Diff. eq. and boundary conditions

[Niles]

1. The problem statement, all variables and given/known data
Hi all.

I am given the following differential equation:

X'' - k*X=0.

I am told that k = -m^2, so the general solution is given by:

X = a*cos(m*x)+b*sin(m*x),

where a and b are constants. I am also given boundary conditions:

1) X(-Pi) = X(Pi)
2) X'(-Pi) = X'(Pi).

To satisfy #1, m must be an integer. But in my book the author states that in order to satisfy #1 and #2, m must be a positive integer. But I don't understand why m must be positive, because cosine is even.

[HallsofIvy]

1. The problem statement, all variables and given/known data
Hi all.

I am given the following differential equation:

X'' - k*X=0.

I am told that k = -m^2, so the general solution is given by:

X = a*cos(m*x)+b*sin(m*x),

where a and b are constants. I am also given boundary conditions:

1) X(-Pi) = X(Pi)
2) X'(-Pi) = X'(Pi).

To satisfy #1, m must be an integer. But in my book the author states that in order to satisfy #1 and #2, m must be a positive integer. But I don't understand why m must be positive, because cosine is even.
I don't know if you are reading that wrong or the author just said it wrong, but, as you say, m does not have to be a positive integer. But X= a cos(mx)+ b sin(mx)= a cos(-mx)+ (-b)sin(-mx) so changing m for -m just changes the sign on the constant b. We can always assume that m is non-negative, since -m would give nothing new, but cannot assume it is positive: taking m= 0 gives X(x)= a which certainly satisfies both differential equation and boundary conditions.

[Niles]

The author says: "For k = -m^2, Nontrivial solutions arise only if m = n for n = 1, 2, 3, ..., and the corresponding solutions are therefore X = a*cos(m*x)+b*sin(m*x).".

I think the reason why we do not want n = 0 is so there are no trivial solutions, i.e. X = const.

[klondike]

Seems to me like an eigenvalue problem with periodic boundary condition.
in this case -k is the eigenvalue, and because k<0, so the soultion

X=a \cos(\sqrt{-k} \space x)+b \sin(\sqrt{-k} \space x)


The periodic boundary condition (heat transfer on a circular ring as a physical example) requires a=\=0 and b=\=0 AND

\begin{*align}
sin (\sqrt{-k} \space \pi)=0
\end{*align}

or

\sqrt{-k} \space \pi =n \pi



note that \sqrt{-k}>0 so n is positive integer.

The author says: "For k = -m^2, Nontrivial solutions arise only if m = n for n = 1, 2, 3, ..., and the corresponding solutions are therefore X = a*cos(m*x)+b*sin(m*x).".

I think the reason why we do not want n = 0 is so there are no trivial solutions, i.e. X = const.