= Solution
With no applied load the free string segments are straight, while in contact $w=f=h-x^2$. By symmetry use contact $[-s,s]$ and a right-hand line through $(1,0)$ tangent at $s$. The <tangency> conditions are $w(s)=h-s^2$ and $w'(s)=-2s$, giving $w=2s(1-x)$ on $[s,1]$ and $h-s^2=2s(1-s)$. Therefore
$$
\boxed{s=1-\sqrt{1-h},\qquad w(x)=\begin{cases}h-x^2,&|x|\leq s,\\2s(1-|x|),&s\leq|x|\leq1.\end{cases}}
$$
The other algebraic root $1+\sqrt{1-h}$ is outside the admissible contact interval except at the coincident endpoint $h=1$.
Verification makes the construction complete: the values and <slopes> match at $\pm s$, the endpoints are zero, and outside contact $w-f=(|x|-s)^2\geq0$. Inside contact $Tw''=-2T\leq0$; outside it $w''=0$, so <unilateral contact complementarity>[complementarity] holds. The minimization argument proves this verified solution is the unique <equilibrium>. At $h=0$ the <contact set of an obstacle problem>[contact set] is the single point $\{0\}$ and $w=0$; at $h=1$ contact fills the whole interval. This is the <parabolic obstacle under a taut string>.
\Image[/past-exam-of-the-elasticity-and-plasticity-course-of-the-university-of-oxford/2008-string-contact.png]
{title=The equilibrium follows the parabolic obstacle in contact and leaves along tangent straight segments}
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