The eikonal equation on an interval has many solutions when endpoint data are omitted

Time-independent profiles have zero test time-derivative. Let $U(x,t)=u(x)$ for a continuous $u:(0,1)\to \mathbb{R}$ and let $\varphi \in C^1(U)$ with $U-\varphi$ having a local extremum at $(x_0,t_0)$. Then, testing along the time line $x=x_0$ and along the space line $t=t_0$, the jet characterisation [F2] gives $p_t=0$ for every $p\in D^{+}U(x_0,t_0)$ or $p\in D^{-}U(x_0,t_0)$: the inequality $u(x_0)-\varphi (x_0,t)\le u(x_0)-\varphi (x_0,t_0)$ near $t_0$ forces ${\mathrm{\partial }}_t\varphi (x_0,t_0)=0$ in the upper case (and dually in the lower case), and the remaining spatial inequality is the one-sided differentiability statement for $u$ at $x_0$. Consequently the viscosity conditions for $U$ reduce to: every upper-test spatial slope $p_x$ satisfies $F(p_x)\le 0$, and every lower-test spatial slope $p_x$ satisfies $F(p_x)\ge 0$.

The three profiles are solutions. For $x_0\mathrm{\ne }\frac{1}{2}$ each $u_i$ is $C^1$ near $x_0$ with $\mathrm{\mid }{u}_i^{\mathrm{\prime }}\mathrm{\mid }=1$, so at any upper or lower test the spatial slope equals $u_i^{\mathrm{\prime }}(x_0)=\pm 1$, and [F1] gives $F(\pm 1)=0$ in both directions. At $x_0=\frac{1}{2}$, only $u_3$ is non-differentiable: writing $h=x-\frac{1}{2}$ we have $u_3(\frac{1}{2}+h)=\frac{1}{2}-\mathrm{\mid }h\mathrm{\mid }$. If $\varphi$ is an upper test with slope $p={\varphi }_x(\frac{1}{2},t_0)$, then $u_3\le \varphi$ near the point gives for $h>0$ the inequality $-h+o(h)\le ph+o(h)$, hence $p\ge -1$, and for $h<0$ the inequality $h+o(\mathrm{\mid }h\mathrm{\mid })\le ph+o(\mathrm{\mid }h\mathrm{\mid })$, hence $p\le 1$; thus every upper-test slope lies in $[-1,1]$ and $F(p)\le 0$. If $\varphi$ were a lower test, $u_3\ge \varphi$ would give for $h>0$ that $p\le -1$ and for $h<0$ that $p\ge 1$, which is impossible; hence there is no lower test at the peak and the supersolution condition is vacuous. By step 1.1 the same reduction applies to $u_1$ and $u_2$ at every point. Hence all three profiles are viscosity solutions of $\mathrm{\mid }{u}^{\mathrm{\prime }}\mathrm{\mid }=1$ on $(0,1)$.

Distinctness and boundary traces. At $x=\frac{1}{4}$ the values are $(u_1,u_2,u_3)=(\frac{1}{4},\frac{3}{4},\frac{1}{4})$ and at $x=\frac{3}{4}$ they are $(\frac{3}{4},\frac{1}{4},\frac{1}{4})$, which distinguishes every pair; also $u_1-u_2=2x-1$ changes sign on $(0,1)$. The continuous extensions to $[0,1]$ have boundary values $u_1(0)=0,u_1(1)=1$; $u_2(0)=1,u_2(1)=0$; and $u_3(0)=u_3(1)=0$. Thus the equation with no prescribed boundary data admits at least three distinct viscosity solutions, and the displayed candidates carry different endpoint traces.