# ODE with nested boundary layers

Problem:

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.

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

Question:

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.

Attempt:

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??

• 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. May 5, 2020 at 19:04
• 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. May 5, 2020 at 19:33

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 $$ε<0.05$$.