I have formally derived a solution to a PDE as a power series

$$u = \sum_{n=0}^\infty \epsilon^n u_n.$$

I would like to show that the radius of convergence for is $\mathbb{R}$. I assume that the easiest way to do that is to show

$$\lim_{n \to \infty}\left|\frac{u_{n+1}}{u_n}\right| = 0.$$

The difficulty in showing this is that the $u_n$ are of the form

$$u_n = \int_\mathbb{R} f_n(x) dx.$$

So far, my best strategy is to introduce

$$g_{n+1} := \frac{f_{n+1}}{f_n}$$

so that

$$\left|\frac{u_{n+1}}{u_n}\right|^2 = \frac{| \int g_{n+1}f_n dx |^2}{|\int f_n dx|^2} \leq \frac{\int | g_{n+1} |^2 dx \cdot \int | f_n |^2 dx}{|\int f_n dx|^2}$$


\begin{align*} \lim_{n \to \infty} |\frac{u_{n+1}}{u_n}|^2 &\leq \limsup_{n \to \infty} \int | g_{n+1} |^2 dx \cdot \lim_{n \to \infty} \frac{ \int | f_n |^2 dx}{|\int f_n dx|^2}\\ &\leq \int \limsup_{n \to \infty} | g_{n+1} |^2 dx \cdot\lim_{n \to \infty} \frac{ \int | f_n |^2 dx}{|\int f_n dx|^2}. \end{align*}

Thus I need to show

$$\limsup_{n \to \infty} | g_{n+1} |^2 = 0 \qquad (A)$$


$$\lim_{n \to \infty} \frac{ \int | f_n |^2 dx}{|\int f_n dx|^2} < \infty \qquad (B)$$

This seems more complicated to me than I think it should be. Intuitively, it seems to me like I should just need to show (A). Anyway, if anybody has a simpler strategy, I would greatly appreciate it.

Additional Information not in my original post

The integrals written above in terms of $x$ in the notation of my paper are

$$\int_\mathbb{R} f_n(\lambda) d\lambda.$$

The specific form of $f_n(\lambda)$ is as follows

$$f_n(\lambda) = \left( \sum_{k=0}^n \frac{e^{t \phi_{\lambda-ik\beta}}} {\prod_{j\neq k}^n (\phi_{\lambda-ik\beta}-\phi_{\lambda-ij\beta})} \right) \left( \prod_{k=0}^{n-1} \chi_{\lambda-ik\beta}\right) (\psi_\lambda, h) \psi_{\lambda}$$


$$\phi_\lambda = \frac{1}{2}a_0^2(-\lambda^2 - i\lambda) - \int_\mathbb{R}\nu_0(dz)(e^{z} - 1 - z)i\lambda + \int_\mathbb{R}\nu_0(dz)(e^{i\lambda z} - 1 - i\lambda z) - c_0$$


$$\chi_\lambda = \frac{1}{2}a_1^2(-\lambda^2 - i \lambda) - \int_\mathbb{R}\nu_1(dz)(e^{z} - 1 - z)i\lambda + \int_\mathbb{R}\nu_1(dz)(e^{i\lambda z} - 1 - i\lambda z) - c_1$$

the $a_i$ and $c_i$ are real positive constants. The $\nu_i$ are Levy measures.

$$\psi_\lambda = e^{i\lambda y} \qquad (\psi_\lambda, h) = \int_\mathbb{R} \psi_\lambda(y) h(y) dy.$$

The domain of $f(\lambda)$, $\psi_\lambda$ and $\chi_\lambda$ are $\mathbb{C}$. For simplicity, assume that $h$ is $L^2(\mathbb{R},dy)$.


2 Answers 2


I think you should offer more information about the mathematical objects you mentionend (e.g. $f_n$ and $u_n$) such like their forms, domain and range.

For example, as you didn't say anything about $f_n$, what if $f_n = constant$, then it is obvious $(A)$ can't be hold.

If these necessary details are omitted, nobody can help you.


Note that $$ \limsup\int |g_n|^2 \leq \int\limsup |g_n|^2 $$ does not hold for arbitrary $g_n$.

  • $\begingroup$ Right. Yes, I should check that there is an integrable function that bounds the $|g_n|^2$ $\endgroup$
    – psyduck
    Jul 15, 2012 at 1:04

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