19 Anisotropic Extension Analysis

This chapter establishes the physical correctness of the anisotropic exchange factor transformation by proving that the final interaction-reflection-scattering matrix \mathbf{K} satisfies the fundamental physical bounds: 0 \leq \mathbf{K}_{ij} \leq \mathbf{F}_{ij} for all elements. These bounds guarantee non-negativity (no negative probabilities) and energy conservation. It is then proven that the row-stochastic normalization preserves the key algebraic identity that ensures global energy balance.

19.1 Physical Bounds on the Interaction Matrix

The construction of \mathbf{K} through the anisotropic parameterization must satisfy strict element-wise bounds to be physically valid.

Theorem 19.1 For the anisotropic formulation with properly bounded scaling coefficient c, the interaction-reflection-scattering matrix \mathbf{K} satisfies:

0 \leq \mathbf{K}_{ij} \leq \mathbf{F}_{ij} \quad \forall \; i,j

The proof proceeds through two lemmas establishing each inequality.

19.1.1 Lower Bound: Non-Negativity

Lemma 19.1 For any element (i,j), the constraint \mathbf{K}_{ij} \geq 0 is satisfied if:

c \leq \frac{b_j}{\Delta_i(1-b_j) - (2\Gamma_{ij}-1)m_j}

whenever the denominator is positive, where:

\Delta_i = \sum_k F_{ik}(2\Gamma_{ik}-1)m_k
m_j = \min(b_j, 1-b_j)

Proof. From Eq. 16.9, it holds that:

\mathbf{K}_{ij} = \frac{\mathbf{K}_{\mathrm{init},ij}}{\mathbf{s}_i} - \mathbf{F}_{ij} \cdot \mathbf{P}_{jj} \left(1 - \frac{1}{\mathbf{s}_i}\right)

Substituting \mathbf{K}_{\mathrm{init},ij} = \mathbf{F}_{ij} \cdot \mathbf{B}_{\mathrm{init},ij} and \mathbf{P}_{jj} = 1-b_j:

\mathbf{K}_{ij} = \mathbf{F}_{ij} \left[\frac{\mathbf{B}_{\mathrm{init},ij}}{\mathbf{s}_i} - (1-b_j)\left(1 - \frac{1}{\mathbf{s}_i}\right)\right]

Simplifying:

\mathbf{K}_{ij} = \mathbf{F}_{ij} \cdot \frac{\mathbf{B}_{\mathrm{init},ij} - (\mathbf{s}_i - 1)(1-b_j)}{\mathbf{s}_i}

Step 1: For \mathbf{K}_{ij} \geq 0 (and \mathbf{s}_i > 0), it is necessary that:

\mathbf{B}_{\mathrm{init},ij} \geq (\mathbf{s}_i - 1)(1-b_j)

Step 2: Express \mathbf{s}_i from Eq. 16.10:

\mathbf{s}_i = \sum_k \mathbf{F}_{ik}(\mathbf{B}_{\mathrm{init},ik} + 1 - b_k)

Substituting \mathbf{B}_{\mathrm{init},ik} = b_k + c(2\Gamma_{ik}-1)m_k:

\mathbf{s}_i = \sum_k \mathbf{F}_{ik}(b_k + c(2\Gamma_{ik}-1)m_k + 1 - b_k)

\mathbf{s}_i = \sum_k \mathbf{F}_{ik}(1 + c(2\Gamma_{ik}-1)m_k)

Since \sum_k \mathbf{F}_{ik} = 1 (row-stochastic):

\mathbf{s}_i = 1 + c\sum_k \mathbf{F}_{ik}(2\Gamma_{ik}-1)m_k = 1 + c\Delta_i

Step 3: Substitute back into the constraint:

b_j + c(2\Gamma_{ij}-1)m_j \geq c\Delta_i(1-b_j)

b_j \geq c[\Delta_i(1-b_j) - (2\Gamma_{ij}-1)m_j]

Step 4: When the term in brackets is positive:

c \leq \frac{b_j}{\Delta_i(1-b_j) - (2\Gamma_{ij}-1)m_j}

When the term in brackets is non-positive, the constraint is automatically satisfied for any c \geq 0. ✓

19.1.2 Upper Bound: Physical Limit

Lemma 19.2 For any element (i,j), the constraint \mathbf{K}_{ij} \leq \mathbf{F}_{ij} is satisfied if:

c \leq \frac{1-b_j}{(2\Gamma_{ij}-1)m_j - \Delta_i(2-b_j)}

whenever the denominator is positive.

Proof. Step 1: From the expression for \mathbf{K}_{ij}, it is necessary that:

\frac{\mathbf{B}_{\mathrm{init},ij} - (\mathbf{s}_i - 1)(1-b_j)}{\mathbf{s}_i} \leq 1

Multiplying by \mathbf{s}_i > 0:

\mathbf{B}_{\mathrm{init},ij} - (\mathbf{s}_i - 1)(1-b_j) \leq \mathbf{s}_i

\mathbf{B}_{\mathrm{init},ij} \leq \mathbf{s}_i + (\mathbf{s}_i - 1)(1-b_j)

\mathbf{B}_{\mathrm{init},ij} \leq \mathbf{s}_i[1 + (1-b_j)] - (1-b_j)

\mathbf{B}_{\mathrm{init},ij} \leq \mathbf{s}_i(2-b_j) - (1-b_j)

Step 2: Substitute \mathbf{s}_i = 1 + c\Delta_i and \mathbf{B}_{\mathrm{init},ij} = b_j + c(2\Gamma_{ij}-1)m_j:

b_j + c(2\Gamma_{ij}-1)m_j \leq (1 + c\Delta_i)(2-b_j) - (1-b_j)

b_j + c(2\Gamma_{ij}-1)m_j \leq (2-b_j) + c\Delta_i(2-b_j) - (1-b_j)

b_j + c(2\Gamma_{ij}-1)m_j \leq 1 + c\Delta_i(2-b_j)

c[(2\Gamma_{ij}-1)m_j - \Delta_i(2-b_j)] \leq 1 - b_j

Step 3: When the term in brackets is positive:

c \leq \frac{1-b_j}{(2\Gamma_{ij}-1)m_j - \Delta_i(2-b_j)}

When the term in brackets is non-positive, the constraint is automatically satisfied for any c \geq 0. ✓

19.2 Energy Conservation

Having established physical bounds on \mathbf{K}, it is now proven that the construction preserves energy conservation through the key algebraic identity.

19.2.1 Preservation of the Fundamental Identity

Theorem 19.2 For the anisotropic formulation with row-stochastic normalization, the fundamental identity holds:

\mathbf{F}(\mathbf{1} - \mathbf{b}) = (\mathbf{I} - \mathbf{K})\mathbf{1}

where \mathbf{b} is the vector of reflection-scattering coefficients.

Proof. Step 1: Analyze the row sums of \mathbf{K}.

From Eq. 16.7 and the row-stochastic property of \mathbf{T} (where \sum_j \mathbf{T}_{ij} = 1):

\sum_j \mathbf{K}_{ij} = \sum_j \mathbf{T}_{ij} - \sum_j \mathbf{F}_{ij} \cdot \mathbf{P}_{jj}

\sum_j \mathbf{K}_{ij} = 1 - \sum_j \mathbf{F}_{ij} \cdot (1-b_j)

Step 2: Express (\mathbf{I} - \mathbf{K})\mathbf{1}.

[(\mathbf{I} - \mathbf{K})\mathbf{1}]_i = 1 - \sum_j \mathbf{K}_{ij}

Substituting from Step 1:

[(\mathbf{I} - \mathbf{K})\mathbf{1}]_i = 1 - \left(1 - \sum_j \mathbf{F}_{ij}(1-b_j)\right)

[(\mathbf{I} - \mathbf{K})\mathbf{1}]_i = \sum_j \mathbf{F}_{ij}(1-b_j) = [\mathbf{F}(\mathbf{1}-\mathbf{b})]_i

Therefore: \mathbf{F}(\mathbf{1} - \mathbf{b}) = (\mathbf{I} - \mathbf{K})\mathbf{1} ✓

Critical Insight

This theorem is the cornerstone of energy conservation. The row-stochastic normalization ensures that despite the full matrix structure of \mathbf{B}_{\mathrm{init}} (rather than column-constant), the fundamental relationship between \mathbf{F}, \mathbf{K}, and \mathbf{b} is preserved.

All subsequent eigenvalue properties from the isotropic case carry over directly to the anisotropic case.

19.2.2 Energy Conservation for Uniform Systems

Theorem 19.3 For a system with uniform reflection-scattering coefficients \mathbf{b} = b\mathbf{1} where b \in [0,1), the system matrix \mathbf{C} satisfies:

\mathbf{1}^T\mathbf{C}\mathbf{j} = 0

for any solution \mathbf{j} of the mixed boundary system \mathbf{M}\mathbf{j} = \mathbf{h}.

Proof. Step 1: From Theorem 19.2, left-multiply by (\mathbf{I}-\mathbf{K})^{-1}:

(\mathbf{I}-\mathbf{K})^{-1}\mathbf{F}(\mathbf{1}-\mathbf{b}) = \mathbf{1}

Left-multiply by \mathbf{P} = (1-b)\mathbf{I} for uniform b:

\mathbf{P}(\mathbf{I}-\mathbf{K})^{-1}\mathbf{F}(\mathbf{1}-\mathbf{b}) = \mathbf{P}\mathbf{1} = (1-b)\mathbf{1}

Step 2: The left side equals \mathbf{A} + \mathbf{R} from Eq. 16.13 and Eq. 16.14:

(\mathbf{A} + \mathbf{R})(\mathbf{1}-\mathbf{b}) = (1-b)\mathbf{1} = (\mathbf{1}-\mathbf{b})

Therefore (\mathbf{1}-\mathbf{b}) is an eigenvector of \mathbf{A}+\mathbf{R} with real eigenvalue 1.

Step 3: Taking the transpose:

(\mathbf{1}-\mathbf{b})^T(\mathbf{A}^T + \mathbf{R}^T) = (\mathbf{1}-\mathbf{b})^T

(\mathbf{1}-\mathbf{b})^T(\mathbf{I} - \mathbf{A}^T - \mathbf{R}^T) = \mathbf{0}^T

(\mathbf{1}-\mathbf{b})^T\mathbf{C} = \mathbf{0}^T

Step 4: For uniform b: (1-b)\mathbf{1}^T\mathbf{C} = \mathbf{0}^T, and since (1-b) > 0:

\mathbf{1}^T\mathbf{C} = \mathbf{0}^T

Therefore: \mathbf{1}^T\mathbf{C}\mathbf{j} = 0 ✓

Equivalence to Isotropic Case

This proof is identical to the isotropic case. The only requirement was establishing Theorem 19.2 for the anisotropic formulation. Once that identity is proven, energy conservation follows automatically.

This demonstrates the elegance of the row-stochastic normalization: it preserves the fundamental algebraic structure regardless of the complexity of \mathbf{\Gamma}.

19.2.3 Energy Conservation for Mixed Boundary Conditions

Theorem 19.4 Consider a mixed boundary system \mathbf{M}\mathbf{j} = \mathbf{h} where:

Rows I of \mathbf{M} are taken from \mathbf{C} with corresponding \mathbf{h}_I = \mathbf{0}
Reflection-scattering coefficients on rows I can be non-uniform: 0 \leq b_i < 1 for all i \in I
Rows J of \mathbf{M} are taken from \mathbf{D} with corresponding \mathbf{h}_J \geq \mathbf{0}
Reflection-scattering coefficients on rows J are uniform: b_j = \gamma < 1 for all j \in J

Then the energy conservation property holds:

\mathbf{1}^T\mathbf{C}\mathbf{j} = 0

Proof. From Theorem 19.3, it holds that (\mathbf{1}-\mathbf{b})^T\mathbf{C} = \mathbf{0}^T.

Partition the system by rows I (known sources) and J (known emissive powers):

\mathbf{1}^T\mathbf{C}\mathbf{j} = \mathbf{1}^T\mathbf{C}_I\mathbf{j} + \mathbf{1}^T\mathbf{C}_J\mathbf{j}

Part 1: For rows I with \mathbf{C}_I\mathbf{j} = \mathbf{0}:

\mathbf{1}^T\mathbf{C}_I\mathbf{j} = \mathbf{1}^T\mathbf{0} = 0

Part 2: For rows J with uniform b_j = \gamma:

From (\mathbf{1}-\mathbf{b})^T\mathbf{C} = \mathbf{0}^T, considering only rows J:

(\mathbf{1}-\gamma\mathbf{1})^T\mathbf{C}_J\mathbf{j} = (1-\gamma)\mathbf{1}^T\mathbf{C}_J\mathbf{j} = 0

Since (1-\gamma) > 0:

\mathbf{1}^T\mathbf{C}_J\mathbf{j} = 0

Conclusion: Combining both parts:

\mathbf{1}^T\mathbf{C}\mathbf{j} = 0 + 0 = 0

Energy conservation holds for mixed boundary conditions. ✓

19.3 Summary

The anisotropic extension preserves all essential physical properties:

Property	Requirement	Status
\mathbf{K} \geq 0	Non-negativity	✓ Theorem 19.1
\mathbf{K} \leq \mathbf{F}	Physical bound	✓ Theorem 19.1
Key identity	Algebraic structure	✓ Theorem 19.2
Energy conservation (uniform)	Global balance	✓ Theorem 19.3
Energy conservation (mixed)	General case	✓ Theorem 19.4

The method provides:

Maximum flexibility: Element-wise control with optimal bounds on c
Physical validity: Guaranteed through proper coefficient selection
Energy conservation: Automatic for uniform and mixed boundary conditions
Computational efficiency: Analytical solution via matrix inversion

The anisotropic formulation represents a true generalization that maintains mathematical rigor while accommodating arbitrary directional distributions within the physical constraint 0 \leq \mathbf{\Gamma}_{ij} \leq 1.

# Anisotropic Extension Analysis {#sec-exchange_fact_anisotropic_extend_analysis} This chapter establishes the physical correctness of the anisotropic exchange factor transformation by proving that the final interaction-reflection-scattering matrix $\mathbf{K}$ satisfies the fundamental physical bounds: $0 \leq \mathbf{K}_{ij} \leq \mathbf{F}_{ij}$ for all elements. These bounds guarantee non-negativity (no negative probabilities) and energy conservation. It is then proven that the row-stochastic normalization preserves the key algebraic identity that ensures global energy balance. ## Physical Bounds on the Interaction Matrix The construction of $\mathbf{K}$ through the anisotropic parameterization must satisfy strict element-wise bounds to be physically valid. ::: {.theorem data-title="Physical Bounds on K" #thm-physical-bounds} For the anisotropic formulation with properly bounded scaling coefficient $c$, the interaction-reflection-scattering matrix $\mathbf{K}$ satisfies: $$ 0 \leq \mathbf{K}_{ij} \leq \mathbf{F}_{ij} \quad \forall \; i,j $$ ::: The proof proceeds through two lemmas establishing each inequality. ### Lower Bound: Non-Negativity ::: {.lemma data-title="Non-Negativity of K" #lem-k-nonnegative} For any element $(i,j)$, the constraint $\mathbf{K}_{ij} \geq 0$ is satisfied if: $$ c \leq \frac{b_j}{\Delta_i(1-b_j) - (2\Gamma_{ij}-1)m_j} $$ whenever the denominator is positive, where: - $\Delta_i = \sum_k F_{ik}(2\Gamma_{ik}-1)m_k$ - $m_j = \min(b_j, 1-b_j)$ ::: ::: {.proof} From @eq-k-simplified, it holds that: $$ \mathbf{K}_{ij} = \frac{\mathbf{K}_{\mathrm{init},ij}}{\mathbf{s}_i} - \mathbf{F}_{ij} \cdot \mathbf{P}_{jj} \left(1 - \frac{1}{\mathbf{s}_i}\right) $$ Substituting $\mathbf{K}_{\mathrm{init},ij} = \mathbf{F}_{ij} \cdot \mathbf{B}_{\mathrm{init},ij}$ and $\mathbf{P}_{jj} = 1-b_j$: $$ \mathbf{K}_{ij} = \mathbf{F}_{ij} \left[\frac{\mathbf{B}_{\mathrm{init},ij}}{\mathbf{s}_i} - (1-b_j)\left(1 - \frac{1}{\mathbf{s}_i}\right)\right] $$ Simplifying: $$ \mathbf{K}_{ij} = \mathbf{F}_{ij} \cdot \frac{\mathbf{B}_{\mathrm{init},ij} - (\mathbf{s}_i - 1)(1-b_j)}{\mathbf{s}_i} $$ **Step 1:** For $\mathbf{K}_{ij} \geq 0$ (and $\mathbf{s}_i > 0$), it is necessary that: $$ \mathbf{B}_{\mathrm{init},ij} \geq (\mathbf{s}_i - 1)(1-b_j) $$ **Step 2:** Express $\mathbf{s}_i$ from @eq-s-expanded: $$ \mathbf{s}_i = \sum_k \mathbf{F}_{ik}(\mathbf{B}_{\mathrm{init},ik} + 1 - b_k) $$ Substituting $\mathbf{B}_{\mathrm{init},ik} = b_k + c(2\Gamma_{ik}-1)m_k$: $$ \mathbf{s}_i = \sum_k \mathbf{F}_{ik}(b_k + c(2\Gamma_{ik}-1)m_k + 1 - b_k) $$ $$ \mathbf{s}_i = \sum_k \mathbf{F}_{ik}(1 + c(2\Gamma_{ik}-1)m_k) $$ Since $\sum_k \mathbf{F}_{ik} = 1$ (row-stochastic): $$ \mathbf{s}_i = 1 + c\sum_k \mathbf{F}_{ik}(2\Gamma_{ik}-1)m_k = 1 + c\Delta_i $$ **Step 3:** Substitute back into the constraint: $$ b_j + c(2\Gamma_{ij}-1)m_j \geq c\Delta_i(1-b_j) $$ $$ b_j \geq c[\Delta_i(1-b_j) - (2\Gamma_{ij}-1)m_j] $$ **Step 4:** When the term in brackets is positive: $$ c \leq \frac{b_j}{\Delta_i(1-b_j) - (2\Gamma_{ij}-1)m_j} $$ When the term in brackets is non-positive, the constraint is automatically satisfied for any $c \geq 0$. ✓ ::: ### Upper Bound: Physical Limit ::: {.lemma data-title="Upper Bound on K" #lem-k-upper} For any element $(i,j)$, the constraint $\mathbf{K}_{ij} \leq \mathbf{F}_{ij}$ is satisfied if: $$ c \leq \frac{1-b_j}{(2\Gamma_{ij}-1)m_j - \Delta_i(2-b_j)} $$ whenever the denominator is positive. ::: ::: {.proof} **Step 1:** From the expression for $\mathbf{K}_{ij}$, it is necessary that: $$ \frac{\mathbf{B}_{\mathrm{init},ij} - (\mathbf{s}_i - 1)(1-b_j)}{\mathbf{s}_i} \leq 1 $$ Multiplying by $\mathbf{s}_i > 0$: $$ \mathbf{B}_{\mathrm{init},ij} - (\mathbf{s}_i - 1)(1-b_j) \leq \mathbf{s}_i $$ $$ \mathbf{B}_{\mathrm{init},ij} \leq \mathbf{s}_i + (\mathbf{s}_i - 1)(1-b_j) $$ $$ \mathbf{B}_{\mathrm{init},ij} \leq \mathbf{s}_i[1 + (1-b_j)] - (1-b_j) $$ $$ \mathbf{B}_{\mathrm{init},ij} \leq \mathbf{s}_i(2-b_j) - (1-b_j) $$ **Step 2:** Substitute $\mathbf{s}_i = 1 + c\Delta_i$ and $\mathbf{B}_{\mathrm{init},ij} = b_j + c(2\Gamma_{ij}-1)m_j$: $$ b_j + c(2\Gamma_{ij}-1)m_j \leq (1 + c\Delta_i)(2-b_j) - (1-b_j) $$ $$ b_j + c(2\Gamma_{ij}-1)m_j \leq (2-b_j) + c\Delta_i(2-b_j) - (1-b_j) $$ $$ b_j + c(2\Gamma_{ij}-1)m_j \leq 1 + c\Delta_i(2-b_j) $$ $$ c[(2\Gamma_{ij}-1)m_j - \Delta_i(2-b_j)] \leq 1 - b_j $$ **Step 3:** When the term in brackets is positive: $$ c \leq \frac{1-b_j}{(2\Gamma_{ij}-1)m_j - \Delta_i(2-b_j)} $$ When the term in brackets is non-positive, the constraint is automatically satisfied for any $c \geq 0$. ✓ ::: ## Energy Conservation Having established physical bounds on $\mathbf{K}$, it is now proven that the construction preserves energy conservation through the key algebraic identity. ### Preservation of the Fundamental Identity ::: {.theorem data-title="Key Identity Preservation" #thm-key-identity} For the anisotropic formulation with row-stochastic normalization, the fundamental identity holds: $$ \mathbf{F}(\mathbf{1} - \mathbf{b}) = (\mathbf{I} - \mathbf{K})\mathbf{1} $$ where $\mathbf{b}$ is the vector of reflection-scattering coefficients. ::: ::: {.proof} **Step 1:** Analyze the row sums of $\mathbf{K}$. From @eq-k-final and the row-stochastic property of $\mathbf{T}$ (where $\sum_j \mathbf{T}_{ij} = 1$): $$ \sum_j \mathbf{K}_{ij} = \sum_j \mathbf{T}_{ij} - \sum_j \mathbf{F}_{ij} \cdot \mathbf{P}_{jj} $$ $$ \sum_j \mathbf{K}_{ij} = 1 - \sum_j \mathbf{F}_{ij} \cdot (1-b_j) $$ **Step 2:** Express $(\mathbf{I} - \mathbf{K})\mathbf{1}$. $$ [(\mathbf{I} - \mathbf{K})\mathbf{1}]_i = 1 - \sum_j \mathbf{K}_{ij} $$ Substituting from Step 1: $$ [(\mathbf{I} - \mathbf{K})\mathbf{1}]_i = 1 - \left(1 - \sum_j \mathbf{F}_{ij}(1-b_j)\right) $$ $$ [(\mathbf{I} - \mathbf{K})\mathbf{1}]_i = \sum_j \mathbf{F}_{ij}(1-b_j) = [\mathbf{F}(\mathbf{1}-\mathbf{b})]_i $$ Therefore: $\mathbf{F}(\mathbf{1} - \mathbf{b}) = (\mathbf{I} - \mathbf{K})\mathbf{1}$ ✓ ::: ::: {.callout-important} ## Critical Insight This theorem is the cornerstone of energy conservation. The row-stochastic normalization ensures that despite the full matrix structure of $\mathbf{B}_{\mathrm{init}}$ (rather than column-constant), the fundamental relationship between $\mathbf{F}$, $\mathbf{K}$, and $\mathbf{b}$ is preserved. All subsequent eigenvalue properties from the isotropic case carry over directly to the anisotropic case. ::: ### Energy Conservation for Uniform Systems ::: {.theorem data-title="Energy Conservation (Uniform Case)" #thm-energy-conservation} For a system with uniform reflection-scattering coefficients $\mathbf{b} = b\mathbf{1}$ where $b \in [0,1)$, the system matrix $\mathbf{C}$ satisfies: $$ \mathbf{1}^T\mathbf{C}\mathbf{j} = 0 $$ for any solution $\mathbf{j}$ of the mixed boundary system $\mathbf{M}\mathbf{j} = \mathbf{h}$. ::: ::: {.proof} **Step 1:** From @thm-key-identity, left-multiply by $(\mathbf{I}-\mathbf{K})^{-1}$: $$ (\mathbf{I}-\mathbf{K})^{-1}\mathbf{F}(\mathbf{1}-\mathbf{b}) = \mathbf{1} $$ Left-multiply by $\mathbf{P} = (1-b)\mathbf{I}$ for uniform $b$: $$ \mathbf{P}(\mathbf{I}-\mathbf{K})^{-1}\mathbf{F}(\mathbf{1}-\mathbf{b}) = \mathbf{P}\mathbf{1} = (1-b)\mathbf{1} $$ **Step 2:** The left side equals $\mathbf{A} + \mathbf{R}$ from @eq-a-matrix and @eq-r-matrix: $$ (\mathbf{A} + \mathbf{R})(\mathbf{1}-\mathbf{b}) = (1-b)\mathbf{1} = (\mathbf{1}-\mathbf{b}) $$ Therefore $(\mathbf{1}-\mathbf{b})$ is an eigenvector of $\mathbf{A}+\mathbf{R}$ with real eigenvalue 1. **Step 3:** Taking the transpose: $$ (\mathbf{1}-\mathbf{b})^T(\mathbf{A}^T + \mathbf{R}^T) = (\mathbf{1}-\mathbf{b})^T $$ $$ (\mathbf{1}-\mathbf{b})^T(\mathbf{I} - \mathbf{A}^T - \mathbf{R}^T) = \mathbf{0}^T $$ $$ (\mathbf{1}-\mathbf{b})^T\mathbf{C} = \mathbf{0}^T $$ **Step 4:** For uniform $b$: $(1-b)\mathbf{1}^T\mathbf{C} = \mathbf{0}^T$, and since $(1-b) > 0$: $$ \mathbf{1}^T\mathbf{C} = \mathbf{0}^T $$ Therefore: $\mathbf{1}^T\mathbf{C}\mathbf{j} = 0$ ✓ ::: ::: {.callout-tip} ## Equivalence to Isotropic Case This proof is identical to the isotropic case. The only requirement was establishing @thm-key-identity for the anisotropic formulation. Once that identity is proven, energy conservation follows automatically. This demonstrates the elegance of the row-stochastic normalization: it preserves the fundamental algebraic structure regardless of the complexity of $\mathbf{\Gamma}$. ::: ### Energy Conservation for Mixed Boundary Conditions ::: {.theorem data-title="Energy Conservation (Mixed Boundary Conditions)" #thm-energy-conservation-mixed} Consider a mixed boundary system $\mathbf{M}\mathbf{j} = \mathbf{h}$ where: - Rows $I$ of $\mathbf{M}$ are taken from $\mathbf{C}$ with corresponding $\mathbf{h}_I = \mathbf{0}$ - Reflection-scattering coefficients on rows $I$ can be non-uniform: $0 \leq b_i < 1$ for all $i \in I$ - Rows $J$ of $\mathbf{M}$ are taken from $\mathbf{D}$ with corresponding $\mathbf{h}_J \geq \mathbf{0}$ - Reflection-scattering coefficients on rows $J$ are uniform: $b_j = \gamma < 1$ for all $j \in J$ Then the energy conservation property holds: $$ \mathbf{1}^T\mathbf{C}\mathbf{j} = 0 $$ ::: ::: {.proof} From @thm-energy-conservation, it holds that $(\mathbf{1}-\mathbf{b})^T\mathbf{C} = \mathbf{0}^T$. Partition the system by rows $I$ (known sources) and $J$ (known emissive powers): $$ \mathbf{1}^T\mathbf{C}\mathbf{j} = \mathbf{1}^T\mathbf{C}_I\mathbf{j} + \mathbf{1}^T\mathbf{C}_J\mathbf{j} $$ **Part 1:** For rows $I$ with $\mathbf{C}_I\mathbf{j} = \mathbf{0}$: $$ \mathbf{1}^T\mathbf{C}_I\mathbf{j} = \mathbf{1}^T\mathbf{0} = 0 $$ **Part 2:** For rows $J$ with uniform $b_j = \gamma$: From $(\mathbf{1}-\mathbf{b})^T\mathbf{C} = \mathbf{0}^T$, considering only rows $J$: $$ (\mathbf{1}-\gamma\mathbf{1})^T\mathbf{C}_J\mathbf{j} = (1-\gamma)\mathbf{1}^T\mathbf{C}_J\mathbf{j} = 0 $$ Since $(1-\gamma) > 0$: $$ \mathbf{1}^T\mathbf{C}_J\mathbf{j} = 0 $$ **Conclusion:** Combining both parts: $$ \mathbf{1}^T\mathbf{C}\mathbf{j} = 0 + 0 = 0 $$ Energy conservation holds for mixed boundary conditions. ✓ ::: ## Summary The anisotropic extension preserves all essential physical properties: | Property | Requirement | Status | |----------|-------------|--------| | $\mathbf{K} \geq 0$ | Non-negativity | ✓ @thm-physical-bounds | | $\mathbf{K} \leq \mathbf{F}$ | Physical bound | ✓ @thm-physical-bounds | | Key identity | Algebraic structure | ✓ @thm-key-identity | | Energy conservation (uniform) | Global balance | ✓ @thm-energy-conservation | | Energy conservation (mixed) | General case | ✓ @thm-energy-conservation-mixed | The method provides: - **Maximum flexibility**: Element-wise control with optimal bounds on $c$ - **Physical validity**: Guaranteed through proper coefficient selection - **Energy conservation**: Automatic for uniform and mixed boundary conditions - **Computational efficiency**: Analytical solution via matrix inversion The anisotropic formulation represents a true generalization that maintains mathematical rigor while accommodating arbitrary directional distributions within the physical constraint $0 \leq \mathbf{\Gamma}_{ij} \leq 1$.