Consider the equation below which is an eigenvalue problem I am studying: $$ \label{left} \lambda(v-v'')+c\left(v''-v+be^\xi v+(1-b)e^\xi v'-e^\xi v''\right)'=0. $$ Restriction on the parameters: $\lambda>0$ and $c>0$. It turns out there are two solutions converging as $\xi\rightarrow -\infty$: $v_1=e^\xi$ and a second solution we denote $v_2={F}(\xi)$. This second solution decays as $e^{\lambda\xi/c}$ as $\xi\rightarrow -\infty$. This rate of decay is simply found by solving the constant coefficient asymptotic system obtained by applying the limit $\xi\rightarrow -\infty$ to the equation. There is also a third solution $v_3=e^{-\xi}$. The point $\xi=0$ is regular singular with indices $r_1=0$, $r_2=1$, and $r_3=-\lambda/c+2-b$. I am interested to associate the series development to the asymptotic behaviors. Since I know explicitly $v_1$ and $v_3$, I only have to deal with $v_2$. To be precise, find the series solution of a solution linearly dependent to $v_1$, that converges to zero as $\xi\rightarrow -\infty$.

More generally, I am interested at a general method on how to connect the series solutions about a regular singular point to the behavior for large value of the independent variable.

  • $\begingroup$ What's the difference between $v_1$ and $v_3$? $\endgroup$ – Yuriy S May 21 at 13:36
  • $\begingroup$ @ Yuriy S Thanks for pointing that out. I corrected my text to say v_3=e^{-\xi}. $\endgroup$ – Stephane May 27 at 17:12

$\lambda(v-v'')+c(v''-v+be^\xi v+(1-b)e^\xi v'-e^\xi v'')'=0$

$\lambda(v-v'')+c((1-e^\xi)v''+(1-b)e^\xi v'+(be^\xi-1)v)'=0$

$\lambda(v-v'')+c((1-e^\xi)v'''-e^\xi v''+(1-b)e^\xi v''+(1-b)e^\xi v'+(be^\xi-1)v'+be^\xi v)=0$







$v=k_1e^\xi+k_2e^{-\xi}+k_3\left(e^\xi\int^\xi e^\frac{(\lambda-c)\xi}{c}(e^\xi-1)^{-\frac{\lambda+bc}{c}}~d\xi-e^{-\xi}\int^\xi e^\frac{(\lambda+c)\xi}{c}(e^\xi-1)^{-\frac{\lambda+bc}{c}}~d\xi\right)$

  • $\begingroup$ Wow thanks. That's really useful. $\endgroup$ – Stephane May 27 at 16:43

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