Suppose $\theta\in[0,1]$ and $s(\cdot)$ is strictly increasing in $\theta$ with $0\leq(0)<s(1)\leq1$ and $s(0)+s(1)>1$
How could I solve the program,
$\max_{s(\cdot)\in C[0,1]}=\int_0^{s^{-1}(1-s(0))} \left(s(0)[s^{-1}(1-s(0))]+\int_0^{\theta} (1-s^{-1}(1-s(x)))dx\right) d \theta +\int_{s^{-1}(1-s(0))}^1 \theta d\theta$
$= s(0)(s^{-1}(1-s(0))^2-\int_0^{s^{-1}(1-s(0))}\int_0^{\theta}s^{-1}(1-s(x))dxd \theta +\frac{1}{2} = s(0)(s^{-1}(1-s(0))^2-\int_0^{s^{-1}(1-s(0))}\left(\int_x^{s^{-1}(1-s(0))}d \theta \right)s^{-1}(1-s(x))dx +\frac{1}{2} = s(0)(s^{-1}(1-s(0))^2-\int_0^{s^{-1}(1-s(0))}[s^{-1}(1-s(0))-x]s^{-1}(1-s(x))dx +\frac{1}{2}$
I continue like this:
Let $g(x)=s^{-1}(1-s(x))$, then $s(x)+s(g(x))=1$. $ s^{-1}(1-s(0))=g(0)$. Because $s(\cdot)$ is strictly increasing, $g(\cdot)$ is strictly decreasing, $s(0)=G(g(0))$ is uniquely determined by $g(\cdot)$. We have
$\max_{g(\cdot)\in C[0,1]} U_i(\theta; s)=G(g(0))[g(0)]^2-\int_0^{g(0)}(g(0)-x)g(x)dx+\frac{1}{2}$
s.t. $s(x)+s(g(x))=1$ and $s(0)=G(g(0))$. Then I don't know how to proceed ?
Thanks.
As a somewhat simple-minded start,
$\begin{align} &\int_0^{s^{-1}(1-s(0))} \left(s(0)[s^{-1}(1-s(0))]+\int_0^{\theta} (1-s^{-1}(1-s(x)))dx\right) d \theta +\int_{s^{-1}(1-s(0))}^1 \theta d\theta\\ &=\int_0^{s^{-1}(1-s(0))} s(0)[s^{-1}(1-s(0))]d \theta +\int_0^{s^{-1}(1-s(0))}\int_0^{\theta} (1-s^{-1}(1-s(x)))dxd \theta +\frac{\theta^2}{2}\big|_{s^{-1}(1-s(0))}^1 \theta d\theta\\ &= s(0)(s^{-1}(1-s(0))^2 +\int_0^{s^{-1}(1-s(0))}\int_0^{\theta} (1-s^{-1}(1-s(x)))dxd \theta +\frac{1-(s^{-1}(1-s(0)) )^2}{2} \\ \end{align} $
The remaining integral can be similarly manipulated as
$\begin{align} \int_0^{s^{-1}(1-s(0))}\int_0^{\theta} (1-s^{-1}(1-s(x)))dxd \theta &=\int_0^{s^{-1}(1-s(0))}\int_0^{\theta}1dxd \theta \\ &\quad -\int_0^{s^{-1}(1-s(0))}\int_0^{\theta}s^{-1}(1-s(x)))dxd \theta\\ &=\int_0^{s^{-1}(1-s(0))}\theta d \theta \\ &\quad -\int_0^{s^{-1}(1-s(0))}\int_0^{\theta}s^{-1}(1-s(x)))dxd \theta\\ &=\frac{(s^{-1}(1-s(0)))^2}{2} \\ &\quad -\int_0^{s^{-1}(1-s(0))}\int_0^{\theta}s^{-1}(1-s(x)))dxd \theta\\ \end{align} $
Combining this, I get
$s(0)(s^{-1}(1-s(0))^2 +\frac{(s^{-1}(1-s(0)))^2}{2} -\int_0^{s^{-1}(1-s(0))}\int_0^{\theta}s^{-1}(1-s(x)))dxd \theta +\frac{1-(s^{-1}(1-s(0)) )^2}{2} \\ =s(0)(s^{-1}(1-s(0))^2 +\frac12 -\int_0^{s^{-1}(1-s(0))}\int_0^{\theta}s^{-1}(1-s(x)))dxd \theta $.
I don't have the slightest idea what to do with this.