Taylor series at a point for which the function isn't defined (perturbation)

[tjackson3]

Homework Statement

This problem arises from the following ODE:

[tex]\epsilon y'' + y' + y = 0, y(0) = \alpha, y(1) = \beta[/tex]

where [itex]0 < x < 1, 0 < \epsilon \ll 1[/itex]

Find the exact solution and expand it in a Taylor series for small [itex]\epsilon[/itex]

Homework Equations

I guess knowing the Taylor series formula would be helpful

The Attempt at a Solution

Using ye olde constant coefficient method, I get that the solution (in non-expanded form) is:

[tex]y(x) = c_1\cosh(rx) + c_2\sinh(rx)[/tex]

where

[tex]r = \frac{-1 \pm \sqrt{1-4\epsilon}}{2\epsilon}[/tex]

(this is real since [itex]\epsilon[/itex] is so small)

Applying the boundary conditions gives that

[tex]y(x) = \alpha\cosh(rx) + \frac{\beta-\alpha\cosh(r)}{\sinh(r)}\sinh(rx)[/tex]

Now the goal is to do a Taylor series not for x, but for epsilon (remember that r is a function of epsilon). Trouble is that this diverges near [itex]\epsilon = 0[/itex], so I don't see how to do it. I tried putting it into Mathematica and got a very crazy answer. I'm not sure it's right, and I'm less sure how to get it.

Thanks! :)

 

[mighty2000]

Thank you for bringing this problem to our attention. It is an interesting and challenging one, and I am happy to help you find the solution.

First, let's start by writing the ODE in a more standard form, by dividing both sides by \epsilon:

y'' + \frac{1}{\epsilon}y' + \frac{1}{\epsilon}y = 0

Next, we can use the method of undetermined coefficients to find the particular solution for this ODE. We assume that the solution has the form y(x) = e^{rx}, and plug it into the ODE:

r^2e^{rx} + \frac{1}{\epsilon}re^{rx} + \frac{1}{\epsilon}e^{rx} = 0

Simplifying and factoring out the common term e^{rx}, we get:

e^{rx} (r^2 + \frac{r}{\epsilon} + \frac{1}{\epsilon}) = 0

Since the exponential function is never equal to zero, the expression in parentheses must be equal to zero. Solving for r, we get:

r = \frac{-1 \pm \sqrt{1-4\epsilon}}{2\epsilon}

Now, we can use these values of r to find the general solution for the ODE:

y(x) = c_1e^{\frac{-1+\sqrt{1-4\epsilon}}{2\epsilon}x} + c_2e^{\frac{-1-\sqrt{1-4\epsilon}}{2\epsilon}x}

Next, we can use the initial conditions to find the values of c_1 and c_2. Plugging in x = 0 and x = 1, we get the following two equations:

c_1 + c_2 = \alpha
c_1e^{\frac{-1+\sqrt{1-4\epsilon}}{2\epsilon}} + c_2e^{\frac{-1-\sqrt{1-4\epsilon}}{2\epsilon}} = \beta

Solving this system of equations, we get:

c_1 = \frac{\alpha e^{\frac{1+\sqrt{1-4\epsilon}}{2\epsilon}} - \beta}{e^{\frac{1+\sqrt{1-4\epsilon}}{2