# Obtain the leading order uniform approximation of the solution

Obtain the leading order uniform approximation of the solution to $$\epsilon y′′-x^2y′-y=0$$.

The boundary conditions are $$y(0)=y(1)=1$$.

Since $$a(x)<0$$ the boundary layer is at $$x=1$$.

The outer solution will be of the form $$y(x; \epsilon) = y_0(x) + \epsilon y_1(x) + ...$$

So the leading order problem is $$-x^2y_0' - y_0 =0$$ and $$y_0(0) = 1$$.

Hence $$y_0 = ke^{1/x}$$.

Here's where the issue arises: to work out $$k$$ I get $$y_0(0)=ke^{1/0}=1$$ using the boundary condition at $$x=0$$, but $$e^{1/0}$$ is obviously undefined. How can I proceed form here?

• I think there is also a boundary layer of width $\sqrt{\epsilon}$ at $x=0$, since for $x~\sim\sqrt{\epsilon}$ you have $\epsilon\sim x^2$, so maybe the first term is important near $x=0$. If you can't satisfy the boundary condition this is often a sign that something is not working in your expansion. – David Dec 3 '18 at 20:16
• Ah I see, that would make sense, thank you! I took the boundary layer to be at x=1 as we were taught that if the coefficient of the y’ term is negative, the boundary layer is at x=1. How can you show that there is also a BL at x=0? – maria1991 Dec 3 '18 at 20:56
• The coefficient is zero at $x=0$ which should make you pause to see if you have a dominant balance near $x=0$. – David Dec 3 '18 at 21:22
• This question asks about the same equation: math.stackexchange.com/q/2034260/131807. – David Dec 4 '18 at 1:43

This is a complicated problem, but if you're careful it does work out. Firstly, let's write out the equation assuming there is a boundary layer or width $$\epsilon^\alpha$$ at $$x_0$$, so let $$X=(x-x_0)/\epsilon^\alpha$$ and $$Y(X)=y(x)$$. We get $$\epsilon^{1-2\alpha}Y_{XX}-\epsilon^{-\alpha}(\epsilon^\alpha X+x_0)^2Y_X-Y=0,$$ expanding $$\epsilon^{1-2\alpha}Y_{XX}-\epsilon^{\alpha}X^2Y_X-2Xx_0Y_X-\epsilon^{-\alpha}x_0^2Y_X-Y=0.$$

Now for dominant balance. If $$x_0\neq0$$, then we would balance $$\epsilon^{1-2\alpha}$$ with $$\epsilon^{-\alpha}$$ (you can check that this is the only dominant balance) to give $$\alpha=1$$. You can leave $$x_0$$ unknown and determine that it must be 1, but since you already know that we'll use the fact and write our leading-order inner equation at $$x=1$$ as $$Y_{XX}-Y_X=0,$$ with boundary condition $$Y(0)=1$$ (since $$x=1$$ corresponds to $$X=0$$). The solution is $$Y(X)=A+Be^X$$ where $$A+B=1$$.

If $$x_0=0$$ then we have a different balance, with $$\alpha=1/2$$. Let $$\mathsf Y(\chi)=y(x)$$ with $$\chi=x/\sqrt{\epsilon}$$ and the leading-order inner equation at $$x=0$$ is $$\mathsf Y_{\chi\chi}-\mathsf Y=0,$$ with $$\mathsf Y(0)=1$$. The solution is $$\mathsf Y=Ce^\chi+De^{-\chi}$$, and since our solution must be bounded as we exit the boundary layer, we need $$A=0$$, and the boundary condition gives $$D=1$$.

The leading order outer solution is, as you found, $$y=ke^{1/x}$$. You can say here that $$k=0$$ since the solution must be bounded as you enter each boundary layer, or do it through asymptotic matching.

To fix the remaining constants $$k$$ and $$A$$ (or $$B$$), we need to match all the parts of the solution. To do this, we need the outer solution to be bounded, so we need $$k=0$$. Alternatively, consider matching the outer solution to the inner solution at $$x=0$$, $$\lim_{\chi\rightarrow\infty}e^{-\chi}=\lim_{x\rightarrow0}ke^{1/x}\Rightarrow0=\lim_{x\rightarrow0}ke^{1/x}\Rightarrow k=0.$$ Then the match between the outer layer and the inner layer at $$x=1$$ gives $$\lim{x\rightarrow1}0=\lim{X\rightarrow-\infty}A+Be^X\Rightarrow A=0$$ and hence $$B=1$$.

So the inner solution at $$x=0$$ is $$\mathsf Y(\chi)=e^{-\chi}$$, the outer solution is $$y(x)=0$$ and the inner solution at $$x=1$$ is $$Y(X)=e^X$$. We can find a uniformly valid approximation by adding the three equations (writing them all in terms of $$x$$) and subtracting off the matching constants (all zero) as, $$y_{unif}(x)=e^{x/\sqrt{\epsilon}}+e^{(x-1)/\epsilon}.$$

This will not be particularly accurate, it's an $$O(\sqrt{\epsilon})$$ approximation, but I did verify that numerically. I've made a plot of the solution for $$\epsilon=0.01$$ first and $$\epsilon=0.02$$ second.