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To illustrate the method we start with linear, finite dimensional ODEs. Consider an ODE for a vector $f(t)=(f_1(t),\,f_2(t),\ldots,f_n(t))^\top\in {\mathbb{C}}^n$: $$\label{ODE:general:n}
  \begin{cases}
   {\frac{{\,\operatorname{d}\!}}{{\,\operatorname{d}\!t}}}{f}= -{\mathbf{C}}f,\quad t\ge 0\,,\\
   f(0)   = f^I \in{\mathbb{C}}^n\,,
  \end{cases}$$ for some (typically non-Hermitian) matrix ${\mathbf{C}}\in{\mathbb{C}}^{n\times n}$. The stability of the steady state $f^0\equiv 0$ is determined by the eigenvalues of matrix ${\mathbf{C}}$:

Let ${\mathbf{C}}\in{\mathbb{C}}^{n\times n}$ and let $\lambda_j$ ($j=1,\ldots,n)$ denote the eigenvalues of ${\mathbf{C}}$ (counted with their multiplicity).

1.  The equilibrium $f^0$ of is stable if and only if (i) $\Re (\lambda_j) \geq 0$ for all $j=1,\ldots,n$; and (ii) all eigenvalues with $\Re(\lambda_j)=0$ are non-defective[^4].

2.  The equilibrium $f^0$ of is asymptotically stable if and only if $\Re (\lambda_j)> 0$ for all $j=1,\ldots,n$.

3.  The equilibrium $f^0$ of is unstable in all other cases.

For positive definite Hermitian matrices ${\mathbf{C}}$, using the Lyapunov functional $\|f\|^2$ in the energy method allows to obtain the sharp decay rate, which is the smallest eigenvalue $\mu$ of ${\mathbf{C}}$: The derivative of $\|f\|^2$ along solutions $f(t)$ of satisfies $$  {\frac{{\,\operatorname{d}\!}}{{\,\operatorname{d}\!t}}}\|f(t)\|^2  = -{\langle {f(t)}\ ,\, {({\mathbf{C}}^* + {\mathbf{C}})f(t)} \rangle} = -2 {\langle {f(t)}\ ,\, {{\mathbf{C}}f(t)} \rangle} \leq -2\mu \|f(t)\|^2 \,,$$ where ${\mathbf{C}}^*$ denotes the Hermitian transpose of ${\mathbf{C}}$. Note that the derivative of $\|f\|^2$ 