Harnack's inequality

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In mathematics, Harnack's inequality is an inequality relating the values of a positive harmonic function at two points, introduced by A. Harnack (1887). J. Serrin (1955) and J. Moser (1961, 1964) generalized Harnack's inequality to solutions of elliptic or parabolic partial differential equations. Perelman's solution of the Poincaré conjecture uses a version of the Harnack inequality, found by R. Hamilton (1993), for the Ricci flow. Harnack's inequality is used to prove Harnack's theorem about the convergence of sequences of harmonic functions. Harnack's inequality can also be used to show the interior regularity of weak solutions of partial differential equations.

The statement[edit]

A harmonic function (green) over a disk (blue) is bounded from above by a function (red) that coincides with the harmonic function at the disk center and approaches infinity towards the disk boundary.

Harnack's inequality applies to a non-negative function f defined on a closed ball in Rn with radius R and centre x0. It states that, if f is continuous on the closed ball and harmonic on its interior, then for any point x with |x - x0| = r < R

\displaystyle{{1-(r/R)\over [1+(r/R)]^{n-1}}f(x_0)\le f(x) \le  {1+(r/R)\over [1-(r/R)]^{n-1}} f(x_0).}

In the plane R2 (n = 2) the inequality can be written:

{R-r\over R+r} f(x_0)\le f(x)\le {R+r\over R-r}f(x_0).

For general domains \Omega in \mathbf{R}^n the inequality can be stated as follows: If \omega is a bounded domain with \bar{\omega} \subset \Omega, then there is a constant C such that

for every twice differentiable, harmonic and nonnegative function u(x). The constant C is independent of u; it depends only on the domains \Omega and \omega.

Proof of Harnack's inequality in a ball[edit]

By Poisson's formula

\displaystyle{f(x) = {1\over \omega_{n-1}} \int_{|y-x_0|=R} {R^2 -r^2\over R|x-y|^n}\cdot f(y)\, dy,}

where ωn − 1 is the area of the unit sphere in Rn and r = |x - x0|.


\displaystyle{R-r \le |x-y| \le R+r,}

the kernel in the integrand satisfies

\displaystyle{{R -r\over R (R+r)^{n-1}} \le {R^2 -r^2\over R|x-y|^n}\le {R+r\over R(R-r)^{n-1}}.}

Harnack's inequality follows by substituting this inequality in the above integral and using the fact that the average of a harmonic function over a sphere equals it value at the center of the sphere:

\displaystyle{f(x_0)={1\over R^{n-1}\omega_{n-1}} \int_{|y-x_0|=R} f(y)\, dy.}

Elliptic partial differential equations[edit]

For elliptic partial differential equations, Harnack's inequality states that the supremum of a positive solution in some connected open region is bounded by some constant times the infimum, possibly with an added term containing a functional norm of the data:

\sup u \le C ( \inf u + ||f||)

The constant depends on the ellipticity of the equation and the connected open region.

Parabolic partial differential equations[edit]

There is a version of Harnack's inequality for linear parabolic PDEs such as heat equation.

Let \mathcal{M} be a smooth domain in \mathbb{R}^n and consider the linear parabolic operator

\mathcal{L}u=\sum_{i,j=1}^n a_{ij}(t,x)\frac{\partial^2 u}{\partial x_i\,\partial x_j}+\sum_{i=1}^n b_i(t,x)\frac{\partial u}{\partial x_i} + c(t,x)u

with smooth and bounded coefficients and a nondegenerate matrix (a_{ij}). Suppose that u(t,x)\in C^2((0,T)\times\mathcal{M}) is a solution of

\frac{\partial u}{\partial t}-\mathcal{L}u\ge0 in (0,T)\times\mathcal{M}

such that

\quad u(t,x)\ge0 in \quad(0,T)\times\mathcal{M}.

Let K be a compact subset of \mathcal{M} and choose \tau\in(0,T). Then there exists a constant \quad C>0 (depending only on K, \tau and the coefficients of \mathcal{L}) such that, for each \quad t\in(\tau,T),

\sup_K u(t-\tau,\cdot)\le C\inf_K u(t,\cdot).\,

See also[edit]