带常源项p-泊松方程的Harnack型估计求证

Let's work through this Harnack-type estimate problem for $p$-harmonic functions step by step. Here's a clear breakdown of the setup, key observations, and proof:

Problem Statement #

Let $B = B_1(0) \subset \mathbb{R}^N$, and let $u \geq 0$ be a solution to the $p$-Laplace equation:
$$-\Delta_p u = 1 \quad \text{in } B$$
Additionally, let $f$ denote the unique solution to the Dirichlet problem:
$$
\begin{cases}
-\Delta_p f = 1 \quad \text{in } B \
f = 0 \quad \text{on } \partial B
\end{cases}
$$
We need to show that in the smaller ball $B_{1/2} = B_{1/2}(0)$, the difference $u - f$ satisfies the following Harnack-type estimate:
$$\frac{1}{C}(u - f)(0) \leq u - f \leq C(u - f)(0)$$
where $C$ is a scale-invariant constant depending only on the space dimension $N$. As noted, since $f$ has an explicit form $f(x) = c\left(1 - |x|^{p/(p-1)}\right)$ (with $c$ chosen to satisfy the PDE), this estimate can be extended to analogous bounds directly for $u$.

Key Observations #

First, let's simplify the problem by defining $v = u - f$. Subtracting the PDEs for $u$ and $f$, we immediately get that $v$ is a homogeneous $p$-harmonic function:
$$-\Delta_p v = -\Delta_p u + \Delta_p f = 1 - 1 = 0 \quad \text{in } B$$
This is critical because $p$-harmonic functions satisfy powerful maximum principles and Harnack inequalities that we can leverage.

By the strong maximum principle for $p$-harmonic functions, if $v$ is non-negative (or non-positive) and attains a zero value anywhere inside $B$, then $v$ is identically zero in $B$. Otherwise, $v$ is strictly positive or strictly negative everywhere in $B$—no sign changes allowed in the interior of the domain.

Proof of the Harnack-Type Estimate #

We split into two cases:

1. Trivial case: $v \equiv 0$
   If $u = f$ everywhere in $B$, the estimate holds trivially (both sides equal zero).

2. Non-trivial case: $v \neq 0$
   Assume $v > 0$ (the case $v < 0$ is symmetric—just replace $v$ with $-v$). Since $v$ is a non-negative $p$-harmonic function in $B$, we can apply the standard Harnack inequality for $p$-harmonic functions:

   For any non-negative $p$-harmonic function in a ball $B_R(x_0)$, there exists a constant $C = C(N)$ (depending only on the space dimension) such that for all $y, z \in B_{R/2}(x_0)$,
   $$v(y) \leq C v(z)$$

   For our problem, take $R = 1$ and $x_0 = 0$. Then $B_{1/2}(0)$ is the smaller ball of radius $1/2$ centered at the origin. For any $x \in B_{1/2}(0)$:

   • Applying the inequality with $y = x$ and $z = 0$ gives $v(x) \leq C v(0)$.
   • Swapping roles (setting $y = 0$ and $z = x$) gives $v(0) \leq C v(x)$, which rearranges to $\frac{1}{C}v(0) \leq v(x)$.

   Combining these two inequalities gives exactly the desired bound:
   $$\frac{1}{C}(u - f)(0) \leq (u - f)(x) \leq C(u - f)(0)$$

   The constant $C$ is scale-invariant because it only depends on the dimension (not the size of the ball), and the estimate holds uniformly across $B_{1/2}$.

As noted, since $f$ has an explicit closed-form expression, we can substitute back to get an analogous estimate directly for $u$:
$$f(x) + \frac{1}{C}(u(0) - f(0)) \leq u(x) \leq f(x) + C(u(0) - f(0))$$

内容的提问来源于stack exchange，提问作者Harish