The Question:

$$\varepsilon y''+f(x)y'+y=0 \qquad y(-1)=0 \qquad y(1)=1$$

where $0<\varepsilon \ll 1$ and $f$ is a given smooth function that is strictly positive with $f(1)=f(-1)=1$.

(i) Determine the location of the boundary layer

(ii) Obtain leading order outer and inner solutions

My Attempt:

(i) By first assuming that $y(x,\varepsilon) \sim y_0(x) + \varepsilon y_1(x)+\cdots$ is of order $O(1)$, I plugged this into the equation to obtain

$$f(x)y_0'(x)+y_0(x)=0 \qquad y_0(-1)=0 \qquad y_0(1)=1$$

at the leading order. Solving this gives

$$y_0(x) = A\exp\biggl(-\int \frac{dx}{f(x)}\biggl)$$

for some constant $A$. If we try to satisfy $y_0(-1)=0$ here, we get $A=0$ which is a contradiction, for then $y$ would not be of order $O(1)$.

It follows that the boundary layer is at $x=-1$.

(ii) So I have already found $y_0$ above, which is indeed the leading order term to the outer solution.

But how do I find the constant $A$ if I don't know what $f$ is?

How do I use the fact that $f(-1)=f(1)=1$?

  • $\begingroup$ If you add limits to your integral, the solution would be $$ y_0 = \exp\left(-\int_1^x \frac{1}{f(t)} dt\right) $$ $\endgroup$ – Dylan May 21 '18 at 3:41
  • $\begingroup$ I don't see how that helps? $\endgroup$ – glowstonetrees May 21 '18 at 12:03
  • $\begingroup$ You asked how to find the constant $A$. That's one way to reduce it $\endgroup$ – Dylan May 21 '18 at 12:06
  • $\begingroup$ Ahhh, I understand what you mean now, thanks $\endgroup$ – glowstonetrees May 21 '18 at 12:29

As an alternative to contemplating inner and outer solutions, the WKB approximation method directs you to consider $y=\exp(S/\delta)$ with accordingly $$ ε(δS''+S'^2)+δfS'+δ^2=0 $$ By the assumption $f>0$, and thus $\min_{x\in[0,1]} f(x)\gg \max(ε,δ)$, the leading terms in magnitude are $εS'^2+δfS'$ and they are in magnitude balance for $ε=δ$. Then the perturbation problem to solve is $$ S'^2+fS'=-ε(S''+1). $$ The series expansion $S=S_0+εS_1+ε^2S_2$ results in (letting $s_k=S_k'$) two solution for the zeroth order equation $s_0^2+fs_0=0$ that lead to two expansions giving approximations to two independent basis solutions. \begin{align} s_0^2+fs_0&=0&:&&s_0&=0&\text{or}&& s_0&=-f(x)& \\ (2s_0+f)s_1&=-(s_0'+1)&:&&s_1&=-\frac1f&|&& s_1&=\frac{-f'+1}f \\ s_1^2+(2s_0+f)s_2&=-s_1'&:&&s_2&=-\frac{1+f'}{f^3}&|&&s_2&=\frac{-f''f+(-2f'+1)(-f'+1)}{f^3} \end{align} etc.

Let $F$ be an anti-derivative of $f$, $G$ of $1/f$, both are monotonously increasing. Select the integration constants so that $F(-1)=0=G(-1)$. Then the two basis solutions using $s_0$ and $s_1$ give the approximation $$ y(x)\approx Ae^{-G(x)}+\frac{B}{f(x)}e^{-\frac1εF(x)+G(x)} $$ The boundary conditions imply \begin{align} 0&=A+B&\implies A&=-B\\ 1&=-Be^{-G(1)}+Be^{-\frac1εF(1)+G(1)}&\implies B&=-e^{G(1)}+\text{very small terms}, \end{align} the last because $\exp(-\frac1εF(1))$ is $O( ε^k)$ for any order $k$. In total $$ y(x)\approx e^{G(1)-G(x)}-\frac1{f(x)}e^{-\frac1εF(x)+G(1)+G(x)}. $$ The first part corresponds to the outer solution, the second the inner solution for the boundary layer at $x=-1$. It is a little more complex than the inner solution $\exp(-\frac1εx+G(1))$ one would get with the direct method.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.