I want to find a composite solution to the boundary value problem:
$$
\epsilon y'' – y' + y^2 = 1, \text{ for }0<x<1 ,\text{ where }y(0) = 1/3,\,y(1) = 1
$$
where $\epsilon\ll 1$.
My approach: I know that I can find a composite solution in four steps:
- Find an outer solution.
- Find a boundary-layer solution.
- Apply matching so that the outer solution and the boundary-layer solution both approximate the same function correctly.
- Find the composite solution, by adding the outer solution and the boundary-layer solution and subtracting the part where they are equal.
Step 1: Outer solution:
I assume that the solution has an expansion in powers of $\epsilon$. So that
$$
y\sim y_0 + \epsilon^\alpha y_1 + \epsilon^\beta y_2 + \ldots
$$
with $0 < \alpha < \beta <\ldots$ If we substitute this into the equation we get
$$
\epsilon(y_0 + \epsilon^\alpha y_1 + \ldots )'' – (y_0 + \epsilon^\alpha y_1 + \ldots)' + (y_0 + \epsilon^\alpha y_1 + \ldots )^2 = 1
$$
If we only look at the $\mathcal{O}(1)$ terms we get
$$
-y_0' + y_0^2 = 1
$$
so that the outer solution becomes
$$
y_0 = \dfrac{1 – e^{2c_1 + 2x}}{e^{2c_1 + 2x} + 1}
$$
Step 2: Boundary layer solution: Let's assume that there is a boundary layer at $x = 0$. I introduce the boundary layer coordinate
$$
\bar{x} = \dfrac{x}{\epsilon^\alpha} \Leftrightarrow \dfrac{d}{dx} = \dfrac{1}{\epsilon^\alpha}\dfrac{d}{d\bar{x}}, \dfrac{d^2}{dx^2} = \dfrac{1}{\epsilon^{2\alpha}}\dfrac{d^2}{d\bar{x}^2}
$$
If we let $Y(\bar{x})$ denote the solution of the problem when using this boundary layer coordinate, the original equation becomes
$$
\epsilon^{1 – 2\alpha}\dfrac{d^2}{d\bar{x}^2}(Y_0 + \epsilon^\gamma Y_1 + \ldots) – \epsilon^{-\alpha}\dfrac{d}{d\bar{x}}(Y_0 + \epsilon^\gamma Y_1 + \ldots) + (Y_0 + \epsilon^\gamma Y_1 + \ldots ) = 1
$$
To balance this equation we need to look at the different terms. We already used the second and third term in part 1 for the outer solution. We can try balancing the first and the second term so that the third term becomes higher order. We need $1 – 2\alpha = -\alpha\Leftrightarrow \alpha = 1$. With $\alpha = 1$ the equation with the boundary layer coordinate becomes
$$
$$\dfrac{1}{\epsilon}\dfrac{d^2}{d\bar{x}^2}(Y_0 + \epsilon^\gamma Y_1 + \ldots ) – \dfrac{1}{\epsilon}\dfrac{d}{d\bar{x}}(Y_0 + \epsilon^\gamma Y_1 + \ldots) + (Y_0 + \epsilon^\gamma Y_1 + \ldots)^2 = 1
$$
If we now look at the order $\mathcal{O}\big(\dfrac{1}{\epsilon}\big)$ terms we get:
$$
Y_0''(\bar{x}) – Y_0'(\bar{x}) = 1, \, Y_0(0) = 1/3
$$
so that $Y_0(\bar{x}) = c_1e^{\bar{x}} + c_2 – \bar{x}$. We need $Y_0(0) = 1/3$ so we have $c_1 + c_2 = 1/3$. If there is a boundary layer at $x = 0$ then we need the outer solution to satisfy the boundary condition at $x = 1$ so that we have $\dfrac{1 – e^{2c_1 + 2}}{e^{2c_1 + 2} + 1} = 1$. This last expression can be rewritten to
\begin{align}
1 – e^{2c_1+2} = e^{2c_1 + 2} + 1\\
\Leftrightarrow -2e^{2c_1 + 2} = 0
\end{align}
Since there is no finite $c_1$ for which this expression holds I should probably look if the outer solution should instead satisfy the boundary condition at $x = 0$. In that case we need to find $c_1$ such that
$$
-2e^{2c_1 + 2} = 1/3
$$
but this results in a complex value for $c_1$ and I don't think that I should end up with such a solution.
Question: What am I doing wrong? How can I find a correct outer solution and a correct boundary layer solution so that I can start with the matching process?
Best Answer
The outer solution with $y(1)=1$ is $y(x)=1$, which you can find as one of the stationary points $y_*=\pm 1 $ of $y'=y^2-1$. In general you should get $$ y(x)=\frac{1-C_1e^{2x}}{1+C_1e^{2x}}. $$
The inner equation is the same independent of the basis point, as the equation is autonomous. You missed to multiply the right side with $ϵ$, which then gives the equation as $Y''(X)-Y'(X)=0$. The solution then is $$Y(X)=C_2e^X+C_3.$$ This is not bounded in direction $X\to+\infty$ so that the boundary layer can only be at the right boundary.
To satisfy the left boundary condition in the outer solution you need $C_1=\frac12$. Thus you need a jump from $C_3=\frac{2-e^2}{2+e^2}$ to $C_2+C_3=1$ on the right boundary.
The combined approximation is then $$ y(x)=\frac{2-e^{2x}}{2+e^{2x}}+\left(1-\frac{2-e^2}{2+e^2}\right)e^{(x-1)/ϵ} $$