# Leading order matching of $\epsilon x^py'' + y' + y = 0$

Question: The function $$y(x)$$ satisfies $$\epsilon x^py'' + y' + y = 0,$$ in $$x\in [0,1]$$, where $$p<1$$, subject to the boundary conditions $$y(0) = 0$$ and $$y(1)=1$$. Find the rescaling for the boundary layer near $$x = 0$$, and obtain the leading order inner and outer approximations, and match them.

My approach: The rescaling for the boundary layer near $$x = 0$$ is given by $$x = \epsilon^{\alpha}x_L$$, where $$x_L>0$$, $$\alpha > 0$$ and $$x_L = \mathrm{ord}(1)$$. Also, in this region, define $$y_L(x_L) = y(\epsilon^{\alpha}x_L) = y(x)$$. In terms of these rescaled variables, the original DE can be rewritten as $$\epsilon^{1+\alpha(p-2)}x_L^p y_L'' + \epsilon^{-\alpha}y'_L + y_L = 0.$$ Now, we know that the first two terms in the above rescaled DE must in balance, which is only possible when $$\alpha = \frac{1}{1-p}$$.

Therefore, the DE can be rewritten as $$x_L^py_L'' + y_L' + \epsilon^{\alpha}y_L = 0,$$ subject to the BC $$y_L(0) = 0$$. The asymptotic expansion of the solution in this layer can be given as $$y_L(x_L) = \sum_{n=0}^{\infty}\epsilon^{n\alpha}y_{L,n\alpha}(x_L).$$ Therefore, at leading order, the DE reduces to the following: $$x_L^py_{L,0}'' + y_{L,0}' = 0,$$ which when solved assuming that $$y_{L,0}(x_L)$$ is strictly increasing in this inner layer leads to $$y_{L,0}(x_L) = A_{L,0}\int e^{{x_L}^{1-p}/(p-1)}dx_L + B_{L,0} = A_{L,0}(1-p)^p\Gamma\left(\frac{1}{1-p}, \frac{x_L^{1-p}}{1-p}\right) + B_{L,0},$$ where $$B_{L,0} = -A_{L,0}(1-p)^p\Gamma\left(\frac{1}{1-p}\right).$$

It is easy to see that the leading order outer solution (assuming no boundary layer other than the one near $$x=0$$) is given by $$y_{\mathrm{out},0}(x) = e^{(1-x)}$$.

Using Van-Dyke's matching rule (following the notations given in Perturbation Methods by E. J. Hinch), we must have $$E_0H_0 y = H_0E_0y$$, where $$H_0E_0y = e^{1-\epsilon^{1/(1-p)}x_L}\approx e$$ and $$E_0H_0y = A_{L,0}(1-p)^p\Gamma\left(\frac{1}{1-p}, \frac{x^{(1-p)}}{\epsilon(1-p)}\right) - A_{L,0}(1-p)^p\Gamma\left(\frac{1}{1-p}\right).$$

I am not sure how to reduce $$E_0H_0y$$ and match it with $$H_0E_0y \approx e$$ and obtain the value of the constant $$A_{L,0}$$. Can someone help me with this? I am looking for a solution using Van-Dyke's matching rule or matching using an intermediate variable (and not using Mathematica). Thanks in advance.

I suggest that you write the solution to $$x_L^py_{L,0}'' + y_{L,0}' = 0$$ satisfying the boundary condition $$y_{L,0}(0)=0$$ as $$y_{L,0}(x_L) = A\int_0^{x_L} \exp\left(-\frac{{\xi}^{1-p}}{1-p}\right)d\xi, \tag{1}$$ which, in terms of the original variables, can be rewritten as $$y(x)=A\int_0^{\epsilon^{-\frac{1}{1-p}}x}\exp\left(-\frac{{\xi}^{1-p}}{1-p}\right)d\xi. \tag{2}$$ For $$0<\epsilon\ll x \ll 1$$, the leading order approximation to $$(2)$$ is $$y(x)\sim A\int_0^{\infty}\exp\left(-\frac{{\xi}^{1-p}}{1-p}\right)d\xi =A(1-p)^{\frac{1}{1-p}}\Gamma\left(1+\frac{1}{1-p}\right). \tag{3}$$ Matching $$(3)$$ with $$y_{\mathrm{out},0}(x) = e^{(1-x)}\sim e$$ if $$0, we obtain $$A=\frac{e(1-p)^{-\frac{1}{p-1}}}{\Gamma\left(1+\frac{1}{1-p}\right)}. \tag{4}$$