Let Ω\documentclass[12pt]{minimal}
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\begin{document}$${\Omega}$$\end{document} a bounded domain in RN\documentclass[12pt]{minimal}
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\begin{document}$${\mathbb{R} ^N }$$\end{document}, and let u∈C1(Ω¯)\documentclass[12pt]{minimal}
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\begin{document}$${u\in C^1 (\overline{\Omega})}$$\end{document} a weak solution of the following overdetermined BVP: -∇(g(|∇u|)|∇u|-1∇u)=f(|x|,u)\documentclass[12pt]{minimal}
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\begin{document}$${-\nabla (g(|\nabla u|)|\nabla u|^{-1}
\nabla u)=f(|x|,u)}$$\end{document}, u>0\documentclass[12pt]{minimal}
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\begin{document}$${ u > 0 }$$\end{document} in Ω\documentclass[12pt]{minimal}
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\begin{document}$${\Omega }$$\end{document} and u=0,|∇u(x)|=λ(|x|)\documentclass[12pt]{minimal}
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\begin{document}$${u=0, \ |\nabla u(x)|
=\lambda (|x|)}$$\end{document} on ∂Ω\documentclass[12pt]{minimal}
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\begin{document}$${\partial \Omega }$$\end{document}, where g∈C([0,+∞)∩C1((0,+∞))\documentclass[12pt]{minimal}
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\begin{document}$${g\in C([0,+\infty
)\cap C^1 ((0,+\infty ) ) }$$\end{document} with g(0)=0\documentclass[12pt]{minimal}
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\begin{document}$${g(0)=0}$$\end{document}, g′(t)>0\documentclass[12pt]{minimal}
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\begin{document}$${g'(t) > 0}$$\end{document} for t>0\documentclass[12pt]{minimal}
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\begin{document}$${t > 0}$$\end{document}, f∈C([0,+∞)×[0,+∞))\documentclass[12pt]{minimal}
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\begin{document}$${f\in C([0,+\infty ) \times [0, +\infty ) )}$$\end{document}, f is nonincreasing in |x|\documentclass[12pt]{minimal}
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\begin{document}$${|x|}$$\end{document}, λ∈C([0,+∞))\documentclass[12pt]{minimal}
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\begin{document}$${\lambda \in C([0, +\infty )) }$$\end{document} and λ\documentclass[12pt]{minimal}
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\begin{document}$${\lambda }$$\end{document} is positive and nondecreasing. We show that Ω\documentclass[12pt]{minimal}
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\begin{document}$${\Omega }$$\end{document} is a ball and u satisfies some “local” kind of symmetry. The proof is based on the method of continuous Steiner symmetrization.