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$x '(t) + g (x (t)) = f (t),quad forall tin mathbb R$ have periodic solution $iff; frac 1T int_0 ^ T f (t) dt in g (mathbb R) $

MathOverflow Asked on November 3, 2021

I have a research work concerning the equation: $$x ‘(t) + g (x (t)) = f (t),quad forall tin mathbb R$$
f and g are defined and continuous in $mathbb R$ and with values ​​in $mathbb R$.
Furthermore f is assumed to be T periodic (There is no initial condition)

First question:

Assume that there is a periodic solution. Using the mean formula, to show that $frac 1T int_0 ^ T f (t) dt in g (mathbb R) $

Second question

Show that this condition : $frac 1T int_0 ^ T f (t) dt in g (mathbb R) $ is sufficient for the existence of a periodic solution

For first question:

$ g circ x $ being continuous, there is a $ c $
between $ 0 $ and $ T $ such as $ frac 1Tint_0 ^ T
g (x (s)) ds = g (x (c)) $
. but $ x $ is a solution
$ T $ periodic, then $ x (T) = x (0) $, and we
will have:

$$frac1T int_0 ^ T f (s) ds = frac 1T int_0 ^ T
(g (x (s)) + x ‘(s)) ds = frac1T int_0 ^ T
g (x (s)) ds = g (x (c)) in g (mathbb R). $$

For second question:

I need help how to use some one of fixed point theorem or any other way to prove that the condition $frac 1T int_0 ^ T f (t) dt in g (mathbb R) $ is sufficient

One Answer

The condition is not sufficient. Take for example the equation $$x'=x^2+sin t$$ with $g(x)=-x^2$ in your notation. Then $$int_0^{2pi} sin t, dt=0=g(0).$$ If $x$ were a $2pi$-periodic solution, then $$ 0=x(2pi)-x(0)=int_0^{2pi} x'(t), dt=int_0^{2pi}(x^2(t)+sin t), dt= int_0^{2pi}x^2(t), dt$$ would imply $xequiv 0$ in $[0, 2pi]$.

Answered by Giorgio Metafune on November 3, 2021

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