Here is a related question which might provide some context: LINK.

Let's consider an oscillator with equation of motion in $n$ dimensions: $$ \frac{d^2}{dt^2} \vec{x} = K \vec{x}. $$

Given that $\vec x=0$ is a stable equilibrium, how can I show that the system will oscillate in sine waves? In other words, how to prove that the system will not behave like $x=\sinh t$? More precisely, how can I show that the $y_i$ in the answer in the link above will be sine waves?

As the answer in the link above suggest, to prove the solution to the DE has sine wave pattern, I need to prove that $K$ is symmetric with negative eigenvalues (see also the comment below the answer).

What about the case that $\frac{d^2}{dt^2} \vec{x} = f(\vec{x})$, where $f(\vec{x})$ doesn't have to be linear, but can be approximated linearly by $K \vec{x}$? If $\vec x=0$ is a stable equilibrium, must the eigenvalues of $K$ be negative OR zero?

I hope I am clear. Please tell me if I am not expressing myself clealy.


In the case that $K\vec{x}$ is exact rather than an approximation, note that if the equilibrium is stable, work done of a virtual displacement $\delta \mathbf x$ must be negative. So, $$ \text{work done}=F.d\propto (K\delta \mathbf x).\delta \mathbf x=\delta \mathbf x^T K\delta \mathbf x<0. $$ So $K$ is negative definite, and all eigenvalues are negative.

If $K$ is just an approximation of $f$, then we may only have $\leq$ instead of $<$. So you can still show that it is non-positive.

But why $K$ must be symmetric?


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