A student has asked me about this exam question (I paraphrase):

- In part a, how is the stable manifold determined?

The stable manifold of the equilibrium 0 of a linear ODE is precisely equal to the union of all (generalized) eigenspaces for eigenvalues that have negative real part. Namely, on these (generalized) eigenspaces, all solution curves converge (exponentially fast) to the equilibrium 0, whereas on the other eigenspaces solutions that start outside the equilbrium never converge to 0. These properties all follow from the explicit solutions of the linear ODE restricted to generalized eigenspaces. In the example at hand the matrix A has eigenvalues 1 and -1 and the eigenvector for -1 is (1,-1). Hence the stable manifold is the line through 0 and (1,-1). - In part d, why does the model answer use the Euler-Lagrange equation rather than the conservation of the Hamiltonian?

Either route is possible. I will show here how we can obtain the answer using the conservation of the Hamiltonian. The Hamiltonain is given by \(H=y'f_{y'}[y]-f[y]=(2y)^2+(y')^2\). The level sets \(H=c\) are ellipses in the \(y-y'\) plane if \(c>0\). We observe that \(c\geq0\). \(H=0\) corresponds to an equilibrium. So let \(c=d^2\). Then we have \(y'=\pm\sqrt{d^2-(2y)^2}\) which can be solved by separation of variables

\[\int dt =\int \sqrt{d^2-(2y)^2}^{-1}.\] Then \[ t+T=\frac{1}{2}\tan^{-1}\left(\frac{2y}{d^2-4y^2}\right),\] where \(T\) is constant. After some algebraic manipulations, one can rewrite this as \(y=\pm d \cos(2(t+T))\). The boundary condition \(y(0)=0\) yields \(T=\pm\frac{\pi}{4}\) so that \(y=\pm d \sin(2t)\) and \(y(1)= 1\) implies that \(y(t)=\frac{\sin(2t)}{\sin(2)}\), in accordance with the model answer.

Clearly in this case, the calculations using the Hamiltonian appear more involved than using the Euler-Lagrange equation. Which route to the answer is most efficient will depend on the example. My advice would be when approaching such a calculation, is to try one route and if it looks tedious, quickly try the other one as well to see if it makes a difference.