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The HJ equation is: $H(u'(x))+1=0, x\in(-1,1)$. And $H:\mathbb{R}\rightarrow\mathbb{R},H(p)=\min_{a\in[-1,1]}ap,p\in\mathbb{R}$.
The question is show that $u(x)=1-|x|,x\in(-1,1)$ is a viscosity solution.
I know that it's a supersolution since there is no element in $D^-u(x)$, but why it is also a subsolution?
I mean for any $\phi(x)$ above $u(x)$, the derivative of it at point x=0 must be in [-1,1], then for any $p\in[-1,1]$, H should have $H\geq-1$, which makes $H+1\geq0$. Then it doesn't satisfy the subsolution definition.
I don't know where I am wrong. Please, any hint?

Einherjar
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  • This is probably trivial, but why does the derivative of $\phi$ above $u$ have to be in $[-1,1]$ at $x=0?$ – Len Dec 30 '23 at 10:43

1 Answers1

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The solution is wrong. Your equation is

$$-|u'(x)| + 1 = 0.$$

The viscosity solution satisfying $u(-1)=0=u(1)$ is $u(x) = |x|-1$. To see this, we only have to check the viscosity solution conditions at $x=0$, since the equation is satisfied classically elsewhere. At $x=0$, there are no test functions touching from above, so $D^+$ is empty. Anything touching from below (i.e. $D^-$) has derivative in the range $[-1,1]$, which yields the supersolution condition $-|u'(x)| + 1\geq 0$.

If you switch the sign of the equation and consider

$$|u'(x)| - 1 = 0$$

then the solution is $u(x)=1-|x|$. This is the peculiar thing about viscosity solutions; the solutions of $H=0$ and $-H=0$ can be different. It is natural if you think about vanishing viscosity, since in one case you regularize by adding $\varepsilon \Delta u$, while in the other case you add $-\varepsilon \Delta u$.

Jeff
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  • Thank you very much for the reply! So you mean that for my equation, u(x)=1-|x| is just a supersolution but not a subsolution and thus not a solution? I am sorry that I am a slow taker, I never majored in math during my life. – Einherjar Dec 20 '22 at 14:34
  • The original question is in Japanese, so maybe I mistake 'check whether' for 'show that'... – Einherjar Dec 20 '22 at 14:37
  • Yes that's right, it's a supersolution but not a subsolution. In fact, it's a general property for your equation that any Lipschitz continuous function that satisfies $|u'(x)|=1$ at all points of differentiability is a viscosity supersolution. The subsolution condition in this case selects the unique viscosity solution. – Jeff Dec 20 '22 at 14:38
  • Oh my, thank you so much! – Einherjar Dec 20 '22 at 14:44
  • @Jeff I've just started dealing with viscosity solutions. Could you please explain why anything touching from below has derivative in [-1,1]? – Len Dec 30 '23 at 10:47
  • Draw a picture to convince yourself of why it is true. Then to prove it, use the finite difference approximation for a derivative, using that $u(x) - \phi(x) \geq u(0) - \phi(0)$ for all $x$, where $\phi$ is the test function and $u(x) = |x|-1$. – Jeff Dec 30 '23 at 16:49
  • @Jeff Thank you for the hint! Do you mean like $\forall h \in \mathbb{R}, u(h) - \phi(h) \geq u(0) - \phi(0) \iff \frac{|h|}{h} = \frac{u(h) - u(0)}{h} \geq \frac{\phi(h) - \phi(0)}{h},$ so $1 \geq \frac{\phi(h) - \phi(0)}{h} \geq -1,$ so taking the limit in $h$ gives $-1 \leq \phi'(0) \leq 1$? – Len Feb 20 '24 at 13:58
  • Yes, exactly like this. – Jeff Feb 21 '24 at 20:20