I'm trying to solve analytically the equation for a thin piece of paper being compressed by both sides on a table. I found this paper trying to solve the same problem doi:10.1088/1742-6596/2012/1/012043 in the end I got to the same equation as the first and have to solve for when it equals zero. What approximation did the author use on this passage? We also use a constraint where the second integral equals L (the length of the paper)
$$ \delta\left[A\int_0^d\frac{y''^2}{(1+y'^2)^{\frac{5}{2}}} \ \text dx + \lambda \int_0^d \sqrt{1+y'^2} \ \text dx\right] \approx \delta\left[A\int_0^d y''^2 \ \text dx + \lambda \int_0^d \sqrt{1 + \frac{1}{2}y'^2} \ \text dx \right] $$
Without using the approximation I got to this differential equation $$ y''\bigg|^d_0 = -\frac{\lambda}{A}(1+y'^2)^{5/2}\sinh^{-1}(y')\bigg|^d_0 - \frac{5(1+y')}{2(1+y'^2)}\Biggr|^d_0-4 $$
I'm trying to solve without the $|^d_0$, I just added it to be consistent.
Goal Determine the ideal shape for an arch bridge, built from a thin sheet of uniform paper. In cross-section the bridge is described by a curve with prescribed endpoints and prescribed total arclength. The endpoints are at equal height $y=0$.
Notation The curvature is defined as $\kappa= \frac{d\theta}{ds}$ where $s$ is arlength and $\theta(s)$ is the angular coordinate of the unit tangent vector $\vec T= \langle\frac{ dx}{ds}, \frac{dy}{ds}\rangle=\langle\cos \theta (s), \sin \theta(s)\rangle$.
Setting up the Nonlinear Model
The ideal shape is that which minimizes the total energy $E$, which is the sum of bending energy and potential energy. The total energy can be written as a linear combination of two fundamental integrals: $E=AI_1 + BI_2$ where $I_1=\int \frac{\kappa^2}{2} ds$ and $I_2= \int y ds$. Here $A$ and $B$ are physical constants. To declutter the calculations that follow, we will prefer to focus on the related energy expression $E/A= I_1 + \frac{B}{A} I_2$ which depends on only the parameter $B/A= C$. (This parameter $C$ is a measure of the relative strength of paper stiffness to weight. It is large when the paper is heavier than it is stiff.)
Note that $I_2=\int (d (sy) - s dy) =-\int s dy$ since $(sy)]_0^L =0$ because $y=0$ at both endpoints. Note further that $\frac{dy}{ds} =\sin \theta$. Thus $I_2= -\int s \sin\theta ds$.
We wish to find critical points of the expression $E/A=I:=( I_1+ \frac{B}{A} I_2)$; that is,
$$I=\int_{s=0}^{s=L} \left[\frac{1}{2}\left(\frac{ d\theta}{ds}\right)^2 - C s \sin \theta \right]\, ds$$
The solution $\theta(s)$ we seek is subject to these two integral constraints:
(i) the total horizontal span of the bridge take the precribed value $J_1= \int dx =\int _{s=0}^{s=L} \frac{dx}{ds} ds =\int_0^L \cos \theta ds$,
(ii) the net vertical displacement across the span of the bridge satisfies $0= J_2=\int_0^L dy= \int_0^L \sin \theta ds$.
Euler-Lagrange Equation
After applying a small endpoint-preserving perturbation of the form $\theta(s)\to \theta(s)+\epsilon (s)$, we obtain the first-order constrained critical point condition $0=\delta (I -\lambda_1 J -\lambda_2 J_2)= \int_{s=0}^{s=L} \{(\frac{ d\theta}{ds}) (\frac{d\epsilon}{ds} ) - [Cs (\cos \theta ) -\lambda_1 \sin \theta + \lambda_2 \cos \theta ] \epsilon(s)\} ds$ from which one deduces (as usual, after integration by parts) that $$\frac { d^2 \theta}{ds^2} =- [C s\cos \theta -\lambda_1 \sin\theta +\lambda_2 \cos \theta]$$
A special limiting case
In the extreme case where $C=0$ the previous equation simplifies to $ \theta''= -\lambda_1 \sin \theta + \lambda _2 \cos \theta$. The RHS can be written in amplitude & phase-shift format as $ -A \sin (\theta+\phi)$ for some constant phase shift $\phi = \arctan{\lambda_2/\lambda_1}$ and $A=\sqrt{\lambda_1^2+\lambda_2^2}$. Thus after setting $ \psi= \theta +\phi$ this reduces to the nonlinear oscillator model for a simple pendulum accelerated by gravity: $\psi''= -A\sin\psi$.
The linearized model
As a first approximation, valid for bridges of modest tilt $\theta$ and curvature $\kappa$, one can use the small angle approximation $\cos \theta \approx 1$ and $\sin \theta \approx \theta$ to obtain the linearized equation $$\theta'' = - Cs - \lambda_1 \theta +\lambda_2$$ After an affine change of the arclength-variable $s$ this linearized equation takes the form of a forced simple harmonic oscillator: $$\theta'' +\lambda_1 \theta =- Cs$$ The general solution of the linearized equation is a sum of a phase-shifted sinuisodal function of $s$ (oscillating with frequency $\sqrt{\lambda_1}$) and the linear function $ms$ where $m= \frac{- C}{\lambda_1}$. That is, $\theta(s) = A \sin (\sqrt{\lambda_1} s+\phi ) + ms$. In this linearized case, the curvature $\kappa= \frac{d\theta}{ds} = m + A\sqrt{\lambda} \cos \sqrt{\lambda_1} s +\phi)$ If the amplitude $A\sqrt{\lambda}$ of the oscillating term is small in comparison to $m$, the curvature is of constant sign. However, if the amplitude of the oscillating term is large compared to $m$, the curvature changes sign; that is the structure has inflection points. In other words, the bridge develops corrugations.
Solutions of the fully nonlinear model Here is an example of a numerical solution for the nonlinear equation. The plotted function is $\kappa(s)$. The sign changes indicate inflection points.
Note that the full shape of the curve can be found by integrating the vector equation $ \frac{ dP}{ds} = T(\theta(s)) =<\cos \theta, \sin \theta> $
And here is expanded Mathematica code that solves for the position vector as well. The picture shows an example of a very floppy piece of paper. (The relevant portion of the solution is that first hump that lies above the line $y=0$.) Note that the solution cannot be written in the format $y= f(x)$ since it curves back on itself. This illustrates that the modeling assumption $y=y(x)$ is generally too restrictive.