# Frobenius series solutions and asymptotic

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.

• What's the difference between $v_1$ and $v_3$? – Yuriy S May 21 '19 at 13:36
• @ Yuriy S Thanks for pointing that out. I corrected my text to say v_3=e^{-\xi}. – Stephane May 27 '19 at 17:12

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

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

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

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

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

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

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

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

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

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

• Wow thanks. That's really useful. – Stephane May 27 '19 at 16:43

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: $$v_2(\xi) = \frac {e^{a \xi}} {2 (a - 1)} \hspace {1px} {_2 \hspace {-1px} F_1}(a - 1, a + b; a; e^\xi) - \frac {e^{a \xi}} {2 (a + 1)} \hspace {1px} {_2 \hspace {-1px} F_1}(a + 1, a + b; a + 2; e^\xi).$$

• Thank you so much! That also helps a lot. – Stephane Sep 9 '19 at 0:15