By Dennis M. Deturck (ed.)
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Extra info for Nonlinear problems in geometry
The proof of the lemma is now complete. 3 Continuous Solutions from Comparison Principle As mentioned above, the maximal subsolution u is not necessarily continuous; hence, its lower semi-continuous envelope u does not coincide necessarily with it. 4 (cf. 13 above). We would get a (continuous viscosity) solution if u D u . On one hand, u is upper semi-continuous by construction and on the other hand u Ä u by definition of the semi-continuous envelopes. 14) if and only if u Ä u in Q. 14) satisfies a comparison principle and that the barriers u˙ satisfy some appropriate inequality on the parabolic boundary.
The fundamental assumptions is that the equation is uniformly elliptic (see below for a definition). For pedagogical purposes, we do not want to prove a theorem for the most general case. x/jDuj @t trace. 25) Assumptions (A) • c is bounded and Lipschitz continuous in Q; • W Q ! e. 29. 25) in an open set Q Rd C1 is H¨older continuous in time and space. 2"/ 1 jx yj2 . Such a result can be found in [CIL92]. 30 (Jensen–Ishii’s Lemma II). Let U and V be two open sets of Rd and I an open interval of R.
4 (cf. 13 above). We would get a (continuous viscosity) solution if u D u . On one hand, u is upper semi-continuous by construction and on the other hand u Ä u by definition of the semi-continuous envelopes. 14) if and only if u Ä u in Q. 14) satisfies a comparison principle and that the barriers u˙ satisfy some appropriate inequality on the parabolic boundary. More precisely, we would like on one hand that Comparison principle. 14) in Q and u Ä v on the parabolic boundary @p Q, then u Ä v in Q.