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Consider the following Sturm Liouville system :

$y'' + \lambda y=0$

$y(-\pi)=y(\pi)$

$y'(-\pi)=y'(\pi)$

I had to find the eigenvalues and corresponding eigenfunctions for this system , well i managed to solve it for $\lambda \gt 0$ and for $\lambda =0$ and got the solution too.

I am a little shaky about the case $\lambda \lt 0$

For this case the general solution turns out to be $y(x)=C_1 e^{kx} + C_2 e^{-kx}$. I then applied the given periodic boundary conditions and obtained a system of two homogeneous equations and found the determinant of their coefficient matrix, which turned out to be zero and so i concluded that only trivial solution exists in this case.

But my book directly states that for $\lambda \lt 0$ the periodic boundary condtions are not satisfied and so no eigenvalues can be found. I don't see, how that's happenening ?

If we substitute the periodic boundary conditions in the general solution, we get,

$y(-\pi) =C_1 e^{-k\pi} + C_2 e^{k\pi}$

$y(\pi) =C_1 e^{k\pi} + C_2 e^{-k\pi}$

So how can we conclude from this, that the first periodic boundary condition is not satisfied, since $C_1$ and $C_2$ are arbitrary constants, can't they have the same value in which case the first periodic boundary condition will be satisfied. I am a little confused here.

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Consider the equation in the form \begin{align} y'' + \lambda^{2} y = 0 \end{align} for which the solution is:


$\lambda^{2} > 0$

$y(x) = A \cos(\lambda x) + B \sin(\lambda x)$. From this it is seen that \begin{align} y(x) &= A \cos(\lambda x) + B \sin(\lambda x) \\ y'(x) &= - \lambda \, A \, \sin(\lambda x) + \lambda \, B \, \cos(\lambda x) \end{align} Applying the condition $y(-\pi) = y(\pi)$ then \begin{align} A \cos(\lambda \pi) - B \sin(\lambda \pi) &= A \cos(\lambda \pi) + B \sin(\lambda \pi)\\ 2 B \, \sin(\lambda \pi) &= 0 \end{align} and $y'(-\pi) = y'(\pi)$ also leads to the same result.


$\lambda = 0$

The solution here is $y(x) = a x + b$. Applying the conditions leads to $y(x) = b$.


$\lambda^{2} < 0$

The differential equation is $y'' - \lambda^{2} y = 0$ for which the solution is \begin{align} y(x) = A \cosh(\lambda x) + B \sinh(\lambda x). \end{align} Applying the first condition yields $2 B \sinh(\lambda \pi) = 0$. This can also be seen in the form $2 i B \sin( i \lambda \pi) = 0$. This leads to $\lambda = - i n$ for $n \geq 0$. Since this involves a complex value then it is not a periodic solution. Another way to view this is that the circular functions have the same form, graphically, but are shifted. Graphing the hyperbolic functions shows that they only behave the same in a small region from $0 \leq x \leq ?$. After this region there is no similarity of functions.

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  • $\begingroup$ how did you get the solution in terms of hyperbolic functions in the last part ? $\endgroup$ – johny Oct 2 '14 at 6:43
  • $\begingroup$ @johny $2 \cosh(x) = e^{x} + e^{-x}$ and $2 \sinh(x) = e^{x} - e^{-x}$. These are the same forms, exponential wise, that you presented. $\endgroup$ – Leucippus Oct 2 '14 at 6:50

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