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

Take a smooth variation $y_{\epsilon }=y+\epsilon \eta$. Differentiating the functional at zero gives the first variationIntegrate the second term once and the third term twice by integration by parts:Choose variations supported away from the endpoints. Stationarity and the fundamental lemma of the calculus of variations forceThis is the Euler-Lagrange equation for a second-derivative functional. With the interior equation imposed, the remaining boundary expression must vanish for every admissible variation.

At each endpoint, a freely varying value $y$ requires $\boxed{F_{y^{\prime }}-(F_{y^{\prime \prime }}{)}^{\prime }=0}$, and a freely varying slope $y^{\prime }$ requires $\boxed{F_{y^{\prime \prime }}=0}$. A fixed value or slope instead makes its corresponding variation zero and supplies no natural condition for that variable. These are the natural boundary conditions for second-derivative functionals, with the value and slope variations treated independently.

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.

The sets of admissible functions are nested:All impose the same left conditions; the right height adds a restriction, and the right slope adds another. Minimizing the same functional over a smaller set cannot lower its minimum. ThereforeThe doubly clamped beam has the greatest minimum energy, and the free-ended beam the smallest. No energy integrals are needed. In fact the inequalities are strict: the positive quadratic bending term makes the minimizer unique on the left-clamped space, and the three solutions in part (b) are different. The relevant principle is monotonicity of a constrained minimum under added constraints.