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Extremise the functional: $ J[y]=\int_0^1 (yy')^2 dx$ subject to the constraint $ \int_0^1 y^2 dx=3, $ And the boundary conditions $y(0)=1$ and $y(1)=2$.

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    I have tried forming $H=(yy')^2+\lambda y^2$ and tried to find the Euler-Lagrange equation of this. However I don't find any answers that are usefull2012-11-25

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This is an isoperimetric problem with $L(y,y')=\int F(y,y')dx=\int (y\,y')^2 dx$ subject to $\int G(y)dx=\int y^2\,dx=3$ The first order condition is $\frac{\partial}{\partial y}\big(F-\lambda G)-\frac{d}{dx}\bigg(\frac{\partial}{\partial y'}\big(F-\lambda G)\bigg)=0$ $\frac{\partial}{\partial y}\big((y\,y')^2-\lambda y^2)-\frac{d}{dx}\bigg(\frac{\partial}{\partial y'}\big((y\,y')^2-\lambda y^2)\bigg)=0$ $2\,y\,y'^2-2\lambda y-\frac{d}{dx}\bigg(2\,y^2\,y'\bigg)=0$ $2\,y\,y'^2-2\lambda y-2y^2y''-4y\,y'^2=0$ $-2y\big(\lambda+y'^2+y\,y''\big)=0$ which can be solved for $y=0$ and $y=\sqrt{\frac{e^{2A}}{\lambda}-\lambda(x+B)^2}=\sqrt{C-\lambda(x+B)^2}$ To satisfy constraint $\int_0^1 y^2\,dx=\int_0^1 \big(C-\lambda(x+B)^2\big)\,dx=C-\frac{\lambda}{3}-B\lambda(1-B)=3$ and boundary conditions $y(0)=\sqrt{C-\lambda(B)^2}=1$ $y(1)=\sqrt{C-\lambda(1+B)^2}=2$ By solving these three equations it follows that $C=4$, $B=-1$ and $\lambda=3$. And the solution $y=\sqrt{3-3(x-1)^2}$

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    Would you please take a look at this variational calculus problem? http://math.stackexchange.com/q/457500/886522013-08-15