Past exam of the short options course of the University of Oxford / 2014 / 5 / b

Here $F_y=\rho g$, $F_{y^{\prime }}=0$, and $F_{y^{\prime \prime }}=K{y}^{\prime \prime }$. The Euler-Lagrange equation for a second-derivative functional is thereforeWrite $q=\rho g/K>0$. After imposing the clamped conditions at zero, the general solution isFree right endpoint. Both endpoint variations are free, so the natural boundary conditions for second-derivative functionals give $y^{\prime \prime }(L)=y^{\prime \prime \prime }(L)=0$. Since $y^{\prime \prime \prime }=-qx+A$ and $y^{\prime \prime }=-q{x}^2/2+Ax+B$, they give $A=qL$, $B=-q{L}^2/2$. HenceRight endpoint with fixed height and free slope. Now $y(L)=0$ is imposed, while the free slope gives $y^{\prime \prime }(L)=0$. These equations yield $A=5qL/8$ and $B=-q{L}^2/8$, soBoth endpoints clamped. The right conditions are $y(L)=y^{\prime }(L)=0$, giving $A=qL/2$, $B=-q{L}^2/12$. ThusEach formula has fourth derivative $-\rho g/K$ and satisfies all four boundary conditions of its case. The three endpoint choices therefore change the integration constants, not the interior differential equation.