Consider the dynamical system

$$\ddot{x}-(a-b x) \dot{x}+x-x^{2}=0, \quad a, b>0 .$$

(a) Show that the fixed point at the origin is an unstable node or focus, and that the fixed point at $x=1$ is a saddle point.

(b) By considering the phase plane $(x, \dot{x})$, or otherwise, show graphically that the maximum value of $x$ for any periodic orbit is less than one.

(c) By writing the system in terms of the variables $x$ and $z=\dot{x}-\left(a x-b x^{2} / 2\right)$, or otherwise, show that for any periodic orbit $\mathcal{C}$

$$\oint_{\mathcal{C}}\left(x-x^{2}\right)\left(2 a x-b x^{2}\right) d t=0$$

Deduce that if $a / b>1 / 2$ there are no periodic orbits.

(d) If $a=b=0$ the system (1) is Hamiltonian and has homoclinic orbit

$$X(t)=\frac{1}{2}\left(3 \tanh ^{2}\left(\frac{t}{2}\right)-1\right)$$

which approaches $X=1$ as $t \rightarrow \pm \infty$. Now suppose that $a, b$ are very small and that we seek the value of $a / b$ corresponding to a periodic orbit very close to $X(t)$. By using equation (3) in equation (2), find an approximation to the largest value of $a / b$ for a periodic orbit when $a, b$ are very small.

[Hint. You may use the fact that $\left.(1-X)=\frac{3}{2} \operatorname{sech}^{2}\left(\frac{t}{2}\right)=3 \frac{d}{d t}\left(\tanh \left(\frac{t}{2}\right)\right)\right]$