We consider the case
B f(x) remains negative for x<0, while it intersects the xaxis once in x>0.
The case ‘f(x) is negative for x>0, while it intersects the xaxis once in x<0’ is treated in exactly the same way.
At first we work in the phaseplane
\{\begin{array}{c}\dot{x}=y,\hfill \\ \dot{y}=f(x)yx\hfill \end{array}
and we give a constructive result based on the methods used in [8].
Theorem 2.1 The phaseplane system (1.2) has exactly one limit cycle, which is stable, provided that f is continuous and

1.
f(x) is monotone decreasing for x<0,

2.
f(x) is monotone increasing for x>0 and has a single zero at a>0,

3.
\underset{x\to +\mathrm{\infty}}{lim}\frac{x}{f(x)}=\alpha <+\mathrm{\infty},
(2.1)

4.
\underset{x\to \mathrm{\infty}}{lim}f(x)=L>1.
(2.2)
Proof We just notice that the continuity of f guarantees the property of uniqueness for the solutions to the Cauchy problem associated to system (1.2), and therefore the trajectories of such a system cannot intersect. Consider the 0isocline of system (1.2), that is, the points in the phaseplane were \dot{y}=0 and \dot{x}\ne 0.
For these points, we have y=\frac{x}{f(x)}. Let Δ be the graph of the function u(x)=\frac{x}{f(x)}. Clearly, we get
\underset{x\to a+}{lim}u(x)=\mathrm{\infty}
(2.3)
and that u(x) is bounded as x\to +\mathrm{\infty}. Consider the slope of the trajectories of system (1.2) given by
{y}^{\mathrm{\prime}}=f(x)\frac{x}{y}.
(2.4)
In x>a, u(x) is negative and the slope at the points below Δ is negative. Hence we can select a point P({x}_{0},u({x}_{0})), with {x}_{0}>a and {x}_{0}a arbitrarily small, such that u({x}_{0}) is smaller than all the other values of u(x) for x>{x}_{0}. Let {\gamma}^{}(P) be a negative semitrajectory which starts at P. Inspection of the slopes shows that such a semitrajectory does not intersect Δ, and therefore is bounded away from the xaxis for x>{x}_{0}. We now consider the positive semitrajectory {\gamma}^{+}(P) which starts at P. At first we observe that being monotone, f(x) has a limit, L\le 0, for x\to \mathrm{\infty}. Hence, for \u03f5>0, there exists {x}_{1}<0 such that L>f(x)>L\u03f5 for every x<{x}_{1}. It is trivial to see that in the strip {x}_{1}<x<{x}_{0} the slope is bounded for y>1, and therefore {\gamma}^{+}(P) intersects the line x={x}_{1} at a point {P}_{1}({x}_{1},{y}_{1}) with {y}_{1}<0, or intersects the xaxis in x<0. Clearly, we must consider the first situation being our goal to reach the xaxis. For x<{x}_{1}, we get
This means that in x<{x}_{1}, Δ lies between the lines \frac{x}{L+\u03f5} and \frac{x}{L} and hence the 0isocline is bounded away from the line \frac{x}{L+\u03f5}. Such a line has the slope \frac{1}{L+\u03f5}. On the other hand, in x<{x}_{1} the slope of {\gamma}^{+}(P) is
f(x)\frac{x}{y}<L+\u03f5\frac{x}{y}<L+\u03f5.
(2.8)
Therefore, if L+\u03f5<\frac{1}{L+\u03f5}, eventually {\gamma}^{+}(P) will intersect the line \frac{x{x}_{1}}{L+\u03f5}\frac{{x}_{1}}{f({x}_{1})} at some \overline{x}<{x}_{1}, and hence also Δ at some \overline{x}<x<{x}_{1}. A straightforward calculation yields that such an inequality holds if L+\u03f5<1, and being ϵ arbitrary, this gives L<1, that is, the value of the limit for x\to \mathrm{\infty} in the assumptions. Now we recall that in x<0 the slope at the points above Δ is negative, and it is easy to see that {\gamma}^{+}(P) eventually will intersect the xaxis in x<0. In fact, any trajectory intersecting the xaxis in x<0 intersects the yaxis in y>0. This is because the slope being bounded for y>1, no vertical asymptote is possible. Then {\gamma}^{+}(P) will intersect the xaxis in x>0. This proves that the trajectory \gamma (P) is winding (see Figure 1). Being f(0)<0, the origin is a source and the existence of a limit cycle is granted by the PoincaréBendixson theorem, while the uniqueness is given by monotonicity properties in virtue of the Massera theorem. □
We observe that if we keep all the sign assumptions and the limits at infinity, but not the monotonicity, clearly the uniqueness is no more granted, but the existence result still holds, and one can relax the assumption
\underset{x\to +\mathrm{\infty}}{lim}\frac{x}{f(x)}=\alpha <+\mathrm{\infty}
(2.9)
with
\underset{x\to +\mathrm{\infty}}{lim}sup\frac{x}{f(x)}=\alpha <+\mathrm{\infty}
(2.10)
as in [8].
Hence we have the following corollary.
Corollary 2.2 The phaseplane system (1.2) has at least a stable limit cycle provided that f is continuous and f(x)(xa)>0, with a>0, and

1.
\underset{x\to \mathrm{\infty}}{lim}f(x)=L>1,
(2.11)

2.
\underset{x\to +\mathrm{\infty}}{lim}sup\frac{x}{f(x)}=\alpha <+\mathrm{\infty}.
(2.12)
As already mentioned, the first result is based on the direct construction of a winding trajectory.
A more powerful approach can be adopted in the Liénard plane, and it is based on the results presented in [9], where the problem of intersection with the vertical isocline y=F(x) is investigated in full for the more general Liénard system
\{\begin{array}{c}\dot{x}=yF(x),\hfill \\ \dot{y}=g(x),\hfill \end{array}
(2.13)
where
F(x)={\int}_{0}^{x}f(t)\phantom{\rule{0.2em}{0ex}}dt,
(2.14)
where as usual xg(x)>0.
The following result holds.
Theorem 2.3 The Liénard system (1.3) has exactly one limit cycle, which is stable, provided that f is continuous and

1.
f(x) is monotone decreasing for x<0,

2.
f(x) is monotone increasing for x>0 and has a single zero at a>0,

3.
\underset{x\to \mathrm{\infty}}{lim}f(x)=L>2,
(2.15)

4.
\underset{x\to +\mathrm{\infty}}{lim}f(x)=M>2.
(2.16)
Proof As already mentioned, the continuity of f guarantees the property of uniqueness for the solutions to the Cauchy problem associated to system (1.3), and therefore the trajectories of such a system cannot intersect.
As the proof is based on different applications of both necessary and sufficient conditions for the intersection of a semitrajectory with the vertical isocline y=F(x) (2.14) given in [9], we are going to introduce the same notation as in [9] and to recall the theorems we are going to use here.
As well as in the phaseplane, for any point P=({x}_{0},{y}_{0})\in {\mathbb{R}}^{2}, we denote by {\gamma}^{+}(P) and {\gamma}^{}(P) the positive, respectively negative, semitrajectory of the Liénard system (2.13) passing through P. With respect to the intersection of {\gamma}^{+}(P) with the vertical isocline (2.14), we observe that by a standard inspection of the direction of the field, it suffices to consider the cases
{x}_{0}\ge 0,\phantom{\rule{2em}{0ex}}{y}_{0}>F({x}_{0})
(2.17)
and
{x}_{0}\le 0,\phantom{\rule{2em}{0ex}}{y}_{0}<F({x}_{0}).
(2.18)
With respect to the intersection of {\gamma}^{}(P) with the vertical isocline (2.14), it suffices to consider the cases
{x}_{0}\ge 0,\phantom{\rule{2em}{0ex}}{y}_{0}<F({x}_{0})
(2.19)
and
{x}_{0}\le 0,\phantom{\rule{2em}{0ex}}{y}_{0}>F({x}_{0}).
(2.20)
As in [9], we define
{F}_{+}(x):=max\{0,F(x)\}
and, finally, we introduce the function
{\mathrm{\Gamma}}_{+}(x):={\int}_{0}^{x}{(1+{F}_{+}(\xi ))}^{1}g(\xi )\phantom{\rule{0.2em}{0ex}}d\xi .
For this particular case, we have the following result.
Theorem 2.4 ([[9], Theorem 2.5.])
If, for every P=({x}_{0},{y}_{0}) verifying (2.19), {\gamma}^{}(P) intersects the curve y=F(x) at (x,F(x)) with x>{x}_{0}, then
\underset{x\to +\mathrm{\infty}}{lim\hspace{0.17em}sup}[4{\mathrm{\Gamma}}_{+}(x)F(x)]=+\mathrm{\infty}.
(2.21)
Theorem 2.5 ([[9], Theorem 2.3.])
If there is a constant
0<\lambda <4
such that
\underset{x\to \mathrm{\infty}}{lim}[\lambda {\mathrm{\Gamma}}_{+}(x)F(x)]=+\mathrm{\infty},
(2.22)
then, for every P=({x}_{0},{y}_{0}) verifying (2.18), {\gamma}^{+}(P) intersects the curve y=F(x) at (x,F(x)) with x<{x}_{0}.
Let us see that hypothesis (2.15) implies that there exists 0<\lambda <4 such that (2.22) is verified. First of all, we observe that in the case we are dealing with {\mathrm{\Gamma}}_{+}(x)={\int}_{0}^{x}{(1+{F}_{+}(\xi ))}^{1}\xi \phantom{\rule{0.2em}{0ex}}d\xi. By hypothesis (2.15), \mathrm{\forall}\u03f5>0 there exists {x}_{1}<0 such that, for x<{x}_{1}, we have f(x)+L<\u03f5. It is not restrictive to suppose L\u03f5>0. This is because we are studying the case L\ge 1, being the result already proved for L<1, in virtue of the previous theorem. Then, for x<{x}_{1}<0, we have
\begin{array}{c}L+\u03f5>f(x)>L\u03f5,\hfill \\ F(x)=F({x}_{1})+{\int}_{x}^{{x}_{1}}f(x)\phantom{\rule{0.2em}{0ex}}dx,\hfill \\ 0<F({x}_{1})+(L\u03f5)({x}_{1}x)<F(x)<F({x}_{1})+(L+\u03f5)({x}_{1}x).\hfill \end{array}
In our assumptions, for x<0, {F}_{+}(x)=F(x). Then
\frac{x}{1+F({x}_{1})+(L+\u03f5)({x}_{1}x)}<\frac{x}{1+{F}_{+}(x)}<\frac{x}{1+F({x}_{1})+(L\u03f5)({x}_{1}x)}.
A straightforward calculation yields
\underset{x\to \mathrm{\infty}}{lim}\frac{x}{1+F({x}_{1})+(L+\u03f5)({x}_{1}x)}=\frac{1}{L+\u03f5},
therefore, choosing \eta >0, we can find {x}_{2}<{x}_{1} such that for x<{x}_{2}<0,
\frac{x}{1+F({x}_{1})+(L+\u03f5)({x}_{1}x)}>\frac{1}{L+\u03f5}\eta .
Then, for x<{x}_{2}<{x}_{1}<0,
{\mathrm{\Gamma}}_{+}(x)={\mathrm{\Gamma}}_{+}({x}_{2})+{\int}_{x}^{{x}_{2}}\frac{x}{1+{F}_{+}(x)}>{\mathrm{\Gamma}}_{+}({x}_{2})+({x}_{2}x)[\frac{1}{L+\u03f5}\eta ].
At last
\begin{array}{rcl}\lambda {\mathrm{\Gamma}}_{+}(x)F(x)& >& \lambda {\mathrm{\Gamma}}_{+}({x}_{2})+\lambda ({x}_{2}x)[\frac{1}{L+\u03f5}\eta ]F({x}_{1})(L+\u03f5)({x}_{1}x)\\ =& x(L+\u03f5\lambda [\frac{1}{L+\u03f5}\eta ])\\ +\{\lambda {\mathrm{\Gamma}}_{+}({x}_{2})+\lambda ({x}_{2})[\frac{1}{L+\u03f5}\eta ]F({x}_{1}){x}_{1}(L+\u03f5)\}.\end{array}
Then limit (2.22) is verified provided that
L+\u03f5\lambda [\frac{1}{L+\u03f5}\eta ]<0.
ϵ and η being arbitrary, the existence of 0<\lambda <4 such that the last inequality holds is ensured by L<2. With a very similar calculation (with {F}_{+}(x)=F(x)F(\alpha ), where α is the nonnull zero of F), hypothesis (2.16) yields to prove that
\underset{x\to +\mathrm{\infty}}{lim\hspace{0.17em}sup}[4{\mathrm{\Gamma}}_{+}(x)F(x)]<+\mathrm{\infty}.
(2.23)
As the necessary condition for the intersection of the negative semitrajectory with the vertical isocline is denied, there exists a point Q=({x}_{0},{y}_{0}) such that {\gamma}^{}(Q) is not intersecting the curve y=F(x) defined in (2.14). By a standard inspection of the direction of the field, {\gamma}^{+}(Q) crosses the negative ysemiaxis and reaches a point P=({x}_{1},{y}_{1}) verifying (2.18). By the existence of 0<\lambda <4 such that the sufficient condition (2.22) holds, the positive semitrajectory {\gamma}^{+}(P) intersects at first the yaxis because again there are no vertical asymptotes and then the curve y=F(x) because the sufficient condition for intersection is fulfilled. By the uniqueness for the solutions to the Cauchy problem associated to system (1.3), trajectories cannot intersect and hence {\gamma}^{+}(Q) is a winding trajectory (see Figure 2). Being f(0)<0, the origin is a source and the existence of a limit cycle is granted by the PoincaréBendixson theorem, while the uniqueness is given by monotonicity properties in virtue of the Massera theorem. □
If we relax the monotonicity assumptions, we lose uniqueness, but existence still holds. Hence we have the following corollary.
Corollary 2.6 The Liénard system (1.3) has at least a stable limit cycle provided that f is continuous and f(x)(xa)>0 with a>0, and

1.
\underset{x\to \mathrm{\infty}}{lim}f(x)=L>2,
(2.24)

2.
\underset{x\to +\mathrm{\infty}}{lim}f(x)=M>2.
(2.25)
Now we consider the case of nonexistence of limit cycles. We consider again the results of [9] and, in particular, the case in which the semitrajectory {\gamma}^{}(P), starting at a point P=({x}_{0},{y}_{0}) verifying (2.19), intersects the vertical isocline in the positive xhalf plane, but {\gamma}^{+}(P) does not intersect the same curve in the negative xhalf plane, and we get the following necessary condition.
Theorem 2.7 The Liénard system (1.3) has no limit cycles provided that f is continuous and

1.
f(x) is monotone decreasing for x<0,

2.
f(x) is monotone increasing for x>0 and has a single zero at a>0,

3.
\underset{x\to \mathrm{\infty}}{lim}f(x)=L<2,
(2.26)

4.
\underset{x\to +\mathrm{\infty}}{lim}f(x)=M<2.
(2.27)
Proof At first we need the following results.
Theorem 2.8 ([[9], Theorem 2.6.])
If there is a constant
0<\lambda <4
such that
\underset{x\to +\mathrm{\infty}}{lim}[\lambda {\mathrm{\Gamma}}_{+}(x)F(x)]=+\mathrm{\infty},
(2.28)
then, for every P=({x}_{0},{y}_{0}) verifying (2.19), {\gamma}^{}(P) intersects the curve y=F(x) at (x,F(x)) with x>{x}_{0}.
Theorem 2.9 ([[9], Theorem 2.2])
If, for every P=({x}_{0},{y}_{0}) verifying (2.18), {\gamma}^{+}(P) intersects the curve y=F(x), then
\underset{x\to \mathrm{\infty}}{lim\hspace{0.17em}sup}[4{\mathrm{\Gamma}}_{+}(x)F(x)]=+\mathrm{\infty}.
(2.29)
As in the previous theorem, the assumptions (2.28) and (2.29) become respectively
\begin{array}{c}\underset{x\to +\mathrm{\infty}}{lim}[\lambda {\int}_{0}^{x}\frac{\xi \phantom{\rule{0.2em}{0ex}}d\xi}{1+F(\xi )F(\alpha )}F(x)]=+\mathrm{\infty},\hfill \\ \underset{x\to \mathrm{\infty}}{lim}[4{\int}_{0}^{x}\frac{\xi \phantom{\rule{0.2em}{0ex}}d\xi}{1+F(\xi )}F(x)]=+\mathrm{\infty},\hfill \end{array}
where α is the nonnull zero of F.
With the same calculation, which we omit for the sake of simplicity, we get that if assumptions hold, we can provide a trajectory which is unwinding and goes to infinity without intersecting the curve y=F(x) for x negative (see Figure 3).
Now assume, by contradiction, that there exists a limit cycle. The monotonicity property of the Massera theorem guarantees that such a limit cycle is unique and stable. Therefore there are trajectories surrounding the limit cycle which are winding and going to such a cycle. But with the same calculation already applied, which we omit for the sake of simplicity, if assumptions (2.26) and (2.27) hold, we can provide a large trajectory which is unwinding and goes to infinity without intersecting the curve y=F(x) for x negative. We can apply again the PoincaréBendixson theorem and get the existence of a second limit cycle which is unstable.
But as already mentioned, the monotonicity properties of the Massera theorem give the uniqueness, and hence in this situation we can conclude that there is no limit cycle and all trajectories are unwinding. □
We observe that there are cases which still are not investigated. For instance, if we consider f(x) such that
\underset{x\to \mathrm{\infty}}{lim}f(x)=L<2\underset{x\to +\mathrm{\infty}}{lim}f(x)=M>2,
we can produce a trajectory which lies entirely below the curve y=F(x) in the Liénard plane (or below the xaxis in the phaseplane) and therefore is not winding or unwinding.
In order to describe in full the possible configurations, we consider the following example.
Example 2.10 Let f be continuous, monotone decreasing for x<0 and monotone increasing for x>0 with a single zero at a>0.
Let f(0)=\alpha <0, and
\underset{x\to \mathrm{\infty}}{lim}f(x)=L=2,\phantom{\rule{2em}{0ex}}\underset{x\to +\mathrm{\infty}}{lim}f(x)=M=2.
Define {f}_{\u03f5}(x)=f(x)+\u03f5 and consider the system
\{\begin{array}{c}\dot{x}=y{F}_{\u03f5}(x),\hfill \\ \dot{y}=x,\hfill \end{array}\text{where}{F}_{\u03f5}(x)={\int}_{0}^{x}{f}_{\u03f5}(t)\phantom{\rule{0.2em}{0ex}}dt.
(2.30)
If \u03f5\in (\mathrm{\infty},0] or [\alpha ,+\mathrm{\infty}), system (2.30) has no limit cycles, while if \u03f5\in (0,\alpha ), system (2.30) has exactly one limit cycle which is stable.
This is because if 2<\u03f5<0, we can apply Theorem 2.7, while if \u03f5\le 2, then {f}_{\u03f5}(x) is negative, and if \u03f5>\alpha, then {f}_{\u03f5}(x) is positive (for \u03f5=\alpha, {f}_{\u03f5}(x) is positive for x\ne 0). And as already mentioned, if {f}_{\u03f5} does not change sign, there is no limit cycle. On the other hand, if \u03f5\in (0,\alpha ), then we can apply Theorem 2.3, and in this case there is a unique limit cycle which is stable.
We can describe the movement of such a cycle. At \u03f5=\alpha, {f}_{\u03f5}(0) changes sign and we have a standard Hopf bifurcation, which gives a small amplitude limit cycle bifurcating from the origin. When ϵ decreases, such a limit cycle enlarges and eventually collapses to a separatrix at \u03f5=0. For a complete description of the bifurcation from a separatrix, we recall [10]. For the bifurcation of limit cycles we recall [11].