Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 58 / 1 / i

The linearization at the zero solution is the self-adjoint differential operator $L=\mu -({\partial }_x^2+K{)}^2$, with domain satisfying both endpoint conditions. The Dirichlet boundary condition for $-{\partial }_x^2$ on $(-1,1)$ gives the orthogonal eigenfunctions $\cos ((n+\frac{1}{2})\pi x)$ and $\sin ((n+1)\pi x)$. Their second derivatives are multiples of themselves, so the additional condition ${\psi }_{xx}(\pm 1)=0$ is automatic. Completeness of the Dirichlet eigenfunctions also shows that these exhaust the spectrum of the polynomial operator $L$. Set $q_j=(j+1)\pi /2$; each eigenfunction has growth rate ${\sigma }_j=\mu -(K-q_j^2{)}^2$. Hence the stationary bifurcation thresholds areEquality of the first two thresholds requires $K$ to be the midpoint of ${\pi }^2/4$ and ${\pi }^2$, since these numbers are distinct. ThereforeAt this point the even and odd lowest eigenfunctions become neutral simultaneously.