Frobenius series solutions and asymptotic

Question

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.

Answer 1

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

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

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

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

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

$c(1-e^\xi)(v''-v)'=(bce^\xi+\lambda)(v''-v)$

$\dfrac{(v''-v)'}{v''-v}=\dfrac{bce^\xi+\lambda}{c(1-e^\xi)}$

$\ln(v''-v)=\dfrac{\lambda\xi}{c}-\dfrac{(\lambda+bc)\ln(e^\xi-1)}{c}+K$

$v''-v=ke^\frac{\lambda\xi}{c}(e^\xi-1)^{-\frac{\lambda+bc}{c}}$

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

Answer 2

I assume you mean the solution linearly dependent on $v_2$. If we make the change of variables $v = e^\xi u$, the equation for $u$ will not contain $u^{(0)}$ (because $u = C$ is a solution), so we get a second-order equation for $u'$. Taking $u' = e^{-2 \xi} w$, we get a first-order equation for $w'$, which has a solution $$w'(\xi) = e^{(a + 1) \xi} (1 - e^\xi)^{-a - b},$$ where $a = \lambda/c$. Taking the binomial expansion and integrating twice gives a convergent asymptotic series $$v_2(\xi) = \sum_{k \geq 0} \frac {(-1)^k} {(k + a - 1) (k + a + 1)} \binom {-a - b} k e^{(k + a) \xi}, \quad \xi < 0,$$ which also has a closed form: $$v_2(\xi) = \frac {e^{a \xi}} {2 (a - 1)} \hspace {1px} {_2 \hspace {-1px} F_1}(a - 1, a + b; a; e^\xi) - \frac {e^{a \xi}} {2 (a + 1)} \hspace {1px} {_2 \hspace {-1px} F_1}(a + 1, a + b; a + 2; e^\xi).$$