Find the eigenvalues and eigenfunctions for $y''+\lambda y=0$ where $y'(1)=0$ and $y'(2)=0$

Question

As stated in the title: Find the eigenvalues and eigenfunctions for $y''+\lambda y=0$, where $y'(1)=0$ and $y'(2)=0$.

So I have already eliminated the cases for $\lambda=0$ and $\lambda<0$ and I'm focused on the case for $\lambda>0$ now. The characteristic equation for this case is $r^2+\lambda=0$, so $r=\pm i \sqrt{\lambda}$. Then the possible solution is $$y(x)=k_1\cos(\sqrt{\lambda}t)+k_2\sin(\sqrt{\lambda}t)$$ and the derivative is $$y'(x)=-k_1\sqrt{\lambda}\sin(\sqrt{\lambda}t) +k_2\sqrt{\lambda}\cos(\sqrt{\lambda}t)$$ Substituting the boundary values into the above equation gives: $$y'(1)=-k_1\sqrt{\lambda}\sin(\sqrt{\lambda}) +k_2\sqrt{\lambda}\cos(\sqrt{\lambda})=0$$ and $$y'(2)=-k_1\sqrt{\lambda}\sin(2\sqrt{\lambda}) +k_2\sqrt{\lambda}\cos(2\sqrt{\lambda})=0$$ I have no idea how to go forward from here. All the other BVP i've done had boundary values (usually one of the BVs IS zero) that led to one of the constants equaling zero or a substitution that was obvious. I'm having trouble seeing an obvious path forward with this problem.

Answer 1

If $y_n(x)=\sin(n\pi x)$, then $y_n^{\prime}(1)=(-1)^n\cdot n\pi$, so that's a problem. Let's go back and eliminate cases. If $\lambda<0$, then the solutions are $$y(x)=k_1\cosh(\sqrt{-\lambda}(x-1))+k_2\sinh(\sqrt{-\lambda}(x-1))$$ $$y^{\prime}(x)=\sqrt{-\lambda}k_1\sinh(\sqrt{-\lambda}(x-1))+\sqrt{-\lambda}k_2\cosh(\sqrt{-\lambda}(x-1))$$ Then $y^{\prime}(1)=\sqrt{-\lambda}k_2=0$ and $$y^{\prime}(2)=\sqrt{-\lambda}k_1\sinh(\sqrt{-\lambda})=0$$ Which has no solution for $\lambda<0$. If $\lambda=0$, then the solutions are $$y(x)=k_1+k_2(x-1)$$ $$y^{\prime}(x)=k_2$$ So applying boundary conditions $y^{\prime}(1)=y^{\prime}(2)=k_2=0$ so we get the solution $y_0(x)=k_1$. Normalizing to unity, this would be $y_0(x)=1$. Now if $\lambda>0$, then the solutions are $$y(x)=k_1\cos(\sqrt{\lambda}(x-1))+k_2\sin(\sqrt{\lambda}(x-1))$$ $$y^{\prime}(x)=-\sqrt{\lambda}k_1\sin(\sqrt{\lambda}(x-1))+\sqrt{\lambda}k_2\cos(\sqrt{\lambda}(x-1))$$ Then $y^{\prime}(1)=\sqrt{\lambda}k_2=0$ and now $$y^{\prime}(2)=-\sqrt{\lambda}k_1\sin(\sqrt{\lambda})=0$$ so $\lambda_n=n^2\pi^2$ and $y_n(x)=k_1\cos(n\pi(x-1))=(-1)^nk_1\cos(n\pi x)$. Normalizing to unity, we have $y_n(x)=\sqrt2\cos(n\pi x)$.

In summary, it would have been easier to have taken functions of $(x-1)$ instead of $x$, and the solutions were cosines, not sines, and there was a constant solution as well.

Answer 2

The system of equations can be rewritten as

$$ \alpha \left( \begin{matrix} -\sin \alpha & \cos \alpha \\ -\sin 2\alpha & \cos 2\alpha \end{matrix} \right) \left( \begin{matrix} k_1 \\ k_2 \end{matrix} \right) = 0 $$

Since your solution vector is in the null space, a solution will exist when the determinant of the coefficient matrix is zero

$$ \left| \begin{matrix} -\sin \alpha & \cos \alpha \\ -\sin 2\alpha & \cos 2\alpha \end{matrix} \right| = 0 $$

Or

$$ \sin 2\alpha \cos \alpha - \sin\alpha \cos2\alpha = \sin(2\alpha -\alpha) = \sin\alpha = 0$$

This gives $\alpha = \sqrt\lambda = n\pi $ as your eigenvalue, and $(\cos\alpha, \sin\alpha)$ as your eigenvector. The general solution is

$$ y(x) = \cos(n\pi)\cos(n\pi x) + \sin(n\pi)\sin(n\pi x) = \cos(n\pi(x-1)) $$