Banach fixed point theorem and interesting integral of arctan!

Question

Let's consider the follwing ecuation: $$x(t)=\int_{0}^{\frac{\pi}{2}} \arctan\left(\dfrac{x(s)}{2}+t\right) ds, $$where $t\in\mathbb{R}$. Using Banach fixed point theorem show that this ecuation has a solution $x(t)\in C[0;\frac{1}{2}]$. I understand the theorem cited above, but I am a bit stroke by the equation. I mean, the solution of it is a contiuous function, right? Thank you for your help!

Answer 1

Let $\Theta :C\left[0, \frac{\pi}{2}\right]\to C\left[0, \frac{\pi}{2}\right],$

$$\Theta (x)(t) =\int_{0}^{\frac{\pi}{2}} \arctan\left(\dfrac{x(s)}{2}+t\right) ds,$$ then we have $$||\Theta (x)-\Theta (y)||_{\infty} =\sup_{0\leq t\leq \frac{\pi}{2}} \left|\int_{0}^{\frac{\pi}{2}} (\arctan\left(\dfrac{x(s)}{2}+t\right)-\arctan\left(\dfrac{y(s)}{2}+t\right)) ds\right|\leq \\\\\leq\sup_{0\leq t\leq\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}} \frac{|x(s)-y(s)|}{2} ds\leq\frac{\pi}{4}||x-y||_{\infty}$$ thus $\Theta $ is contraction and by Banach Fixed Point Theorem $$\Theta (x) =x $$ for some $x\in C\left[0, \frac{\pi}{2}\right].$

Answer 2

Note that

$$x(t)-y(t)=\int_{0}^{\frac{\pi}{2}} \bigg(\arctan\left(\dfrac{x(s)}{2}+t\right)- \arctan\left(\dfrac{y(s)}{2}+t\right) \bigg)\,ds =\int_{0}^{\frac{\pi}{2}}\left(\int_{\frac{y(s)}{2}+t}^{\frac{x(s)}{2}+t}\frac{dw}{1+w^2}\right)\,ds, $$ and hence $$ \lvert x(t)-y(t)\rvert\le \int_{0}^{\frac{\pi}{2}}\left|\frac{y(s)}{2}+t-\frac{y(s)}{2}-t\right|\,ds=\frac{1}{2}\int_0^{\pi/2}\lvert x(s)-y(s)\rvert\,ds \le \frac{\pi}{4}\max_{s}\lvert x(s)-y(s)\rvert. $$ Finally, $\frac{\pi}{4}<1 $.