University of Florida/Egm4313/s12.team11.perez.gp/R3.2

Problem Statement
Developing the second homogeneous solution for the case of double real root as a limiting case of distinct roots.

Given
Consider two distinct roots of the form:

$$ \lambda_1 = x \! $$ and $$ \lambda_2 = x + \epsilon \! $$

(where $$ \epsilon \! $$ is perturbation).

Given
Find the homogeneous L2-ODE-CC having the above distinct roots.

Solution
$$ (\lambda - \lambda_1)(\lambda - (\lambda_1))=0 \! $$

$$ (\lambda - x)(\lambda - (x + \epsilon))=0 \! $$

$$ \lambda^2 - \lambda x - \lambda \epsilon - \lambda x + x^2 + x \epsilon =0 \! $$

$$ \lambda^2 - \lambda (2x+\epsilon) + x (x+\epsilon )=0 \! $$

$$ \therefore y'' - y'(2\lambda +\epsilon) + y \lambda (\lambda + \epsilon) = 0 \! $$ (1)

Given
Show that $$ \frac{e^{(\lambda +\epsilon) x} - e^{\lambda x}}{\epsilon } \! $$ is a homogeneous solution. (2)

Solution
Let's find the corresponding derivatives:

$$ y = \frac{e^{(\lambda+\epsilon) x} - e^{\lambda x}}{\epsilon} \! $$

$$ y' = \frac{(\lambda +\epsilon) e^{\lambda + \epsilon x}-\lambda e^{\lambda x}}{\epsilon } \! $$

$$ y'' = \frac{(\lambda +\epsilon)^2 e^{(\lambda + \epsilon) x}-\lambda^2 e^{\lambda x}}{\epsilon} \! $$

If we now take these three equations and plug them into the homogeneous L2-ODE-CC (1), we get:

$$ e^{(\lambda + \epsilon)x} (\lambda^2+2\lambda \epsilon + \epsilon^2 - 2\lambda^2-2\lambda \epsilon - \lambda \epsilon -\epsilon^2 + \lambda^2 + \epsilon \lambda)+e^{\lambda x}(- \lambda^2+2\lambda^2+\lambda \epsilon - \lambda ^2 - \lambda \epsilon) = 0 \! $$

$$ \therefore 0 \equiv 0. \! $$

Since the left and right hand sides of the equation are zero, the solution is in fact a homogeneous equation.

Given
Find the limit of the homogeneous solution in (2) as epsilon approaches zero (think l'Hopital's Rule).

Solution
Using l'Hopital's Rule,

$$ \lim_{\epsilon \rightarrow 0}\frac{e^{(\lambda +\epsilon) x }-e^{\lambda x}}{\epsilon} = \frac{0}{0} \! $$

(this is an indeterminate form).

L'Hopital's Rule states that we can divide this function into two functions, $$ f(\epsilon) \! $$ and $$ g(\epsilon) \! $$, and then find their derivatives and attempt to find the limit of $$ \frac{f'(\epsilon)}{g'(\epsilon)} \! $$. If a limit exists for this, then a limit exists for our original function.

$$ \lim_{\epsilon \rightarrow 0} \frac{f'(\epsilon)}{g'(\epsilon)} = \lim_{\epsilon \rightarrow 0} \frac{xe^{x(\lambda + \epsilon)}}{1}\! $$

$$ = \frac{x e^{(\lambda+0)x}}{1} \! $$

$$ = xe^{x \lambda}. \! $$

Given
Take the derivative of $$ e^{\lambda x} $$ with respect to lambda.

Solution
Taking the derivative with respect to lambda, we find that:

$$ \frac{d(e^{\lambda x})}{d\lambda} = xe^{\lambda x} $$.

It is important to remember that we must hold $$ x \! $$ as a constant when finding this derivative.

Given
Compare the results in parts (3) and (4), and relate to the result by using variation of parameters

Solution
Taking a closer look at Parts 3 and 4 of this problem, we discover that they're in fact equal:

$$ \frac{d(e^{\lambda x})}{d\lambda }=\lim_{e\rightarrow 0}\frac{e^{(\lambda +\epsilon )x} - e^{\lambda x}} {\epsilon} = xe^{\lambda x} \! $$

Given
Numerical experiment: Compute (2) setting lambda equal to 5 and epsilon equal to 0.001, and compare to the value obtained from the exact second homogeneous solution.

Solution
After performing these calculations, from (2) we get 148.478.

And from the exact second homogeneous solution, we get 200.05.