# viscosity solution vs. weak solution

I am confused between the two. Is one a subset of the other or they are the same/completely different notions? Say I have an euqaiton $u_t=\mathcal{L}u$ for an elliptic operator $\mathcal{L}$ with bad coefficients so that it doesn't have a strong solution, but I can find a weak solution. What is the relation to viscosity then?

-
What do you mean by weak solution, here? Viscosity solutions do solve PDEs (for suitable definition of solving) without necessarily being differentiable, but they are not obtained by integration by parts. –  Tommi Brander Jul 6 '12 at 10:10
that's what I am trying to understand. Weak solution by the definition, I take a test function $v\in C_c^{\infty}$, integrate over domain $D$ and obtain $$\int_Du_tvdx=\int_D\mathcal{L}uvdx$$then the function $u\in H_0^1$ s.t. satisfies that is a weak solution. As treat them these weak solutions are the ones lacking smoothness which is required to use them directly in original pde. What is the intuition behind viscosity? I see two words being used in a similar context but what the difference then? –  Medan Jul 6 '12 at 12:29
For nonlinear PDEs like a Hamilton-Jacobi equation or a hyperbolic systems of conservation laws, a weak solution need not be unique. For a Hamilton-Jacobi equation, the viscosity solution is a special weak solution which turns out to exist and be unique. For hyperbolic systems of conservation laws, a similar concept is the entropy solution, which can be shown to exist and be unique in a number of "sufficiently well understood" situations. –  Thomas Klimpel Jul 11 '12 at 7:41
arxiv.org/abs/math/9207212 may be relevant. –  Willie Wong Sep 11 '12 at 12:54

The viscosity solutions was first introduced in the context of the Hamilton-Jacobi equation by the vanishing viscosity method. It would be difficult to apply the notion of distributional weak solution in this context, because the derivatives occur inside a nonlinear function. Further complications would arise because a distributional weak solution is not necessarily unique (for degenerate elliptic equations or nonlinear first order equations). Because of the nonlinearity, a Hamilton-Jacobi equations often doesn't have a classical solution, even if the Hamiltonian is an analytic function.

The viscosity solution was later generalized to degenerate elliptic equations, where the vanishing viscosity method itself no longer works. In case the equation is linear in the derivatives, both distributional weak solutions and viscosity solutions are defined, and we can investigate their relationship. My guess would be that the viscosity solutions turns out to also be a weak solutions in this context, but that there will in general be also weak solutions which are not viscosity solutions.

-

In the elliptic case, these two notions are equivalent. The proof (which is not trivial at all) can be found in the paper

H. Ishii, "On the equivalence of two notions of weak solutions, viscosity solutions and distribution solutions", Funkcial Ekvac. Ser. Int. 38 (1) (1995) 101–120.(pdf)

Probably the same proof can be extended to the parabolic case (but this is only a guess on my part).

-
ok, great, because sometimes it's confusing reading two names for the same concept. The same about "weak" derivative and "distributional" derivative! However, when the two are different? At least intuitively, how to understand the difference between the two, what does it depend on, on regularity? –  Medan Jul 7 '12 at 23:21
I asked a question about your guess, here: mathoverflow.net/questions/157433/… no answer so far :( I only need it for linear equations! –  Lost1 Feb 17 '14 at 16:01
@Lost1 Sorry for the late reply; actually I believe that this paper directly applies to the parabolic case as well, since those can be seen as a special case of degenerate elliptic equations (and there is no assumption of strict ellipticity in Ishii's paper) –  pgassiat Feb 24 '14 at 18:46
@Lost1 Re-check your computations, $G$ satisfies properness if and only if $F$ does :) –  pgassiat Feb 25 '14 at 16:22
@Lost1 Yes absolutely (for second order equations). But don't take my word for it, check it yourself, it is easy enough ! By the way, for the fact that one can require arbitrary smoothness for test functions in the definition, a complete proof can be found in the book by Y. Giga, "Surface evolution equations, a level set approach", Proposition 2.2.3. –  pgassiat Feb 27 '14 at 15:52