Boundary Value Problem and ODE: How to Rescale and Solve for Inner Solutions?

[fionamb83]

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

Consider the boundary 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 boundary 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.
Thank you in advance!

 

[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.

 

[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 ("$\infty$").