Consider the equation

$$\varepsilon^3 \frac{d^2y}{dx^2} + 2x^3 \frac{dy}{dx} - 4\varepsilon y = 2x^3 \qquad \qquad y(0) = a \;, \; y(1)=b$$

in the limit as $\varepsilon \rightarrow 0^+$, where $0<a<b-1$.

Assuming that there are nested boundary layers at $x=0$, determine the thickness of the boundary layers and the leading-order additive composite solution.


My attempt is shown below. Basically, I don't understand how you are meant to match the different solutions in the various domains.

Would be grateful if someone could explain this to me, or give me some hints.


Firstly, the leading term $y_0$ of the outer solution satisfies

$$2x^3 \frac{dy_0}{dx} =2x^3 \qquad \qquad y_0(1)=b$$

This is easily solved to give $\color{red}{y_0(x) = x+b-1}$.

Now suppose we scale the equation with $x=\varepsilon ^\alpha X$ where $\alpha>0$ and $X = \mathcal O(1)$. The equation becomes

$$\varepsilon^{3-2\alpha} \frac{d^2y}{dX^2} + 2\varepsilon^{2\alpha}X^3 \frac{dy}{dX} - 4\varepsilon y = 2\varepsilon^{3\alpha}X^3$$

The possible leading order balances come from $\alpha = 1$ and $\alpha = 1/2$.

When $\alpha = 1/2$ (i.e. $x = \varepsilon^{1/2} X$), the leading order term $Y_0$ in this layer should satisfy

$$2X^3 \frac{dY_0}{dX} - 4Y_0 = 0 \qquad \qquad Y_0(X=0) = a$$

The general solution is $\color{red}{Y_0(X) = A\exp (-1/X^2)}$, which cannot satisfy the boundary condition, since $a>0$. So I suppose this is why we need another boundary layer.

Scaling instead with $x = \varepsilon \tilde X$, the leading order solution $\tilde Y_0$ in this layer satisfies

$$\frac{d^2\tilde Y_0}{d\tilde X^2} - 4\tilde Y_0=0\qquad \qquad \tilde Y_0(\tilde X = 0) = a$$

The solution is $\color{red}{\tilde Y_0(\tilde X) = \tilde A\sinh(2\tilde X)} \color{green}{+a\cosh(2\tilde X)}$.

$\color{blue}{\text{Now the problem is, how do I match these three solutions?}}$

From what I have learnt, I need to do something like

$$\lim_{x\rightarrow 0^+} y_0(x) = \lim_{X\rightarrow +\infty}Y_0(X)$$

to obtain $\color{blue}{A= b-1?}$ But if I similarly try to do

$$\lim_{X\rightarrow 0^+}Y_0(X) = \lim_{\tilde X\rightarrow +\infty}\tilde Y_0(\tilde X)$$

it doesn't work, because $\tilde Y_0$ is not bounded??

  • $\begingroup$ How did you get to the third solution? I would rather get $\tilde Y_0(\tilde X)=\tilde A\exp(-2\tilde X)$ as the solution that is bounded to the right. $\endgroup$ May 5, 2020 at 19:04
  • $\begingroup$ Right, I solved it incorrectly. However, given the boundary condition, shouldn't it be $$\tilde Y_0(\tilde X) = \tilde A \sinh (2\tilde X) + a \cosh (2\tilde X)$$ instead? It is still unbounded. $\endgroup$ May 5, 2020 at 19:33

2 Answers 2


You have three solutions to work with, the outer solution $$ y_0(x) = x+B $$ the "wide" inner solution $$ Y_0(X) = C_1\exp\left(-\frac{1}{X^2}\right),\quad X=\frac{x}{\sqrt{\epsilon}} $$ and the "narrow" interior solution $$ \tilde Y_0(\tilde X) = D_1\exp\left(-2\tilde X\right)+D_2\exp\left(2\tilde X\right),\quad\tilde X=\frac{x}{\epsilon}. $$

Immediately you can say $B=b-1$, $D_2=0$ and $D_1=a$.

There's no need to use $\cosh$ and $\sinh$, they're just combinations of $\exp(2X)$ and $\exp(-2X)$ anyway. The whole reason you set $D_2=0$ is to ensure that $\tilde Y_0(\tilde X)$ is bounded as $\tilde X\to\infty$, but $\cosh$ and $\sinh$ are both unbounded.

You have two places where you need to do asymptotic matching (to find constant or verify things match), which you identified correctly, $$ \lim_{X\to\infty}Y_0(X)=\lim_{x\to0^+}y_0(x) \Rightarrow C_1=b-1,\qquad(\star)$$ and $$ \lim_{X\to0^+}Y_0(X)=\lim_{\tilde X\to\infty}\tilde Y_0(\tilde X) \Rightarrow 0=0 \qquad(\dagger),$$ so the inner solutions are exponentially small in the transition region.

You can combine all this to get a uniform approximation, $$y(x) = \underbrace{\left[x+b-1\right]}_\text{outer}+\underbrace{\left[(b-1)\exp\left(-\frac{\epsilon}{x^2}\right)\right]}_\text{wide inner}+\underbrace{\left[a\exp\left(-\frac{2X}{\epsilon}\right)\right]}_\text{narrow inner}-\underbrace{\left[b-1\right]}_\text{matching constant $(\star)$}-\underbrace{\left[0\right]}_\text{matching constant $(\dagger)$}+O(\epsilon^{1/2}),$$ that is, $$y(x) = x+(b-1)\exp\left(-\frac{\epsilon}{x^2}\right)+a\exp\left(-\frac{2\color{red}{x}}{\epsilon}\right)+O(\epsilon^{1/2}),$$


Illustration of the solution above: A numerical solution of the boundary value problem is plotted against the approximation to show that indeed the 3 parts do fit together that way for sufficiently small $ε$, visually for $ε<0.05$.

enter image description here


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