ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)

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Positive admittance removes energy

Physical scope

A local sign check for the impedance boundary condition in the finite-energy model, using an orthonormal tangent basis on a flat face. It records outward power and the corresponding sign of the energy balance; it is not an outgoing-wave model or an experiment.

Example

Use the boundary geometry of the preceding finite-energy model, with $Γ = \partial Ω$ . On a flat boundary face with outward normal $n = e_{z}$ , take a right-handed tangent basis $e_{x}, e_{y}$ , with $e_{z} \times e_{x} = e_{y}$ , and prescribe tangential boundary data $E_{τ} = E_{0} e_{x}$ for $E_{0} \in R$ . For surface admittance $Y \geq 0$ , the boundary condition $H_{τ} = Y (n \times E_{τ})$ gives $H_{τ} = Y E_{0} e_{y}$ and outward power

$n \cdot (E_{τ} \times H_{τ}) = Y E_{0}^{2} \geq 0.$

If the sign in the boundary law is reversed, $H_{τ} = - Y (n \times E_{τ})$ , the outward power is $- Y E_{0}^{2}$ ; it is strictly inward when $Y E_{0}^{2} > 0$ .

For the conditional energy consequence, suppose an actual finite-energy model solution exists on a nonempty open interval $I$ and bounded Lipschitz polyhedron $Ω$ : $E, H \in C^{1} (I; L^{2} (Ω; R^{3})) \cap C (I; H (curl; Ω))$ , $J \in C (I; L^{2} (Ω; R^{3}))$ , the two evolution equations of the preceding boundary model hold as $L^{2}$ identities, and tangential boundary fields $E_{τ}, H_{τ} \in L^{2} (Γ \times I; R^{3})$ satisfy its Green compatibility identity. Take $Y \in L^{\infty} (Γ)$ , $Y \geq 0$ , as in that model. If the reversed law $H_{τ} = - Y (n \times E_{τ})$ holds almost everywhere on all of $Γ \times I$ , then the model energy $W (t) = \frac{1}{2} \int_{Ω} (E \cdot ε E + H \cdot μ H) d λ_{3}$ obeys

$W (t) - W (s) = - \int_{s}^{t} \int_{Ω} E \cdot J d λ_{3} d r + \int_{s}^{t} \int_{Γ} Y ∣ E_{τ} ∣^{2} d S d r (s < t) .$

In particular, if $J = 0$ and the last integral is positive, the stored energy increases. This conclusion is conditional on existence of a solution satisfying the reversed law.

Statement references

Finite-energy PEC and impedance boundary formulation

Energy balance and zero-data forward uniqueness for admissible weak Maxwell fields

Source convention

Chen's boundary relation is $n \times H - λ E_{t} = g$ . Here $E_{t}$ denotes the tangential electric component, identified with $E_{τ}$ , and $g$ is prescribed tangential boundary data. For comparison with the homogeneous law of this example, set $g = 0$ and use $H_{τ} = Y (n \times E_{τ})$ . Since $∣ n ∣ = 1$ and $n \cdot E_{τ} = 0$ , the right-handed cross product gives $n \times H_{τ} = Y n \times (n \times E_{τ}) = - Y E_{τ}$ . Thus our law corresponds to Chen's homogeneous relation with $λ = - Y$ . This is a conversion of conventions; the cited passage does not stipulate a sign for $λ$ or supply a passivity claim.

Diagram

The drawing is a projection of the tangent plane: $e_{z}$ points out of the page, and the right-handed convention is $e_{x} \times e_{y} = e_{z}$ . The arrows depict the representative case $E_{0} > 0$ and $Y > 0$ . If $E_{0} < 0$ while $Y > 0$ , both tangential arrows reverse; if $E_{0} = 0$ , both vectors vanish; and if $Y = 0$ , $H_{τ}$ vanishes. The verification below supplies the sign calculation.

\begin{tikzpicture}[>=stealth,scale=1.1]
  \draw[->] (-0.2,0) -- (2.8,0) node[below] {$e_x$};
  \draw[->] (0,-0.2) -- (0,2.5) node[left] {$e_y$};
  \draw[->,thick,blue] (0,0) -- (1.7,0) node[below=4pt] {$E_\tau=E_0e_x$};
  \draw[->,thick,red] (0,0) -- (0,1.7) node[left=4pt] {$H_\tau=YE_0e_y$};
  \node at (2.2,1.4) {$n=e_z\ \odot$};
\end{tikzpicture}

Facts & Assumptions

Given: The impedance boundary law, $n = e_{z}$ , $E_{τ} = E_{0} e_{x}$ , an amplitude $E_{0}$ in $V m^{- 1}$ , and an admittance $Y \geq 0$ in siemens. The outward power density is measured in $W m^{- 2}$ .

[A1]

For the conditional energy conclusion only, the finite-energy model solution class has $E, H \in C^{1} (I; L^{2} (Ω; R^{3})) \cap C (I; H (curl; Ω))$ , $J \in C (I; L^{2} (Ω; R^{3}))$ , the displayed evolution equations as $L^{2}$ identities, and a tangential $L^{2} (Γ \times I)$ pair satisfying the Green compatibility identity in the model definition. The reversed boundary law is an additional assumption on all of $Γ \times I$ ; it is not part of the model's original nonnegative-admittance branch. Finite-energy PEC and impedance boundary formulation

[F1]

In the impedance branch, $H_{τ} = Y (n \times E_{τ})$ , and the outward boundary power is the scalar $n \cdot (E_{τ} \times H_{τ})$ . Finite-energy PEC and impedance boundary formulation

[F2]

For an admissible impedance solution, the energy identity includes the boundary term $- \int_{\partial Ω} Y ∣ E_{τ} ∣^{2} d S$ in the change of field energy. Energy balance and zero-data forward uniqueness for admissible weak Maxwell fields

[F3]

The cross product is the bilinear right-handed coordinate product of The cross product in $R^{3}$ . For a unit normal $n$ and tangential vector $e$ with $e \cdot n = 0$ , substituting that coordinate formula gives $n \cdot (e \times (n \times e)) = ∣ n ∣^{2} ∣ e ∣^{2} - (e \cdot n)^{2} = ∣ e ∣^{2}$ ; in the displayed tangent basis, $e_{z} \times e_{x} = e_{y}$ and $e_{x} \times e_{y} = e_{z}$ . The dot product and norm are those of The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ .

Verification

Given: The values and conventions of the Example.

1.1givenF1F3algebra

Apply the impedance law [F1]: $H_{τ} = Y (e_{z} \times E_{0} e_{x}) = Y E_{0} e_{y}$ by $e_{z} \times e_{x} = e_{y}$ from [F3].

2.1step 1.1F3givenalgebra

Using bilinearity and $e_{x} \times e_{y} = e_{z}$ from [F3], $n \cdot (E_{τ} \times H_{τ}) = e_{z} \cdot (E_{0} e_{x} \times Y E_{0} e_{y}) = Y E_{0}^{2},$ which is nonnegative because $Y \geq 0$ and $E_{0}^{2} \geq 0$ . Replacing the boundary law by its negative changes $H_{τ}$ to $- Y E_{0} e_{y}$ , so the same calculation gives $- Y E_{0}^{2}$ , strictly negative when $Y E_{0}^{2} > 0$ .

3.1A1F2F3step 2.1algebra∎

For the conditional solution in [A1], differentiate the bounded symmetric quadratic form $W$ in the real $L^{2}$ norm and substitute the evolution equations. For almost every $t$ , this gives $W^{'} (t) = (curl H, E)_{L^{2}} - (J, E)_{L^{2}} - (H, curl E)_{L^{2}}$ . The Green compatibility identity then yields $W^{'} (t) = - (J, E)_{L^{2}} - \int_{Γ} n \cdot (E_{τ} \times H_{τ}) d S$ . Under the reversed law, [F3] gives $n \cdot (E_{τ} \times H_{τ}) = - Y ∣ E_{τ} ∣^{2}$ , so integration in time gives the stated positive boundary term. For the original law the same calculation agrees with [F2]. The boundary sign conclusion uses the Green identity and assumes an admissible solution; the example itself asserts no solution existence.