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Let us consider the fully nonlinear problem $$ \begin{cases} F(x,u,Du,D^2 u) = 0 & \text{ in } \Omega \\ u=0 & \text{ in } \partial \Omega \end{cases} $$ Suppose that we know that the solution $u_n$ to $$ \begin{cases} F_n(x,u_n,Du_n,D^2 u_n) = 0 & \text{ in } \Omega \\ u_n=0 & \text{ in } \partial \Omega \end{cases} $$ satisfies $u_n \to u$ uniformly on compact subsets of $\Omega$.

In this setup, is it generally true that the solution of $$ \begin{cases} \partial_t u_n + F_n(x,u_n,Du_n,D^2 u_n) = 0 & \text{ in } \Omega\times(0,\infty) \\ u_n=0 & \text{ in } \partial \Omega\times(0,\infty) \\ u_n(\cdot,0) = g & \text{ in } \Omega \end{cases} $$ converges to the one of $$ \begin{cases} \partial_t u + F(x,u,Du,D^2 u) = 0 & \text{ in } \Omega\times(0,\infty) \\ u=0 & \text{ in } \partial \Omega\times(0,\infty) \\ u(\cdot,0) = g & \text{ in } \Omega \end{cases} $$ or not? If not, under which additional conditions is it true?

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  • $\begingroup$ Hi and welcome to the MathOverflow. Are you talking about a sequence of solutions to finer discretizations of the original differential operator or about a numerable sequence of solutions to some smooth approximations to it. I am asking this because I don't understand the notation $D u_n$: if it were a finite difference approximation, then the divided difference operator $\Delta_{h_n} u_n$ (and its higher order versions) should appear instead of the derivative. $\endgroup$ Jul 8, 2022 at 8:57

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This is not, in general, true. This is because your hypothesis only says something about the zero set of $F$, but the parabolic equation needs to see the entire function (all level sets). For instance, I could define $F(D^2u) = f(-\Delta u)$ where $f:\mathbb{R} \to \mathbb{R}$ is any strictly increasing, but nonlinear, function satisfying $f(0) = 0$. Then clearly set of solutions of $F(D^2u) = 0$ are just the harmonic functions. However, the parabolic equations will definitely not have the same set of solutions. Thus, even if we strengthen your hypothesis to allow for any boundary condition (not just zero), and dispense with the sequence and talk about two operators, it still wouldn't be enough.

However, if you strengthen your hypothesis to allow for any general right-hand side function $h\in C(\overline{\Omega})$, instead of just zero, then in that case the answer is yes: this will imply convergence of the parabolic problems. A direct way to see this is to approximate the parabolic problem by a discrete-in-time problem which then reduces to solving a sequence of elliptic problems.

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