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For the equation $ F(x,u,u',u'') = -au''-1 =0$ for $ x\in (0,2)$ with $ u(0) = 0 = u(2) $ and $a(x)$ is $1$ for $x\in (0,1)$ and $2$ for $x\in [1,2)$. Need to show that the function $$ u(x) = \begin{cases} -x^2/2 + 5x/6 &\mbox{if } x \in (0,1] \\ -x^2/4 + 5x/12+1/6 & \mbox{if } x\in(1,2) \end{cases} $$ is not a viscosity solution. Here we observe $ u(1) = 1/3 $ from both sides, thus I need to show there exists $ \phi \in C^2(0,2)$ such that $ \phi(x) < u(x) $ for $ x\neq 1$ and $\phi(1) = 1/3 = u(1)$ with $ F(1,\phi(1),\phi'(1))<0$. I can not seem to be able to find such a $ \phi$. Any hints or solutions would be extremely helpful. Thank you.

smiley06
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  • Is it as intended that in the rhs of $F(x,u,u') = -au''-1 =0$ there are no arguments of $F$ but $u''$ is present which is not an argument? – Andrew Mar 20 '15 at 18:52
  • @Andrew edited.....thanx – smiley06 Mar 21 '15 at 20:57
  • Take $\phi$ such that $\phi(1) = 1/3$, $-1/6 <\phi'(1)< -1/12$ and $\phi''(1)=0$. – Ben Derrett Mar 21 '15 at 22:53
  • You can directly add some viscosity to the equation ($-au''-\epsilon u' -1=0$), solve it, then take the $\lim_{\epsilon\rightarrow\infty}$ of the solution. You recover the "nice" solution with $u'(1)$ also continuous, a property the solution above does not have. If you can prove uniqueness of the viscosity solution for this equation, then that would prove that the above cannot be a viscosity solution. Proving uniqueness of the viscosity solution for this equation should boil down to showing that $F$ is continuous some sense (Lipschitz?). – rajb245 Mar 25 '15 at 18:39
  • Another way: can you prove (or are you allowed to assume) that classical solutions are viscosity solutions? This with uniqueness also precludes the above from being a viscosity solution. – rajb245 Mar 25 '15 at 18:45
  • @smiley06 A function $\phi$ with behaviour near 1 as per my comment and fast decay elsewhere has the properties you need. – Ben Derrett Mar 27 '15 at 07:49

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