0
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ "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. $\endgroup$ May 12 at 13:09
  • $\begingroup$ From 2. $x(t) = \lambda\exp{i(k_+ + k_- t)} = \lambda\exp{ik_+ t}\exp{ik_- t}$ $\endgroup$
    – niobium
    May 12 at 13:14
  • $\begingroup$ 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($e_2$) $\endgroup$
    – niobium
    May 12 at 13:27
  • 1
    $\begingroup$ 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. $\endgroup$ May 12 at 13:29
  • $\begingroup$ Well in this video -> youtube.com/… the teacher does it $\endgroup$
    – niobium
    May 12 at 13:45

1 Answer 1

1
$\begingroup$

Ah, so these are solutions to two different but related problems

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 $k = 0$.

$\endgroup$
1
  • $\begingroup$ But the second case is still an Ordinary Differential Equation? I think from the "basis" point of view these are the same. By basis I mean linear algebra. If you think there is a difference btw the two could you highlight in what extent they differ please? $\endgroup$
    – niobium
    May 12 at 16:55

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.