# Proving that any solution to the differential equation of an oscillator can be written as a sum of sinusoids.

Suppose you have a differential equation with n distinct functions of $$t$$ where

$$\frac{d^2x_1}{dt^2}=k_{11}x_1+...k_{1n}x_n$$

.

.

.

$$\frac{d^2x_n}{dt^2}=k_{n1}x_1+...k_{nn}x_n$$

I want to show that any set of solutions of this differential equation $$(x_1,x_2,...,x_n)$$

can be written as a linear combination of solutions of the form $$(e^{iw_1t},...,e^{iw_1t}), (e^{iw_2t},...e^{iw_2t}), ...,(e^{iw_mt},...e^{iw_mt})$$ where each $$w_j$$ is a real number.

i.e. I want to know why the motion of any oscillator can be written as a linear combination of its normal modes. I would also appreciate it if you could tell me if a proof of this fact has to do with eigenvalues and eigenvectors in general.

• Was my answer helpful to you? If not let me know where you need additional clarification. – Spencer Apr 22 at 18:14

The system of differential equations you wrote could be written as,

$$\frac{d^2}{dt^2} \left[\begin{array}{c} x_1 \\ \vdots \\ x_n \end{array}\right] = \left[\begin{array}{ccc} k_{11} & \cdots & k_{1n} \\ \vdots & \ddots & \vdots \\ k_{1n} & \cdots & k_{nn} \end{array}\right] \left[ \begin{array}{c} x_1 \\ \vdots \\ x_n\end{array}\right]$$

$$\frac{d^2}{dt^2} \vec{x} = K \vec{x}$$

The matrix $$K=[k_{ij}]$$ acts on the indices of the functions $$x_1,\dots,x_n$$.

We will suppose that $$K$$ is diagonalizable with eigenvalues $$\lambda_1, \dots, \lambda_n$$.

Let $$\Lambda$$ be the diagonal form of $$K$$.

Let $$V$$ be the matrix of eigenvectors of $$K$$. Note that $$\Lambda = V^{-1} K V$$.

We can now write the system of differential equations as,

$$\frac{d^2}{dt^2} \vec{x} = V \Lambda V^{-1} \vec{x}$$

$$V^{-1}\frac{d^2}{dt^2} \vec{x} = \Lambda V^{-1} \vec{x}$$

$$\frac{d^2}{dt^2} V^{-1}\vec{x} = \Lambda V^{-1} \vec{x}$$

Let $$\vec{y} = V^{-1} \vec{x}$$, then we have $$\frac{d^2}{dt^2} \vec{y} = \Lambda \vec{y}$$. This corresponds to the following system of equations.

$$\frac{d^2 y_1}{dt^2} = \lambda_1 y_1$$ $$\frac{d^2 y_2}{dt^2} = \lambda_2 y_2$$ $$\vdots$$ $$\frac{d^2 y_n}{dt^2} = \lambda_n y_n$$

Clearly the solutions are of the form,

$$y_j(t) = C_1 e^{\sqrt{\lambda_j}\ t} + C_2 e^{-\sqrt{\lambda_j}\ t},$$

to obtain the $$x_j$$'s we just multiply by the $$V$$ matrix.

$$x_j(t) = \sum_i V_{ji} y_i(t)$$

Whether or not the solutions are oscillators depends on whether the eigenvalues are positive, negative, or complex. In physical applications it wouldn't be uncommon for $$K$$ to be a symmetric matrix with negative eigenvalues.

• In fact, you can show that if the $x_i$ coordinates are displacements from a stable equilibrium position, and the equations of motion given are exact (as opposed to a linearized approximation to a more complicated set of ODEs), then $K$ is symmetric with negative eigenvalues. – Michael Seifert Apr 22 at 13:59