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I am working with cauchy problem of the form $$ ( - \partial_t + A^\delta) u^\delta = 0 , \qquad u^\delta(0,x) = h(x), $$ where the domain of $u^\delta$ is $[0,\infty) \times \mathbb{R}$. The operator $A^\delta$ can be written as follows: $$ A^\delta = A_0 + \delta e^{b x} A_1 , $$ where $b<0$ and $\delta \geq 0$. I have found a formal solution to the above problem by assuming that $u^\delta$ has a power series expansion in $\delta$. That is $$ u^\delta = \sum_{n=0}^\infty \delta^n u_n .$$ Plugging the expansion into the Cauchy problem yields $$ ( - \partial_t + A_0) u_0 = 0 , \qquad u_0(0,x) = h(x) , $$ and $$ ( - \partial_t + A_1) u_n = -A_1 u_{n-1} , \qquad u_n(0,x) = 0 $$ I can solve these equations explicity for a given $u_n$. But the solution for each $u_n$ is written as an integral (a Fourier Transform). And it is not easy to check if my series solution converges.

So, it would be very nice if I had some sort of theorem that would tell me that my assumption -- that $u^\delta$ can be expanded power series in $\delta$ -- is justified. If I had such a theorem, then I wouldn't need to check convergence of the sum.

So my question is: When can I assume that $u^\delta$ has a power series expansion in $\delta$? Specifically, is this a valid assumption in my problem? And, more generally, are there conditions I can check in the future to know if a perturbation can be treated as regular?

If it helps, both $A_0$ and $A_1$ are infinitesimal generators of Levy processes. i.e. $$ A_i = \mu_i \partial x + \frac{1}{2} \sigma_i^2 \partial_x^2 + \int_{\mathbb{R}} \nu_i(dz) ( e^{z \partial_x} - 1 - z \partial_x )$$ where $\mu_i$ and $\sigma_i$ and constants, $\nu_i(dz)$ is Levy measure and $e^{z \partial_x} f(x) = f(x+z)$ for any $C^\infty$ function.

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seems the recurrence relation linking $u_n$ with $u_{n-1}$ has an $A_0$ on the left side instead of $A_1$. From this relation, if you can show that the operator mapping $u_{n-1}$ to $u_n$ is a contraction from an appropriate space to itself, then the series is justified. –  Francois Monard Aug 6 '12 at 19:11
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