Existence of continuous function using fixed point theorem

Question

Using the fixed point theorem, the problem asked me to prove there is a continuous function $x(t):[0,1]\to\mathbb{R}$ (maybe unique??)satisfying the equation $$x(t)=\frac{1}{3}\int_0^t s^2\sin(x(s))ds $$

What I did (bounding in $[0,1]$): $$|x(t)-x(v)|=\frac{1}{3}\left|\int_v^t s^2\sin(x(s))ds\right|\leq \frac{1}{3}|t-v| $$ so $x$ is Lipschitz and uniformly continuous in $[0,1]$. Indeed, $x'(t)=\frac{1}{3} t^2\sin(x(t))\geq 0$ this is because $\sin(\cdot)$ is positive in $[0,1]$ so this means $x$ non decreasing $x(t)\leq x(1)\leq \frac{1}{3}$. Accordingly to fixed point theorem if $x:X\to X$, and $d(x(t),x(v))\leq \lambda d(x,y)$ then there is a unique $x(t_0)=t_0$. But how does this imply the existence of the solution? for my problem and its uniqueness?

Answer 1

Apply Fixed Point Theorem to the space $C[0,1]$ of continuous functions on $[0,1]$ with the sup norm. Define $F(x)(t)=\frac1 3 \int_0^{t} s^{2} \sin (x(s))ds$. Then you get $\|F(x)-F(y)\|\leq \frac 1 3 \|x-y\|$. Hence there exists $x\in C[0,1]$ such that $F(x)=x$ and this gives the solution to the given integral equation.

Answer 2

Let $F:C[0,1] \to C[0,1]$ be defined by $(Fx)(t) = {1 \over 3}\int_0^t s^2 \sin x(s) ds $ then you have shown that $\|Fx\| \le { 1\over 3} \|x\|$ and so is a contraction. Hence there exists a unique fixed point satisfying $Fx = x$.