## 2 Lossless Transmission Circuit Equations with Terminal Conditions
### Transmission Line Circuit Equation
\right)=\frac{1}{\sqrt{2}}\left( \begin{array}{c}\cos(-\frac {\pi}{4})\ -Z\sin(-\frac{\pi}{4})\\ \sin(-\frac{\pi}{4})\ Z\cos(-\frac{\pi}{4})\end{array} \right)\left( \begin{array} []{c}v\\ i\end{array} \right). \tag{8}\]  
Equation (8) denotes that the \((\psi,\phi)\)-coordinate system is identical with the \(-\pi/4\) rotation of \((v,i)\)-coordinate system. Substituting Equation (7) into Equations (3) and (4), we rewrite the boundary conditions as follows:  
\[\psi(\tau)-\phi(\tau) = E-R\frac{(\psi(\tau)+\phi(\tau))}{Z}, \tag{9}\] \[\frac{\psi(\tau-1)+\phi(\tau)}{Z} = f_{k}((\psi(\tau-1)-\phi(\tau)),\zeta,\eta,\zeta), \tag{10}\]  
where the time is shifted \(1/2\) back in (10) and \((k=N,T)\). The following Equations (9) and (10) provide the boundary conditions of Figure 6. Equations (9) and (10) also provide the boundary conditions of Figure 5 as follows. At one side terminal, we have the following: (i) a dc bias voltage source in series with 3 SPWL resistor (N) or (T): \(\xi\
eq 0\) and (ii) 3 SPWL resistor (N) or (T): \(\xi=0\). At another side terminal, we have the following: (1) short: \(E=0\), \(R=0\); (2) open: \(E=0\), \(R\to\infty\); (3) a load resistor: \(E=0\), \(R\
eq 0\); (4) a dc bias voltage source: \(E\
eq 0\), \(R=0\); or (5) a dc bias voltage source in series with a load resistor: \(E\
eq 0\), \(R\
eq 0\). Since \(f_{k}(v,\zeta,\eta,\xi)\) is a three segment piecewise linear function, we can identify Equation (11) with \(f_{k}(v,\zeta,\eta,\xi)\) for \(v\in I_{kw}\subset\mathcal{R}\), \(w=1,2,3\):  
\[ \begin{array}{l}m_{kw}v+n_{kw},\ v\in I_{kw}\subset\mathcal{R},\ m_{kw},n_{kw }\in\mathcal{R},\ w=1,2,3,\\ I_{kw}\subset\mathcal{R},\ :\text{with segment of three domain intervals},\end{array}  \tag{11}\]  
where \(m_{kw}:(\zeta,\eta,\xi)\to\mathcal{R}\) and \(n_{kw}:(\zeta,\eta,\xi)\to\mathcal{R}\). Here, from Equations (9) and (10), we briefly show how to derive an explicit piecewise linear 1-D map of \(\psi(\tau)\) and \(\psi(\tau-1)\). First of all, substitute Equation (11) into \(f_{k}(v,\zeta,\eta,\xi)\) at the right hand side of Equation (10). We obtain Equations (12) and (13). Now since Equations (9) and (12) are simultaneous linear equations, we solve Equations (9) and (12) as Equation (14). We also solve Equations (12) and (13) as Equation (15). \[\frac{\psi(\tau-1)+\phi(\tau)}{Z}=m_{kw}(\psi(\tau-1)-\phi(\tau))+n_{kw}, \tag{12}\]  
\[\psi(\tau-1)-\phi(\tau)\in I_{kw}\subset\mathcal{R}. \tag{13}\]  
\[\psi(\tau)=\frac{1-\frac{\beta}{2}}{1+\frac{\beta}{2}}\left(\frac{m_{kw}-\frac {1}{2}}{m_{kw}+\frac{1}{2}}\psi(\tau-1)-\frac{n_{kw}}{m_{kw}+\frac{1}{2}}+ \frac{E}{1-\frac{\beta}{Z}}\right), \tag{14}\]  
\[\psi(\tau-1)\in\frac{m_{kw}+\frac{1}{Z}}{\frac{\beta}{2}}\left(I_{kw}+\frac{n _{kw}}{m_{kw}+\frac{1}{Z}}\right)\subset\mathcal{R}, \tag{15}\]  
In the following, using the similar procedure from Equations (9)-(15), we provide Theorem 1 and Corollary 1 such that \(\phi(\tau)\) and \(\psi(\tau)\) are explicitly determined by 1-D maps of \(\psi(\tau-1)\) with the iterative time 1. We also provide Corollary 2 such that \(v(\tau+\frac{1}{2},\frac{1}{2})\) is explicitly determined by the1-D map of \(v(\tau-\frac{1}{2},\frac{1}{2})\). **Theorem 1**. : _If \(1>\frac{\tau}{Z}>\zeta>0\) and [PERSON]'s 3 SPWL Resistor \((k=N)\) is \(\eta>-1\) or TD 3 SPWL Resistor \((k=T)\) is \(\eta>1\), then \(\chi_{k}\), \((k=N,T)\): \(\mathcal{R}\to\mathcal{R}\) is defined by the following:_  
\[\phi(\tau)=\chi_{k}(\psi(\tau-1))=\left\{ \begin{array}{ll}\chi_{k1}(\psi( \tau-1))&,\ \psi(\tau-1)<\psi_{k1},\\ \chi_{k2}(\psi(\tau-1))&,\ \psi_{k1}\leq\psi(\tau-1)\leq\psi_{k2},\\ \chi_{k3}(\psi(\tau-1))&,\ \psi_{k2}<\psi(\tau-1)\end{array} \right.