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I found that for the following problem \begin{cases} -\Delta_p u = 1,&x\in B_1(0)\\ u = 0,\quad &x\in\partial B_1(0) \end{cases} where $B_1(0)$ is the unitary ball of $\mathbb{R}^N$ and $\Delta_p u = \operatorname{div}(\|\nabla u\|^{p-2}\nabla u) $, there exists a weak solution of the form: $ u(x) = C_{N,p} (1-\|x\|^{\frac{p}{p-1}})$ with $C_{N,p}\in\mathbb{R}$ a constant depending just on the dimension of the ball $N$ and the value of $p>2$.

The problem is that I cannot find the constant $C_{N,p},$ and I need it for the case $N=3$.

Eilif
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Dadeslam
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1 Answers1

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Assume the given solution $u$ and note that it is smooth almost everywhere.

Compute the partial derivative (define $C:=C_{N,p}$) \begin{align} \partial_i u &= -C \partial_i |x|^\frac{p}{p-1} \\ &= -C \frac{p}{p-1} |x|^{\frac{p}{p-1}-1} \partial_i |x| \\ &= -C \frac{p}{p-1} |x|^{\frac{p}{p-1}-1} \frac{x_i}{|x|} \\ &= -C \frac{p}{p-1} |x|^{\frac{1}{p-1}-1} x_i \,. \end{align} Gather for the gradient \begin{align} \nabla u &= -C \frac{p}{p-1} |x|^{\frac{1}{p-1}-1} x \,. \end{align} The norm of the gradient (assume $C>0$) \begin{align} | \nabla u | &= C \frac{p}{p-1} |x|^\frac{1}{p-1} \,. \end{align} The factor needed for the $p$-Laplacian \begin{align} | \nabla u |^{p-2} \nabla u &= \left( C \frac{p}{p-1} |x|^\frac{1}{p-1} \right)^{p-2} \left( -C \frac{p}{p-1} |x|^{\frac{1}{p-1}-1} x \right) \\ &= - \left( C \frac{p}{p-1} \right)^{p-1} |x|^{\frac{p-2}{p-1}+\frac{1}{p-1}-1} x \\ &= - \left( C \frac{p}{p-1} \right)^{p-1} x \,. \end{align} For the $p$-Laplacian we need to compute the divergence of the above computed factor \begin{align} \Delta_p u &= \nabla \cdot \left( | \nabla u |^{p-2} \nabla u \right) \\ &= \nabla \cdot \left( - \left( C \frac{p}{p-1} \right)^{p-1} x \right) \\ &= -N \left( C \frac{p}{p-1} \right)^{p-1} \\ &\overset{!}{=} -1 \\ \implies C &= \frac{p-1}{p} N^\frac{1}{1-p} = C_{N,p} \,. \end{align}

Note, that there are no special cases for $p>2$.

jack
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