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So, for the ODE \begin{align} x' = -x^{3} + \sin t, \end{align} we can show that there exists a $2\pi$ periodic solution. To do this, we denote by \begin{align} x(t,\alpha) \end{align} The solution $x(t)$ of the ODE such that $x(0) = \alpha$. Then, let $\alpha \in [-2,2]$. Consider the function $f(\alpha) = x(2\pi, \alpha)$ \begin{align} x > 1 &\Rightarrow x' < 0 \\ x < 1 &\Rightarrow x' > 0. \end{align}

Thus, solutions beginning in $[-2,2]$ stay there and we can use Brouwer's fixed point theorem to show that there exists a fixed point of $f$. Therefore there is some $\alpha^{*}$ such that $f(\alpha^{*}) = \alpha^{*}$, which represents the periodic solution $x(t,\alpha)$.

Now, my question: Is this $2\pi$- periodic solution unique? Are there other $2\pi$ periodic solutions of this ODE? I have started by trying to subtract two periodic solutions from one another but this hasn't taken me anywhere productive.

Arctic Char
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DRich
  • 799

2 Answers2

2

Let $v(t,α)=\frac{\partial x}{\partial α}(t,α)$, then

  • $v(0,α)=1$ as $x(0,α)=α$ and
  • the $α$ derivative of the ODE gives $$ v'(t,α)=-3x(t,α)^2v(t,α) $$ which means that $v$ is falling towards zero, moreover $$ v(2π,α)=\exp\left(-3\int_0^{2π}x(t,α)^2\,dt\right) $$ and as there are no solutions that are constant zero, $0<v(2π,α)<1$.

By the usual contraction argument this excludes the existence of multiple fixed points of $f$, thus proving the uniqueness of the periodic solution.

A recent duplicate to this question with the same ODE asked also after the uniqueness after a not so recent duplicate only asked after the existence. I felt that this more specific question was the more appropriate place to answer.

Lutz Lehmann
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Not quite an answer, more of a suggestion. Since $x$ is differentiable and periodic, if it happens to be periodic, it has a Fourier series. Let's pretend it is a trigonometric polynomial. The ODE tells you that there are some conditions on the coefficients, which you can write down explicitly, and now you can attempt to solve the equations one by one. What happens when you try this?

Igor Rivin
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