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Let $A:=[a,b]$ be a closed interval in $\mathbb{R}$. Let $F(x,p,q,r)$ be a function from $[a,b]\times \mathbb{R} \times \mathbb{R} \times\mathbb{R} \rightarrow \mathbb{R}$ describing a second order ODE. We assume that $F$ is continuous on $[a,b]$ and satisfies all the nice properties we need for viscosity solutions.

Let our DE be -

$F(x,v(x), v'(x), v''(x)) = 0$ (1)

Suppose $v$ is a viscosity solution to $F(x,v(x),v'(x),v''(x))=0$ on $(a,b]$ i.e. it is both a subsolution and supersolution. Moreover, I know that $v$ is continuous on $[c,d] \supset [a,b]$ Then, can I say that $v$ is also a viscosity solution to Equation (1) on $[a,b]$?

The reason why I think this should be true is the following. Suppose my claim is false. In particular, let's assume that $v$ is not a supersolution of $(1)$. Then, there exists a $C^2$ function $h$ such that $v - h$ is minimized at $a$ and $F(a, v(a), h'(a),h''(a)) < 0$.

But then, since $v$ is continuous and $h$ is $C^2$, I should get a function that is "very similar" to $h$, say $\bar{h}$, such that $v-\bar{h}$ is minimized at some $a + \epsilon$ and $F(a+\epsilon, v(a+\epsilon), \bar{h}'(a+\epsilon),\bar{h}''(a+\epsilon)) < 0$ Therefore, $v$ could not have been a supersolution on $(a,b)$ in the first place.

Is it absolutely bizarre to expect such a thing? And if I can get it, what are the nice conditions I would need on $v$ besides continuity?

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In general, no. For example, $v(x)=x$ is a viscosity solution of $v'(x)=1$ on the interval $(0,1)$, but not on $[0,1)$ ($v - mx$ has a local max at $x=0$ relative to $[0,1)$ for any $m\geq 1$, so the subsolution property does not hold at $x=0$).

The issue is that you have a great degree of freedom in choosing test functions at the boundary of the domain, so classical solutions up to the boundary may not be viscosity solutions as well. This is covered in great detail in section 7.A "Boundary conditions in the viscosity sense" of the User's Guide

http://arxiv.org/abs/math/9207212

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  • $\begingroup$ Wow. Awesome. Thanks a lot. Saw this only today. :) $\endgroup$ – avk255 Oct 27 '15 at 17:55

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