Analyzing a Difference Equation

[Theia]

Hello!

I started to study a differential equation, and turned the problem into a difference equation ($$a_0, a_1 \in \mathbb{R}, a_2 = 0, a_3 = \frac{a_0 + a_1}{6}$$)

$$a_{v+2} = \frac{v^2a_v + a_{v-1}}{(v+1)(v+2)}$$, where $$v \ge 2$$.

The numbers $$a_v$$ are coefficients of a serie solution to the original differential equation.

But now the question: I'd like to prove if the serie converges or not (i.e. is it usefull at all). But how is this done in this case? As far as I can see, the difference equation has not err... a nice solution and so all I have is the difference equation. Sometimes I've seen, someone takes a limit of an equation and assumes $$a_{\infty} = a$$ and then substituting this into the equation. But this method clearly fails in this case. Are there another methods?

Thank you! ^^

 

[Ackbach]

Hmm, well one thing you could say, if you were trying to take the limit as $v\to\infty$, is that
\begin{align*}
\lim_{v\to\infty}a_{v+2}&=\lim_{v\to\infty}\frac{v^2a_v+a_{v-1}}{v^2+3v+2} \\
&=\lim_{v\to\infty}\frac{a_v+\frac{a_{v-1}}{v^2}}{1+\frac{3}{v}+\frac{2}{v^2}}.
\end{align*}
From this, we see that $a_{v+2}\to a_v$. This isn't a proof of convergence, but I think I'd be shocked if it didn't.

 

[I like Serena]

Hmm, well one thing you could say, if you were trying to take the limit as $v\to\infty$, is that
\begin{align*}
\lim_{v\to\infty}a_{v+2}&=\lim_{v\to\infty}\frac{v^2a_v+a_{v-1}}{v^2+3v+2} \\
&=\lim_{v\to\infty}\frac{a_v+\frac{a_{v-1}}{v^2}}{1+\frac{3}{v}+\frac{2}{v^2}}.
\end{align*}
From this, we see that $a_{v+2}\to a_v$. This isn't a proof of convergence, but I think I'd be shocked if it didn't.

I'm afraid that doesn't quite fly.
Consider for instance $a_v = \ln v$ that satisfies the criterion $a_{v+2}\to a_v$. Successive terms get arbitrarily close to each other... but the sequence still does not converge.

Anyway, if we pick for instance $a_0=a_1=0$, the sequence does converge.
I'm not sure yet if it holds for the general case though.

 

[Theia]

*sigh* I've tried a ratio test to attack on this. Just to see that it is inconclusive... Also trying to write e.g. $$a_{v+4}$$ in terms of lower indices seems not to work and all I got was that $$a_{v+4} \to a_{v+2}$$, which we already know.

To make a numerical study, I coded a program that computes the coefficients until one of them is smaller than $$10^{-9}$$. If I got everything right, it looks like that needs about million terms (plus some hundreds of thousands terms depending on initial values)... But still, it is not saying anything does it really converge or not... Maybe this needs quite a tricky way to be handled properly. (Sadface)

 

[Theia]

... Nice. Post just disappeared. -.- Anyway, if one writes $$a_v = \alpha_va_0 + \beta_va_1$$, then one obtains two sequences pure numbers, namely

$$\alpha_0 = 1, \alpha_1 = \alpha_2 = 0, \alpha_3 = \frac{1}{6}, \ldots$$ and $$\beta_0 = 0, \beta_1 = 1, \beta_2 = 0, \beta_3 = \frac{1}{6}, \ldots$$.

Then one can compute more easily the continuation (the same original equation applies to both sequences). But does this help or not, I have no idea.

 

[Theia]

*phew* After some err... months(?!) I finally got an answer whether the serie - mentioned in the first post - converges or not. The answer is that there is some sort of radius of convergence around 0 where the serie converges and it is the solution for the ode I was studying (well... I'm still).

Shortly, I was/am studying ode $$y'' = y\sin x$$. Near $$x = 0$$ I derived another form of the ode, namely $$(1 - t^2)u'' -tu' - tu =0$$ ($$\ ' = \tfrac{d}{dt}$$), which clearly has singular points at $$t = \pm 1$$ (with the fact that substitution $$t = \sin x$$ was used), there is no point to try to search values other than $$|x| \le 1$$. But some numerical computations I made, gave me also a hint that the serie will break earlier, about at $$x \approx \sin 1$$. (Up to about 9 million terms used.)

Being such, I think this question is answered and case closed. I'll see what I can and want to do with the ode. Thank you again all the helpers! ^^