# 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.

P.S. This is posted on a different thread. A mistake on my part. Not sure how to delete it so please could someone tell me!

2. May 10, 2012

### HallsofIvy

Why do you call this a "boundary value problem" when there is no boundary condition given?

If a condition is that y(0)= e, yes, that would be correct.

Yes, that's good.

??? at X= 0, y(0)= A+ B but if you are given no specific value of y there, it gives you nothing. If you are taking y(0)= 0, then you would have A= -B, not A= B.

Why? If you do have A= B, then it would be y= A(e^{-X}+ 1). If A= -B, it would be y= A(e^{-X}- 1)
Now, as X goes to infinity (x is not infinitesmal) this would give y= -A and you want to match that to y= e^{1- x} which is e at x= 0.