# Equivalent circuit for linear full-wave problem
Assumptions:
1. domain is linear: $\sigma(\boldsymbol{x})=\sigma$, $\varepsilon(\boldsymbol{x})=\varepsilon$
2. domain has 2 complementary subdomain $\Omega_1$ with constant and linear $\sigma_1,\varepsilon_1$ and $\Omega_2$ with $\sigma_2,\varepsilon_2$.
## 1. Maxwell equations
Ampere's law: $$\nabla \times \boldsymbol{H} = \boldsymbol{J} + \frac{\partial \boldsymbol{D}}{\partial t} \tag{1.1}$$
Faraday's law: $$\nabla \times \boldsymbol{E} = - \frac{\partial \boldsymbol{B}}{\partial t} \tag{1.2}$$
Gauss's law: $$\nabla \cdot \boldsymbol{B} = 0 \tag{1.3}$$
electric Gauss law: $$\nabla \cdot \boldsymbol{D} = \rho \tag{1.4}$$

## 2. Circuit


General KCL: $$\frac{v_{2n}}{R_{2n}^{\Omega_1}} + C_{2n}^{\Omega_1}\frac{dv_{2n}}{dt} + \frac{v_{2n}}{R_{2n}^{\Omega_2}} + C_{2n}^{\Omega_2}\frac{dv_{2n}}{dt} - \frac{v_{2n+2}}{R_{2n+2}^{\Omega_1}} - C_{2n+2}^{\Omega_1}\frac{dv_{2n+2}}{dt} - \frac{v_{2n+2}}{R_{2n+2}^{\Omega_2}} - C_{2n+2}^{\Omega_2}\frac{dv_{2n+2}}{dt} =i_{2n+1} \tag{2.1}$$
General KVL: $$v_{2n}=L_{2n-1}\frac{di_{2n-1}}{dt} - L_{2n+1}\frac{di_{2n+1}}{dt} \tag{2.2}$$

## 3. Field Decomposition
$$\boldsymbol{E}(\boldsymbol{x},t) = \sum_{n=0}^{\infty} v_{2n}(t) \boldsymbol{E}_{2n}(\boldsymbol{x}) = \sum_{n=0}^{\infty} v_{2n}(t) \boldsymbol{E}_{2n}^{\Omega_1}(\boldsymbol{x}) + \sum_{n=0}^{\infty} v_{2n}(t) \boldsymbol{E}_{2n}^{\Omega_2}(\boldsymbol{x}) \tag{3.1}$$
$$\boldsymbol{H}(\boldsymbol{x},t) = \sum_{n=0}^{\infty} i_{2n+1}(t) \boldsymbol{H}_{2n+1}(\boldsymbol{x}) = \sum_{n=0}^{\infty} i_{2n+1}(t) \boldsymbol{H}_{2n+1}^{\Omega_1}(\boldsymbol{x}) + \sum_{n=0}^{\infty} i_{2n+1}(t) \boldsymbol{H}_{2n+1}^{\Omega_2}(\boldsymbol{x}) \tag{3.2}$$
$$\boldsymbol{D}(\boldsymbol{x},t) = \sum_{n=0}^{\infty} v_{2n}(t) \boldsymbol{D}_{2n}(\boldsymbol{x}) = \sum_{n=0}^{\infty} v_{2n}(t) \boldsymbol{D}_{2n}^{\Omega_1}(\boldsymbol{x}) + \sum_{n=0}^{\infty} v_{2n}(t) \boldsymbol{D}_{2n}^{\Omega_2}(\boldsymbol{x}) \tag{3.3}$$

## 4. Energies
$$\begin{align} 2P(t)=\left( \boldsymbol{E}(t),\sigma \boldsymbol{E}(t) \right)_\Omega &=
\left( \sum_{n=0}^{\infty} v_{2n}(t) \boldsymbol{E}_{2n}(\boldsymbol{x}),\sigma \sum_{m=0}^{\infty} v_{2m}(t) \boldsymbol{E}_{2m}(\boldsymbol{x}) \right)_\Omega
\\
&=\left( \sum_{n=0}^{\infty} v_{2n}(t) \boldsymbol{E}_{2n}^{\Omega_1}(\boldsymbol{x}),\sigma_1 \sum_{m=0}^{\infty} v_{2m}(t) \boldsymbol{E}_{2m}^{\Omega_1}(\boldsymbol{x}) \right)_{\Omega_1} + \left( \sum_{n=0}^{\infty} v_{2n}(t) \boldsymbol{E}_{2n}^{\Omega_2}(\boldsymbol{x}),\sigma_2 \sum_{m=0}^{\infty} v_{2m}(t) \boldsymbol{E}_{2m}^{\Omega_2}(\boldsymbol{x}) \right)_{\Omega_2}
\\
&=\sum_{n=0}^{\infty}\frac{v_{2n}^2(t)}{R_{2n}^{\Omega_1}} + \sum_{n=0}^{\infty}\frac{v_{2n}^2(t)}{R_{2n}^{\Omega_2}} \tag{4.1} \end{align}$$

$$2W_m(t)=\left( \boldsymbol{H}(t),\mu \boldsymbol{H}(t) \right)_\Omega =
\left( \sum_{n=0}^{\infty} i_{2n+1}(t) \boldsymbol{H}_{2n+1}(\boldsymbol{x}),\mu \sum_{m=0}^{\infty} i_{2m+1}(t) \boldsymbol{H}_{2m+1}(\boldsymbol{x}) \right)_\Omega = \sum_{n=0}^{\infty}L_{2n+1}i_{2n+1}^2(t) \tag{4.2}$$

$$\begin{align}2W_e(t)=\left( \boldsymbol{D}(t),\varepsilon \boldsymbol{D}(t) \right)_\Omega &=
\left( \sum_{n=0}^{\infty} v_{2n}(t) \boldsymbol{D}_{2n}(\boldsymbol{x}),\varepsilon \sum_{m=0}^{\infty} v_{2m}(t) \boldsymbol{D}_{2m}(\boldsymbol{x}) \right)_\Omega
\\
&= \left( \sum_{n=0}^{\infty} v_{2n}(t) \boldsymbol{D}_{2n}^{\Omega_1}(\boldsymbol{x}),\varepsilon_1 \sum_{m=0}^{\infty} v_{2m}(t) \boldsymbol{D}_{2m}^{\Omega_1}(\boldsymbol{x}) \right)_{\Omega_1} + \left( \sum_{n=0}^{\infty} v_{2n}(t) \boldsymbol{D}_{2n}^{\Omega_2}(\boldsymbol{x}),\varepsilon_2 \sum_{m=0}^{\infty} v_{2m}(t) \boldsymbol{D}_{2m}^{\Omega_2}(\boldsymbol{x}) \right)_{\Omega_2}
\\
&= \sum_{n=0}^{\infty}C_{2n}^{\Omega_1}v_{2n}^2 + \sum_{n=0}^{\infty}C_{2n}^{\Omega_2}v_{2n}^2 \tag{4.3} \end{align}$$

The orthogonality applies that:
$$\left( \boldsymbol{E}_{2n}^{\Omega_1}, \sigma_1\boldsymbol{E}_{2m}^{\Omega_1} \right)_{\Omega_1} = \frac{\delta_{mn}}{R_{2n}^{\Omega_1}}, \quad \left( \boldsymbol{E}_{2n}^{\Omega_2}, \sigma_2\boldsymbol{E}_{2m}^{\Omega_2} \right)_{\Omega_2} = \frac{\delta_{mn}}{R_{2n}^{\Omega_2}} \tag{4.4}$$
$$\left( \boldsymbol{H}_{2n+1}, \mu\boldsymbol{H}_{2m+1} \right)_\Omega = L_{2n+1}\delta_{mn} \tag{4.5}$$
$$\left( \boldsymbol{D}_{2n}^{\Omega_1}, \varepsilon_1\boldsymbol{D}_{2m}^{\Omega_1} \right)_{\Omega_1} = C_{2n}^{\Omega_1}\delta_{mn}, \quad \left( \boldsymbol{D}_{2n}^{\Omega_2}, \varepsilon_2\boldsymbol{D}_{2m}^{\Omega_2} \right)_{\Omega_2} = C_{2n}^{\Omega_2}\delta_{mn} \tag{4.6}$$

From (4.4) and (4.6) and having $\boldsymbol{D}=\varepsilon \boldsymbol{E}$ in mind:
$$C_{2n}^{\Omega_1}\delta_{mn} = \left( \boldsymbol{D}_{2n}^{\Omega_1}, \varepsilon_1\boldsymbol{D}_{2m}^{\Omega_1} \right)_{\Omega_1} = \left( \boldsymbol{E}_{2n}^{\Omega_1}, \varepsilon_1\boldsymbol{E}_{2m}^{\Omega_1} \right)_{\Omega_1} = \frac{\varepsilon_1}{\sigma_1}\left( \boldsymbol{E}_{2n}^{\Omega_1}, \sigma_1\boldsymbol{E}_{2m}^{\Omega_1} \right)_{\Omega_1} = \frac{\varepsilon_1}{\sigma_1} \frac{\delta_{mn}}{R_{2n}^{\Omega_1}} \tag{4.7} $$
Since $\frac{\varepsilon_1}{\sigma_1}=cte$ then
$$C_{2n}^{\Omega_1} = \frac{\varepsilon_1}{\sigma_1} \frac{1}{R_{2n}^{\Omega_1}}, \quad C_{2n}^{\Omega_2} = \frac{\varepsilon_2}{\sigma_2} \frac{1}{R_{2n}^{\Omega_2}} \tag{4.8}$$

## 5. Modes
### 5.1 Ampere's law + general KCL
Substitution of decomposed fields (section 3) into Ampere's Law yield to
$$\nabla \times \sum_{n=0}^{\infty} i_{2n+1} \boldsymbol{H}_{2n+1} = \sigma \sum_{n=0}^{\infty} v_{2n} \boldsymbol{E}_{2n} + \frac{d}{dt}\sum_{n=0}^{\infty} v_{2n} \boldsymbol{D}_{2n} \tag{5.1}$$
which upon using the general KCL expressed in (2.2) on the left-hand side of (5.1) yields
$$\nabla \times \sum_{n=0}^{\infty} \left( \frac{v_{2n}}{R_{2n}} + C_{2n}\frac{dv_{2n}}{dt} - \frac{v_{2n+2}}{R_{2n+2}} - C_{2n+2}\frac{dv_{2n+2}}{dt} \right)\boldsymbol{H}_{2n+1} = \sigma \sum_{n=0}^{\infty} v_{2n} \boldsymbol{E}_{2n} + \frac{d}{dt}\sum_{n=0}^{\infty} v_{2n+1} \boldsymbol{D}_{2n+1} \tag{5.2}$$
Upon Introducing $\boldsymbol{H}_{-1}=0$ the left hand side of (5.2) can be rewritten as follows
$$\begin{aligned}
\nabla \times \sum_{n=0}^{\infty} \left( \frac{v_{2n}}{R_{2n}} + C_{2n}\frac{dv_{2n}}{dt} - \frac{v_{2n+2}}{R_{2n+2}} - C_{2n+2}\frac{dv_{2n+2}}{dt} \right)\boldsymbol{H}_{2n+1} &=
\nabla \times \left[ (\frac{v_0}{R_0}+C_0\frac{dv_0}{dt}-\frac{v_2}{R_2}-C_2\frac{dv_2}{dt})\boldsymbol{H}_1 + (\frac{v_2}{R_2}+C_2\frac{dv_2}{dt}-\frac{v_4}{R_4}-C_4\frac{dv_4}{dt})\boldsymbol{H}_3 +\cdots \right] \\
&=\nabla \times \left[ (\frac{v_0}{R_0}+C_0\frac{dv_0}{dt})(\boldsymbol{H}_1-\boldsymbol{H}_{-1}) + (\frac{v_2}{R_2}+C_2\frac{dv_2}{dt})(\boldsymbol{H}_3-\boldsymbol{H}_1) + \cdots \right]\\
&=\nabla \times \sum_{n=0}^{\infty} \frac{v_{2n}}{R_{2n}}(\boldsymbol{H}_{2n+1}-\boldsymbol{H}_{2n-1}) + \nabla\times \sum_{n=0}^{\infty}C_{2n}\frac{dv_{2n}}{dt}(\boldsymbol{H}_{2n+1}-\boldsymbol{H}_{2n-1})
\end{aligned} \tag{5.3}$$

Reconsidering the (5.2) with the simplified left hand side in (5.3) provides
$$\nabla \times \sum_{n=0}^{\infty} \frac{v_{2n}}{R_{2n}}(\boldsymbol{H}_{2n+1}-\boldsymbol{H}_{2n-1}) + \nabla\times \sum_{n=0}^{\infty}C_{2n}\frac{dv_{2n}}{dt}(\boldsymbol{H}_{2n+1}-\boldsymbol{H}_{2n-1})
=\sigma \sum_{n=0}^{\infty} v_{2n} \boldsymbol{E}_{2n} + \frac{d}{dt}\sum_{n=0}^{\infty} v_{2n} \boldsymbol{D}_{2n} \tag{5.4}$$
(5.4) constitutes the two of the recurrence relations
$$\nabla \times (\boldsymbol{H}_{2n+1}-\boldsymbol{H}_{2n-1})= R_{2n}\sigma \boldsymbol{E}_{2n} \tag{5.5}$$
$$\nabla \times (\boldsymbol{H}_{2n+1}-\boldsymbol{H}_{2n-1})= \frac{1}{C_{2n}}\boldsymbol{D}_{2n} \tag{5.6}$$

### 5.2 Faraday's law + general KVL
Substitution of decomposed fields (section 3) into Faraday's Law yield to
$$\nabla \times \sum_{n=0}^{\infty} v_{2n} \boldsymbol{E}_{2n} = -\mu \frac{d}{dt}\sum_{n=0}^{\infty} i_{L(2n+1)} \boldsymbol{H}_{2n+1} \tag{5.7}$$
then the general KVL comes to simplify the left hand side
$$\nabla \times \sum_{n=0}^{\infty} \left( L_{2n-1}\frac{di_{L(2n-1)}}{dt} - L_{2n+1}\frac{di_{L(2n+1)}}{dt} \right) \boldsymbol{E}_{2n} = -\mu \frac{d}{dt}\sum_{n=0}^{\infty} i_{L(2n+1)} \boldsymbol{H}_{2n+1} \tag{5.8}$$
Since $\boldsymbol{H}_{-1}=0$ then $L_{-1}=0$. The reft hand side of (5.8) is rewritten as
$$\begin{aligned}
\nabla \times \sum_{n=0}^{\infty} \left( L_{2n-1}\frac{di_{L(2n-1)}}{dt} - L_{2n+1}\frac{di_{L(2n+1)}}{dt} \right) \boldsymbol{E}_{2n} &=\nabla \times \sum_{n=0}^{\infty} \left[ (L_{-1}\frac{di_{L_{-1}}}{dt} - L_1\frac{di_{L_1}}{dt}) \boldsymbol{E}_0 + (L_1\frac{di_{L_1}}{dt} - L_3\frac{di_{L_3}}{dt}) \boldsymbol{E}_2 + \cdots \right] \\
&=\nabla \times \sum_{n=0}^{\infty} \left[ L_1\frac{di_{L_1}}{dt}(\boldsymbol{E}_2-\boldsymbol{E}_0) + L_3\frac{di_{L_3}}{dt}(\boldsymbol{E}_4-\boldsymbol{E}_2) + \cdots \right] \\
&=\nabla \times \sum_{n=0}^{\infty}L_{2n+1}\frac{di_{L(2n+1)}}{dt}(\boldsymbol{E}_{2n+2}-\boldsymbol{E}_{2n})
\end{aligned} \tag{5.9}$$
Now the (5.8) becomes
$$\nabla \times \sum_{n=0}^{\infty}L_{2n+1}\frac{di_{L(2n+1)}}{dt}(\boldsymbol{E}_{2n+2}-\boldsymbol{E}_{2n})
= -\mu \sum_{n=0}^{\infty} \frac{di_{L(2n+1)}}{dt} \boldsymbol{H}_{2n+1} \tag{5.10}$$
The (5.10) translates into
$$\nabla \times (\boldsymbol{E}_{2n+2}-\boldsymbol{E}_{2n}) = -\frac{1}{L_{2n+1}}\mu \boldsymbol{H}_{2n+1} \tag{5.11}$$