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


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:


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:


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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.