Lotka-Volterra: is stability analysis done on both equations separately or to their sum? Lotka-Volterra: is stability analysis done on both equations separately or to their sum?
So if the systems are e.g. notated as:
$$u_t=u(v-1)$$
$$v_t=v(1-u)$$
then would one do stability analysis for $u_t$ and then for $v_t$ or would one consider 
$$u_t+v_t=0$$
or
$$u_t=v_t=0 \implies u_t-v_t=0$$
and then do the analysis?
 A: Neither considering the variables separately or analyzing the sum variable $u_t + v_t$ will resolve the stability issues for the system
$u_t = u(v - 1), \tag 1$
$v_t = v(1 - u). \tag 2$
Considering the variables separately fails for the reason that $u$ and $v$ are coupled in the system (1)-(2); the time evolution of $u(t)$ affects the time evolution of $v(t)$ and vice-versa; we can't determine one without the other.  And while it is legitamite to form the sum $u_t + v_t$, in the absence of a second variable such as perhaps $u(t)$, $v(t)$, or $u_t - v_t$, the system is incomplete and will lack sufficient information to determine its evolution.  Furthermore, it is not clear that any useful simplification will be had by re-writing (1)-(2) in terms of the variable $u_t + v_t$; one would have to present a specific equation for $u_t + v_t$ to resolve such a question.
One needs to consider (1)-(2) as a single entity in the vector variable
$\mathbf r(u, v) = \begin{pmatrix} u \\ v \end{pmatrix}; \tag 3$
in so doing, we may present the right-hand sides of (1)-(2) as the vector field
$\mathbf X(u, v) = \begin{pmatrix} u(v - 1) \\ v(1 - u) \end{pmatrix}; \tag 4$
the given system is then written in vector form
$\dot{\mathbf r}(u, v) = \mathbf X(u, v).  \tag 5$
We may find the equilibria of (1)-(2), (5) by solving
$u_t = u(v - 1) = 0, \tag 6$
$v_t = v(1 - u) = 0 \tag 7$
simultaneously for $u$, $v$; we see that if either 
$u = 0 \; \text{or} \; v = 0 \tag 8$
then
$u = v = 0; \tag 9$
also, from (6) and (7),
$u \ne 0 \Longrightarrow v = 1 \Longrightarrow u = 1; \tag{10}$
thus another equilibrium point is
$u = v = 1; \tag{11}$
it is easy to see there are no others.
We next linearize the system about the equilibria $(0, 0)$, $(1, 1)$; the Jacobean matrix of first derivatives is
$J_{\mathbf X}(u, v) = \begin{bmatrix} u_{t, u} & u_{t, v} \\ v_{t, u} & v_{t, v} \end{bmatrix}, \tag{12}$
where I have introduced the shorthand notation
$u_{t, u} = \dfrac{\partial u_t}{\partial u}, \tag{13}$
and so forth; then
$J_{\mathbf X}(u, v) = \begin{bmatrix} v - 1 & u \\ -v & 1 - u \end{bmatrix}; \tag{14}$
at $(0, 0)$,
$J_{\mathbf X}(0, 0) = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}; \tag{15}$
the eigenvalues of this matrix are clearly $\pm 1$; $(0, 0)$ is thus a saddle point, an equilibrium with both stable an unstable trajectories nearby, hence unstable as a critical point of (1)-(2); as for $(1, 1)$ we have
$J_{\mathbf X}(1, 1) = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}, \tag{16}$
the eigenvalues of which are easily seen to be $\pm i$; we have in $(1, 1)$ a center, indicating the possible presence of periodic orbits surrounding $(1, 1)$, though the eigenvalues alone are not decisive in this case.
The preceding calculations show what is done in a typical, first order, elementary stability analysis for the system (1)-(2), (5).  At a more advanced level, we might address the stability of non-equilibrium orbits, that is, whether trajectories near a given one $(u(t), v(t))$ converge to, or diverge from, it.  But such an undertaking, as well as the determination of the existence of truly periodic orbits, is a more advanced and involved undertaking which will be reserved for perhaps future posts on the subject.
