 work I'm going to present to you today is the first step of a big project which consists of building a model and numerical simulations to study vesicular release at neural synapses. So the first step is about asymptotic analysis of the mean first passage time of a renown particle to a small hole. Okay so this the I will first present a short overview of synaptic transmission at chemical synapses very short no problem and then I will show you the asymptotic analysis of narrow escape time. So the synapse maybe you everybody knows what it is is a structure not very well known today in the is the functioning that permit to transmit an electrical signal from one neuron the presynaptic neuron to another the post-synaptic so when an electrical signal arrives an action potential in red here it triggers the opening of the voltage gated calcium channels which are the blue channels on the on this picture and this will trigger the entry of calcium ions within the terminal the orange dots and those calcium ions can go to bind to specific receptors localized on docked vesicles the docked vesicle are the one which are just opposed to the membrane and if they find those docked vesicle and the specific receptors on those docked vesicle it will trigger the release of neurotransmitter within the synaptic cleft and then it will they can bind to the pink receptors specific receptor for neurotransmitters and the chemical cascade and everything will trigger the will create an electrical signal in the post-synaptic neuron so it's a very complex phenomenon I'm just today interested in what's happening in the presynaptic part when calcium calcium ions enter through the calcium channel and have to find the very small receptors located at the docked vesicle okay so just a zoom on this part so the dot vesicle are really attached to the membrane through the snare complex which is the complex of protein and on the snare complex you have the receptors for the calcium so calcium enters through the blue channels then they diffuse and if few calcium like 4 to 8 find the receptors then the synaptic vesicle will fuse with the membrane the membrane of the vesicle with the membrane and then everything will start for the transmission so the model here that we did is calcium ions are bronion particles that diffuse freely in the terminal we know that particles that calcium ions are charged particles but we will neglect this at the first step and the docked vesicle is just a sphere tangent to the plane okay here you have the sphere in green and the tangent plane in blue and the binding on the snare complex is when the bronion particle will find the red cylinder you can find here so this is our model of these ions finding the complex find the small cylinder here which is the center of the cylinder is the center of the in the tangential point between the vesicle and the plane okay so we are now interested in the time and in the further step of the project the probability that an ion starting from this channel will find this small cylinder okay so I will present now the asymptotic analysis we did for this narrow escape time so the narrow escape problem we have here is the following a brilliant particle is described by a stochastic equation x dot equals square root of 2d w dot d is a diffusion coefficient of the particle and w dot is white noise and the first time to it did domain here on megabar which represent the domain we are interested in through the small hole in red the cylinder in red in this picture is given by this so the first time the particle find the domain okay the particle is reflected everywhere on the border of the domain except at the red cylinder as it's absorbed so we are interested in the mean first passage time to this cylinder which is u of x and we know it's very well known it's dinkin that this mean first passage time follow the solution of the mixed boundary value problem which is here d laplacian u equal minus one d is diffusion coefficient d u dn equals 0 is a classical reflection at the boundaries and u equals 0 is a classical absorption absorption at the red cylinder here so we are going to to try to find the asymptotic analysis of the of this time okay so the the the first tough we can look to see when we look at this this domain that it's invariant rotation around the axis delta here which is very easy to see and according to the equation we have we know that this problem is independent of theta if we consider the cynical coordinates around delta so we can just study an equivalent problem in domain omega which is a 2d domain the projection of the domain on theta equals 0 for example so this on this domain you see the blue is going to the blue the the red sphere is here the red the arc of circle the small cylinder is now just segment okay and the equation becomes the following equation where you recognize here the laplacian in cylindrical coordinates very simply okay so I'm now in a 2d domain with a 2d problem which is equivalent to the problem we had in 3d you notice as well that I put the diffusion coefficient on the right and that we have reflection on the blue black and green boundaries and absorption at the red segment we are interested in the asymptotics of u when epsilon which is the length of this segment goes to zero very small goes to zero is very small in front of the rest of the boundary so the problem we have here is that we have a singularity because when epsilon goes to zero the point here is not regular so to desingularize this singularity we we will have to okay the problem of this singularity is that it prevents us to use a classical method which are for example the green function on this domain because we don't know the order of the green function in epsilon and desingularity this it depends on the domain and we don't have access to this the other method which is very classical is matching asymptotics so we try to find a solution near the cusp near the hole we try to find a solution in the domain we try to match and then we can obtain something but here because of the geometry is not possible as well so the idea is to desingularize the singularity and we are using conformal maps so here because we have tangent circles we will use the inverse map which comes very easily when you start to think about it because it will transform these domains into the main tilde omega we have here because the inverse map will transform the sphere tangent in zero into parallel straight lines you notice that the small hole here in red is going to a straight line here and you know that inverse maps are putting spheres circled into straight lines and straight line into circles so in reality the the hole here is not a straight line it's a knock of circle part of the circle but the radius of this circle is 1 over 2 r a which is here the the abscissa of the hole and if you maybe don't remember because I didn't tell you this before r a is in all those quote of epsilon okay it's a simple computation so sorry this mean that this at the first order we can consider that this part is a straight line okay the first order is straight line so putting setting v of st equal u of r vd we can now consider the equation in the map domain which is much simpler domain because at the first order it's like rectangular domain the equation is much more complicated because it captures the geometry of the problem in omega but still the domain now is much easier and we get this this equation we have the reflection dv dn again on the blue and black and green boundaries and the absorption on the red boundary so to solve this equation we will now use a scaling property sorry a scaling method so we put zeta equal s square root of 2r epsilon square root of tilde epsilon okay which means that which means that we are going we are taking your our interest is in the region very close to the hole okay so very close to the red boundary so by setting this using the scaling we can put big y equals of psi t equals v of sts very simple and then we can do a regular expansion of big y in power of epsilon tilde tilde and injecting this regular expansion into the equation and expanding the equation now we obtain here the asymptotic of the equation in power of epsilon and we are we will now take care of the terms of leading of much interest leading order terms of the equation the first one is in one of a tilde epsilon square then the second one is one of epsilon etc etc so using this expansion now we are looking because we are looking for asymptotic analysis we are looking of the leading order term looking for the other term sorry we will first consider the first term which is the t the second derivative of y0 over t equals 0 so using this equation and the boundary condition we have we have on the blue and green boundaries we can show very easily that y0 is not is independent of t okay this is what we get from the first term the second term the second order term gives us this equation here and integrating over t again we can show that y1 is as well independent of t it's simple computation and using this we obtain a complete an equation for y0 in zeta and using the boundary condition now at the whole which state that in zero in zeta equal 1 solution equals to 0 we obtain the form of the leading order term of the solution big y0 near the whole going back into our first parameters we have that v of st take this form okay so it's equals to 0 near the whole this solution is an approximation is not an approximation it's the leader order term of the solution near the whole okay our problem is we have to compute a so usually you use the rest of the parameters but this approximation of the solution is not good far from the whole we cannot use the reflection the last last boundary condition we get so to compute a we will use the divergence theorem which state that the integral of bar omega of laplacian u it's equal to the integral of the border very classical equation so this the laplacian u equals minus tilde the volume of tilde omega over d just comes from the equation d laplacian u equal minus 1 and the other part because we have reflection everywhere except at the whole say that this integral on the border is exactly the integral of the solution at the whole it's good because we have a solution at the whole so we can then using our first order term of the solution at the whole give this equation and then find a and we now have another first order term the leading order term of the solution in epsilon for the the whole domain so to get toe to get the mean first passage time sorry we just put the value of a in v of st and then consider when s is equal to zero when we are far from the whole s equals to 1 at the whole equals to 0 for the whole and we get the mean first passage time for a Bernan particle to the small hole and in the boundary layer we also have the leading order of the mean first passage time and we know as well the way is behaves okay so we computed the leading order term of the mean first passage time to a small ribbon located between two tangents first and we could show that this mean first pass time is constant outside of the boundary layer as we just noticed and is well this tells us that this mean first pass time is well approximated by a Poisson process so this is a result we already know so it's mean that we can we can consider that the arrival of this particle to the small hole is Poissonian and the first approximation the law of arrival is an exponential this is very interesting because using this analysis in further development of the project that I don't have time to present here we could compute as well the probability when the particles start inside the boundary layer because here it's we have the mean first passage time to go to the whole when you are far from the hole because we set s equals to 0 so it means we are far what's happening when you are close the problem is still a problem so using the same kind of analysis what we have here we could compute the probability to reach the cylinder before escaping the domain okay it's the same kind of computation and using this we could build a model of the active zone which is the place where vesicle are located and the calcium channel are located as well to investigate the influence of the position of the channels on the crowding of vesicle or not crowding of vesicle etc etc on the release probability which is the probability to have the fusion so the probability to have four ions that are coming to this vesicle or not and using this we could build a model of the prescine optic terminal which bigger with the whole stuff we have over there and the fact that this mean first pass time is Poissonian is really what is important here in our study because if you want to model the prescine optic terminal you have a bronyon particle you you can do bronyon simulations if you want it's very simple not that simple because the domain is not that fun but okay you can do it but you will wait for hours because of the very small hole you have here and because of the very long time you have to find this small hole the fact that it's a Poisson process allows you to use simply Gillespie algorithm to say okay it's Poissonian I can at least use Markov chain for what's happening because it's constant it doesn't depend on the position so it doesn't depend on the time and it's what we did in the following of this project and I wanted to show you this because it's a very first step but it's everything is is going after that because we could show that the mean first pass time is independent of the position and that this time is Poissonian time so this work was published in the Siam multicore modeling simulations so I want to thank just my PhD supervisor David Altman and my lab mates and the organizer of this conference for giving me the opportunity to present my work today thank you I was just wondering what do you mean by it's well approximated by Poisson process I mean the first passage time is approximately exponentially because of this because it doesn't depend on the position in fact so it doesn't depend when you start so it's yeah you have the mean and the variance is the same and it's well approximated there any other questions if not let's send the speaker again