Why are all solutions to this system of pendulum differential equations a linear combination of the two given solutions?

Question

I am currently trying to do a lab report for a coupled pendulums experiment in which we find the following linear system of second order differential equations (describing the position as a function of time of the two masses): $$\dfrac{\text{d}^2}{\text{d}t^2}\begin{pmatrix}x_1\\x_2\\\end{pmatrix} = \begin{pmatrix}-\omega_0^2-k & & k\\k & & -\omega_0^2-k\\\end{pmatrix}\begin{pmatrix}x_1\\x_2\\\end{pmatrix}$$

And the lab notes states that all solutions are linear combinations of the following two solutions $$\begin{pmatrix}x_1\\x_2\\\end{pmatrix} = A_{sym}\begin{pmatrix}1\\1\\\end{pmatrix}\cos{(\omega_{sym}t+\phi_{sym})}$$

$$\begin{pmatrix}x_1\\x_2\\\end{pmatrix} = A_{asym}\begin{pmatrix}1\\-1\\\end{pmatrix}\cos{(\omega_{asym}t+\phi_{asym})}$$

where $\omega_{sym} = \omega_0$ and $\omega_{asym} = \sqrt{\omega_0^2+2k}\\$

I can understand intuitively why the two solutions span all solutions but I can't really make sense of it mathematically. I think my confusion comes from the fact that they are spanning a space of functions rather than finite vectors. Can somebody please explain fully in a simple way why the two solutions span all solutions?

Answer 1

Recall from linear algebra that given an operator $A$ acting on vectors $\vec{v}$ of $\mathbb{R}^n$, the equation $$A\vec{v}=0$$ defines the kernel or null-space of the operator $A$. This is the subspace of all vectors that are mapped to 0 by such operator and its dimension correspond to $n-\textrm{Rank}(A)$. This is the case for finite dimensional spaces but is generalized to vector spaces of functions and almost the same things apply.

So here you have a differential operator $$A = \begin{pmatrix} \frac{d}{dt^2} + \omega_0^2 + k^2 & k \\ k & \frac{d}{dt^2} + \omega_0^2 + k^2 \end{pmatrix}$$ which is acting on some space space of functions $V$ (I will not go down in detail into which space is this but, assume "nice" functions). As you can see the the operator is actually acting on the direct product of two such spaces, $V^2=V\times V$, that is it is acting on two copies of some space of functions, so vectors look like $\vec{v} = (f_1(t),f_2(t))$ and we are solving for \begin{equation}A\vec{v}=0 \tag{1} \label{eq:nullspace} \end{equation} as before. Now under the same philosophy of the usual linear algebra case, the most that can happen is that the operator sends everything to zero, in which case the null-space will have dimension $2n$ where $n$ would be now the dimension of $V$ and could be infinite. However, from differential equations theory we now that second order ODE's will need to be supplied with two boundary conditions, and this is exactly what is related to the freedom of "directions" you have in this null space. So if you think about the solutions as directions in a vector space, this is telling you that there is at most two directions in which you can "move" or change the conditions while still satisfying $\eqref{eq:nullspace}$. This would happen to each copy separately if there were no off-diagonal terms, but since here they are not zero, what they do is couple the equations and thus force a particular relation between the components, given by the eigenvalues. At the end of the day the important thing is to realize that all boils down to the dimension of a null-space of an operator in a space of functions and this is limited by the order of the ODE's involved and the size of the system.