# Behaviour of solutions to ODE near singular points

I am having trouble understanding how to classify what happens to solutions of an ODE near singular points. For example;

I have a question that is about the ODE given by;

$$(x^2-36)y''+(6-x)y'+(x^2+12x+36)y=0$$

And the question asks about the behaviour of the solutions near each of the singular points.

I found that the singular points were $x_1=6$ and $x_2=-6$

And I found that by taking limits, that both point were regular because of the finite limits,

$$\lim_{x \to 6} (x-6)\frac{6-x}{x^2-36}=0$$ and $$\lim_{x \to 6} \frac{(x-6)^{2}(x^2+12x+36}{x^2-36}$$ 0

and similarly $$\lim_{x \to -6}(x+6)^{2} \frac{6-x}{x^2-36}$$ 0 and $$\lim_{x \to -6} \frac{(x+6)^{2}(x^2+12x+36)}{x^2-36}=0$$

So what can I infer from this at all? I see that it has two real roots, but what can this tell me about the solution's behaviour? Can someone please take some time to help.

Do I just need to look at the Euler equation corresponding to the case of two real distinct roots?

By the way, I do not think the answer is supposed to take any extreme amounts of work, it is from a past online assignment, i,e That is, I have what the answer is already, but I want to understand it. I really do appreciate the answers given but they are definitely beyond my level and beyond what we are even supposed to consider. We only have covered things such as that behaviour of points is similar to the behaviour of the solutions of the associated euler, and things of that level, etc. that is , the answer provided by the user a couple days ago seems to have left the impression that the question is answered but unfortunately it really is of no help to me even though I do appreciate the effort done

Another example of the type of question is to do the same thing but say for a different form such as ,

$$(x^2-4x-21)^{2}y''+(x^2-9)y'-xy=0$$

For which I found the two singular points to be $x_o=-3$, a regular singular point and $x_1=7$, an irregular singular point. And I am wanting to know the same things about the solutions in regard to $x_o$, for example, how can I know if all solutions remain bounded near it? or if all non zero solutions are unbounded, etc? Should I just look at the limits as x goes to the solutions of the euler equation as above? Or is there something else going on too? I have been trying to understand it for a while and I just can't seem to figure it out. Thanks a lot

PS: I am really still trying to figure this out, I have an exam coming up want to learn this.

• I think the answer to your question relies entirely on the distinction between a regular singularity and an irregular singularity. For instance, all the solutions of the differential equation will remain bounded near a regular singular point whereas it doesn't for an irregular singular point. I cannot provide a rigorous justification of this right now, but maybe I can think of it and let you know if I get to something. Cheers! – Dmoreno May 29 '15 at 4:54
• Some of your last edits were a bit insubstantial. It would be good if you could constrain the cosmetic edits to only be done together with material edits. – Daniel Fischer May 29 '15 at 8:50
• Okay, will do . – Quality May 29 '15 at 15:08
• @Dmoreno Sir that could very possibly be exactly the answer I am looking for. So, if it is true that all solutions remained bounded near regular singular points, then that answers the question, if you want to fill in any details and such and leave an answer I will gladly award the bounty as well – Quality May 29 '15 at 16:31
• @Quality, the problem is that I'm not completely sure about my statement. Consider, for example, the Bessel differential equation, where $x=0$ is a regular singular point. One of the solutions, $Y_{\nu}(x)$ is diverging at $x=0$ whereas the other one is bounded, $J_{\nu}(x)$ but I think this is a particular case. – Dmoreno May 29 '15 at 16:35

It seems for some reason, you haven't learn the characteristic exponents for a regular singular point.

Consider following $2^{nd}$ order ODE which has a regular singular point at $x = 0$.

$$x^2 y''(x) + xp(x) y'(x) + q(x) y(x) = 0\ \quad\text{ where }\quad \begin{cases} p(z) &= \sum\limits_{k=0}^\infty p_k z^k\\ q(z) &= \sum\limits_{k=0}^\infty q_k z^k\\ \end{cases} \;\;\text{ for } |z| < \rho.$$ For simplicity, we will assume all coefficients $p_k, q_k$ are real. From these coefficients, we can construct a quadratic equation

$$r(r-1) + p_0 r + q_0 = 0$$

This is known as the indicial equation for the ODE. Let $r_1, r_2$ be the two roots of indicial equations. They are known as the characteristic exponents and the convention is labeling them to make $\Re(r_1) \ge \Re(r_2)$.

Using these two exponents, we can say something about the solutions of the ODE near $0$.

• If $r_1, r_2$ are complex or $r_1, r_2$ are real but $r_1 - r_2 \notin \mathbb{N}$, then the ODE has two independent solutions of the form: $$y_1(x) = x^{r_1} \sum_{k=0}^\infty a_k x^k \quad\text{ and }\quad y_2(x) = x^{r_2} \sum_{k=0}^\infty b_k x^k$$

• Otherwise, $r_1, r_2$ are real but $r_1 - r_2 \in \mathbb{N}$, the two linear independent solutions has the form $$y_1(x) = x^{r_1} \sum_{k=0}^\infty a_k x^k \quad\text{ and }\quad y_2(x) = K y_1(x)\log x + x^{r_2} \sum_{k=0}^\infty b_k x^k$$ If $r_1 = r_2$, $K \ne 0$ and we can choose it to be $1$. If $r_1 > r_2$, $K$ can be zero, can be non-zero.

Apply this to our ODE

\begin{align} & (x^2-36)y''+(6-x)y'+(x^2+12x+36)y=0\\ \iff & y'' - \frac{1}{x+6} y' + \frac{x+6}{x-6} y = 0 \end{align} It has two regular singular points at $x = -6$ and $6$.

• At $x = -6$, the indicial equation is $$r(r-1) - r = 0 \quad\implies\quad r_1 = 2, r_2 = 0$$ So the two solutions of the ODE have following forms: \left\{\begin{align} y_1(x) &= (x+6)^2 \sum_{k=0}^\infty a_k (x+6)^k\\ y_2(x) &= K y_1(x)\log(x+6) + \sum_{k=0}^\infty b_k (x+6)^k \end{align}\right. Independent of whether $K = 0$ or not, the $\log(x+6)$ term in $y_2(x)$ will be suppressed by the $(x+6)^2$ factor in $y_1(x)$. As a result, both solutions $y_1(x), y_2(x)$ and hence all solutions of the ODE will be bounded near $x = -6$.

• At $x = 6$, the indicial equation is $$r(r-1) = 0 \quad\implies\quad r_1 = 1, r_2 = 0$$ The two solutions there have following forms: \left\{\begin{align} y_1(x) &= (x-6) \sum_{k=0}^\infty a_k (x-6)^k\\ y_2(x) &= K y_1(x)\log(x-6) + \sum_{k=0}^\infty b_k (x-6)^k \end{align}\right. Similar to the $x = -6$ case, the $\log(x-6)$ term in $y_2(x)$ get suppressed by the $(x-6)$ factor in $y_1(x)$. As a result, all solutions to the ODE is bounded near $x = 6$.

For the other ODE \begin{align} &(x^2-4x-21)^{2}y''+(x^2-9)y'-xy=0\\ \iff & y'' + \frac{x-3}{(x-7)^2(x+3)} y' + \frac{x}{(x-7)^2(x+3)^2} y = 0 \end{align} It has a regular singular point at $x = -3$ and an irregular one at $x = 7$. I have no idea how to deal with the irregular singular point. For the regular singular point at $x = -3$, the corresponding indicial equation is

$$r(r-1) - \frac{3}{50} r -\frac{3}{100} = 0 \quad\implies\quad \begin{cases} r_1 &= \frac{53+\sqrt{3109}}{100} \approx +1.087584074378026\\ r_2 &= \frac{53-\sqrt{3109}}{100} \approx -0.027584074378026 \end{cases}$$ Since $r_1, r_2$ are real but $r_1 - r_2$ isn't an integer, the two solutions has the from

\left\{\begin{align} y_1(x) &= (x+3)^{r_1} \sum_{k=0}^\infty a_k(x+3)^k\\ y_2(x) &= (x+3)^{r_2} \sum_{k=0}^\infty b_k(x+3)^k \end{align}\right. Since $r_1 > 0$ while $r_2 < 0$, $y_1(x)$ is bounded near $x = -3$ while $y_2(x)$ is unbounded there.

• Thank you!! That really helped. Sir, if you please have any time do you think you could take a look at my question I posted recently called Looking for help in regard to series solutions with ordinary points? I think it can be very simple to answer but I can't figure it out. Thanks a lot again – Quality May 31 '15 at 1:52

I will analyze $x>6$ and $x=6$ is an irregular singular so we can use what is called a WKB Approximation. If you have an equation of the form:

$$y''=Qy$$

then the asymptotic solution of that equation will be of the form:

$$y=\frac{exp(\pm \int\sqrt Q d x)}{Q^{1/4}}$$

Lets try to approximate the equation with initial conditions $y(0)=1$ and $y'(0)=0$. Rewriting the original equation:

$$y''-\frac{1}{x+6}y'+\frac{x+6}{x-6}y=0$$

This equation can be turned into $y''=Qy$. Substitute $y=uv$

$$u''v+uv''+2u'v'-\frac{1}{x+6}u'v-\frac{1}{x+6}uv'+\frac{x+6}{x-6}uv=0$$

I want to get rid of $u'$, extract all the $u'$ terms and make them $0$ (I have the right to do it because I choose what $v$ and $u$ is):

$$+2u'v'-\frac{1}{x+6}u'v=0$$

$$v=\sqrt{x+6}$$

The rest of the equation becomes:

$$u''=\frac{4 x^3+72 x^2+429 x+882}{4 (x-6) (x+6)^2}u$$

Which means

$$Q=\frac{4 x^3+72 x^2+429 x+882}{4 (x-6) (x+6)^2}$$

$$\sqrt Q=i \sqrt{\frac{4 x^3+72 x^2+429 x+882}{4 (6-x) (x+6)^2}}$$

I have turned $x-6$ into $6-x$ If possible, always try to make $Q$ positive by extracting the $i$ out. We don't need to worry about $Q^{1/4}$ because the constant will be absorbed by the arbitrary constant of the linear equation.

What we have as a result is:

$$u=C_{\pm}\frac{\exp\bigg(\pm i \int^x_7 \sqrt{\frac{4 t^3+72 t^2+429 t+882}{4 (t-6) (t+6)^2}} dt\bigg)}{\sqrt{\frac{4 x^3+72 x^2+429 x+882}{4 (x-6) (x+6)^2}}}$$

We can write complex exponential as (cos and sin) or (sin + phase). So it becomes:

$$u=C\frac{\sin\bigg(\int^x_7 \sqrt{\frac{4 t^3+72 t^2+429 t+882}{4 (t-6) (t+6)^2}} dt + \phi\bigg)}{\sqrt{\frac{4 x^3+72 x^2+429 x+882}{4 (x-6) (x+6)^2}}}$$

Combining with $v$

$$y=C\sqrt{x+6}\frac{\sin\bigg(\int^x_7 \sqrt{\frac{4 t^3+72 t^2+429 t+882}{4 (t-6) (t+6)^2}} dt + \phi\bigg)}{\sqrt{\frac{4 x^3+72 x^2+429 x+882}{4 (x-6) (x+6)^2}}}$$

But how good is it? Let's compare it with the numeric solution of the actual equation. Using the initial conditions picked above, the solution becomes:

$$y=0.528063 \sqrt{x+6} \frac{\sin\bigg(\int^x_7 \sqrt{\frac{4 t^3+72 t^2+429 t+882}{4 (t-6) (t+6)^2}} dt -0.444830\bigg)}{\sqrt{\frac{4 x^3+72 x^2+429 x+882}{4 (x-6) (x+6)^2}}}$$

This is the comparison between the numerical solution: enter link description here

Here the third picture blue is the approximation and orange is the numerical solution. They completely overlap.

You may think the solution does not look nice, but for many cases, it will look VERY clean (especially if you can actually integrate $Q$)

This probably wasn't something that you wanted to learn but I enjoy this so I'll post it anyways.

• Wow well thanks a lot, for the effort. But do you think you know a more less rigorous approach? The options for the answers are between, "All solutions remain bounded near $x_1$ or atleast one is bounded and one is unbounded or all non zero are unbounded? and same for x_2, so is there a way to see this? – Quality May 22 '15 at 21:31
• By the way, I updated the question, It seems that I made a mistake when I took the limits as I used x and x^2 instead of (x-6) and (x-6)^2 for calculating of the points were regular ( which they are). I don't know if this changes anything – Quality May 27 '15 at 18:52
• @Quality I wouldn't want to say something absolute, but I can say that for some initial conditions, the solution will blow up at $+6$ Consider Bessel $J_0$ and $Y_0$, only $Y_0$ blows up. – grdgfgr May 28 '15 at 9:52
• I posted the picture of the answer given – Quality May 29 '15 at 0:30
• @Quality Numerical methods can be wrong sometimes, but I would count on it this time. i.imgur.com/3VA6GSa.png – grdgfgr May 29 '15 at 4:45