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Consider the initial value problem $$\begin{cases} u_t^\epsilon + H(x,t,u^\epsilon,\nabla_x u^\epsilon) = \epsilon\Delta_x u^\epsilon & \text{ in } \mathbb{R}^n \times (0,\infty)\\ u^\epsilon = g &\text{ in } \mathbb{R}^n \times \{t= 0\}\end{cases}, $$ for $\epsilon>0$.

Can you point out one (some) reference(s) for a complete proof of a theorem roughly to the effect that, under adequate hypothesis on $H$ and $g$, this Cauchy problem admits a unique $C^{2,1}(\mathbb{R}^n \times [0,\infty))$ solution (that is $C^2$ with respect to $x$ and $C^1$ with respect to $t$)?

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  • $\begingroup$ See Sect. 6 of the Crandall, Ishii, Lions survey ams.org/journals/bull/1992-27-01/S0273-0979-1992-00266-5/… $\endgroup$
    – Liviu Nicolaescu
    Commented Oct 30, 2016 at 11:13
  • $\begingroup$ For $\epsilon>0$ this is a quasilinear parabolic equation and existence for such equations is known under certain conditions on $H$ (stronger than continuity). Theorem 6.5 in the above survey gives a sufficient condition for convergence. $\endgroup$
    – Liviu Nicolaescu
    Commented Oct 30, 2016 at 15:49
  • $\begingroup$ Crandall, Michael G.; Lions, Pierre-Louis Viscosity solutions of Hamilton-Jacobi equations. Trans. Amer. Math. Soc. 277 (1983), no. 1, 1–42. $\endgroup$
    – Liviu Nicolaescu
    Commented Oct 30, 2016 at 18:07
  • $\begingroup$ Read the survey of Crandall Lions and Ishii and check the references there $\endgroup$
    – Liviu Nicolaescu
    Commented Oct 30, 2016 at 20:50
  • $\begingroup$ The statement you want is not true. Consider the linear case $H=H(x,t)$. For some continuous functions $H$ the bounded solution does not belong to the class $C^{2,1}_{x,t}(\mathbb{R}^n \times \{t>0\})$. $\endgroup$
    – Andrew
    Commented Nov 9, 2016 at 16:35

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