Capillary problem for singular degenerate parabolic equations on a half space (Q1374483)
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scientific article; zbMATH DE number 1095808
| Language | Label | Description | Also known as |
|---|---|---|---|
| English | Capillary problem for singular degenerate parabolic equations on a half space |
scientific article; zbMATH DE number 1095808 |
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Capillary problem for singular degenerate parabolic equations on a half space (English)
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10 December 1997
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The paper is devoted to the study of solutions to a geometric evolution of an interface \(\{\Gamma_t\}_{t \geq 0}\) in a fixed domain \(\Omega\) under the condition that the geometric boundary \(b\Gamma_t = \partial \Omega \cap \bar {\Gamma_t}\) of \(\Gamma_t\) touches the boundary \(\partial \Omega\) with a given angle. Intrinsically, \(\Gamma_t\) means a surface dividing two regions \(D^{+}_{t}, D^{-}_{t}\) in \(\Omega\) occupied in the time \(t\) by two phases of the studied materials. In the paper, the interface \(\Gamma_t\) is looked for as the zero level set of a function \(u\) which is positive on \(D^{+}\) and negative on \(D^{-}\). If \(u \in C^2\) and \(\nabla u \neq 0\) near \(\Gamma_t\), the problem can be formulated as follows: \[ \partial_t u + F(\nabla u, \nabla^2 u) = 0\;\;\text{on} \;\Omega \cap \Gamma_t,\qquad\langle \nu (x), \nabla u\rangle - k |\nabla u |= 0\;\;\text{on} b\Gamma_t \] with \(F\) given, \(k = \cos \theta\) (\(\theta\) being the angle between \(\Gamma_t\) and \(\partial \Omega \)) and \(\nu(x)\) the outer unit normal of \(\partial \Omega\) at a point \(x\). The crucial step in the proof is a comparison principle for viscosity sub- and super-solutions. It allows to construct a unique viscosity solution for any continuous initial value \(a\) (constant outside a compact subset of \(\Omega \)). For simplicity reasons, only the case of \(\Omega\) equal to a halfspace is considered here. The general case is treated in a forthcoming paper.
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capillary problem
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viscosity solutions
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level-set approach
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comparison principle
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0.9017185
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0.8916062
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0.8853923
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0.8849715
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0.88396305
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0.88079906
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0.8801055
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