2
$\begingroup$

As extension of this question, I was wondering what would be the Euler-Lagrange equations associated with the functional

$$ E(u) = \frac{1}{2}\int_{\gamma} \lVert \nabla u \rVert^2 ds $$

The difference is that the integral is not defined in a region but on a curve, and the gradient $\nabla u$ in this case is defined by

$$ \nabla u = \begin{pmatrix} u_x(x,y) \\ u_y(x,y) \end{pmatrix} = \begin{pmatrix} u_x(x(t),y(t)) \\ u_y(x(t),y(t)) \end{pmatrix} $$

However the curve $\gamma$ is fixed, it's not the unknown, what I want to find is a function $u$ (or pixel values) defined on that curve.

So the argument of the integral is not $\mathcal{L}(x,y,u,u_x,u_y)$ but $\mathcal{L}(t,x,y,u,u_x,u_y)$ where $t$ is a the curve parameter and $x = x(t),y = y(t)$ are known functions of $u$, but I'm totally confused how to derive the EL equations in this case.

The clue I had was to substitute the gradient with the directional derivative, but I want to consider the whole region sorrounding the curve $\gamma$ which I would be missing.

Here $ds = \sqrt{x'^2 + y'^2} dt$.

Can anyone help me out working out the math?

Just an observation, it might be possible that maybe what I want is actually minimizing the functional

$$ E(u) = \frac{1}{2} \int_{\gamma} | \left\langle \nabla u, d \gamma \right\rangle | $$

Because my functional depends on the gradient defined on the curve, so I guess directional derivative is more natural.

Thank you

Maybe I have made a small progress, first of all I observe that if $\gamma = \gamma(t), t \in [a,b]$ I have

$$ E(u) = \frac{1}{2} \int_a^b | \left\langle \nabla u, d\gamma \right\rangle | = \frac{1}{2} \int_a^b | du | = \frac{1}{2} \int_{a}^{b} | u' | dt = \int_{a}^b \mathcal{L}(u') dt $$

Now I can apply the EL equations for single variable, single function

$$ \frac{d \mathcal{L}}{du} - \frac{d}{dt} \frac{d \mathcal{L}}{du'} = 0 $$

Using distributional derivatives and the fact that $\mathcal{L} = \mathcal{L}(u')$ I get

$$ 0 = \frac{d \mathcal{L}}{du} - \frac{d}{dt} \frac{d \mathcal{L}}{du'} = - \frac{d}{dt} \frac{d \mathcal{L}}{du'} = -\frac{1}{2}\frac{d}{dt} sgn(u') = - \delta(u') u'' \Rightarrow \delta(u') u'' = 0 $$

Therefore with appropriate boundary conditions I need to solve

$$ \delta(u') u'' = 0 $$

However now since I have the dirac delta in the equation I'm not sure how I can get rid of it.

$\endgroup$
  • $\begingroup$ Yes, your last formula seems geometrically much more natural than your first formula. $\endgroup$ – Qmechanic Oct 27 '18 at 16:20
  • $\begingroup$ But is there a specific form of the EL equations? $\endgroup$ – user8469759 Oct 27 '18 at 16:23
  • $\begingroup$ I think this paper (citeseerx.ist.psu.edu/viewdoc/…) might be related (section 3), though the problem seems kind of the other way around i.e. given the gradient find an optimal curve such that. $\endgroup$ – user8469759 Oct 29 '18 at 10:15
  • $\begingroup$ Consider to provide a permalink/DOI number to paper if possible. $\endgroup$ – Qmechanic Oct 29 '18 at 15:43
0
$\begingroup$

I believe there're two ways to set up this problem (which I guess would provide similar results anyway).

Suppose $\gamma$ is a simple parametric open curve (with uniform parametrization), we want to find $u$ such that the integral

$$ E(u) = \frac{1}{2} \int_{\gamma} | \left\langle \nabla u, \hat{t} \right\rangle | ds $$

is minimized, here $\hat{t} = \hat{t}(s)$ denotes the unit tangent vector in $\gamma$ at coordinate $s$, we observe that

$$\left\langle \nabla u, \hat{t} \right\rangle = \frac{du}{ds}$$

(because of the total derivative). Therefore the energy function might be rewritten as

$$ E(u) = \frac{1}{2} \int_{\gamma} | \left\langle \nabla u, \hat{t} \right\rangle | ds = \frac{1}{2} \int_{a}^{b} |u'| ds = \int_{a}^{b} \mathcal{L}(u')ds $$

the Euler Lagrange equations are given by

$$ 0 = \frac{\partial \mathcal{L}}{\partial u} - \frac{d}{ds}\frac{\partial \mathcal{L}}{\partial u'} = - \frac{1}{2} \frac{d}{ds} sign(u') = - \frac{1}{2} \frac{d}{ds} sign\left(\left\langle \nabla u, \hat{t} \right\rangle \right) $$

such equation with appropriate boundary conditions allow to solve the problem.

A similar problem is to minimize the integral

$$ E(u) = \frac{1}{2} \int_{\gamma} \left| \left\langle \nabla u, \hat{t} \right\rangle \right|^2 ds = \frac{1}{2} \int_a^b \left| u' \right|^2 ds = \int_{a}^{b} \mathcal{L}(u') ds $$

The Euler lagrange equations in this case are given by

$$ -\frac{d}{ds} \left\langle \nabla u,\hat{t} \right\rangle = 0 $$

And with appropriate boundary conditions even in this case we can solve the problem. Boundary conditions are given by some conditions fixed at the extremes of the curve, namely $\gamma(a)$ and $\gamma(b)$.

Implementing the second case should be very straight forward since it would involve setting up a linear system with appropriate boundary conditions, the directional derivative can be computed using the sobel operators multiplied by the weights given by the tangent unit vector $\hat{t}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.