I am trying to extremise the functional $\int{[y + \frac{1}{2}y^2 - \frac{1}{2}(y^{''})^2]}dy$ and so from Euler-Lagrange I get the differential equation
$1 + y + y^{(4)} = 0$ and I have no idea how to solve it. It's supposed to be an easy question so there must be a trick.
I have the initial conditions $y(0) = -1, y'(0) = 0, y(\pi) = \cosh(\pi), y'(\pi) = \sinh(\pi)$.
Just to check, I'm using the E-L equation, $\frac{\partial f}{\partial y} - \frac{d}{dx} \frac{\partial f}{\partial y'} + \frac{d^2}{dx^2} \frac{\partial f}{\partial y^{''}} = 0$
Can someone help?
Thanks
I think your equation should be $1+y-y^{(4)}=0$, but in any case it's a linear differential equation with constant coefficients, so you get the particular solution $y_p(x)=-1$ and solve the homogeneous differential equation $$y_h^{(4)}-y_h=0$$ The trial function is $y=e^{rx}$ leading to the characteristic equation $r^4-1=(r+1)(r-1)(r+i)(r-i)=0$, so the general solution is $$y(x)=-1+c_1\cos x+c_2\sin x+c_3\cosh x+c_4\sinh x$$ Applying initial conditions, $y(0)=-1=-1+c_1+c_3$, $y^{\prime}(0)=0=c_2+c_4$, $y(\pi)=\cosh(\pi)=-1-c_1+c_3\cosh\pi+c_4\sinh\pi$, $y^{\prime}(\pi)=\sinh\pi=-c_2+c_3\sinh\pi+c_4\cosh\pi$, we should be able to get a solution.
EDIT The solution turns out to be very simple, as might have been expected from the initial conditions: $c_3=1=-c_1$, $c_4=0=-c_2$.