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Consider singular perturbation problem $$\epsilon \left[\frac{d}{dx}\left(h^3p\frac{dp}{dx}\right)\right]=\frac{d}{dx}(hp)$$ $$p(0)=p(1)=1$$ where $h(x)$ is a positive smooth function with $h(0)\ne h(1)$. I trying to construct the approximate solution using matched asymptotic method. Numerical solution suggests that there exist boundary layer near $x=1$. Observe that the differential equation can be written as

$$\epsilon\left[3h^2h'pp'+h^3(p')^2+h^3pp''\right]=h'p+hp'.$$

For the outer solution I obtained $$p_{\mathrm{out},0}(x)=\frac{h(0)}{h(x)}.$$ This solution is consistent with numerical result. For the boundary layer, using $\hat{x}=(x-1)/\epsilon$, we obtain $$\frac{d}{dx}p(\hat{x})=\frac{1}{\epsilon}p'(\hat{x}),\quad \frac{d^2}{dx^2}p(\hat{x})=\frac{1}{\epsilon^2}p''(\hat{x})$$ Substituting into differential equation $$\epsilon 3h^2h'p(\hat{x})p'(\hat{x})+h^3(p'(\hat{x}))^2+h^3p(\hat{x})p''(\hat{x})=\epsilon h'p(\hat{x})+hp'(\hat{x}).$$ Expanding and taking the leading order $$h^2\left[(p_0'(\hat{x}))^2+p_0(\hat{x})p_0''(\hat{x})\right]=p_0'(\hat{x})$$ Since $h(x)$ is abstract I don't know how to proceed.

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  • $\begingroup$ How did you come up with the boundary layer equation? Can you put some working in? $\endgroup$ – David Apr 7 '16 at 0:11
  • $\begingroup$ I made mistake. I have corrected it. $\endgroup$ – Sukan Apr 7 '16 at 1:38
  • $\begingroup$ I think you should also have $h'(x)=h'(\hat x)/\epsilon$, just like $p'(x)=p'(\hat x)/\epsilon$. I don't know how to solve the resulting equation though. $\endgroup$ – David Apr 7 '16 at 1:42
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First, observe that you can integrate the original equation once with respect to $x$ to obtain \begin{equation} \epsilon h^3 p\,p' = h\,p + c. \tag{1} \end{equation} If you look for a solution where the left boundary condition is satisfied in the slow coordinate $x$, you obtain as an outer solution \begin{equation} p(x) = -\frac{c}{h(x)} = \frac{h(0)}{h(x)}. \end{equation} In the fast coordinate $\xi = \frac{x}{\epsilon}$ with $\frac{\text{d}p}{\text{d} \xi} = \dot{p}$, equation $(1)$ is \begin{equation} h^3 p\,\dot{p} = h\,p + c. \tag{2} \end{equation} As $h$ is positive, we can rewrite $(2)$ as \begin{equation} p\,\dot{p} = \frac{p}{h^2} + \frac{c}{h^3}. \tag{3} \end{equation} Equation $(3)$ is an Abel (differential) equation of the second kind. For general $h$, this equation does not have an explicit solution. However, for specific choices of $h$, the equation can be solved once it is brought into canonical form, see here.

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