Vector Fields Flows and Lie Derivatives: Examples

The curve ${\gamma }_{x_0}(t)=x_0+tv$ satisfies ${\gamma }_{x_0}(0)=x_0$ and ${\gamma }_{x_0}^{\mathrm{\prime }}(t)=v=X({\gamma }_{x_0}(t))$, so it is an integral curve of $X$ through $x_0$.

The formula is defined for every $t\in \mathbb{R}$, so the flow exists globally and [L1] shows that $X$ is complete. The time-$t$ map is the translation ${\mathrm{\Phi }}_t(x)=x+tv$.

The curve ${\gamma }_{x_0}(t)=e^t{x}_0$ satisfies ${\gamma }_{x_0}(0)=x_0$ and ${\gamma }_{x_0}^{\mathrm{\prime }}(t)=e^t{x}_0=X({\gamma }_{x_0}(t))$, so it is the integral curve through $x_0$.

Since $e^t{x}_0$ is defined for all $t\in \mathbb{R}$, the flow is global. By [L1], the radial vector field is complete, and its flow maps are the dilations ${\mathrm{\Phi }}_t(x)=e^{t}x$.

Differentiating the displayed formula gives $\frac{d}{dt}{\mathrm{\Phi }}_t(x,y)=(-x\sin t-y\cos t,x\cos t-y\sin t)=X({\mathrm{\Phi }}_t(x,y))\mathrm{.}$ Also ${\mathrm{\Phi }}_0(x,y)=(x,y)$, so ${\mathrm{\Phi }}_t(x,y)$ is the integral curve through $(x,y)$.

The formula is defined for all $t\in \mathbb{R}$, so the flow is global. Therefore [L1] implies that the rotation field is complete.

The curve ${\gamma }_{x_0}(t)=x_0\mathrm{/}(1-x_{0}t)$ satisfies ${\gamma }_{x_0}(0)=x_0$ and ${\gamma }_{x_0}^{\mathrm{\prime }}(t)={\gamma }_{x_0}(t{)}^2$, so it is the integral curve through $x_0$.

For $x_0>0$, the denominator vanishes at $t=1\mathrm{/}{x}_0$, so this integral curve is not defined for all real times. Hence [L1] shows that $X$ is not complete.

Apply [L1] to $[-1,1]\subset (-2,2)$ to obtain the required smooth bump. Put $f(x)=\chi (x)x^2$. It is smooth, equals $x^2$ on $[-1,1]$, and vanishes off $[-2,2]$. Hence $\mathrm{\mid }f\mathrm{\mid }\le 4$. Its derivative is continuous and zero off $[-2,2]$, so [L4] bounds $\mathrm{\mid }{f}^{\mathrm{\prime }}\mathrm{\mid }$ globally. By [L3], $f$ is Lipschitz.

Fix any initial value $x(0)=a$ and take its maximal solution $x:(\alpha ,\beta )\to \mathbb{R}$ from [L2]. Since $\mathrm{\mid }{x}^{\mathrm{\prime }}\mathrm{\mid }=\mathrm{\mid }f(x)\mathrm{\mid }\le 4$, [L3] on the interval between $0$ and $t$ gives $\mathrm{\mid }x(t)-a\mathrm{\mid }\le 4\mathrm{\mid }t\mathrm{\mid }$. If $\beta <\mathrm{\infty }$, set $t_j=\beta (1-1\mathrm{/}(j+1))$ for $j\ge 1$. These times increase to $\beta$, and the graph points lie in the compact rectangle $[0,\beta ]\times [-R,R]$, where $R=\mathrm{\mid }a\mathrm{\mid }+4\beta +1$. The ODE domain is all of ${\mathbb{R}}^2$, so [L5] extends the solution past $\beta$, contrary to maximality. Therefore $\beta =\mathrm{\infty }$. Reflecting time applies the same argument to $-f$ and gives $\alpha =-\mathrm{\infty }$. This uses an explicitly defined sequence and no choice assumption.

Every integral curve of $Y$ is thus global, so $Y$ is complete. In contrast, $x(t)=1\mathrm{/}(1-t)$ solves $x^{\mathrm{\prime }}=x^2$ with $x(0)=1$ for $t<1$ and diverges as $t\uparrow 1$, so the original field is incomplete. This verifies both the local agreement and the claimed change in completeness.

For every smooth $h:\mathbb{R}\to \mathbb{R}$, the commutator definition [L1] gives $[X,Y]h=X(x{h}^{\mathrm{\prime }})-Y(h^{\mathrm{\prime }})=(x{h}^{\mathrm{\prime }}{)}^{\mathrm{\prime }}-x{h}^{\mathrm{\prime }\mathrm{\prime }}=h^{\mathrm{\prime }}\mathrm{.}$ Thus $[X,Y]$ acts exactly as $d\mathrm{/}dx$ on all smooth functions.

Hence $[X,Y]=d\mathrm{/}dx$, so the Lie bracket is nonzero even though $X$ is constant.

The explicit integral curves give the flows ${\mathrm{\Phi }}_t^X(x,y)=(x+t,y)$ and ${\mathrm{\Phi }}_s^Y(x,y)=(x,y+s)$.

Their compositions satisfy ${\mathrm{\Phi }}_t^X({\mathrm{\Phi }}_s^Y(x,y))=(x+t,y+s)={\mathrm{\Phi }}_s^Y({\mathrm{\Phi }}_t^X(x,y))\mathrm{.}$ This agrees with [L1] and [L2].

At $0$, both pairs have the same point values: $X_0=X_0^{\mathrm{\prime }}=d\mathrm{/}dx{\mathrm{\mid }}_0$ and $Y_0=Y_0^{\mathrm{\prime }}=0$.

For every smooth $h:\mathbb{R}\to \mathbb{R}$, [L1] gives $[X,Y]h=(x{h}^{\mathrm{\prime }}{)}^{\mathrm{\prime }}-x{h}^{\mathrm{\prime }\mathrm{\prime }}=h^{\mathrm{\prime }}$ and $[X^{\mathrm{\prime }},Y^{\mathrm{\prime }}]h=0$. Thus $[X,Y]=d\mathrm{/}dx$ and $[X^{\mathrm{\prime }},Y^{\mathrm{\prime }}]=0$, whence $[X,Y{]}_0=d\mathrm{/}dx{\mathrm{\mid }}_0\mathrm{\ne }0=[X^{\mathrm{\prime }},Y^{\mathrm{\prime }}{]}_0\mathrm{.}$

Hence equal point values do not determine the Lie bracket value at a point, giving the required counterexample.

Differentiating the proposed formula yields $\frac{d}{dt}{\mathrm{\Psi }}_{t,s}(x)=a(t)=X_t({\mathrm{\Psi }}_{t,s}(x)),$ and clearly ${\mathrm{\Psi }}_{s,s}(x)=x$. So the formula solves the initial-value problem.

Because $I$ is an interval, the segment between any $s,t\in I$ stays in $I$, so the integral and the displayed map are defined for every $s,t\in I$. Thus this is the global evolution operator extending the local one from [L1]. Moreover, ${\mathrm{\Psi }}_{r,t}({\mathrm{\Psi }}_{t,s}(x))=x+{\int }_s^{t}a+{\int }_t^{r}a=x+{\int }_s^{r}a={\mathrm{\Psi }}_{r,s}(x),$ matching [L2].

Compute $X(u)=1,X(v)=X(y{e}^{-x})=y(-e^{-x})+e^{-x}X(y)=-y{e}^{-x}+e^{-x}y=0.$

Step 1.1 means that in the coordinates $(u,v)$ the field differentiates the first coordinate by $1$ and the second by $0$, so it is exactly $\mathrm{\partial }\mathrm{/}\mathrm{\partial }u$. This is an explicit flow-box chart, as predicted by [L1].