# symmetry of a solution of an ordinary differential equation

Let $\gamma$ be a symmetry of the ordinary differential equation  (http://planetmath.org/SymmetryOfAnOrdinaryDifferentialEquation) and $x_{0}$ be a steady state solution of $\dot{x}=f(x)$. If

 $\gamma x_{0}=x_{0}$

then $\gamma$ is called a symmetry of the solution of $x_{0}$.

Let $\gamma$ be a symmetry of the ordinary differential equation and $x_{0}(t)$ be a periodic solution of $\dot{x}=f(x)$. If

 $\gamma x_{0}(t-t_{0})=x_{0}(t)$

for a certain $t_{0}$ then $(\gamma,t_{0})$ is called a symmetry of the periodic solution of $x_{0}(t)$.

lemma: If $\gamma$ is a symmetry of the ordinary differential equation and let $x_{0}(t)$ be a solution(either steady state or periodic) of $\dot{x}=f(x)$. Then $\gamma x_{0}(t)$ is a solution of $\dot{x}=f(x)$.
proof: If $x_{0}(t)$ is a solution of $\frac{dx}{dt}=f(x)$ implies $\frac{dx_{0}(t)}{dt}=f(x_{0}(t))$. Let’s now verify that $\gamma x_{0}(t)$ is a solution, with a substitution into $\frac{dx}{dt}=f(x)$. The left hand side of the equation becomes $\frac{d\gamma x_{0}(t)}{dt}=\gamma\frac{dx_{0}(t)}{dt}$ and the right hand side of the equation becomes $f(\gamma x_{0}(t))=\gamma f(x_{0}(t))$ since $\gamma$ is a symmetry of the differential equation. Therefore we have that the left hand side equals the right hand side since $\frac{dx_{0}(t)}{dt}=f(x_{0}(t))$. qed

## References

• GSS Golubitsky, Martin. Stewart, Ian. Schaeffer, G. David: Singularities and Groups in Bifurcation Theory (Volume II). Springer-Verlag, New York, 1988.
Title symmetry of a solution of an ordinary differential equation SymmetryOfASolutionOfAnOrdinaryDifferentialEquation 2013-03-22 13:42:26 2013-03-22 13:42:26 Daume (40) Daume (40) 11 Daume (40) Definition msc 34-00 symmetry of a periodic solution solution of an ordinary differential equation