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I was studying the proof of the following theorem

Theorem. Let $x_0$ be a real number and suppose that the coefficients $a(x)$, $b(x)$ in $$L[y](x)=y^{''}(x)+a(x)y^{'}(x)+b(x)y(x)\tag{1}$$ have convergent power series expansions in powers of $x-x_0$ in some interval as follows $$a(x)=\sum_{n=0}^{\infty}a_n(x-x_0)^n, \quad b(x)=\sum_{n=0}^{\infty}b_n(x-x_0)^n, \quad |x-x_0| \lt r_0, \quad r_0 \gt 0\tag{2}$$ Then there exists a nontrivial solution $\phi$ of the problem $$L[y](x)=0 \tag{3}$$ with a power series expansion $$\phi(x)=\sum_{n=0}^{\infty}c_n(x-x_0)^n\tag{4}$$ convergent for $|x-x_0|<\rho$ where $\rho \ge r_0$.

Proof Outline

The proof is straight forward. It considers the power series expansions of the coefficients in $(2)$ which are convergent for $|x-x_0| \lt r_0,\,r_0 \gt 0$. Then it puts the series solution $(4)$ into the ODE mentioned in $(3)$ and obtains a recurrence relation for the $c_i$s such that the ODE is satisfied. Then it proves that the series solution $(4)$ whose coefficients obey the aforementioned recurrence relation will converge in the interval $|x-x_0| \lt r_0$.

The following links show the original proof if you want to check it out.

Proof, Page 1
Proof, Page 2
Proof, Page 3


Question

According to the proof, it can be concluded that the solution will converge for $\rho=r_0$.

The proof does not mention anything about the last sentence of the theorem that $\rho \ge r_0$ which is saying that $r_0$ is lower bound for radius of convergence of the solution.

I just cannot understand that how the radius of convergence of the solution can be $\rho \gt r_0$. I think having $\rho>r_0$ is meaningless as the coefficients of ODE can just be replaced by their power series only in $|x-x_0|<r_0$.

Can someone shed some light on this?

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  • $\begingroup$ There seems to be some parts missing. It is a linear, homogeneous ODE and no initial value given. So obviously $y(x)\equiv 0$ is a solution. This solution is btw. converging on $\mathbb{R}$. So you could skip the whole given proof. And this seems also to answer your question. In the worst case, the radius of convergence might be $r_{0}$ but it can be larger or even be infinity. $\endgroup$ – Alex Jul 8 '16 at 22:58
  • $\begingroup$ @Alex: Thanks for attention. No parts is missing. :) You don't need initial values to prove the things mentioned in the theorem. You are forcing me to say what if we exclude the trivial solution $y=0$!? :) $\endgroup$ – H. R. Jul 8 '16 at 23:03
  • $\begingroup$ Yeah, of course the interesting part is the non-trivial solution. I just wanted to make the point that if this is really the Theorem without excluding the trivial solution, there was nothing to prove :-)! I will think about what you asked... $\endgroup$ – Alex Jul 8 '16 at 23:07
  • $\begingroup$ Ok, if I get the idea right $c_{i}$ is a function of $(a_{n})$ and $(b_{n})$. So given the $a_{n}$ could one not just choose the $b_{n}$ such that the $\phi(x)$ is converging in a larger radius of convergence buy just making many of the $c_{i}$ zero? This would be recurrence 9.9 in the files you gave. $\endgroup$ – Alex Jul 8 '16 at 23:27
  • $\begingroup$ And again to be even more precise: The author does not speak of the convergence radius of the involved functions but just claims that both $a(x)$ and $b(x)$ are convergent AT least in that given interval. So both series could converge also in $\mathbb{R}$ and of course then the power series of the solution would also converge on $\mathbb{R}$. So it is just not very sharp. $\endgroup$ – Alex Jul 8 '16 at 23:36
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I think, I have found an example for you. Suppose $a(x)=\frac{1}{1+x^{2}}$ and $b(x)=-1-\frac{1}{1+x^{2}}$. Then, obviously, both power series have radius of convergence $1$ with $x_{0}=0$ but as solution you can find \begin{equation} y(x)=\exp(x). \end{equation} So there is at least one solution which has greater radius of convergence, (even infinity) and for that both $a(x)$ and $b(x)$ can still have the same radius of convergence.

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  • $\begingroup$ (+1) Really nice example. So, what is the point that we are missing in the analysis? I mean this would show up in the proof of theorem some where! We may spot the point if you take the whole computations with this example. $\endgroup$ – H. R. Jul 9 '16 at 12:10
  • $\begingroup$ Are you there Alex? :) $\endgroup$ – H. R. Jul 9 '16 at 21:43
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    $\begingroup$ Now I am. Yeah, I would expect that if you compute the $c_{i}$ that you get they will decrease sufficiently fast, such that you get a larger existence interval. Maybe you can use the formula 9.9 and assume that $c_{0}=c_{1}=1$ (to get exactly the exponential function) and then you would just have to clarify that the recurrence indeed give $c_{k}=\tfrac{1}{k!}$. If you cannot make it, let me know, I will try it then. $\endgroup$ – Alex Jul 9 '16 at 23:20
  • $\begingroup$ I thought about it, I think having $\rho>r_0$ is meaningless as the coefficients of ODE can just be replaced by their power series only in $|x-x_0|<r_0$. But sometimes, luckily, the solution we obtained formally satisfies the ODE while we have not replaced the coefficients by their power series. $\endgroup$ – H. R. Jul 12 '16 at 0:33
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    $\begingroup$ Isn't it more that in principle the solution does not have to be analytic to satisfy the differential equation. So for the example I gave above we have still functions with are $C^{\infty}$ on $\mathbb{R}$ but not analytical. So there might be a solution which satisfies the ODE. It might also be reasonable to reconsider that the function I gave are not analytical on the whole $\mathbb{R}$ but except $|x|=1$ they can locally been developed into a power series. So clustering the whole $\mathbb{R}$ with different discs of convergence could do the trick, if the solution remains bounded.... $\endgroup$ – Alex Jul 12 '16 at 3:12
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The followings are other examples that show there may exist nontrivial solutions which are analytic at $x=0$ with their radius of convergence being greater than that of coefficients of the ODE.

$$\begin{array}{rcll} y''+ \dfrac{x}{1-x} y'-\dfrac{1}{1-x} y=0, & & y=x \\ y''- 2\dfrac{x}{1-x^2} y'+ \dfrac{n(n+1)}{1-x^2} y=0, & & y=P_n(x), & \text{Legendre Differential Equation} \end{array}$$

There is also a nice example in the answer by @Alex. However, in all of the examples, the radius of convergence of the solution we are interested in is $\infty$. I couldn't find some example which the radius of convergence of solution is greater than that of coefficients but it is not $\infty$.

I think any arguments about $ρ>r_0$, in such a proof, is meaningless as the coefficients of ODE can just be replaced by their power series only in $|x−x_0|<r_0$. But sometimes, luckily, the solution we obtained formally by this method satisfies the ODE while we have not replaced the coefficients by their power series. In this case, what just matters is the radius of convergence of the solution itself.

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