= Solution
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> $\cos((n+\tfrac12)\pi x)$ and $\sin((n+1)\pi x)$. Their second derivatives are multiples of themselves, so the additional condition $\psi_{xx}(\pm1)=0$ 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 are
$$
\boxed{\mu_j(K)=\left(K-\frac{(j+1)^2\pi^2}{4}\right)^2.}
$$
Equality of the first two thresholds requires $K$ to be the midpoint of $\pi^2/4$ and $\pi^2$, since these numbers are distinct. Therefore
$$
\boxed{K^*=\frac{5\pi^2}{8},\qquad \mu^*=\frac{9\pi^4}{64}.}
$$
At this point the even and odd lowest <eigenfunctions> become neutral simultaneously.
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