# 2nd order LDE basis of solutions

Consider the LDE $$x''+k^2 x = 0$$

Then depending on how we define $$k$$ we have the solutions

1. $$x(t)=\lambda_1 \exp{ikt} + \lambda_2 \exp{(-ikt)}$$
2. $$x(t) = \lambda \exp{ik_nt}$$

In 1. we have defined k as the positive square root of $$k^2$$

In 2. we have defined $$k_n$$ as either the positive or the negative square root of $$k^2$$

As 1. and 2. must be the same (i.e. the solution to the LDE), I would like to know how this can be the same in terms of basis?

I tried to go from 2. to 1. using $$k_n = k_+ + k_-$$ where $$k_+$$ and $$k_-$$ are the $$k_n$$ restricted to positive or negative values but then I go with the product of 2 exponentials which is not at all the same formula as in 1.

• "I tried to go from 2. to 1. using..." Could you reproduce that work here? It would likely help determine where your confusion is coming from. May 12 at 13:09
• From 2. $x(t) = \lambda\exp{i(k_+ + k_- t)} = \lambda\exp{ik_+ t}\exp{ik_- t}$ May 12 at 13:14
• I think I got it: it's like saying vect($e_1$,$e_2$) (where $e_i$'s are the canonical basis vectors) or vect($e_1$) + vect($e_2$) May 12 at 13:27
• I would stick to the first formulation, as it is definitely correct. I'm still not sure what you're getting at with the second one, but you shouldn't end up with products like that. May 12 at 13:29
• Well in this video -> youtube.com/… the teacher does it May 12 at 13:45

In the first case, we're just solving the equation $$x'' + k^2x = 0$$ which has for a general solution $$x(t) = \lambda_1 e^{ikt} + \lambda_2e^{-ikt}$$, as you said.
In the second case, we're solving an eigenvalue problem. That is, we want to know the values $$k$$ such that the following boundary-value problem $$x'' - k^2 x = 0\;\;\;\;;\;\;\;\; x(0) = x(L), x'(0) = x'(L)$$ has a solution other than $$x(t) = 0$$. In this case, we need an integer number of periods of $$e^{ikt}$$ to fit into the interval $$[0,L]$$ to satisfy the boundary conditions. (Or that $$x(t)$$ is constant, i.e., $$k=0$$). This will be satisfied exactly when $$kL = 2n\pi$$ for some integer $$n$$. Thus, for each positive integer $$n$$, we have $$k_n = 2n\pi/L$$, and there will be a solution $$x_n = \lambda_{n+}e^{ik_nt} + \lambda_{n-}e^{-ik_n t}.$$ Now we only defined $$k_n$$ for $$n > 0$$. However, the formula for $$k_n$$ still makes sense for $$n \le 0$$, where $$k_n = -k_{-n}$$. So instead of writing the above, we can write $$x_n = \lambda_ne^{ik_n t}$$ with the understanding that $$x_n$$ and $$x_{-n}$$ are the two linearly independent solutions that fit $$n$$ periods into $$[0,L]$$ and $$x_0$$ is the constant solution with $$k = 0$$.