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Completing squares reveals the geometry of the sheared quartic double-well gradient flow:The critical points satisfy $2(x+y)=0$ and $4x^3-2x+2y=0$, so they are $(0,0)$, $(1,-1)$ and $(-1,1)$. The Hessian matrix isAt the origin its determinant is $-8$, so $(0,0)$ is a saddle point, with $V=0$. At either other critical point, the determinant is $16$ and the upper-left entry is $10$, so it is a positive-definite matrix. They are strict local minima; the completed squares show that both are global minima, with $V=-1$.

A contour at level $c$ obeysThere are no contours for $c<-1$, two isolated minimum points for $c=-1$, and two separate closed ovals for $-1<c<0$. At $c=0$ the two lobes meet at the saddle point; near the origin their tangents are $y=(-1\pm \sqrt{2})x$. For $c>0$ one closed contour surrounds both minima.

Along the gradient flow, the gradient-flow dissipation identity isAt the stated initial point, $V(1,-1/2)=-3/4$. Its trajectory stays in the invariant sublevel set $V\le -3/4$. This set is compact because $V$ is a coercive function. It also excludes every point with $x=0$, where $V=y^2\ge 0$; continuity therefore keeps the trajectory in its positive-$x$ component.

To justify its limit rather than merely infer it from the sketch, integrate the dissipation identity: ${\int }_0^{\infty }\Vert \nabla V{\Vert }^{2}dt$ is finite. On this compact set the gradient and its time derivative are bounded, so $\Vert \nabla V{\Vert }^2$ is uniformly continuous. A nonnegative uniformly continuous function with finite integral tends to zero: otherwise separated intervals of a fixed positive height and width would force an infinite integral. Every accumulation point is consequently a critical point. In the positive-$x$ component below this energy level the only one is $(1,-1)$. Compactness then implies convergence, andThis is a compact gradient-flow trapping criterion; the drawn trajectory illustrates, rather than replaces, the convergence proof.

Along the downward gradient flow, the gradient-flow dissipation identity, the chain rule and the defining property of the Riemannian gradient giveA smooth vector field on a compact manifold is complete, so the trajectory exists for every $t\ge 0$. In particular, its function value never increases.

If the trajectory is still in $C_{[a,b]}$ at time $t$, its monotone function value has stayed in $[a,b]$ throughout $[0,t]$. ConsequentlyThus a uniform exit bound isThe trajectory is outside the closed band for every $t>T$. Once it leaves below level $a$, monotonicity prevents its return. The argument also covers the zero-width band $a=b$.

For the total occupation-time bound for a gradient flow, let $m={\min }_{M}f$ and $F={\max }_{M}f$, which exist by compactness. For every $R>0$, the same gradient flow identity givesTaking $R\to \infty$, by monotone convergence theorem, gives the total occupation-time boundThis allows arbitrarily many visits to $C_{\epsilon }$; it bounds their total duration. Neither of these two estimates needs the Morse function hypothesis.

For exponential convergence of a Morse gradient flow, first observe that a limiting point $z=x_{\infty }$ is a critical point. Indeed, writing ${\varphi }_s$ for the flow and using continuous dependence on initial data,Differentiation at $s=0$ gives $\nabla f(z)=0$.

We use the Morse lemma: near a nondegenerate critical point of Morse index $\lambda$, there are coordinates $(u,v)\in {\mathbb{R}}^{\lambda }\times {\mathbb{R}}^{n-\lambda }$ in whichChoose a Riemannian metric equal to the Euclidean metric on a smaller ball in these coordinates. Such a global Riemannian metric exists by a partition of unity: patch this metric to any background metric outside a slightly larger ball. One can do this simultaneously at every critical point, since a Morse function on a compact manifold has only finitely many critical points. The metric is chosen before forming its gradient flow; changing the metric can change trajectories.

For this metric, any trajectory converging to $z$ eventually remains in the smaller coordinate ball. From some time $t_0$ onward its equations are exactlyConvergence forces $u(t_0)=0$, and thenThe straight radial segment stays in the coordinate ball, so its length bounds the Riemannian distance from above:Increase $C$ beyond both $e^{2t_0}|v(t_0)|$ and ${\max }_{0\le t\le t_0}{e}^{2t}d({\gamma }_x(t),z)$. Then the requested strict estimate holds for all $t\ge 0$:This proof uses the Morse lemma, existence and uniqueness for smooth ordinary differential equations, continuous dependence on initial data, and the existence of a partition of unity. The expanding coordinates explain why convergence to a saddle occurs only along its stable manifold.

For a counterexample, use an arc coordinate $x$ around a point of the circle, its flat local metric, and a globally smooth function that equals $x^4$ on this arc. Such a function can be made by a smooth cutoff, agreeing with a positive constant outside a larger arc. For $x_0>0$ sufficiently small the downward gradient flow remains in the arc and isIt converges to the degenerate critical point $0$, and its Riemannian distance from $0$ is $x(t)$ for sufficiently large $t$. For every $k>0$, $e^{kt}x(t)\to \infty$, so no exponential bound is possible. This counterexample is on a compact manifold as well.

The completed-square form gives global minima at $(1,-1)$ and $(-1,1)$, both with value $-1$, and a saddle point at $(0,0)$ with value zero. Negative levels split into two closed components; the zero contour joins their lobes at the saddle. The gradient flow $\dot{x}=-V_x,\dot{y}=-V_y$ decreases $V$ by the gradient-flow dissipation identity. A negative initial level traps each trajectory on its original side of $x=0$, because $V(0,y)\ge 0$.

If $f$ is bounded along a downward gradient flow, the total time spent where $\Vert {\nabla }_{g}f\Vert \ge \epsilon$ is at most $(\sup f-\inf f)/{\epsilon }^2$. The gradient-flow dissipation identity proves this by integrating the indicator of that region. It bounds all visits together, without requiring the visited region to be connected.