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A Beginners Guide to

the Theory of Viscosity Solutions

Shigeaki Koike

August 27, 2010

2nd edition

26 August 2010

Department of Mathematics, Saitama University, 255 Shimo-Okubo, Sakura, Saitama,338-8570 JAPAN

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Preface for the 2nd edition

The first version of this text has been out of print since 2009. It seems thatthere is no hope to publish the second version from the publisher. Therefore,I have decided to put the files of the second version in my Home Page.

Although I corrected many errors in the first version, there must be somemistakes in this version. I would be glad if the reader would kindly informme errors and typos etc.

Acknowledgment for the 2nd edition

I would like to thank Professors H. Ishii, K. Ishii, T. Nagasawa, M. Ohta,T. Ohtsuka, and graduate students, T. Imai, K. Kohsaka, H. Mitake, S.Nakagawa, T. Nozokido for pointing out numerous errors in the First edition.

26 August 2010 Shigeaki Koike

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Preface

This book was originally written in Japanese for undergraduate studentsin the Department of Mathematics of Saitama University. In fact, the firsthand-written draft was prepared for a series of lectures on the viscosity so-lution theory for undergraduate students in Ehime University and HokkaidoUniversity.

The aim here is to present a brief introduction to the theory of viscositysolutions for students who have knowledge on Advanced Calculus (i.e. differ-entiation and integration on functions of several-variables) and hopefully, alittle on Lebesgue Integration and Functional Analysis. Since this is written

for undergraduate students who are not necessarily excellent, I try to giveeasy proofs throughout this book. Thus, if you do not feel any difficultyto read Users guide [6], you should try to read that one.

I also try not only to show the viscosity solution theory but also to men-tion some related classical results.

Our plan of this book is as follows: We begin with our motivation insection 1. Section 2 introduces the definition of viscosity solutions and theirproperties. In section 3, we first show classical comparison principles andthen, extend them to viscosity solutions of first- and second-order PDEs,separately. We establish two kinds of existence results via Perrons methodand representation formulas for Bellman and Isaacs equations in section 4.

We discuss boundary value problems for viscosity solutions in sections 5.Section 6 is a short introduction to the Lp-viscosity solution theory, on whichwe have an excellent book [4].

In Appendix, which is the hardest part, we give proofs of fundamentalpropositions.

In order to learn more on viscosity solutions, I give a list of books:A popular survey paper [6] by Crandall-Ishii-Lions on the theory of viscos-

ity solutions of second-order, degenerate elliptic PDEs is still a good choicefor undergraduate students to learn first. However, to my experience, itseems a bit hard for average undergraduate students to understand.

Bardi-Capuzzo Dolcettas book [1] contains lots of information on viscos-ity solutions for first-order PDEs (Hamilton-Jacobi equations) while Fleming-Soners [10] complements topics on second-order (degenerate) elliptic PDEswith applications in stochastic control problems.

Barles book [2] is also nice to learn his original techniques and Frenchlanguage simultaneously !

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It has been informed that Ishii would write a book [15] in Japanese on

viscosity solutions in the near future, which must be more advanced thanthis.

For an important application via the viscosity solution theory, we referto Gigas [12] on curvature flow equations. Also, I recommend the reader toconsult Lecture Notes [3] (Bardi-Crandall-Evans-Soner-Souganidis) not onlyfor various applications but also for a friendly introduction by Crandall,who first introduced the notion of viscosity solutions with P.-L. Lions in early80s.

If the reader is interested in section 6, I recommend him/her to attackCaffarelli-Cabres [4].

As a general PDE theory, although there are so many books on PDEs, Ionly refer to my favorite ones; Gilbarg-Trudingers [13] and Evans [8]. Alsoas a textbook for undergraduate students, Han-Lins short lecture notes [14]is a good choice.

Since this is a text-book, we do not refer the reader to original papersunless those are not mentioned in the books in our references.

Acknowledgment

First of all, I would like to express my thanks to Professors H. Morimotoand Y. Tonegawa for giving me the opportunity to have a series of lectures intheir universities. I would also like to thank Professors K. Ishii, T. Nagasawa,and a graduate student, K. Nakagawa, for their suggestions on the first draft.

I wish to express my gratitude to Professor H. Ishii for having given meenormous supply of ideas since 1980.

I also wish to thank the reviewer for several important suggestions.My final thanks go to Professor T. Ozawa for recommending me to publish

this manuscript. He kindly suggested me to change the original Japanese title(A secret club on viscosity solutions).

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Contents

1 Introduction 11.1 From classical solutions to weak solutions 21.2 Typical examples of weak solutions 41.2.1 Burgers equation 41.2.2 Hamilton-Jacobi equation 6

2 Definition 92.1 Vanishing viscosity method 92.2 Equivalent definitions 17

3 Comparison principle 23

3.1 Classical comparison principle 243.1.1 Degenerate elliptic PDEs 243.1.2 Uniformly elliptic PDEs 24

3.2 Comparison principle for first-order PDEs 263.3 Extension to second-order PDEs 313.3.1 Degenerate elliptic PDEs 333.3.2 Remarks on the structure condition 353.3.3 Uniformly elliptic PDEs 38

4 Existence results 404.1 Perrons method 40

4.2 Representation formulas 454.2.1 Bellman equation 464.2.2 Isaacs equation 51

4.3 Stability 58

5 Generalized boundary value problems 625.1 Dirichlet problem 655.2 State constraint problem 685.3 Neumann problem 725.4 Growth condition at |x| 75

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6 Lp-viscosity solutions 78

6.1 A brief history 786.2 Definition and basic facts 816.3 Harnack inequality 846.3.1 Linear growth 856.3.2 Quadratic growth 87

6.4 Holder continuity estimates 896.5 Existence result 92

7 Appendix 957.1 Proof of Ishiis lemma 957.2 Proof of the ABP maximum principle 100

7.3 Proof of existence results for Pucci equations 1057.4 Proof of the weak Harnack inequlity 1087.5 Proof of the local maximum principle 113

References 118

Notation Index 120Index 121

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1 Introduction

Throughout this book, we will work in (except in sections 4.2 and 5.4),where

Rn is open and bounded.We denote by , the standard inner product in Rn, and set |x| =

x, x for x Rn. We use the standard notion of open balls: For r > 0and x Rn,

Br(x) := {y Rn | |x y| < r}, and Br := Br(0).For a function u :

R, we denote its gradient and Hessian matrix at

x , respectively, by

Du(x) :=

u(x)x1

...u(x)xn

,

D2u(x) :=

2u(x)x2

1

j-th 2u(x)x1xn

......

...

i-th 2u(x)xixj

......

..

.

..

.2u(x)xnx1

2u(x)x2n

.

Also, Sn denotes the set of all real-valued n n symmetric matrices. Notethat if u C2(), then D2u(x) Sn for x .

We recall the standard ordering in Sn:

X Y X , Y , for Rn.We will also use the following notion in sections 6 and 7: For =t

(1, . . . , n), =t (1, . . . , n) Rn, we denote by the n n matrix

whose (i, j)-entry is ij for 1 i, j n;

=

11 j-th 1n...

......

i-th ij ......

......

n1 nn

.

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We are concerned with general second-order partial differential equations

(PDEs for short):

F(x, u(x), Du(x), D2u(x)) = 0 in . (1.1)

We suppose (except in several sections) that

F : R Rn Sn R is continuous

with respect to all variables.

1.1 From classical solutions to weak solutions

As the first example of PDEs, we present the Laplace equation:

u = 0 in . (1.2)

Here, we define u := trace(D2u). In the literature of the viscosity solutiontheory, we prefer to have the minus sign in front of .

Of course, since we do not require any boundary condition yet, all poly-nomials of degree one are solutions of (1.2). In many textbooks (particularlythose for engineers), under certain boundary condition, we learn how to solve(1.2) when has some special shapes such as cubes, balls, the half-space or

the whole space Rn. Here, solve means that we find an explicit formula ofu using elementary functions such as polynomials, trigonometric ones, etc.

However, the study on (1.2) in such special domains is not applicablebecause, for instance, solutions of equation (1.2) represent the density of agas in a bottle, which is neither a ball nor a cube.

Unfortunately, in general domains, it seems impossible to find formulas forsolutions u with elementary functions. Moreover, in order to cover problemsarising in physics, engineering and finance, we will have to study more generaland complicated PDEs than (1.2). Thus, we have to deal with general PDEs(1.1) in general domains.

If we give up having formulas for solutions of (1.1), how do we investigatePDEs (1.1) ? In other words, what is the right question in the study of PDEs?

In the literature of the PDE theory, the most basic questions are as fol-lows:

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(1) Existence: Does there exist a solution ?

(2) Uniqueness: Is it the only solution ?(3) Stability: If the PDE changes a little,

does the solution change a little ?

The importance of the existence of solutions is trivial since, otherwise,the study on the PDE could be useless.

To explain the significance of the uniqueness of solutions, let us remem-ber the reason why we study the PDE. Usually, we discuss PDEs or theirsolutions to understand some specific phenomena in nature, engineerings oreconomics etc. Particularly, people working in applications want to knowhow the solution looks like, moves, behaves etc. For this purpose, it mightbe powerful to use numerical computations. However, numerical analysisonly shows us an approximate shapes, movements, etc. Thus, if thereare more than one solution, we do not know which is approximated by thenumerical solution.

Also, if the stability of solutions fails, we could not predict what will hap-pen from the numerical experiments even though the uniqueness of solutionsholds true.

Now, let us come back to the most essential question:

What is the solution of a PDE ?

For example, it is natural to call a function u : R a solution of(1.1) if there exist the first and second derivatives, Du(x) and D2u(x), forall x , and (1.1) is satisfied at each x when we plug them in the lefthand side of (1.1). Such a function u will be called a classical solution of(1.1).

However, unfortunately, it is difficult to seek for a classical solution be-cause we have to verify that it is sufficiently differentiable and that it satisfiesthe equality (1.1) simultaneously.

Instead of finding a classical solution directly, we have decided to choose

the following strategy:(A) Find a candidate of the classical solution,(B) Check the differentiability of the candidate.

In the standard books, the candidate of a classical solution is called aweak solution; if the weak solution has the first and second derivatives, then

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it becomes a classical solution. In the literature, showing the differentiability

of solutions is called the study on the regularity of those.Thus, with these terminologies, we may rewrite the above with mathe-

matical terms:

(A) Existence of weak solutions,(B) Regularity of weak solutions.

However, when we cannot expect classical solutions of a PDE to exist,what is the right candidate of solutions ?

We will call a function the candidate of solutions of a PDE if it is aunique and stable weak solution under a suitable setting. In section 2,

we will define such a candidate named viscosity solutions for a large classof PDEs, and in the proceeding sections, we will extend the definition tomore general (possibly discontinuous) functions and PDEs.

In the next subsection, we show a brief history on weak solutions toremind what was known before the birth of viscosity solutions.

1.2 Typical examples of weak solutions

In this subsection, we give two typical examples of PDEs to derive two kindsof weak solutions which are unique and stable.

1.2.1 Burgers equation

We consider Burgers equation, which is a model PDE in Fluid Mechanics:

u

t+

1

2

(u2)

x= 0 in R (0, ) (1.3)

under the initial condition:

u(x, 0) = g(x) for x R, (1.4)where g is a given function.

In general, we cannot find classical solutions of (1.3)-(1.4) even if g issmooth enough. See [8] for instance.In order to look for the appropriate notion of weak solutions, we first

introduce a function space C10 (R [0, )) as a test function space:

C10(R [0, )) :=

C1(R [0, )) there is K > 0 such thatsupp [K, K] [0, K]

.

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Here and later, we denote by supp the following set:

supp := {(x, t) R [0, ) | (x, t) = 0}.Suppose that u satisfies (1.3). Multiplying (1.3) by C10(R [0, ))

and then, using integration by parts, we have

R

0

u

t+

u2

2

x

(x, t)dtdx +

R

u(x, 0)(x, 0)dx = 0.

Since there are no derivatives of u in the above, this equality makes sense ifu K>0L1((K, K)(0, K)). Hence, we may adapt the following propertyas the definition of weak solutions of (1.3)-(1.4).

R

0

u

t+

u2

2

x

(x, t)dtdx +

R

g(x)(x, 0)dx = 0

for all C10(R [0, )).We often call this a weak solution in the distribution sense. As you no-

ticed, we derive this notion by an essential use of integration by parts. Wesay that a PDE is in divergence form when we can adapt the notion ofweak solutions in the distribution sense. When the PDE is not in divergence

form, we say that it is in nondivergence form.We note that the solution of (1.3) may have singularities even though theinitial value g belongs to C by an observation via characteristic method.From the definition of weak solutions, we can derive the so-called Rankine-Hugoniot condition on the set of singularities.

On the other hand, unfortunately, we cannot show the uniqueness of weaksolutions of (1.3)-(1.4) in general while we know the famous Lax-Oleinikformula (see [8] for instance), which is the expected solution.

In order to obtain the uniqueness of weak solutions, for the definition, weadd the following property (called entropy condition) which holds for theexpected solution given by the Lax-Oleinik formula: There is C > 0 such

that

u(x + z, t) u(x, t) Czt

for all (x,t,z) R (0, ) (0, ). We call u an entropy solution of (1.3)if it is a weak solution satisfying this inequality. It is also known that sucha weak solution has a certain stability property.

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We note that this entropy solution satisfies the above mentioned impor-

tant properties; existence, uniqueness and stability. Thus, it must be aright definition for weak solutions of (1.3)-(1.4).

1.2.2 Hamilton-Jacobi equations

Next, we shall consider general Hamilton-Jacobi equations, which arise inOptimal Control and Classical Mechanics:

u

t+ H(Du) = 0 in (x, t) Rn (0, ) (1.5)

under the same initial condition (1.4).In this example, we suppose that H : Rn R is convex, i.e.

H(p + (1 )q) H(p) + (1 )H(q) (1.6)for all p, q Rn, [0, 1].

Remark. Since a convex function is locally Lipschitz continuous in general,we do not need to assume the continuity of H.

Example. In Classical Mechanics, we often call this H a Hamiltonian.As a simple example of H, we have H(p) =

|p|2.

Notice that we cannot adapt the weak solution in the distribution sensefor (1.5) since we cannot use the integration by parts.

We next introduce the Lagrangian L : Rn R defined byL(q) = sup

pRn{p, q H(p)}.

When H(p) = |p|2, it is easy to verify that the maximum is attained in theright hand side of the above.

It is surprising that we have a neat formula for the expected solution

(called Hopf-Lax formula) presented by

u(x, t) = minyRn

tL

x y

t

+ g(y)

. (1.7)

More precisely, it is shown that the right hand side of (1.7) is differentiableand satisfies (1.5) almost everywhere.

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Thus, we could call u a weak solution of (1.5)-(1.4) when u satisfies (1.5)

almost everywhere. However, if we decide to use this notion as a weak solu-tion, the uniqueness of those fails in general. We will see an example in thenext section.

As was shown for Burgers equation, in order to say that the uniqueweak solution is given by (1.7), we have to add one more property for thedefinition of weak solutions: There is C > 0 such that

u(x + z, t) 2u(x, t) + u(x z, t) C|z|2 (1.8)for all x, z R, t > 0. This is called the semi-concavity of u.

We note that (1.8) is a hypothesis on the one-sided bound of second

derivatives of functions u.In 60s, Kruzkov showed that the limit function of approximate solutionsby the vanishing viscosity method (see the next section) has this property(1.8) when H is convex. He named u a generalized solution of (1.5) whenit satisfies (1.5) almost everywhere and (1.8).

To my knowledge, between Kruzkovs works and the birth of viscositysolutions, there had been no big progress in the study of first-order PDEs innondivergence form.

Remark. The convexity (1.6) is a natural hypothesis when we consideronly optimal control problems where one person intends to minimize somecosts (energy in terms of Physics). However, when we treat game prob-lems (one person wants to minimize costs while the other tries to maximizethem), we meet non-convex and non-concave (i.e. fully nonlinear)

Hamiltonians. See section 4.2.

In this book, since we are concerned with viscosity solutions of PDEs innondivergence form, for which the integration by parts argument cannot beused to define the notion of weak solutions in the distribution sense, we shallgive typical examples of such PDEs.

Example. (Bellman and Isaacs equations)We first give Bellman equations and Isaacs equations, which arise in

(stochastic) optimal control problems and differential games, respectively.As will be seen, those are extensions of linear PDEs.

Let A and B be sets of parameters. For instance, we suppose A and Bare (compact) subsets in Rm (for some m 1). For a A, b B, x ,

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r

R, p =t(p1, . . . , pn)

Rn, and X = (Xij)

Sn, we set

La(x,r,p,X) := trace(A(x, a)X) + g(x, a), p + c(x, a)r,La,b(x,r,p,X) := trace(A(x,a,b)X) + g(x,a,b), p + c(x,a,b)r.

Here A(, a), A(, a , b), g(, a), g(, a , b), c(, a) and c(, a , b) are given functionsfor (a, b) A B.

For inhomogeneous terms, we consider functions f(, a) and f(, a , b) in for a A and b B.

We call the following PDEs Bellman equations:

supaA{La(x, u(x), Du(x), D2u(x)) f(x, a)} = 0 for x . (1.9)

Notice that the supremum over A is taken at each point x .Taking account of one more parameter set B, we call the following PDEs

Isaacs equations:

supaA

infbB

{La,b(x, u(x), Du(x), D2u(x)) f(x,a,b)} = 0 for x (1.10)

and

infbB

supaA

{La,b(x, u(x), Du(x), D2u(x)) f(x,a,b)} = 0 for x . (1.10)

Example. (Quasi-linear equations)We say that a PDE is quasi-linear if the coefficients of D2u contains u

or Du. Although we will not study quasilinear PDEs in this book, we givesome of those which are in nondivergence form.

We first give the PDE of mean curvature type:

F(x,p,X) := |p|2trace(X) Xp,p

.

Notice that this F is independent of x-variables. We refer to [12] for appli-cations where this kind of operators appears.

Next, we show a relatively new one called L-Laplacian:F(x,p,X) := Xp,p.

Again, this F does not contain x-variables. We refer to Jensens work [16],where he first studied the PDE D2uDu, Du = 0 in via the viscositysolution approach.

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2 Definition

In this section, we derive the definition of viscosity solutions of (1.1) via thevanishing viscosity method.

We also give some basic properties of viscosity solutions and equivalentdefinitions using semi-jets.

2.1 Vanishing viscosity method

When the notion of viscosity solutions was born, in order to explain thereason why we need it, many speakers started in their talks by giving thefollowing typical example called the eikonal equation:

|Du|2 = 1 in . (2.1)

We seek C1 functions satisfying (2.1) under the Dirichlet condition:

u(x) = 0 for x . (2.2)

However, since there is no classical solution of (2.1)-(2.2) (showing the non-existence of classical solutions is a good exercise), we intend to derive areasonable definition of weak solutions of (2.1).

In fact, we expect that the following function (the distance from )would be the unique solution of this problem (see Fig 2.1):

u(x) = dist(x, ) := infy

|x y|.

Fig 2.1

1

1

10

y

y = u(x)

x

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If we consider the case when n = 1 and = (

1, 1), then the expected

solution is given by

u(x) = 1 |x| for x [1, 1]. (2.3)

Since this function is C except at x = 0, we could decide to call u a weaksolution of (2.1) if it satisfies (2.1) in except at finite points.

Fig 2.2

1

1

1 0 x

y

However, even in the above simple case of (2.1), we know that there areinfinitely many such weak solutions of (2.1) (see Fig 2.2); for example, u isthe weak solution and

u(x) =

x + 1 for x [1, 12

),x for x [1

2, 12

),x 1 for x [1

2, 1],

...etc.

Now, in order to look for an appropriate notion of weak solutions, weintroduce the so-called vanishing viscosity method; for > 0, we considerthe following PDE as an approximate equation of (2.1) when n = 1 and = (1, 1):

u + (u

)

2 = 1 in (1, 1),u(1) = 0.

(2.4)

The first term, u , in the left hand side of (2.4) is called the vanishingviscosity term (when n = 1) as 0.

By an elementary calculation, we can find a unique smooth function uin the following manner: We first note that if a classical solution of (2.4)

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exists, then it is unique. Thus, we may suppose that u(0) = 0 by symmetry.

Setting v = u, we first solve the ODE: v + v2 = 1 in (1, 1),v(0) = 0.

(2.5)

It is easy to see that the solution of (2.5) is given by

v(x) = tanh

x

.

Hence, we can find u by

u(x) = logcosh

x

cosh

1

= log

ex + e

x

e1

+ e1

.

It is a good exercise to show that u converges to the function in (2.3)uniformly in [1, 1].

Remark. Since u(x) := u(x) is the solution ofu + (u)2 = 1 in (1, 1),

u(

1) = 0,

we have u(x) := lim0 u(x) = u(x). Thus, if we replace u by +u,then the limit function would be different in general.

To define weak solutions, we adapt the properties which hold for the(uniform) limit of approximate solutions of PDEs with the minus vanishingviscosity term.

Let us come back to general second-order PDEs:

F(x,u,Du,D2u) = 0 in . (2.6)

We shall use the following definition of classical solutions:

Definition. We call u : R a classical subsolution (resp.,supersolution, solution) of (2.6) ifu C2() and

F(x, u(x), Du(x), D2u(x)) 0 (resp., 0, = 0) in .

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Remark. IfF does not depend on X-variables (i.e. F(x,u,Du) = 0; first-

order PDEs), we only suppose u C1() in the above in place ofu C2().Throughout this text, we also suppose the following monotonicity condi-

tion with respect to X-variables:

Definition. We say that F is (degenerate) elliptic if

F(x,r,p,X) F(x,r,p,Y)

for all x , r R, p Rn, X , Y Sn provided X Y. (2.7)

We notice that if F does not depend on X-variables (i.e. F = 0 is thefirst-order PDE), then F is automatically elliptic.

We also note that the left hand side F(x,r,p,X) = trace(X) of theLaplace equation (1.2) is elliptic.

We will derive properties which hold true for the (uniform) limit (as +0) of solutions of

u + F(x,u,Du,D2u) = 0 in ( > 0). (2.8)

Note that since trace(X) + F(x,r,p,X) is uniformly elliptic (see insection 3 for the definition) provided that F is elliptic, it is easier to solve

(2.8) than (2.6) in practice. See [13] for instance.

Proposition 2.1. Assume that F is elliptic. Let u C2() C()be a classical subsolution (resp., supersolution) of (2.8). If u converges tou C() (as 0) uniformly in any compact sets K , then, for any C2(), we have

F(x, u(x), D(x), D2(x)) 0 (resp., 0)

provided that u attains its maximum (resp., minimum) at x .

Remark. When F does not depend on X-variables, we only need to sup-pose and u to be in C

1() as before.

Proof. We only give a proof of the assertion for subsolutions since theother one can be shown in a symmetric way.

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Suppose that u

attains its maximum at x

for

C2(). Setting

(y) := (y) + |y x|4 for small > 0, we see that(u )(x) > (u )(y) for y \ {x}.

(This tiny technique to replace a maximum point by a strict one will appearin Proposition 2.2.)

Let x be a point such that (u )(x) = max(u ). Notethat x also depends on > 0.

Since u converges to u uniformly in Br(x) and x is the unique maximumpoint of u , we note that lim0 x = x. Thus, we see that x forsmall > 0. Notice that if we argue by instead of, the limit ofx mightdiffer from x.

Thus, at x , we have

u(x) + F(x, u(x), Du(x), D2u(x)) 0.

Since D(u )(x) = 0 and D2(u )(x) 0, in view of ellipticity, wehave

(x) + F(x, u(x), D(x), D2(x)) 0.Sending 0 in the above, we have

F(x, u(x), D(x), D2(x)) 0.Since D(x) = D(x) and D

2(x) = D2(x), we conclude the proof. 2

Definition. We call u : R a viscosity subsolution (resp.,supersolution) of (2.6) if, for any C2(),

F(x, u(x), D(x), D2(x)) 0 (resp., 0)

provided that u attains its maximum (resp., minimum) at x .We call u :

R a viscosity solution of (2.6) if it is both a viscosity

sub- and supersolution of (2.6).

Remark. Here, we have given the definition to general functions butwe will often suppose that they are (semi-)continuous in Theorems etc.

In fact, in our propositions in sections 2.1, we will suppose that viscositysub- and supersolutions are continuous.

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However, all the proposition in section 2.1 can be proved by replacing up-

per and lower semi-continuity for viscosity subsolutions and supersolutions,respectively.

We will introduce general viscosity solutions in section 3.3.

Notation. In order to memorize the correct inequality, we will oftensay that u is a viscosity subsolution (resp., supersolution) of

F(x,u,Du,D2u) 0 (resp., 0) in

if it is a viscosity subsolution (resp., supersolution) of (2.6).

Proposition 2.2. For u : R, the following (1) and (2) are equiva-lent:

(1) u is a viscosity subsolution (resp., supersolution) of (2.6),(2) if 0 = (u )(x) > (u )(x) (resp., < (u )(x))

for C2(), x and x \ {x},then F(x, (x), D(x), D2(x)) 0 (resp., 0).

Fig 2.3

graph ofgraph of

graph of () + (u )(x)

xx

graph ofu

Proof. The implication (1) (2) is trivial.For the opposite implication in the subsolution case, suppose that u

attains a maximum at x . Set

(x) = (x) + |x x|4 + (u )(x).

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See Fig 2.3. Since 0 = (u

)(x) > (u

)(x) for x

\ {x

}, (2) gives

F(x, (x), D(x), D2(x)) 0,

which implies the assertion. 2

By the next proposition, we recognize that viscosity solutions are rightcandidates of weak solutions when F is elliptic.

Proposition 2.3. Assume that F is elliptic. A function u : Ris a classical subsolution (resp., supersolution) of (2.6) if and only if it is aviscosity subsolution (resp., supersolution) of (2.6) and u

C2().

Proof. Suppose that u is a viscosity subsolution of (2.6) and u C2().Taking u, we see that u attains its maximum at any points x .Thus, the definition of viscosity subsolutions yields

F(x, u(x), Du(x), D2u(x)) 0 for x .

On the contrary, suppose that u C2() is a classical subsolution of(2.6).

Fix any C2(). Assuming that u takes its maximum at x ,we have D(u )(x) = 0 and D2(u )(x) 0.Hence, in view of ellipticity, we have

0 F(x, u(x), Du(x), D2u(x)) F(x, u(x), D(x), D2(x)). 2

We introduce the sets of upper and lower semi-continuous functions: ForK Rn,

USC(K) := {u : K R | u is upper semi-continuous in K},

and

LSC(K) := {u : K R | u is lower semi-continuous in K}.

Remark. Throughout this book, we use the following maximum principlefor semi-continuous functions:

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An upper semi-continuous function in a compact set attains its maximum.

We give the following lemma which will be used without mentioning it.Since the proof is a bit technical, the reader may skip it over first.

Proposition 2.4. Assume that u U SC() (resp., u LSC()) is a viscos-ity subsolution (resp., supersolution) of(2.6) in . Assume also that infKu > (resp., supKu < ) for any compact sets K .

Then, for any open set , u is a viscosity subsolution (resp., supersolution)of (2.6) in .

Proof. We only show the assertion for subsolutions since the other can be shownsimilarly.

For C2(), by Proposition 2.2, we suppose that for some x ,0 = (u )(x) > (u )(y) for all y \ {x}.

Choose r > 0 such that B3r(x) , and set s := infB2r(x)(u ) > 0.We recall the standard mollifier C(Rn) and supp B: First, we set

(x) := C0e

1

|x|21 for |x| < 1,

0 for |x| 1,where C0 := B1 e

1

|x|21 dx1

.

Next, we define by

(x) :=1

n

x

.

Notice thatRn dx = 1 for > 0.

Now, setting (x) := (x) =Rn (y)(x y)dy C(Rn), where

(x) :=

(x) for x Br(x),2s + sup

u for x \ Br(x),

we easily verify that for 0 < < r,

max\B2r(x)

(u ) 2s.

On the other hand, since we have

(u )(x) =B(x)

(x y)((x) (y))dy,

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there is 0 > 0 such that

(u )(x) s for (0, 0).Therefore, we can choose x B2r(x) such that max(u ) = (u )(x).

Thus, taking a subsequence of {x}>0 denoted by x again (if necessary),we find z B2r(x) such that lim0 x = z. Hence, sending 0, we havemaxB2r(x)(u ) = (u )(z), which gives z = x

By the definition, we have

F(x, u(x), D(x), D2(x)) 0.

Note that lim0 u(x) = u(x). Therefore, sending 0 in the above, we concludethe proof. 2

2.2 Equivalent definitions

We present equivalent definitions of viscosity solutions. However, since wewill need those in the proof of uniqueness for second-order PDEs,

the reader may postpone this subsection until section 3.3.

First, we introduce semi-jets of functions u : R at x by

J2,+u(x) :=

(p, X) R

n Sn

u(y) u(x) + p, y x+

1

2X(y

x), y

x

+o(|y x|2) as y x

and

J2,u(x) :=

(p, X) R

n Sn

u(y) u(x) + p, y x+

1

2X(y x), y x

+o(|y x|2) as y x

.

Note that J2,u(x) = J2,+(u)(x).

Remark. We do not impose any continuity for u in these definitions.

We recall the notion of small order o in the above: For k 1,f(x) o(|x|k) (resp., o(|x|k)) as x 0

there is C([0, ), [0, )) such that (0) = 0, andsup

xBr\{0}

f(x)

|x|k (r)

resp., infxBr\{0}

f(x)

|x|k (|x|)

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In the next proposition, we give some basic properties of semi-jets: (1)

is a relation between semi-jets and classical derivatives, and (2) means thatsemi-jets are defined in dense sets of .

Proposition 2.5. For u : R, we have the following:(1) If J2,+u(x) J2,u(x) = , then Du(x) and D2u(x) exist and,

J2,+u(x) J2,u(x) = {(Du(x), D2u(x))}.(2) If u USC() (resp., u LSC()), then

= x

xk

such that J2,+u(xk)

=

, limk

xk = xresp., =

x

xk such that J2,u(xk) = , limk

xk = x

.

Proof. The proof of (1) is a direct consequence from the definition.We give a proof of the assertion (2) only for J2,+.Fix x and choose r > 0 so that Br(x) . For > 0, we can choose

x Br(x) such that u(x) 1|x x|2 = maxyBr(x)(u(y) 1|y x|2).Since |x x|2 (maxBr(x) u(x)), we see that x converges to x Br(x)as 0. Thus, we may suppose that x Br(x) for small > 0. Hence, wehave

u(y) u(x) +1

(|y x|2 |x x|2) for all y Br(x).It is easy to check that (2(x x)/, 21I) J2,+u(x). 2

We next introduce a sort of closure of semi-jets:

J2,

u(x) :=

(p, X) Rn Sn

xk and (pk, Xk) J2,u(xk)

such that (xk, u(xk), pk, Xk) (x, u(x), p , X ) as k

.

Proposition 2.6. For u : R, the following (1), (2), (3) are equiva-lent.

(1) u is a viscosity subsolution (resp., supersolution) of (2.6).(2) For x and (p, X) J2,+u(x) (resp., J2,u(x)),

we have F(x, u(x), p , X ) 0 (resp., 0).(3) For x and (p, X) J2,+u(x) (resp., J2,u(x)),

we have F(x, u(x), p , X ) 0 (resp., 0).

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Proof. Again, we give a proof of the assertion only for subsolutions.Step 1: (2) = (3). For x and (p, X) J2,+u(x), we can find (pk, Xk)

J2,+u(xk) with xk such that limk(xk, u(xk), pk, Xk) = (x, u(x), p , X )and

F(xk, u(xk), pk, Xk) 0,which implies (3) by sending k .

Step 2: (3) = (1). For C2(), suppose also (u)(x) = max(u).Thus, the Taylor expansion of at x gives

u(y)

u(x)+D(x), y

x

+1

2D2(x)(y

x), y

x

+o(|x

y|2) as y

x.

Thus, we have (D(x), D2(x)) J2,+u(x) J2,+u(x).Step 3: (1) = (2). For (p, X) J2,+u(x) (x ), we can find nonde-

creasing, continuous : [0, ) [0, ) such that (0) = 0 and

u(y) u(x) + p, y x + 12X(y x), y x + |y x|2(|y x|) (2.9)

as y x. In fact, by the definition of o, we find 0 C([0, ), [0, )) suchthat 0(0) = 0, and

0(r) supyBr(x)\{x}

1

|x y|2

u(y) u(x) p, y x 12X(y x), y x

,

we verify that (r) := sup0tr 0(t) satisfies (2.9).Now, we define by

(y) := p, y x + 12X(y x), y x + (|x y|),

where

(t) :=3t

t2s

s (r)dr

ds t2

(t).It is easy to check that

(D(x), D2(x)) = (p, X) and (u )(x) (u )(y) for y .

Therefore, we conclude the proof. 2

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Remark. In view of the proof of Step 3, we verify that for x

,

J2,+u(x) =

(D(x), D2(x)) Rn Sn

C2() such that u

attains its maximum at x

,

J2,u(x) =

(D(x), D2(x)) Rn Sn

C2() such that u

attains its minimum at x

.

Thus, we intuitively know J2,u(x) from their graph.

Example. Consider the function u C([1, 1]) in (2.3). From the graphbelow, we may conclude that J2,u(0) =

, and J2,+u(0) = (

{1

} [0,

))

({1} [0, )) ((1, 1) R). See Fig 2.4.1 and 2.4.2.We omit how to obtain J2,u(0) of this and the next examples.We shall examine J2, for discontinuous functions. For instance, consider

the Heaviside function:

u(x) :=

1 for x 0,0 for x < 0.

We see that J2,u(0) = and J2,+u(0) = ({0} [0, )) ((0, ) R). SeeFig 2.5.

Fig 2.4.1

y = (x)

y = x + 1y

y = u(x)

x

In order to deal with boundary value problems in section 5, we preparesome notations: For a set K Rn, which is not necessarily open, we define

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Fig 2.4.2

y = (x)

y

y = u(x)

x

y = ax + 1 (|a| < 1)

Fig 2.5

y = (x)y

y = u(x)

x

1

0

y = ax + 1(a > 0)

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semi-jets of u : K

R at x

K by

J2,+K u(x) :=

(p, X) R

n Sn

u(y) u(x) + p, y x+

1

2X(y x), y x

+o(|y x|2) as y K x

,

J2,K u(x) :=

(p, X) R

n Sn

u(y) u(x) + p, y x+

1

2X(y x), y x

+o(|y x|2) as y K x

,

and

J2,K u(x) :=

(p, X) Rn Sn

xk K and (pk, Xk) J2,K u(xk)

such that (xk, u(xk), pk, Xk) (x, u(x), p , X ) as k

.

Remark. It is obvious to verify that

x = J2, u(x) = J2, u(x) and J2, u(x) = J

2, u(x).

For x , we shall simply write J2,u(x) (resp., J2,u(x)) for J2, u(x) =J2,

u(x) (resp, J2, u(x) = J

2, u(x)).

Example. Consider u(x) 0 in K := [0, 1]. It is easy to observe thatJ2,+u(x) = J2,+K u(x) = {0} [0, ) provided x (0, 1). It is also easy toverify that

J2,+K u(0) = ({0} [0, )) ((0, ) R),and

J2,K u(0) = ({0} (, 0]) ((, 0) R).We finally give some properties of J2,

and J

2, . Since the proof is easy,

we omit it.

Proposition 2.7. For u : R, C2() and x , we haveJ2,

(u + )(x) = (D(x), D2(x)) + J2,

u(x)

andJ2, (u + )(x) = (D(x), D

2(x)) + J2, u(x).

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3 Comparison principle

In this section, we discuss the comparison principle, which implies the unique-ness of viscosity solutions when their values on coincide (i.e. under theDirichlet boundary condition). In the study of the viscosity solution theory,the comparison principle has been the main issue because the uniqueness ofviscosity solutions is harder to prove than existence and stability of them.

First, we recall some classical comparison principles and then, showhow to modify the proof to a modern viscosity version.

In this section, the comparison principle roughly means that

Comparison principle

viscosity subsolution uviscosity supersolution v

u v on

= u v in

Modifying our proofs of comparison theorems below, we obtain a slightlystronger assertion than the above one:

viscosity subsolution uviscosity supersolution v

= max

(u v) = max

(u v)

We remark that the comparison principle implies the uniqueness of (con-tinuous) viscosity solutions under the Dirichlet boundary condition:

Uniqueness for the Dirichlet problem

viscosity solutions u and vu = v on

= u = v in

Proof of the comparison principle implies the uniqueness.Since u (resp., v) and v (resp., u), respectively, are a viscosity subsolution

and supersolution, by u = v on , the comparison principle yields u v(resp., v u) in . 2

In this section, we mainly deal with the following PDE instead of (2.6).

u + F(x,Du,D2u) = 0 in , (3.1)

where we suppose that 0, (3.2)

andF : Rn Sn R is continuous. (3.3)

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3.1 Classical comparison principle

In this subsection, we show that if one of viscosity sub- and supersolutionsis a classical one, then the comparison principle holds true. We call this theclassical comparison principle.

3.1.1 Degenerate elliptic PDEs

We first consider the case when F is (degenerate) elliptic and > 0.

Proposition 3.1. Assume that > 0 and (3.3) hold. Assume alsothat F is elliptic. Let u USC() (resp., v LSC()) be a viscositysubsolution (resp., supersolution) of (3.1) and v LSC() C2() (resp.,u USC() C2()) a classical supersolution (resp., subsolution) of (3.1).

If u v on , then u v in .

Proof. We only prove the assertion when u is a viscosity subsolution of(3.1) since the other one can be shown similarly.

Set max(u v) =: and choose x such that (u v)(x) = .Suppose that > 0 and then, we will get a contradiction. We note that

x because u v on .Thus, the definition of u and v respectively yields

u(x) + F(x,Dv(x), D2v(x)) 0 v(x) + F(x,Dv(x), D2v(x)).

Hence, by these inequalities, we have

= (u v)(x) 0,

which contradicts > 0. 2

3.1.2 Uniformly elliptic PDEs

Next, we present the comparison principle when = 0 but F is uniformly

elliptic in the following sense. Notice that if > 0 and F is uniformly ellip-tic, then Proposition 3.1 yields Proposition 3.3 below because our uniformellipticity implies (degenerate) ellipticity.

Throughout this book, we freeze the uniform ellipticity constants:

0 < .

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With these constants, we introduce the Puccis operators: For X

Sn,

P+(X) := max{trace(AX) | I A I for A Sn},

P(X) := min{trace(AX) | I A I for A Sn}.We give some properties ofP. We omit the proof since it is elementary.

Proposition 3.2. For X, Y Sn, we have the following:(1) P+(X) = P(X),(2) P(X) = P(X) for 0,

(3) P+

is convex, P is concave,(4)

P(X) + P(Y) P(X+ Y) P(X) + P+(Y) P+(X+ Y) P+(X) + P+(Y).

Definition. We say that F : Rn Sn R is uniformly elliptic(with the uniform ellipticity constants 0 < ) if

P(X Y) F(x,p,X) F(x,p,Y) P+(X Y)

for x , p Rn, and X, Y Sn.

We also suppose the following continuity on F with respect to p Rn:There is > 0 such that

|F(x,p,X) F(x, p, X)| |p p| (3.4)

for x , p , p Rn, and X Sn.

Proposition 3.3. Assume that (3.2), (3.3) and (3.4) hold. Assume alsothat F is uniformly elliptic. Let u USC() (resp., v LSC()) be aviscosity subsolution (resp., supersolution) of(3.1) and v LSC() C2()(resp., u

USC()

C2()) a classical supersolution (resp., subsolution) of

(3.1).If u v on , then u v in .

Proof. We give a proof only when u is a viscosity subsolution and v aclassical supersolution of (3.1).

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Suppose that max(u

v) =: > 0. Then, we will get a contradiction

again.For > 0, we set (x) = e

x1 , where := max{( + 1)/,+ 1} > 0.We next choose > 0 so small that

maxx

ex1 2

Let x be the point such that (u v + )(x) = max(u v + ) .By the choice of > 0, since u v on , we see that x .

From the definition of viscosity subsolutions, we have

u(x) + F(x, D(v )(x), D2

(v )(x)) 0.By the uniform ellipticity and (3.4), we have

u(x) + F(x,Dv(x), D2v(x)) + P(D2(x)) |D(x)| 0.Noting that |D(x)| ex1 and P(D2(x)) 2ex1 , we have

u(x) + F(x,Dv(x), D2v(x)) + ( )ex1 0. (3.5)Since v is a classical supersolution of (3.1), by (3.5) and ( + 1)/, wehave

(u v)(x) + ex1 0.Hence, we have

( (x)) ex1,which gives a contradiction because + 1. 2

3.2 Comparison principle for first-order PDEs

In this subsection, without assuming that one of viscosity sub- and supersolu-tions is a classical one, we establish the comparison principle when F in (3.1)does not depend on D2u; first-order PDEs. We will study the comparisonprinciple for second-order ones in the next subsection.

In the viscosity solution theory, Theorem 3.4 below was the first surprisingresult.

Here, instead of (3.1), we shall consider the following PDE:

u + H(x,Du) = 0 in . (3.6)

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We shall suppose that

> 0, (3.7)

and that there is a continuous function H : [0, ) [0, ) such thatH(0) = 0 and

|H(x, p) H(y, p)| H(|x y|(1 + |p|)) for x, y and p Rn. (3.8)In what follows, we will call H in (3.8) a modulus of continuity. For

notational simplicity, we use the following notation:

M := { : [0, ) [0, ) | () is continuous, (0) = 0}.

Theorem 3.4. Assume that (3.7) and (3.8) hold. Let u U SC() andv LSC() be a viscosity sub- and supersolution of (3.6), respectively.


Proof. Suppose max(u v) =: > 0 as usual. Then, we will get acontradiction.

Notice that since both u and v may not be differentiable, we cannot usethe same argument as in Proposition 3.1.

Now, we present the most important idea in the theory of viscosity solu-tions to overcome this difficulty.

Setting (x, y) := u(x) v(y) (2)1|x y|2 for > 0, we choose(x, y) such that

(x, y) = maxx,y

(x, y).

Noting that (x, y) maxx (x, x) = , we have|x y|2

2 u(x) v(y) . (3.9)

Since is compact, we can find x, y

, and k > 0 such that limk

k = 0

and limk(xk , yk) = (x, y).We shall simply write for k (i.e. in what follows, 0 means that

k 0 when k ).Setting M := max u min v, by (3.9), we have

|x y|2 2M 0 (as 0).

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Thus, we have x = y.

Since (3.9) again implies

0 lim inf0

|x y|22

lim sup0

|x y|22

lim sup0

(u(x) v(y)) (u v)(x) 0,

we have

lim0

|x y|2

= 0. (3.10)

Moreover, since (u

v)(x) = > 0, we have x

from the assumption

u v on . Thus, for small > 0, we may suppose that (x, y) .Furthermore, ignoring the left hand side in (3.9), we have

lim inf0

(u(x) v(y)). (3.11)

Taking (x) := v(y) + (2)1|x y|2, we see that u attains its

maximum at x . Hence, from the definition of viscosity subsolutions, wehave

u(x) + H

x,x y

0.

On the other hand, taking (y) := u(x)

(2)1

|y

x

|2, we see that

v attains its minimum at y . Thus, from the definition of viscositysupersolutions, we have

v(y) + H

y,x y

0.

The above two inequalities yield

(u(x) v(y)) H|x y| + |x y|

2

.

Sending

0 in the above together with (3.10) and (3.11), we have

0,

which is a contradiction. 2

Remark. In the above proof, we could show that lim0 u(x) = u(x) andlim0 v(y) = v(x) although we do not need this fact. In fact, by (3.9), wehave

v(y) u(x) ,

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which implies that

v(x) lim inf0

v(y) lim inf0

u(x) lim sup0

u(x) u(x) ,

and

v(x) lim inf0

v(y) lim sup0

v(x) lim sup0

u(x) u(x) .

Hence, since all the inequalities become the equalities, we have

u(x) = lim inf0 u

(x) = lim sup0

u(x) and v(x) = lim inf0 v

(y) = lim sup0

v(y).

We remark here that we cannot apply Theorem 3.4 to the eikonal equation(2.1) because we have to suppose > 0 in the above proof.

We shall modify the above proof so that the comparison principle forviscosity solutions of (2.1) holds.

To simplify our hypotheses, we shall consider the following PDE:

H(x,Du) f(x) = 0 in . (3.12)Here, we suppose that H has homogeneous degree > 0 with respect to thesecond variable; there is > 0 such that

H(x,p) = H(x, p) for x , p Rn and > 0. (3.13)To recover the lack of assumption > 0, we suppose the positivity of f C(); there is > 0 such that

minx

f(x) =: > 0. (3.14)

Example. When H(x, p) = |p|2 (i.e. = 2) and f(x) 1 ( i.e. = 1),equation (3.12) becomes (2.1).

The second comparison principle for first-order PDEs is as follows:

Theorem 3.5. Assume that (3.8), (3.13) and (3.14) hold. Let u USC() and v LSC() be a viscosity sub- and supersolution of (3.12),respectively.


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Proof. Suppose that max(u

v) =: > 0 as usual. Then, we will get a

contradiction.If we choose (0, 1) so that

(1 )max

u 2

,

then we easily verify that

max

(u v) =: 2

.

We note that for any z

such that (u

v)(z) = , we may suppose

z . In fact, otherwise (i.e. z ), if we further suppose that < 1 isclose to 1 so that (1 )min v /4, then the assumption (u v on) implies

2 = u(z) v(z) ( 1)v(z)

4,

which is a contradiction. For simplicity, we shall omit writing the dependenceon for and (x, y) below.

At this stage, we shall use the idea in the proof of Theorem 3.4: Considerthe mapping : R defined by

(x, y) := u(x) v(y) |x

y

|2

2 .

Choose (x, y) such that maxx,y (x, y) = (x, y). Notethat (x, y) /2.

As in the proof of Theorem 3.4, we may suppose that lim0(x, y) =(x, y) for some (x, y) (by taking a subsequence if necessary). Also,we easily see that

|x y|22

u(x) v(y) M := max

u min

v. (3.15)

Thus, sending 0, we have x = y. Hence, (3.15) implies that u(x) v(x) = , which yields x because of the choice of . Thus, we see that(x, y) for small > 0.

Moreover, (3.15) again implies

lim0

|x y|2

= 0. (3.16)

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Now, taking (x) := (v(y)+(2)1

|x

y

|2)/, we see that u

attains

its maximum at x . Thus, we have

H

x,

x y

f(x).

Hence, by (3.13), we have

H

x,x y

f(x). (3.17)

On the other hand, taking (y) = u(x) (2)1|y x|2, we see thatv

attains its minimum at y

. Thus, we have

H

y,x y

f(y). (3.18)

Combining (3.18) with (3.17), we have

f(y) f(x) H

y,x y

H

x,

x y

H|x y|

1 +

|x y|

.

Sending

0 in the above with (3.16), we have

(1 )f(x) 0,which contradicts (3.14). 2

3.3 Extension to second-order PDEs

In this subsection, assuming a key lemma, we will present the comparisonprinciple for fully nonlinear, second-order, (degenerate) elliptic PDEs (3.1).

We first remark that the argument of the proof of the comparison principle

for first-order PDEs cannot be applied at least immediately.Let us have a look at the difficulty. Consider the following simple PDE:

u u = 0, (3.19)where > 0. As one can guess, if the argument does not work for thiseasiest PDE, then it must be hopeless for general PDEs.

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However, we emphasize that the same argument as in the proof of The-

orem 3.4 does not work. In fact, let u USC() and v LSC() be aviscosity sub- and supersolution of (3.19), respectively, such that u v on. Setting (x, y) := u(x) v(y) (2)1|x y|2 as usual, we choose(x, y) so that maxx,y (x, y) = (x, y) > 0 as before.

We may suppose that (x, y) converges to (x, x) (as 0) forsome x such that (u v)(x) > 0. From the definitions of u and v, wehave

u(x) n

0 v(y) + n

.

Hence, we only have

(u(x) v(y)) 2n

,which does not give any contradiction as 0.

How can we go beyond this difficulty ?

In 1983, P.-L. Lions first obtained the uniqueness of viscosity solutionsfor elliptic PDEs arising in stochastic optimal control problems (i.e. Bell-man equations; F is convex in (Du,D2u)). However, his argument heavilydepends on stochastic representation of viscosity solutions as value func-tions. Moreover, it seems hard to extend the result to Isaacs equations; F

is fully nonlinear.The breakthrough was done by Jensen in 1988 in case when the coeffi-

cients on the second derivatives of the PDE are constant. His argument reliespurely on real-analysis and can work even for fully nonlinear PDEs.

Then, Ishii in 1989 extended Jensens result to enable us to apply toelliptic PDEs with variable coefficients. We present here the so-called Ishiislemma, which will be proved in Appendix.

Lemma 3.6. (Ishiis lemma) Let u and w be in USC(). For C2( ), let (x, y) be a point such that

maxx,y

(u(x) + w(y) (x, y)) = u(x) + w(y) (x, y).

Then, for each > 1, there are X = X(), Y = Y() Sn such that

(Dx(x, y), X) J2,+ u(x), (Dy(x, y), Y) J2,+

w(y),

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and

( + A) I 00 I X 00 Y A + 1A2,where A = D2(x, y) S2n.

Remark. We note that if we suppose that u, w C2() and (x, y) in the hypothesis, then we easily have

X = D2u(x), Y = D2w(y), and

X 00 Y

A.

Thus, the last matrix inequality means that when u and w are only contin-

uous, we get some error term 1

A2

, where > 1 will be large.We also note that for (x, y) := |x y|2/(2), we have

A := D2(x, y) =1

I I

I I

and A = 2

. (3.20)

For the last identity, since

A2 := sup

A

xy

, A

xy

|x|2 + |y|2 = 1

,

the triangle inequality yields A2 = 22 sup{|x y|2 | |x|2 + |y|2 = 1} 4/

2

. On the other hand, taking x = y (i.e. |x|2

= 1/2) in the supremumof the definition of A2 in the above, we have A2 4/2.

Remark. The other way to show the above identity, we may use the factthat for B Sn, in general,

B = max{|k| | k is the eigen-value of B}.

3.3.1 Degenerate elliptic PDEs

Now, we give our hypotheses on F, which is called the structure condition.

Structure conditionThere is an F M such that if X, Y Sn and > 1 satisfy

3

I 00 I

X 00 Y

3

I I

I I

,

then F(y, (x y), Y) F(x, (x y), X) F(|x y|(1 + |x y|)) for x, y .

(3.21)

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In section 3.3.2, we will see that if F satisfies (3.21), then it is elliptic.

We first prove the comparison principle when (3.21) holds for F usingthis lemma. Afterward, we will explain why assumption (3.21) is reasonable.

Theorem 3.7. Assume that > 0 and (3.21) hold. Let u USC()and v LSC() be a viscosity sub- and supersolution of (3.1), respectively.


Proof. Suppose that max(u v) =: > 0 as usual. Then, we will get acontradiction.

Again, for > 0, consider the mapping : R defined by

(x, y) = u(x) v(y) 12

|x y|2.

Let (x, y) be a point such that maxx,y (x, y) = (x, y) . As in the proof of Theorem 3.4, we may suppose that

lim0

(x, y) = (x, x) for some x (i.e. x, y for small > 0).

Moreover, since we have (u v)(x) = ,

lim0 |x

y

|2

= 0, (3.22)

and lim inf

0(u(x) v(y)). (3.23)

In view of Lemma 3.6 (taking w := v, := 1/, (x, y) = |xy|2/(2))and its Remark, we find X, Y Sn such that

x y

, X

J2,+u(x),

x y

, Y

J2,v(y),

and 3

I 00 I

X 00 Y

3

I I

I I

.

Thus, the equivalent definition in Proposition 2.6 implies that

u(x) + F

x,x y

, X

0 v(y) + F

y,

x y

, Y

.

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Hence, by virtue of our assumption (3.21), we have

(u(x) v(y)) F|x y| + |x y|

2

. (3.24)

Taking the limit infimum, as 0, together with (3.22) and (3.23) in theabove, we have

0,which is a contradiction. 2

3.3.2 Remarks on the structure condition

In order to ensure that assumption (3.21) is reasonable, we first present someexamples. For this purpose, we consider the Isaacs equation as in section1.2.2.

F(x,p,X) := supaA

infbB

{La,b(x,p,X) f(x,a,b)},

where

La,b(x,p,X) := trace(A(x,a,b)X) + g(x,a,b), p for (a, b) A B.

If we suppose that A and B are compact sets in Rm (for some m 1),and that the coefficients in the above and f(, a , b) satisfy the hypothesesbelow, then F satisfies (3.21).

(1) M1 > 0 and ij(, a , b) : R such that Aij(x,a,b) =mk=1

ik(x,a,b)jk(x,a,b), and |jk(x,a,b) jk(y,a,b)| M1|x y|for x, y , i , j = 1, . . . , n , k = 1, . . . , m, a A, b B,

(2) M2 > 0 such that |gi(x,a,b) gi(y, a , b)| M2|x y| for x, y ,i = 1, . . . , n , a A, b B,

(3) f M such that

|f(x,a,b)

f(y, a , b)

| f(

|x

y|) for x, y

, a

A, b

B.

We shall show (3.21) only when

F(x,p,X) := n

i,j=1

mk=1

ik(x,a,b)jk(x,a,b)Xij

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for a fixed (a, b)

A

B because we can modify the proof below to general

F.Thus, we shall omit writing indices a and b.To verify assumption (3.21), we choose X, Y Sn such that

X 00 Y

3

I I

I I

.

Setting k =t(1k(x), . . . , nk(x)) and k =

t(1k(y), . . . , nk(y)) for anyfixed k {1, 2, . . . , m}, we have

X 00 Y

k

k

, k

k 3

I

I

I I k

k

, k

k

= 3|k k|2 3nM21 |x y|2.

Therefore, taking the summation over k {1, . . . , m}, we have

F(y, (x y), Y) F(x, (x y), X) n

i,j=1

(Aij(y)Yij + Aij(x)Xij)

=mk=1

(Y k, k + Xk, k)

3mnM21 |x y|

2

.2

We next give other reasons why (3.21) is a suitable assumption. Thereader can skip the proof of the following proposition if he/she feels that theabove reason is enough to adapt (3.21).

Proposition 3.8. (1) (3.21) implies ellipticity.(2) Assume that F is uniformly elliptic. If M satisfies that supr0 (r)/(r +1) < , and

|F(x,p,X) F(y ,p,X)| (|x y|(X + |p| + 1)) (3.25)

for x, y , p Rn, X Sn, then (3.21) holds for F.

Proof. For a proof of (1), we refer to Remark 3.4 in [6].For the readers convenience, we give a proof of (2) which is essentially used

in a paper by Ishii-Lions (1990). Let X, Y Sn satisfy the matrix inequality in(3.21). Note that X Y.

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Multiplying I II I to the last matrix inequality from both sides, wehave

X Y X + YX+ Y X Y

12

0 00 I

.

Thus, multiplying

s

for s R and , Rn with || = || = 1, we see that

0 (12 (X Y), )s2 2(X + Y), s (X Y), .

Hence, we have

|(X + Y), |2 |(X Y), |(12 + |(X Y), |),which implies

X + Y X Y1/2(12 + X Y)1/2.Thus, we have

X 12

(X Y + X + Y) X Y1/2(6 + X Y)1/2.

Since X Y (i.e. the eigen-values of X Y are non-positive), we see that

F(y ,p,X)

F(y ,p,Y) P

(X

Y)

X

Y

. (3.26)

For the last inequality, we recall Remark after Lemma 3.6.Since we may suppose is concave, for any fixed > 0, there is M > 0 such

that (r) + Mr and (r) = inf>0( + Mr) for r 0. By (3.25) and (3.26),since X 3 and Y 3, we have

F(y,p,Y) F(x,p,X) + M|x y|(|p| + 1) + sup

0t6

M|x y|t1/2(6 + t)1/2 t

.

Noting that

M|x y|t1/2

(6 + t)1/2

t 3

M2 |x y|

2

,

we haveF(y, (x y), Y) F(x, (x y), X)

+ M|x y|(|x y| + 1) + 31M2 |x y|2,which implies the assertion by taking the infimum over > 0. 2

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3.3.3 Uniformly elliptic PDEs

We shall give a comparison result corresponding to Proposition 3.3; F isuniformly elliptic and 0.

Theorem 3.9. Assume that (3.2), (3.3), (3.4) and (3.21) hold. Assumealso that F is uniformly elliptic. Let u USC() and v LSC() be aviscosity sub- and supersolution of (3.1), respectively.


Remark. As in Proposition 3.3, we may suppose = 0.

Proof. Suppose that max(u v) =: > 0.Setting := ( + 1)/, we choose > 0 so that

maxx

ex1 2

.

We then set := maxx(u(x) v(x) + ex1) > 0.Putting (x, y) := (2)1|x y|2 ex1 , we let (x, y) be the

maximum point of u(x) v(y) (x, y) over .By the compactness of , we may suppose that (x, y) (x, y)

as

0 (taking a subsequence if necessary). Since u(x)

v(y)

(x, y),we have |xy|2 2(max umin v +21) and moreover, x = y. Hence,we have

u(x) v(x) + ex1 ,which implies x because of our choice of . Thus, we may suppose that(x, y) for small > 0. Moreover, as before, we see that

lim0

|x y|2

= 0. (3.27)

Applying Lemma 3.6 to u(x) := u(x)+ex1 and

v(y), we find X, Y

Sn

such that ((x y)/,X) J2,+

u(x), ((x y)/,Y) J2,v(y), and

3

I OO I

X OO Y

3

I I

I I

.

We shall simply write x and y for x and y, respectively.

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Note that Proposition 2.7 implies

x y

ex1e1, X 2ex1I1

J2,+u(x),

where e1 Rn and I1 Sn are given by

e1 :=

10...0

and I1 :=

1 0 00 0 0...

......

0 0 0

.

Setting r := ex1

, from the definition of u and v, we have

0 F

y,x y

, Y

F

x,

x y

re1, X rI1

.

In view of the uniform ellipticity and (3.4), we have

0 r + rP+(I1) + F

y,x y

, Y

F

x,

x y

, X

.

Hence, by (3.21) and the definition ofP+, we have

0 r( ) + F|x y| + |x y|2

,

which together with (3.27) yields 0 ex1(). This is a contradictionbecause of our choice of > 0. 2

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4 Existence results

In this section, we present some existence results for viscosity solutions ofsecond-order (degenerate) elliptic PDEs.

We first present a convenient existence result via Perrons method, whichwas established by Ishii in 1987.

Next, for Bellman and Isaacs equations, we give representation formulasfor viscosity solutions. From the dynamic programming principle below, wewill realize how natural the definition of viscosity solutions is.

4.1 Perrons method

In order to introduce Perrons method, we need the notion of viscosity solu-tions for semi-continuous functions.

Definition. For any function u : R, we denote the upper andlower semi-continuous envelope of u by u and u, respectively, which aredefined by

u(x) = lim0

supyB(x)

u(y) and u(x) = lim0

infyB(x)

u(y).

We give some elementary properties for u and u without proofs.

Proposition 4.1. For u : R, we have(1) u(x) u(x) u(x) for x ,(2) u(x) = (u)(x) for x ,(3) u(resp., u) is upper (resp., lower) semi-continuous in , i.e.

lim supyx

u(y) u(x), (resp., lim infyx u(y) u(x)) for x ,

(4) if u is upper (resp., lower) semi-continuous in ,then u(x) = u(x) (resp., u(x) = u(x)) for x .

With these notations, we give our definition of viscosity solutions of

F(x,u,Du,D2u) = 0 in . (4.1)

Definition. We call u : R a viscosity subsolution (resp., superso-lution) of (4.1) ifu (resp., u) is a viscosity subsolution (resp., supersolution)of (4.1).

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We call u :

R a viscosity solution of (4.1) if it is both a viscosity

sub- and supersolution of (4.1).

Remark. We note that we supposed that viscosity sub- and supersolu-tions are, respectively, upper and lower semi-continuous in our comparisonprinciple in section 3. Adapting the above new definition, we omit the semi-continuity for viscosity sub- and supersolutions in Propositions 3.1, 3.3 andTheorems 3.4, 3.5, 3.7, 3.9.

In what follows,

we use the above definition.

Remark. We remark that the comparison principle Theorem 3.7 impliesthe continuity of viscosity solutions.

Continuity of viscosity solutions

viscosity solution usatisfies u = u on

= u C()

Proof of the continuity of u. Since u and u are, respectively, a viscositysubsolution and a viscosity supersolution and u u on , Theorem 3.7yields u u in . Because u u u in , we have u = u = u in ;u

C(). 2

We first show that the point-wise supremum (resp., infimum) of viscos-ity subsolutions (resp., supersolution) becomes a viscosity subsolution (resp.,supersolution).

Theorem 4.2. Let Sbe a non-empty set of upper (resp., lower) semi-continuous viscosity subsolutions (resp., supersolutions) of (4.1).

Set u(x) := supvSv(x) (resp., u(x) := infvSv(x)). If supxK |u(x)| < for any compact sets K , then u is a viscosity subsolution (resp.,supersolution) of (4.1).

Proof. We only give a proof for subsolutions since the other can be provedin a symmetric way.

For x , we suppose that 0 = (u)(x) > (u)(x) for x \ {x}and C2(). We shall show that

F(x, (x), D(x), D2(x)) 0. (4.2)

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Let r > 0 be such that B2r(x)

. We can find s > 0 such that

maxBr(x)

(u ) s. (4.3)

We choose xk Br(x) such that limk xk = x, u(x) k1 u(xk)and |(xk) (x)| < 1/k. Moreover, we select upper semi-continuous uk Ssuch that uk(xk) + k

1 u(xk).By (4.3), for 3/k < s, we have

maxBr(x)

(uk ) < (uk )(xk).

Thus, for large k > 3/s, there is yk Br(x) such that uk attains itsmaximum over Br(x) at yk. Hence, we haveF(yk, uk(yk), D(yk), D

2(yk)) 0. (4.4)Taking a subsequence if necessary, we may suppose z := limk yk. Since

(u )(x) (uk )(xk) + 3k

(uk )(yk) + 3k

(u )(yk) + 3k

by the upper semi-continuity of u, we have

(u

)(x) (u

)(z),

which yields z = x, and moreover, limk uk(yk) = u(x) = (x). Therefore,sending k in (4.4), by the continuity of F, we obtain (4.2). 2

Our first existence result is as follows.

Theorem 4.3. Assume that F is elliptic. Assume also that there area viscosity subsolution U SC() Lloc() and a viscosity supersolution LSC() Lloc() of (4.1) such that

in .Then, u(x) := supvSv(x) (resp., u(x) = infwSw(x)) is a viscosity solu-

tion of (4.1), where

S:=

v USC() v is a viscosity subsolutionof (4.1) such that v in

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resp., S:= w LSC() w is a viscosity supersolution

of (4.1) such that w in .Sketch of proof. We only give a proof for u since the other can be shown

in a symmetric way.First of all, we notice that S = since S.Due to Theorem 4.2, we know that u is a viscosity subsolution of (4.1).

Thus, we only need to show that it is a viscosity supersolution of (4.1).Assume that u LSC(). Assuming that 0 = (u )(x) < (u )(x)

for x \ {x} and C2(), we shall show that

F(x, (x), D(x), D2(x))

0.

Suppose that this conclusion fails; there is > 0 such that

F(x, (x), D(x), D2(x)) 2.

Hence, there is r > 0 such that

F(x, (x) + t,D(x), D2(x)) for x Br(x) and |t| r. (4.5)First, we claim that (x) < (x). Indeed, otherwise, since u in

,

attains its minimum at x

. See Fig 4.1.

Fig 4.1

y = (x)

y = u(x)

y = (x)

xx

y = (x)

Hence, from the definition of supersolution , we get a contradiction to(4.5) for x = x and t = 0.

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We may suppose that (x) < (x) since, otherwise, = = at x.

Setting 3 := (x) u(x) > 0, from the lower and upper semi-continuity of and , respectively, we may choose s (0, r] such that

(x) + (x) + 2 (x) for x B2s(x).Moreover, we can choose (0, s) and 0 (0, min{ , r}) such that

(x) + 20 u(x) for x Bs+(x) \ Bs(x).If we can define a function w Ssuch that w(x) > u(x), then we finish

our proof because of the maximality of u at each point.Now, we set

w(x) :=

max{u(x), (x) + 0} in Bs(x),u(x) in \ Bs(x).See Fig 4.2.

Fig 4.2

y = (x)

y = u(x)

y = (x)

y = w(x)

xx

y = (x) + 0

3

It suffices to show that w S. Because of our choice of 0, s > 0, it iseasy to see w in . Thus, we only need to show that w is a viscositysubsolution of (4.1).

To this end, we suppose that (w )(x) (w )(z) = 0 for x ,and then we will get

F(z, w(z), D(z), D2(z)) 0. (4.6)Ifz \Bs(x) =: , by Proposition 2.4, then u attains its maximum

at z , we get (4.6).

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If z

Bs(x), then (4.6) holds again since w = u in Bs+(x)

\Bs(x).

It remains to show (4.6) when z Bs(x). Since + 0 is a viscositysubsolution of (4.1) in Bs(x), Theorem 4.2 with := Bs(x) yields (4.6). 2

Correct proof, which the reader may skip first. Since we do not suppose thatu LSC() here, we have to work with u.

Suppose that 0 = (u)(x) < (u)(x) for x \{x} for some C2(),x , > 0 and

F(x, (x), D(x), D2(x)) 2.Hence, we get (4.5) even in this case.

We also show that the w defined in the above is a viscosity subsolution of (4.1).

It only remains to check that sup(w u) > 0.In fact, choosing xk B1/k(x) such that

u(x) +1

k u(xk),

we easily verify that if 1/k min{0/2, s} and |(x)(xk)| < 0/2, then we have

w(xk) (xk) + 0 > (x) + 02

= u(x) +02

u(xk). 2

4.2 Representation formula

In this subsection, for given Bellman and Isaacs equations, we present theexpected solutions, which are called value functions. In fact, via the dy-namic programming principle for the value functions, we verify that they areviscosity solutions of the corresponding PDEs.

Although this subsection is very important to learn how the notion ofviscosity solutions is the right one from a view point of applications in optimalcontrol and games,

if the reader is more interested in the PDE theory than these applications,he/she may skip this subsection.

We shall restrict ourselves to

investigate the formulas only for first-order PDEs

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because in order to extend the results below to second-order ones, we need

to introduce some terminologies from stochastic analysis. However, this istoo much for this thin book.

As will be seen, we study the minimization of functionals associated withordinary differential equations (ODEs for short), which is called a deter-ministic optimal control problem. When we adapt stochastic differentialequations instead of ODEs, those are called stochastic optimal controlproblems. We refer to [10] for the later.

Moreover, to avoid mentioning the boundary condition, we will work onthe whole domain Rn.

Throughout this subsection, we also suppose (3.7); > 0.

4.2.1 Bellman equation

We fix a control set A Rm for some m N. We define A byA := { : [0, ) A | () is measurable}.

For x Rn and A, we denote by X(; x, ) the solution ofX(t) = g(X(t), (t)) for t > 0,

X(0) = x,(4.7)

where we will impose a sufficient condition on continuous functions g : Rn

A Rn so that (4.7) is uniquely solvable.For given f : Rn A R, under suitable assumptions (see (4.8) below),

we define the cost functional for X(; x, ):J(x, ) :=

0

etf(X(t; x, ), (t))dt.

Here, > 0 is called a discount factor, which indicates that the right handside of the above is finite.

Now, we shall consider the optimal cost functional, which is called thevalue function in the optimal control problem;

u(x) := infA J(x, ) for x Rn

.

Theorem 4.4. (Dynamic Programming Principle) Assume that

(1) supaA

f(, a)L(Rn) + g(, a)W1,(Rn)

< ,

(2) supaA

|f(x, a) f(y, a)| f(|x y|) for x, y Rn, (4.8)

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where f

M.

For any T > 0, we have

u(x) = infA

T0

etf(X(t; x, ), (t))dt + eTu(X(T; x, ))

.

Proof. For fixed T > 0, we denote by v(x) the right hand side of theabove.

Step 1: u(x) v(x). Fix any > 0, and choose A such that

u(x) +

0


Setting x = X(T; x, ) and A by (t) = (T + t) for t 0, we have0

etf(X(t; x, ), (t))dt =T0

etf(X(t; x, ), (t))dt

+eT0


Here and later, without mentioning, we use the fact that

X(t + T; x, ) = X(t; x, ) for T > 0, t 0 and A,

where(t) := (t + T) (t 0) and x := X(T; x, ).

Indeed, the above relation holds true because of the uniqueness of solutionsof (4.7) under assumptions (4.8). See Fig 4.3.

Thus, taking the infimum in the second term of the right hand side of theabove among A, we have

u(x) + T0

etf(X(t; x, ), (t))dt + eTu(x),

which implies one-sided inequality by taking the infimum over A since > 0is arbitrary.

Step 2: u(x) v(x). Fix > 0 again, and choose A such that

v(x) + T0

etf(X(t; x, ), (t))dt + eTu(x),

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Fig 4.3

X(t; x, )

X(t + T; x, ) = X(t; x, )

t = 0

xx

t = T

x = X(T; x, )

where x := X(T; x, ). We next choose 1 A such that

u(x) + 0

etf(X(t; x, 1), 1(t))dt.

Now, setting

0(t) :=

(t) for t [0, T),1(t T) for t T,

we see thatv(x) + 2

0

etf(X(t; x, 0), 0(t))dt,

which gives the opposite inequality by taking the infimum over 0 A since > 0 is arbitrary again. 2

Now, we give an existence result for Bellman equations.

Theorem 4.5. Assume that (4.8) holds. Then, u is a viscosity solutionof

supaA

{u g(x, a), Du f(x, a)} = 0 in Rn. (4.9)

Sketch of proof. In Steps 1 and 2, we give a proof when u USC(Rn

)and u LSC(Rn), respectively.Step 1: Subsolution property. Fix C1(Rn), and suppose that 0 =

(u )(x) (u )(x) for some x Rn and any x Rn.Fix any a0 A, and set 0(t) := a0 for t 0 so that 0 A.

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For small s > 0, in view of Theorem 4.4, we have

(x) es(X(s; x, 0)) u(x) esu(X(s; x, 0))

s0

etf(X(t; x, 0), a0)dt.

Setting X(t) := X(t; x, 0) for simplicity, by (4.7), we see that

et{(X(t)) g(X(t), 0), D(X(t))} = ddt

et(X(t))

. (4.10)

Hence, we have

0 s

0et{(X(t)) g(X(t), a0), D(X(t)) f(X(t), a0)}dt.

Therefore, dividing the above by s > 0, and then sending s 0, we have0 (x) g(x, a0), D(x) f(x, a0),

which implies the desired inequality of the definition by taking the supremumover A.

Step 2: Supersolution property. To show that u is a viscosity supersolu-tion, we argue by contradiction.

Suppose that there are x

Rn, > 0 and

C1(Rn) such that 0 =(u )(x) (u )(x) for x Rn, and that

supaA

{(x) g(x, a), D(x) f(x, a)} 2.

Thus, we can find > 0 such that

supaA

{(x) g(x, a), D(x) f(x, a)} for x B(x). (4.11)

By assumption (4.8) for g, setting t0 := /(supaA g(, a)L(Rn)+1) > 0,we easily see that

|X(t; x, ) x| t0

|X(s; x, )|ds for t [0, t0] and A.

Hence, by setting X(t) := X(t; x, ) for any fixed A, (4.11) yields(X(t)) g(X(t), (t)), D(X(t)) f(X(t), (t)) (4.12)

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for t

[0, t0]. Since (4.10) holds for in place of 0, multiplying e

t in

(4.12), and then integrating it over [0, t0], we obtain

(x) et0(X(t0)) t00

etf(X(t), (t))dt

(1 et0).

Thus, setting 0 = (1 et0)/ > 0, which is independent of A, wehave

u(x) t00

etf(X(t), (t))dt + et0u(X(t0)) 0.Therefore, taking the infimum over A, we get a contradiction to Theorem4.4. 2

Correct proof, which the reader may skip first.Step 1: Subsolution property. Assume that there are x Rn, > 0 and

C1(Rn) such that 0 = (u )(x) (u )(x) for x Rn and that

supaA

{(x) g(x, a), D(x) f(x, a)} 2.

In view of (4.8), there are a0 A and r > 0 such that

(x) g(x, a0), D(x) f(x, a0) for x B2r(x). (4.13)

For large k 1, we can choose xk B1/k(x) such that u(x) u(xk) + k1

and |(x) (xk)| < 1/k. We will only use k such that 1/k r.Setting 0(t) := a0, we note that Xk(t) := X(t; xk, 0) B2r(x) for t [0, t0]

with some t0 > 0 and for large k.On the other hand, by Theorem 4.4, we have

u(xk) t00

etf(Xk(t), a0)dt + et0u(Xk(t0)).

Thus, we have

(xk)

2

k (x)

1

k u(xk)

t0

0

etf(Xk(t), a0)dt + et0(Xk(t0)).

Hence, by (4.13) as in Step 1 of Sketch of proof, we see that

2k

t00

et{f(Xk(t), a0) + g(Xk(t), a0), D(Xk(t)) (Xk(t))}dt

(1 et0),

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which is a contradiction for large k.

Step 2: Supersolution property. Assume that there are x Rn, > 0 and C1(Rn) such that 0 = (u )(x) (u )(x) for x Rn and that

supaA

{(x) g(x, a), D(x) f(x, a)} 2.

In view of (4.8), there is r > 0 such that

(x) g(x, a), D(x) f(x, a) for x B2r(x) and a A. (4.14)

For large k 1, we can choose xk B1/k(x) such that u(x) u(xk) k1and |(x) (xk)| < 1/k. In view of (4.8), there is t0 > 0 such that

Xk(t; xk, ) B2r(x) for all k 1r

, A and t [0, t0].

Now, we select k A such that

u(xk) +1

k

t00

etf(X(t; xk, k), k(t))dt + et0u(X(t0; xk, k)).

Setting Xk(t) := X(t; xk , k), we have

(xk) +3

k (x) + 2

k u(xk) + 1

k

t00

etf(Xk(t), k(t))dt + et0(Xk(t)).

Hence, we have

3

k

t00

et{g(Xk(t), k(t)), D(Xk(t)) + f(Xk(t), k(t)) (Xk(t))}dt.

Putting (4.14) with k in the above, we have

3

k

t00

etdt,

which is a contradiction for large k 1. 2

4.2.2 Isaacs equation

In this subsection, we study fully nonlinear PDEs (i.e. p Rn F(x, p) isneither convex nor concave) arising in differential games.

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We are given continuous functions f : Rn

A

B

R and g : Rn

A B Rn such that

(1) sup(a,b)AB

f(, a , b)L(Rn) + g(, a , b)W1,(Rn)

< ,

(2) sup(a,b)AB

|f(x,a,b) f(y, a , b)| f(|x y|) for x, y Rn, (4.15)

where f M.Under (4.15), we shall consider Isaacs equations:

supaA

infbB

{u g(x,a,b), Du f(x,a,b)} = 0 in Rn, (4.16)

and

infbB

supaA

{u g(x,a,b), Du f(x,a,b)} = 0 in Rn. (4.16)

As in the previous subsection, we shall derive the expected solution.We first introduce some notations: While we will use the same notion A

as before, we set

B := { : [0, ) B | () is measurable}.

Next, we introduce the so-called sets of non-anticipating strategies:

:=

: A B

for any T > 0, if 1 and 2 A satisfy

that 1(t) = 2(t) for a.a. t (0, T),then [1](t) = [2](t) for a.a. t (0, T)

and

:=

: B A

for any T > 0, if 1 and 2 B satisfythat 1(t) = 2(t) for a.a. t (0, T),

then [1](t) = [2](t) for a.a. t (0, T)

.

Using these notations, we will consider maximizing-minimizing problemsof the following cost functional: For A, B, and x Rn,

J(x,,) :=0

etf(X(t; x,,), (t), (t))dt,

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where X(

; x,,) is the (unique) solutions of

X(t) = g(X(t), (t), (t)) for t > 0,X(0) = x.

(4.17)

The expected solutions for (4.16) and (4.16), respectively, are given by

u(x) = sup

infA

0

etf(X(t; x,,[]), (t), [](t))dt,

andv(x) = inf

supB

0

etf(X(t; x, [], ), [](t), (t))dt.

We call u and v upper and lower value functions of this differential game,respectively. In fact, under appropriate hypotheses, we expect that v u,which cannot be proved easily. To show v u, we first observe that u and vare, respectively, viscosity solutions of (4.16) and (4.16). Noting that

supaA

infbB

{rg(x,a,b), pf(x,a,b)} infbB

supaA

{rg(x,a,b), pf(x,a,b)}

for (x,r,p) RnRRn, we see that u (resp., v) is a viscosity supersolution(resp., subsolution) of (4.16) (resp., (4.16)). Thus, the standard comparisonprinciple implies v u in Rn (under suitable growth condition at |x| for u and v).

We shall only deal with u since the corresponding results for v can beobtained in a symmetric way.

To show that u is a viscosity solution of the Isaacs equation (4.16), we firstestablish the dynamic programming principle as in the previous subsection:

Theorem 4.6. (Dynamic Programming Principle) Assume that (4.15)hold. Then, for T > 0, we have

u(x) = sup

infA

T0

etf(X(t; x,,[]), (t), [](t))dt

+eTu(X(T; x,,[]))

.

Proof. For a fixed T > 0, we denote by w(x) the right hand side of theabove.

Step 1: u(x) w(x). For any > 0, we choose such that

u(x) infA

0

etf(X(t; x,,[]), (t), [](t))dt =: I.

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For any fixed 0

A, we define the mapping

T0 :

A Aby

T0[] :=

0(t) for t [0, T),(t T) for t [T, ) for A.

Thus, for any A, we have

I T0

etf(X(t; x, 0, [0]), 0(t), [0](t))dt

+T

etf(X(t; x, T0[], [T0[]]), T0[](t), [T0[]](t))dt=: I1 + I

2 .

We next define by[](t) := [T0[]](t + T) for t 0 and A.

Note that belongs to .Setting x := X(T; x, 0, [0]), we have

I2 = eT

0

etf(X(t; x,, []), (t), [](t))dt.

Taking the infimum over A, we have

u(x) I1 + eT infA

0

etf(X(t; x,, []), (t), [](t))dt

=: I1 + I2 .

Since I2 eTu(x), we have

u(x) I1 + eTu(x),

which implies u(x) w(x) by taking the infimum over 0 A and then,the supremum over . Therefore, we get the one-sided inequality since > 0is arbitrary.

Step 2: u(x) w(x). For > 0, we choose 1 such that

w(x) infA

T0

etf(X(t; x,,1 []), (t), 1 [](t))dt

+eTu(X(T; x,,1 []))

.

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For any fixed 0

A, setting x = X(T; x, 0,

1 [0]), we have

w(x) T0

etf(X(t; x, 0, 1 [0]), 0(t), 1 [0](t))dt + e

Tu(x).

Next, we choose 2 such that

u(x) infA

0

etf(X(t; x,,2 []), (t), 2 [](t))dt. =: I.

For A, we define the mapping T1 : A A by

T1[](t) := (t + T) for t

0.

Thus, we have

I 0

etf(X(t; x, T1[0], 2 [T1[0]]), T1[0](t), 2 [T1[0]](t))dt =: I .

Now, for A, setting

[](t) :=

1 [](t) for t [0, T),2 [T1[]](t T) for t [T, ),

and X(t) := X(t; x,

T1[0],

2 [

T1[0]]), we have

I =T

e(tT)f(X(t T), T1[0](t T), 2 [T1[0]](t T))dt= eT

T

etf(X(t T), 0(t), [0](t))dt.

Since

X(t; x, 0, [0]) =

X(t; x, 0,

1 [0]) for t [0, T),

X(t T) for t [T, ),we have

w(x) 2 0

etf(X(t; x, 0, [0]), 0(t), [0](t))dt.

Since 0 is arbitrary, we have

w(x) 2 infA

0

etf(X(t; x,, []), (t), [](t))dt,

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which yields the assertion by taking the supremum over and then, by

sending 0. 2Now, we shall verify that the value function u is a viscosity solution of

(4.16).Since we only give a sketch of proofs, one can skip the following theorem.

For a correct proof, we refer to [1], originally by Evans-Souganidis (1984).

Theorem 4.7. Assume that (4.15) holds.(1) Then, u is a viscosity subsolution of (4.16).(2) Assume also the following properties:

(i) A Rm is compact for some integer m 1.(ii) there is an A M such that

|f(x,a,b) f(x, a, b)| + |g(x,a,b) g(x, a, b)| A(|a a|)for x Rn, a,a A and b B.

(4.18)

Then, u is a viscosity supersolution of (4.16).

Remark. To show that v is a viscosity subsolution of (4.16), instead of (4.18),we need to suppose the following hypotheses:

(i) B Rm is compact for some integer m 1.(ii) there is an B

Msuch that

|f(x,a,b) f(x,a,b)| + |g(x,a,b) g(x,a,b)| B(|b b|)for x Rn, b , b B and a A,

(4.18)

while to verify that v is a viscosity supersolution of (4.16), we only need (4.15).

Sketch of proof. We shall only prove the assertion assuming that u U SC(Rn)and u LSC(Rn) in Step 1 and 2, respectively.

To give a correct proof without the semi-continuity assumption, we need a bitcareful analysis similar to the proof for Bellman equations. We omit the correctproof here.

Step 1: Subsolution property. Suppose that the subsolution property fails; there

are x Rn

, > 0 and C1

(Rn

) such that 0 = (u )(x) (u )(y) (for ally Rn) and

supaA

infbB

{u(x) g(x,a,b), D(x) f(x,a,b)} 3.

We note that X(; x,,[]) are uniformly continuous for any (, ) A in view of (4.15).

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Thus, we can choose that a0 A such that

infbB

{(x) g(x, a0, b), D(x) f(x, a0, b)} 2.

For any , setting 0(t) = a0 for t 0, we simply write X() forX(; x, 0, [0]). Thus, we find small t0 > 0 such that

(X(t)) g(X(t), a0, [0](t)), D(X(t)) f(X(t), a0, [0](t))

for t [0, t0]. Multiplying et in the above and then, integrating it over [0, t0],we have

(1 et0

) t0

0 d

dt

et

(X(t))

+ et

f(X(t), a0, [0](t))

dt

= (x) et0(X(t0)) t00

etf(X(t), a0, [0](t))dt.

Hence, we have

u(x)

(1 et0) t00

etf(X(t), a0, [0](t))dt + et0u(X(t0)) =: I.

Taking the infimum over A, we have

I

infA

t0

0

etf(X(t; x,,[]), (t), [](t))dt

+et0u(X(t0; x,,[])) .

Therefore, since is arbitrary, we have

u(x)

(1 et0) sup

infA

t00

etf(X(t; x,,[]), (t), [](t))dt

+et0u(X(t0; x,,[]))

,

which contradicts Theorem 4.6.Step 2: Supersolution property. Suppose that the supersolution property fails;

there are x Rn, > 0 and C1(Rn) such that 0 = (u )(x) (u )(y)for y

Rn, and

supaA

infbB

{u(x) g(x,a,b

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