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Which curve joining the points (0,0) and (1,0) minimizes the integral $$ J[y]=\int_0^1(y'')^2 dx. $$ subject to the condition $$ \int_0^{1} (y')^2dx=1, $$ if $y(0)=0$, $y'(0)=\alpha$ and $y(1)=0$.

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Where are you stuck? What have you tried? Do you know the Euler-Lagrange equation you need to solve? –  Antonio Vargas Nov 25 '12 at 21:35
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Let's assume that $y'=u$ then $$J[y'']=\int(y'')^2 dx\Rightarrow J[u']=\int F(u') dx=\int(u')^2 dx$$ subject to $$\int_0^{1} (y')^2dx\Rightarrow \int G(u)dx\int (u)^2dx=1$$ The first order condition is $$\frac{\partial}{\partial u}\big(F-\lambda G)-\frac{d}{dx}\bigg(\frac{\partial}{\partial u'}\big(F-\lambda G)\bigg)=0$$ $$-2\lambda u-\frac{d}{dx}\bigg(2u'\bigg)=0$$ $$-2\lambda u-2u''=0$$ where it can be solved $$u(x)=A\cos(x\sqrt{\lambda})+B\sin(x\sqrt{\lambda})$$ since $y'=u$ $$y(x)=\frac{A}{\sqrt{\lambda}}\sin(x\sqrt{\lambda})-\frac{B}{\sqrt{\lambda}}\cos(x\sqrt{\lambda})+F=C\sin(x\,E)-D\cos(x\,E)+F$$ To satisfy the constraint $$\int_0^{1} (y')^2dx=\int_0^{1} \bigg(C\,E\,\cos(x\,E)+D\,E\sin(x\,E)\bigg)^2dx=1$$ $$\Rightarrow D-D\cos(E)+C\sin(E)=1$$ and the boundary conditions $$y(0)=-D+F=0$$ $$y(1)=C\sin(\,E)-D\cos(\,E)+F=0$$ $$y'(0)=C\,E=\alpha$$

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