Question

Determine the periodic solutions, if any, of the system $$ \frac{d x}{d t}=y+\frac{x}{\sqrt{x^{2}+y^{2}}}\left(x^{2}+y^{2}-2\right), \quad \frac{d y}{d t}=-x+\frac{y}{\sqrt{x^{2}+y^{2}}}\left(x^{2}+y^{2}-2\right) $$

Short answer

The periodic solutions of the given ODE system are: r(t) = √2, θ(t) = -t + c.

Step-by-step solution

Rewrite the ODE system in polar coordinates

Using the polar coordinate transformation \(x = r\cos\theta\), and \(y = r\sin\theta\), rewrite the given ODEs in terms of \(r\) and \(\theta\). To do this, we first differentiate \(x\) and \(y\) with respect to time \(t\) and in terms of \(r\) and \(\theta\): $$ \frac{dx}{dt} = \frac{dr}{dt}\cos\theta - r\sin\theta\frac{d\theta}{dt} $$ $$ \frac{dy}{dt} = \frac{dr}{dt}\sin\theta + r\cos\theta\frac{d\theta}{dt} $$ Now, substitute the given ODEs into these equations to form a system of equations involving \(\frac{dr}{dt}\) and \(\frac{d\theta}{dt}\).

Isolate \(\frac{dr}{dt}\) and \(\frac{d\theta}{dt}\)

Solve for \(\frac{dr}{dt}\) and \(\frac{d\theta}{dt}\) to obtain the following ODE system in polar coordinates: $$ \frac{dr}{dt} = r\left(1 - \frac{2}{r^2}\right), $$ $$ \frac{d\theta}{dt} = -1 $$

Solve the ODE for \(r\)

The \(\frac{dr}{dt}\) ODE can be separated and solved for \(r(t)\): $$ \int\frac{dr}{r\left(1-\frac{2}{r^2}\right)} = \int dt. $$ By solving this integral, we find that: $$ r^2(t) = k e^{2t} + 2, $$ where \(k\) is an integration constant.

Solve the ODE for \(\theta\)

The \(\frac{d\theta}{dt}\) ODE is already separated, so it's easy to find the solution for \(\theta(t)\): $$ \int d\theta = -\int dt \quad \Rightarrow \quad \theta(t) = -t + c, $$ where \(c\) is another integration constant.

Identify periodic solutions

Since the ODEs are linear and autonomous, periodic solutions occur when \(\frac{dr}{dt} = 0\). This happens when the constant term in parentheses is zero: $$ 1-\frac{2}{r^2} = 0. $$ The solution to this equation is \(r^2 = 2\), meaning that \(r = \sqrt{2}\), which corresponds to a periodic solution. Therefore, the periodic solutions of the given ODE system are given by: $$ \boldsymbol{r(t) = \sqrt{2}, \quad \theta(t) = -t + c}. $$