### Continuous time stochastic model for neurite branching (van Elburg 2011)

Accession:129071
"In this paper we introduce a continuous time stochastic neurite branching model closely related to the discrete time stochastic BES-model. The discrete time BES-model is underlying current attempts to simulate cortical development, but is difficult to analyze. The new continuous time formulation facilitates analytical treatment thus allowing us to examine the structure of the model more closely. ..."
Reference:
1 . van Elburg R (2011) Stochastic Continuous Time Neurite Branching Models with Tree and Segment Dependent Rates Journal of Theoretical Biology 276(1):159-173 [PubMed]
Model Information (Click on a link to find other models with that property)
 Model Type: Axon; Dendrite; Brain Region(s)/Organism: Cell Type(s): Channel(s): Gap Junctions: Receptor(s): Gene(s): Transmitter(s): Simulation Environment: C or C++ program; MATLAB; Model Concept(s): Development; Implementer(s): van Elburg, Ronald A.J. [R.van.Elburg at ai.rug.nl];
 / ContinuousTimeDendriticBranchingModel figures mat-files mexhandle readme.html cBEModel.m mexSDependenceCalculator.cpp mexSDependenceCalculator.mexglx ObjectHandle.h plotBEModelCurves.m SDependenceCalculation.m SDependenceCalculator.cpp SDependenceCalculator.h SDependenciesPlot.m
% plotBEModeCurves

b=1; % basal branching rate
% The parameter E giving the dependence of branch rate on the number of terminal segments will varied
scrsz = get(0,'ScreenSize');
figure(1)
set(gcf,'Position',[1, 1 , scrsz(4) , scrsz(4) ])
clf
hold on

nplotrange=1:5:101;

for E=0:0.25:1
% Calculate numerical solutions
[T,Y,mun,sigman,validPoints]=cBEModel(b,E);

% Plot top row: temporal development for the probabilities $p(n,t)$ for
% $n=1,6,11,...,101$ and three different values of $E$: $E=0$ (left),
% $E=\frac{1}{2}$ (middle) and $E=1$ (right) in all cases lower
% $n$-values lead to earlier $p(n,t)$-peaks, while late peaking traces
% might peak outside the window shown.

subplot(3,3,1)
if(E==0)
plot(T(validPoints),Y(validPoints,nplotrange))
end
axis([0,4,0,0.2])

subplot(3,3,2)
if(E==0.5)
plot(T(validPoints),Y(validPoints,nplotrange))
end
axis([0,25,0,0.2])

subplot(3,3,3)
if(E==1)
plot(T(validPoints),Y(validPoints,nplotrange))
end
axis([0,25,0,0.2])

% Plot middle row: expectation value and variance for the number of terminal
% segments for three different values of $E$ (matching those in the top
% row) calculated using the first 1000 $p(n,t)$'s while keeping
% $p_{high}< 10^{-6}$.
subplot(3,3,4)
hold on
if(E==0)
plot(T(validPoints),mun(validPoints),'k-')
plot(T(validPoints),sigman(validPoints),'k--')
end
axis([0,4,0,30])

subplot(3,3,5)
hold on
if(E==0.5)
plot(T(validPoints),mun(validPoints),'k-')
plot(T(validPoints),sigman(validPoints),'k--')
end
axis([0,10,0,30])

subplot(3,3,6)
hold on
if(E==1)
plot(T(validPoints),mun(validPoints),'k-')
plot(T(validPoints),sigman(validPoints),'k--')
legend('\mu(n)','\sigma(n )','p_{high}10^7 ','Location','NorthWest')
end
axis([0,25,0,30])

% Plot bottom row left: comparison expectation value (solid grey lines) with
% mean field prediction (dashed black lines) for five different $E$
% values: $E= 0,0.25, 0.5, 0.75, 1$; at $E$-values $0, 1$ the mean
% field solutions coincide with the exact solution.
subplot(3,2,5)
hold on
plot(T(validPoints),mun(validPoints),'k-')
if(E==0)
plot(T(validPoints),exp(T(validPoints)),'r--')
else
plot(T(validPoints),(E*T(validPoints)+1).^(1/E),'r--')
legend('\mu(n)','\mu_{MF}(n)','Location','SouthEast')
end
axis([0,25,0,30])

% Plot bottom row right: comparison of mean field solution and numerical
% results; after an initial growth of the relative error for
% intermediate values of $E$ , i.e.  $E= 0.25, 0.5, 0.75$, the
% relative error attenuates. The standard deviation and mean for  $E=0$
% and $E=1$ correspond within the numerical error with the exact
% solutions.
markers=['.';'+';'d';'o';'-'];

subplot(3,2,6)
hold on

if(E==0 )
plot(T(validPoints),(exp(T(validPoints))./mun(validPoints)'),markers(1+round(E/0.25)))
else
plot(T(validPoints),((E*T(validPoints)+1).^(1/E)./mun(validPoints)'),markers(1+round(E/0.25)))
end

legend('E=0','E=0.25','E=0.50','E=0.75','E=1','Location','SouthEast')
axis([0,25,0.99,1.1])

end

%% Save figure to file (pdf)

figure(1)
saveas(gcf,'figures/BEModel.pdf')