TITLE: Why don't these ODEs produce the same result?
QUESTION [40 upvotes]: I am relatively new to differential equations, and the following problem is confusing me. Consider, for example, the ODE $x'+x=0$ such that $x(0)=1$. This has solution $x(t)=e^{-t}$. But consider an $\epsilon>0$ and the ODE.
$$\epsilon x'' + x' + x = 0$$
subject to $x(0)=1$. In the limit as $\epsilon\to 0$ the above equation becomes $x'+x=0$. But no matter how small we make epsilon, the solution will never look like $e^{-t}$, and in fact, is unique up to a constant. While mathematically I think I can see what is going on - the first equation only has one linearly independent solution, while the second has two - physically I cannot understand how this can be.

REPLY [7 votes]: As epiliam points out your initial equation $x' + x = 0$ with $x(0)=1$ has unique solution $x = e^{-t}$, and the perturbed equation, $\epsilon x'' + x' +x = 0$ has a family of solutions of the form
$$
x = A e^{\lambda_1 t} + (1-A) e^{\lambda_2 t}
$$
where $\lambda_1,\lambda_2 = \frac{-1 \pm \sqrt{1 - 4\epsilon}}{2\epsilon}$
To find a version of these that closely matches the original solution we can look at adding a first derivative condition that matches the original. The natural choice would be $x'(0) = -1$. The expression for the first derivative is:
$$
x' = A \lambda_1 e^{\lambda_1 t} + (1-A) \lambda_2 e^{\lambda_2 t}
$$
Solving $x'(0) = A \lambda_1 + (1-A) \lambda_2 = -1$ gives
$$
A = \frac{\lambda_2 + 1}{\lambda_2 - \lambda_1}
$$
If we now plot these curves we see that these do in fact resemble $e^{-t}$ even for relatively large values of $\epsilon$. The following graph shows $\epsilon=0.1$

This correspondence gets better with smaller epsilon as we might expect. It is easier to see this if we look at the log-x vs t graph

We can see this is getting closer and closer to $e^{-t}$ by graphing the difference (again in log format):