Where linearization runs out
The last two guides handed you a sharp tool and an honest warning. Build the Jacobian, read its eigenvalues, and at a hyperbolic equilibrium the Hartman-Grobman theorem guarantees the linear verdict really is the local truth. But that guarantee evaporates the instant an eigenvalue lands on the imaginary axis — the borderline, non-hyperbolic case — where the discarded curved terms decide everything and linearization simply shrugs.
The undamped pendulum is the canonical headache. Its bottom rest point has Jacobian eigenvalues +/- i — a pure center, zero real part — so linearization predicts endless circles but cannot say whether the real nonlinear orbits spiral slowly in, slowly out, or close up exactly. We need a method that does not lean on the linearization at all. That method is Lyapunov's direct method, and its central idea is almost embarrassingly physical: watch the energy.
The energy you already trust
Think of a marble in a bowl. Wherever you place it, total mechanical energy — height plus motion — can only stay level or drip away to friction; it can never spontaneously climb. The marble therefore cannot wander uphill out of the bowl, and if there is any friction it must settle at the bottom. Notice we concluded all of that without ever solving for the marble's path. We reasoned purely about how one scalar quantity, the energy, changes in time.
Lyapunov's leap was to realize the marble's energy is not sacred — what mattered was only its shape and its trend. Any scalar function V(x, y) that is bowl-shaped around an equilibrium and never increases as the system flows will trap trajectories exactly the same way. Such a V is called a Lyapunov function, and it need not be a real physical energy at all; it can be any clever expression we invent, as long as it has the two right properties.
Two conditions, and the chain-rule trick
Spell out 'bowl-shaped' and 'never increases' precisely. First, positive definiteness: V should be zero exactly at the equilibrium and strictly positive everywhere nearby — that is the bowl, with its single low point right at the rest state. Second, the trend condition on V's rate of change along trajectories, the quantity written V' or dV/dt. We want this to be non-positive, so V can only sit still or slide downhill as time runs.
Here is the move that makes the whole method work without solving anything. By the multivariable chain rule, the rate of change of V along a trajectory is V' = (dV/dx) x' + (dV/dy) y'. But the system itself tells us x' = f(x, y) and y' = g(x, y) — so we just substitute the vector field directly in. The unknown x(t) never appears. We compute V' purely from the formula for V and the formula for the field, evaluate its sign, and read off the system's fate.
conditions on V(x,y) near an equilibrium (x*, y*):
V(x*, y*) = 0 and V > 0 elsewhere nearby (positive definite: a bowl)
V' = (dV/dx) x' + (dV/dy) y' (chain rule)
= (dV/dx) f(x,y) + (dV/dy) g(x,y) (substitute the field, no x(t) needed)
V' <= 0 ==> stable (energy never grows)
V' < 0 ==> asymptotically stable (energy strictly drains away)
Reading the verdict
Now the two cases. If V' <= 0 throughout a neighbourhood, trajectories can never climb to a higher bowl-level, so they stay trapped near the equilibrium: that is Lyapunov stability — start close, stay close. If the trend is strictly stronger, V' < 0 everywhere except at the equilibrium itself, then the value keeps strictly draining and trajectories must spiral all the way down to the rest point: that is asymptotic stability — start close, and actually arrive.
There is a mirror version for proving the opposite. If you can find a function that is positive definite yet whose V' is strictly positive — energy that is forced to grow — then trajectories are driven away and the equilibrium is unstable. This is sometimes called a Chetaev-type argument, and it lets the same accounting prove instability just as cleanly as it proves stability. One scalar function, and you have settled the question either way.
When V' only touches zero: LaSalle
A subtlety hides in the damped pendulum. With friction, V' is negative whenever the bob is moving, but it momentarily equals zero at every instant the velocity y passes through zero — at the top of each swing. So V' is not strictly negative; it merely touches zero along the way. The plain asymptotic-stability test asks for V' < 0 everywhere but the equilibrium, and this V fails that letter of the law even though the pendulum obviously does coast to rest.
The LaSalle invariance principle repairs exactly this gap. It says: let V' <= 0, then look at the set where V' = 0, and ask which whole trajectories could live inside that set forever. The system must end up on the largest such invariant piece. For the damped pendulum, V' = 0 needs y = 0 at all times; but if the bob stays at y = 0 forever it cannot be swinging, which forces it to the bottom rest point. The only trajectory trapped in the zero-set is the equilibrium itself — so LaSalle promotes mere stability back up to genuine asymptotic stability.
The honest catch: finding V
Be candid about the price of this power. The method gives no recipe for finding a Lyapunov function — that is its one real weakness. You must guess a candidate V, then check the two conditions; if V' comes out the wrong sign, your guess told you nothing and you simply try another. Physical energy is the natural first guess for mechanical and circuit systems; quadratic forms like V = a x^2 + b y^2 are the workhorse near a generic equilibrium, with the constants a and b tuned to make V' come out non-positive.
Two more honest limits. The method is one-directional: failing to find a working V does not prove instability — perhaps you just have not been clever enough. And its conclusions are usually local, valid only on the neighbourhood where your conditions actually hold; claiming global stability demands that the bowl and the downhill trend reach everywhere, which is a much stronger thing to verify. Within those bounds, though, the reward is remarkable: a full verdict on stability — including the very borderline cases where the Jacobian falls silent — extracted from one well-chosen scalar function and a single application of the chain rule, with the unsolvable system never solved.