Research article Special Issues

The vanishing discount problem for monotone systems of Hamilton-Jacobi equations: a counterexample to the full convergence

  • In recent years there has been intense interest in the vanishing discount problem for Hamilton-Jacobi equations. In the case of the scalar equation, B. Ziliotto has recently given an example of the Hamilton-Jacobi equation having non-convex Hamiltonian in the gradient variable, for which the full convergence of the solutions does not hold as the discount factor tends to zero. We give here an explicit example of nonlinear monotone systems of Hamilton-Jacobi equations having convex Hamiltonians in the gradient variable, for which the full convergence of the solutions fails as the discount factor goes to zero.

    Citation: Hitoshi Ishii. The vanishing discount problem for monotone systems of Hamilton-Jacobi equations: a counterexample to the full convergence[J]. Mathematics in Engineering, 2023, 5(4): 1-10. doi: 10.3934/mine.2023072

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  • In recent years there has been intense interest in the vanishing discount problem for Hamilton-Jacobi equations. In the case of the scalar equation, B. Ziliotto has recently given an example of the Hamilton-Jacobi equation having non-convex Hamiltonian in the gradient variable, for which the full convergence of the solutions does not hold as the discount factor tends to zero. We give here an explicit example of nonlinear monotone systems of Hamilton-Jacobi equations having convex Hamiltonians in the gradient variable, for which the full convergence of the solutions fails as the discount factor goes to zero.



    We consider the system of Hamilton-Jacobi equations

    {λu1(x)+H1(Du1(x))+B1(u1(x),u2(x))=0  in Tn,λu2(x)+H2(Du2(x))+B2(u1(x),u2(x))=0  in Tn, (1.1)

    where λ>0 is a given constant, the functions Hi:RnR and Bi:R2R, with i=1,2, are given continuous functions, and Tn denotes the n-dimensional flat torus Rn/Zn.

    In a recent paper [6], the authors have investigated the vanishing discount problem for a nonlinear monotone system of Hamilton-Jacobi equations

    {λu1(x)+G1(x,Du1(x),u1(x),u2(x),,um(x))=0  in Tn,λu1(x)+G1(x,Du1(x),u1(x),u2(x)λum(x)+Gm(x,Dum(x),u1(x),u2(x),,um(x))=0  in Tn, (1.2)

    and established under some hypotheses on the GiC(Tn×Rn×Rm) that, when uλ=(uλ,1,,uλ,m)C(Tn)m denoting the (viscosity) solution of (1.2), the whole family {uλ}λ>0 converges in C(Tn)m to some u0C(Tn)m as λ0+. The constant λ>0 in the above system is the so-called discount factor.

    The hypotheses on the system are the convexity, coercivity, and monotonicity of the Gi as well as the solvability of (1.2), with λ=0. Here the convexity of Gi is meant that the functions Rn×Rm(p,u)Gi(x,p,u) are convex. We refer to [6] for the precise statement of the hypotheses.

    Prior to work [6], there have been many contributions to the question about the whole family convergence (in other words, the full convergence) under the vanishing discount, which we refer to [1,3,4,6,8,9,10] and the references therein.

    In the case of the scalar equation, B. Ziliotto [11] has recently shown an example of the Hamilton-Jacobi equation having non-convex Hamiltonian in the gradient variable for which the full convergence does not hold. In Ziliotto's approach, the first step is to find a system of two algebraic equations

    {λu+f(uv)=0,λv+g(vu)=0, (1.3)

    with two unknowns u,vR and with a parameter λ>0 as the discount factor, for which the solutions (uλ,vλ) stay bounded and fail to fully converge as λ0+. Here, an "algebraic" equation is meant not to be a functional equation. The second step is to interpolate the two values uλ and vλ to get a function of xT1 which satisfies a scalar non-convex Hamilton-Jacobi equation in T1.

    In the first step above, Ziliotto constructs f,g based on a game-theoretical and computational argument, and the formula for f,g is of the minimax type and not quite explicit. In [5], the author has reexamined the system given by Ziliotto, with a slight generality, as a counterexample for the full convergence in the vanishing discount.

    Our purpose in this paper is to present a system (1.3), with an explicit formula for f,g, for which the solution (uλ,vλ) does not fully converge to a single point in R2. A straightforward consequence is that (1.1), with B1(u1,u2)=f(u1u2) and B2(u1,u2)=g(u2u1), has a solution given by

    (uλ,1(x),uλ.2(x))=(uλ,vλ)   for xTn,

    under the assumption that Hi(x,0)=0 for all xTn, and therefore, gives an example of a discounted system of Hamilton-Jacobi equations, the solution of which fails to satisfy the full convergence as the discount factor goes to zero.

    The paper consists of two sections. This introduction is followed by Section 2, the final section, which is divided into three subsections. The main results are stated in the first subsection of Section 2, the functions f,g, the key elements of (1.3), are contstructed in the second subsection, and the final subsection provides the proof of the main results.

    Our main focus is now the system

    {λu+f(uv)=0λv+g(vu)=0, (2.1)

    where f,gC(R,R) are nondecreasing functions, to be constructed, and λ>0 is a constant, to be sent to zero. Notice that (2.1) above is referred as (1.3) in the previous section.

    We remark that, due to the monotonicity assumption on f,g, the mapping (u,v)(f(uv),g(vu)),R2R2 is monotone. Recall that, by definition, a mapping (u,v)(B1(u,v),B2(u,v)),R2R2 is monotone if, whenever (u1,v1),(u2,v2)R2 satisfy u1u2v1v2 (resp., v1v2u1u2), we have B1(u1,v1)B1(u2,v2) (resp., B2(u1,v1)B2(u2,v2)).

    Our main results are stated as follows.

    Theorem 1. There exist two increasing functions f,gC(R,R) having the properties (a)–(c):

    (a) For any λ>0 there exists a unique solution (uλ,vλ)R2 to (2.1),

    (b) the family of the solutions (uλ,vλ) to (2.1), with λ>0, is bounded in R2,

    (c) the family {(uλ,vλ)}λ>0 does not converge as λ0+.

    It should be noted that, as mentioned in the introduction, the above theorem has been somewhat implicitly established by Ziliotto [11]. In this note, we are interested in a simple and easy approach to finding functions f,g having the properties (a)–(c) in Theorem 1.

    The following is an immediate consequence of the above theorem.

    Corollary 2. Let HiC(Rn,R), i=1,2, satisfy H1(0)=H2(0)=0. Let f,gC(R,R) be the functions given by Theorem 1, and set B1(u1,u2)=f(u1u2) and B2(u1,u2)=g(u2u1) for all (u1,u2)R2. For any λ>0, let (uλ,1,uλ,2) be the (viscosity) solution of (1.1). Then, the functions uλ,i are constants, the family of the points (uλ,1,uλ,2) in R2 is bounded, and it does not converge as λ0+.

    Notice that the convexity of Hi in the above corollary is irrelevant, and, for example, one may take Hi(p)=|p|2 for iI, which are convex functions.

    We remark that a claim similar to Corollary 2 is valid when one replaces Hi(p) by degenerate elliptic operators Fi(x,p,M) as far as Fi(x,0,0)=0, where M is the variable corresponding to the Hessian matrices of unknown functions. (See [2] for an overview on the viscosity solution approach to fully nonlinear degenerate elliptic equations.)

    If f,g are given and (u,v)R2 is a solution of (2.1), then w:=uv satisfies

    λw+f(w)g(w)=0. (2.2)

    Set

    h(r)=f(r)g(r)   for rR, (2.3)

    which defines a continuous and nondecreasing function on R.

    To build a triple of functions f,g,h, we need to find two of them in view of the relation (2.3). We begin by defining function h.

    For this, we discuss a simple geometry on xy-plane as depicted in Figure 1 below. Fix 0<k1<k2. The line y=12k2+k1(x+12) has slope k1 and crosses the lines x=1 and y=k2x at P:=(1,12(k1+k2)) and Q:=(12,12k2), respectively, while the line y=k2x meets the lines x=1 and x=12 at R:=(1,k2) and Q=(12,12k2), respectively.

    Figure 1.  Graph of ψ.

    Choose k>0 so that 12(k1+k2)<k<k2. The line y=kx crosses the line y=12k2+k1(x+12) at a point S:=(x,y) in the open line segment between the points P=(12,12(k1+k2)) and Q=(12,12k2). The line connecting R=(1,k2) and S=(x,y) can be represented by y=k2+k+(x+1), with k+:=y+k2x+1>k2.

    We set

    ψ(x)={k2x for x(,1][1/2,),min{k2+k+(x+1),12k2+k1(x+12)}   for x(1,12).

    It is clear that ψC(R) and increasing on R. The building blocks of the graph y=ψ(x) are three lines whose slopes are k1<k2<k+. Hence, if x1>x2, then ψ(x1)ψ(x2)k1(x1x2), that is, the function xψ(x)k1x is nondecreasing on R.

    Next, we set for jN,

    ψj(x)=2jψ(2jx)   for xR.

    It is clear that for all jN, ψjC(R), the function xψj(x)k1x is nondecreasing on R, and

    ψj(x){>k2x   for all x(2j,2j1),=k2x   otherwise.

    We set

    η(x)=maxjNψj(x)   for xR.

    It is clear that ηC(R) and xη(x)k1x is nondecreasing on R. Moreover, we see that

    η(x)=k2x   for all x(,12][0,),

    and that if 2j<x<2j1 and jN,

    η(x)=ψj(x)>k2x.

    Note that the point S=(x,y) is on the graph y=ψ(x) and, hence, that for any jN, the point (2jx,2jy) is on the graph y=η(x). Similarly, since the point S=(x,y) is on the graph y=kx and for any jN, the point (2jx,2jy) is on the graph y=kx. Also, for any jN, the point (2j,k22j) lies on the graphs y=η(x) and y=k2x.

    Fix any d1 and define hC(R) by

    h(x)=η(xd).

    For the function h defined above, we consider the problem

    λz+h(z)=0. (2.4)

    Lemma 3. For any λ0, there exists a unique solution zλR of (2.4).

    Proof. Fix λ0. The function xh(x)+λx is increasing on R and satisfies

    limx(h(x)+λx)=   and   limx(h(x)+λx)=.

    Hence, there is a unique solution of (2.4).

    For any λ0, we denote by zλ the unique solution of (2.4). Since h(d)=0, it is clear that z0=d.

    For later use, observe that if λ>0, k>0, and (z,w)R2 is the point of the intersection of two lines y=λx and y=k(xd), then w=λz=k(zd) and

    z=kdk+λ. (2.5)

    Lemma 4. There are sequences {μj} and {νj} of positive numbers converging to zero such that

    zμj=k2dk2+μj  and  zνj=kdk+νj.

    Proof. Let jN. Since (2j,k22j) is on the intersection of the graphs y=k2x and y=η(x), it follows that (2j+d,k22j) is on the intersection of the graphs y=k2(xd) and y=h(x). Set

    μj=k22jd2j, (2.6)

    and note that μj>0 and that

    μj(d2j)=k22j,

    which says that the point (d2j,k22j) is on the line y=μjx. Combining the above with

    k22j=h(d2j)

    shows that d2j is the unique solution of (2.4). Also, since (d2j,μj(d2j))=(d2j,k22j) is on the line y=k2(xd), we find by (2.5) that

    zμj=k2dk2+μj.

    Similarly, since (2jx,2jy) is on the intersection of the graphs y=kx and y=η(x), we deduce that if we set

    νj:=2jyd+2jx=2j|y|d2j|x|, (2.7)

    then

    zνj=kdk+νj.

    It is obvious by (2.6) and (2.7) that the sequences {μj}jN and {νj}jN are decreasing and converge to zero.

    We fix k0(0,k1) and define f,gC(R) by f(x)=k0(xd) and

    g(x)=f(x)h(x).

    It is easily checked that g(x)(k1k0)x is nondecreasing on R, which implies that g is increasing on R, and that h(x)=f(x)g(x) for all xR. We note that

    f(d)=h(d)=g(d)=0. (2.8)

    We fix f,g,h as above, and consider the system (2.1).

    Lemma 5. Let λ>0. There exists a unique solution of (2.1).

    The validity of the above lemma is well-known, but for the reader's convenience, we provide a proof of the lemma above.

    Proof. By choice of f,g, the functions f,g are nondecreasing on R. We show first the comparison claim: if (u1,v1),(u2,v2)R2 satisfy

    λu1+f(u1v1)0,λv1+g(v1u1)0, (2.9)
    λu2+f(u2v2)0,λv2+g(v2u2)0, (2.10)

    then u1u2 and v1v2. Indeed, contrary to this, we suppose that max{u1u2,v1v2}>0. For instance, if max{u1u2,v1v2}=u1u2, then we have u1v1u2v2 and u1>u2, and moreover

    0λu1+f(u1v1)λu1+f(u2v2)>λu2+f(u2v2),

    yielding a contradiction. The other case when max{u1u2,v1v2}=v1v2, we find a contradiction, 0>λv2+g(v2u2), proving the comparison.

    From the comparison claim, the uniqueness of the solutions of (2.1) follows readily.

    Next, we may choose a constant C>0 so large that (u1,v1)=(C,C) and (u2,v2)=(C,C) satisfy (2.9) and (2.10), respectively. We write S for the set of all (u1,u2)R2 such that (2.9) hold. Note that (C,C)S and that for any (u,v)S, uC and vC. We set

    u=sup{u:(u,v)S  for some v},v=sup{v:(u,v)S  for some u}.

    It follows that Cu,vC. We can choose sequences

    {(u1n,v1n)}nN,{(u2n,v2n)}nNS

    such that {u1n},{v2n} are nondecreasing,

    limnu1n=u   and   limnv2n=v.

    Observe that for all nN, u2nu, v1nv, and

    0λu1n+f(u1nv1n)λu1n+f(u1nv),

    which yields, in the limit as n,

    0λu+f(uv).

    Similarly, we obtain 0λv+g(vu). Hence, we find that (u,v)S.

    We claim that (u,v) is a solution of (2.1). Otherwise, we have

    0>λu+f(uv)   or   0>λv+g(vu).

    For instance, if the former of the above inequalities holds, we can choose ε>0, by the continuity of f, so that

    0>λ(u+ε)+f(u+εv).

    Since (u,v)S, we have

    0λv+g(vu)λv+g(vuε).

    Accordingly, we find that (u+ε,v)S, which contradicts the definition of u. Similarly, if 0>λv+g(vu), then we can choose δ>0 so that (u,v+δ)S, which is a contradiction. Thus, we conclude that (u,v) is a solution of (2.1).

    Theorem 6. For any λ>0, let (uλ,vλ) denote the unique solution of (2.1). Let {μj},{νj} be the sequences of positive numbers from Lemma 2.4. Then

    limjuμj=k0dk2  and  limjuνj=k0dk.

    In particular,

    lim infλ0uλk0dk2<k0dklim supλ0uλ.

    With our choice of f,g, the family of solutions (uλ,vλ) of (2.1), with λ>0, does not converge as λ0.

    Proof. If we set zλ=uλvλ, then zλ satisfies (2.4). By Lemma 4, we find that

    zμj=k2dk2+μj   and   zνj=kdk+νj.

    Since uλ satisfies

    0=λuλ+f(zλ)=λuλ+k0(zλd),

    we find that

    uμj=k0(zμjd)μj=k0dμj(k2k2+μj1)=k0dμjμjk2+μj=k0dk2+μj,

    which shows that

    limjuμj=k0dk2.

    A parallel computation shows that

    limjuνj=k0dk.

    Recalling that 0<k<k2, we conclude that

    lim infλ0uλk0dk2<k0dklim supλ0uλ.

    We remark that, since

    limλ0zλ=d   and   vλ=uλzλ,
    limjvμj=k0dk2d   and   limjvνj=k0dkd.

    We give the proof of Theorem 1.

    Proof of Theorem 1. Assertions (a) and (c) are consequences of Lemma 5 and Theorem 6, respectively.

    Recall (2.8). That is, we have f(d)=h(d)=g(d)=0. Setting (u2,v2)=(d,0), we compute that for any λ>0,

    λu2+f(u2v2)>f(d)=0   and   λv2+g(v2u2)=g(d)=0.

    By the comparison claim, proved in the proof of Lemma 5, we find that uλd and vλ0 for any λ>0. Similarly, setting (u1,v1)=(0,d), we find that for any λ>0,

    λu1+f(u1v1)=f(d)=0   and   λv1+g(v1u1)g(v1u1)=g(d)=0,

    which shows by the comparison claim that uλ0 and vλd for any λ>0. Thus, the sequence {(uλ,vλ)}λ>0 is bounded in R2, which proves assertion (b).

    Proof of Corollary 2. For any λ>0, let (uλ,vλ)R2 be the unique solution of (2.1). Since H1(0)=H2(0)=0, it is clear that the constant function (uλ,1(x),uλ,2(x)):=(uλ,vλ) is a classical solution of (1.1). By a classical uniqueness result (see, for instance, [7,Theorem 4.7]), (uλ,1,uλ,2) is a unique viscosity solution of (1.1). The rest of the claims in Corollary 2 is an immediate consequence of Theorem 1.

    Some remarks are in order. (ⅰ) Following [11], we may use Theorem 6 as the primary cornerstone for building a scalar Hamilton-Jacobi equation, for which the vanishing discount problem fails to have the full convergence as the discount factor goes to zero.

    (ⅱ) In the construction of the functions f,gC(R,R) in Theorem 6, the author has chosen d to satisfy d1, but, in fact, one may choose any d>0. In the proof, the core step is to find the function h(x)=f(x)g(x), with the properties: (a) the function xh(x)εx is nondecreasing on R for some ε>0 and (b) the curve y=h(x), with x<d, meets the lines y=p(xd) and y=q(xd), respectively, at Pj and Qj for all jN, where p,q,d are positive constants such that ε<p<q, and the sequences {Pj}jN,{Qj}jN converge to the point (d,0). Obviously, such a function h is never left-differentiable at x=d nor convex in any neighborhood of x=d. Because of this, it seems difficult to select f,gC(R,R) in Theorem 1, both smooth everywhere. In the proof of Theorem 6, we have chosen ε=k0, p=k, q=k2, Pj=(uνj,k(uνjd)), and Qj=(uμj,k2(uμjd))

    Another possible choice of h among many other ways is the following. Define first η:RR by η(x)=x(sin(log|x|)+2) if x0, and η(0)=0 (see Figure 2). Fix d>0 and set h(x)=η(xd) for xR. we remark that ηC(R{0}) and hC(R{d}). Note that if x0,

    η(x)=sin(log|x|)+cos(log|x|)+2[22,2+2],
    Figure 2.  Graph of η (slightly deformed).

    and that if we set xj=exp(2πj) and ξj=exp(2πj+π2), jN, then

    η(xj)=2xj   and   η(ξj)=3ξj.

    The points Pj:=(xj+d,2xj) are on the intersection of two curves y=h(x) and y=2(xd), while the points Qj:=(d+ξj,3ξj) are on the intersection of y=h(x) and y=3(xd). Moreover, limPj=limQj=(d,0).

    The author would like to thank the anonymous referees for their careful reading and useful suggestions. He was supported in part by the JSPS Grants KAKENHI No. 16H03948, No. 20K03688, No. 20H01817, and No. 21H00717.

    The author declares no conflict of interest.



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    2. Julie Clutterbuck, Jiakun Liu, Preface to the Special Issue: Nonlinear PDEs and geometric analysis – Dedicated to Neil Trudinger on the occasion of his 80th birthday, 2023, 5, 2640-3501, 1, 10.3934/mine.2023095
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