Let $f:\mathbb{R}\times\mathbb{R^n} \rightarrow \mathbb{R^n}$ with $f$ a continuous function. Let $\tau>0$ such that $\forall(t,x) \in \mathbb{R}\times\mathbb{R^n}, f(t+\tau,x) = f(t,x)$. And we are given the fact that $f$ is Lipschitz for $x\in [0,\tau]\times\mathbb{R^n}$. Then, the following Cauchy problem, $x'=f(t,x), x(t_0)=x_0$ for some $(t_0,x_0)$.
For my problem I'm considering $x'= f(t,x)=\sin^2({t+||x||^2})g(x)$, with $(t,x)\in \mathbb{R}\times\mathbb{R^n}$ and $g:\mathbb{R^n}\rightarrow \mathbb{R^n}$ Lipschitz. And I'm trying to prove that if $\phi$ is a solution, then $\forall n\in \mathbb{N}$ $\phi_n(t)=\phi(t+n\pi)$ is also a solution of the problem.
I think I just need to prove that $f(t,x)$ is a Lipschitz function, so that a solution exists and from uniqueness we get that $\phi=\phi_n$, therefore $\phi_n$ is also a solution. So I know that the sine function is Lipschitz then sine squared is too, $g(x)$ is Lipschitz so I believe that I just need the $||x||^2$ to be Lipschitz for everything to work out. But I'm not quite sure if it is or how to prove that.
thanks in advance for any help!
A solution of an ODE is a function (with its domain) that satisfies the ODE, the equation, in every point. If present, the initial condition too.
This means that one (impossible) solution method for ODE is to iterate through all admissible functions (sufficiently continuously differentiable, domain and range compatible with the domain of the ODE) and test them. The existence theorems just say that this method has at least one valid result.
Here you have given the situation where one no further qualified solution $\phi$ is given and a class of functions $\phi_n$ (with shifted domain) is constructed from it. You are to apply the above test process to these new functions.
So insert the new functions into the ODE, if they are solutions, it should be true that $$ \phi_n'(t)=f(t,\phi_n(t)). $$ This remains equivalent when the definition of $\phi_n$ is inserted. Note that a shift in the argument gives the trivial inner derivative $1$. $$ \phi'(t+n\pi)=f(t,\phi(t+n\pi)). $$ Now apply the periodicity of $f$ to make its time argument compatible with the arguments of $\phi$. $$ \phi'(t+n\pi)=f(t+n\pi,\phi(t+n\pi)). $$ This now is true as $\phi$ is a solution per assumption.