Harnack's inequality

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 every point x with |x − x₀| = r < R,

${\displaystyle {\frac {1-(r/R)}{[1+(r/R)]^{n-1}}}f(x_{0})\leq f(x)\leq {1+(r/R) \over [1-(r/R)]^{n-1}}f(x_{0}).}$

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

${\displaystyle {R-r \over R+r}f(x_{0})\leq f(x)\leq {R+r \over R-r}f(x_{0}).}$

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

${\displaystyle \sup _{x\in \omega }u(x)\leq C\inf _{x\in \omega }u(x)}$

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

By Poisson's formula

${\displaystyle f(x)={\frac {1}{\omega _{n-1}}}\int _{|y-x_{0}|=R}{\frac {R^{2}-r^{2}}{R|x-y|^{n}}}\cdot f(y)\,dy,}$

where ω_{n − 1} is the area of the unit sphere in Rⁿ and r = |x − x₀|.

Since

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

the kernel in the integrand satisfies

${\displaystyle {\frac {R-r}{R(R+r)^{n-1}}}\leq {\frac {R^{2}-r^{2}}{R|x-y|^{n}}}\leq {\frac {R+r}{R(R-r)^{n-1}}}.}$

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 its value at the center of the sphere:

${\displaystyle f(x_{0})={\frac {1}{R^{n-1}\omega _{n-1}}}\int _{|y-x_{0}|=R}f(y)\,dy.}$

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:

${\displaystyle \sup u\leq C(\inf u+\|f\|)}$

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

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

Let ${\displaystyle {\mathcal {M}}}$ be a smooth (bounded) domain in ${\displaystyle \mathbb {R} ^{n}}$ and consider the linear elliptic operator

${\displaystyle {\mathcal {L}}u=\sum _{i,j=1}^{n}a_{ij}(t,x){\frac {\partial ^{2}u}{\partial x_{i}\,\partial x_{j}}}+\sum _{i=1}^{n}b_{i}(t,x){\frac {\partial u}{\partial x_{i}}}+c(t,x)u}$

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

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

such that

${\displaystyle \quad u(t,x)\geq 0{\text{ in }}(0,T)\times {\mathcal {M}}.}$

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

${\displaystyle \sup _{K}u(t-\tau ,\cdot )\leq C\inf _{K}u(t,\cdot ).}$

• Harnack's theorem
• Harmonic function

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