# Boundary Value problem and ODE

1. May 10, 2012

### fionamb83

Hi, I'm not sure if this is on the right thread but here goes. It's a perturbation type problem.

Consider the boundry value problem

$$\epsilon y'' + y' + y = 0$$
Show that if $$\epsilon = 0$$ the first order constant coefficient equation has
the solution
$$y_{outer} (x) = e^{1-x}$$
I have done this fine.
Find a suitable rescaling $$X = x/\delta$$ so that the highest derivative is important and balances another term and find the solutions $y_{inner}(X)$ of
this equation, containing one free parameter, satisfying the boundary
condition at $$x = X = 0$$

So I am at the rescaling part and solved the differential equation (after neglecting the δy part of the full equation)
$$\frac{d^2y}{dX^2} + \frac{dy}{dX} = 0$$
yielding $$y = Ae^{-X} + B$$

imposing the boundry condition $$x = X = 0$$

gives $$A = B$$

so is $$y_{inner} = Ae^{-X}$$ ??
I think I covered that the highest derivative is important (Although again I was unsure about the wording)

When I continue on I think I either have this part wrong or the matching is wrong as I am not getting the right answer. I am supposed to be doing intermediate scaling. If anyone has any comments about this that would also be much appreciated.

Last edited: May 10, 2012
2. May 11, 2012

### hunt_mat

You have yet to do the matching, the matching will yield the value for A.

However the rest of your work seems okay.

3. May 11, 2012

### pasmith

You haven't actually told us the second condition that $y$ is supposed to satisfy along with $y(0) = 0$, but assuming your $y_{outer}$ is correct it must be $y(1) = 1$.

You have the correct ODE for the inner solution, but you haven't solved it correctly.

The equation for the inner solution is 2nd order, and you have two boundary conditions. The first is $y_{inner}(0) = 0$, which requires $A + B = 0$. The second, which comes from matching with the outer solution, is $y_{inner}(X) \to y(1) = 1$ as $X \to \infty$. Since $e^{-X} \to 0$ as $X \to \infty$, we must have $B = 1$.

The idea behind the second condition is that $x = \epsilon X$, so $x = 1$ corresponds to $X = 1/\epsilon$; since $\epsilon$ is small ("0"), $X$ must be large ("$\infty$").

Last edited: May 11, 2012