Why is linear independence of harmonic oscillator solutions important?

Question

The equation of motion for the harmonic oscillator (mass on spring model)

$$\frac{d^2x}{dt^2} + \omega_0^2 x = 0$$

with $\omega_0^2 = D/m$, $D$ and $m$ being the force constant of the spring and the mass, has the solution

$$x=ce^{\lambda t}$$

where $c$ and $\lambda$ are a constant and a parameter. Inserting $x$ leads to

$$\lambda_1 = +i\omega_0$$

and

$$\lambda_2 = -i\omega_0$$

and so the solutions are

$$x_1(t) = c_1 e^{i \omega_0 t}$$

and

$$x_2(t) = c_2 e^{-i \omega_0 t}.$$

In my book, I now read "...these solutions are linearly independent for $\omega_0 \neq 0$."

What does this mean (i.e. how can I see this) and why is it important?

Answer 1

They would be linearly dependent if and only if there exist complex numbers $\alpha$ and $\beta$ such that

$\alpha x_{1}(t) + \beta x_{2}(t) = 0 \forall t$

Clearly, if $\omega_{0}=0$ then this is the case for $\alpha = 1$ and $\beta = -c_{1}/c_{2}$. So then they are linearly dependent. However, if $\omega_{0}\neq0$, you can't find a combination of $\alpha$ and $\beta$ that fulfills this requirement for all $t$. The importance lies in the fact that

(1) Any linear combination $\gamma x_{1}(t) + \delta x_{2}(t)$ of the two functions is also a solution. (Just plug the linear combination into the equation to see this.)

(2) These are the only solutions. Namely, if you would find a solution $y(t)$ you could always write it as a combination of $x_{1}$ and $x_{2}$. So these are the only solution to care about, all the dynamics of the system is contained in them.

Answer 2

It means, that you can't produce one solution out of a linear combination of the other. This is important because if you could gain a solution produced by superposition of other known solutions, it is actually not a new solution. It's information is already stored in the other known solution and therefor not relevant. So here the author wants to tell you: These are two different solutions.

Answer 3

It is like $\sin(\omega t)$ cannot be reduced to $\cos(\omega t),$ i.e., obtained by multiplying by a coefficient, because they are different functions - they are not proportional to each other.

Existence of two independent solutions is also like existence of two independent integration constants for a second order differential equation.

Using both of them guarantees that you can cover a general case and miss nothing.

Answer 4

Linearly independent means that the equation $$\lambda_1x_1(t) + \lambda_2 x_2(t) =0$$ has for unique solution $\lambda_1=\lambda_2=0$ for $\omega_0 \ne 0$. Otherwise, your equation being linear you can see (and it maybe is the easiest view) the solution space as a vector space, of which your $x_1(t)$ and $x_2(t)$ are one basis.