Triangulation sensing

The mathematical formulation consists in considering diffusing cues that have to bind to N narrow windows located on the surface of a three dimensional shaped object, typically a ball (in dimension 3) or a disk in dimension 2. M windows ranging between 10 and 50 are located on a ball B(a) of radius a. Individual cue molecules are described as Brownian particles. The cues are released from a source at position ${\displaystyle x_{0}}$ outside the ball. The triangulation sensing method consists in reconstructing the source position from the steady-state fluxes estimated at each narrow window for fast binding, i.e. the probability density function has an absorbing boundary condition at the windows.[4]

The inner solution is constructed near each small window[5] by scaling the arclength s and the distance to the boundary ${\displaystyle \eta }$ by ${\displaystyle {\bar {\eta }}={\frac {\eta }{\epsilon }}}$ and ${\displaystyle {\bar {s}}={\frac {s}{\epsilon }}}$, so that the inner problem reduces to the classical two-dimensional Laplace equation[6]

${\displaystyle \Delta w=0{\hbox{ in }}\mathbb {R} _{+}^{2},{\frac {\partial w}{\partial n}}=0{\hbox{ for }}|{\bar {s}}|>{\frac {1}{2}},{\bar {\eta }}=0,w({\bar {s}},{\bar {\eta }})=0{\hbox{ for }}|{\bar {s}}|<{\frac {1}{2}},{\bar {\eta }}=0.}$

The far field behavior for ${\displaystyle |x|\rightarrow \infty }$ and for each hole i=1, 2 is ${\displaystyle w_{i}(x)\approx A_{i}\{\log |x-x_{i}|-\log \epsilon \}+o(1),}$ where A_i is the flux ${\displaystyle A_{i}={\frac {2}{\pi }}\int _{0}^{1/2}{\frac {\partial w(0,{\bar {s}})}{\partial {\bar {\eta }}}}d{\bar {s}}.}$

The general solution of equation for n=2 is obtained from the outer solution of the external Neumann-Green's function

${\displaystyle -\Delta _{x}G(x,y)=\delta (x-y){\hbox{ for }}x\in \,\mathbb {R} _{+}^{2}}$

${\displaystyle {\frac {\partial G}{\partial n_{x}}}(x,x_{0})=0{\hbox{ for }}x\in {\partial \mathbb {R} _{+}^{2}}.}$

Given for ${\displaystyle x,x_{0}\in \mathbb {R} _{+}^{2}}$ by ${\displaystyle G(x,x_{0})={\frac {-1}{2\pi }}\left(\ln |x-x_{0}|+\ln \left|x-{\bar {x_{0}}}\right|\right),}$ where ${\displaystyle {\bar {x_{0}}}}$is the symmetric image of ${\displaystyle x_{0}}$through the boundary axis 0z.

The uniform solution is the sum of inner and outer solution (Neuman-Green's function) ${\displaystyle P(x,x_{0})=G(x,x_{0})+A_{1}\{\log |x-x_{1}|-\log \epsilon \}+A_{2}\{\log |x-x_{2}|-\log \epsilon \}+C,}$ where ${\displaystyle A_{1},A_{2},C}$ are constants to be determined.

To that purpose, the behavior of the solution near each point ${\displaystyle x_{i}}$ is crutial. In the boundary layer, we get

${\displaystyle P(x,y)\approx A_{i}\{\log |x-x_{i}|-\log \epsilon \}.}$

Using this condition on each window, we obtain the two conditions:

${\displaystyle G(x_{1},x_{0})+A_{2}\{\log |x_{1}-x_{2}|-\log \epsilon \}+C=0}$

${\displaystyle G(x_{2},x_{0})+A_{1}\{\log |x_{2}-x_{1}|-\log \epsilon \}+C=0.}$

Due to the recursion property of the Brownian motion in dimension 2, there are no fluxes at infinity, thus the conservation of flux gives: ${\displaystyle \int _{\partial \Omega _{1}}{\frac {\partial P(x,y)}{\partial n}}dS_{x}+\int _{\partial \Omega _{2}}{\frac {\partial P(x,y)}{\partial n}}dS_{x}=-1.}$ In the limit of two well separated windows (${\displaystyle |x_{1}-x_{2}|\gg 1)}$, using the condition for the flux, we get for each window i=1,2, ${\displaystyle \int _{\partial \Omega _{i}}{\frac {\partial P(x,y))}{\partial n}}dS_{x}=-\pi A_{i}}$ (the minus sign is due to the outer normal orientation), thus ${\displaystyle \pi A_{1}+\pi A_{2}=1.}$ We finally obtain the system

${\displaystyle {\begin{cases}{\frac {G(x_{1},x_{0})-G(x_{2},x_{0})}{\{\log |x_{1}-x_{2}|-\log \epsilon \}}}+(A_{2}-A_{1})=0\\\\A_{1}+A_{2}={\frac {1}{\pi }}.\end{cases}}}$

To conclude, the absorbing probabilities are given by

${\displaystyle {\begin{cases}P_{1}=\pi A_{1}={\frac {1}{2}}+{\frac {\pi }{2}}{\frac {G(x_{2},x_{0})-G(x_{1},x_{0})}{\{\log |x_{1}-x_{2}|-\log \epsilon \}}}={\frac {1}{2}}-{\frac {1}{4}}{\frac {\ln {\frac {|x_{2}-x_{0}|\left|x_{2}-{\bar {x_{0}}}\right|}{|x_{1}-x_{0}|\left|x_{1}-{\bar {x_{0}}}\right|}}}{\{\log |x_{2}-x_{1}|-\log \epsilon \}}}.\\\\P_{2}=\pi A_{2}={\frac {1}{2}}+{\frac {\pi }{2}}{\frac {G(x_{1},x_{0})-G(x_{2},x_{0})}{\{\log |x_{1}-x_{2}|-\log \epsilon \}}}={\frac {1}{2}}-{\frac {1}{4}}{\frac {\ln {\frac {|x_{1}-x_{0}|\left|x_{1}-{\bar {x_{0}}}\right|}{|x_{2}-x_{0}|\left|x_{2}-{\bar {x_{0}}}\right|}}}{\{\log |x_{1}-x_{2}|-\log \epsilon \}}}.\end{cases}}}$

These probabilities precisely depend on the source position ${\displaystyle x_{0}}$ and the relative position of the two windows. When one of the splitting probabilities (either ${\displaystyle P_{1}}$ or ${\displaystyle P_{2}}$) is known and fixed in [0,1], recovering the position of the source requires inverting the previous equations. For ${\displaystyle P_{2}=\alpha \in [0,1],}$ the position ${\displaystyle x_{0}}$ lies on the curve

${\displaystyle S_{source}=\{x_{0}{\hbox{ such that}}\,{\frac {|x_{1}-x_{0}|\left|x_{1}-{\bar {x_{0}}}\right|}{|x_{2}-x_{0}|\left|x_{2}-{\bar {x_{0}}}\right|}}=\exp \left((4\alpha -2)\{\log |x_{1}-x_{2}|-\log \epsilon \}\right)\}.}$

To conclude knowing the splitting probability between two windows is not enough to recover the exact the source ${\displaystyle x_{0}}$ , because it lies on one dimensional curve solution. However the direction can be obtained by simply checking which one of the two probability is the highest.

To reconstruct the location of a source from the measured fluxes, at least three windows are needed. With two windows only, a source located on the line perpendicular to one of the connecting windows would, for example generate the same splitting probability ${\displaystyle P_{1}=P_{2}}$, leading to a one dimensional curve degeneracy for the reconstructed source positions ${\displaystyle x_{0}}$.

Reconstructing the source location ${\displaystyle x_{0}}$ from the splitting probabilities,[7] requires to invert a system equation. Starting from the general solution as given above, we get for three windows:

${\displaystyle P(x,x_{0})=G(x,x_{0})+A_{1}\{\log |x-x_{1}|-\log \epsilon \}+A_{2}\{\log |x-x_{2}|-\log \epsilon \}+A_{3}\{\log |x-x_{3}|-\log \epsilon \}+C,}$

where ${\displaystyle A_{1},A_{2},A_{3},C}$ are constants to be determined. The three absorbing boundary conditions for${\displaystyle P(x,x_{0})}$leads to the system of equations

${\displaystyle G(x_{1},x_{0})+A_{2}\{\log |x_{1}-x_{2}|-\log \epsilon \}+A_{3}\{\log |x_{1}-x_{3}|-\log \epsilon \}+C=0}$

${\displaystyle G(x_{2},x_{0})+A_{1}\{\log |x_{2}-x_{1}|-\log \epsilon \}+A_{3}\{\log |x_{2}-x_{3}|-\log \epsilon \}+C=0}$

${\displaystyle G(x_{3},x_{0})+A_{1}\{\log |x_{1}-x_{3}|-\log \epsilon \}+A_{2}\{\log |x_{2}-x_{3}|-\log \epsilon \}+C=0,}$

with the normalization condition for the fluxes ${\displaystyle \pi A_{1}+\pi A_{2}+\pi A_{3}=1.}$ With ${\displaystyle \Delta _{123}=\left(\log {\frac {d_{13}d_{12}}{d_{32}\epsilon }}\right)^{2}-4\log {\frac {d_{12}}{\epsilon }}\log {\frac {d_{13}}{\epsilon }}}$, and using the notations ${\displaystyle d_{ij}=|x_{i}-x_{j}|}$, we obtain the solution

${\displaystyle A_{1}={\frac {1}{\pi }}-A_{2}-A_{3}.}$

${\displaystyle A_{2}={\frac {\log {\frac {d_{13}d_{12}}{d_{32}\epsilon }}(G_{30}-G_{10}+{\frac {1}{\pi }}\log {\frac {d_{13}}{\epsilon }})-(G_{10}-G_{20}+{\frac {1}{\pi }}\log {\frac {d_{12}}{\epsilon }})\log {\frac {d_{13}^{2}}{\epsilon ^{2}}})}{\Delta _{123}}}.}$

${\displaystyle A_{3}={\frac {\log {\frac {d_{13}d_{12}}{d_{32}\epsilon }}(G_{20}-G_{10}+{\frac {1}{\pi }}\log {\frac {d_{12}}{\epsilon }})-(G_{10}-G_{30}+{\frac {1}{\pi }}\log {\frac {d_{13}}{\epsilon }})\log {\frac {d_{12}^{2}}{\epsilon ^{2}}})}{\Delta _{123}}}}$

To conclude, we can specify the flux values ${\displaystyle \alpha >0}$ and ${\displaystyle \beta >0}$ such that ${\displaystyle \alpha +\beta <1}$, with ${\displaystyle \alpha =\pi A_{1}=\int _{\partial \Omega _{1}}{\frac {\partial P(x,y))}{\partial n}}dS_{x}}$and ${\displaystyle \beta =\pi A_{2}=\int _{\partial \Omega _{2}}{\frac {\partial P(x,y))}{\partial n}}dS_{x}.}$

To conclude, when two fluxes are know, the flux coefficients ${\displaystyle A_{1},A_{2},A_{3}}$can be computed and the position ${\displaystyle x_{0}}$ can be reconstructed from the three equations.[8][9]

With three windows in a x-y plane in the confirguration ${\displaystyle x_{1}=(0,0,0),x_{2}=(d,0,0)\,{\hbox{and}}\,x_{3}=(e,f,0)}$ (window 1 is at the origin and window 2 is on the x-axis). Using the leading order from the expansion of the fluxes, the location of the source is the solution of the three non-linear equations

${\displaystyle \gamma _{1}^{2}=(x_{0}^{(1)})^{2}+(x_{0}^{(2)})^{2}+(x_{0}^{(3)})^{2}}$

${\displaystyle \gamma _{2}^{2}=(d-x_{0}^{(1)})^{2}+(x_{0}^{(2)})^{2}+(x_{0}^{(3)})^{2}}$

${\displaystyle \gamma _{3}^{2}=(e-x_{0}^{(1)})^{2}+(f-x_{0}^{(2)})^{2}+(x_{0}^{(3)})^{2},}$

where ${\displaystyle \gamma _{i}={\frac {2\epsilon }{\pi \Phi _{i}}}}$. Solving for the coordinates of ${\displaystyle x_{0}}$ and requiring that ${\displaystyle x_{0}}$.x >0, leads to the analytical solution

${\displaystyle x_{0}^{(1)}={\frac {d^{2}+\gamma _{1}-\gamma _{2}}{2d}}}$

${\displaystyle x_{0}^{(2)}={\frac {1}{2df}}\left[d(e^{2}+f^{2}+\gamma _{1}-\gamma _{3})-e(d^{2}+\gamma _{1}-\gamma _{2})\right]}$

${\displaystyle x_{0}^{(3)}={\frac {1}{2df}}\left[(e^{2}+f^{2})(\{\gamma _{1}-\gamma _{2}\}^{2}-d^{4})+2de(e^{2}+f^{2}+\gamma _{1}-\gamma _{3})(d^{2}+\gamma _{1}-\gamma _{2})\right.\left.-d^{2}(e^{4}+f^{4}+[\gamma _{1}-\gamma _{3}]^{2}+2e^{2}[f^{2}+2\gamma _{1}-\gamma _{2}-\gamma _{3}]-2f^{2}[\gamma _{2}+\gamma _{3}])\right]^{1/2}.}$Thus the position can be recovered from the steady-state fluxes.[8][9][10]

In general, with N windows, no explicit inverse is easily available, hence numerical procedures are used to find the position of the source ${\displaystyle x_{0}}$ at any order in ${\displaystyle \epsilon }$. To find the position of the source ${\displaystyle x_{0}}$ from the measured fluxes ${\displaystyle \Phi _{i},i=1...N,}$we need to invert a system of equations. Each of these equations describe a non-planar surface ${\displaystyle S_{i}}$ in three dimensions, corresponding to window i and intersecting the half-plane (in the case of the windows located on the half-plane) or the unit ball (in the case of the windows located on the ball). Each pair of surfaces ${\displaystyle S_{i}}$and ${\displaystyle S_{j}}$ intersect, forming three-dimensional curves ${\displaystyle C_{ij}}$and all of these curves intersect at the location of the source ${\displaystyle x_{0}}$ . In case of N>3 windows, the system is overdetermined and we can simply choose any combination k, l and m of three fluxes from the N available windows. Any combination lead to the same source position.