Find the extremes of the functional $J=\int_0^{\pi}(2yz-2y^2+(y^\prime)^2-(z^\prime)^2)dx$

Question

Find the extremes of the functional

$$J=\int_0^{\pi}(2yz-2y^2+(y^\prime)^2-(z^\prime)^2)dx,$$ under the boundary conditions $y(0)=1,$ $y(\pi)=1,$ $z(0)=0$ y $z(\pi)=1$

I tried to solve this exercise with the function $F(x,y,y^\prime,z^\prime)=2yz-2y^2+(y^\prime)^2-(z^\prime)^2$ and the Euler-Langrange equations: $$\frac{\partial F}{\partial y}-\frac{d}{dx}{\{\frac{\partial F}{\partial y^\prime}}\}=0\;\;\;y \;\;\;\frac{\partial F}{\partial z}-\frac{d}{dx}{\{\frac{\partial F}{\partial z^\prime}}\}=0$$

and then I deduced that $$\frac{d^2y}{dx^2}=z-2y\;\;\;y\;\;\;\frac{d^2z}{dx^2}=-y$$

What else can I do? How do I apply the conditions $y(0)=1,$ $y(\pi)=1,$ $z(0)=0$ y $z(\pi)=1$?

Thanks you.

Answer 1

$$\frac{d^2y}{dx^2}=z-2y\\ \frac{d^2z}{dx^2}=-y$$ Substract both equations, you can deduce that: $$y''-z''=-(y-z)$$ $$u''+u=0$$ Where $u=y-z$ and initial conditions :$u(0)=1,u(\pi)=0$

Your initial conditions are incompatible: $$u=c_1 \cos x +c_2 \sin x$$ $$u(0)=1 \implies c_1=1$$ But $$u(\pi)=0 \implies c_1=0$$

Answer 2

This is a second order linear system, so you can solve it either with matrices, or by plugging one equation into the other. For example, taking two derivatives of the first equation and plugging in the second equation gives:

$y''''=z''-2y''=-y-2y'',$

which you can solve with the usual characteristic roots (you'll get some combination of exponentials). You'll get 4 constants which you can solve for with your boundary conditions. Can you finish it from here?