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:

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

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:

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 ${u(x)-(-\sqrt{1-x^2})}$:

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

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.

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:

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

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

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

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.

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.