My question is quite general, but let me explain it with a specific example. Suppose I have a PDE: $$\frac{d^2}{d\rho^2}f(\rho)+\frac{3}{\rho}\frac{d}{d\rho}f(\rho)+U[f(\rho)]=0,\tag{1}$$ where $\rho\in R$, $f(\rho)$ is a real-valued function and $U[f]$ is an arbitrary real-valued function of $f$ when viewing $f$ as a number in $R$. Suppose we obtain the solution of Eq.(1) for a specific boundary condition, denoted by $f_1(\rho)$.
Now I want to continue the Eq.(1) through $\rho\rightarrow i\tilde\rho$ with $\tilde\rho\in R$ and obtain $$-\frac{d^2}{d\tilde{\rho}^2}f(\tilde{\rho})-\frac{3}{\tilde{\rho}}\frac{d}{d\tilde{\rho}}f(\tilde{\rho})+U[f(\tilde{\rho})]=0.\tag{2}$$ And we would like to solve Eq.(2) with boundary condition $f(\tilde{\rho}=0)=f_1(\rho=0)$. Denote the solution by $f_2(\tilde\rho)$. My question is will $$f_2(\tilde\rho)=f_1(i\tilde\rho)\ ?$$ That is, can we obtain the solution to the continued PDE via the direct continuation of the solution to the `before-continued' PDE?
My superficial guess is no. But if yes, can we give a proof on that?