I have a question about a boundary value problem. It may sort of look intuitive at first, but it's sort of strange for me.

$$ BC: y(0) = 0, \ y(1) +y'(1) = 0\\ y''(x)+\lambda y(x) = 0 $$

So I solved the differential equation and got the general solution as: $$ y = c_1 cos(\sqrt\lambda x) + c_2sin(\sqrt\lambda x) $$ Applied the first boundary condition and got that: $$c_1 = 0$$

Then I applied the second boundary condition and got: $$ c_2(sin(\sqrt\lambda) + \sqrt\lambda cos(\sqrt\lambda )) = 0 $$

The only thing I could really say is that either: $$ c_2 = 0 $$ or $$ sin(\sqrt\lambda) + \sqrt\lambda cos(\sqrt\lambda ) = 0 $$

The problem is, I can't really see how to find the eigenvalues for this problem. It doesn't look very obvious to me, any sort of hints?


First, lets write the equation in the form $\sqrt{\lambda} = -\tan(\sqrt{\lambda})$. Here's a graph of the left and right side of the equation, together with their points of intersection:

enter image description here

From this graph, it's fairly clear that there are countably many eigenvalues converging to $\infty$ in absolute value, as guaranteed by the Sturm-Liousville theory. Furthermore, they asymptotically approach the vertical asymptotes, which occur when $\cos(\sqrt{\lambda})=0$.

They cannot be solved for explicitly, but you can find them numerically. The first few are: ${0., 4.11586, 24.1393, 63.6591, 122.889, 201.851}$.


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