Well, I'm struggling with this, and thought that posting what I'm doing could help.

I have the transport equation:

\(\displaystyle \hat \Omega \cdot \nabla I(\mathbf{r},\hat \Omega)+\mu_t I(\mathbf{r},\hat \Omega)-\mu_s\int d\hat \Omega' p(\hat \Omega \cdot \hat \Omega') I(\mathbf{r},\hat \Omega')=0\)

with prescribed boundary conditions on the inner surface of the domain \(\displaystyle V\) with boundary \(\displaystyle \partial V\): \(\displaystyle I(\mathbf{r},\hat \Omega)=f(\hat \Omega_r \cdot \hat n)I(\mathbf{r},\hat \Omega_r)\) for \(\displaystyle (\mathbf{r},\hat \Omega) \in \Gamma_{-}\), \(\displaystyle \Gamma_{\pm}=\{(\mathbf{r},\hat \Omega)| \mathbf{r}\in \partial V, \hat \Omega \cdot n \gtrless 0 \}\)

Here \(\displaystyle \hat n=\hat n(\mathbf{r})\) is the normal to the surface at the boundary, and \(\displaystyle f(\hat \Omega' \cdot \hat n)\) is the Fresnel coefficient. This are Fresnel boundary conditions. Part of the radiation is reflected at the boundary according to this boundary condition, and \(\displaystyle \hat \Omega_r=\hat R \hat \Omega\), being \(\displaystyle \hat R=\hat R(\hat n) \) a reflection operator

I want to derive the adjoint equation for this problem, i.e., if I define the transport operator

\(\displaystyle \mathcal{T}I=\hat \Omega \cdot \nabla I(\mathbf{r},\hat \Omega)+\mu_t I(\mathbf{r},\hat \Omega)-\mu_s\int d\hat \Omega' p(\hat \Omega \cdot \hat \Omega') I(\mathbf{r},\hat \Omega')\),

Then I should have for the adjoint operator that:

\(\displaystyle \langle \psi, \mathcal{T}I \rangle=\langle I, \mathcal{T}^{\dagger} \psi \rangle\)

This is what I've done:

\(\displaystyle \langle \psi, \mathcal{T}I \rangle=\int d\mathbf{r}\int d \hat \Omega \psi(\mathbf{r},\hat \Omega) \left( \hat \Omega \cdot \nabla I(\mathbf{r},\hat \Omega)+\mu_t I(\mathbf{r},\hat \Omega)-\mu_s\int d\hat \Omega' p(\hat \Omega \cdot \hat \Omega') I(\mathbf{r},\hat \Omega') \right )\)

After some manipulations it can be shown that:

\(\displaystyle \langle \psi, \mathcal{T}I \rangle=\int d\mathbf{r}\int d \hat \Omega I(\mathbf{r},\hat \Omega) \left( - \hat \Omega \cdot \nabla \psi(\mathbf{r},\hat \Omega)+\mu_t \psi(\mathbf{r},\hat \Omega)-\mu_s\int d\hat \Omega' p(\hat \Omega \cdot \hat \Omega') \psi(\mathbf{r},\hat \Omega') + \oint_{\partial V} d\hat \Omega' \hat n \cdot \hat \Omega' \psi(\mathbf{r},\hat \Omega') I(\mathbf{r},\hat \Omega') \right)\)

In order to obtain the adjoint equation, the last integral should vanish. I should impose boundary conditions in \(\displaystyle \psi\) that make this surface integral vanish. However, I haven't been able to do it. My intuition says that I should have the boundary conditions \(\displaystyle \psi(\mathbf{r},\hat \Omega)=f(\hat \Omega_r \cdot \hat n)\psi(\mathbf{r},\hat \Omega_r)\) for \(\displaystyle (\mathbf{r},\hat \Omega) \in \Gamma_{+}\), this is based on some physical interpretation of the adjoint problem, I'm not sure this is correct neither, so I'm trying to prove it.

The boundary integral can be written:

\(\displaystyle \oint_{\partial V} d\hat \Omega' \hat n \cdot \hat \Omega' \psi(\mathbf{r},\hat \Omega') I(\mathbf{r},\hat \Omega') = \int_{\Gamma_{+}} d\hat \Omega' \hat n \cdot \hat \Omega' \psi(\mathbf{r},\hat \Omega') I(\mathbf{r},\hat \Omega') +\int_{\Gamma_{-}} d\hat \Omega' \hat n \cdot \hat \Omega' \psi(\mathbf{r},\hat \Omega') I(\mathbf{r},\hat \Omega')\)

Given the boundary condition for the transport problem, one also has that: \(\displaystyle \int_{\Gamma_{-}} d\hat \Omega' \hat n \cdot \hat \Omega' \psi(\mathbf{r},\hat \Omega') I(\mathbf{r},\hat \Omega')=\int_{\Gamma_{-}} d\hat \Omega' \hat n \cdot \hat \Omega' \psi(\mathbf{r},\hat \Omega') f(\hat \Omega'_r \cdot \hat n ) I(\mathbf{r},\hat \Omega'_r)\)

I thought of defining a reflection operator that acts on the scalars \(\displaystyle \psi(\mathbf{r},\hat \Omega)\) and \(\displaystyle I(\mathbf{r},\hat \Omega)\) such that:

\(\displaystyle \hat R \psi(\mathbf{r},\hat \Omega)=\psi(\mathbf{r},\hat \Omega_r)\).

One would have for the surface integral that:

\(\displaystyle \hat R \left( \oint_{\partial V} d\hat \Omega' \hat n \cdot \hat \Omega' \psi(\mathbf{r},\hat \Omega') I(\mathbf{r},\hat \Omega') \right)=\oint_{\partial V} d\hat \Omega' \hat n \cdot \hat \Omega' \psi(\mathbf{r},\hat \Omega') I(\mathbf{r},\hat \Omega')\)

That is, the reflection operator would not change the value of the integral, because the integral is evaluated for every \(\displaystyle \hat \Omega\), and reflecting it only would change the order on how the variables are swept. However, the definition I'm giving is vague, because in here I am applying to everything inside the parenthesis, before the integral operator. Now it would be crucial to know if I can commute the reflection operator with the integral operator, and act over the scalars \(\displaystyle I,\psi\).

I've tried by applying this operator to the integral to obtain the boundary condition for the adjoint problem, and I've found a result, but I don't if what is did is right, because I haven't been rigorous with the application of the reflection operator, and on how it should act. I don't know if I am being consistent when I apply it.

I have the transport equation:

\(\displaystyle \hat \Omega \cdot \nabla I(\mathbf{r},\hat \Omega)+\mu_t I(\mathbf{r},\hat \Omega)-\mu_s\int d\hat \Omega' p(\hat \Omega \cdot \hat \Omega') I(\mathbf{r},\hat \Omega')=0\)

with prescribed boundary conditions on the inner surface of the domain \(\displaystyle V\) with boundary \(\displaystyle \partial V\): \(\displaystyle I(\mathbf{r},\hat \Omega)=f(\hat \Omega_r \cdot \hat n)I(\mathbf{r},\hat \Omega_r)\) for \(\displaystyle (\mathbf{r},\hat \Omega) \in \Gamma_{-}\), \(\displaystyle \Gamma_{\pm}=\{(\mathbf{r},\hat \Omega)| \mathbf{r}\in \partial V, \hat \Omega \cdot n \gtrless 0 \}\)

Here \(\displaystyle \hat n=\hat n(\mathbf{r})\) is the normal to the surface at the boundary, and \(\displaystyle f(\hat \Omega' \cdot \hat n)\) is the Fresnel coefficient. This are Fresnel boundary conditions. Part of the radiation is reflected at the boundary according to this boundary condition, and \(\displaystyle \hat \Omega_r=\hat R \hat \Omega\), being \(\displaystyle \hat R=\hat R(\hat n) \) a reflection operator

I want to derive the adjoint equation for this problem, i.e., if I define the transport operator

\(\displaystyle \mathcal{T}I=\hat \Omega \cdot \nabla I(\mathbf{r},\hat \Omega)+\mu_t I(\mathbf{r},\hat \Omega)-\mu_s\int d\hat \Omega' p(\hat \Omega \cdot \hat \Omega') I(\mathbf{r},\hat \Omega')\),

Then I should have for the adjoint operator that:

\(\displaystyle \langle \psi, \mathcal{T}I \rangle=\langle I, \mathcal{T}^{\dagger} \psi \rangle\)

This is what I've done:

\(\displaystyle \langle \psi, \mathcal{T}I \rangle=\int d\mathbf{r}\int d \hat \Omega \psi(\mathbf{r},\hat \Omega) \left( \hat \Omega \cdot \nabla I(\mathbf{r},\hat \Omega)+\mu_t I(\mathbf{r},\hat \Omega)-\mu_s\int d\hat \Omega' p(\hat \Omega \cdot \hat \Omega') I(\mathbf{r},\hat \Omega') \right )\)

After some manipulations it can be shown that:

\(\displaystyle \langle \psi, \mathcal{T}I \rangle=\int d\mathbf{r}\int d \hat \Omega I(\mathbf{r},\hat \Omega) \left( - \hat \Omega \cdot \nabla \psi(\mathbf{r},\hat \Omega)+\mu_t \psi(\mathbf{r},\hat \Omega)-\mu_s\int d\hat \Omega' p(\hat \Omega \cdot \hat \Omega') \psi(\mathbf{r},\hat \Omega') + \oint_{\partial V} d\hat \Omega' \hat n \cdot \hat \Omega' \psi(\mathbf{r},\hat \Omega') I(\mathbf{r},\hat \Omega') \right)\)

In order to obtain the adjoint equation, the last integral should vanish. I should impose boundary conditions in \(\displaystyle \psi\) that make this surface integral vanish. However, I haven't been able to do it. My intuition says that I should have the boundary conditions \(\displaystyle \psi(\mathbf{r},\hat \Omega)=f(\hat \Omega_r \cdot \hat n)\psi(\mathbf{r},\hat \Omega_r)\) for \(\displaystyle (\mathbf{r},\hat \Omega) \in \Gamma_{+}\), this is based on some physical interpretation of the adjoint problem, I'm not sure this is correct neither, so I'm trying to prove it.

The boundary integral can be written:

\(\displaystyle \oint_{\partial V} d\hat \Omega' \hat n \cdot \hat \Omega' \psi(\mathbf{r},\hat \Omega') I(\mathbf{r},\hat \Omega') = \int_{\Gamma_{+}} d\hat \Omega' \hat n \cdot \hat \Omega' \psi(\mathbf{r},\hat \Omega') I(\mathbf{r},\hat \Omega') +\int_{\Gamma_{-}} d\hat \Omega' \hat n \cdot \hat \Omega' \psi(\mathbf{r},\hat \Omega') I(\mathbf{r},\hat \Omega')\)

Given the boundary condition for the transport problem, one also has that: \(\displaystyle \int_{\Gamma_{-}} d\hat \Omega' \hat n \cdot \hat \Omega' \psi(\mathbf{r},\hat \Omega') I(\mathbf{r},\hat \Omega')=\int_{\Gamma_{-}} d\hat \Omega' \hat n \cdot \hat \Omega' \psi(\mathbf{r},\hat \Omega') f(\hat \Omega'_r \cdot \hat n ) I(\mathbf{r},\hat \Omega'_r)\)

I thought of defining a reflection operator that acts on the scalars \(\displaystyle \psi(\mathbf{r},\hat \Omega)\) and \(\displaystyle I(\mathbf{r},\hat \Omega)\) such that:

\(\displaystyle \hat R \psi(\mathbf{r},\hat \Omega)=\psi(\mathbf{r},\hat \Omega_r)\).

One would have for the surface integral that:

\(\displaystyle \hat R \left( \oint_{\partial V} d\hat \Omega' \hat n \cdot \hat \Omega' \psi(\mathbf{r},\hat \Omega') I(\mathbf{r},\hat \Omega') \right)=\oint_{\partial V} d\hat \Omega' \hat n \cdot \hat \Omega' \psi(\mathbf{r},\hat \Omega') I(\mathbf{r},\hat \Omega')\)

That is, the reflection operator would not change the value of the integral, because the integral is evaluated for every \(\displaystyle \hat \Omega\), and reflecting it only would change the order on how the variables are swept. However, the definition I'm giving is vague, because in here I am applying to everything inside the parenthesis, before the integral operator. Now it would be crucial to know if I can commute the reflection operator with the integral operator, and act over the scalars \(\displaystyle I,\psi\).

I've tried by applying this operator to the integral to obtain the boundary condition for the adjoint problem, and I've found a result, but I don't if what is did is right, because I haven't been rigorous with the application of the reflection operator, and on how it should act. I don't know if I am being consistent when I apply it.

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