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Let us consider the following parameterised complex-valued second order differential equations, and $u(x,\lambda)$ be the solution for $$ u''+u'-i\lambda V(x)u=0, \, x\in [0,1], $$ What is the relationship between $u(x,\lambda_1)$ and $u(x,\lambda_2)$? Is that possible to construct $u(x,\lambda_1)$ from $u(x,1)$? Do we have to impose some assumptions on $V(x)$?

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    $\begingroup$ The solution? You need initial or boundary conditions to specify a particular solution. Which one are you interested in? $\endgroup$ Commented Feb 19, 2016 at 18:26

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Presumably $V(x)$ is a continuous function of $x$.

Formally write $$u(x,\lambda) = \sum_{j=0}^\infty u_j(x) (\lambda - 1)^j$$ The coefficients of each power of $\lambda-1$ in the differential equation give you a sequence of d.e.'s for the $u_j$'s: $u_0'' + u_0' - i V u_0$ (so that $u(x,1) = u_0(x)$ is a solution of your d.e. for $\lambda = 1$), and $u_j'' + u_j' - i V u_j = i V u_{j-1}$ for $j \ge 1$. Thus $u_j$ is a solution of the inhomogeneous linear equation corresponding to your homogeneous equation, with forcing term depending on $u_{j-1}$.

Of course, to get a particular solution you need to specify initial or boundary conditions, and you have not done so. Convergence or divergence of the series may depend on those conditions.
If you specify initial conditions that are fixed or entire as functions of $\lambda$, then $u(x,\lambda)$ should be an entire function of $\lambda$, and the power series will always converge.

On the other hand, a solution satisfying specified boundary conditions might fail to exist for certain complex $\lambda$, and presumably this will be reflected in lack of convergence of the power series.

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