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This is a soft question. For my background, I am an undergraduate math major just finishing my first course in PDE. In my PDE class, we have been learning about viscosity solutions.

I am following the definitions in these notes

We say a continuous function $u : [0, T ] \times \Omega → \mathbb{R}$ is a subsolution ( alt supersolution) of the equation $u_t + H(x,t,u,\nabla u) =0$, if every time there exists a $C^1$ function $\varphi$ for which there is a point $(t_0, x_0) \in (0, T ] \times \Omega$ and $r>0$, such that $\varphi (t_0, x_0) = u(t_0, x_0)$, $\varphi (t,x)≥u(t,x)$ (alt $≤$) for all $(t,x) \in(t_0 −r,t_0]×B_r(x_0)$, then $\varphi_t +H(t,x,\varphi,\nabla \varphi)≤0$ (alt $≥0$).

A continuous function $u$ is a viscosity solution in $(0, T ] × Ω$ when it is both a subsolution and a supersolution.

Evans also discusses these solutions in Chapter 10 of his book on PDE. Evans motivates viscosity solutions by looking at a sequence of functions $u^\varepsilon$ that are solutions to $u_t + H(x,t,u,\nabla u) - \varepsilon \Delta u =0$. This equation has smooth solutions, because the Laplacian regularizes the Hamilton-Jacobi equation. Often we can assure that the sequence $u^\varepsilon$ is bounded and equicontinuous, giving some locally uniformly convergent subsequence, and these turn out to be viscosity solutions. (Though not all viscosity solutions arise in this way)

I am also aware that viscosity solutions can be defined for somewhat more general non-linear PDEs.

My questions are the following:

  • Why are such generalized solutions interesting? Are they only interesting because we can prove the existence and uniqueness of such solutions? Or do viscosity solutions to Hamilton-Jacobi equations actually model anything?

  • Are these solutions ever useful in applications? This may be just a more specific form of the above question.

  • Why should the notion of viscosity solution for Hamilton-Jacobi equations be the "right" definition of generalized solution?

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    $\begingroup$ I can see I've been down voted twice now. If anyone has suggestions on how I could improve the question or what's wrong with it, please let me know. I am quite interested in its answer. $\endgroup$
    – msm
    Jun 7, 2017 at 14:27

2 Answers 2

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The method of vanishing viscosity is not a particularly good way to think about viscosity solutions. The reason viscosity solutions are so useful is that they satisfy a maximum principle (or rather a comparison principle), which gives them very nice uniqueness, existence, and stability results (making it easy to pass to limits, for instance). For these reasons, viscosity solutions are almost always the correct notion of solution in applications, which include things like optimal control theory and geometric flows like mean curvature motion, to name a few.

Viscosity solutions apply to fully nonlinear equations for which no other notion of solution is available (e.g., classical solutions do not exist, and weak distributional solutions do not make sense). Without viscosity solutions, we would have no way to rigorously make sense of a very broad class of PDEs. I could go on, but others have written more elegant answers to this question elsewhere:

https://mathoverflow.net/questions/59449/why-are-viscosity-solutions-useful-solutions

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  • $\begingroup$ "viscosity solutions are almost always the correct notion of solution" until another notion comes along, no? $\endgroup$
    – PatrickT
    Sep 20, 2020 at 0:38
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    $\begingroup$ New notions of solutions to PDEs come about to address classes of PDEs where existing notions do not give existence or uniqueness, or for which existing notions do not describe the physically correct solution. For the reasons I mentioned above (mainly the strong stability of viscosity solutions), for any PDE where the viscosity solution exists and is unique, it is extremely unlikely another notion of solution would be needed or would be useful. Of course, new notions will eventually come along for other classes of PDEs that we do not fully understand (e.g., Navier-Stokes equations). $\endgroup$
    – Jeff
    Sep 20, 2020 at 20:41
  • $\begingroup$ How about for the economics? Is the notion always correct? $\endgroup$ Sep 16, 2021 at 3:49
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I learned this from the book "Calculus of Variations and Optimal Control Theory: A Concise Introduction" by Daniel Liberzon (2012), in section 5.2. This book is excellent in many ways.

Pontryagin's maximum principle is more general (and much more difficult to prove) than the HJB equation because the latter requires the value function to be differentiable with respect to x (and additionally that this gradient is further differentiable with respect to time). In many control problems, the solution is a value function with some points where a classical derivative doesn't exist.

However, the notion of differentiation can be extended to super-differentials and sub-differentials. These are linear functions with a slope such that the target function's graph lies entirely above or below the slope, near the point in question. Something like this:

sub-differential

This leads to the notion of viscosity super-solution and sub-solutions. For example if the PDE is of the form: $$F(x, v(x), \nabla v(x)) = 0$$ where $v$ is the value function, the viscosity sub-solution is defined by: $$F(x,V(x),\xi) \le 0 \quad \forall \xi\in D^+v(x), \forall x $$ where $\xi \in D^+$, the set of all super-differentials at point $x$.

A viscosity solution is both a viscosity sub-solution and super-solution.

The name 'viscosity' comes from a variation of the standard PDE which describes viscuous fluids: $$F(x, v(x), \nabla v(x)) = \epsilon \Delta v(x) $$ as $\epsilon \rightarrow 0$, where $\Delta$ is the Laplacian. It can be shown that the $v$ that solves the above equation is both a sub-solution and a super-solution.

See Liberzon's book $\S5.2$ for more details.

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