Local strict singular characteristics II: existence for stationary equations on ${\mathbb {R}}^2$

Abstract

The notion of strict singular characteristics is important in the wellposedness issue of singular dynamics on the cut locus of the viscosity solutions. We provide an intuitive and rigorous proof of the existence of the strict singular characteristics of Hamilton–Jacobi equation $H(x,Du(x),u(x))=0$ in two dimensional case. We also proved if ${\textbf{x}}$ is a strict singular characteristic, then we really have the right-differentiability of ${\textbf{x}}$ and the right-continuity of $\dot{{\textbf{x}}}^+(t)$ for every t. Such a strict singular characteristic must give a selection $p(t)\in D^+u({\textbf{x}}(t))$ such that $p(t)=\arg \min _{p\in D^+u({\textbf{x}}(t))}H({\textbf{x}}(t),p,u({\textbf{x}}(t)))$.

Appendix A. Rescaling of a lipschitz curve

In this section, we suppose $\tau _0>0$, $\gamma :[0,\tau _0]\rightarrow {\mathbb {R}}^n$ is a Lipschitz curve with Lipschitz constant $\kappa$. For $t\in [0,\tau _0]$, we define

$$\begin{aligned} \tau _1(t)&=\inf \{\tau \in [0,t]:\gamma (r)=\gamma (t),\ \forall r\in [\tau ,t]\},\\ \tau _2(t)&=\sup \{\tau \in [t,\tau _0]:\gamma (r)=\gamma (t),\ \forall r\in [t,\tau ]\}. \end{aligned}$$

Obviously, we have $0\leqslant \tau _1(t)\leqslant t\leqslant \tau _2(t)\leqslant \tau _0$. For $0\leqslant t_1\leqslant t_2\leqslant \tau _0$, we define the length of $\gamma \Big \vert _{[t_1,t_2]}$ as follows

$$\begin{aligned} l(t_1,t_2)=\sup \left\{ \sum _{i=1}^{k}|\gamma (r_i)-\gamma (r_{i-1})|:k\in {\mathbb {N}},\ t_1=r_0\leqslant r_1\leqslant \cdots \leqslant r_k=t_2\right\} . \end{aligned}$$

We can easily obtain $0\leqslant l(t_1,t_2)\leqslant \kappa (t_2-t_1)$.

Lemma A.1

(1):

For any $t_1,t_2\in [0,\tau _0]$, we have

$$\begin{aligned}{}[\tau _1(t_1),\tau _2(t_1)]\cap [\tau _1(t_2),\tau _2(t_2)]\ne \emptyset \Longleftrightarrow [\tau _1(t_1),\tau _2(t_1)]=[\tau _1(t_2),\tau _2(t_2)] \end{aligned}$$

(2):

$l(t_1,t_3)=l(t_1,t_2)+l(t_2,t_3)$, $\forall 0\leqslant t_1\leqslant t_2\leqslant t_3\leqslant \tau _0$.

(3):

For any $t_1,t_2\in [0,\tau _0]$, we have

$$\begin{aligned} l(0,t_1)=l(0,t_2)\Longleftrightarrow [\tau _1(t_1),\tau _2(t_1)]=[\tau _1(t_2),\tau _2(t_2)]. \end{aligned}$$

(4):

If $t,t_i\in [0,\tau _0]$ satisfies $l(0,t_i)>l(0,t)$, $i\in {\mathbb {N}}$ and $\lim _{i\rightarrow \infty }l(0,t_i)=l(0,t)$, then

$$\begin{aligned} t_i>\tau _2(t),\ \forall i\in {\mathbb {N}},\qquad \lim _{i\rightarrow \infty }t_i=\tau _2(t). \end{aligned}$$

Proof

The proofs of (1) and (2) are immediate. To prove (3), without loss of generality, we assume $t_1\leqslant t_2$. Then

$$\begin{aligned} l(0,t_1)=l(0,t_2)&\Longleftrightarrow l(t_1,t_2)=0 \Longleftrightarrow \gamma (t)=\gamma (t_1),\ \forall t\in [t_1,t_2]\\&\Longleftrightarrow t_2\in [\tau _1(t_1),\tau _2(t_1)] \Longleftrightarrow [\tau _1(t_1),\tau _2(t_1)]=[\tau _1(t_2),\tau _2(t_2)], \end{aligned}$$

where the first equivalence is from (2) and the last is from (1).

Finally, it follows from (3) that $l(0,\tau _2(t))=l(0,t)<l(0,t_i)$, $i\in {\mathbb {N}}$, which implies $t_i>\tau _2(t)$, $i\in {\mathbb {N}}$. If there exists ${\tilde{t}}>\tau _2(t)$ and a subsequence $\{t_{i_k}\}$ of $\{t_i\}$ such that $t_{i_k}\geqslant {\tilde{t}}$, $\forall k\in {\mathbb {N}}$, then by (3) we have $l(t,{\tilde{t}})=l(0,{\tilde{t}})-l(0,t)>0$ and

$$\begin{aligned} l(0,t_{i_k})\geqslant l(0,{\tilde{t}})=l(0,t)+l(t,{\tilde{t}}),\qquad \forall k\in {\mathbb {N}}, \end{aligned}$$

which leads to a contradiction to $\lim _{i\rightarrow \infty }l(0,t_i)=l(0,t)$. Therefore, we have $\lim _{i\rightarrow \infty }t_i=\tau _2(t)$. This completes the proof of (4). $\square$

Now, we define a curve $\gamma _l:[0,l(0,\tau _0)]\rightarrow {\mathbb {R}}^n$, $\gamma _l(s)=\gamma (t)$, where $t\in [0,\tau _0]$ satisfies $l(0,t)=s$. Due to Lemma A.1 (3), the curve $\gamma _l$ is well defined.

Lemma A.2

$\gamma _l$ is a Lipschitz curve with constant 1 and satisfies

$$\begin{aligned} |{\dot{\gamma }}_l(s)|=1,\qquad a.e.\ s\in [0,l(0,\tau _0)]. \end{aligned}$$

(A.1)

Proof

For any $0\leqslant s_1\leqslant s_2\leqslant l(0,\tau _0)$, there holds

$$\begin{aligned} |\gamma _l(s_2)-\gamma _l(s_1)|=|\gamma (t_2)-\gamma (t_1)|\leqslant l(t_1,t_2)=l(0,t_2)-l(0,t_1)=s_2-s_1, \end{aligned}$$

where $t_i\in [0,\tau _0]$ satisfies $l(0,t_i)=s_i$, $i=1,2$. Thus, $\gamma _l$ is a Lipschitz curve with constant 1 and

$$\begin{aligned} |{\dot{\gamma }}_l(s)|\leqslant 1,\qquad a.e.\ s\in [0,l(0,\tau _0)]. \end{aligned}$$

(A.2)

On the other hand, for any $0=t_0\leqslant t_1\leqslant \cdots \leqslant t_k=\tau _0$, we have

$$\begin{aligned} \sum _{i=1}^{k}|\gamma (t_i)-\gamma (t_{i-1})|&=\sum _{i=1}^{k}|\gamma _l(l(0,t_i))-\gamma _l(l(0,t_{i-1}))|=\sum _{i=1}^{k}|\int _{l(0,t_{i-1})}^{l(0,t_i)}{\dot{\gamma }}_l(s)\ ds|\\&\leqslant \sum _{i=1}^{k}\int _{l(0,t_{i-1})}^{l(0,t_i)}|{\dot{\gamma }}_l(s)|\ ds=\int _{0}^{l(0,\tau _0)}|{\dot{\gamma }}_l(s)|\ ds. \end{aligned}$$

It follows that $l(0,\tau _0)\leqslant \int _{0}^{l(0,\tau _0)}|{\dot{\gamma }}_l(s)|\ ds$, that is, $\int _{0}^{l(0,\tau _0)}1-|{\dot{\gamma }}_l(s)|\ ds\leqslant 0$. Combing the inequality with (A.2), (A.1) follows. $\square$

Lemma A.3

Suppose $\gamma :[0,\tau _0]\rightarrow {\mathbb {R}}^n$ is a Lipschitz curve, $f:{\mathbb {R}}^n\rightarrow {\mathbb {R}}^n$ is a map.

(1):

If for any $t\in [0,\tau _0]$ with $\tau _2(t)<\tau _0$, there holds $\lim _{r\rightarrow \tau _2(t)^+}f(\gamma (r))=f(\gamma (t))$, then we have

$$\begin{aligned} \lim _{r\rightarrow s+}f(\gamma _l(r))=f(\gamma _l(s)),\qquad \forall s\in [0,l(0,\tau _0)). \end{aligned}$$

(2):

Under the assumption of (1), if for any $t\in [0,\tau _0]$ with $\tau _2(t)<\tau _0$, there holds

$$\begin{aligned} \lim _{r\rightarrow \tau _2(t)^+}\frac{\gamma (r)-\gamma (t)}{|\gamma (r)-\gamma (t)|}=f(\gamma (t)), \end{aligned}$$

we have

$$\begin{aligned} {\dot{\gamma }}_l^+(s)=f(\gamma _l(s)),\qquad \forall s\in [0,l(0,\tau _0)). \end{aligned}$$

Proof

Set $s\in [0,l(0,\tau _0))$. For any $r_i\rightarrow s^+$, $i\rightarrow \infty$, choose $t,t_i\in [0,\tau _0]$, $i\in {\mathbb {N}}$ such that $l(0,t)=s$, $l(0,t_i)=r_i$, $i\in {\mathbb {N}}$. Lemma A.1 (4) implies $t_i\rightarrow \tau _2(t)^+$, $i\rightarrow \infty$. Thus,

$$\begin{aligned} \lim _{i\rightarrow \infty }f(\gamma _l(r_i))=\lim _{i\rightarrow \infty }f(\gamma (t_i))=f(\gamma (t))=f(\gamma _l(s)). \end{aligned}$$

It follows that $\lim _{r\rightarrow s+}f(\gamma _l(r))=f(\gamma _l(s))$, and (1) holds.

Now, we turn to the proof of (2). First, for almost all $s\in [0,l(0,\tau _0))$, we have

$$\begin{aligned} {\dot{\gamma }}_l(s)=\lim _{r\rightarrow s+}\frac{\gamma _l(r)-\gamma _l(s)}{r-s}=\lim _{r\rightarrow s+}\frac{\gamma (t_r)-\gamma (t)}{|\gamma (t_r)-\gamma (t)|}\cdot \frac{|\gamma _l(r)-\gamma _l(s)|}{r-s}, \end{aligned}$$

where $t_r$ and t satisfy $l(0,t_r)=r$ and $l(0,t)=s$. Lemma A.2 implies

$$\begin{aligned} \lim _{r\rightarrow s+}\frac{|\gamma _l(r)-\gamma _l(s)|}{r-s}=|{\dot{\gamma }}_l(s)|=1. \end{aligned}$$

It follows from Lemma A.1 (4) that

$$\begin{aligned} \lim _{r\rightarrow s+}\frac{\gamma (t_r)-\gamma (t)}{|\gamma (t_r)-\gamma (t)|}=\lim _{r\rightarrow \tau _2(t)^+}\frac{\gamma (r)-\gamma (t)}{|\gamma (r)-\gamma (t)|}=f(\gamma (t))=f(\gamma _l(s)). \end{aligned}$$

Therefore, ${\dot{\gamma }}_l(s)=f(\gamma _l(s))$. Now, for any $s\in [0,l(0,\tau _0))$, we have

$$\begin{aligned} {\dot{\gamma }}^+_l(s)&=\lim _{r\rightarrow s+}\frac{\gamma _l(r)-\gamma _l(s)}{r-s}=\lim _{r\rightarrow s+}\frac{1}{r-s}\int _{s}^{r}{\dot{\gamma }}_l(\tau )\ d\tau \\&=\lim _{r\rightarrow s+}\frac{1}{r-s}\int _{s}^{r}f(\gamma _l(\tau ))\ d\tau =f(\gamma _l(s)), \end{aligned}$$

where the last equality is from (1). $\square$

We say a function $g:{\mathbb {R}}\rightarrow {\mathbb {R}}$ is b-increasing with $b>0$ if

$$\begin{aligned} g(t_2) \geqslant g(t_1)+b(t_{2}-t_{1}),\qquad \forall t_1 \leqslant t_2. \end{aligned}$$

Lemma A.4

Suppose $a>0$, $0<b_1\leqslant b_2<+\infty$, $f:[0,a]\rightarrow [b_1,b_2]$ is right-continuous. Then there exists a Lipschitz function $x:[0,\frac{a}{b_2}]\rightarrow {\mathbb {R}}$ such that it is a solution of

$$\begin{aligned} {\left\{ \begin{array}{ll} {\dot{x}}^+(t)=f(x(t)),\qquad t\in [0,\frac{a}{b_2})\\ x(0)=0 \end{array}\right. } \end{aligned}$$

Proof

For $m\in {\mathbb {N}}$, we define $x_m:[0,\frac{a}{b_2}]\rightarrow {\mathbb {R}}$ as follows:

$$\begin{aligned} x_m(t)= {\left\{ \begin{array}{ll} f(0)t,\qquad t\in [0,\frac{1}{m}\frac{a}{b_2}]\\ x_m(\frac{i-1}{m}\frac{a}{b_2})+f(x_m(\frac{i-1}{m}\frac{a}{b_2}))(t-\frac{i-1}{m}\frac{a}{b_2}),\quad t\in [\frac{i-1}{m}\frac{a}{b_2},\frac{i}{m}\frac{a}{b_2}],2\leqslant i\leqslant m. \end{array}\right. } \end{aligned}$$

It is easy to verify that $x_m$ are all $b_1$-increasing, $b_2$-Lipschitz functions on $[0,\frac{a}{b_2}]$, and $x_m(0)=0$. By Alzela-Ascoli theorem, there exists a subsequence $\{x_{m_k}\}$ and an $x:[0,\frac{a}{b_2}]\rightarrow {\mathbb {R}}$ such that $x_{m_k}$ converges uniformly to x on $[0,\frac{a}{b_2}]$. x is also $b_1$-increasing, $b_2$-Lipschitz on $[0,\frac{a}{b_2}]$, and $x(0)=0$. So we have

$$\begin{aligned} 0\leqslant x(t)<a,\qquad \forall t\in [0,\frac{a}{b_2}). \end{aligned}$$

Fix any $t\in [0,\frac{a}{b_2})$. For any $\varepsilon >0$, by the right-continuity of f, there exists $\delta >0$ such that

$$\begin{aligned} |f(y)-f(x(t))|<\varepsilon ,\qquad \forall x(t)\leqslant y\leqslant x(t)+\delta . \end{aligned}$$

Since $x_{m_k}$ converges uniformly to x, there exists $n_0\in {\mathbb {N}}$ such that $\Vert x_{m_k}-x\Vert <\frac{1}{2}\delta$ for all $k\geqslant n_0$. Thus, we have

$$\begin{aligned} x_{m_k}(s)&<x(s)+\frac{1}{2}\delta \leqslant x(t)+(s-t)b_2+\frac{1}{2}\delta \leqslant x(t)+\frac{\delta }{2b_2}b_2+\frac{1}{2}\delta \\&=x(t)+\delta ,\qquad \forall k\geqslant n_0,t\leqslant s\leqslant t+\frac{\delta }{2b_2}. \end{aligned}$$

Furthermore, choosing any $t<{\tilde{t}}<s\leqslant t+\frac{\delta }{2b_2}$, by the uniform convergence of $x_{m_k}$, there exists $n_1\in {\mathbb {N}}$ such that $\Vert x_{m_k}-x\Vert <\frac{b_1}{2}({\tilde{t}}-t)$ for all $k\geqslant n_1$. Therefore, when $k\geqslant \max \{n_0,n_1\}$, $\frac{t+{\tilde{t}}}{2}\leqslant \tau \leqslant s$, we have

$$\begin{aligned} x_{m_k}(\tau )&\geqslant x_{m_k}\left( \frac{t+{\tilde{t}}}{2}\right) >x\left( \frac{t+{\tilde{t}}}{2}\right) -\frac{b_1}{2}({\tilde{t}}-t)\\&\geqslant x(t)+\frac{b_1}{2}({\tilde{t}}-t)-\frac{b_1}{2}({\tilde{t}}-t)=x(t), \end{aligned}$$

and $x_{m_k}(\tau )\leqslant x(t)+\delta$. This implies $|f(x_{m_k}(\tau ))-f(x(t))|<\varepsilon$. Combing this with the definition of $x_m$, it is not hard to obtain that

$$\begin{aligned} \frac{x(s)-x({\tilde{t}})}{s-{\tilde{t}}}=\lim _{k\rightarrow \infty }\frac{x_{m_k}(s)-x_{m_k}({\tilde{t}})}{s-{\tilde{t}}}\in (f(x(t))-\varepsilon ,f(x(t))+\varepsilon ). \end{aligned}$$

Since ${\tilde{t}}$ is arbitrary, it follows that

$$\begin{aligned} \frac{x(s)-x(t)}{s-t}=\lim _{{\tilde{t}}\rightarrow t^+}\frac{x(s)-x({\tilde{t}})}{s-{\tilde{t}}}\in (f(x(t))-\varepsilon ,f(x(t))+\varepsilon ). \end{aligned}$$

Finally, we have

$$\begin{aligned} {\dot{x}}^+(t)=\lim _{s\rightarrow t^+}\frac{x(s)-x(t)}{s-t}=f(x(t)). \end{aligned}$$

This completes the proof. $\square$