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It shall be an old story in PDE. I am looking for a sufficient condition of Dirichlet problem for the existence of the unique viscosity solution of the equation in the form of $$\inf_{a \in [-1,1]} \{b(a) \cdot D u(x) + \frac 1 2 tr (A(a) D^{2} u(x) )\} + f(x) = 0, \ \forall x\in O$$ with boundary condition $u = g$ on $\partial O$.

In particular, I want to explain the following two simple examples.

[ex1] There is no solution for $$- u' + u - 1 = 0, \ \forall x\in (-1,1) \hbox{ with } u (\pm 1) = 0.$$ Although $u(x) = 1 - e^{-1+x}$ is the unique solution of generalized Dirichlet problem according to the generalized definition of section 7 of [User's Guide by Crandall, Ishii, and Lions], it does not meet boundary data in the strong sense. END

[ex2] One can check $u(x) = 1 - e^{-1+|x|}$ is the solution of $$- \inf_{b\in [-1,1]} b u' + u - 1 = 0, \ \forall x\in (-1,1) \hbox{ with } u (\pm 1) = 0.$$ This equation also satisfies comparison principle, and hence it has unique solution. END

[ex3] As of discussion by @Willis Wong below, there is no solution for $$- \inf_{b\in [-1,1]} b u' + u + 1 = 0, \ \forall x\in (-1,1) \hbox{ with } u (\pm 1) = 0.$$

Indeed, all equations satisfy comparison principle. But first and last one have no solution, while second has unique one. According to Perron's method, the existence holds if one can construct sub and supersolution, but the sufficient condition of the existence is still left vague from Perron's method.

[Q.] Is there any reference which can explicitly explain why the [ex1] and [ex3] have no solution but [ex2] does? It's more desirable if the sufficient condition works for the general equation with the second order term in the above.

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  • $\begingroup$ Regarding your tag choice: [ex1] is not elliptic in any way you look at it. The principal part of [ex2] is in fact just a complicated way to write $|u'|$ which I guess may be thought of as being elliptic due to coercivity? (Incidentally, do you really mean $-\inf bu'$? Maybe you want it without the leading minus sign?) $\endgroup$ Commented Jun 28, 2016 at 15:45
  • $\begingroup$ @WillieWong I am not familiar in PDE, so I am not confident. Anyway, by the definition of [User's guide], it may be regarded as elliptic, but degenerate. In [ex2], I do really mean $- \inf b u'$, which is indeed $|u'|$ as you mentioned. But, the result is the same if the $\inf$ is over $[-1, 2]$, by only changing a bit on explicit form of the solution. Thanks. $\endgroup$
    – kenneth
    Commented Jun 28, 2016 at 15:58
  • $\begingroup$ @WillieWong I am also curious about the case without leading minus sign. I believe the unique solvability shall be true? $\endgroup$
    – kenneth
    Commented Jun 28, 2016 at 16:02
  • $\begingroup$ Sorry, you are right about the minus sign. Mea culpa (I dropped a minus sign in checking your computations.) // I'll let someone more familiar with the method answer the technical part of your question. $\endgroup$ Commented Jun 28, 2016 at 16:03
  • $\begingroup$ With the other sign the second equation has no (continuous) solutions. $|u'| \geq 0$ but by boundary data assumption $u-1 < 0$ in a neighborhood of $\pm1$. (In fact that's why I queried in my first comment; I've mistakenly thought you were in this case.) $\endgroup$ Commented Jun 28, 2016 at 16:09

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