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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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  • $\begingroup$ What have you tried? Could you form a precise question instead of just stating a problem from a book or homework? $\endgroup$ – icurays1 Nov 25 '12 at 18:01
  • $\begingroup$ 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 usefull $\endgroup$ – user45503 Nov 25 '12 at 18:04
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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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