Past exam of the mathematical methods II course of the University of Oxford / 2013 resit / 11 / b / Solution

The stationarity equations are $4x^3-16y=0$ and $4y^3-16x=0$. Thus $y=x^3/4$ and $x=y^3/4$, so $x(x^8-256)=0$. The real solutions are $(0,0)$, $(2,2)$ and $(-2,-2)$. The Hessian matrix isAt the origin its determinant is $-256$, giving a saddle. At either nonzero stationary point it has eigenvalues $32,64$, so both are strict minima, with value $-32$:They are also global minima, because $f=(x^2-y^2{)}^2+2(xy-4{)}^2-32\ge -32$, with equality exactly at those two points. This is the special coefficient choice in a quartic double well with bilinear coupling.