**4. Keller–Dykhne Duality**

Circuit duality can be extended for a continuous system. As with circuits, the effective response of a continuous system is associated with that of the dual one; thus, a self-dual response is automatically guaranteed. Such a constraint can induce critical behaviors of self-dual systems. Here, these topics are reviewed. Next, we introduce differential forms to clearly extract the structure of the duality in a continuous system. The correspondence between the dualities in continuous and discrete systems is formulated through discretization of continuous fields. In this section, we always use the right-hand vector product, in other words, *A*, *B*, and *A* × *B* obey the right-hand rule.

#### *4.1. Two-Dimensional Resistive Sheets*

Consider an electric field *<sup>E</sup>*(*<sup>x</sup>*, *y*) and current density *<sup>K</sup>*(*<sup>x</sup>*, *y*) on a two-dimensional resistive sheet located at *z* = 0 with a sheet conductance *<sup>G</sup>*(*<sup>x</sup>*, *y*). Here, we assume that the fields inside the thin sheet are uniform along *z* and omit *z*-dependency for the fields. Thus, we treat the fields as two-dimensional vector fields independent of *z*. An example configuration is shown in Figure 24a. The physical dimensions of *E*, *K*, and *G* are V/m, A/m, and 1/Ω, respectively. From KVL and KCL, we obtain

$$
\nabla \times \mathbf{E} = \mathbf{0},
\tag{28}
$$

$$
\nabla \cdot \mathbf{K} = 0.\tag{29}
$$

**Figure 24.** (**a**) two-dimensional resistive sheet with boundary conditions; (**b**) solution for a sheet with a constant sheet conductance *<sup>G</sup>*(*<sup>x</sup>*, *y*) = *G*ref. Black lines represent the current flow. The potential is shown as a color map with isopotential gray contours.

Note that these equations can be directly obtained from Maxwell's equations; by omitting the time-derivative terms for steady states, we can obtain Equations (28) and (29) from Faraday's law and the law of charge conservation. Ohm's law is given by

$$\mathbf{K} = \mathbf{G} \mathbf{E},\tag{30}$$

where *G* is a conductance and generally a 2 × 2 matrix. For a metallic electrode, the boundary condition is written as

$$
\mathfrak{n} \times \mathcal{E} = 0,
$$

where *n* is the unit vector normal to the boundary. For an open boundary, the boundary condition is given by

$$
\mathbf{u} \cdot \mathbf{K} = 0.\tag{32}
$$

From Equation (28), the electric field is represented by

$$E = -\nabla \varphi \tag{33}$$

 with a potential *ϕ*(*<sup>x</sup>*, *y*). Combining Equation (33) with Equations (29) and (30), we obtain

$$
\nabla \cdot G \nabla \varphi = 0. \tag{34}
$$

The boundary of the *i*-th electrode is specified by the Dirichlet boundary condition

$$
\varphi = \varphi\_{\dot{\nu}} \tag{35}
$$

which is constant along the boundary. The open boundary is given by the Neumann boundary condition

$$
\mathfrak{n} \cdot \mathbb{G} \,\nabla \varphi = 0. \tag{36}
$$

For a simplified system with a constant scalar conductance, we have the Laplace equation

$$
\nabla^2 \varphi = 0\tag{37}
$$

from Equation (34). An example solution of the Laplace equation calculated by COMSOL Multiphysics-R is shown in Figure 24b.

#### *4.2. Duality in Laplace Equation*

In this subsection, we discuss duality in Laplace equations [76]. The solution of the Laplace equation is called a *harmonic function*. A harmonic function can be considered as a part of a holomorphic function. To see this fact, consider a holomorphic function *w*(*z*) = *u*(*z*) + j*v*(*z*) with *z* = *x* + j*y*, *u*(*z*) = Re[*w*(*z*)], and *v*(*z*) = Im[*w*(*z*)]. The holomorphism leads to the Cauchy–Riemann equations:

$$\frac{\partial \mu}{\partial x} = \frac{\partial v}{\partial y'}\tag{38}$$

$$\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}.\tag{39}$$

These equations can be expressed as

$$
\begin{bmatrix}
\frac{\partial \upsilon}{\partial x} \\
\frac{\partial \upsilon}{\partial y}
\end{bmatrix} = J \begin{bmatrix}
\frac{\partial \mu}{\partial x} \\
\frac{\partial \mu}{\partial y}
\end{bmatrix} \tag{40}
$$

with *J* = -0 −1 1 0 , which induces counterclockwise 90◦ rotation. As we differentiate Equations (38) and (39) along *x* and *y*, respectively, and combine the results, we obtain

$$
\nabla^2 u = 0,\tag{41}
$$

which states that *u* is harmonic. Similarly, *v* is also harmonic. Here, *v* is called a *harmonic conjugate* of *u*.

If *u* is given, how can we obtain its harmonic conjugate *v*? Focusing on *<sup>v</sup>*(*<sup>x</sup>*, *y*) = *<sup>v</sup>*(*<sup>x</sup>*0, *y*0) + path:(*<sup>x</sup>*0,*y*0)→(*<sup>x</sup>*,*y*)(∇*v*) · d*r* with Equation (40), we have

$$\upsilon(\mathbf{x}, y) = \upsilon(\mathbf{x}\_0, y\_0) + \int\_{\text{path:}(\mathbf{x}\_0, y\_0) \to (\mathbf{x}, y)} (f \nabla u) \cdot \mathbf{d} \mathbf{r}\_\prime \tag{42}$$

where *r* = [*x y*]<sup>T</sup> and (*<sup>x</sup>*0, *y*0) is a fixed point. We have assumed that the considered region is simply connected to define Equation (42). When we consider a small displacement Δ*r* along a line *<sup>v</sup>*(*<sup>x</sup>*, *y*) = const., we have (*J*∇*u*) · Δ*r* = 0, which leads to ∇*<sup>u</sup>* <sup>Δ</sup>*<sup>r</sup>*. Therefore, *u* and *v* constitute an orthogonal coordinate around the point ∇*u* = 0, as shown in Figure 25a. In this subsection, the operation of taking the harmonic conjugate is treated as a duality transformation.

**Figure 25.** (**a**) holomorphic function *w*(*z*) = *u*(*z*) + j*v*(*z*) defines an orthogonal coordinate around a point with d*w*/d*z* = 0; (**b**) harmonic potential *ϕ* and the lines of force −∇*ϕ*; (**c**) harmonic conjugate *ψ* for *ϕ* and the lines of force −∇*ψ*.

The above result induces duality for the potential problem of the Laplace equation. Let *ψ* be the harmonic conjugate of a harmonic potential *ϕ*. The relation between *ϕ* and *ψ* is depicted in Figure 25b,c. Now, we come back to resistive sheet problems. The current stream lines and isopotential contours are replaced with each other under the harmonic conjugate as shown in the simplest example of Figure 26. Furthermore, we can see that the harmonic conjugate interchanges the Dirichlet and the Neumann boundary conditions because of the 90◦ rotation.

**Figure 26.** Current and potential distributions for (**a**) original and (**b**) its dual resistive sheets with a uniform conductance.
