91Ó°ÊÓ

A system is conservative if its potential energy depends only on the position of the particles and satisfies the force acting on them is derived from a potential function. Mathematically, this means that the force can be written as the gradient of some scalar potential function: \(\mathbf{f}(\mathbf{x}) = -\nabla U(\mathbf{x})\).

We have a given system of the form \(\ddot{\mathbf{x}}=\mathbf{f}(\mathbf{x}), \mathbf{x} \in E^{2}\). Here, \(\mathbf{x}\) is a position vector in a two-dimensional Euclidean space, \(E^2\), while the dots represent second-order derivatives with respect to time, and \(\mathbf{f}(\mathbf{x})\) represents the force acting on the particle. Now, let's find a system that is not conservative.

Consider a system with a dampened oscillator, where the force is given by \(\mathbf{f}(\mathbf{x}) = -k \mathbf{x} - c \dot{\mathbf{x}}\). The first term (\(-k \mathbf{x}\)) is a standard linear spring force, while the second term (\(-c \dot{\mathbf{x}}\)) is a damping force proportional to the velocity of the particle. The damping force is a non-conservative force because it does not conserve total mechanical energy. This means that we cannot write this force as the gradient of a scalar potential function.

Now, we can rewrite the dampened oscillator system in the form of the given problem. Let the position vector \(\mathbf{x}=(x(t), y(t))^T\). The force acting on the particle is given by: \(\mathbf{f}(\mathbf{x}) = \begin{pmatrix}-kx - cx'\\ -ky - cy'\end{pmatrix}\), where \((x', y')^T = \dot{\mathbf{x}}\) is the particle's velocity. The second-order equation for the system is: \(\ddot{\mathbf{x}} = \mathbf{f}(\mathbf{x})\), \(\begin{pmatrix}x''\\ y''\end{pmatrix} = \begin{pmatrix}-kx - cx'\\ -ky - cy'\end{pmatrix}\). Finally, we have found an example of a system of the form \(\ddot{\mathbf{x}}=\mathbf{f}(\mathbf{x}), \mathbf{x} \in E^{2}\), which is not conservative, given by the dampened oscillator system with damping force.