 Solution

Let $\eta$ be a smooth variation with $\eta ={\eta }^{\prime }=0$ at both endpoints, and consider $y_{ϵ}=y+ϵ\eta$. stationary point gives the first variationIntegrate the middle term once by parts and the last term twice. All boundary terms vanish because both $\eta$ and ${\eta }^{\prime }$ vanish there. Assuming the derivatives needed for the classical formula, the given fundamental lemma of the calculus of variations yieldsWithout assuming those classical derivatives initially, the same calculation with compactly supported variations states this equation weakly; sufficient regularity then gives the classical equation.

For the specific integrand, $F_y=-2y$, $F_{y^{\prime }}=0$, $F_{y^{\prime \prime }}=2y^{\prime \prime }$, so the equation is $y^{(4)}-y=0$. Its weak solutions here are smooth, and its general classical solution isPut $\ell =\pi /2$, $c=\cosh \ell$, $s=\sinh \ell$. The conditions at zero give $C=-A$ and $D=1-B$. Those at $\ell$ giveUsing $c^2-s^2=1$, solve to obtain $A=c/2$, $B=(1-s)/2$, $C=-c/2$, $D=(1+s)/2$. Thus an extremal isSubstitution checks all four boundary values. The additive term $x^2$ in the integrand contributes no variation in $y$, so it does not alter this equation.