Gradient Like Vector Fields and Morse Trajectories — Examples

Solving the separated equations gives $u(t)=e^{2t}{u}_0$ and $v(t)=e^{-2t}{v}_0$. Thus the forward-stable coordinate disk is $\{u=0\}$ and the backward-unstable disk is $\{v=0\}$.

Directly, $\frac{d}{dt}f(u(t),v(t))=-4\mathrm{\mid }u(t){\mathrm{\mid }}^2-4\mathrm{\mid }v(t){\mathrm{\mid }}^2$, which is strictly negative away from the origin.

The tangential projection of $e_z$ is $e_z-z(x,y,z)$, so the displayed vector field is the negative gradient. Along it, $\dot{z}=z^2-1<0$ whenever $\mathrm{\mid }z\mathrm{\mid }<1$.

The longitude is constant along the flow, while $z$ decreases from $1$ to $-1$. Therefore every nonpolar longitude gives a meridian from north to south, a one-parameter family before quotienting by time translation.

Differentiating gives $df=-\sin \theta d\theta -\sin \varphi d\varphi$, hence the displayed negative-gradient equations. Their zeros give exactly the four listed critical points, and the diagonal Hessian $\mathrm{diag}(-\cos \theta ,-\cos \varphi )$ gives the stated indices.

The circles $\theta \in \{0,\pi \}$ and $\varphi \in \{0,\pi \}$ are invariant and supply the coordinate-circle separatrices. On either component of $S^1\setminus \{0,\pi \}$, each nonconstant coordinate trajectory has backward limit $0$ and forward limit $\pi$ (it increases on $(0,\pi )$ and decreases on $(\pi ,2\pi )$ in the displayed coordinate). Thus the trajectories off those circles run from the maximum to the minimum in four arc-choice components. In each component, quotienting the two integration constants by common time translation leaves a one-parameter family.

The positive-gradient equations are $\dot{u}=-2u$ and $\dot{v}=2v$, so $u(t)=e^{-2t}{u}_0$ and $v(t)=e^{2t}{v}_0$.

Thus the forward-stable space is $\{v=0\}$ of index dimension and the backward-unstable space is $\{u=0\}$ of coindex dimension. The labels are reversed from the descending convention, refuting the assertion.

Since $\mathrm{grad}f=-x^2{\mathrm{\partial }}_x$, the negative-gradient field is $x^2{\mathrm{\partial }}_x$ and its equation is $\dot{x}=x^2$.

For $x_0>0$, the solution is $x(t)=x_0\mathrm{/}(1-x_{0}t)$, which tends to $+\mathrm{\infty }$ as $t\uparrow 1\mathrm{/}{x}_0$. Hence its maximal interval has a finite positive endpoint and the field is incomplete.