second order – Calculus VII

Imagine a circular (or cylindrical, in cross-section) object being supported by an elastic string. Like this:

To actually compute the equilibrium mass-string configuration, I would have to take some values for the mass of the object and for the resistance of the string. Instead, I simply chose the position of the object: it is the unit circle with center at ${(0,0)}$ . It remains to find the equilibrium shape of the string. The shape is described by equation ${y=u(x)}$ where ${u}$ minimizes the appropriate energy functional subject to boundary conditions ${u(-2)=0=u(2)}$ and the obstacle ${u(x)\le -\sqrt{1-x^2}}$ . The functional could be the length

$\displaystyle L(u) = \int_{-2}^2 \sqrt{1+u'(x)^2}\,dx$

or its quadratization

$\displaystyle E(u) = \frac12 \int_{-2}^2 u'(x)^2 \,dx$

The second one is nicer because it yields linear Euler-Lagrange equation/inequality. Indeed, the obstacle permits one-sided variations ${u+\varphi}$ with ${ \varphi\le 0}$ smooth and compactly supported. The linear term of ${E(u+\varphi)}$ is ${\int u'\varphi'}$ , which after integration by parts becomes ${-\int u'' \varphi}$ . Since the minimizer satisfies ${E(u+\varphi)-E(u)\ge 0}$ , the conclusion is ${\int u'' \varphi \le 0 }$ whenever ${\varphi\le 0}$ . Therefore, ${u''\ge 0}$ everywhere (at least in the sense of distributions), which means ${u}$ is a convex function. In the parts where the string is free, we can do variation of either sign and obtain ${u''=0}$ ; that is, ${u}$ is an affine function there.

The convexity of ${u}$ in the part where it touches the obstacle is consistent with the shape of the obstacle: the string can assume the same shape as the obstacle.

The function ${u}$ can now be determined geometrically: the only way the function can come off the circle, stay convex, and meet the boundary condition is by leaving the circle along the tangents that pass through the endpoint ${(\pm 2,0)}$ . This is the function pictured above. Its derivative is continuous: Lipschitz continuous, to be precise.

First derivative is Lipschitz continuous

The second derivative does not exist at the transition points. Still, the minimizer has a higher degree of regularity (Lipschitz continuous derivative) than a generic element of the function space in which minimization takes place (square-integrable derivative).

As a bonus, the minimizer of energy ${E}$ turns out to minimize the length ${L}$ as well.

All in all, this was an easy problem. Next post will be on its fourth-order version.

Tag: second order

Second order obstacle problem