The roots are complex numbers when \(b^2 - 4ac \lt 0\text{.}\) Here is where I need a little bit of arithmetic for complex numbers to understand the solutions (as well as the use of Euler’s formula). I can factor \(\imath\) out of the square root to get a real root.
\begin{equation*}
r = \frac{-b \pm \sqrt{b^2-4ac}}{2a} = \frac{-b}{2a} \pm
\imath \frac{\sqrt{4ac-b^2}}{2a}
\end{equation*}
The roots are a pair of complex numbers \(x \pm y \imath\) with the real part \(x = \frac{-b}{2a}\) and imaginary part \(y = \frac{ \sqrt{4ac-b^2}}{2a}\text{.}\) They are a conjugate pair (each is the conjugate of the other). This is expected behaviour: the complex roots to a real quadratic always come as a conjugate pair. To write this more succinctly, I’ll define two new constants.
\begin{align*}
\amp \alpha = \frac{-b}{2a} \amp \amp \beta =
\frac{\sqrt{4ac-b^2}}{2a}
\end{align*}
Then the complex roots are described via these constants: \(\alpha \pm \imath \beta\text{.}\) The solutions to the DE are
\begin{equation*}
e^{(\alpha \pm \imath \beta)t} = e^{\alpha t} e^{\pm \imath
\beta t}\text{.}
\end{equation*}
What do these complex functions means? The
\(e^{\alpha t}\) term is fine, it is just a real exponential. The complex exponential is understood through Euler’s formula (
Proposition 1.5.4 in
Section 1.5).
\begin{equation*}
e^{\alpha t} e^{\pm \imath \beta t} = e^{\alpha t} ( \cos
(\beta t) \pm \imath \sin (\beta t))
\end{equation*}
These solutions are still problematic because they involve complex numbers. I am trying to solve a real system with real coefficient and I want real solutions. To find them, I need to take clever linear combinations (over \(\CC\text{!}\)) of the two solutions.
\begin{align*}
\frac{1}{2} e^{\alpha t}(\cos \beta t + \imath \sin \beta t)
+ \frac{1}{2} e^{\alpha t}(\cos \beta t - \imath \sin \beta
t) \amp = e^{\alpha t} \cos \beta t\\
\frac{1}{2\imath } e^{\alpha t}(\cos \beta t + \imath \sin
\beta t) - \frac{1}{2\imath} e^{\alpha t}(\cos \beta t -
\imath \sin \beta t) \amp = e^{\alpha t} \sin \beta
\end{align*}
On the basis of these linear combinations, I conclude that the following two functions are the linearly independent real solutions.
\begin{align*}
\amp y_1 = e^{\alpha t} \sin \beta t \amp \amp y_2 =
e^{\alpha t} \cos \beta t
\end{align*}
The general, real-valued solutions are linear combinations of the two linearly independent solutions.
\begin{equation*}
y = A e^{\alpha t} \sin \beta t + B e^{\alpha t} \cos \beta t
\end{equation*}
The complex numbers, via Euler’s formula, product sine and cosine instead of exponential growth or decay. In a harmonic system, this behaviour happens when \(b^2 \lt 4ac\text{,}\) that is, when friction is small enough. If there is some friction, but not too much, harmonic systems have the oscillations that we expect. The real exponential term will be decay (\(\alpha \lt 0\) in harmonic systems since all the coefficient are positive). The total result is oscillations with exponentially declining amplitude. Such harmonic systems are called underdamped.
