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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

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  • $\begingroup$ Your functional has no explicit $x$-dependence, so were it not for the $y''$-dependence you could simply appeal to the Beltrami identity (Wikipedia). That obviously won't work directly here, but the derivation of said identity in the Wikipeda link can perhaps be adapted to the case at hand. $\endgroup$ – Semiclassical May 5 '16 at 20:34
  • $\begingroup$ You got a sign wrong, it seems. The E-L equation becomes $1+y-y^{(4)}=0$ instead. $\endgroup$ – Ivo Terek May 5 '16 at 20:36
  • $\begingroup$ Just in case you are interested, your equation is related to Bernoulli-Euler beam equation, which is a classical way to describe the bending of beams at small frequencies. $\endgroup$ – Yuriy S May 5 '16 at 21:11
  • $\begingroup$ I was too lazy to figure out whether or not my comment above was actually valid, so I posted a question re: generalizing the Beltrami identity here. $\endgroup$ – Semiclassical May 5 '16 at 21:24
  • $\begingroup$ Thanks, it's nice to have some background. This is a past exam question for my course, so I don't get any context for it $\endgroup$ – ThanksABundle May 6 '16 at 12:47
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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: $c_3=1=-c_1$, $c_4=0=-c_2$.

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