Consider the differential equation $$\theta '' + \lambda \sin ( \theta ) = 0, \quad \text{ for } 0 < x < 1 ,$$ in which $ \theta ' (0) = \theta ' (1) = 0$. For a homework question, I am asked to find the solutions to this equations that bifurcate from the equilibrium solution $\theta_{s} = 0$. This problem is related to a physical situation in which an initially straight rod is subjected to an axial load $\lambda$. The variable $\theta(x)$ is the angle the tangent to the rod makes with the horizontal at the spatial position $x$.
I find it hard to get started with this exercise. I tried reasoning that, if $\theta_{s} = 0$, then surely $\theta_{ss} = 0$. So we are left with the equation $\lambda \sin( \theta ) = 0$. I could also expand the sine into a Taylor series, but that doesn't seem to be of much help.
According to the solution manual, there should be a supercritical pitchfork when $\lambda_{n} = (n \pi )^{2} $ and $ \theta_{n} = 2 (2 \epsilon)^{1/2} \cos(n \pi x) / n \pi $.
I don't have the faintest idea how they arrived at this solution. Do you know how I should approach this problem?
Well this is how no less than Leonhard Euler himself solved the same physical problem. For sufficiently small deviations from the trivial equilibrium $\theta = 0$ the sine function can be approximated by the linear term of ist Taylor expansion $sin (x) \sim x$, yielding the ODE $$ \theta´´ + \lambda \theta = 0$$ The general solution is $$ \theta =A cos(\sqrt\lambda x) + Bsin(\sqrt\lambda x) $$ where the constants of integration $A$ and $B$ are determined via the boundary conditions $$ \theta´(0) = 0 $$ implies $B=0$. The second condition translates into $$-A sin (\sqrt \lambda ) \lambda = 0$$ There are two ways to satisfy the latter, either $A=0$, which yields the trivial equilibrium configuration, or $$ \sqrt \lambda = n\pi $$ for $n= 1, 2, 3, ...$, which coincides with the expression for the critical $\lambda$ you gave.