Team:TU Delft-Leiden/Modeling/Curli

Curli Module

introduction curli model, explain three different modules

summary of the conclusions

Gene Level Modeling

We will start with the modeling of the expression of curli on the gene level. Proteins that are dedicated to the curli formation are CsgA/B/D/E/F/G [1]. CsgA is the main building block of the curli. When produced, this protein is secreted out of the cell by the CsgEFG complex. In the absence of CsgB, there is no curli formation, since the CsgA proteins remain unpolymerized. CsgB is the starting block of the curli fibrils and connect the cell membrane to the first CsgA protein in the curli fibril. Once CsgB is located on the outside of the cell surface, the CsgA can polymerize onto the starting curli fibril.
In the constructs we made in the wet lab, CsgA is continuously being produced. However, in our constructs the CsgB gene is placed under the control of a landmine promoter, activated by either TNT or DNT reference to landmine. So, when the cells get induced by TNT or DNT, CsgB protein production will get started and CsgA will already be present in the system, as CsgA is continuously being produced.

Extensive Gene Level Modeling

to be written, low priority

Simplified Gene Level Modeling

Though the model described above, providing that all rates are known, has a more accurate (though still simplified) representation of the curli assembly system, we have chosen to decrease the complexity further to the bare essentials, as most of the production rates cannot be found in literature. Measuring the accurate rates in the wet lab is, within the scope of this project, infeasible and therefore, we constructed a model that only includes the rate limiting step of the system as this will mostly determine the dynamics of the system.
First of all, we investigated if the diffusion of the CsgA and CsgB proteins to their final destination is the rate limiting step in curli formation. From the literature and the wet lab, we know that the system response to the induction by TNT or DNT is in the order of hours font color="red">[reference]. If diffusion is the rate limiting step, it would mean that CsgA and CsgB proteins would pile up inside and outside the cell, because it takes a long time for them to travel to their final destination, the end of a growing curli fibril and the outer membrane, respectively. A quick calculation shows that after one second, the displacement of a spherical particle with radius \(r = \ 10 \ nm\) is 6.6 μm due to Brownian motion in liquid water at room temperature using equation 1; many times the bacterial radius! Hence, we conclude that diffusion is not rate limiting [4].

$$ \bar{x}^2 = \ \frac{k_b T t}{3 \pi \eta r} \tag{1}$$

What we do expect to be the rate limiting step for curli formation is the large amount of CsgA and CsgB proteins that have to be produced. Hence, we expect the production rate of one of these proteins to be the rate limiting step. Instead of including the intermediate steps, we have implemented the production of the CsgA and CsgB proteins with one reaction and associated production rate each. These rates have to be measured in the lab. We will use the following system of equations:

$$ \emptyset \xrightarrow{p_{A}} \ CsgA_{free} \tag{2} $$ $$ \emptyset \xrightarrow{p_{B}} \ CsgB \tag{3} $$ $$ CsgA_{free} + \ CsgB \xrightarrow{k} \ CsgA_{curli} + \ CsgB \tag{4} $$

Reactions 2 and 3 represent the production of CsgA and CsgB proteins, respectively. Equation 4 represents the growing of a curli fibril, where a curli fibril reacts with a free CsgA protein to become part of the curli. In reality, this reaction only happens at the end of the curli fibrils. In our model, we assume a homogeneous concentration of all the substances and we cannot discriminate between curli subunits. It is theoretically possible to model the system as an infinite amount of possible reactions that can take place to increase a curli fibril with length i to length i+1 at rate k [7]. However, we are merely interested in the growth rates of the curli, since the distribution of the curli length will follow from the model at the cell level. Therefore, we decided to model the growing of curli at the gene level as reaction 4. We assume that each CsgB protein is the start of a curli fibrils, thus the concentration of CsgB equals the concentration of curli. We can do this, because we showed that the diffusion of CsgA and CsgB proteins to their final destination is not the rate limiting step. Therefore, nearly all the CsgB proteins will be the beginning of a curli fibril in reality and our assumption is valid.
So, in reaction 4 we let a free CsgA protein react with a curli fibril to a CsgA protein that is part of that curli and the curli itself again, as it is immediatily again availible for the next reaction with a free CsgA protein to grow even more. Therefore, curli growth is dependent on the rate k and the concentration of \(CsgA_{free}\) and CsgB.

Writing reactions 2-4 into differential equations results in:

$$ \frac{d}{dt} [CsgA_{free}] = \ p_{A} - \ k [CsgA_{free}][CsgB] \tag{5.1} $$ $$ \frac{d}{dt} [CsgB] = \ p_{B} \tag{5.2} $$ $$ \frac{d}{dt} [CsgA_{curli}] = \ k [CsgA_{free}][CsgB] \tag{5.3} $$

Fortunately, this system can be solved analytically. To do this, we need the initial conditions. Say the CsgB promoter is activated at \(t= \ 0\). At this time there are no curli present, so \([CsgB]|_{t=0} = \ [CsgA_{curli}]|_{t=0}= \ 0\). However, the CsgA promoter is continuously active, so we expect to have an initial concentration \(A_0\) of free CsgA proteins at time \(t= \ 0\).

The solution to equation 5.2 is trivial:

$$ [CsgB] = \ p_B t \tag{6}$$

Substituting this into equation 5.1 results in:

$$ \frac{d}{dt} [CsgA_{free}] = \ p_{A} - \ K p_B [CsgA]t \tag{7} $$

It can easily be proven that a first order differential equation of the form

$$ y(t)' + \ f(t)y(t) = \ g(t) $$

has a solution of the form

$$ y(t) = \ e^{-F(t)} \int{g(t) e^{F(t)} dt} + \ y_0 e^{-F(t)} $$

where \(F(t)= \int{f(t) dt}\). In our case, \(f(t) = \ k p_B t\) and \(g(t) = \ p_A\). This yields equation 8.

$$ [CsgA_{free}] = \ p_A e^{\frac{-k \ p_B t^2}{2}} \int{e^{\frac{k \ p_B t^2}{2}} dt} + \ C_{1} e^{\frac{-k \ p_B t^2}{2}} = \ p_A e^{\frac{-k \ p_B t^2}{2}} \int_{0}^{t}{e^{\frac{k \ p_B \tau^2}{2}} d\tau} + \ C_{2} e^{\frac{-k \ p_B t^2}{2}} \tag{8} $$

One with a keen eye may recognize the Dawson function (equation 9):

$$ D_+ (x) = \ e^{-x^2 } \int_{0}^x{e^{y^2} dy} \tag{9} $$

As in our case, \(x^2 = \ k p_B t^2 \) and \(y^2 = k p_B \tau^2 \) and equation 10 obtained.

$$ [CsgA_{free}] = \ \frac{p_A D_+ (t\sqrt{\frac{k \ p_B}{2}})}{\sqrt{\frac{k \ p_B}{2}}} + \ C_{2} e^{\frac{-k \ p_B t^2}{2}} \tag{10}$$

Using the boundary condition \([CsgA_{free}]|_{t=0}= \ A_0\), the expression for the concentration of free CsgA proteins becomes:

$$ [CsgA_{free}] = \ \frac{p_A D_+ (t\sqrt{\frac{k \ p_B}{2}})}{\sqrt{\frac{k \ p_B}{2}}} + \ A_0 e^{\frac{-k \ p_B t^2}{2}} \tag{11}$$

Now, we can fill in equations 11 and 6 into equation 5.3, which gives us equation 12.

$$ \frac{d}{dt} [CsgA_{curli}] = \ k p_B t \frac{p_A D_+ (t\sqrt{\frac{k \ p_B}{2}})}{\sqrt{\frac{k \ p_B}{2}}} + \ A_0 e^{\frac{-k \ p_B t^2}{2}} \tag{12} $$

For the parameters \(p_{A}\), \(p_{B}\), \(k\) and \(A_0\), we have estimated the following values explain:

Table 1: Parameters used to obtain quantitative results from the analytical solution for the curli production.

Parameters	Value	Unit
\(\boldsymbol{p_{A}}\)	\(1.0 \cdot 10^{-10}\)	\(\frac{1}{Ms}\)
\(\boldsymbol{p_{B}}\)	\(3.4 \cdot 10^{-12}\)	\(\frac{M}{s}\)
\(\boldsymbol{k}\)	\(4.0 \cdot 10^{4}\)	\(\frac{1}{Ms}\)
\(\boldsymbol{A_0}\)	\(6.0 \cdot 10^{-6}\)	\(M\)

Plotting equation 12 with the parameter values in table 1 yields the graph shown in figure 1. insert caption

Figure 1 shows a steady production of CsgB. \(CsgA_{curli}\) concentration at \(t= \ 0\) is zero as expected, since there is no CsgB at that point. In the next few hours, \(CsgA_{curli}\) concentration peaks. We think that this is due to the high concentration of \(CsgA_{free}\) that is present at \(t= \ 0\). In figure 2, curli growth as function of time is plotted for different initial concentrations of \(CsgA_{free}\). insert caption

We conclude the following from figure 2:
Firstly, as expected, curli growth stabilizes to a rate equal to \(p_{A}\) after approximately 2 hours, independent of the initial concentration of \(CsgA_{free}\), \(A_0\).
Secondly, increasing the initial concentration of \(CsgA_{free}\), \(A_0\), increases the height of the peak. Even with zero initial \(CsgA_{free}\) concentration, a small peak can be found at one hour. This is a consequence of \(CsgA_{free}\) build-up when the CsgB concentration is still very small.
Thirdly, during the first two hours, few CsgB proteins are present in the system. We therefore expect that the length of the curli fibrils that started in the first few hours are much longer than the fibrils that started at later times.

write some conclusion about rate limiting step and which information we gained from this level and what will be used in the subsequent levels

Cell Level Modeling

Now that the growth rate of curli and production of CsgB protein as function of time is obtained, the conductivity as a function of time can be computed. The relevant length scale is the cell length, or the micrometre scale. The approach we used for this is relatively simple:

• We discretize the amount of curli subunits (\(CsgA_{curli}\) in the gene level model) and CsgB proteins that have to be added for each time step.
• At each time step, we add more curli subunits to growing curli fibrils. Also, we add more new curli fibrils to the model.
• From the density of the curli fibrils around the cell as a function of the radius, we calculate the conductive radius of the cell.

Discretization of Gene Level Model

We have discretized equations 5.2 and 12 in N time steps. These give the expected number of new CsgB proteins and curli subunits for each time step, as we plotted the solution of these two equations in figures 1 and 2. From these figures we determine the expected number of new CsgB proteins and curli subunits for each time step. However, a fundamental assumption in deterministic modeling is that the concentration is continuous. In reality, the amount of added curli subunits is discrete, since we cannot add half a curli subunit.
Furthermore, in the gene level model we did not take into account the statistical variation of gene transcription and adding of curli subunits; sometimes less and some times more curli subunits are added with respect to the expected value. To include this in the cell level model, we drew the amount of new curli subunits from a Poisson distribution where λ equals the expected amount of added subunits.
So, for each time step we now have \(B_n\) new CsgB proteins and \(C_n\) new curli subunits, where \(C_n\) varies for each time step, as it is drawn from a Poisson distribution. A assumption of this distribution is that the time at which a new curli subunit is added, is uncorrelated to the time at which the previous curli subunit was added, we think this is a fair assumption. Note that the cell level model we made, accounts for the stochasticity of adding curli subunits, but not for the stochasticity of gene expression, so for the production of CsgB protein. The value \(B_n\) and the Poisson distribution are determined from figures 1 and 2. say something about the time steps, how much time represents each step and determine Bn and Cn from figures

Building the Curli Fibrils

Firstly, \(B_n\) CsgB proteins are added to our model that mark the starting points for new curli fibrils. These new curli fibrils are located at random points on a sphere with radius r, which represents the cell. The radius r is chosen such that the volume of the cell is\(\ \sim 1.1 \ \mu m^3\) [5]. A CsgB protein is modeled by a line of length 4 nm that points radially outward, perpendicular to the cell surface [source]. In reality, the distribution of CsgB on the cell surface is not uniformly distributed [6]. However, we assumed uniformly distributed CsgB to keep our model prehensile. This is a point that may be used to further improve the model.

Next, \(C_n\), which is drawn from the Poisson distribution, where λ equals the expected amount of added curli subunits, new curli subunits are added to curli fibrils by repeating the following process \(C_n\) times:

• Firstly, a random curli fibril is selected, e.g. curli number k. A curli fibril is represented by a 3 (the x, y and z coordinates) by l+1 matrix, where l is the amount of curli subunits of the curli fibril and the origin is chosen to be the center of the sphere. Thus, by storing the ending coordinates of each curli subunit, we know the starting and end coordinates of each curli subunit. The curli subunits are modeled by a line of length 4 nm [source].
• Secondly, the polar angle in spherical coordinates of the last curli subunit is computed, \(\theta_{1}\).
• Thirdly, the new curli subunit has a small angular deviation with respect to the previous one. This polar angle \(\theta_{2}\) is chosen from a Gaussian distribution with parameters N(0,σ). σ is chosen such that the persistence length, the distance over which a fibril has bend by \(90^{\circ}\) and has ‘lost’ its directional information, is 4 µm. The azimuthal angle ϕ is completely random between 0 and 2π radians, and chosen from an uniform distribution.
• Fourthly, for the new curli subunit for which we determined \(\theta_{2}\) and ϕ, the polar angle is determined to be \(\theta_{1} + \theta_{2}\). We now know the length of the new curli subunit (4 nm), its polar angle and its azimuthal angle. Subsequently, we add it to the previous curli subunit of the fibril and calculate the ending coordinate of the added curli subunit from its length, polar angle and azimuthal angle and the ending coordinate of the previous curli subunit. This calculated ending coordinate of the added curli subunit is stored in the matrix that represents the curli fibril.

The angular deviation σ is a critical parameter in our model. Increasing this value increases the flexibility of our curli, where decreasing this value increases the stiffness of the curli. This is shown in figure 3. If the length of one subunit is 4 nm and the total persistence length is 4 µm, then \(\sigma = \ 3.47^{\circ}\). Furthermore, we think that it is justified to add the curli subunits one at a time to a random curli. We expect no discrimination of the CsgA proteins for binding to a large or small curli or one that has recently gotten a new curli subunit. figure caption

An illustrative view of what our cell looks like during the adding of curli subunits is shown in figure 4. This figure is created when just a few curli were added (\( \sim 1/2 \ hour\)). A similar figure after \(t = \ 10 \ hr\) would look like a fuzzy ball of curli. figure caption

Now that we have a model of a cell with growing curli, we want to extract relevant data for the colony level modeling. Ideally, the resistance as function of radius and time would be calculated by looking at connections between the curli fibrils. However, this requires insight of the behavior of the curli on the nanoscopic scale. For instance, what is the conductivity of a single curli fibril with gold nanoparticles and what is the critical distance between the fibrils that make them connect? After an extensive literature study, we have decided to simplify this model. Furthermore, when interactions between the curli fibrils have to be taken into account, the model gets too computationally expensive.

[write something about the part where we tried the percolation on this level], low priority

To have a reasonable computational time, we decided to extract our parameters for the colony level modeling from the curli density around the cell. Figure 5 contains a histogram with the amount of curli subunits as a function of the cell radius after 10 hours. [add figure] Note how no curli are found below the actual cell radius. It can be seen from the figure that there is a large peak, followed by a plateau. When this histogram is observed in time, you would notice that at first large curli are being created. Figure 6 shows the length of all curli after 10 hours. figure caption

Looking back at figures 1 and 2, the fact that, as can be seen in figure 6, the first curli are much longer that the later ones, can be explained by the fact that there is relatively large curli growth in the beginning, because few CsgB have been produced and therefore, only a few curli fibrils are available for CsgA proteins. After a couple of hours there are more CsgB proteins, thus more curli fibrils, but CsgA protein production does not increase. Therefore, the ratio [CsgA]/[curli fibrils] is much smaller than in the beginning and each curli will grow much slower. A consequence of this is that the ‘newer’ curli fibrils are much shorter. The first histogram of figure 7 shows the amount of curli as function of radius after a growth time of 10 hour. The second histogram shows the curli density as function of radius after a growth time of 10 hour. [add figure]

Conductive Radius of the Cell

We think that a reasonable approximation of the conductivity is the density of the curli around the cell as a function of the radius (see figure 8 in the middle, the blue line). When the density is higher, there are more gold particles, thus higher conductivity. In our simplest model we say that there is a critical density \(\rho_{crit}\) of curli that is needed to have conductivity. The density \(\rho_{curli}\) decreases as function of the radius. The largest radius where \(\rho_{curli} > \rho_{crit}\), we call the conductive radius \(r_{cond}\). With only this simple approximation we can calculate some interesting properties of our system: the time at which we expect percolation to happen and the resistivity of our system. Though this approximation seems to be rather arbitrary, we do have some reasoning for this: [add caption]

• First of all, the goal of this parameter is to get information about our system that will be calculated in colony level modeling. We use this parameter in colony level modeling to find connections between cells. To have a continuous path from one electrode to the other electrode, we must have a lot of cells that are connected to each other. In order to know when cells are connected to each other, we have to assume that everything at a certain radius from the cell is conductive; for this radius we use the critical density \(\rho_{crit}\). However, for this to be true the fibrils on one side of the cell must be connected to the fibrils on the other side. The Percolation Theory prescribes that this is a sharp transition as a function of the density, so we can choose \(\rho_{crit}\) in such a way that we are very sure that everything at \(\rho_{crit}\) from the cell is conductive.
• While the precise value of \(\rho_{crit}\) may be unknown and should be measured, we think that we can still get plenty of information about the qualitative behaviour of our system in advance. Figure 8 at the bottom shows the conductive radius \(r_{cond}\) as function of time using \(\rho_{crit}\) as shown in figure 8 in the middle as the red line. Increasing or decreasing \(\rho_{crit}\) would result in a similar \(r_{cond}\) as function of time. Hence, the qualitative behaviour is preserved.
• Due to the simplifications that we made in order to be able to model our system, we cannot include interactions or cluster forming between the curli themselves. Using \(\rho_{crit}\), we have an elegant way to filter out modeling errors.

Note that figure 8 at the bottom shows a discrete increase of the conductive radius. This is because \(\rho_{curli}\) is dependent on the radius, and because we run a computer simulation, the radius is therefore discretized. Moreover, if we rerun the script to calculate the conductive radius, we expect some variation in the curve, as \(\rho_{curli}\) is subject to stochasticity. Therefore, we have repeated our simulations on the cell level many times in order to get statistically valid results for the mean and standard deviation of \(r_{cond}\).
When we ran our simulation 100 times, we got the results displayed in figure 9. Figure 9 displays the curli density at \(\ t= \ 2 \ hours\) for all cells in the left figure. The orange line represents the average of the simulations. It can be concluded that the intercellular variation is relatively small. This makes sense, since the relative deviation of stochastic processes decreases with the sample size. In the right figure, the mean and standard deviation of the curli density as a function of the radius is shown. insert caption

Another interesting figure to look at, is the density for different times, shown in figure 10. This figure shows that, corresponding with what we have seen previously, \(\rho_{curli}\) decreases as a function of the radius. Also, it decreases faster as a function of the radius in the first two hours. After two hours, we can see that the curli density increases only for small r, as mainly short curli are added to the system. This agrees with our previous results. insert caption

We use the data from figure 10 to make a first approximation for \(\rho_{crit}\). However, the exact value of \(\rho_{crit}\) has to be measured. We set \(\rho_{crit}\) equal to 1% of the maximum \(\rho_{curli}\) observed in figure 9. So, we set \(\rho_{crit} = \ 10^3 \ \frac{number \ of \ curli}{\mu m^{2}}\). The following figure is obtained. insert caption and check if rho_crit value is correct

Again, the orange line represents the mean conductive radius. A sharp increase in the conductive radius can be observed for \(t < 1 \ hour\), and after \(t = \ 1 \ hour\) the conductive radius increases slowly. The cellular variation in the second regime is relatively large, as all the lines except the orange line in figure 11 represent single simulations of our system. Different values of \(\rho_{crit}\) result in different characteristic curves for \(r_{cond}\) as shown in figure 12. insert caption

From figure 12, we conclude that low values of \(\rho_{crit}\) result in a sharp increase of \(r_{cond}\) followed by a steady, slow increase of \(r_{cond}\) in time. During the steady, slow increase of \(r_{cond}\) in time, the cellular variation is relatively large. For high values of \(\rho_{crit}\), there is a delayed sharp increase of \(r_{cond}\) and less cellular variation.

make a fit for \(\rho_{curli}\)
from this fit determine \(\rho_{crit}\) (transition value between the two different fits)
plot \(r_{cond}\) for this found value of \(\rho_{crit}\)

Colony Level Modeling

Percolation

still has to be written

Resistivity

To get a quantitative measure for the conductivity between two cells, we will quantify the overlap of two conducting spheres, where we assumed that the conducting spheres represent cells surrounded by curli filaments. To do this, we will subdivide the overlapping region in infinitesimal volumes \(dV\). The infinitesimal conductivity of such an infinitesimal volume is given by:

$$ d \sigma (y) = \ \frac{\rho_1}{r_1} dV \frac{\rho_2}{r_2} dV \tag{}$$

The factor \( 1/r \) is introduced to account for the conductivity of the wires itself. Further away from the cell, the wires need a longer distance to go to the cell. Since we want to know the strongness of the connection between the cells we have to include this factor. For a straight line this inversely proportional to the distance. For a single curli fibril, this relation does not hold. However, we assume that the curli density is high, thus there are many connections between the curli themselves. Then there is a pathway from the origing to \( r \) roughly proportional to the distance from the cell. To find the total conductivity, we integrate on both sides. To account for the fact that both volume elements \(dV\) are the same, we make use of the Dirac-delta function \(\delta_3\) [source]. This gives us the following:

$$ \sigma (y) = \int{ \frac{\rho_1(\vec{r_1})\rho_2(\vec{r_2})}{r_1 r_2}\delta_3(\vec{r_2}-f(\vec{r_1}))d^3\vec{r_1}d^3\vec{r_2}} \tag{} $$

The Dirac delta allows us to remove the \(\vec{r_2}\) dependence by expressing these in \(\vec{r_1}\). The still undetermined relation between \(\vec{r_1}\) and \(\vec{r_2}\) is given by \(\vec{r_2} = f(\vec{r_1})\). Applying this removes one of the two volume integrations. Using spherical coordinates, the resulting single volume integration can be written as:

$$ \sigma (y) = \int_{r_0}^{r_{max}} \int_0^{\theta_{max}(r)} \int_0^{2\pi} \rho(r_1)\rho_2(f(r_1))\frac{r_1}{r_2} \sin(\theta_1) d\phi_1 d\theta_1 dr_1 \tag{} $$

Here we have made use of the fact that the density \(\rho\) is only dependent on \(r\) and not on \(\phi\) and \(\theta \). The integral over \(phi\) is trivial and gives us a multiplication factor of \(2 \pi\):

$$ \sigma (y) = \ 2 \pi \int_{r_0}^{r_{max}} \int_0^{\theta_{max}(r)} \rho(r_1)\rho_2(f(r_1))\frac{r_1}{r_2} \sin(\theta_1) d\theta_1 dr_1 \tag{} $$

Now that we have reduced our integration to two dimensions, we will work out \(f(\vec{r_1})\). To do this, we introduce the vector from the origin of cell 1 to the origin of cell 2, \(\vec{y}\). This allows us to express \(\vec{r_2}\) in terms of \(\vec{y}\) and \(\vec{r_1}\):

$$ \vec{r_2} = \ \vec{y} - \ \vec{r_1} = \begin{bmatrix}y \\0\\ \end{bmatrix} - \begin{bmatrix} r_1 \cos(\theta_1) \\r_1 \sin(\theta_1)\\ \end{bmatrix} \tag{} $$

Now it is straightforward to express \(r_2\) in terms of \(y\), \(r_1\) and \(\theta_1\):

$$ r_2 = \ |\vec{r_2}| = \ \sqrt{(y - r_1 \cos(\theta_1))^2 + \ r_1^2 \sin^2(\theta_1)} \tag{} $$

Plugging this in yields the following integral:

$$ \sigma (y) = \ 2 \pi \int_{r_0}^{r_{max}} \int_0^{\theta_{max}(r)} \frac{\rho(r_1)\rho_2 \left( \sqrt{(y - r_1 \cos(\theta_1))^2 + r_1^2 \sin^2(\theta_1)}\right) r_1 \sin(\theta_1)}{ \sqrt{(y - r_1 \cos(\theta_1))^2 + r_1^2 \sin^2(\theta_1)}} d\theta_1 dr_1 \tag{} $$

We will now have a closer look at the boundary values for \(r\) and \(\theta\). We want to integrate over the entire volume. Therefore, \( \theta(max) = \pi \) and \( r_{max}=\infty \). Here we have approximated our cells as points in space. Hence \( r_0 =0 \).

We will now use the previously [link] found fact that the curli density can be described as:

$$ \rho(r) = \ C_{1}e^{-\frac{r}{C_{2}}} \tag{} $$

Plugging the boundary values and our expression for \(\rho(r)\), we find the following expression for the conductivity between two cells:

$$ \sigma (y) = \ 2 \pi C_{1}^2 \int_{0}^{\infty} \int_0^{\pi} \frac{e^{-\frac{r_1}{C_{2}}} e^{-\frac{ \sqrt{(y - \ r_1 \cos(\theta_1))^2 + \ r_1^2 \sin^2(\theta_1)}}{C_{2}}} r_1 \sin(\theta_1)}{\sqrt{(y - r_1 \cos(\theta_1))^2 + r_1^2 \sin^2(\theta_1)}} d\theta_1 dr_1 \tag{} $$

This integral looks very complicated, but don't panic! It can algebraically be simplified with some clever substitutions. We can rewrite this integral by moving all terms independent of \( \theta \). Furthermore, using that \( sin^2 (\theta_1) + cos^2\theta_1 = 1 \) we get.

$$ \sigma (y) = \ 2 \pi C_{1}^2 \int_{0}^{\infty} r_1 e^{-\frac{r_1}{C_{2}}} \int_0^{\pi} \frac{e^{-\frac{ \sqrt{y^2+r_1^2-2yr_1 cos( \theta_1 ) }}{C_{2}}} \sin(\theta_1)}{ \sqrt{y^2+r_1^2-2yr_1 cos( \theta_1 ) }} d\theta_1 dr_1 \tag{} $$

Now we must recognize that we can substitute \( x= cos(\theta_1) \) such that \( dx = -\sin(\theta_1) d\theta_1 \). This results in:

$$ \sigma (y) = - \ 2 \pi C_{1}^2 \int_{0}^{\infty} r_1 e^{-\frac{r_1}{C_{2}}} \int_0^{\pi} \frac{e^{-\frac{ \sqrt{y^2+r_1^2-2yr_1 x }}{C_{2}}}}{\sqrt{y^2+r_1^2-2yr_1 x }} dx dr_1 \tag{} $$

In the second integral we recognize something of the form \( \int \frac{e^{-\sqrt{a+bx}}}{C_2\sqrt{a+bx}} dx \) with \( a= \frac{y^2+r_1^2}{C^2_2} \) and \(b=-\frac{2yr_1}{C^2_2} \). Substituting \( h= \sqrt{a+bx} \) with \( dx= \frac{2h}{b} dh \) yields.

$$ \frac{2}{bC_2} \int e^{-h} dh= \frac{-2}{bC_2} e^{-h}$$

Now h, a and b x can be resubtituted:

$$ \frac{-2}{bC_2} e^{-h}=\frac{-2}{bC_2} e^{-\sqrt{a+bx}}= \frac{C_2}{yr_1} e^{-\frac{\sqrt{y^2+r_1^2-2yr_1cos(\theta_1)}}{C_2} }$$

Filling in the upper limit, makes

$$\frac{C_2 e^{-\frac{y+r_1}{C_2} }}{yr_1} $$

And for the lower boundary this results in:

$$ \frac{C_2 e^{-\frac{|y-r_1|}{C_2} }}{yr_1}$$ Hence, the entire integral now becomes $$ \sigma (y) = \frac{ 2 \pi C_{1}^2 C_2 }{y} \int_{0}^{\infty} e^{-\frac{|y-r_1|+r_1}{C_2} } - e^{-\frac{y+2r_1}{C_2} } dr_1 \tag{} $$

Solving the second integral is fairly easy:

$$ \sigma (y) = \frac{ 2 \pi C_{1}^2 C_2 }{y} \int_{0}^{\infty} e^{-\frac{|y-r_1|+r_1}{C_2} }-e^{-\frac{y+2r_1}{C_2} } dr_1 = \sigma (y) = \frac{ 2 \pi C_{1}^2 C_2 }{y} \left( \int_{0}^{y} e^{-\frac{y}{C_2}} dr_1 +\int_{y}^{\infty} e^{-\frac{2r_1-y}{C_2}} dr_1 -e^{\frac{-y}{c_2}}\int_0^{\infty} e^{-\frac{2r_1}{C_2} } dr_1 \right) \tag{} $$

Which brings us to the final end result: $$ \sigma (y) = \frac{ 2 \pi C_{1}^2 C_2 \left( C_2 + y \right)}{y} e^{\frac{-y}{c_2}} \tag{} $$

solving this integral numerically, storing values in table
giving the conductivity as weights to the adjacency matrix
from these weights (conductivity) determine if cells are connected or not
from the resulting adjacency matrix, calculate the resistivity of the system using graph theory

References

still has to be made