Euler-Lagrange equation for a second-derivative functional ($F_y-(F_{y^{\prime }}{)}^{\prime }+(F_{y^{\prime \prime }}{)}^{\prime \prime }=0$)

For a sufficiently smooth functional $I[y]={\int }_a^{b}F(x,y,y^{\prime },y^{\prime \prime })dx$, the first variation in direction $\eta$ is $\int (F_y\eta +F_{y^{\prime }}{\eta }^{\prime }+F_{y^{\prime \prime }}{\eta }^{\prime \prime })dx$. Twice applying integration by parts gives the interior coefficient $F_y-(F_{y^{\prime }}{)}^{\prime }+(F_{y^{\prime \prime }}{)}^{\prime \prime }$ and boundary expression $[(F_{y^{\prime }}-(F_{y^{\prime \prime }}{)}^{\prime })\eta +F_{y^{\prime \prime }}{\eta }^{\prime }{]}_a^b$. Variations supported in the interior and the fundamental lemma of the calculus of variations force the displayed equation. The remaining boundary terms give the natural boundary conditions for second-derivative functionals.

With values $y(0)=y(\pi /2)=y^{\prime }(\pi /2)=0$ and $y^{\prime }(0)=1$, the higher-order Euler-Lagrange equation is $y^{(4)}-y=0$. Write $c=\cosh (\pi /2)$ and $s=\sinh (\pi /2)$. Solving the four linear boundary equations gives $y=[c(\cos x-\cosh x)+(1-s)\sin x+(1+s)\sinh x]/2$. The added function $x^2$ does not affect the first variation.

For a smooth functional $\int F(x,y,y^{\prime },y^{\prime \prime })dx$, let $P=F_{y^{\prime }}$ and $Q=F_{y^{\prime \prime }}$. Along an extremal, the Euler-Lagrange equation for a second-derivative functional is $F_y-P^{\prime }+Q^{\prime \prime }=0$. Differentiating $H=y^{\prime \prime }Q+y^{\prime }(P-Q^{\prime })-F$ by the chain rule givesConsequently $H$ is constant whenever $F$ has no explicit dependence on $x$. This extends the Beltrami identity to a second-derivative variational problem. If $F$ is independent of $y^{\prime \prime }$, the expression reduces to $y^{\prime }{F}_{y^{\prime }}-F$.

At an endpoint of a second-derivative variational problem, a free value $y$ gives the first displayed condition, while a free slope $y^{\prime }$ gives the second. Each condition is required only when its corresponding endpoint variation is free; a prescribed value or slope instead forces that variation to vanish. For $F=K(y^{\prime \prime }{)}^2/2+qy$, the two conditions become $y^{\prime \prime \prime }=0$ and $y^{\prime \prime }=0$, respectively. A clamped beam has both value and slope fixed, a pinned beam fixes value but leaves slope free, and a free end leaves both free.