2
$\begingroup$

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:

  1. Find an outer solution.
  2. Find a boundary-layer solution.
  3. Apply matching so that the outer solution and the boundary-layer solution both approximate the same function correctly.
  4. 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?

$\endgroup$
1
$\begingroup$

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 $$ y(x)=\frac{2-e^{2x}}{2+e^{2x}}+\left(1-\frac{2-e^2}{2+e^2}\right)e^{(x-1)/ϵ} $$

enter image description here

$\endgroup$

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.