Consider a glider flying at velocity $\bf u$ (giving $u = |\bf u| $ on its speedometer) at an angle $θ ∈ (−π, π]$ to the horizon. Positive angle $θ ∈ (0, π/2)$ coincides with the glider’s nose tilted upwards relative to the horizon, whereas for $θ ∈ (−π, −π/2) ∪ (π/2, π)$ the plane is upside down. Its trajectory is governed approximately by the dynamical system $$\dot{u} = − \sin θ − bu^2$$ $$u \dot{θ} = − \cos θ + u^2$$ Here, the trigonometric terms are due to gravity whereas the quadratic terms result from the hydrodynamic drag and lift experienced by the glider. The control parameter b is the drag coefficient.
Zero drag ($b = 0$):
$(a)$ For the special (ideal) case of zero drag, show that $3u \cos θ − u ^3$ does not vary with time, $t$. Hence, explain why this system is conservative.
$(b)$ Calculate the fixed points and using the result in $(a)$ only, determine the phase portrait in the immediate vicinity of the fixed points. Do not conduct a linear stability analysis. Hint: A Taylor expansion may help.
$(c)$ Find the curves corresponding to where the quantity in $(a)$ vanishes.
Hello everyone. I'm struggling to understand this question. Where does the result "$3u \cos θ − u^3$" come from? Also how would you show that it does not vary with time? Should I be solving the differential equations? I don't know how to attempt this question, mainly because I can't see the relevance of "$3u \cos θ − u^3$" to this question. Any help would be appreciated. Thanks in advance.
For $b=0$ we have the system \begin{align} \dot{u} &= -\sin(\theta) \\ u \dot{\theta} &= -\cos(\theta) + u^2 \end{align} We now check the time derivative of the given term \begin{align} (d/dt) (3u\cos(\theta)-u^3) &= 3 \dot{u} \cos(\theta) - 3u\sin(\theta)\dot{\theta} -3u^2 \dot{u} \\ &= 3 (\underbrace{-\sin(\theta)}_{\dot{u}})\cos(\theta) -3 (\underbrace{-\cos(\theta) + u^2}_{u\dot{\theta}})\sin(\theta) - 3 u^2 (-\sin(\theta)) \\ &= 0 \end{align}