I was reading "A deterministic map of Waddington's epigenetic landscape for cell fate specification", where the authors derive the quasi-potential landscape for the system.
Quoting the paper:
\begin{equation} \Delta V_q = \frac{\partial V_q}{\partial x} \cdot \Delta x + \frac{\partial V_q}{\partial y} \cdot \Delta y \end{equation} \begin{equation} = -\frac{dx}{dt} \cdot \Delta x - \frac{dy}{dt} \cdot \Delta y \qquad (4) \end{equation} where $\Delta x$ and $\Delta y$ are sufficiently small increments along the trajectory such that $\frac{dx}{dt}$ and $\frac{dy}{dt}$ can be assumed to remain unchanged over the interval $[(x, x+\Delta x); (y, y+\Delta y)]$. The quantities $\Delta x$ and $\Delta y$ are obtained as the products $\frac{dx}{dt} \cdot \Delta t$ and $\frac{dy}{dt} \cdot \Delta t$, respectively, where $\Delta t$ is the time increment.
The change in the quasi-potential, $\Delta V_q$, can be rewritten from Eq. 4 as: \begin{equation} \Delta V_q = -\frac{dx}{dt} \cdot \left( \frac{dx}{dt} \cdot \Delta t \right) - \frac{dy}{dt} \cdot \left( \frac{dy}{dt} \cdot \Delta t \right) \end{equation} \begin{equation} = -\left[ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 \right] \cdot \Delta t \qquad (5)\end{equation}
For positive increments in time $\Delta t$, $\Delta V_q$ is thus always negative along an evolving trajectory, ensuring that trajectories flow “downhill” along a putative “quasi-potential surface”. Stable steady states of the system ($\frac{dx}{dt} = 0; \frac{dy}{dt} = 0$) would correspond to local minima on this quasi-potential surface, given that at these states $\Delta V_q = 0$ (per Eq. 5).
As they also state that the initial value for any trajectory, $V_q = 0$, then the quasi-potential should be zero at the steady states, and negative otherwise. Thus steady states should be (local) maxima in the landscape instead of minima.
Also in their landscape plots, they have positive values for the quasi-potential, which should not be possible given $\Delta V_q$ always negative.
I also found another paper using the same approach, titled "Exploiting delayed transitions to sustain semiarid ecosystems after catastrophic shifts", which in its Supporting Information gives the the algorithm to compute the quasi-potential following the same approach, but the formula is not really equivalent
$U(x(0), y(0)) = \left( \sum_{t=1}^{T} x(t) - x(t - 1) \right)^2 + \left( \sum_{t=1}^{T} y(t) - y(t - 1) \right)^2$
as it does not have the minus sign, and squares the differences along the trajectory instead. So in this case steady states should indeed be minima but $\Delta V_q$ will be always positive along an evolving trajectory.
I really feel like I missed something...