I have a paper which lies on a flat surface. The paper is fixed on one side and the opposite side can slide in the direction of the opposing side. As side end slides toward the other, a "bump" forms. I want to know what the solution is to the shape of this bump. There must be some standard solution for this case.

EDIT: Luboš Motl did provide a really nice answer for the case when the paper is assumed to lie flat against the surface at the ends (also with the assumption that the bump is small). I also found this interesting paper on a similar topic.


A very natural and interesting question.

If the paper stays nearly flat, we may use the linearized approximation. I will neglect the gravitational potential energy because it's probably much smaller than the bending energy (note that the shape of the bent paper is almost the same when the desk is vertical as when it is horizontal).

The paper doesn't want to bend much, so the energy contains a term that punishes the second derivatives of $y$ (the curvature) $$E = K \int_0^L {\rm d}x\,(y'')^2$$ where $y'\equiv dy/dx$, and so on. We want to keep the length of the paper fixed - it's prescribed by the way how you attach it at the end points. In the linear approximation, the length is a linear function of $$D = \int_0^L {\rm d} x\, (y')^2 $$ which may be seen, if you need it, by a Taylor expansion of the exact expression for the length of the graph, $\int (1+y^{\prime 2})^{1/2}{\rm d}x$. Add it with a Lagrange multiplier, requiring $$\delta (E+\lambda D) = 0.$$ I think that the resulting equations say $$ y'''' = \frac{\lambda}{K} y''.$$ Note that in the Euler-Lagrange equations, all the primes "clump". If you write $Y=y''$, it says that the second derivative of $Y$ is proportional to $Y$ itself. The physically relevant solution is $$ Y = Y_0 \cos (2\pi x n / L) $$ where only $n=1$ is possible if there is a table underneath the paper. Consequently, $$ y = y_0 [1-\cos (2\pi x / L)],$$ too. I added the term $1$ as an integration constant (while the other is zero) to guarantee that $y=0$ and $y'=0$ at the boundaries.

So what you get in practice is one wave of a cosine, from one minimum to the next one.

A more detailed discussion would be needed to eliminate the other cosine-like function with the same frequency, the sine, as well as linear functions, and to discuss whether the exponentially increasing/decreasing solutions may ever be relevant. Well, it wouldn't be too difficult. A more general solution for the same (cosine-like) sign of $\lambda$ would be $$ y = y_0 [1-\cos (\phi_0+ 2\pi x N / L)] + Ax^2+B,$$ and the four conditions $y=0$, $y'=0$ at $x=0,L$ as well as the fixed length of the paper would imply $N=1$, $\phi_0=0$, $A=0$, $B=0$, as well as the right normalization $y_0$. As suggested above, the conditions would have other solutions where $N=2,3,4\dots$ but these solutions wouldn't satisfy $y\geq 0$ for $0\leq x \leq L$, so the paper couldn't arrange itself above the table. If there were a hole in the table, these higher harmonics (several periods of the wave containing) solutions would be possible - but I guess that they would be unstable.

I am not sure whether I would be able to solve it analytically if the angle $y'$ were not infinitesimal. It seems clear that the exact solution for significant angles is not just a cosine: the bent paper tends to resemble a balloon - for which $y(x)$ is not even single-valued - when there's too much redundant paper in the middle.

On the other hand, if you sharply bend the paper at $x=0,L$ so that $y'$ may be arbitrary at these two points, the paper in between will be bent as an arc of a circle - see another answer below - and this result is accurate even if $y'$ is of order one. Note that in the linearized approximation, the arc gives $y$ being a quadratic function of $x$ so that $y''=0$ still solves the fourth-order equation above.

  • $\begingroup$ Hey, great to see that my intuition still works well 20 years later... ;-) $\endgroup$ – ysap May 16 '11 at 16:43
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    $\begingroup$ Dear ysap, excellent to hear about it. But is it really intuition, or just plain good old memory that remembers someone else's intuition or, more likely, calculation? ;-) $\endgroup$ – Luboš Motl May 16 '11 at 16:44
  • $\begingroup$ Well, I guess it's both. Honestly, this specific course was one of my least favorites (to say the least) while studying Aerospace Eng. What I did vaguely remembered is that the this problem is modeled by a simple diff. eqn., and that the shape of the buckled beam is an eigenfunction of that eqn., which is harmonic. Testing it with an actual sheet of paper, I assured myself that it is a full cycle of the harmony (due to the table B.C., as you described), and not just half of it. $\endgroup$ – ysap May 16 '11 at 16:52
  • $\begingroup$ @Lubos Pardon my ignorance, but I did not understand the statement "The paper doesn't want to bend much, so the energy contains a term that punishes the second derivatives of y (the curvature) " $\endgroup$ – yayu May 16 '11 at 16:58
  • $\begingroup$ Dear @yayu, the $(y')^2$ term vanishes if the paper is straight. If it is curved at a given point $x$, the term is nonzero. It's quadratic because you may really imagine that there are springs on both sides of the paper, trying to keep is straight, and the $(y')^2$ term generalizes the $kx^2/2$ potential energy of a spring. $\endgroup$ – Luboš Motl May 16 '11 at 19:09

You gave some boundary contraints, but not enough to solve this problem without some assumptions on the part of the answerer. The missing piece is: For each end; is the end...

1.) constrained to lie flat against the table, or

2.) allowed to pivot up.

If the answer is 1 for either or both ends, the answer is non-trivial.

If the answer is 2 for BOTH ends, and the weight of the paper is non-negligible, then, again the answer is non-trivial.

If the answer is 2 for BOTH ends -AND- the weight of the paper is negligible, then the paper will bend in the arc of an almost perfect circle. With the distace between two ends = paper height, the circle radius is infinity. With the distance between the ends = 0, the radius is paper height divided by (2*Pi); and the arc is the complete circle.

EDIT, based on fearlesscoder's, Lubos' and Ysap's comments:

If the paper is constrained to lie in the plain of the table at the ends (dy/dx = 0), then the answer could indeed be (or approximate) a complete cycle of a cosine wave (with an offset in the vertical axis); as long as the ends are constrained to not approach closer than L/8 to each other, where L is the length of the sheet. Not sure what to do as the ends of the paper get really close to one another: the cosine thingy breaks down in that region.

It would not, in any case (where the ends are constrained to lie flat), resemble a circle, nor a segment of a circle, nor a quadratic nor a Gaussian curve (which is the first thing that popped to mind when I realized exactly what boundary conditions the OP meant).

I would make the assumption that the resistance of the sheet to bending is proportional to the radius of curvature (((d^2)x)/(dy^2) = ((d^2)y)/(dx^2)). I would make the peak of the bump at x=0. I would have to put in a lot of thought getting to the point where I can find out if y=cos(x) is a sol'n to that problem.

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    $\begingroup$ I think that the effect of the paper weight can be eliminated (at least, reduced to a great extent) by considering a vertical setup, where the paper is bound to a vertical surface and is pressed from both sides. Other than that, real life solutions are never trivial, but simple, and pretty accurate models do exist frequently. @Lubos' answer develops such a model and solution that show the shape to be harmonic, and not circular. $\endgroup$ – ysap May 16 '11 at 17:58
  • $\begingroup$ Dear @Vintage, I agree that if one is allowed to sharply bend the paper at $x=0$ and $x=L$, then the paper in between will make an arc of a circle, or - in my approximation - $y=ax^2+bx+c$. This is a sort of trivial solution. I think that the OP meant 1) that the paper is never sharply bent and it continues as flat plane on both end points. $\endgroup$ – Luboš Motl May 16 '11 at 20:35
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    $\begingroup$ never mind the hand waving, how about proving this mathematically. $\endgroup$ – Larry Harson May 17 '11 at 11:05
  • $\begingroup$ @ user2146 - No, thank you. As I said, the sol'n is non-trivial. $\endgroup$ – Vintage May 17 '11 at 17:06
  • $\begingroup$ Sorry about the lack of boundary constraints, your 1.) was the one I had in mind, though the other one is interesting as well. $\endgroup$ – Krumelur May 17 '11 at 17:17

I'm quite sure that the idealized shape is one cycle of a cosine. This is due to the differential equations governing the buckling of thin structures (beams).

However, there has been many, many years since I studied this kind of stuff, so I don't have reference at hand.

  • $\begingroup$ What if you have a shape that does not support bending moments (zero thickness paper)? It will still form a shape due to the conservation of mass, but the EI component will not be there... Interesting! $\endgroup$ – John Alexiou May 16 '11 at 17:54
  • $\begingroup$ @ja72 - I am not sure, but I think that the simple model assumes zero thickness. $\endgroup$ – ysap May 16 '11 at 18:01

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