Convergence & rates for Hamilton–Jacobi equations with Kirchoff junction conditions

Abstract

We investigate rates of convergence for two approximation schemes of time-independent and time-dependent Hamilton–Jacobi equations with Kirchoff junction conditions. We analyze the vanishing viscosity limit and monotone finite-difference schemes. Following recent work of Lions and Souganidis, we impose no convexity assumptions on the Hamiltonians. For stationary Hamilton–Jacobi equations, we obtain the classical $\epsilon ^{\frac{1}{2}}$ rate, while we obtain an $\epsilon ^{\frac{1}{7}}$ rate for approximations of the Cauchy problem. In addition, we present a number of new techniques of independent interest, including a quantified comparison proof for the Cauchy problem and an equivalent definition of the Kirchoff junction condition.

Appendices

Appendix A: Reformulated Kirchoff condition

We present an equivalent definition of viscosity solutions of Kirchoff problems for the general problem:

$$\begin{aligned} \left\{ \begin{array}{l l} F_{i}(x,u,u_{x_{i}}) = 0 &{} \text {in} \, \, I_{i} \\ {\sum }_{i = 1}^{K} u_{x_{i}} = B &{} \text {on} \, \, \{0\} \end{array} \right. \end{aligned}$$

(82)

where $F_{i} : \overline{I_{i}} \times {\mathbb {R}} \times {\mathbb {R}} \rightarrow {\mathbb {R}}$ is continuous for each i. We start with sub-solutions:

Proposition 10

$u \in USC({\mathcal {I}})$ is a sub-solution of (82) if and only if for each $\varphi \in C^{2}({\mathcal {I}})$ and any local maximum $x_{0}$ of $u - \varphi$, the following inequalities are satisfied:

$$\begin{aligned} \left\{ \begin{array}{l l} F_{i}(x_{0},u(x_{0}),\varphi _{x_{i}}(x_{0})) \le 0 &{} \text {if} \, \, x_{0} \in I_{i} \\ \min _{i, {\tilde{\theta }} \in [0,1]} F_{i}\left( 0,u(0),\varphi _{x_{i}}(0) + {\tilde{\theta }} \left( {\sum }_{j = 1}^{K} \varphi _{x_{i}}(0) - B\right) ^{-}\right) \le 0 &{} \text {if} \, \, x_{0} = 0 \end{array} \right. \end{aligned}$$

In fact, the proposition only requires $\varphi \in C^{1}({\mathcal {I}})$, but we will not expand on that here.

Before proceeding, we need to recall the definitions of first-order differential sub-jets and super-jets. Given an upper semi-continuous function $u : {\mathcal {I}} \rightarrow {\mathbb {R}}$ and an $x \in \overline{I_{i}}$, we say that $p \in J^{+}_{i}u(x)$ if and only if

$$\begin{aligned} u(y) \le u(x) + p(y - x) + o(|y - x|) \quad \text {if} \, \, y \in \overline{I_{i}}, \end{aligned}$$

where $\lim _{y \rightarrow x} \frac{o(|y - x|)}{|y - x|} = 0$. Similarly, given a lower semi-continuous function $v : {\mathcal {I}} \rightarrow {\mathbb {R}}$ and an $x \in \overline{I_{i}}$, we say that $q \in J^{-}_{i}v(x)$ if and only if

$$\begin{aligned} v(y) \ge v(x) + q(y - x) + o(|y-x|) \quad \text {if} \, \, y \in \overline{I_{i}}. \end{aligned}$$

Note that there is a $\varphi \in C^{2}({\mathcal {I}})$ such that $u - \varphi$ has a local maximum at 0 if and only if $\xi _{i} := \varphi _{x_{i}}(0) \in J_{i}^{+}u(0)$ for each i. Therefore, in what follows, we often work with K-tuples $(\xi _{1},\dots ,\xi _{K})$ instead of test functions.

As in the Neumann problem, Proposition 10 rests on the next lemma.

Lemma 9

Fix $i \in \{1,\dots ,K\}$ and assume that u is an upper semi-continuous sub-solution of

$$\begin{aligned} F_{i}(x,u,u_{x_{i}}) = 0 \quad \text {in} \, \, I_{i}. \end{aligned}$$

Let $\xi _{i} \in J_{i}^{+}u(0)$ and set $\lambda _{i,0} = \sup \{\lambda \ge 0 \, \mid \, \xi _{i} + \lambda \in J^{+}_{i}u(0) \}$. If $\lambda _{i,0} < \infty$, then

$$\begin{aligned} F_{i}(0,u(0),\xi _{i} + \lambda _{i,0}) \le 0. \end{aligned}$$

Proof

The proof is the same as the one in [18, Lemma 3]. For the sake of completeness, we reproduce it here. Since $J_{i}^{+}u(0)$ is closed, we can pick $\psi \in C^{2}(\overline{I_{i}})$ such that $u - \psi$ has a strict local maximum at 0 in $\overline{I_{i}}$, $u(0) = \psi (0)$, and $\psi _{x_{i}}(0) = \xi _{i} + \lambda _{i,0}$. Fix $\delta >0$, set $\mu (\delta ) = \psi (-\delta _{i}) - u(-\delta _{i})$ and $\alpha (\delta ) = \min \{\delta ,\frac{\mu (\delta )}{2 \delta }\}$, and let $x_{\delta }$ be a maximum of $u - \psi - \alpha (\delta )x$ in $\overline{I^{\delta }_{i}} \subseteq \overline{I_{i}}$.

Observe that $x_{\delta } \ne 0$. Indeed, if $x_{\delta } = 0$, then

$$\begin{aligned} \xi _{i} + \lambda _{i,0} + \alpha (\delta ) =\psi _{x_{i}}(0) + \alpha (\delta ) \in J^{+}u_{i}(0), \end{aligned}$$

contradicting the definition of $\lambda _{i,0}$.

Additionally, $x_{\delta } \ne -\delta _{i}$ since

$$\begin{aligned} u(-\delta _{i}) - \psi (-\delta _{i}) + \alpha (\delta )\delta&\le \frac{u(-\delta _{i}) - \psi (-\delta _{i})}{2} \\&< 0 = u(0) - \psi (0) < u(x_{\delta }) - \psi (x_{\delta }) - \alpha (\delta ) x_{\delta }. \end{aligned}$$

Thus, $x_{\delta } \in (0,\delta )$ and the sub-solution property of u gives

$$\begin{aligned} F_{i}(x_{\delta },u(x_{\delta }),\psi _{x_{i}}(x_{\delta }) + \alpha (\delta )) \le 0. \end{aligned}$$

From the inequality $u(0) - \psi (0) < u(x_{\delta }) - \psi (x_{\delta }) - \alpha (\delta ) x_{\delta }$, the upper semi-continuity of u implies $u(x_{\delta }) \rightarrow u(0)$. Therefore, observing that

$$\begin{aligned} \lim _{\delta \rightarrow 0^{+}} (x_{\delta },\alpha (\delta ),u(x_{\delta })) = (0,0,u(0)), \end{aligned}$$

the result follows. $\square$

In the proof of Proposition 10 we will use the following fact about sub-solutions of (82). In fact, the corresponding property of the Neumann problem is actually embedded in the definition in [18].

Proposition 11

Suppose u is a sub-solution of (82), $\varphi \in C^{2}({\mathcal {I}})$, and $u - \varphi$ has a local maximum at 0. If ${\sum }_{i = 1}^{K} \varphi _{x_{i}}(0) \ge B$, then $\min _{i} F_{i}(0,u(0),\varphi _{x_{i}}(0)) \le 0$.

Proof

By the definition of sub-solution, it suffices to consider the case when ${\sum }_{i = 1}^{K} \varphi _{x_{i}}(0) = B$. Define $(\xi _{1},\dots ,\xi _{K})$ by $\xi _{i} = \varphi _{x_{i}}(0)$ and let $(\lambda _{1,0},\dots ,\lambda _{K,0})$ be defined as in Lemma 9. If $\lambda _{j,0} = 0$ for some j, then Lemma 9 implies $F_{j}(0,u(0),\xi _{j}) \le 0$. On the other hand, if $\min _{i}\lambda _{i,0} > 0$, then, for small enough $\delta > 0$, $\xi _{i} + \delta \in J_{i}^{+}u(0)$ holds, no matter the choice of i. From ${\sum }_{i = 1}^{K} (\xi _{i} + \delta ) = B + \delta K$ and the sub-solution property, we find $\min _{i}F_{i}(0,u(0),\xi _{i} + \delta ) \le 0$. We conclude by sending $\delta \rightarrow 0^{+}$. $\square$

We continue with the

Proof of Proposition 10

Since one direction is immediate, here we prove only the “only if” statement.

Suppose $(\xi _{1},\dots ,\xi _{K})$ is a K-tuple satisfying $\xi _{i} \in J^{+}_{i}u(0)$ for each i.

In what follows, we use the notation in the statement of Lemma 9. If there is a $j \in \{1,2,\dots ,K\}$ such that $\left( {\sum }_{i = 1}^{K} \xi _{i} -B\right) ^{-} < \lambda _{j,0}$, then let ${\tilde{\xi }}_{k} = \xi _{k}$ if $k \ne j$ and ${\tilde{\xi }}_{j} = \xi _{j} + \left( {\sum }_{i = 1}^{K} \xi _{i} -B\right) ^{-}$. For each i, ${\tilde{\xi }}_{i} \in J_{i}^{+} u(0)$ and

$$\begin{aligned} \sum _{i = 1}^{K} {\tilde{\xi }}_{i} = \left( \sum _{i = 1}^{K} \xi _{i}\right) + \left( \sum _{j = 1}^{K} \xi _{j} - B\right) ^{-} \ge B. \end{aligned}$$

(83)

Thus, Proposition 11 implies $\min _{i} F_{i} \left( 0,u(0), {\tilde{\xi }}_{i} \right) \le 0$. From this and the definition of $({\tilde{\xi }}_{1},\dots ,{\tilde{\xi }}_{K})$, we conclude

$$\begin{aligned} \min _{j} \min _{{\tilde{\theta }} \in [0,1]} F_{j}\left( 0,u(0), \xi _{j} + {\tilde{\theta }} \left( \sum _{i = 1}^{K} \xi _{i} -B\right) ^{-} \right) \le 0. \end{aligned}$$

It only remains to consider the case when $\lambda _{j,0} \le \left( {\sum }_{i = 1}^{K} \xi _{i}-B\right) ^{-}$ independently of the choice of j. In this case, we can fix $({\tilde{\theta }}_{1},{\tilde{\theta }}_{2},\dots ,{\tilde{\theta }}_{K}) \in [0,1]^{K}$ such that $\lambda _{j,0} = {\tilde{\theta }}_{j} \left( {\sum }_{i = 1}^{K} \xi _{i} - B\right) ^{-}$. Therefore, Lemma 9 yields that, for each j,

$$\begin{aligned} \min _{{\tilde{\theta }} \in [0,1]} F_{j} \left( 0,u(0), \xi _{j} + {\tilde{\theta }} \left( \sum _{i = 1}^{K} \xi _{i} - B\right) ^{-} \right) \le F_{j}(0,u(0),\xi _{j} + \lambda _{j,0}) \le 0. \end{aligned}$$

$\square$

The result for super-solutions is stated next. Since the proof is so similar, we omit the details.

Proposition 12

A function $v \in \text {LSC}({\mathcal {I}})$ is a viscosity super-solution of (82) if and only if for each $\varphi \in C^{2}({\mathcal {I}})$ and any local minimum $x_{0}$ of $u - \varphi$, the following inequalities are satisfied:

$$\begin{aligned} \left\{ \begin{array}{l l} F_{i}(x_{0},v(x_{0}),\varphi _{x_{i}}(x_{0})) \ge 0 &{} \text {if} \, \, x_{0} \in I_{i}\\ \max _{i, {\tilde{\theta }} \in [0,1]} F_{i}\left( 0,v(0),\varphi _{x_{i}}(0) - {\tilde{\theta }} \left( {\sum }_{j = 1}^{K} \varphi _{x_{i}}(0) -B \right) ^{+}\right) \ge 0 &{} \text {if} \, \, x_{0} = 0 \end{array} \right. \end{aligned}$$

There is an analogous reformulation of time-dependent equations like (2). We do not prove it here since we have no immediate use for it and it does not simplify the uniqueness proof presented in Sect. 3.

Appendix B: Dimensionality reduction lemma

In this section, we show how to obtain time-independent equations from those in which time-derivatives do not appear. The following result implies Proposition 3:

Lemma 10

Assume that, for each i, $F_{i} : [0,T] \times {\mathcal {I}} \times {\mathbb {R}}^{2} \rightarrow {\mathbb {R}}$ is a continuous function, and fix $B \in {\mathbb {R}}$ and $\delta > 0$. Let the upper semi-continuous function $u : \bigcup _{i = 1}^{K} \overline{I_{i}^{\delta }} \times [0,T] \rightarrow {\mathbb {R}}$ be a sub-solution of

$$\begin{aligned} \left\{ \begin{array}{l l} F_{i}(t,x,u,u_{x_{i}}) = 0 &{} \text {in} \, \, I_{i}^{\delta } \times (0,T) \\ {\sum }_{i = 1}^{K} u_{x_{i}} = B &{} \text {on} \, \, \{0\} \times (0,T) \end{array} \right. \end{aligned}$$

(84)

For each $t_{0} \in (0,T]$, the function $u(\cdot ,t_{0}) : \bigcup _{i = 1}^{K} \overline{I_{i}^{\delta }} \rightarrow {\mathbb {R}}$ is a sub-solution of

$$\begin{aligned} \left\{ \begin{array}{l l} F_{i}(t_{0},x,u(\cdot ,t_{0}),u_{x_{i}}(\cdot ,t_{0})) = 0 &{} \text {in} \, \, I_{i}^{\delta } \\ {\sum }_{i = 1}^{K} u_{x_{i}}(\cdot ,t_{0}) = B &{} \text {on} \, \, \{0\} \end{array} \right. \end{aligned}$$

We remark that a version of Lemma 10 for super-solutions follows from it by replacing u by $-u$.

Proof

Fix a $t_{0} \in (0,T]$. Given $\varphi \in C^{2}({\mathcal {I}})$, suppose $u(\cdot ,t_{0})- \varphi$ has a strict global maximum in $\bigcup _{i = 1}^{K} \overline{I_{i}^{\delta }}$ at $x_{0} \in \bigcup _{i = 1}^{K} I_{i}^{\delta }$. We consider only the case when $x_{0} = 0$, the other case being slightly easier.

For each $\epsilon > 0$, let $\Phi _{\epsilon } : \bigcup _{i = 1}^{K} \overline{I_{i}^{\delta }} \times [0,T] \rightarrow {\mathbb {R}}$ be the function given by

$$\begin{aligned} \Phi _{\epsilon }(x,t) = u(x,t) - \varphi (x) - \frac{(t- t_{0})^{2}}{2 \epsilon }. \end{aligned}$$

Write $\Phi _{\epsilon }(x,t) = u(x,t) - \Psi _{\epsilon }(x,t)$ and note that $\Psi _{\epsilon } \in C^{2,1}({\mathcal {I}} \times [0,T])$. Let $(x_{\epsilon }, t_{\epsilon })$ denote a maximum point of $\Phi _{\epsilon }$ in its domain. Since 0 is a strict global maximum of $u(\cdot ,t_{0}) - \varphi$, it follows that $t_{\epsilon } \rightarrow t_{0}$, $x_{\epsilon } \rightarrow 0$, and $u(x_{\epsilon },t_{\epsilon }) \rightarrow u(0,t_{0})$ as $\epsilon \rightarrow 0^{+}$. Fix $\epsilon _{1} > 0$ such that $t_{\epsilon } > 0$ if $\epsilon \in (0,\epsilon _{1})$. If there is a sequence $\epsilon _{n} \rightarrow 0$ such that $x_{\epsilon _{n}} \in I_{j}$ for some j and each n, then we immediately obtain

$$\begin{aligned} F_{j}(t_{\epsilon _{n}},x_{\epsilon _{n}},u(x_{\epsilon _{j}},t_{\epsilon _{j}}),\varphi _{x_{j}}(x_{\epsilon _{n}})) = F_{j}(t_{\epsilon _{n}}, x_{\epsilon _{n}}, u(x_{\epsilon _{j}},t_{\epsilon _{j}}),\Psi _{x_{j}}(x_{\epsilon _{n}},t_{\epsilon _{n}})) \le 0. \end{aligned}$$

Sending $n \rightarrow \infty$, we recover $F_{j}(t_{0},0,u(x_{0},t_{0}),\varphi _{x_{j}}(0)) \le 0$.

It remains to consider the case when there is an $\epsilon _{2} > 0$ such that $x_{\epsilon } = 0$ for all $\epsilon \in (0,\epsilon _{2})$. Fix such an $\epsilon$. For each $j \in \{1,2,\dots ,K\}$, the map

$$\begin{aligned} (x,t) \mapsto u(x,t) - \varphi (x) - \frac{(t - t_{0})^{2}}{2 \epsilon } \end{aligned}$$

defined in $\overline{I_{j}^{\delta }} \times [0,T]$ has a local maximum at $(0,t_{\epsilon })$. Thus,

$$\begin{aligned} \min \left\{ \min _{i} F_{i}(t_{\epsilon },0,u(0,t_{\epsilon }),\varphi _{x_{i}}(0)), \sum _{i = 1}^{K} \varphi _{x_{i}}(0) - B \right\} \le 0. \end{aligned}$$

We conclude by sending $\epsilon \rightarrow 0^{+}$ and appealing to continuity of the functions $F_{1},\dots ,F_{K}$. $\square$

Appendix C: Existence of solutions of the Cauchy problems

In this section, we prove the existence of solutions of (2) and (4). The main results proved herein follow:

Theorem 11

If $u_{0} \in \text {UC} \left( {\mathcal {I}} \right)$, then there is a $u \in \text {UC}({\mathcal {I}} \times [0,T])$ solving (2). If, in addition, $u_{0} \in \text {Lip}({\mathcal {I}})$, then $u \in \text {Lip}({\mathcal {I}} \times [0,T])$, and $\text {Lip}(u)$ depends on $u_{0}$ only through $\text {Lip}(u_{0})$.

Theorem 12

Fix $\epsilon > 0$. If $u_{0} \in \text {UC} \left( {\mathcal {I}} \right)$, then there is a $u^{\epsilon } \in \text {UC}({\mathcal {I}} \times [0,T])$ solving (4). Moreover, if $[u_{0}]_{1} + [u_{0}]_{2} < \infty$, then there is a $C > 0$ depending only on $[u_{0}]_{1} + \epsilon [u_{0}]_{2}$ such that $\text {Lip}(u^{\epsilon }) \le C$.

We prove Theorem 11 by sending $\epsilon \rightarrow 0^{+}$ in Theorem 12. Therefore, the main thrust of this section is the proof of Theorem 12 and associated estimates.

The proof of Theorem 12 is divided into three steps. First, we use the estimate proved by von Below in [22] and Schaefer’s fixed point theorem to obtain solutions of (4) when $u_{0}$ is a smooth function satisfying some compatibility conditions. Next, we prove Lipschitz estimates when the initial data is sufficiently regular. Finally, we approximate arbitrary initial data by smooth data and use the comparison principle to pass to the limit.

Recall that in Remark 8 above we observed that an alternative proof of Theorem 11 can be obtained using the finite-difference scheme (6) and the method of half-relaxed limits.

1.1 C.1: Existence for regular data

Here we obtain solutions of (4) using a priori Hölder estimates for linear parabolic equations on networks and Schaefer’s fixed point theorem.

To begin with, for a given $R > 0$, we let $\{{\tilde{H}}_{1}^{(R)},\dots ,{\tilde{H}}_{K}^{(R)}\}$ take the form

$$\begin{aligned} {\tilde{H}}_{i}^{(R)}(t,x,p) = \psi ^{(R)}(p) H_{i}(t,x,p) + (1 - \psi ^{(R)}(p)) R, \end{aligned}$$

where $\psi ^{(R)} : {\mathbb {R}} \rightarrow [0,1]$ is a smooth cut-off function satisfying $\psi ^{(R)}(p) = 1$ if $|p| \le \frac{R}{2}$ and $\psi ^{(R)}(p) = 0$ if $|p| \ge R$. Notice that $\{{\tilde{H}}_{1}^{(R)},\dots ,{\tilde{H}}_{K}^{(R)}\}$ are bounded functions on their respective domains, and the assumptions (7) and (9) continue to hold.

The result is stated next:

Proposition 13

Suppose $a > 0$ and $u_{0} \in C^{3}({\mathcal {I}})$ satisfies, for each $i \in \{1,2,\dots ,K\}$,

$$\begin{aligned} \epsilon u_{0,x_{i}x_{i}}(0) - H_{i}(0,0,u_{0,x_{i}}(0))&= \epsilon u_{0,x_{1}x_{1}}(0) - H_{1}(0,0,u_{0,x_{1}}(0)) \end{aligned}$$

(85)

$$\begin{aligned} \sum _{i = 1}^{K} u_{0,x_{i}}(0)&= 0 \end{aligned}$$

(86)

$$\begin{aligned} _{1} + [u_{0}]_{2} + [u_{0}]_{3}&< \infty \end{aligned}$$

(87)

Assume, in addition, that $R \ge 2 [u_{0}]_{1}$. Then there is a viscosity solution $u^{(a)}: \bigcup _{i = 1}^{K} \overline{I^{a}_{i}} \times [0,T] \rightarrow {\mathbb {R}}$ of the following equation:

$$\begin{aligned} \left\{ \begin{array}{l l} u^{(a)}_{t} - \epsilon u^{(a)}_{x_{i}x_{i}} + {\tilde{H}}^{(R)}_{i}(t,x,u^{(a)}_{x_{i}}) = 0 &{} \text {in} \, \, I_{i}^{a} \times (0,T) \\ {\sum }_{i = 1}^{K} u^{(a)}_{x_{i}} = 0 &{} \text {on} \, \, \{0\} \times (0,T) \\ u^{(a)} = u_{0} &{} \text {on} \, \, {\bigcup }_{i = 1}^{K} \overline{I_{i}^{a}} \times \{0\} \\ u^{(a)} = \beta _{i} &{} \text {on} \, \, \{-a_{i}\} \times (0,T) \end{array} \right. \end{aligned}$$

where the functions $\beta _{1},\dots ,\beta _{K} : [0,T] \rightarrow {\mathbb {R}}$ are given by

$$\begin{aligned} \beta _{i}(t) = u_{0}(-a_{i}) + \left( \epsilon u_{0,x_{i}x_{i}}(-a_{i}) - {\tilde{H}}^{(R)}_{i}(0,-a_{i},u_{0,x_{i}}(-a_{i}))\right) t. \end{aligned}$$

(88)

For each $i \in \{1,2,\dots ,K\}$, the functions $u^{(a)}$, $u^{(a)}_{t}$, $u^{(a)}_{x_{i}}$, and $u^{(a)}_{x_{i}x_{i}}$ are Hölder continuous in $\overline{I_{i}^{a}} \times [0,T]$.

A similar result has been obtained in [2] starting with weak solutions in $L^{p}$ spaces.

Our proof of Proposition 13 follows the same general outline presented in [17, Chapter 5]. As in the fixed point arguments contained there, the next remark will play a significant role here. For a proof, see, for example, [17, Lemma 3.1].

Remark 9

Suppose $I \subseteq {\mathbb {R}}$ is an open interval and $u : {\overline{I}} \times [0,T] \rightarrow {\mathbb {R}}$ is twice continuously differentiable in space and once continuously differentiable in time. Then $u_{x}$ is $\frac{1}{2}$-Hölder continuous in time with a Hölder constant that only depends on I and the suprema of $|u_{t}|$ and $|u_{xx}|$.

It will be convenient in what follows to use the semi-norms $[\cdot ]_{\alpha }$ and $[\cdot ]_{1 + \alpha }$ on functions with domain $\bigcup _{i = 1}^{K} \overline{I_{i}^{a}} \times [0,T]$, abusing the notation somewhat. By this, we mean the semi-norms as defined in Sect. 1.5, but with $\overline{I_{i}}$ replaced everywhere in the definitions with $\overline{I_{i}^{a}}$.

Proof of Proposition 13

First, for each $\alpha \in (0,1)$, define a norm on functions $v : \bigcup _{i = 1}^{K} \overline{I_{i}^{a}} \times [0,T] \rightarrow {\mathbb {R}}$ by

$$\begin{aligned} \Vert v\Vert _{\alpha } = [v]_{0} + [v]_{\alpha } + \max _{i} \, [v_{x_{i}}]_{i,0} + [v]_{1 + \alpha }. \end{aligned}$$

Let $V_{\alpha }$ be the Banach space of functions v with $\Vert v\Vert _{\alpha } < \infty$. We will find the solution as the fixed point of a certain operator on $V_{\alpha }$.

Fix $\alpha \in (0,1)$. We claim we can define a compact, continuous operator $T : V_{\alpha } \rightarrow V_{\alpha }$ so that $u = T(v)$ solves the equation

$$\begin{aligned} \left\{ \begin{array}{l l} u_{t} - \epsilon u_{x_{i}x_{i}} + {\tilde{H}}^{(R)}_{i}(x,t,v_{x_{i}}) = 0 &{} \text {in} \, \, I_{i}^{a} \times (0,T) \\ {\sum }_{i = 1}^{K} u_{x_{i}} = 0 &{} \text {on} \, \, \{0\} \times (0,T) \\ u = u_{0} &{} \text {on} \, \, {\bigcup }_{i = 1}^{K} \overline{I_{i}^{a}} \times \{0\} \\ u = \beta _{i} &{} \text {on} \, \, \{-a\} \times (0,T) \end{array} \right. \end{aligned}$$

(89)

Indeed, since $[v]_{1 + \alpha } < \infty$ and ${\tilde{H}}^{(R)}_{i}(0,0,u_{0,x_{i}}(0)) = H_{i}(0,0,u_{0,x_{i}}(0))$ for all i by the choice of R, the compatibility conditions (85), (86), and (88) imply the result of [22] is applicable. In particular, a bounded solution u of (89) exists and the functions $u_{t},u_{x_{1}x_{1}},\dots ,u_{x_{K}x_{K}}$ are bounded and continuous. Thus, Remark 9 implies $[u]_{2} < \infty$, and $u \in V_{\alpha }$ follows. Arguing as in Theorem 9, we see that u is uniquely determined. Therefore, T is well-defined.

We claim that T is compact and continuous. Suppose $(v_{n})_{n \in {\mathbb {N}}} \subseteq V_{\alpha }$ and $\Vert v_{n}\Vert _{\alpha } \le C$ independently of n. Let $u^{(n)} = T(v_{n})$. The main result of [22] implies $u^{(n)},u^{(n)}_{t},u^{(n)}_{x_{1}x_{1}},\dots ,u^{(n)}_{x_{K}x_{K}}$ are bounded continuous functions with bounds depending on $(v_{n})_{n \in {\mathbb {N}}}$ only through the constant C. Thus, Remark 9 implies $[u^{(n)}]_{1} + [u^{(n)}]_{2}$ is uniformly bounded. Since $\alpha < 1$ and $\{u^{(n)}(\cdot ,0)\}_{n \in {\mathbb {N}}} = \{u_{0}\}$, it follows that $(u^{(n)})_{n \in {\mathbb {N}}}$ is relatively compact in $V_{\alpha }$. Therefore, by definition, T is compact. Since solutions of (89) in $V_{\alpha }$ are unique, if, in addition, $v_{n} \rightarrow v$ in $V_{\alpha }$, then it is straightforward to check that T(v) is the only possible subsequential limit point of $(u^{(n)})_{n \in {\mathbb {N}}}$. In particular, $u^{(n)} \rightarrow T(v)$ in $V_{\alpha }$, which proves that T is continuous.

Finally, we check the hypotheses of Schaefer’s fixed point theorem (cf. [12, Section 9.2.2]). Recall that we need to find a $C >0$ such that if $u \in V_{\alpha }$ satisfies $u= \sigma T(u)$ for some $\sigma \in [0,1]$, then $\Vert u\Vert _{\alpha } \le C$. Indeed, arguing as in Proposition 14 below, we see that $u_{t}$ is bounded independently of $\sigma$. Since $\{{\tilde{H}}_{1}^{(R)},\dots ,{\tilde{H}}_{K}^{(R)}\}$ are bounded functions, the equation implies $u_{x_{i}x_{i}}$ is also bounded independently of $\sigma$ and i. From this, we obtain a bound on $[u]_{2}$ by Remark 9. Finally, the regularity of $u_{0}$ and the uniform bound on $[u]_{2}$ gives a bound on $\max _{i} \, [u_{x_{i}}]_{i,0} + [u]_{1 + \alpha }$, and this together with the uniform bound on $u_{t}$ provides one for $[u]_{0} + [u]_{\alpha }$. It follows that $\Vert u\Vert _{\alpha }$ is bounded independently of $\sigma$.

By Schaefer’s theorem, we conclude there is a $u^{(a)} \in V_{\alpha }$ such that $T(u^{(a)}) = u^{(a)}$. The regularity of $u^{(a)}$ and its derivatives follows directly from the result of [22]. $\square$

1.2 C.2: A priori bounds

In the previous subsection, we showed that smooth initial data have smooth solutions, provided certain compatibility conditions are satisfied. Now we prove some a priori estimates satisfied by these solutions.

We start with a bound on the time derivative, which follows from (9) and the maximum principle.

Proposition 14

Let $a> 0$. If $u_{0}$ and $R > 0$ satisfy the hypotheses of Proposition 13, and if $u^{(a)}$ is the solution obtained therein, then

$$\begin{aligned} u^{(a)}_{t}(x,0) = \epsilon u_{0,x_{i}x_{i}}(x) - {\tilde{H}}^{(R)}_{i}(x,0,u_{0,x_{i}}(x)) \quad \text {if} \, \, x \in \overline{I_{i}^{a}}, \, \, i \in \{1,2,\dots ,K\}, \end{aligned}$$

and, for each $(x,t) \in {\mathcal {I}} \times [0,T]$,

$$\begin{aligned} |u^{(a)}_{t}(x,t)| \le [u^{(a)}_{t}(\cdot ,0)]_{0} + Dt. \end{aligned}$$

(90)

Proof

The claim concerning $u^{(a)}_{t}(\cdot ,0)$ follows from the regularity established in Proposition 13.

Given $\zeta \in (0,T)$, define $v^{\zeta } : \bigcup _{i = 1}^{K} \overline{I_{i}^{a}} \times [0,T - \zeta ] \rightarrow {\mathbb {R}}$ by $v^{\zeta }(x,t) = \frac{u^{(a)}(x,t + \zeta ) - u^{(a)}(x,t)}{\zeta }$. An immediate computation shows $v^{\zeta }$ is a classical solution of a linear parabolic equation of the following form:

$$\begin{aligned} \left\{ \begin{array}{l l} v^{\zeta }_{t} - \epsilon v^{\zeta }_{x_{i}x_{i}} + b_{i}^{\zeta }(x,t) v^{\zeta }_{x_{i}} - g_{i}^{\zeta }(x,t) = 0 &{} \text {in} \, \, I^{a}_{i} \times (0,T - \zeta ) \\ {\sum }_{i = 1}^{K} v^{\zeta }_{x_{i}} = 0 &{} \text {on} \, \, \{0\} \times (0,T - \zeta ) \\ v^{\zeta } = u^{(a)}_{t}(\cdot ,0) &{} \text {on} \, \, \{-a_{i}\} \times (0,T - \zeta ) \end{array} \right. \end{aligned}$$

Notice that $\{b_{1}^{\zeta },\dots ,b_{K}^{\zeta }\}$ and $\{g_{1}^{\zeta },\dots ,g_{K}^{\zeta }\}$ are bounded functions by (7) and (9). Specifically, the functions $\{g_{1}^{\zeta },\dots ,g_{K}^{\zeta }\}$ are bounded above and below by D and $-D$, respectively, independently of the choice of $\zeta$.

We claim that if $(x,t) \in \bigcup _{i = 1}^{K} \overline{I_{i}^{a}} \times [0,T-\zeta ]$, then

$$\begin{aligned} v^{\zeta }(x,t) - Dt \le \sup \left\{ v^{\zeta }(x,0) \, \mid \, x \in \bigcup _{i = 1}^{K} \overline{I_{i}^{a}}\right\} . \end{aligned}$$

(91)

To see this, fix $K > 0$ strictly greater than the suprema of the functions $\{|b_{1}^{\zeta }|,\dots ,|b_{K}^{\zeta }|\}$ and notice that if $\delta > 0$, then the function $(x,t) \mapsto v^{\zeta }(x,t) - \delta x - (D + K\delta )t$ cannot attain its maximum in $\bigcup _{i = 1}^{K} (\overline{I^{a}_{i}} \setminus \{-a_{i}\})\times (0,T - \zeta ]$. Recalling that $v^{\zeta }$ is constant on $\bigcup _{i = 1}^{K} \{-a_{i}\} \times [0,T - \zeta ]$ and sending $\delta \rightarrow 0^{+}$, we recover (91).

Notice that for each $\zeta ' \in (0,T)$, the Hölder continuity of $u^{(a)}_{t}$ implies $v^{\zeta } \rightarrow u^{(a)}_{t}$ uniformly in $\bigcup _{i = 1}^{K} \overline{I_{i}^{a}} \times [0,T - \zeta ']$ as $\zeta \rightarrow 0^{+}$. Thus, after sending $\zeta \rightarrow 0^{+}$ in (91), we find $u^{(a)}_{t}(x,t) \le [u^{(a)}_{t}(\cdot ,0)]_{0} + Dt$. To see that $u^{(a)}_{t}(x,t) \ge -( [u^{(a)}_{t}(\cdot ,0)]_{0} + Dt)$, we repeat the previous argument, replacing $v^{\zeta }$ by $-v^{\zeta }$.

$\square$

Next, we leverage the bound on the time derivative to obtain a matching bound on the first order space derivatives.

Proposition 15

If $u^{(a)}$ is the solution obtained in Proposition 13 and we define $C_{1} = [u_{t}(\cdot ,0)]_{0} + DT$, then there is an $L > 0$ independent of a, depending on $u_{0}$ only through $[u_{0}]_{1} + \epsilon [u_{0}]_{2}$, and such that if $a > 2\left( 2C_{1} + 1\right) T$ and $R \ge 2KL$, then

$$\begin{aligned} |u^{(a)}(x,t) - u^{(a)}(y,t)| \le KLd(x,y) \quad \text {if} \, \, d(x,0), d(y,0) \le \frac{a}{2}, \, \, t \in [0,T]. \end{aligned}$$

(92)

Proof

First, let $L_{0} = [u_{0}]_{1} + \epsilon [u_{0}]_{2}$. By (8), there is an $L_{1} \ge 1$ such that

$$\begin{aligned} -(C_{1} + 1) + H_{i}(t,x,p) \ge 1 \quad \text {if} \, \, |p| \ge L_{1}, \, \, i \in \{1,2,\dots ,K\}. \end{aligned}$$

Let $L_{2} = L_{0} + L_{1}$. Notice that since $C_{1}$ depends on $u_{0}$ only through $[u_{0}]_{1} + \epsilon [u_{0}]_{2}$, it follows that $L_{2}$ depends on $u_{0}$ only through that quantity. Assume in what follows that $R \ge 2K(3L_{2} + 1)$.

Fix $i \in \{1,2,\dots ,K\}$ and $(x,t) \in \overline{I_{i}^{a}} \times (0,T)$ such that $d(x,0) \le \frac{a}{2}$. Define the test function $\varphi : {\mathcal {I}} \rightarrow {\mathbb {R}}$ exactly as in (17), but with $u^{(a)}(x,t)$ in place of u(x) and $3L_{2} + 1$ in place of L. Finally, define $w : {\mathcal {I}} \times [0,t] \rightarrow {\mathbb {R}}$ by

$$\begin{aligned} w(y,s) = \varphi (y) + (C_{1} + 1)(t - s). \end{aligned}$$

We claim that the function $(y,s) \mapsto u^{(a)}(y,s) - w(y,s)$ defined in $\bigcup _{i =1}^{K} \overline{I_{i}^{a}} \times [0,t]$ is maximized at (x, t). First, note that $u^{(a)}(x,t) - w(x,t) = 0$. Moreover, if $s < t$, then the inequality $[u_{t}^{(a)}]_{0} \le C_{1}$ implies

$$\begin{aligned} u^{(a)}(x,s) - w(x,s) = u^{(a)}(x,s) - u^{(a)}(x,t) - (C_{1} + 1)(t - s) \le - (t -s) < 0. \end{aligned}$$

Therefore, the maximum does not occur at a point of the form (x, s), where $s \in [0,t)$.

If (y, s) is the maximum of $u^{(a)} - w$ in $\bigcup _{i = 1}^{K} \overline{I^{a}_{i}} \times [0,t]$, $y \in \bigcup _{i = 1}^{K} I_{i}^{a} \setminus \{x\}$, and $s \in (0,t]$, then, in view of the choice of R, the equation yields

$$\begin{aligned} -(C_{1} + 1) + H_{i}(t,x, u (3L_{2} + 1)) \le 0 \quad \text {for some} \, \, u \in \{-1,1, K,-K\}, \end{aligned}$$

contradicting the choice of $L_{1}$. We get a contradiction similarly in the case when $y = 0$ and $s \in (0,t]$.

If (y, s) is the maximum, $y \ne x$, and $s = 0$, then the inequalities $[u_{0}]_{1} \le L_{2}$ and $[u^{(a)}_{t}]_{0} \le C_{1}$ yield the following

$$\begin{aligned} 0 \le u^{(a)}(y,0) - w(y,0) \le u_{0}(y) - u_{0}(x) - (3L_{2}+1) d(x,y) < 0, \end{aligned}$$

which is a contradiction.

Finally, if (y, s) is the maximum and $y = -a_{j}$ for some j, then the assumption $d(x,0) \le \frac{a}{2}$, the inequalities $[u_{0}]_{1} \le L_{2}$ and $[u^{(a)}_{t}]_{0} \le C_{1}$, and the assumption $a > 2\left( 2C_{1} + 1\right) T$ all come together to imply the following:

$$\begin{aligned} u^{(a)}(-a_{j},s)&\ge w(-a_{j},s) \\&= \varphi (-a_{j}) + (C_{1} + 1)(t - s) \\&\ge u^{(a)}(x,t) + (3L_{2} + 1)\left( \frac{a}{2}\right) + (C_{1} + 1)(t - s) \\&\ge \left( u_{0}(x) - u_{0}(-a_{j})\right) - C_{1}(t + s) + u^{(a)}(-a_{j},s) + (3L_{2} + 1) \left( \frac{a}{2}\right) \\&\quad \quad + (C_{1} + 1)(t - s) \\&\ge -L_{2} \left( \frac{3a}{2}\right) - (2C_{1} + 1) T + (3L_{2} + 1) \left( \frac{a}{2}\right) + u^{(a)}(-a_{j},s) \\&> u^{(a)}(-a_{j},s), \end{aligned}$$

which is another contradiction. Therefore, the function $(y,s) \mapsto u^{(a)}(y,s) - w(y,s)$ is maximized in $\bigcup _{i = 1}^{K} \overline{I_{i}^{a}} \times [0,t]$ at the point (x, t).

Thus, restricting to points $(y,s) = (y,t)$, we find

$$\begin{aligned} u^{(a)}(y,t) - u^{(a)}(x,t) \le K (3L_{2} + 1) |x - y| \quad \text {if} \, \, y \in \overline{I_{j}}. \end{aligned}$$

After setting $L = 3L_{2} + 1$, we conclude that (92) holds. $\square$

1.3 C.3: Viscosity solutions

Now we prove Theorems 11 and 12.

To prove these, we need to ensure that we can approximate the initial datum with a regular function that satisfies the compatibility conditions (85) and (86). That is the purpose of the next two results.

Lemma 11

Suppose $p : (-\infty ,0] \rightarrow {\mathbb {R}}$ is a thrice continuously differentiable function for which there is a constant $C_{p} > 0$ such that, for each $x \in (-\infty ,0]$,

$$\begin{aligned} |p'(x)| + |p''(x)| \le C_{p} \end{aligned}$$

and $\sup \left\{ |p'''(x)| \, \mid \, x \in (-\infty ,0]\right\} < \infty$. Let $b \in {\mathbb {R}}$. There is a universal constant $C'_{p} > 0$ depending only on $C_{p}$ and b such that if $\zeta > 0$, then there is a thrice continuously differentiable function $p_{\zeta } : (-\infty ,0] \rightarrow {\mathbb {R}}$ such that $p_{\zeta }(0) = p(0)$, $p_{\zeta }'(0) = p'(0)$, $p_{\zeta }''(0) = b$, $\sup \left\{ |p_{\zeta }'''(x)| \, \mid \, x \in (-\infty ,0]\right\} < \infty$, and, for each $x \in (-\infty ,0]$,

$$\begin{aligned} |p_{\zeta }'(x)| + |p_{\zeta }''(x)|&\le C_{p}'\\ |p_{\zeta }(x) - p(x)|&\le C_{p}' \zeta ^{2} \end{aligned}$$

Proof

Given $\zeta > 0$, choose a smooth function $\varphi _{\zeta } : (-\infty ,0] \rightarrow {\mathbb {R}}$ such that

$$\begin{aligned} \varphi _{\zeta }(x) = 0 \quad \text {if} \, \, x \in (-\infty ,-2\zeta ]&, \quad \varphi _{\zeta }(x) = 1 \quad \text {if} \, \, x \in [-\zeta ,0], \\ \max \left\{ \zeta |\varphi _{\zeta }'(x)|, \zeta ^{2} |\varphi _{\zeta }''(x)| \right\}&\le C_{0} \quad \text {if} \, \, x \in (-\infty ,0], \end{aligned}$$

where $C_{0} \ge 1$ is a universal constant independent of $\zeta$, p, $C_{p}$, and b.

Define $q(x) = p(0) + p'(0)x + \frac{b x^{2}}{2}$ and then let $p_{\zeta } : (-\infty ,0] \rightarrow {\mathbb {R}}$ be given by

$$\begin{aligned} p_{\zeta }(x) = (1 - \varphi _{\zeta }(x))p(x) + \varphi _{\zeta }(x)q(x). \end{aligned}$$

The choice of $\varphi _{\zeta }$ implies $p_{\zeta }(0) = p(0)$, $p_{\zeta }'(0) = p'(0)$, and $p_{\zeta }''(0) = b$. Moreover, $\sup \left\{ |p'''_{\zeta }(x)| \, \mid \, x \in (-\infty ,0]\right\} < \infty$ holds.

Differentiating $p_{\zeta }$, we find

$$\begin{aligned} p_{\zeta }'(x)&= (1 - \varphi _{\zeta }(x))p'(x) + \varphi _{\zeta }(x) q'(x) + \varphi _{\zeta }'(x)(q(x) - p(x)), \\ p_{\zeta }''(x)&= (1 - \varphi _{\zeta }(x))p''(x) + \varphi _{\zeta }(x) q''(x) + 2\varphi _{\zeta }'(x)(q'(x) - p'(x)) \\&\qquad + \varphi _{\zeta }''(x)(q(x) - p(x)). \end{aligned}$$

Thus, the regularity of p and the definition of $\varphi _{\zeta }$ imply the desired bounds by Taylor expansion at 0. $\square$

Now we use Lemma 11 to show how to approximate a $C^{3}({\mathcal {I}})$ function by one that satisfies the compatibility conditions.

Proposition 16

Suppose $u_{0} \in C^{3}({\mathcal {I}})$ satisfies ${\sum }_{i = 1}^{K} u_{0,x_{i}}(0) = 0$ and $[u_{0}]_{1} + [u_{0}]_{2} + [u_{0}]_{3} < \infty$. Then there is a universal constant $C' > 0$ depending only on $[u_{0}]_{1} + [u_{0}]_{2}$ such that for all $\zeta > 0$, there is a $u_{0}^{\zeta } \in C^{3}({\mathcal {I}})$ satisfying the following conditions:

• (i) $[u_{0}^{\zeta }]_{1} + [u_{0}^{\zeta }]_{2} + [u_{0}^{\zeta }]_{3} < \infty$

• (ii) For each $i \in \{1,2,\dots ,K\}$,

  $$\begin{aligned} u^{\zeta }_{0,x_{i}}(0)&= u_{0,x_{i}}(0) \\ -\epsilon u^{\zeta }_{0,x_{i}x_{i}}(0) + H_{i}(0,0,u^{\zeta }_{0,x_{i}}(0))&= - \epsilon u^{\zeta }_{0,x_{1}x_{1}}(0) + H_{1}(0,0,u^{\zeta }_{0,x_{1}}(0)) \\ \end{aligned}$$

• (iii) For each $i \in \{1,2,\dots ,K\}$ and each $x \in \overline{I_{i}}$,

  $$\begin{aligned} |u^{\zeta }_{0,x_{i}}(x)| + |u^{\zeta }_{0,x_{i}x_{i}}(x)|&\le C' \end{aligned}$$

  (93)
  $$\begin{aligned} |u^{\zeta }_{0}(x) - u_{0}(x)|&\le C'\zeta ^{2} \end{aligned}$$

  (94)

Proof

Define $\{b_{1},\dots ,b_{K}\}$ by $b_{1} = 1$ and

$$\begin{aligned} b_{i} = \epsilon ^{-1}\left( \epsilon - H_{1}(0,0,u_{0,x_{1}}(0)) + H_{i}(0,0,u_{0,x_{i}}(0))\right) . \end{aligned}$$

(95)

Notice that this immediately implies $\{b_{1},\dots ,b_{K}\}$ satisfy, for each i,

$$\begin{aligned} -\epsilon b_{i} + H{i}(0,0,u_{0,x_{i}}(0)) = -\epsilon b_{1} + H_{1}(0,0,u_{0,x_{1}}(0)). \end{aligned}$$

(96)

Now apply Lemma 11 to obtain functions $\{\psi ^{\zeta ,(1)},\dots ,\psi ^{\zeta ,(K)}\}$ and a constant $C' > 0$ so that, for each $i \in \{1,2,\dots ,K\}$, $\psi ^{\zeta ,(i)}$ has domain $(-\infty ,0)$ and the following relations hold:

$$\begin{aligned} \sup \left\{ |\psi ^{\zeta ,(i)}_{x}(x)| + |\psi ^{\zeta ,(i)}_{xx}(x)| \, \mid \, x \le 0 \right\}&\le C' \end{aligned}$$

(97)

$$\begin{aligned} \psi ^{\zeta ,(i)}(0)&= u_{0}(0) \end{aligned}$$

(98)

$$\begin{aligned} \psi ^{\zeta ,(i)}_{x}(0)&= u_{0,x_{i}}(0) \end{aligned}$$

(99)

$$\begin{aligned} \psi ^{\zeta ,(i)}_{xx}(0)&= b_{i} \end{aligned}$$

(100)

$$\begin{aligned} \sup \left\{ |\psi ^{\zeta ,(i)}(x) - u_{0}(x)| \, \mid \, x \in \overline{I_{i}}\right\}&\le C' \zeta ^{2} \end{aligned}$$

(101)

By construction, $\{\psi ^{\zeta ,(1)},\dots ,\psi ^{\zeta ,(K)}\}$ come together to form a function $u^{\zeta }_{0} \in C^{3}({\mathcal {I}})$ with the desired properties. $\square$

In the proof that follows, we will not use the $\epsilon$ superscript to denote solutions of (4). Since we are only dealing with (4) and not (2) in the proof, we hope this will not cause too much confusion.

Proof of Theorem 12

First, assume $u_{0} \in C^{3}({\mathcal {I}})$ and $[u_{0}]_{1} + [u_{0}]_{2} + [u_{0}]_{3} < \infty$. For $\zeta > 0$ sufficiently small, let $u^{\zeta }_{0}$ be the function obtained from Proposition 16, and fix $R \ge 2C'$, where $C'$ is the constant defined in the proposition. For each $a > 0$, let $u^{(a),\zeta }$ be the solution of (89) with initial datum $u_{0}^{\zeta }$ obtained in Proposition 13.

By Propositions 14 and 15 and the uniform bound (93), there are constants $B, L, a_{0} > 0$, all independent of $\zeta$, such that if $a \ge a_{0}$ and $R \ge 2KL$, then $[u_{t}^{(a),\zeta }]_{0} \le B$ and (92) holds with $u^{(a)} = u^{(a),\zeta }$. Henceforth, assume $R \ge 2KL$.

The estimates obtained in the previous paragraph imply we can fix a sequence $(a_{n})_{n \in {\mathbb {N}}} \subseteq [a_{0},\infty )$ and a function $u^{\zeta } : {\mathcal {I}} \times [0,T] \rightarrow {\mathbb {R}}$ such that $\lim _{n \rightarrow \infty } a_{n} = \infty$ and $u^{\zeta } = \lim _{n \rightarrow \infty } u^{(a_{n}),\zeta }$ locally uniformly in ${\mathcal {I}} \times [0,T]$. The local uniform convergence and the stability of viscosity solutions together imply $u^{\zeta }$ is a solution of (4) with Hamiltonians $\{{\tilde{H}}^{(R)}_{1},\dots ,{\tilde{H}}^{(R)}_{K}\}$ and initial datum $u_{0}^{\zeta }$.

Since $\lim _{n \rightarrow \infty } a_{n} = \infty$, (92) shows that $u^{\zeta }$ satisfies $\text {Lip}(u^{\zeta }(\cdot ,t)) \le KL$ for all $t \in [0,T]$. Thus, as ${\tilde{H}}^{(R)}_{i}(t,x,p) = H_{i}(t,x,p)$ for all $|p| \le KL$, it follows that $u^{\zeta }$ is actually a solution of (4) with the Hamiltonians $\{H_{1},\dots ,H_{K}\}$. By Theorem 9, we deduce that the limit is unique and, in fact, $u^{\zeta } = \lim _{a \rightarrow \infty } u^{(a),\zeta }$ locally uniformly in ${\mathcal {I}} \times [0,T]$.

Finally, we send $\zeta \rightarrow 0^{+}$. Since $u^{\zeta }_{0} \rightarrow u_{0}$ uniformly in ${\mathcal {I}}$ as $\zeta \rightarrow 0^{+}$, Remark 2 implies $(u^{\zeta })_{\zeta > 0}$ is uniformly Cauchy in ${\mathcal {I}} \times [0,T]$. In particular, $u = \lim _{\zeta \rightarrow 0^{+}} u^{\zeta }$ exists uniformly in ${\mathcal {I}} \times [0,T]$ and the stability of viscosity solutions implies u solves (4) with initial datum $u_{0}$.

Since L and B were independent of $\zeta$, the uniform convergence $u^{\zeta } \rightarrow u$ implies $\text {Lip}(u) \le B + L$.

To remove the $C^{3}({\mathcal {I}})$ assumption, we argue by approximation. That is, if $u_{0} \in C^{1}({\mathcal {I}})$ and $[u_{0}]_{1} + [u_{0}]_{2} < \infty$, we obtain the solution u of (4) and show that it is in $\text {Lip}({\mathcal {I}} \times [0,T])$ by approximating $u_{0}$ with functions $(u_{0,n}) \subseteq C^{3}({\mathcal {I}})$ such that $u_{0,n} \rightarrow u_{0}$ uniformly in ${\mathcal {I}}$ and $\sup \left\{ [u_{0,n}]_{1} + [u_{0,n}]_{2} \, \mid \, n \in {\mathbb {N}} \right\} < \infty$. Since the proof that it is possible to do this is very similar to some of the arguments presented in Appendix C.5, we omit it.

Finally, if $u_{0} \in UC({\mathcal {I}})$, then, arguing as in Remark 10 below, we can fix a sequence $(u_{0}^{(n)})_{n \in {\mathbb {N}}} \subseteq C^{1}({\mathcal {I}})$ satisfying $[u_{0}^{(n)}]_{1} + [u_{0}^{(n)}]_{2} < \infty$ for each n and such that $u_{0}^{(n)} \rightarrow u_{0}$ uniformly in ${\mathcal {I}}$ as $n \rightarrow \infty$. By the previous step, we can let $u^{(n)}$ be the solution of (4) with initial datum $u_{0}^{(n)}$, and Remark 2 shows that $(u^{(n)})_{n \in {\mathbb {N}}}$ is uniformly Cauchy in ${\mathcal {I}} \times [0,T]$. Therefore, as before, the limit $u = \lim _{n \rightarrow \infty } u^{(n)}$ is a continuous viscosity solution of (4). In fact, $u \in UC({\mathcal {I}} \times [0,T])$, being the uniform limit of such functions. $\square$

1.4 C.4: Existence of solutions of (2)

Finally, we establish the existence of solutions of (2). Here, as in the error analysis, we invoke Proposition 17.

Proof of Theorem 11

First, assume $u_{0} \in \text {Lip}({\mathcal {I}})$. By Proposition 17 below, there is a family $(v_{0}^{\epsilon })_{\epsilon > 0} \subseteq C^{1} \left( {\mathcal {I}} \right)$ such that $\lim _{\epsilon \rightarrow 0^{+}} [v_{0}^{\epsilon } - u_{0}]_{0} =0$ and $\sup \{[v_{0}^{\epsilon }]_{1} + \epsilon [v_{0}^{\epsilon }]_{2} \, \mid \, \epsilon > 0\} \le C'$, where $C'$ only depends on $\text {Lip}(u_{0})$.

For each $\epsilon > 0$, let $v^{\epsilon }$ be the solution of (4) with initial datum $v_{0}^{\epsilon }$. Since $[v_{0}^{\epsilon }]_{1} + \epsilon [v_{0}^{\epsilon }]_{2}$ is bounded, Theorem 12 implies there is an $L' > 0$ depending on $C'$, but not $\epsilon$, such that $\text {Lip}(v^{\epsilon }) \le L'$.

In view of the uniform Lipschitz estimate, we can fix $(\epsilon _{n})_{n \in {\mathbb {N}}}$ and a function $u : {\mathcal {I}} \times [0,T] \rightarrow {\mathbb {R}}$ such that $\lim _{n \rightarrow \infty } \epsilon _{n} = 0$ and $u = \lim _{n \rightarrow \infty } v^{\epsilon _{n}}$. By the stability of viscosity solutions, u solves (2) with initial datum $u_{0}$. In fact, Theorem 8 implies u is independent of the choice of subsequence, and, thus, $u = \lim _{\epsilon \rightarrow 0^{+}} v^{\epsilon }$. Moreover, $\text {Lip}(u) \le L'$.

In general, if $u_{0} \in UC({\mathcal {I}})$, then there is a sequence $(u_{0}^{(n)})_{n \in {\mathbb {N}}} \subseteq \text {Lip}({\mathcal {I}})$ such that $u_{0}^{(n)} \rightarrow u_{0}$ uniformly in ${\mathcal {I}}$ as $n \rightarrow \infty$. (See Remark 10.) Let $u^{(n)}$ denote the solution of (2) with initial datum $u_{0}^{(n)}$. By Remark 2, $(u^{(n)})_{n \in {\mathbb {N}}}$ is uniformly Cauchy in ${\mathcal {I}} \times [0,T]$. Invoking stability of viscosity solutions, we conclude that the limit $u = \lim _{n \rightarrow \infty } u^{(n)}$ is a solution of (2) with initial datum $u_{0}$. Moreover, as a uniform limit of such functions, $u \in UC({\mathcal {I}} \times [0,T])$. $\square$

1.5 C.5: A useful approximation result

In the error analysis of Sect. 6, we used the following result:

Proposition 17

Let $u_{0} \in \text {Lip}\left( {\mathcal {I}} \right)$. For each $\epsilon >0$, there is a $v_{0}^{\epsilon } \in C^{1} \left( {\mathcal {I}}\right)$ and a universal constant $C > 0$ such that:

$$\begin{aligned}{}[v_{0}^{\epsilon } - u_{0}]_{0}&\le \text {Lip}(u_{0})\epsilon \\ [v_{0}^{\epsilon }]_{1}&\le C \text {Lip}(u_{0}) \\ [v_{0}^{\epsilon }]_{2}&\le C \epsilon ^{-1} \text {Lip}(u_{0}) \end{aligned}$$

Moreover, $v_{0}^{\epsilon }$ can be chosen in such a way that both $v_{0}^{\epsilon }(0) = u_{0}(0)$ and ${\sum }_{i = 1}^{K} v^{\epsilon }_{0,x_{i}}(0) = 0$.

The same method used to prove Proposition 17 below can be used to establish more general approximation results for functions on ${\mathcal {I}}$ with varying degrees of regularity. We will not expound on those here. However, since we use an approximation result for functions in $\text {UC}({\mathcal {I}})$ in the proof of Theorems 11 and 12, we include its statement as a remark:

Remark 10

Arguing as in the proof that follows, we can show that if $u_{0} \in \text {UC}({\mathcal {I}})$, then there is a sequence of functions $(u^{(n)}_{0})_{n \in {\mathbb {N}}} \subseteq C^{2}({\mathcal {I}})$ satisfying $[u^{(n)}_{0}]_{1} + [u^{(n)}_{0}]_{2} < \infty$, ${\sum }_{i = 1}^{K} u_{0,x_{i}}^{(n)}(0) = 0$, and such that

$$\begin{aligned}{}[u_{0}^{(n)} - u_{0}]_{0} \le \omega (2 n^{-1}), \end{aligned}$$

where $\omega$ is the modulus of continuity of $u_{0}$ in ${\mathcal {I}}$.

Proof of Proposition 17

First, given $\epsilon > 0$, let $\varphi _{\epsilon }$ be as in the proof of Lemma 11. Additionally, let $\rho : {\mathbb {R}} \rightarrow [0,\infty )$ be a smooth symmetric function supported in $(-1,1)$ and satisfying $\int _{-\infty }^{\infty } \rho (x) \, dx = 1$.

Define ${\tilde{\psi }}_{i}^{\epsilon } : \overline{I_{i}} \rightarrow {\mathbb {R}}$ by ${\tilde{\psi }}_{i}^{\epsilon }(x) = \epsilon ^{-1} \int _{-\infty }^{\infty } u_{0,i}(y) \rho (\epsilon ^{-1}(x - y)) \, dy$, where $u_{0,i}$ is given by $u_{0,i}(x) = u_{0}(x_{i})$ if $x < 0$ and $u_{0,i}(x) = u_{0}(0)$, otherwise. Recall the following well-known properties of ${\tilde{\psi }}_{i}^{\epsilon }$:

$$\begin{aligned} \sup \left\{ |{\tilde{\psi }}_{i}^{\epsilon }(x) - u_{0}(x)| \, \mid \, x \in I_{i} \right\}&\le \text {Lip}(u_{0})\epsilon \\ \sup \left\{ |{\tilde{\psi }}_{i,x_{i}}^{\epsilon }(x)| \, \mid \, x \in I_{i} \right\}&\le \text {Lip}(u_{0}) \\ \sup \left\{ |{\tilde{\psi }}_{i,x_{i}x_{i}}^{\epsilon }(x)| \, \mid \, x \in I_{i} \right\}&\le C \text {Lip}(u_{0}) \epsilon ^{-1} \end{aligned}$$

We proceed by combining $\{{\tilde{\psi }}^{\epsilon }_{1},\dots ,{\tilde{\psi }}^{\epsilon }_{K}\}$ into a function on ${\mathcal {I}}$.

Define $v_{0}^{\epsilon } : {\mathcal {I}} \rightarrow {\mathbb {R}}$ by

$$\begin{aligned} v_{0}^{\epsilon }(x) = (1 - \varphi _{\epsilon }(x)) {\tilde{\psi }}^{\epsilon }_{i}(x) + \varphi _{\epsilon }(x) u_{0}(0) \quad \text {if} \, \, x \in \overline{I_{i}}, \, \, i \in \{1,2,\dots ,K\}. \end{aligned}$$

Observe that $\min \{|\varphi _{\epsilon }'(x)|, |\varphi _{\epsilon }''(x)|\} > 0$ only if $x \in [-2\epsilon , \epsilon ]$. Moreover, for such x, the following inequality holds:

$$\begin{aligned} |{\tilde{\psi }}^{\epsilon }_{i}(x) - u_{0}(0)| \le |{\tilde{\psi }}^{\epsilon }_{i}(x) - u_{0}(x)| + |u_{0}(x) - u_{0}(0)| \le 3 \text {Lip}(u_{0}) \epsilon . \end{aligned}$$

Therefore, we can argue as in Lemma 11 to see that $v_{0}^{\epsilon }$ satisfies the required estimates.

Finally, $v_{0}^{\epsilon }(x) = u_{0}(0)$ if $x \in \bigcup _{i = 1}^{K} \overline{I_{i}^{\epsilon }}$ so ${\sum }_{i = 1}^{K} v^{\epsilon }_{0,x_{i}}(0) = 0$. $\square$

Appendix D: Time-dependent finite-difference schemes

In this section, we show that the finite-difference scheme approximating (2) is monotone provided a CFL-type condition is satisfied. We also establish the required regularity properties of the solution.

We begin by introducing the necessary terminology. A function $V : {\mathcal {J}} \times S \rightarrow {\mathbb {R}}$ is said to be a sub-solution of the scheme (74) if it satisfies the system of inequalities obtained by replacing all equal signs with $\le$. Analogously, a function W on the same domain is called a super-solution of the scheme (74) if it satisfies the system of inequalities obtained by replacing all equal signs with $\ge$. As in the stationary case, the scheme is monotone when sub- and super-solutions obey a discrete maximum principle. This is made precise in the following definition.

Definition 2

The finite-difference scheme (74) is called monotone if the following two criteria hold:

(i) If $V,\chi : {\mathcal {J}} \times \{0,1,\dots ,N\} \rightarrow {\mathbb {R}}$, V is a sub-solution of (74), and $V - \chi$ has a global maximum at (m, s) with $s > 0$ and $m \in J_{i}$, then

$$\begin{aligned} \frac{\chi (m,s) - \chi (m,s - 1)}{\Delta t} + F_{i}(D^{+}\chi (m,s - 1),D^{-}\chi (m,s - 1)) \le f_{i}((s-1)\Delta t, -m \Delta x) \end{aligned}$$

if $m \ne 0$, and

$$\begin{aligned} \sum _{i = 1}^{K} (\chi (0,s) - \chi (1_{i},s)) \le 0, \quad \text {otherwise}. \end{aligned}$$

(ii) If $W,\chi : {\mathcal {J}} \times \{0,1,\dots ,N\} \rightarrow {\mathbb {R}}$, W is a super-solution of (74), and $W - \chi$ has a global minimum at (m, s) with $s > 0$ and $m \in J_{i}$, then

$$\begin{aligned} \frac{\chi (m,s) - \chi (m,s - 1)}{\Delta t} + F_{i}(D^{+}\chi (m,s - 1),D^{-}\chi (m,s - 1)) \ge f_{i}((s-1)\Delta t, -m \Delta x) \end{aligned}$$

if $m \ne 0$, and

$$\begin{aligned} \sum _{i = 1}^{K} (\chi (0,s) - \chi (1_{i},s)) \ge 0, \quad \text {otherwise}. \end{aligned}$$

As in the time-independent setting, when we use the term “monotone” in reference to (74), we always mean it in the sense of the previous definition.

The error analysis of (74) uses a discrete version of Lipschitz continuity. Specifically, given a function $U : {\mathcal {J}} \times S \rightarrow {\mathbb {R}}$, we say that U is Lipschitz if

$$\begin{aligned} \text {Lip}(U) := \sup \left\{ |U(m,s) - U(k,r)| \, \mid \, \frac{d(-m\Delta x, -k \Delta x)}{\Delta x} + |s - r| \le 1 \right\} < \infty . \end{aligned}$$

The following result gives sufficient conditions under which the scheme (74) is monotone and the solution is Lipschitz. Recall that $L_{G}$ is a uniform bound on the Lipschitz constants of the numerical Hamiltonians $G_{1},\dots ,G_{K}$, and $L_{c}$ is the cut-off in assumption (48).

Proposition 18

There is an ${\tilde{L}}_{c} > 0$ depending only on $\text {Lip}(u_{0})$, D, $L_{G}$, $L_{2}$, and T such that if (78) holds and $L_{c} \ge {\tilde{L}}_{c}$, then the finite-difference scheme (74) is monotone and the solution U of (74) satisfies $\text {Lip}(U) \le {\tilde{L}}_{c} \Delta x$.

Proof

From (78), we see that $\epsilon \ge 2L_{G} \Delta x$ and $\frac{\Delta x}{\Delta t} - \frac{ \epsilon }{\Delta x} - 2L_{G} \ge 0$. From this, it follows that the expression

$$\begin{aligned} \frac{\chi (k,s) - \chi (k,s-1)}{\Delta t} + F_{i}(D^{+}\chi (k,s-1),D^{-}\chi (k,s-1)) \end{aligned}$$

is non-increasing in the variables $\chi (k,s-1)$, $\chi (k+1,s-1)$, and $\chi (k-1,s-1)$. We leave it to the reader to verify that this implies (74) is monotone according to Definition 2.

To see that U is Lipschitz, we argue as in the continuum case. To start with, define $V : {\mathcal {J}} \times (S \setminus \{N\}) \rightarrow {\mathbb {R}}$ by $V(k,s) = \frac{U(k,s + 1) - U(k,s)}{\Delta t}$. Observe that if $s \in S \setminus \{N,N - 1\}$ and $k \in {\mathcal {J}} \setminus \{0\}$, then

$$\begin{aligned} D_{t}V(k,s) + B_{i}^{+}(k,s) D^{+}V(k,s) + B_{i}^{-}(k,s) D^{-}V(k,s) - D_{t} \Gamma (k,s) = 0, \end{aligned}$$

(102)

where the coefficients of the equation are defined as follows:

$$\begin{aligned} B_{i}^{+}(k,s)&= \frac{F_{i}(D^{+}U(k,s + 1),D^{-}U(k,s+1)) - F_{i}(D^{+}U(k,s),D^{-}U(k,s+1))}{D^{+}U(k,s + 1) - D^{+}U(k,s)}, \\ B_{i}^{-}(k,s)&= \frac{F_{i}(D^{+}U(k,s),D^{-}U(k,s+1)) - F_{i}(D^{+}U(k,s),D^{-}U(k,s))}{D^{-}U(k,s+1) - D^{-}U(k,s)}, \\ \Gamma (k,s)&= f_{i}(s \Delta t, -k\Delta x). \end{aligned}$$

The discussion in the previous paragraph implies $B_{i}^{+} \le 0$ and $B_{i}^{-} \ge 0$ pointwise in ${\mathcal {J}} \times (S \setminus \{N\})$.

In addition to (102), V satisfies ${\sum }_{i = 1}^{K} D^{+}V(1_{i},s) = 0$ if $s \in S \setminus \{1,N\}$. Notice that if we define a scheme using (102) and this discrete Kirchoff condition, then the signs of $B_{i}^{+}$ and $B_{i}^{-}$ imply it is monotone in ${\mathcal {J}} \times (S \setminus \{N\})$ in the sense of Definition 2.

By (9), $|D_{t}\Gamma | \le D$ pointwise in ${\mathcal {J}} \times (S \setminus \{N\})$. Therefore, using monotonicity and arguing as in Proposition 14, we find that if $(k,s) \in {\mathcal {J}} \times (S \setminus \{N\})$, then

$$\begin{aligned} |V(k,s)| \le \sup \left\{ |V(k,0)| \, \mid \, k \in {\mathcal {J}} \right\} + DT. \end{aligned}$$

In particular, since V(k, 0) is determined by $u_{0}$, there is a constant $C_{0}$ depending only on $\text {Lip}(u_{0})$ such that $|V| \le C_{0} + DT$ pointwise. Notice that, by (78), $C_{0}$ can be chosen independent of $\Delta x$ and $\epsilon$, though it does depend on $L_{2}$.

Now we show that the finite differences $D^{+}U$ and $D^{-}U$ are uniformly bounded. Indeed, if we fix $s \in S \setminus \{N\}$, then the function $m \mapsto U(m,s)$ defined in ${\mathcal {J}}$ satisfies the stationary finite difference equation

$$\begin{aligned} V(m,s) + F_{i}(D^{+}U(m,s),D^{-}U(m,s)) = f_{i}(s\Delta t,-m\Delta x) \quad \text {in} \, \, J_{i}. \end{aligned}$$

(103)

Since V is uniformly bounded and the assumption $\epsilon \ge 2 L_{G} \Delta x$ implies the difference equation (103) is monotone, we can argue exactly as in Theorem 10 to see that there is an ${\tilde{L}}_{c} > 0$ depending only on $C_{0}$ and D, but not on s, such that if $L_{c} \ge {\tilde{L}}_{c}$, then $\text {Lip}(U(\cdot ,s)) \le {\tilde{L}}_{c}\Delta x$.

The bound we obtained through the equation only applies if $s < N$. To get a bound at $s = N$, observe that the assumption $\frac{\Delta x}{\Delta t} \ge 2 L_{G}$ implies

$$\begin{aligned} |U(k + 1,N) - U(k,N)|&\le |U(k + 1,N) - U(k+1,N-1)| \\&\quad + |U(k+1,N-1)- U(k,N-1)| \\&\quad + |U(k,N-1) - U(k,N)| \\&\le 2(C_{0} + DT)\Delta t + {\tilde{L}}_{c}\Delta x \\&\le \left( \frac{C_{0} + DT}{L_{G}} + {\tilde{L}}_{c} \right) \Delta x \end{aligned}$$

Thus, making ${\tilde{L}}_{c}$ larger if necessary, we can assume that $\text {Lip}(U(\cdot ,s)) \le {\tilde{L}}_{c}$ independently of $s \in S$. Making ${\tilde{L}}_{c}$ larger again, we can assume that ${\tilde{L}}_{c} \ge \frac{C_{0} + DT}{L_{G}}$ and, thus, $|V| \le {\tilde{L}}_{c}L_{G}$ pointwise. From this and the assumption that $\frac{\Delta t}{\Delta x} \le L_{G}^{-1}$, we conclude that $\text {Lip}(U) \le {\tilde{L}}_{c}\Delta x$ on ${\mathcal {J}} \times S$. $\square$

Appendix E: Proof of Theorem 7

In this section, we take on the hardest step in the comparison results presented above. In order to apply [20, Lemma 3.1], we need to understand, roughly speaking, the extent to which the equation “sees” the differentiability (or lack thereof) of a sub-solution or super-solution at the junction.

In what follows, given $u : (-\infty ,0] \rightarrow {\mathbb {R}}$ and $x \in (-\infty ,0]$, we define $J^{+}u(x)$ to be the set of all $p \in {\mathbb {R}}$ such that

$$\begin{aligned} u(y) \le u(x) + p(y - x) + o(|y -x|) \quad \text {as} \, \, y \rightarrow x. \end{aligned}$$

$J^{-}u(x)$ is defined by $J^{-}u(x) = -J^{+}(-u)(x)$.

Notice that this is analogous to the definitions in Appendix A. In particular, given $x \in \overline{I_{i}}$ and $u : {\mathcal {I}} \rightarrow {\mathbb {R}}$, if $u_{i} : (-\infty ,0] \rightarrow {\mathbb {R}}$ is defined by restricting u to ${\overline{I}}_{i}$, then $J^{+}_{i}u(x) = J^{+}u_{i}(x)$ and $J^{-}_{i}u(x) = J^{-}u_{i}(x)$.

Lemma 12

If $u : (-\infty ,0] \rightarrow {\mathbb {R}}$ is upper semi-continuous continuous and $u_{x}(0)$ exists, then there are sequences $(x_{n}^{+})_{n \in {\mathbb {N}}} \subseteq (-\infty ,0)$, $(p_{n}^{+})_{n \in {\mathbb {N}}} \subseteq {\mathbb {R}}$ such that

• (a) $p_{n}^{+} \in J^{+}u(x_{n}^{+})$ for each $n \in {\mathbb {N}}$

• (b) $\lim _{n \rightarrow \infty } p_{n}^{+} = u_{x}(0)$

• (c) $\lim _{n \rightarrow \infty } x_{n}^{+} = 0$

• (d) $\lim _{n \rightarrow \infty } u(x_{n}^{+}) = u(0)$

Similarly, if $v : (-\infty ,0] \rightarrow {\mathbb {R}}$ is lower semi-continuous and $v_{x}(0)$ exists, then there are sequences $(x_{n}^{-})_{n \in {\mathbb {N}}} \subseteq (-\infty ,0)$ and $(q_{n}^{-})_{n \in {\mathbb {N}}} \subseteq {\mathbb {R}}$ such that

• (a) $q_{n}^{-} \in J^{-}v(x_{n}^{-})$ for each $n \in {\mathbb {N}}$

• (b) $\lim _{n \rightarrow \infty } q_{n}^{-} = v_{x}(0)$

• (c) $\lim _{n \rightarrow \infty } x_{n}^{-} = 0$

• (d) $\lim _{n \rightarrow \infty } v(x_{n}^{-}) = v(0)$

Proof

Regarding $(x_{n}^{+},p_{n}^{+})$, this follows from the proof of Lemma 9 and the fact that, in this case, $J^{+}u(0) = (-\infty ,u_{x}(0)]$. To obtain the sequences $(x_{n}^{-},p_{n}^{-})$, use the fact that $-v$ is upper semi-continuous and $J^{+}(-v)(x) = -J^{-}v(x)$.

$\square$

When the solution is not differentiable at the junction, Lemma 12 is replaced by the following one:

Lemma 13

Suppose $u : (-\infty ,0] \rightarrow {\mathbb {R}}$ is continuous and $u_{x}(0)$ does not exist. Let $p^{+} = \limsup _{x \rightarrow 0^{-}} \frac{u(x) - u(0)}{x}$ and $p^{-} = \liminf _{x \rightarrow 0^{-}} \frac{u(x) - u(0)}{x}$. If $p \in (p^{-},p^{+})$, then there is a sequence $(x_{n}^{+})_{n \in {\mathbb {N}}} \subseteq (-\infty ,0)$ such that

• (a) $p \in J^{+}u(x_{n}^{+})$ for all $n \in {\mathbb {N}}$

• (b) $\lim _{n \rightarrow \infty } x_{n}^{+} = 0$

• (c) $\lim _{n \rightarrow \infty } u(x_{n}^{+}) = u(0)$

Similarly, suppose $v : (-\infty ,0] \rightarrow {\mathbb {R}}$ is continuous and $v_{x}(0)$ does not exist. Let $q^{+} = \limsup _{x \rightarrow 0^{-}} \frac{v(x) - v(0)}{x}$ and $q^{-} = \liminf _{x \rightarrow 0^{-}} \frac{v(x) - v(0)}{x}$. If $q \in (q^{-},q^{+})$, then there is a sequence $(x_{n}^{-})_{n \in {\mathbb {N}}} \subseteq (-\infty ,0)$ such that

• (a) $q \in J^{-}v(x_{n}^{-})$ for all $n \in {\mathbb {N}}$

• (b) $\lim _{n \rightarrow \infty } x_{n}^{-} = 0$

• (c) $\lim _{n \rightarrow \infty } v(x_{n}^{-}) = v(0)$

Proof

We only provide the arguments in the upper semi-continuous case since the lower semi-continuous case follows by a transformation as in the previous lemma.

First, observe that since $p^{+} > p^{-}$, u crosses the line $x \mapsto px$ infinitely often as $x \rightarrow 0^{-}$. Therefore, there is a sequence $(y_{n})_{n \in {\mathbb {N}}} \subseteq (-\infty ,0)$ such that

• (i) $y_{n} < y_{n + 1}$ for all $n \in {\mathbb {N}}$,

• (ii) $\lim _{n \rightarrow \infty } y_{n} = 0$,

• (iii) $\frac{u(y_{n}) - u(0)}{y_{n}} \le p$ for all $n \in {\mathbb {N}}$, and

• (iv) For all $n \in {\mathbb {N}}$, $y \mapsto u(y) - u(0) - py$ has a positive maximum in $[y_{n},y_{n + 1}]$.

For each $n \in {\mathbb {N}}$, let $x_{n}^{+}$ be a point in $[y_{n},y_{n + 1}]$ where $y \mapsto u(y) - u(0) - py$ is maximized. Notice that (iii) and (iv) imply $x_{n}^{+} \in (y_{n},y_{n + 1})$. Therefore, $p \in J^{+}u(x_{n}^{+})$. Moreover, $\lim _{n \rightarrow \infty } x_{n}^{+} = 0$, and, thus, by assumption, $\lim _{n \rightarrow \infty } u(x_{n}^{+}) = u(0)$. $\square$

Finally, we have the ingredients necessary to establish Theorem 7. For the sake of clarity, we begin by boiling Lemmas 12 and 13 down into the form we will use in the proof.

Proposition 19

Fix $i \in \{1,2,\dots ,K\}$. Suppose that $u : (-\infty ,0] \rightarrow {\mathbb {R}}$ is a continuous sub-solution (resp. super-solution) of $u + H_{i}(x,u_{x}) = 0$ in $(-\infty ,0)$. Let $p^{+} = \limsup _{x \rightarrow 0^{-}} \frac{u(x) - u(0)}{x}$ and $p^{-} = \liminf _{x \rightarrow 0^{-}} \frac{u(x) - u(0)}{x}$. If $p \in (p^{-},p^{+})$, then $u(0) + H_{i}(0,p) \le 0$ (resp. $\ge 0$).

If $|p^{+}| < \infty$ (resp. $|p^{-}| < \infty$), then the conclusion holds with $p = p^{+}$ (resp. $p = p^{-}$) as well.

Proof

We only provide the arguments when u is a sub-solution since the super-solution case follows in the same way.

Fix $p \in (p^{-},p^{+})$. Notice that Lemmas 12 and 13 together imply that there is a sequence $(x_{n},p_{n})_{n \in {\mathbb {N}}} \subseteq (-\infty ,0) \times {\mathbb {R}}$ such that $p_{n} \in J^{+}u(x_{n})$ independently of $n \in {\mathbb {N}}$ and $\lim _{n \rightarrow \infty } (x_{n},p_{n},u(x_{n})) = (0,p,u(0))$. Since $x_{n} < 0$, we can invoke the sub-solution property to find

$$\begin{aligned} u(x_{n}) + H_{i}(x_{n},p_{n}) \le 0, \end{aligned}$$

which, upon sending $n \rightarrow \infty$, becomes $u(0) + H_{i}(0,p) \le 0$.

If $|p^{+}| < \infty$, then $u(0) + H_{i}(0,p^{+}) \le 0$ follows from the continuity of $p \mapsto H_{i}(0,p)$. The same can be said if $|p^{-}| < \infty$. $\square$

The proof of Theorem 7 is now an application of Proposition 19 and Remark 1:

Proof of Theorem 7

We will only give the details for sub-solutions. In addition to $(p_{1}^{+},\dots ,p_{K}^{+})$, let us also define $(p_{1}^{-},\dots ,p_{K}^{-})$ by

$$\begin{aligned} p_{i}^{-} = \liminf _{I_{i} \ni x \rightarrow 0} \frac{u(x) - u(0)}{x}. \end{aligned}$$

Proposition 19 implies (i) directly. Additionally, it shows that if ${\tilde{p}}_{i} \ge p_{i}^{-}$ for some $i \in \{1,2,\dots ,K\}$, then (12) in (ii) holds.

It only remains to establish (ii) in the case when ${\tilde{p}}_{i} < p_{i}^{-}$ for all $i \in \{1,2,\dots ,K\}$. Remark 1 implies that, in this case, if $\varphi : {\mathcal {I}} \rightarrow {\mathbb {R}}$ is given by

$$\begin{aligned} \varphi (x) = u(0) + {\tilde{p}}_{i}x \quad \text {if} \, \, x \in \overline{I_{i}}, \end{aligned}$$

then $u - \varphi$ has a local maximum at 0. Therefore, since u is a sub-solution, we find

$$\begin{aligned} \min \left\{ \sum _{i = 1}^{K} {\tilde{p}}_{i}, u(0) + \min _{i} H_{i}(0,{\tilde{p}}_{i}) \right\} \le 0. \end{aligned}$$

Thus, (12) holds, as claimed. $\square$