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The Green's function is defined for a linear differential operator $L$ as the solution of the equation $LG = \delta$, where $\delta$ is Dirac's delta function. A direct consequence of the definition of $G$ is that the solution of the problem $Lu = f$ is the convolution $G*f$, where $G$ is the Green's function.

I am interested to know if there exists a solution to the equation: $$ \Delta_p G = \delta $$ in some bounded domain $\Omega$ with some boundary condition, where $\Delta_p$ is the $p$-Laplacian defined by: $$ \Delta_p u = div (|\nabla u|^{p-2}\nabla u), $$ with $p\neq 2$ (the case $p=2$ is the Laplacian). I know that I won't be able to build solutions of the problem $\Delta_p u = f$ by the convolution $G*f$, because $\Delta_p$ is nonlinear.

I did not find any paper about this problem, so I think that maybe it is a very difficult problem or maybe it is well known that it does not exist a solution. I would appreciate if you enlighten me on this issue.

korina
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1 Answers1

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Let $\omega_d = |\mathbb{S}^{d-1}| = \frac{2\,\pi^{d/2}}{\Gamma(d/2)}$ and $p>\frac{d-1}{d+1}$ verifying $p\neq d-1$. One solution of the equation $$ \Delta_p u = \delta_0 $$ in $\mathbb{R}^d$ is $$ u = \tfrac{p}{p+1-d} \frac{1}{\omega_d^\frac{1}{p}\,|x|^{\frac{(d-1)}{p}-1}}. $$ As you say, this is not useful to solve the equation with an other right-hand side, since $\Delta_p$ is not linear. Therefore, I would not call it the Green function.

Remark: When $p=d-1$, the same procedure will give $u = C\,\ln(|x|)$.

Proof: For such a function $u$, indeed, one has $$ ∇u = \frac{x}{\omega_d^\frac{1}{p}\,|x|^{\frac{d-1}{p}+1}} $$ so that $$ |∇u|^p = \frac{1}{\omega_d\,|x|^{d-1}} = \left|\frac{x}{\omega_d\,|x|^{d}}\right| = |∇G_1| $$ where $G_1 = \frac{-1}{(d-2)\,\omega_d\,|x|^{d-2}}$ is the solution of the Laplace equation $\Delta G_1 = \delta_0$. Therefore, since $|∇u|^{p-1}∇u$ and $∇G_1$ are parallel, of same direction and of same norm, we deduce $|∇u|^{p-1}∇u = ∇G_1$, so that $$ \Delta_p u = \mathrm{div}(|∇u|^{p-1}∇u) = \mathrm{div}(∇G_1) = \Delta G_1 = \delta_0. $$

LL 3.14
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