Unique solution of $ 4f'(t)+\sin f(t) + \int_{\frac{t}{2}}^{t} (1 + y^2(s)) \sin(s)ds=0 $

Question

Let $C([0,1],[-1,1])=\{f:[0,1] \to [-1,1] | f $ continuous$ \}$ the unit ball of space $C([0,1])$. Show that there exists only one $f \in C([0,1],[-1,1])$ such that:

$ \displaystyle 4f'(t)+\sin (f(t)) + \int_{\frac{t}{2}}^{t} (1 + f^2(s)) \sin(s)ds = 0 \;\;\forall t \in [0,1], f(0)=0.$

I already proved $C([0,1])$ complete, and I know Banach Fix Point Theorem. Probably I can use it here, but I can't find a useful contraction.

Thanks.

Answer 1

Hint: Look at the operator

$$T(f)(x) = -\frac{1}{4} \int_{0}^{x}\left( \sin(f(t)) + \int_{t/2}^{t} (1+f^2 (s))\sin(s) \mathrm{d}s \right)\mathrm{d}t.$$ Note that if $f \in C([0,1],[-1,1])$ with $f(0) = 0$ is a fixed point of this operator, $f$ solves your equation.