# Resonance and Repeated Factors

When studying Calculus, I asked this question about why non-repeated and repeated factors are handled differently in partial fraction decomposition. As I work Laplace Transform problems, I'm noticing repeated factors occur when there is resonance between an ODE's forcing term and associated homogeneous solution. This makes at least half of a bit of sense, since the algebraic step of a Laplace Transform problem involves dividing everything by the characteristic polynomial, which is related to the associated homogeneous solution. Perhaps the denominator of a forcing term's Laplace Transform is related to the homogeneous linear ODE which that forcing term would solve, creating this overlap in the resonant case.

Is it accurate to say that repeated factors in the s-domain occur if, only if, or if and only if the ODE has resonance? If so, is my linked question about the additional term in a repeated-factor partial fraction decomposition nothing more than the s-domain framing of the common question of why an additional factor of $$t$$ appears in the particular solution to a resonant linear ODE?

Technically "resonance" doesn't 100% make sense in the general situation. For example,

$$y'-y=e^t$$

is not really "resonant" because $$y'-y=0$$ was already growing anyway. Or more dramatically,

$$y'+y=e^{-t}$$

is not really "resonant" either because it doesn't even grow at all. Resonance is more about equations like

$$y''+y=\sin(t)$$

which were not growing without the forcing but are suddenly growing when the forcing is included.

But the change in the algebraic form of the solution can indeed be traced to an algebraic difference in the Laplace transforms; for example, with $$y'-y=e^{at},y(0)=0$$ you have $$sY-Y=\frac{1}{s-a}$$ or $$Y=\frac{1}{(s-a)(s-1)}$$ and the form of the solution is changed depending on whether $$a=1$$ or $$a \neq 1$$.

In short I'd say the answer to your second question is "yes", while the answer to your first question is "it depends on what you really mean by "has resonance"".

Really I would say the right intuition for the connection between the "resonant" and "non-resonant" cases is gained by considering a problem with a parameter and a fixed initial condition, like

$$y'-y=e^{at},y(0)=0.$$

The solution to this problem for $$a \neq 1$$ is $$\frac{e^{at}-e^t}{a-1}$$, and lo and behold that as $$a \to 1$$ the solution continuously goes to $$te^t$$, where this extra factor of $$t$$ just comes from differentiating with respect to $$a$$. You can come at this from the other side by looking at the integrating factor method:

$$(e^{-t} y)'=e^{(a-1)t}$$

where the $$t$$ appears as the integral of $$1$$ when $$a=1$$. Note that this integrating factor method can still be applied in the higher order constant coefficient case, it just gets called the "annihilator method" instead.

This perspective dodges what I think is really bothering you, which is that there is a sort of "algebraic discontinuity" when we compare, e.g., $$te^t + Ce^t$$ to $$\frac{e^{at}}{a-1} + Ce^t$$ without taking into account that the $$C$$ in one and the $$C$$ in the other are drastically different when the initial condition is the same in both cases. (Indeed if, say, we parametrize by $$y(0)=y_0$$, then in the first case $$C=y_0$$ and in the second case $$C=y_0-\frac{1}{a-1}$$.)

Laplace Transform is just another way of solving the differential equations.

Thus if you have resonance in your solution you will get it with or without Laplace Transform.

Repeated eigenvalues in general does not result in resonance.

For example $$y''+2y'+y=e^{-t}$$ has repeated eigenvalues but not resonance due to the fact that the exponential decay factor dominates the polynomial part.

For your second question the answer is yes because the repeated eigenvalues translate to repeated factors in the denominator of F(s).