I want to solve the problem: Find the curve satisfies following conditions.

  1. Minimize the functional $J$
  2. The coordinates of the start/end points are given
  3. Direction(tangential vector) at the start/end points are given
  4. The length of the curve is $L$

I parameterized the curve using $\theta(s)$. $s$ is the length from the start point, and $\theta$ is the angle of the tangential vector at $s$.

The functional $J$ is defined as following. $$ J[\theta] = \int_0^L f(\theta, \theta'; s)ds \qquad \left(\theta'\equiv \frac{d\theta}{ds}\right) $$

The curve $(x(s), y(s))$ can be writen using $\theta$ $$ x(s) = \int_0^s \cos(\theta(s))ds \\ y(s) = \int_0^s \sin(\theta(s))ds $$

Then, the second condition becomes following isoperimetric constraints. (set the start point as origin.) $$ x(L) = \int_0^L \cos(\theta)ds = p_x\\ y(L) = \int_0^L \sin(\theta)ds = p_y $$

condition 3, and 4 is already considered by the definitions.

Therefore the problem is equivalent to the euler lagrange equation with two isoperimetric constraints.

I wanted to verify this solution, so took it to my acquaintance. He said that the solution is wrong, but I can't understand.

Is there any wrong part?

  • $\begingroup$ Could you please clarify why conditons 3 and 4 woud already be considered by the definitions? $\endgroup$ – An aedonist Sep 26 '17 at 9:16
  • $\begingroup$ @Anaedonist direction at start/end point is equivalent to \theta(0) and \theta(L). It specifies solution of the Euler-Lagrange equation. $\endgroup$ – damhiya Sep 26 '17 at 9:24
  • $\begingroup$ @Anaedonist The length of the curve is \int \sqrt (dx^2 + dy^2) = \int \sqrt(x'^2+y'^2)ds = \int \sqrt(cos^2 \theta + sin^2 \theta)ds = \int ds = L $\endgroup$ – damhiya Sep 26 '17 at 9:33

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