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

Our main results are stated as follows.

It should be noted that, as mentioned in the introduction, the above theorem has been somewhat implicitly established by Ziliotto [Zi]. 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.

Notice that the convexity of $H_{i}$ in the above corollary is irrelevant, and, for example, one may take $H_{i}(p)=|p|^{2}$ for $i\in\mathbb{I}$, which are convex functions.

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

If $f,g$ are given and $(u,v)\in\mathbb{R}^{2}$ is a solution of (2), then $w:=u-v$ satisfies

Set

which defines a continuous and nondecreasing function on $\mathbb{R}$.

To build a triple of functions $f,g,h$, we need to find two of them in view of the relation (4). We begin by defining function $h$.

For this, we discuss a simple geometry on $xy$-plane as depicted in Fig. 1 below. Fix $0<k_{1}<k_{2}$. The line $y=-\frac{1}{2}k_{2}+k_{1}(x+\frac{1}{2})$ has slope $k_{1}$ and crosses the lines $x=-1$ and $y=k_{2}x$ at $\mathrm{P}:=(-1,-\frac{1}{2}(k_{1}+k_{2}))$ and $\mathrm{Q}:=(-\frac{1}{2},-\frac{1}{2}k_{2})$, respectively, while the line $y=k_{2}x$ meets the lines $x=-1$ and $x=-\frac{1}{2}$ at $\mathrm{R}:=(-1,-k_{2})$ and $\mathrm{Q}=(-\frac{1}{2},-\frac{1}{2}k_{2})$, respectively.

Choose $k^{*}>0$ so that $\frac{1}{2}(k_{1}+k_{2})<k^{*}<k_{2}$. The line $y=k^{*}x$ crosses the line $y=-\frac{1}{2}k_{2}+k_{1}(x+\frac{1}{2})$ at a point $\mathrm{S}:=(x^{*},y^{*})$ in the open line segment between the points $\mathrm{P}=(-\frac{1}{2},-\frac{1}{2}(k_{1}+k_{2}))$ and $\mathrm{Q}=(-\frac{1}{2},-\frac{1}{2}k_{2})$. The line connecting $\mathrm{R}=(-1,-k_{2})$ and $\mathrm{S}=(x^{*},y^{*})$ can be represented by $y=-k_{2}+k^{+}(x+1)$, with $k^{+}:=\frac{y^{*}+k_{2}}{x^{*}+1}>k_{2}$.

We set

It is clear that $\psi\in C(\mathbb{R})$ and increasing on $\mathbb{R}$. The building blocks of the graph $y=\psi(x)$ are three lines whose slpoes are $k_{1}<k_{2}<k^{+}$. Hence, if $x_{1}>x_{2}$, then $\psi(x_{1})-\psi(x_{2})\geq k_{1}(x_{1}-x_{2})$, that is, the function $x\mapsto\psi(x)-k_{1}x$ is nondecreasing on $\mathbb{R}$.

Next, we set for $j\in\mathbb{N}$,

It is clear that for all $j\in\mathbb{N}$, $\psi_{j}\in C(\mathbb{R})$, the function $x\mapsto\psi_{j}(x)-k_{1}x$ is nondecreasing on $\mathbb{R}$, and

We set

It is clear that $\eta\in C(\mathbb{R})$ and $x\mapsto\eta(x)-k_{1}x$ is nondecreasing on $\mathbb{R}$. Moreover, we see that

and that if $-2^{-j}<x<-2^{-j-1}$ and $j\in\mathbb{N}$,

Note that the point $\mathrm{S}=(x^{*},y^{*})$ is on the graph $y=\psi(x)$ and, hence, that for any $j\in\mathbb{N}$, the point $(2^{-j}x^{*},2^{-j}y^{*})$ is on the graph $y=\eta(x)$. Similarly, since the point $\mathrm{S}=(x^{*},y^{*})$ is on the graph $y=k^{*}x$ and for any $j\in\mathbb{N}$, the point $(2^{-j}x^{*},2^{-j}y^{*})$ is on the graph $y=k^{*}x$. Also, for any $j\in\mathbb{N}$, the point $(-2^{-j},-k_{2}2^{-j})$ lies on the graphs $y=\eta(x)$ and $y=k_{2}x$.

Fix any $d\geq 1$ and define $h\in C(\mathbb{R})$ by

For the function $h$ defined above, we consider the problem

For any $\lambda\geq 0$, we denote by $z_{\lambda}$ the unique solution of (5). Since $h(d)=0$, it is clear that $z_{0}=d$.

We fix $k_{0}\in(0,k_{1})$ and define $f,g\in C(\mathbb{R})$ by $f(x)=k_{0}(x-d)$ and

It is easily checked that $g(x)-(k_{1}-k_{0})x$ is nondecreasing on $\mathbb{R}$, which implies that $g$ is increasing on $\mathbb{R}$, and that $h(x)=f(x)-g(-x)$ for all $x\in\mathbb{R}$. We note that

We fix $f,g,h$ as above, and consider the system (2).

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

With our choice of $f,g$, the family of solutions $(u_{\lambda},v_{\lambda})$ of (2), with $\lambda>0$, does not converge as $\lambda\to 0$.

We remark that, since

We give the proof of Theorem 1.

Some remarks are in order. (i) Following [Zi], we may use Theorem 7 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 $f,g\in C(\mathbb{R},\mathbb{R})$ in Theorem 7, the author has chosen $d$ to satisfy $d\geq 1$, but one may choose any $d>0$. In the process, the crucial step is to find the function $h(x)=f(x)-g(-x)$, with the properties: (a) the function $x\mapsto h(x)-\varepsilon x$ is nondecreasing on $\mathbb{R}$ for some $\varepsilon>0$ and (b) the curve $y=h(x)$, with $x<d$, meets the lines $y=p(x-d)$ and $y=q(x-d)$, respectively, at $P_{n}$ and $Q_{n}$ for all $n\in\mathbb{N}$, where $p,q,d$ are positive constants such that $\varepsilon<p<q$, and the sequences $\{P_{n}\}_{n\in\mathbb{N}},\,\{Q_{n}\}_{n\in\mathbb{N}}$ converge to the point $(d,0)$. Thus, a possible choice of $h$ among many other ways is the following. Define first $\eta\,:\,\mathbb{R}\to\mathbb{R}$ by $\eta(x)=x(\sin(\log|x|)+2)$ if $x\not=0$, and $\eta(0)=0$ (see Fig. 2). Fix $d>0$ and set $h(x)=\eta(x-d)$ for $x\in\mathbb{R}$. Note that if $x\not=0$,

and that if we set $x_{n}=-\exp(-2\pi n)$ and $\xi_{n}=-\exp\left(-2\pi n+\frac{\pi}{2}\right)$, $n\in\mathbb{N}$, then

The points $P_{n}:=(x_{n}+d,2x_{n})$ are on the intersection of two curves $y=h(x)$ and $y=2(x-d)$, while the points $Q_{n}:=(d+\xi_{n},3\xi_{n})$ are on the intersection of $y=h(x)$ and $y=3(x-d)$. Moreover, $\lim P_{n}=\lim Q_{n}=(d,0)$.