This is exercise (c) from Introduction to Perturbation Methods on page $101$. $$\epsilon y'' = yy' – y^3 \text{ for } 0 < x < 1, \; y(0 ) = \frac{3}{5}, y(1) = -\frac{2}{3}$$
The outer solution: $y_0 = \frac{1}{c-x}$ where $c$ is some constant.
Immediate observation: if $c \in [0,1]$, then a singularity exists and an interior layer exists at c. Otherwise, the function is monotonically increase and cannot satisfy both boundary conditions. So a boundary layer exists and I am not sure if this implies an interior layer necessarily exists.
I check boundary layers.
At $x = 0$? Set $\overline{x} = \frac{x}{\epsilon^{\alpha}}$ and balancing yields $\alpha = 1$. The corresponding solution involves $\tan(\cdot)$ and so the limit does not exists as $\overline{x} \to \infty$ and thus matching is not possible.
At $x = 1$? Set $\frac{1-x}{\epsilon^{\alpha}}$ and similarly we get a solution which does not allow matching to occur.
So no boundary layers… now what about the interior layer? Set $\overline{x} = \frac{x – x_0}{\epsilon^{\alpha}}$ where $x_0 \in [0,1]$. This also yields a solution in terms of $\tan(\cdot)$ so I am not sure what is going on as this doesn't allow matching to occur.
Help.
Best Answer
At issue is the crossing of the $x$ axis, as the first outer solution $y(x)=\frac1{c-x}$ can not deliver that, and the second $y=0$ is not very compatible with the bounded variants of the inner solution.
As you found out, the inner equation is after a first integration $Y'(X)=\frac12(Y(X)^2+C)$, $X=x_0+ϵX$. For $C>0$ this gives a tangent that is not finite in both directions. For $C=0$ the hyperbolic function is bounded to one side, thus suitable for boundary layers at the interval boundaries. For $C=-A^2<0$ the solutions contain the hyperbolic tangent that falls from $A>0$ to $-A$ from $-\infty$ to $\infty$. Checking the combinations only this can be used as an inner layer.
For the given boundary values the inner layer applies with outer solutions $$ y_1(x)=\frac3{5-3x}, ~~~ y_2(x)=-\frac2{1+2x} $$ For the inner layer we need $y_1(x_0)=A=-y_2(x_0)$ or $\frac53-x_0=\frac12+x_0\implies x_0=\frac7{12}$, $A=\frac{12}{13}$.
For the inner solution set $Y=A\tanh(U)$ gives $A(1-Y^2)U'=\frac{A^2}2(Y^2-1)\implies U'=-\frac{A}2$ $$ y_{inner}(x)=-A\tanh\left(\frac{A}{2ϵ}(x-x_0)\right) $$