Fourth order obstacle problem

Having solved the obstacle problem for a string, let us turn to a more difficult one, in which an elastic string is replaced with an elastic rod (or plate, if we are looking at a cross-section). Elastic rods resist bending much in the same way that strings don’t. This can be modeled by minimizing the bending energy

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

subject to the boundary conditions {u(-2)=0=u(2)}, {u'(-2)=0=u'(2)}, and the same obstacle as before: {u(x)\le -\sqrt{1-x^2}}. The boundary conditions for {u'} mean that the rod is clamped on both ends.

As before, the obstacle permits one-sided variations {u+\varphi} with { \varphi\le 0} smooth and compactly supported. The linear term of {B(u+\varphi)} is {\int u''\varphi''}, which after double integration by parts becomes {\int u^{(4)} \varphi}. Since the minimizer satisfies {E(u+\varphi)-E(u)\ge 0}, the conclusion is {\int u^{(4)} \varphi \ge 0 } whenever {\varphi\le 0}. Therefore, {u^{(4)}\le 0} everywhere, at least in the sense of distributions. In the parts where the rod does not touch the obstacle, we can do variation of either sign and obtain {u^{(4)}=0}; that is, {u} is a cubic polynomial there.

So far everything looks similar to the previous post. But the fourth derivative of the obstacle function {-\sqrt{1-x^2}} is {3(4x^2+1)/(1-x^2)^{7/2}}, which is positive. Since the minimizer {u} must satisfy {u^{(4)}\le 0}, it cannot assume the shape of the obstacle. The contact can happen only at isolated points.

Therefore, {u} is a cubic spline with knots at the contact points and at {\pm 2}. The distributional derivative {u^{(4)}} consists of negative point masses placed at the contact points. Integrating twice, we find that {u''} is a piecewise affine concave function; in particular it is continuous. The minimizer will be {C^2}-smooth in {(-2,2)}.

How many contact points are there? If only one, then by symmetry it must be at {x=0}, and the only three-knot cubic spline that satisfies the boundary conditions and passes through {(0,-1)} with zero derivative is {(-1/4)(1+|x|)(2-|x|)^2}. But it does not stay below the obstacle:

Three-knot spline fails the obstacle condition
Three-knot spline fails the obstacle condition

With a smaller circle, or a longer bar, the one-contact (three-knot) spline would work. For example, on {[-3,3]}:

With a longer bar, a three-knot spline solves the problem
With a longer bar, a three-knot spline solves the problem

But with our parameters we look for two contact points. By symmetry, the middle piece of the spline must be of the form {q(x)=cx^2+d}. The other two will be {p(x)=(ax+b)(2-x)^2} and {p(-x)}, also by symmetry and to satisfy the boundary conditions at {\pm 2}. At the positive knot {x_0} the following must hold:

\displaystyle p(x_0)=q(x_0)=-\sqrt{1-x_0^2}, \quad p'(x_0)=q'(x_0)=\frac{-x_0}{\sqrt{1-x_0^2}}, \quad p''(x_0)=q''(x_0)

where the last condition comes from the fact that {u''} is concave and therefore continuous. With five equations and five unknowns, Maple finds solutions in closed form. One of them has {x_0=0}, {a=b=-1/4} as above, and is not what we want. The other has {x_0=(\sqrt{10}-2)/3\approx 0.3874} and coefficients such as {a=-\frac{1}{3645}\sqrt{505875+164940\sqrt{10}}}. Ugly, but it works:

Minimizer
Minimizer

This time, the bar does stay below the obstacle, touching it only at two points. The amount by which it comes off the obstacle in the middle is very small. Here is the difference {u(x)-(-\sqrt{1-x^2})}:

Deviation from the obstacle
Deviation from the obstacle

And this is the second derivative {u''}.

Second derivative is Lipschitz continuous
Second derivative is Lipschitz continuous

Again, the minimizer has a higher degree of regularity (Lipschitz continuous second derivative) than a generic element of the function space in which minimization takes place (square-integrable second derivative).

If the rod is made shorter (and the obstacle stays the same), the two-contact nature of the solution becomes more pronounced.

Shorter rod
Shorter rod

Assuming the rod stays in one piece, of course.

Second order obstacle problem

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

Obstacle problem
Obstacle problem

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
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.

Inner-variational equations

It’s been a while since the last time I posted a post-colloquium post. This one is based on a colloquium given by me, but don’t worry: none of my results are included here.

As a warm-up, consider the (trivial) problem of finding the shortest path between two points a,b\in\mathbb R^n. The naive approach is to minimize the length L(f)=\int_0^1 |f\,'(t)|\,dt among all maps f\colon [0,1]\to \mathbb R^n that are sufficiently smooth and satisfy the boundary conditions f(0)=a and f(1)=b. This turns out to be a bad idea: L(f) is neither strictly convex nor differentiable, and the set of minimizing maps is huge, containing some rather nonsmooth specimen.

The right approach is to minimize the energy E(f)=\int_0^1 |f\,'(t)|^2\,dt. While this functional is not immediately related to length for general maps, it is not hard to see that for minimizing maps f we have E(f)=L(f)^2. Indeed, consider performing the inner variation \widetilde{f}(t)=f(t+\epsilon \eta(t)) where \eta\colon [0,1]\to\mathbb R is smooth and vanishes at the endpoints. Expanding the inequality E(\widetilde{f})\ge E(f), we arrive at \int_0^1 |f\,'(t)|^2\eta'(t)\,dt=0, which after integration by parts yields \frac{d}{dt}|f\,'(t)|^2=0. Thus, minimization of energy enforces constant-speed parametrization, and since for constant-speed maps we have E(f)=L(f)^2, the geometric nature of the variational problem has not been lost.

As a side remark, the inner-variational equation \frac{d}{dt}|f\,'(t)|^2=0 could be written as f\,''\cdot f\,'=0, which is a nonlinear second-order equation.

For comparison, try the first variation \widetilde{f}(t)=f(t)+\epsilon \eta(t) where \eta\colon [0,1]\to\mathbb R^n is smooth and vanishes at the endpoints. Expanding the inequality E(\widetilde{f})\ge E(f), we arrive at \int_0^1 f\,'(t)\cdot \eta'(t)\,dt=0, which after integration by parts yields f\,''\equiv 0, a linear second-order equation. This Euler-Lagrange equation immediately tells us what the minimizing map f is: there is only one linear map with the given boundary values. Obviously, f\,''\equiv 0 is a much stronger statement that f\,''\cdot f\,'=0. However, if f+\epsilon \eta is not admissible for some geometric reason (e.g., the curve must avoid an obstacle), the Euler-Lagrange equation may not be available.

Moving one dimension up, consider the problem of parameterizing a given simply-connected domain \Omega\subset \mathbb C. Now we are to minimize the energy of diffeomorphisms f\colon \mathbb D\to\Omega which is defined, as before, to be the sum of squares of derivatives. (Here \mathbb D is the unit disk.) In complex notation, E(f)=\iint_{\mathbb D}(|f_z|^2+|f_{\bar z}|^2). For definiteness assume f is sense-preserving, that is |f_z|\ge |f_{\bar z}|. Minimizers ought to be conformal maps onto \Omega, but how can we see this from variational equations?

The Euler-Lagrange equation that we get from E(f)\le E(f+\epsilon\eta) turns out to be the Laplace equation \Delta f=0. This is much weaker than the Cauchy-Riemann equation f_{\bar z}=0 that we expect. One problem is that \eta must vanish on the boundary: otherwise f+\epsilon\eta will violate the geometric constraint. We could try to move the values of f in the direction tangent to \partial\Omega, but since does not necessarily make sense since the boundary of \Omega could be something like the von Koch snowflake. And of course, it is not at all clear why the minimum of E must be attained by a diffeomorphism. If the class of maps is expanded to include suitable limits of diffeomorphisms, then it’s no longer clear (actually, not true) that f+\epsilon\eta belongs to the same class. All things considered, the approach via the first variation does not appear promising.

Let’s try the inner variation instead. For small \epsilon the map z\mapsto z+\epsilon\eta is a diffeomorphism of \mathbb D, hence its composition with f is as good a candidate as f itself. Furthermore, since the inner variation deals with the model domain \mathbb D and not with the generic domain \Omega, it is easy to allow modification of boundary values: \eta should be tangent to \partial \mathbb D, i.e., \mathrm{Re}\,\bar z\eta(z) should vanish on the boundary. It takes a bit of computation to turn E(f(z+\epsilon \eta(z))) into something manageable, but the final outcome is remarkably simple. The inner-variational equation says that the function \varphi:=f_z\overline{f_{\bar z}} is holomorphic in \mathbb D and z^2 \varphi is real on the boundary \partial \mathbb D. (Technically speaking, \varphi\, dz^2 must be a real holomorphic quadratic differential.) What can we conclude from this? To begin with, the maximum principle implies that z^2 \varphi is a constant function. And since it vanishes at z=0, the inevitable conclusion is \varphi\equiv 0. Recalling the sense-preserving constraint |f_z|\ge |f_{\bar z}|, we arrive at f_{\bar z}\equiv 0, the desired Cauchy-Riemann equation.

Executive summary

  • Suppose we are to minimize some quantity (called “energy”) that depends on function (or a map) f
  • We can consider applying the first variation f+ \epsilon\eta of the inner variation f\circ (\mathrm{id}+\epsilon \eta). Here \eta is a small perturbation which is applied differently: in the first case it changes the values of f, in the second it shuffles them around.
  • Inner variation applies even in the presence of geometric constraints that make first variation illegal. One example of such constraint is “f must be injective”.
  • Inner-variational equations are quite different from the Euler-Lagrange equations. Even for simple quadratic functionals they are nonlinear.
  • Inner-variational equations are useful because they tell us something about the maps of minimal energy.