Unique solution of parametrized functional

Show that for $\lambda \in [0,\pi)$, the functional $$J_{\lambda}(y) = \frac{1}{2}\int_0^1y'(x)^2 - \lambda^2y(x)^2dx$$ has a unique minimum $y \equiv 0$ over $y \in C^1[0,1]$ and $y(0) = y(1) = 0$.

Solution: Note the Euler-Lagrange equation gives us $2y'' + 2\lambda^2y = 0$. This is a second-order differential equation whose characteristic equation $2r^2 + 2\lambda^2 = 0$ has roots $r_{1,2} = \pm i\lambda$. Hence, we have the solution $y(x) = c_1\cos(\lambda x) + c_2\sin(\lambda x)$. Applying the boundary conditions, \begin{align*} y(0) &= c_1 = 0 \\ y(1) &= c_1\cos(\lambda) + c_2\sin(\lambda) = c_2\sin(\lambda) = 0. \end{align*} I had this solved in my head but I can't put it back together. How do I finish this to show the claim?

Well, if $\lambda = 0$, then $y(x) = c_2 \sin \lambda = c_2 \sin 0 = 0$, good. If $0 < \lambda < \pi$, then $\sin \lambda \neq 0$ and $$c_2 \sin \lambda = 0 \implies c_2 = 0,$$hence $y(x) = 0$. Umm, done?
• What happens if $\lambda$ attains $\pi$? – clocktower Feb 27 '17 at 23:43
• Same thing if $\lambda = 0$, since $\sin \pi = 0$. But the problem states $\lambda \in [0,\pi)$, so we don't have to worry about that. – Ivo Terek Feb 27 '17 at 23:44