Functionals

Figure 1: Illustration of a rope or chain hanging from two anchor points. The differential length on the rope is \(ds=\sqrt{dx^2+dy^2}\). The shape of the rope will be such that its potential energy is minimized.

The potential energy of the differential length \(ds\) is \(d g\, y(x) ds\), where \(d\) is the mass density of the rope, and \(g\) is the gravitational acceleration. We can compute the potential energy by integrating over the length to get v=d g \(\propto\int_a^by(x) ds\), where \(ds=\sqrt{dx^2+dy^2}=\sqrt{1+y'^2}dx\). This is also how \(y'\equiv\frac{\partial y}{dx}\) naturally shows up is such problems.

Both ends fixed

Figure 2: The blue curve \(y(x)\), which is unknown, gives the minimum value for the functional \(v\). The orange curve represents random deformations around \(y(x)\). The variation \(\alpha \eta(x)\) must vanish at the end points since the value \(y\) at the end points are fixed.

Keeping the boundary conditions in Eq. (4) in mind, let us calculate Eq. (3): \[\begin{eqnarray} \frac{d v(\alpha)}{d\alpha}\bigg \vert_{\alpha=0}&=&\int^{x_1}_{x_0} \frac{d}{d\alpha}\mathscr{L}(x,y+\alpha \eta,y'+\alpha \eta'(x))\bigg \vert_{\alpha=0} dx=\int^{x_1}_{x_0}\left[\frac{\partial}{\partial y}\mathscr{L}(x,y,y') \eta +\frac{\partial}{\partial y'}\mathscr{L}(x,y,y') \frac{d\eta}{dx}\right]dx\nonumber\\ &=&\int^{x_1}_{x_0}\left[\frac{\partial}{\partial y}\mathscr{L}(x,y,y') \eta +\frac{d}{dx}\left(\frac{\partial}{\partial y'}\mathscr{L}(x,y,y')\eta\right)- \frac{d}{dx}\left(\frac{\partial}{\partial y'}\mathscr{L}(x,y,y')\right)\eta\right]dx\nonumber\\ &=&\int^{x_1}_{x_0}\left[\frac{\partial\mathscr{L}}{\partial y} -\frac{d}{dx}\left(\frac{\partial\mathscr{L}}{\partial y'}\right)\right]\eta dx +\frac{\partial\mathscr{L}}{\partial y'}\eta\bigg \vert_{x_0}^{x_1}=\int^{x_1}_{x_0}\left[\frac{\partial\mathscr{L}}{\partial y} -\frac{d}{dx}\left(\frac{\partial\mathscr{L}}{\partial y'}\right)\right]\eta dx, \tag{5} \end{eqnarray}\] where the boundary terms become \(0\) due to the constraints in Eq. (4). Since \(\eta\) is an arbitrary function, in order to set this equation to \(0\), we require the following:

\[\begin{eqnarray} \frac{\partial\mathscr{L}}{\partial y} -\frac{d}{dx}\left(\frac{\partial\mathscr{L}}{\partial y'}\right)=0. \tag{6} \end{eqnarray}\] Equation (6) is known as the Euler-Lagrange equation and the function \(\mathscr{L}\) is called the Lagrangian. In the case of the hanging rope, we have \[\begin{eqnarray} \mathscr{L}=d g y \sqrt{1 +y'^2}. \tag{7} \end{eqnarray}\] We can plug this into Eq. (6) and solve the resulting differential equation for \(y(x)\). However, it is easy to see that it will be a second order differential equation. It won’t be too hard to solve, but we can do better than that. The crucial observation is that \(\mathscr{L}\) has no explicit \(x\) dependence, i.e., \(\frac{\partial\mathscr{L}}{\partial x}=0\). This means the total derivative of \(\mathscr{L}\) can be written as: \[\begin{eqnarray} \frac{d\mathscr{L}}{dx}=\frac{\partial\mathscr{L}}{\partial x}+\frac{\partial\mathscr{L}}{\partial y} y'+\frac{\partial\mathscr{L}}{\partial y'} y''=\cancel{\frac{\partial\mathscr{L}}{\partial x}}+\frac{\partial\mathscr{L}}{\partial y} y'+\frac{\partial\mathscr{L}}{\partial y'} y''=\frac{\partial\mathscr{L}}{\partial y} y'+\frac{\partial\mathscr{L}}{\partial y'} y''. \tag{8} \end{eqnarray}\] We can create these terms out of the Eq. (6) if we multiply it with \(y'\): \[\begin{eqnarray} \frac{d\mathscr{L}}{\partial y}y' -\frac{d}{dx}\left(\frac{d\mathscr{L}}{\partial y'}\right)y'&=&\frac{d\mathscr{L}}{\partial y}y' -\frac{d}{dx}\left(\frac{d\mathscr{L}}{\partial y'}y'\right)+\frac{d\mathscr{L}}{\partial y''}y''=\frac{d}{dx}\left( \mathscr{L}-\frac{d\mathscr{L}}{\partial y'}y'\right)=0, \tag{9} \end{eqnarray}\] which means that \[\begin{eqnarray} \mathscr{L}-\frac{d\mathscr{L}}{\partial y'}y'=C. \tag{10} \end{eqnarray}\] This reduces the order of the differential equation from two to one! Inserting the expression for \(\mathscr{L}\) into Eq. (7) we get: \[\begin{eqnarray} y \sqrt{1 +y'^2}-\frac{y y'2}{ \sqrt{1 +y'^2}}=\frac{y}{\sqrt{1 +y'^2}}=C. \tag{10} \end{eqnarray}\] It is best to solve the equation in a parametric form by defining \(y'=\sinh t\) \[\begin{eqnarray} y=C\sqrt{1 +y'^2}=C\sqrt{1 +\sinh^2 t}=C\cosh t. \tag{11} \end{eqnarray}\] We can extract \(x(t)\) as follows: \[\begin{eqnarray} \frac{dy}{dt}=C\sinh t=\frac{dy}{dx}\frac{dx}{dt}=\sinh t\frac{dx}{dt} \rightarrow \frac{dx}{dt}=C, \tag{12} \end{eqnarray}\] which gives \(x=C t+D\). We can eliminate \(t\) in favor of \(x\): \(t=\frac{x-D}{C}\) and put it back it \(y(t)\) to get \[\begin{eqnarray} y(x)=C \cosh \left(\frac{x-D}{C}\right). \tag{13} \end{eqnarray}\] \(C\) and \(D\) are the integration constants and they can be fixed by requiring that \(y(x_0)=y_0\) and \(y(x_0)=y_1\), i.e., the anchor points are fixed.

If you think about this practically, you will notice a problem. You have a rope and you can decide on the anchor points as you wish. That completely fixes all the constants. How about the length of the rope, though? A longer rope will definitely have a different shape than that of a shorter one. There should have been another parameter in our solution so that it can be adjusted to give the correct length. That is why we have to introduce a Lagrange multiplier to address variation problems with constraints. In this case the constraint is that the solution should give the correct length: \(\int_{x_0}^{x_0} ds=L\), \(L\) being the length of the rope. In order to enforce this requirement we revise \(\mathscr{L}\) to \(\mathscr{L}-\lambda \left(\int_{x_0}^{x_0} ds -L\right)\) where \(\lambda\) is the Lagrange parameter. The new Lagrangian can be written as¹ \[\begin{eqnarray} \mathscr{L}=d g (y-\lambda) \sqrt{1 +y'^2}, \tag{14} \end{eqnarray}\] Note that we don’t have to solve the differential equation all over again since the new term just shifts \(y\). Therefore, the final solution is simply a shifted version of previous one: \[\begin{eqnarray} y(x)=\lambda+C \cosh \left(\frac{x-D}{C}\right). \tag{15} \end{eqnarray}\] This makes more sense now: we have a solution with \(3\) free parameters and we have \(3\) conditions [\(2\) end points and the length]. Imposing the conditions we will get a unique solution. Let’s do that by first calculating the length: \[\begin{eqnarray} L&=&\int_{x_0}^{x_0} ds =\int_{x_0}^{x_0} \sqrt{1 +y'^2} dx=\int_{x_0}^{x_0} \sqrt{1 +\sinh^2 \left(\frac{x-D}{C}\right)} dx=\int_{x_0}^{x_0} \cosh \left(\frac{x-D}{C}\right) dx\nonumber\\ &=&C\left[ \sinh \left(\frac{x_0-D}{C}\right)-\sinh \left(\frac{x_0-D}{C}\right)\right]. \tag{16} \end{eqnarray}\] Along with the end point requirements, we have the following conditions: \[\begin{eqnarray} y(x_0)=\lambda+C \cosh \left(\frac{x_0-D}{C}\right)&\equiv& y_0\nonumber\\ y(x_0)=\lambda+C \cosh \left(\frac{x_0-D}{C}\right)&\equiv& y_1\nonumber\\ C\left[ \sinh \left(\frac{x_0-D}{C}\right)-\sinh \left(\frac{x_0-D}{C}\right)\right]&=&L. \tag{17} \end{eqnarray}\] In principle, these equations can be solved numerically, but we can simplify them a bit. Taking the difference of two lines gives: \[\begin{eqnarray} \frac{y_1-y_0}{C}=\cosh \left(\frac{x_0-D}{C}\right)- \cosh \left(\frac{x_0-D}{C}\right). \tag{18} \end{eqnarray}\] Take its square and subtract the square of the third line: \[\begin{eqnarray} \frac{(y_1-y_0)^2}{C^2} -\frac{L^2}{C^2}&=&\cosh^2 \left(\frac{x_0-D}{C}\right)- 2 \cosh \left(\frac{x_0-D}{C}\right)\cosh \left(\frac{x_0-D}{C}\right)+\cosh^2 \left(\frac{x_0-D}{C}\right)\nonumber\\ &&-\sinh^2 \left(\frac{x_0-D}{C}\right)+2 \sinh \left(\frac{x_0-D}{C}\right)\sinh \left(\frac{x_0-D}{C}\right)-\sinh^2 \left(\frac{x_0-D}{C}\right)\nonumber\\ &=&2\left[1-\cosh\left( \frac{x_0-x_0}{C}\right) \right], \tag{18} \end{eqnarray}\] which still needs to be solved numerically. What we accomplished by going through the algebra was that we reduced the problem from three equation with three unknows to one equation with one unknown, \(C\). Once we solve for \(C\), we can insert it back into Eq. (18) to get \(D\), and finally we can compute \(\lambda\).

One end sliding

Figure 3: The blue curve \(y(x)\), which is unknown, gives the minimum value for the functional \(v\). The orange curve represents random deformations around \(y(x)\). The variation \(\alpha \eta(x)\) does not necessarily vanish at the end points.

Figure 4: A closer look at the variation at the end point \(x_1\). The displacement we are looking for is \(\delta y\bigg \vert_{x_1}\), and it is the vertical distance between the curves at \(x_1\).

As seen from the geometry in Fig. 4, we can write the following equation \[\begin{eqnarray} \delta y\bigg \vert_{x_1}=\delta y_1- y'(x_1)\delta x_1. \tag{23} \end{eqnarray}\] Putting this back in Eq. (22) we get \[\begin{eqnarray} \left[\mathscr{L} -y'\frac{\partial\mathscr{L}}{\partial y'}\right]_{x_1}\delta x_1+\frac{\partial\mathscr{L}}{\partial y'}\bigg \vert_{x_1}\delta y_1=0. \tag{24} \end{eqnarray}\]

If \(\delta x_1\) and \(\delta y_1\) are independent, we need to set the two terms in Eq. (24) to \(0\) individually. However, in most physical problems, the end point is constrained to move on a curve \(y_1=\varphi(x_1)\). In such cases we will have \(\frac{\delta y_1}{\delta x_1}=\varphi'(x_1)\), and Eq. (24) simplifies to \[\begin{eqnarray} \left[\mathscr{L} +\left(\varphi'-y'\right)\frac{\partial\mathscr{L}}{\partial y'}\right]_{x_1}\delta x_1=0, \tag{25} \end{eqnarray}\] which is known as the transversality condition. For the specific case of \(\mathscr{L}\) in Eq. (14) we get

\[\begin{eqnarray} \left[\frac{(y-\lambda)(1+\varphi' y') }{\sqrt{1 +y'^2}}\right]_{x_1}=0. \tag{26} \end{eqnarray}\] Assuming \(y-\lambda\neq0\) at the boundary, the only way to satisfy this equation would require \(\varphi' y'\vert_{x_1}=-1\), that is \(y\) should be orthogonal to the curve \(\varphi\). This is really neat. It all boils down this: the curve will still be a catenary, but when it hits the boundary, it should be perpendicular to it. For example, if you have a vertical post, the chain will be parallel to the ground at the post!

Image processing with Python

This is not that much of a processing: we just want to load the image an overlay a \(\cosh\) curve to see if it fits the rope shape.

  # Fitting a cosh curve on a rope in a photo
  #  06/17/2021, tetraquark@gmail.com
  # tetraquark.netlify.app
  
  from PIL import Image
  import math
  imgid=   'featured.jpg'
  img = Image.open(imgid)
  
  pixels = img.load()
  
  
  leftpostx=550
  leftposty=1280
  
  rightpostx=3100
  rightposty=1350
  
  midx=(leftpostx+rightpostx)/2 +50
  midy=1720
  
  offset=750
  catparameter= img.size[1]-midy +offset
  
  
  def catenary(c,x):
      return c* math.cosh(x/c)
  
  for i in range(img.size[0]):
      for j in range(img.size[1]):
          catpoint=img.size[1]-catenary(catparameter,(i-midx))+offset
  
          if  abs(j-catpoint ) <5 and i>leftpostx and i<rightpostx:
              pixels[i, j] = (0, 255,255)
  
          if ((i-leftpostx)**2+(j-leftposty)**2)**0.5 <20:
              pixels[i, j] = (255, 0, 0)
          if  ((i-rightpostx)**2+(j-rightposty)**2)**0.5 <20:
              pixels[i, j] = (0, 255,0)
          if  ((i-midx)**2+(j-midy)**2)**0.5 <20:
              pixels[i, j] = (0, 0,255)
  img.show()
  img.save('marked_'+imgid)

Figure 8: Drawing a \(\cosh\) curve onto the hanging rope in the original image results in a very good fit.

A test set up

How about the sliding end case? We probably won’t find this out in the wild, so I have to build a test rig. I have a pile of metal shafts that I have been pulling off from dead printers. They are very polished and have low friction. I also found a plastic cylinder that fits perfectly on the shaft, and it slides easily. The curve indeed hits the shaft at \(90^{\circ}\), and I find it very cool!

Figure 9: If one of the ends can freely slide, it will settle down to the point for which the chain leaves from the shaft at an angle of \(90^{\circ}\). The curious cat is for scale.