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I have the following ODE defined on $D_x = [-1,1]$:

$$y''(x)=-k^2y(x).$$

Prom the physical problem, I know that the solution is non-zero and that $y(x)$ is a even function [$y(-x)=y(x)$] vanishing at $x=1$.

The general solution to the previous ODE is given by:

$$y(x)=a\sin(kx)+b\cos(kx).$$

I tried to solve this problem by using two different boundary values.

Method 1: $y(1)=0$ and because $y(x)$ is a even function, we can state $y'(0)=0$. Using the general solution we get:

$$0=a\sin(k)+b\cos(k)$$ $$0=ak$$

  • Case 1 $a=0$: We obtain $0=b\cos(k)$.
    • Case 1a $b=0$: Leads to trivial zero solution.
    • Case 1b $\cos(k)=0$: Leads to $k=\pi/2+\pi l$, in which $l \in \mathbb{Z}$.
  • Case 2 $a\neq 0$: this implies $k=0$ which leads to $b=0$, hence the trivial zero solution.

So only Case 1b gave a meaningful result given by $$y(x)=\sum_{l=-\infty}^{\infty}b_l\cos((\pi/2+\pi l)x)$$

Method 2: As $y(x)$ vanishes on the boundaries I can also use $y(1)=y(-1)=0$ as boundary condition. Again, using the general solution it is possible to obtain:

$$0=a\sin(k)+b\cos(k)$$ $$0=-a\sin(k)+b\cos(k).$$

From here there are two possible ways to solve this problem.

Procedure 1: Adding both equations:

$$0=2b\cos(k).$$

Case 1: $b=0$, implies $0=a\sin(k)$ Case 1a $a=0$: Leads to trivial zero soltuion. Case 1b $a\neq 0$: Leads to $k=\pi l$, hence the general solution $$y(x)=\sum_{l=1}^{\infty}a_l\sin(\pi l x)$$

Procedure 2: Subtracting both equations: $$0=2a\sin(k)$$

Case 2 $a=0$: Implies $0=b\cos(k)$

Case 2a $b=0$: Implies trivial zero solution.

Case 2b $b\neq 0$: Implies $k=\pi/2+\pi l$, in which $l\in \mathbb{Z}$. So the general solution is given by $$y(x)=\sum_{l=-\infty}^{\infty}b_l\cos((\pi/2+\pi l)x)$$

My Question: How is it possible that two different sets of boundary conditions, descibing the same physical $y(x)$ lead to different results? And how is it possible that for the second method different ways of solving the system lead to different solutions? How can I know which boundary conditions I need to pick, in order to get the "right" solution? I would be glad if someone could point out a mistake that I made or explain me why this happened.

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  • $\begingroup$ Maybe it's because I'm on my smartphone... But what is the difference between the last equation in 1b and 2a? $\endgroup$ – N74 Nov 2 '16 at 21:35
  • $\begingroup$ One solution of method 2 is the same as for method 1. But the other solution is different. $\endgroup$ – MrYouMath Nov 2 '16 at 21:44
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    $\begingroup$ The first procedure in method 2 gives you functions which are not even, so they are not solutions to your problem. And you missed case 2, $b \ne 0$, which gives you the same solutions as the other methods. $\endgroup$ – Lukas Geyer Nov 6 '16 at 20:21
  • $\begingroup$ Your solution $y(x) = \sum b_l \cos(\cdots)$ is not a solution to $y'' = -k^2 y$. Each of these solutions you have added are solutions to that equation for a different value of $k$. $\endgroup$ – Winther Nov 12 '16 at 20:59
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Assuming that $y(1) = y(-1)$ does not guarantee that the solution is even. There are plenty for function satisfying this without being even. So in Method 2 your total solution will be a sum of the solutions found in Procedure 1 and Procedure 2. However the solutions from Procedure 1 does not fulfil the requirement that the solution is even so they have to be discarded (i.e. $a_l = 0$ for all $l$).

In this case it does not matter which method you use as they both yield the correct result in the end (once you keep only the even solutions).

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