The problem is to find all eigenvalues and eigenfunctions for the following SL system.

$u'' + \lambda u = 0, x \in [a,b]$

$u'(a) = u'(b) = 0$

I know the general idea of how to do these problems and can do them for more simple boundary conditions say $u(0) = u(L) = 0$

I know the general solution to the ODE is:

$$u(x) = A\cos(\sqrt{\lambda}x) +B\sin(\sqrt{\lambda}x)$$, (do i need the constants for eigenvalue problems?)

My issue is with how to apply to boundary conditions to get it in the form given in the solution.

The ansatz given in the solution (without explanation) says "Explicit left boundary condition implies $u(x) = \cos(\sqrt{\lambda}(x-a))$.

I can see how this solution obviously satisfies $u'(a) = 0$ but I'm not sure how you would get there from:

$u'(a) = -\sqrt{\lambda}A\sin(\sqrt{\lambda}a) + \sqrt{\lambda}B\cos(\sqrt{\lambda}a) = 0$

They also drop the constant in the ansatz and I understand thats because any multiple of an eigenfunction is an eigenfunction so may aswell make that 1, but I'm also confused about when and where the constants should be disregarded(and if they should be included anywhere at all).

Thanks in advance!


If you want a solution with $u'(a)=0$, then you may safely assume $u(a)=1$ because $u'(a)=u(a)=0$ will force $u\equiv 0$. Any non-trivial solution will be a multiple of the solution where $u'(a)=0,u(a)=1$, and that solution is $$ u(x) = \cos(\sqrt{\lambda}(x-a)). $$ To complete the solution, set $u'(b)=0$, which gives $$ \sin(\sqrt{\lambda}(b-a))\sqrt{\lambda}=0. $$ $\lambda=0$ does give a solution $u(x)=1$. For $\lambda\ne 0$, then solutions are $$ \sqrt{\lambda}(b-a)=\pi,2\pi,3\pi,\cdots \\ \lambda = \frac{n^2\pi^2}{(b-a)^2},\;\;\; n=1,2,3,\cdots. $$ The corresponding non-trivial eigenfunctions are non-zero scalar multiples of $$ u_n(x)=\cos\left(\frac{n\pi(x-a)}{b-a}\right),\;\;\; n=0,1,2,3,\cdots. $$

  • $\begingroup$ Ah genius thankyou. One question, what constitutes the jump from $u'(a) = 0, u(a) = 1$ to that cos solution? Clearly it works but is there a specific way to get there. $\endgroup$ – iCaird Jun 2 '20 at 23:13
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    $\begingroup$ @iCaird : Seeing it once, you'll be able to reconstruct. The solutions are translation invariant, and you adjust to get $0$ at a specific point Then you normalize to match the value of the derivative at that point. $\endgroup$ – Disintegrating By Parts Jun 2 '20 at 23:41

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