= Solution
Seek an <even function> with isolated contact at $x=0$. On $0\leq x\leq L$, the contact and reflection conditions impose $w(0)=-H$ and $w'(0)=0$, while the free interval still has $w''''=0$. Write
$$
w(x)=-H+A_2x^2+A_3x^3.
$$
The two end <boundary conditions> give
$$
A_2L^2+A_3L^3=H,\qquad
2A_2L+3A_3L^2=\alpha,
$$
and therefore the full <displacement field> is
$$
\boxed{w(x)=-H+\frac{3H-\alpha L}{L^2}x^2
+\frac{\alpha L-2H}{L^3}|x|^3.}
$$
The <jump condition> at zero gives the upward <point force>
$$
\boxed{R_0=B[w''']_0=\frac{12B}{L^3}(\alpha L-2H).}
$$
It is nonnegative precisely when $\alpha\geq2H/L$. In that range $A_3\geq0$. Nonpenetration near zero requires $A_2\geq0$, equivalently $\alpha\leq3H/L$. Conversely, when both inequalities hold,
$$
w(x)+H=x^2(A_2+A_3|x|)\geq0,
$$
with strict inequality away from zero. Thus this is an admissible isolated-contact solution throughout
$$
\boxed{\frac{2H}{L}\leq\alpha\leq\frac{3H}{L}.}
$$
At the upper endpoint the contact <curvature> $w''(0)=2A_2$ vanishes, and the profile remains above the obstacle for $x\ne0$ because $A_3>0$. Increasing $\alpha$ further would make $A_2<0$, forcing penetration near zero. An interval of contact must then develop.
The admissible solution is unique within this linear model: the bending <elastic energy> $B\int(w'')^2/2$ is strictly convex on the <convex set> of admissible profiles with the prescribed end data. Integration by parts shows that a solution with nonnegative contact <point forces> satisfies the corresponding minimum condition, since every admissible variation lifts a point already on the obstacle. Reflection therefore justifies the assumed <even function> also after contact.
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