The question asks to find a global approximation to the leading order of $\epsilon$.

$\epsilon y'' + xy' + \epsilon y =0$, with boundary conditions $y(0)=1,y(1)=-1$.

I assumed it's a boundary layer problem, and tried to work with both $x=0$ and $x=1$. But did not work out...

(new edit below...)

I start by trying boundary layer at x=0 and x=1 normal. Outer solution is $y=-1$, for inner solution, do the substitution of $x = \epsilon X$, $Y=y(x/\epsilon)$, then $\frac{d^2 Y}{d X^2} + \epsilon \frac{dY}{dX} + \epsilon ^ 2 Y = 0$,which when take $\epsilon \rightarrow 0$, becomes $\frac{d^2 Y}{d X^2} = 0$, and gives you an inner solution of $Ax/\epsilon +B$. I can't match this with inner solution...

However if I do boundary layer at $x=1$, the outer solution at $x=0$ would just be $y=1$, for inner solution, I did substitution of $1-x = \epsilon X$, the equation then becomes $\frac{d^2 Y}{d X^2} - \frac{dY}{dX} =0$, which will give the inner solution $y = A + B exp(\frac{1-x}{\epsilon}$. I can't match this with outer solution either, since when $\frac{1-x}{\epsilon} \rightarrow \infty$, inner solution would blow up.

Somehow I feel I am missing something, even the outer solution might not be something that simple. since when $\epsilon \rightarrow 0$, the equation becomes $y'=0$, but in that case $\epsilon$ terms aren't negligible any more.

  • $\begingroup$ Could you give a little more detail about how it did not work out? Which step are you stuck on? (For example, do you know the width of the boundary layer? Have you found the inner and outer solutions? Are you having trouble with the matching?) $\endgroup$ Commented Jan 18, 2014 at 7:35
  • $\begingroup$ I added new detail.. $\endgroup$
    – initial_D
    Commented Jan 18, 2014 at 13:50

1 Answer 1


The outer solution $y_{\text{outer}}(x) = -1$ is fine, but it turns out that in this problem the width of the boundary layer isn't $\approx \epsilon$ as you have assumed with your substitution $x = \epsilon X$. To determine the correct width, instead substitute $x = \epsilon^{\alpha} X$, where $\alpha$ is a constant to be determined. The equation

$$ \epsilon y'' + xy' + \epsilon y = 0 $$

then becomes

$$ \epsilon^{1-2\alpha} \frac{d^2 Y}{dX^2} + X \frac{dY}{dX} + \epsilon Y = 0. $$

The third term, $\epsilon Y$, being order $\epsilon$, is certainly smaller than the second term, $X \frac{dY}{dX}$, which is order $1$. Therefore the third term must not be involved in the dominant balance. This only leaves the first two terms, which are balanced when $\alpha = 1/2$. The resulting equation is then, to leading order,

$$ \frac{d^2 Y}{dX^2} + X \frac{dY}{dX} = 0. $$

After multiplying by the integrating factor $e^{X^2/2}$ we can write the equation as

$$ \frac{d}{dX} \left(e^{X^2/2} \frac{dY}{dX}\right) = 0, $$

and integrating yields

$$ \frac{dY}{dX} = A e^{-X^2/2}. $$

Integrating once more we see that

$$ \newcommand{erf}{\operatorname{erf}} Y(X) = A\erf\left(\frac{X}{\sqrt{2}}\right) + B, $$

where $\erf$ is the error function. To satisfy the boundary condition $y(0) = 1$ we must then take $B = 1$. To match this with the outer solution we need

$$ \lim_{X \to \infty} Y(X) = y_{\text{outer}}(0) = -1, $$

which is satisfied when $A = -2$.

We now have the inner and outer solutions, namely

$$ y_{\text{outer}}(x) = -1, \\ y_{\text{inner}}(x) = 1 - 2\erf\left(\frac{x}{\sqrt{2\epsilon}}\right). $$

The matched solution is therefore

$$ y(x) \sim y_{\text{inner}}(x) + y_{\text{outer}}(x) - y_{\text{outer}}(0) = 1 - 2\erf\left(\frac{x}{\sqrt{2\epsilon}}\right). $$

Here's a plot which shows the actual solution in blue and this approximate solution in purple for $\epsilon = 1/50$.

enter image description here

  • $\begingroup$ Note that the result $\alpha = 1/2$ implies that the width of the boundary layer is $\approx \sqrt{\epsilon}$. $\endgroup$ Commented Jan 18, 2014 at 17:41
  • $\begingroup$ Aww, thanks, that definitely makes sense. I still have doubt regarding the outer solution, if $y'=0$, is the other terms still negligible? Another problem I always have is to determine the boundary layer. How do you see from the problem that the boundary layer is at x=0? $\endgroup$
    – initial_D
    Commented Jan 19, 2014 at 6:25
  • $\begingroup$ Well, in the outer region the actual solution $y(x,\epsilon)$ isn't exactly equal to $-1$, so it's not exactly true that $y' = 0$ for the actual outer solution. If you try substituting $y(x,\epsilon) = -1$ into the ODE you'll find that it leaves behind a term of order $\epsilon$, so in a sense it is true that $y' = 0$ before you include effects of order $\epsilon$ to cancel out what remains. (...) $\endgroup$ Commented Jan 19, 2014 at 17:02
  • $\begingroup$ (...) Think of it this way: the $y_{\text{outer}}(x)$ approximation you found is leading order in the sense that, if you try a Poincaré expansion like $$y(x,\epsilon) \sim \sum_{n=0}^{P} \epsilon^n y_n(x),$$ you would find that $y_0(x) = -1$ and $y_1(x) = \log x$, so that $y(x,\epsilon) \sim -1 + \epsilon \log x$. Indeed, $$\lim_{\epsilon \to 0^+} \frac{y(x,\epsilon) + 1}{\epsilon} = \log x$$ pointwise on the interval $(0,1)$. This $\epsilon \log x$ term is what cancels out any effects of order $\epsilon$ in the equation (and in turn leaves behind effects of order $\epsilon^2$). $\endgroup$ Commented Jan 19, 2014 at 17:02
  • $\begingroup$ As for the location of the boundary layer, I'm not really sure how to determine it's location a priori. Usually one tries some changes of variables to see how the equation looks when stretched one way or the other. What characterizes a boundary layer is a term which decays--not necessarily exponentially--when leaving the boundary region, so if you don't find one of those then you haven't found a boundary layer. (...) $\endgroup$ Commented Jan 19, 2014 at 17:02

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