin(\beta \ln x)\big] $$

In this case, $y(x\to0^+)$ is bounded if $\alpha \ge 0$, or $a \le 1$ and $b > \frac{(a-1)^2}{4}$. You actually have two different behaviors here

  • If $\alpha > 0$ then $y(0^+)\to 0$

  • If $\alpha = 0$ then $y(0^+)$ does not exist but always oscillates between $[-1,1]$

Case 3: If $\Delta > 0$, you have two distinct real roots, and the general solution $$ y(x) = c_1x^{r_1} + c_2x^{r_2} $$

For $y(0^+)$ to be finite, both roots need to be non-negative, but only one can be zero, i.e. $r_2 > r_1 \ge 0$.

From Vieta's formulas:

\begin{align} r_1 + r_2 &= 1-a > 0 \\ r_1r_2 &= b \ge 0 \end{align}

or equivalently $a < 1$ and $0 \le b < \frac{(a-1)^2}{4}$

Altogether: Almost all $(a,b)$ pairs in the solution space $a \le 1$, $b \ge 0$ will the satisfy the given condition, with the exception of the point $(1,0)$. In set notation, this is

$$ (a,b) \in \{(\infty,1] \times [0,\infty)\} \backslash (1,0) $$


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