Liouville's formula

Statement

Let $I$ be an open interval, let $A : I \to \mathbb{R}^{n \times n}$ be continuous, and let $\Phi : I \to \mathbb{R}^{n \times n}$ be a differentiable function with $\Phi'(t) = A(t)\, \Phi(t)$ for each $t \in I$. Then for each $t_0, t \in I$, $$\det \Phi(t) = \det \Phi(t_0)\, \exp\left( \int_{t_0}^{t} \operatorname{tr} A(s)\, ds \right),$$ where $\operatorname{tr} A(s)$ is the trace of $A(s)$.

Remarks

So $\det \Phi(t)$ is zero at each point of $I$ or at no point of $I$. This shows that a fundamental matrix is invertible everywhere if it is invertible at one point. For a constant matrix $A$, the formula gives $\det e^{tA} = e^{t \operatorname{tr} A}$. For the system that comes from $y'' + p y' + q y = 0$, the trace is $-p$, and the formula is Abel's identity.

Explanations

L1 human

The determinant of $\Phi$ is the volume spanned by $n$ solutions. The formula says how this volume changes in time: its rate of growth is the trace of $A$.