I have attached pg. 275 and pg. 276 of [BS91]. My concern is with the claim (2.7) on pg. 276. To prove this claim, I require the following additional assumption, which is not made by the authors:

Assumption: $u^{\rho}$ is upper semicontinuous for each $\rho$.

Remark: In the supersolution case, the analogue of the above assumption is that $u^\rho$ is lower semicontinuous for each $\rho$.

My proof of (2.7) under the additional assumption above: Let $(\rho_n,x_n)_n$ be a sequence with $\rho_n\rightarrow 0$, $x_n\rightarrow x_0$, and $u^{\rho_n}(x_n)\rightarrow \overline{u}(x_0)$. Then, by the assumption above, for each $n$, there exists $y_n$ such that $$(u^{\rho_n}-\varphi)(y_n) = \sup_{ \overline{B(x_0;r)} } \{ u^{\rho_n} - \varphi \}.$$ Extract a subsequence of $(\rho_n,x_n,y_n)_n$ converging to some $(0,x_0,\hat{y})$ and relabel it, with a slight abuse of notation, $(\rho_n,x_n,y_n)_n$. It follows that

\begin{multline*} 0=(\bar{u}-\varphi)(x_{0})=\lim_{n\rightarrow\infty}(u^{\rho_{n}}-\varphi)(x_{n})\leq\limsup_{n\rightarrow\infty} (u^{\rho_{n}}-\varphi)(y_{n}) \\ \leq\limsup_{\substack{\rho\rightarrow0\\ y\rightarrow \hat{y} } }(u^{\rho}-\varphi)(y)=(\overline{u}-\varphi)(\hat{y}). \end{multline*} Because it was assumed that $x_0$ is a strict local maximum of $\overline{u}-\varphi$, we get $\hat{y}=x_0$. This proves (2.7), with $(\rho_n,y_n)$ being the desired sequence.

My problem: I cannot see how to remove the assumption. While the assumption can be relaxed, as far as I can tell, we need to impose something to ensure that the maximum is attained in the proof below. Am I mistaken?

[BS91] Barles, Guy, and Panagiotis E. Souganidis. "Convergence of approximation schemes for fully nonlinear second order equations." Asymptotic analysis 4.3 (1991): 271-283.

  • Try looking instead for an approximate maximizer when $u^\rho$ is not continuous. In your proof of (2.7) choose $y_n$ satisfying $u^{\rho_n}(y_n) - \phi(y_n) + \frac{1}{n} \geq u^{\rho_n}(x) - \phi(x)$ for all $x \in \overline{B(x_0,r)}$. – Jeff Nov 6 '16 at 7:38
  • @Jeff: I did try that, and this is what happens: let $y_{n}$ be as suggested. Using the argument above, we get $y_{n}\rightarrow x_{0}$. However, now we have $u^{\rho_{n}}\leq\varphi+\xi_{n}^{\prime}$ where $\xi_{n}^{\prime}=u^{\rho_{n}}(y_{n})-\varphi(y_{n})+\frac{1}{n}$ (compare this with $\xi_{n}$ in the original proof). Namely, we can no longer establish $S(\rho_{n},y_{n},\varphi(y_{n})+\xi_{n}^{\prime},\varphi+\xi_{n}^{\prime})\leq0$ since $\varphi(y_{n})+\xi_{n}^{\prime}=u^{\rho_{n}}(y_{n})+\frac{1}{n}\neq u^{\rho_{n}}(y_{n})$. – parsiad Nov 6 '16 at 13:54
  • Right, but you do have $S(\rho_n,y_n,\phi(y_n)+\xi_n'-\frac{1}{n},\phi+\xi'_n) \leq 0$. Then the proof goes through with a minor change to the consistency condition. – Jeff Nov 6 '16 at 16:38
  • It's possible this is a small oversight; I cannot speak for the authors and don't know what they had in mind. Note that if you're working with a finite difference scheme, you can restrict the domain of $u^\rho$ to the discrete grid and this issue won't come up. – Jeff Nov 6 '16 at 16:42
  • Let us continue this discussion in chat. – parsiad Nov 6 '16 at 17:25
up vote 2 down vote accepted

After checking the arguments with @Jeff, I am willing to conclude that the issue in the question above is an extremely minor error in the original paper [BS91]. Both Jeff and I came up with "fixes" for the case of $u^{\rho}$ not continuous. Both are detailed below.

Update: I had the chance to speak to Professor Barles about this, who was kind enough to suggest a way to improve my fix. The result is below.

Parsiad and Professor Barles' fix

Redefine the notion of an approximation scheme as a pair of functions $$\underline{S},\overline{S}:(0,\infty)\times\overline{\Omega}\times\mathbb{R}\times B(\overline{\Omega})\rightarrow\mathbb{R}.$$ For fixed $\rho$, a solution of this approximation scheme is a bounded function $u^{\rho}:\overline{\Omega}\rightarrow\mathbb{R}$ such that $$ \underline{S}(\rho,x,(u^{\rho})^{*}(x),(u^{\rho})^{*})\leq0\leq \overline{S}(\rho,x,(u^{\rho})_{*}(x),(u^{\rho})_{*})\text{ for all }x\in\overline{\Omega}. $$

Remark: Intuitively, $(u^{\rho})^*$ and $(u^{\rho})_*$ are simply upper and lower semicontinuous envelopes of the solution $u^{\rho}$ of an approximation scheme in the sense of [BS91].

Monotonicity is extended in the obvious way: an approximation scheme $(\underline{S},\overline{S})$ is monotone if for each $S\in\{\underline{S},\overline{S}\}$, \begin{gathered} S(\rho,x,t,u)\leq S(\rho,x,t,v)\text{ if }u\geq v\\ \text{ for all }\rho>0\text{, }x\in\overline{\Omega}\text{, }t\in\mathbb{R}\text{, and }u,v\in B(\overline{\Omega}). \end{gathered} Stability is also extended in the obvious way: an approximation scheme $(\underline{S},\overline{S})$ is stable if \begin{gathered} \text{for all }\rho>0\text{, there exists a solution }u^{\rho}\in B(\overline{\Omega})\text{,}\\ \text{with }u^{\rho}\text{ bounded independent of }\rho. \end{gathered} The consistency requirement is modified slightly. In particular, we require that $\overline{S}$ (resp. $\underline{S}$) only satisfy (2.4a) (resp. (2.4b)) of [BS91]. That is, an approximation scheme $(\underline{S},\overline{S})$ is consistent if for all $x\in\overline{\Omega}$ and $\varphi\in C_{b}^{\infty}(\overline{\Omega})$, \begin{align*} \limsup_{\substack{\rho\rightarrow0\\ y\rightarrow x\\ \xi\rightarrow0 } }\frac{\overline{S}(\rho,y,\varphi(y)+\xi,\varphi+\xi)}{\rho} & \leq F^{*}(D^{2}\varphi(x),D\varphi(x),\varphi(x),x)\\ \text{and }\liminf_{\substack{\rho\rightarrow0\\ y\rightarrow x\\ \xi\rightarrow0 } }\frac{\underline{S}(\rho,y,\varphi(y)+\xi,\varphi+\xi)}{\rho} & \geq F_{*}(D^{2}\varphi(x),D\varphi(x),\varphi(x),x). \end{align*} Proposition: Let $(\underline{S},\overline{S})$ be an approximation scheme that is monotone, stable, and consistent (in the sense above). Assume $F=0$ satisfies a strong comparison principle (i.e., (2.5) of [BS91]). Then, as $\rho\rightarrow0$, the solution $u^{\rho}$ of the scheme converges locally uniformly to the unique viscosity solution of $F=0$ (in the sense of [BS91]).

Proof: With the setup above, the proof is essentially identical to that of [BS91] (and avoids the issue of being unable to take extrema).

Jeff's fix

Proposition: Theorem 2.1 of [BS91] holds if we replace (2.4) of [BS91] by the following modified consistency requirement \begin{align*} \limsup_{\substack{\rho\rightarrow0\\ y\rightarrow x\\ \xi\rightarrow0 } }\frac{S(\rho,y,\varphi(y)+\xi+e^{-1/\rho},\varphi+\xi)}{\rho} & \leq F^{*}(D^{2}\varphi(x),D\varphi(x),\varphi(x),x) & \text{(2.4a}\text{)}\phantom{.}\\ \text{and }\liminf_{\substack{\rho\rightarrow0\\ y\rightarrow x\\ \xi\rightarrow0 } }\frac{S(\rho,y,\varphi(y)+\xi-e^{-1/\rho},\varphi+\xi)}{\rho} & \geq F_{*}(D^{2}\varphi(x),D\varphi(x),\varphi(x),x) & \text{(2.4b}\text{)}. \end{align*} Proof: While the claim (2.7) cannot be proved, we can instead establish the existence of a sequence $(\rho_{n},y_{n})$ such that $$ \rho_{n}\rightarrow0\text{, }y_{n}\rightarrow x_{0}\text{, and }(u^{\rho_{n}}-\varphi)(y_{n})+e^{-1/\rho_{n}}\geq(u^{\rho_{n}}-\varphi)(\cdot). $$

To see this, let $(\rho_{n},x_{n})_{n}$ be a sequence with $\rho_{n}\rightarrow0$, $x_{n}\rightarrow x_{0}$, and $u^{\rho_{n}}(x_{n})\rightarrow\overline{u}(x_{0})$. Then, for each $n$, there exists $y_{n}$ such that $$ (u^{\rho_{n}}-\varphi)(y_{n})+e^{-1/\rho_{n}}\geq\sup_{\overline{B(x_{0};r)}}\left\{ u^{\rho_{n}}-\varphi\right\} . $$ Extract a subsequence of $(\rho_{n},x_{n},y_{n})_{n}$ converging to some $(0,x_{0},\hat{y})$ and relabel it, with a slight abuse of notation, $(\rho_{n},x_{n},y_{n})_{n}$. It follows that \begin{multline*} 0=(\overline{u}-\varphi)(x_{0})=\lim_{n\rightarrow\infty}(u^{\rho_{n}}-\varphi)(x_{n})\leq\limsup_{n\rightarrow\infty}\left\{ (u^{\rho_{n}}-\varphi)(y_{n})+e^{-1/\rho_{n}}\right\} \\ \leq\limsup_{\substack{\rho\rightarrow0\\ y\rightarrow\hat{y} } }\left\{ (u^{\rho}-\varphi)(y)+e^{-1/\rho}\right\} =\limsup_{\substack{\rho\rightarrow0\\ y\rightarrow\hat{y} } }(u^{\rho}-\varphi)(y)=(\overline{u}-\varphi)(\hat{y}). \end{multline*} Because it was assumed that $x_{0}$ is a strict local maximum of $\overline{u}-\varphi$, we get $\hat{y}=x_{0}$, as desired.

Now, the remainder of the proof in [BS91] is modified as follows. Let $\xi_{n}=(u^{\rho_{n}}-\varphi)(y_{n})+e^{-1/\rho_{n}}$. Then, $u^{\rho_{n}}\leq\varphi+\xi_{n}$ and hence $$ 0=S(\rho_{n},y_{n},u^{\rho_{n}}(y_{n}),u^{\rho_{n}})\geq S(\rho_{n},y_{n},\varphi(y_{n})+\xi_{n}-e^{-1/\rho_{n}},\varphi+\xi_{n}). $$ The desired result now follows by dividing by $\rho$, taking limit inferiors, and applying the consistency condition. $\square$

Remark: the exponential decay is chosen as it does not cause any problems in creating approximation schemes, practically speaking. For example, consider the approximation $$\frac{u^\rho(x+\rho) - u^\rho(x)}{\rho}.$$ If we substitute the corresponding entities in the revised consistency requirement, we get $$\frac{\varphi(x+\rho) - \varphi(x) \pm e^{-1/\rho}}{\rho} = \varphi^\prime(x) + \mathcal{O}(\rho)$$ by a Taylor approximation.

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