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






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.

  • $\begingroup$ Was my answer helpful to you? If not let me know where you need additional clarification. $\endgroup$ – 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.

  • 1
    $\begingroup$ 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. $\endgroup$ – Michael Seifert Apr 22 at 13:59

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