Verification of the Existence of Normal Coordinates for Small Vibrations

Statement of the Problem #

The treatment of first-order small vibrations near the stable equilibrium position of a multi-degree-of-freedom system (e.g., with \(n\) degrees of freedom) involves solving the system of differential equations

$$ \bm{\ddot{x}} = -\mathbf{A} \bm{x} \tag{1} $$

The solution can be formally written as

$$ \bm{x}=\bm{x_1}\exp\left(i\mathbf{A^\frac{1}{2}}\ t \right) + \bm{x_2}\exp\left(-i \mathbf{A^\frac{1}{2}}\ t \right) \tag{2} $$

Assuming \(\mathbf{A}\) is diagonalizable, we can set

$$ \mathbf{A} = \mathbf{P}\mathbf{\Lambda}\mathbf{P^{-1}} \tag{3} $$

Using the Taylor expansion, equation (2) transforms into

$$ \bm{x}=\bm{x_1}\mathbf{P}\exp\left(i\mathbf{\Lambda^\frac{1}{2}}\ t \right)\mathbf{P^{-1}} + \bm{x_2}\mathbf{P}\exp\left(- i \mathbf{\Lambda^\frac{1}{2}}\ t \right)\mathbf{P^{-1}} \tag{4} $$

where

$$ \exp(i\mathbf{\Lambda^{\frac{1}{2}}}\ t) = \begin{bmatrix} \exp(i\lambda_1^{\frac{1}{2}}\ t) & 0 & 0 & \\ 0 & \exp(i\lambda_2^{\frac{1}{2}}\ t) & 0 & \cdots \\ 0 & 0 & \exp(i\lambda_3^{\frac{1}{2}}\ t) & \\ & \cdots & & \end{bmatrix} \tag{5} $$

Thus, equation (4) represents the solution to the system of differential equations.

Another more common approach found in many textbooks is to introduce a trial solution for (1)

$$ \bm{x} = \bm{x_0} \cos (\omega t + \varphi) \tag{6} $$

Substituting yields

$$ (\mathbf{A}-\omega^2\mathbf{I})\bm{x_0}=0 \tag{7} $$

For non-trivial solutions \(\bm{x_0}\), we require

$$ \det(\mathbf{A} - \mathbf{\omega^2} \mathbf{I}) = 0 \tag{8} $$

This solves for the normal mode frequencies \(\omega_i\) and the corresponding eigenvectors \(\bm{x_i}\) \((i=1,2,\cdots, n)\). The solution to (1) is then

$$ \bm{x}=\sum_i \ \bm{x_i} \cos (\omega_i t + \varphi_i) \tag{9} $$

In many mechanics textbooks, the following is stated without proof

1. The eigenvalues \(\lambda_i\) of the coefficient matrix \(\mathbf{A}\) are positive (\(\lambda_i >0\)).
2. \(\mathbf{A}\) is diagonalizable, meaning its geometric multiplicity equals \(n\).

This implies that for first-order small vibrations near the stable equilibrium position of a multi-degree-of-freedom system, there exist normal coordinates equal in number to its degrees of freedom.

\(\lambda_i >0\) means the motion described is oscillatory (normal modes). The geometric multiplicity of \(\mathbf{A}\) equaling \(n\) implies the number of normal coordinates matches the system’s degrees of freedom. Hence the statement: there exist normal coordinates equal in number to its degrees of freedom.

This article aims to discuss the existence of these conditions.

Similarly, under the alternative approach, the statement ‘solves for the normal mode frequencies \(\omega_i\) and the corresponding eigenvectors \(\bm{x_i}\) \((i=1,2,\cdots, n)\)’ is not self-evident. Potential issues are

1. \(\omega_i\) being imaginary ( \(\omega_i^2 < 0\) ).
2. The geometric multiplicity of \(\mathbf{A}\) being less than \(n\). Even over \(\mathbb{C}\), \(n\) linearly independent eigenvectors \(\bm{x_i}\) may not exist if there are degenerate eigenvalues. Consequently, \(\bm{x}\) could have particular solutions not expressible as trigonometric functions (or finite sums thereof).

These two points are equivalent to the two points above. We need to discuss whether such situations can occur.

Existence of Normal Coordinates #

For first-order small vibrations near the stable equilibrium position of a multi-degree-of-freedom system, the equations of motion can be written as

$$ \mathbf{M}\bm{\ddot{x}}+\mathbf{K}\bm{x}=0 \tag{10} $$

where the matrices \(\mathbf{M}\) and \(\mathbf{K}\) are both real, symmetric, and positive definite. The coefficient matrix \(\mathbf{A}\) in (1) can then be expressed as

$$ \mathbf{A} = \mathbf{M^{-1}K} \tag{11} $$

Given that \(\mathbf{M}\) is positive definite, this can be rewritten as

$$ \mathbf{A} = \mathbf{M^{-\frac{1}{2}}\ M^{-\frac{1}{2}}\ K\ M^{-\frac{1}{2}}\ M^{\frac{1}{2}}} \tag{12} $$

Here, \(\mathbf{M^{-\frac{1}{2}}\ K\ M^{-\frac{1}{2}}}\) is a real symmetric matrix. Therefore, it can be diagonalized

$$ \mathbf{M^{-\frac{1}{2}}\ K\ M^{-\frac{1}{2}}} = \mathbf{P\Lambda P^{-1}} \tag{13} $$

Substituting back gives

$$ \mathbf{A} = \mathbf{M^{-\frac{1}{2}} P\Lambda P^{-1} M^{\frac{1}{2}}} \tag{14} $$

This demonstrates that

$$ \boxed{\text{\textbf{The coefficient matrix \(\mathbf{A}\) is diagonalizable}}} \notag $$

Consider the eigenvalue equation for \(\mathbf{A}\)

$$ \mathbf{A}\bm{x} = \mathbf{M^{-1}K}\bm{x} = \lambda \bm{x} \tag{15} $$

which is equivalent to the generalized eigenvalue problem

$$ \mathbf{K}\bm{x} = \lambda\ \mathbf{M}\bm{x} \tag{16} $$

Left-multiplying both sides by \(\bm{x^T}\) yields

$$ \bm{x^T} \mathbf{K} \bm{x} = \lambda\ \bm{x^T} \mathbf{M}\bm{x} \tag{17} $$

Since \(\mathbf{M}\) and \(\mathbf{K}\) are positive definite matrices, we have

$$ \bm{x^T} \mathbf{K} \bm{x} > 0 \quad \text{and} \quad \bm{x^T} \mathbf{M}\bm{x} > 0 \quad \text{(for } \bm{x} \neq \bm{0}\text{)} \tag{18} $$

Therefore

$$ \lambda = \frac{\bm{x^T} \mathbf{K} \bm{x}}{\bm{x^T} \mathbf{M}\bm{x}} > 0 \tag{19} $$

This proves that

$$ \boxed{\text{\textbf{The eigenvalues \(\lambda\) of the coefficient matrix \(\mathbf{A}\) are positive}}} \notag $$

That means, for first-order small vibrations near the stable equilibrium position of a multi-degree-of-freedom system, there exist normal coordinates equal in number to its degrees of freedom.