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Light Transport in Tissue


Formal solution of $\varphi _d(\mathbf {r})$ in terms of Green's functions

Define $\varphi_d(r)$ as a solution to the inhomogeneous Helmholtz equation (Equation (4.77))

\begin{displaymath}
\nabla^2\varphi_d(\mathbf{r}) -\kappa_d^2\varphi_d(\mathbf{r})=S(\mathbf{r})
\end{displaymath} (4.109)

where

\begin{displaymath}
S(\mathbf{r})=S_0\pi F_0(\rho)\exp(-\zeta)\qquad S_0=-3(1-r_s)a'(1+g'-a'g')
\end{displaymath}

and

\begin{displaymath}
\kappa_d^2=3(1-a')(1-a'g')
\end{displaymath}

subject to the following inhomogeneous mixed boundary conditions
\begin{displaymath}
\varphi_d(\mathbf{r})-A_{\mathrm{top}}h'{\partial \varphi_d(...
...-A_{\mathrm{top}}Q'(\mathbf{r})\qquad\mathrm{at}\qquad \zeta=0
\end{displaymath} (4.110)

and
\begin{displaymath}
\varphi_d(\mathbf{r})+A_{\mathrm{bottom}}h'{\partial \varphi...
...thrm{bottom}}Q'(\mathbf{r})\qquad\mathrm{at}\qquad \zeta=\tau'
\end{displaymath} (4.111)

with

\begin{displaymath}
Q'(\mathbf{r})=Q_0\pi F_0(\rho)\exp(-\zeta)
\qquad
Q_0=3h'a'g'(1-r_s)
\end{displaymath}

Let G(r;r') be a Green's function solution to the homogeneous Helmholtz equation
\begin{displaymath}
\nabla^2G(\mathbf{r};\mathbf{r}') -\kappa_d^2G(\mathbf{r};\mathbf{r}')=-\delta(\mathbf{r}-\mathbf{r}')
\end{displaymath} (4.112)

subject to the following homogeneous mixed boundary conditions
\begin{displaymath}
G(\mathbf{r};\mathbf{r}')-A_{\mathrm{top}}h'{\partial G(\mat...
...}')\over \partial \zeta} = 0
\qquad\mathrm{at}\qquad
\zeta=0
\end{displaymath} (4.113)


\begin{displaymath}
G(\mathbf{r};\mathbf{r}')+A_{\mathrm{bottom}}h'{\partial G(\...
...')\over \partial\zeta}= 0
\qquad\mathrm{at}\qquad
\zeta=\tau'
\end{displaymath} (4.114)

The solution to Equations (4.109)-(4.111) can be found using Green's second identity [51],
\begin{displaymath}
\int_{\mathrm{volume}}(u\nabla^2v-v\nabla^2u)\,dV=
\int_{\ma...
...(u{\partial v\over\partial n}-v{\partial u\over\partial n}\,dS
\end{displaymath} (4.115)

Where n is the outward normal to the surface enclosing the volume of integration on the left hand side of the equation. If u=G(r;r') and $v=\varphi_d(\mathbf{r})$, then Equation (4.115) becomes
    $\displaystyle \int_{\mathrm{volume}}
\Big(G(\mathbf{r};\mathbf{r}')\nabla^2\var...
...\mathbf{r}')-\varphi_d(\mathbf{r}')\nabla^2G(\mathbf{r};\mathbf{r}')\Big)\,dV'=$  
    $\displaystyle \qquad\qquad\int_{\mathrm{surface}}\Bigg(G(\mathbf{r};\mathbf{r}'...
...hi_d(\mathbf{r}'){\partial G(\mathbf{r};\mathbf{r}')\over\partial n}\Bigg)\,dS'$ (4.116)

Adding and subtracting $G(\mathbf{r};\mathbf{r}')\kappa_d^2\varphi_d(\mathbf{r})$ to the left hand side of Equation (4.116) yields
$\displaystyle \hbox{L.H.S.}$ = $\displaystyle \int_{\mathrm{volume}}
\Big(G(\mathbf{r};\mathbf{r}')[\nabla^2\varphi_d(\mathbf{r}') -\kappa_d^2\varphi_d(\mathbf{r}')]$  
  - $\displaystyle \varphi_d(\mathbf{r}')[\nabla^2G(\mathbf{r};\mathbf{r}')-\kappa_d^2G(\mathbf{r};\mathbf{r}')]\Big)\,dV'$ (4.117)

Using Equations (4.109) and (4.112) to simplify the bracketed quantities reduces the LHS to
\begin{displaymath}
\hbox{L.H.S.}=\int_{\mathrm{volume}}G(\mathbf{r};\mathbf{r}')S(\mathbf{r}')\,dV'-\varphi_d(\mathbf{r})
\end{displaymath} (4.118)

The surface integral on the right hand side of Equation (4.115) can be rewritten with the stipulation that on the top surface of the slab
\begin{displaymath}
{\partial\over\partial n} = -{\partial\over\partial\zeta'}
\end{displaymath} (4.119)

and on the bottom surface
\begin{displaymath}
{\partial\over\partial n} = {\partial\over\partial\zeta'}
\end{displaymath} (4.120)

because $\zeta$ increases with depth in the slab and n is an outward normal to the slab. Upon substitution of Equations (4.119) and (4.120) into the R.H.S. of Equation (4.116),
$\displaystyle \hbox{R.H.S.}=$ - $\displaystyle \int_{\zeta'=0}\Bigg(
G(\mathbf{r};\mathbf{r}'){\partial \varphi_...
...\mathbf{r}'){\partial G(\mathbf{r};\mathbf{r}')\over\partial \zeta'}\Bigg)\,dS'$ (4.121)
  + $\displaystyle \int_{\zeta'=\tau'}\Bigg(
G(\mathbf{r};\mathbf{r}'){\partial \var...
...\mathbf{r}'){\partial G(\mathbf{r};\mathbf{r}')\over\partial \zeta'}\Bigg)\,dS'$  

This equation simplifies using the boundary conditions (4.110),(4.111),(4.113) and (4.114)
\begin{displaymath}
\hbox{R.H.S.}={1\over h'}\int_{\zeta'=\tau'}G(\mathbf{r};\ma...
...'}\int_{\zeta'=0}G(\mathbf{r};\mathbf{r}')Q'(\mathbf{r}')\,dS'
\end{displaymath} (4.122)

Equating Equations (4.118) and (4.122) results in an expression for $\varphi _d(\mathbf {r})$ in terms of the Green's function G(r;r')
$\displaystyle \varphi_d(\mathbf{r})$ = $\displaystyle -\int_{\mathrm{volume}} G(\mathbf{r};\mathbf{r}')S(\mathbf{r}')\,dV'
+{1\over h'}\int_{\zeta'=\tau'}G(\mathbf{r};\mathbf{r}')Q'(\mathbf{r}')\,dS'$  
  - $\displaystyle {1\over h'}\int_{\zeta'=0}G(\mathbf{r};\mathbf{r}')Q'(\mathbf{r}')\,dS'$ (4.123)

The volume integral accounts for the inhomogeneous (source) term in the Helmholtz Equation (4.109), and the last two integrals arise from inhomogeneous boundary conditions at the top and bottom surfaces of the slab.

S. A. Prahl."Light Transport in Tissue," PhD thesis, University of Texas at Austin, 1988.