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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 $y^{\prime }{F}_{y^{\prime }}-F$.

Let $P=F_{y^{\prime }}$ and $Q=F_{y^{\prime \prime }}$. For a smooth variation $y+\epsilon \eta$, the first variation isHere the last expression follows by two integrations by parts. Taking variations supported in the interior and applying the fundamental lemma of the calculus of variations gives the Euler-Lagrange equation for a second-derivative functional:If both endpoint values and slopes are free, their independent variations force the natural boundary conditions for second-derivative functionalsIf an endpoint value or slope is fixed instead, its corresponding variation vanishes and the associated natural condition is not required.

For $H=y^{\prime \prime }Q-y^{\prime }(Q^{\prime }-P)-F$, differentiate using the chain rule:The Euler-Lagrange equation cancels the first term. Consequently $H^{\prime }=-F_x$, and if there is no explicit $x$ dependence,This is a Beltrami identity for a second-derivative functional.

Let $P=F_{y^{\prime }}$ and $Q=F_{y^{\prime \prime }}$. For a smooth variation $y+\epsilon \eta$, the first variation isHere the last expression follows by two integrations by parts. Taking variations supported in the interior and applying the fundamental lemma of the calculus of variations gives the Euler-Lagrange equation for a second-derivative functional:If both endpoint values and slopes are free, their independent variations force the natural boundary conditions for second-derivative functionalsIf an endpoint value or slope is fixed instead, its corresponding variation vanishes and the associated natural condition is not required.

For $H=y^{\prime \prime }Q-y^{\prime }(Q^{\prime }-P)-F$, differentiate using the chain rule:The Euler-Lagrange equation cancels the first term. Consequently $H^{\prime }=-F_x$, and if there is no explicit $x$ dependence,This is a Beltrami identity for a second-derivative functional.

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.