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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.

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  • $\begingroup$ Rewrite one equation for k1 as function of k2 and use as input to the other equation. Then solve that equation for k1 and use the solution to find k2. Quite straightforward. It is possible that the solution is not very elegant but perhaps you can apply some trigonometric identities to rewrite the solution. $\endgroup$ – nluigi Apr 10 '16 at 17:53
  • $\begingroup$ I tried the substitution you mentioned already and i couldn't get it to work out to a solution. I let $k_2=k_1\tan(\sqrt{\lambda})$ then when i plug it into the other equation i end up with $k_1\sqrt{\lambda}(\cos(2\sqrt{\lambda})\tan(\sqrt{\lambda})-\sin(2\sqrt{\lambda}))=0$. Since we are looking for a non-trivial solution for $k_1$ and since $\lambda\neq 0$ we must have that $(\cos(2\sqrt{\lambda})\tan(\sqrt{\lambda})=\sin(2\sqrt{\lambda})$ or $\tan(\sqrt{\lambda})=\tan(2\sqrt{\lambda})$ which is not true. Sooo, i think i'm misunderstanding something. $\endgroup$ – Iron Charioteer Apr 10 '16 at 18:24
  • $\begingroup$ likely because the boundary conditions make this an illposed problem. You give two zero gradients at different locations but the solution is periodic so how do you determine which (or both) of the bcs are minima or maxima? You need a BC in terms of y. $\endgroup$ – nluigi Apr 10 '16 at 18:42
  • $\begingroup$ So you're saying that we don't have enough information to solve this problem? It wouldn't be the first time my prof. has given us a problem that he hasn't made sure there is a solution for and/or given the wrong information. $\endgroup$ – Iron Charioteer Apr 10 '16 at 18:49
  • $\begingroup$ I'm on my mobile so it's difficult to work it out at the moment but I would say so. Sorry. Maybe that is what he is after for you to find out? Are you sure you read the problem statement correctly? $\endgroup$ – nluigi Apr 10 '16 at 18:52
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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

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

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  • $\begingroup$ Well you proved me wrong! I don't mind dredging up an old post if it teaches me something new. +1 $\endgroup$ – nluigi Oct 13 '17 at 21:18
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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.

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