Minimisation functional problem $$ \min_{y} \int _ {(-\beta,0)}^A\sqrt{(\dot{y}(x))^2+1} dx \mbox{ such that } x^2+y(x)^2\geq R^2$$ Scaling everything, take $R=1, \beta >1$

Starting at $(-\beta,0)$ find the path with minimum distance taking you to $A=(x_0,y_0), y_0 \geq 0 \mbox{ wlog }$ without going through the interior of the unit circle.

Let $ \alpha $ be the angle between the x axis and the line segment joining $ (-\beta,0)$ with $ A $. If $ \sin \alpha \geq 1/ \beta $, then the curve is that line segment.

You can try the problem with a square, a rhombus any compact set you like. In fact:

Generalisation: find a curve that minimises the distance between two points $A(x_A,y_A)$ and $ B(x_B,y_B) $ that doesn't go through some obstacles. The obstacles are open bounded sets $B_i$, with the curve lying in $ \mathbb{R} ^2 \setminus \cup_{i=1}^{\infty} B_i$.

My difficulty: I am not familiar with inequality constraints. Any advice on how to tackle these problems either analytically or computationally would be appreciated.

  • 4
    $\begingroup$ Draw a line through the two points. If it avoids the circle, you're done. Else, the shortest path lies on the side of the circle on which the line passes the centre; it consists of the tangents from the points to the circle, up to the point of tangency, and the arc on the circle between the points of tangency. $\endgroup$ – joriki Jul 22 '18 at 16:45
  • $\begingroup$ You need a $\ge$ in the inequality constraint or you won't attain the minimum described in the above comment $\endgroup$ – Calvin Khor Jul 22 '18 at 16:54
  • $\begingroup$ I edited my question @Calvin Khor $\endgroup$ – George Aliatimis Jul 22 '18 at 17:15
  • $\begingroup$ Tightening a string seems to be a good way of dealing with those problems. Can it be made more rigorous? $\endgroup$ – George Aliatimis Jul 22 '18 at 17:17

Considering the problem

$$ \min_{y} \int _ {(-\beta,0)}^A\left(\sqrt{(\dot{y}(x))^2+1}\right) dx \mbox{ such that } x^2+y^2\ge R^2 $$

we will try to give an informal introduction to the handling of this problem with the help of slack variables $\epsilon(x)$ and lagrange multipliers $\lambda_1(x),\lambda_2(x),\lambda_3(x)$

First, the inequality is converted into an equivalent equality with the contribution of the slack variable

$$ x^2+y^2 \ge R^2 \equiv x^2+y^2-R^2-\epsilon^2(x) = 0 $$

and then the main functional reads

$$ F(x,z,y,y',\epsilon,\epsilon')=\sqrt{z^2+1}+\lambda_1(x^2+y^2-R^2-\epsilon^2)+\lambda_2(2x+2y -2\epsilon \epsilon'z)+\lambda_3(z-y') $$

Note the inclusion as restriction also the derivative of $x^2+y^2-R^2-\epsilon^2=0$

Now applying the Euler-Lagrange conditions for stationarity we have

$$ F_y-\left(F_{y'}\right)'=2\lambda_1y+2\lambda_2 z+\lambda_3'=0\\ F_z-\left(F_{z'}\right)'=2\lambda_2y+\frac{z}{\sqrt{z^2+1}}+\lambda_3 = 0\\ F_{\epsilon}-\left(F_{\epsilon'}\right)'=2\epsilon(\lambda'_2-\lambda_1)=0 $$

Assuming $\epsilon(x) \ne 0$ we have $\lambda_1=\lambda'_2$ and the equations reduce to

$$ (2\lambda_2 y)'+\lambda'_3=0\Rightarrow 2\lambda_2 y +\lambda_3 = C_1 $$

and then

$$ 2\lambda_2 y +\lambda_3 = C_1\\ 2\lambda_2y+\frac{y'}{\sqrt{(y')^2+1}}+\lambda_3 = 0 $$


$$ C_1 = \frac{y'}{\sqrt{(y')^2+1}} $$

with solution

$$ y = \pm \frac{C_1 x}{\sqrt{1-C_1^2}}+C_2 $$

Concluding, for $\epsilon(x)\ne 0\;$ the solutions are $y = y^0_i+m_i(x-x_0^i)\;$ for $i\in \{1,2\}$

The solution for $\epsilon(x) = 0\;$ involves the angular circle segment.

  • $\begingroup$ @GeorgeAliatimis I am applying the variations to a family of functions $y_*$. The first integral can be presented as two boundary conditions which satisfy the line functions. This is only a non rigorous presentation, handling the functional tools and the calculus of variations. It is not my purpose to develop a full theory about the theme. Regards. $\endgroup$ – Cesareo Jul 22 '18 at 19:40
  • $\begingroup$ @GeorgeAliatimis I hope now it looks better. I am intending to introduce some ideas about how to handle state inequality restrictions. $\endgroup$ – Cesareo Jul 26 '18 at 10:20

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