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Parameterize the initial line by $(s,2)$ and use characteristic time $\tau$. The characteristic equations for a transport equation integrate toHere $dU/d\tau =U^2$ and $U(0)=h$. Since $\tau =\frac{1}{4}\log (x_2/2)$, the solution isDirect differentiation gives $u_{x_1}=0$, $u_{x_2}=u^2/(4x_2)$, verifying the equation and the initial values. The map $(s,\tau )\mapsto (x_1,x_2)$ has determinant $8e^{6\tau }\ne 0$, so every point of this strip is reached exactly once.

At the upper boundary $u\to +\infty$, a finite-time blow-up of an ordinary differential equation. At the lower boundary $u\to 0$, but $u_{x_2}=u^2/(4x_2)$ diverges as $x_2\downarrow 0$, so no $C^1$ continuation through that boundary is possible. Thus the displayed strip is the maximal connected domain containing the initial line on which this classical Cauchy solution exists.

For $r>0$, differentiate $r^2=x^2+y^2$ and use the angular derivative $(x\dot{y}-y\dot{x})/r^2$. The polar coordinates satisfyHence the rotation is clockwise. The origin is an unstable equilibrium, while $r=1$ is a limit cycle of period $2\pi$. Writing $u=r^2$ gives the logistic differential equation $\dot{u}=2u(1-u)$. For $r(t_0)=r_0>0$,All nonzero solutions tend to the unit-circle attracting limit cycle as $t\to \infty$, spiralling outward for $r_0<1$ and inward for $r_0>1$. For $0<r_0<1$ the solution exists for every real time and approaches the origin as $t\to -\infty$; for $r_0=1$ it remains periodic in both time directions. For $r_0>1$, the denominator first vanishes atThis finite-time blow-up of an ordinary differential equation occurs backward in time. Such exterior solutions have no $t\to -\infty$ behavior: their maximal time interval is $(t_{\ast },\infty )$. The zero solution exists for all time and has no defined polar angle.

For $x\ne \pm 1$, eliminating time gives the separable differential equationand henceDefine the possible limit at $x=s\in \{-1,1\}$ for a trajectory starting off the boundary lines byThe $k<0$ set is empty because $|y|\to \infty$, so there is no finite accumulation point. Since $\dot{x}<0$ on $(-1,1)$ and $\dot{x}>0$ outside $[-1,1]$, the complete classification off the boundary lines isFor $x_0>1$ the solution has forward finite-time blow-up, and for $x_0<-1$ it has backward finite-time blow-up, explaining the other empty entries under the usual infinite-time definitions.

It remains to treat $x_0=s=\pm 1$. Here $x(t)=s$ and $y(t)=y_0{e}^{kst}$. If $y_0=0$, both limit sets equal $\{(s,0)\}$. If $y_0\ne 0$, thenThis includes all boundary cases in both $k$ and the initial point.

The substitution $y=xv$ gives $y^{\prime }=v+x{v}^{\prime }$. Cancelling $x^{3}v$ on both sides of the differential equation, for $x>0$, givesThe initial condition is $v(1)=1$. This selects a positive solution locally, so separation of variables givesUsing $v(1)=1$ gives $C=-1/2$, and continuity selects the positive square root:The real solution containing $x=1$ has maximal positive domain $0<x<\sqrt{e}$. Its denominator tends to zero and $y\to +\infty$ as $x\uparrow \sqrt{e}$, a finite-time blow-up of an ordinary differential equation in the independent variable $x$. Thus the source's wording “for $x>0$” needs this domain restriction; the prescribed initial value admits no real solution on the whole positive half-line. The division by $v^3$ loses the separate solution $v=0$, but that solution does not meet the initial condition.

The planar system $\dot{x}=y+(1-x^2-y^2)x$, $\dot{y}=-x+(1-x^2-y^2)y$ has polar coordinates $\dot{r}=r(1-r^2)$, $\dot{\theta }=-1$. Every nonzero orbit approaches the clockwise unit-circle limit cycle, of period $2\pi$. For $r(t_0)=r_0>0$,Interior orbits approach the origin backward in time, but exterior orbits have backward finite-time blow-up of an ordinary differential equation at $t=t_0+\frac{1}{2}\log (1-r_0^{-2})$. Thus forward convergence to a limit cycle does not imply existence for all negative times.