Asymptotic Analysis of the Narrow Escape Problem in Dendritic Spine Shaped Domain: Three Dimension
AAsymptotic Analysis of the Narrow Escape Problem in DendriticSpine Shaped Domain: Three Dimension
Hyundae Lee ∗ Xiaofei Li † Yuliang Wang ‡ September 20, 2018
Abstract
This paper deals with the three-dimensional narrow escape problem in dendritic spine shapeddomain, which is composed of a relatively big head and a thin neck. The narrow escape problemis to compute the mean first passage time of Brownian particles traveling from inside the head tothe end of the neck. The original model is to solve a mixed Dirichlet-Neumann boundary valueproblem for the Poisson equation in the composite domain, and is computationally challenging.In this paper we seek to transfer the original problem to a mixed Robin-Neumann boundaryvalue problem by dropping the thin neck part, and rigorously derive the asymptotic expansionof the mean first passage time with high order terms. This study is a nontrivial generalizationof the work in [14], where a two-dimensional analogue domain is considered.
The narrow escape problem (NEP) in diffusion theory, which goes back to Lord Rayleigh [18],is to calculate the mean first passage time (MFPT) of a Brownian particle to a small absorbingwindow on the otherwise reflecting boundary of a bounded domain. NEP has recently attractedsignificant attention from the point of view of mathematical and numerical modeling due to itsrelevance in molecular biology and biophysics. The small absorbing window often represents asmall target on a cellular membrane, such as a protein channel, which is a target for ions [8], areceptor for neurotransmitter molecules in a neuronal synapse [5], a narrow neck in the neuronalspine, which is a target for calcium ions [13], and so on. A main concern for NEP is to derive anasymptotic expansion of the MFPT when the size of the small absorbing window tends to zero.There have been several significant works deriving the leading-order and higher-order terms of theasymptotic expansions of MFPT for regular and singular domains in two and three dimensions[1, 2, 3, 6, 9, 10, 11, 19, 20].In this paper we focus on the MFPT of calcium ion in the three-dimensional dendritic spineshape domain. For many neurons in the mammalian brain, the postsynaptic terminal of an ex-citatory synapse is found in a specialized structure protruding from the dendritic draft, knownas a dendritic spine (Figure 1(a)). Dendritic spines function as biochemical compartments thatregulate the duration and spread of postsynaptic calcium fluxes produced by glutamatergic neuro-transmission at the synapses [16]. Calcium ions are ubiquitous signaling molecules that accumulate ∗ Department of Mathematics, Inha University, Incheon 402-751, Korea ([email protected]). † Department of Mathematics, South University of Science and Technology of China, Shenzhen, China ([email protected]). Corresponding author. ‡ Department of Mathematics, Hong Kong Baptist University, Kowloon Tong, Hong Kong SAR ([email protected]). a r X i v : . [ m a t h - ph ] F e b pine neck Spine head dendritic (a) Ω h Ω n Γ " @ Ω a spine head spine neck (b) Figure 1: (a).Dendritic spine morphology. (b).The modeling shape of dendritic spine Ω = Ω h ∪ Ω n with a spherical spine head Ω h and a cylindrical spine neck Ω n .in the cytoplasm in response to diverse classes of stimuli and, in turn, regulate many aspects of cellfunction [7]. The rate of diffusional escape from spines through their narrow neck, which we callMFPT, is one factor that regulates the retention time of calcium ions in dendritic spines.Most spines have a bulbous head, and a thin neck that connects the head of the spine to theshaft of the dendrite (Figure 1). Spines are separated from their parent dendrites by this thin neckand compartmentalize calcium during synaptic stimulation. It has been given that many organellesinside the spine head do not affect the nature of the random motion of ions, mainly due to theirlarge size relative to that of ions [12]. So we assume the environment inside the spine is homogenous,which means the movement of the calcium ions is pure diffusion.The NEP can be mathematically formulated in the following way. Let Ω h be a bounded simplyconnected domain in R . Let Ω n be a cylinder with length L and radius r = O ( ε ) which is muchsmaller than the length. Connecting this cylinder with Ω h , we have the geometry for the spine(Figure 1(b)). The connection part between Ω h and Ω n is a small interface which is denoted byΓ ε . Let Ω = Ω h ∪ Ω n denote the domain for the whole spine. Suppose that the boundary ∂ Ω isdecomposed into the reflecting part ∂ Ω r and the absorbing part ∂ Ω a , where ∂ Ω a is the end of thethin cylinder neck. We assume that the area of ∂ Ω a , | ∂ Ω a | = O ( ε ) is much smaller than the areaof the whole boundary. The NEP is to calculate the MFPT u which is the unique solution to thefollowing boundary value problem, ∆ u = − , in Ω ,∂u∂ν = 0 , on ∂ Ω r ,u = 0 , on ∂ Ω a , (1)where ν is the outer unit normal to ∂ Ω. The asymptotic analysis for NEP arises in deriving theasymptotic expansion of u as ε →
0, from which one can estimate the escape time of the calciumions. In [15, 4], the first order of the asymptotic expansion has been obtained numerically as u ≈ | Ω h | Lπε , (2)where | Ω h | denotes the volume of the spine head. In this study, we derive higher order asymptoticsolution to (1) by means of the Neumann-Robin model which is proposed in [14] to deal with thenarrow escape time in a two-dimensional analogue domain.2n the Robin-Neumann model the solution to the original boundary value problem (1) in thesingular domain Ω is approximated by the solution to the following boundary value problem in thesmooth domain Ω h : ∆ u ε = − , in Ω h ,∂u ε ∂ν = 0 , on ∂ Ω r ,∂u ε ∂ν + αu ε = β, on Γ ε := ∂ Ω h \ ∂ Ω r , (3)where Ω h is the spine head of Ω mentioned in Figure 1, and Γ ε := ∂ Ω h \ ∂ Ω r is the connection partbetween the spine head and the spine neck. Here α > β > α < α for some a priori constant α > ε is sufficiently small so that αε (cid:28) u ε to (3) as u ε ( x ) ≈ | Ω h | παε + | Ω h | Mπ ε + βα − | Ω h | π | x − x ∗ | for x ∈ Ω h and away from Γ ε , where M is a computable constant, x ∗ is a fixed point in Γ ε .This study is organized as follows. In section 2, we review the Neumann function for theLaplacian in R , which is a major tool for our study. In section 3, we derive the asymptoticsolution for Robin-Neumann model. In section 4, we apply the Robin-Neumann boundary modelto approximate the MFPT of calcium ion in dendritic spine. Numerical experiments are also givenin this section to confirm the theoretical results. This study ends with a short conclusion in section5. R Let Ω be a bounded domain in R with C smooth boundary ∂ Ω, and let N ( x, z ) be the Neumannfunction for − ∆ in Ω with a given z ∈ Ω. That is, N ( x, z ) is the solution to the boundary valueproblem ∆ x N ( x, z ) = − δ z , x ∈ Ω ,∂N∂ν x = − | ∂ Ω | , x ∈ ∂ Ω , (cid:90) ∂ Ω N ( x, z ) dσ ( x ) = 0 , where ν is the outer unit normal to the boundary ∂ Ω.If z ∈ Ω, then N ( x, z ) can be written in the form N ( x, z ) = 14 π | x − z | + R Ω ( x, z ) , x ∈ Ω , where R Ω ( x, z ) has weaker singularity than 1 / | x − z | and solves the boundary value problem − ∆ x R ∂ Ω ( x, z ) = 0 , x ∈ Ω ,∂R Ω ∂ν x (cid:12)(cid:12)(cid:12) x ∈ ∂ Ω = − | ∂ Ω | + 14 π (cid:104) x − z, ν x (cid:105)| x − z | , x ∈ ∂ Ω . where (cid:104)· , ·(cid:105) denotes the inner product in R . 3f z ∈ ∂ Ω, then Neumann function on the boundary is denoted by N ∂ Ω and can be written as N ∂ Ω ( x, z ) = 12 π | x − z | + R ∂ Ω ( x, z ) , x ∈ Ω , z ∈ ∂ Ω , (4)where R ∂ Ω ( x, z ) has weaker singularity than 1 / | x − z | and solves the boundary value problem ∆ x R ∂ Ω ( x, z ) = 0 , x ∈ Ω ,∂R ∂ Ω ∂ν x (cid:12)(cid:12)(cid:12) x ∈ ∂ Ω = − | ∂ Ω | + 12 π (cid:104) x − z, ν x (cid:105)| x − z | , x ∈ ∂ Ω , z ∈ ∂ Ω . The structure of R ∂ Ω is given in [17] as R ∂ Ω ( x, z ) = − π H ( z ) ln | x − z | + v ∂ Ω ( x, z ) , (5)where z ∈ ∂ Ω, x ∈ Ω ∪ ∂ Ω, where H ( z ) denotes the mean curvature of ∂ Ω at z , and v ∂ Ω is abounded function. The goal in this section is to derive the asymptotic expansion of u ε to (3) as ε →
0. For simplicitywe assume the connection part Γ ε lies in a plane. The general case where Γ ε is curved can behandled with minor modifications. Theorem 3.1.
The solution u ε to the boundary value problem (3) has the following asymptoticexpansion, u ε ( x ) = | Ω h | παε + | Ω h | Mπ ε + βα − | Ω h | π | x − x ∗ | + Φ( x ) + O ( α ) , where M is a constant given by M = (cid:90) Γ (cid:90) Γ π | x − z | dxdz, (6) x ∗ is a fixed point in Γ ε , and Φ( x ) is a bounded function depending only on Ω h . The remainder O ( α ) is uniform in x ∈ Ω h satisfying dist ( x, Γ ε ) ≥ c for some constant c > . Proof. By integrating the first equation in (3) over Ω h using the divergence theorem we get thecompatibility condition (cid:90) Γ ε ∂u ε ∂ν dσ = −| Ω h | . (7)Let us define g ( x ) by g ( x ) = (cid:90) Ω h N ( x, z ) dz, x ∈ Ω h , which is seen to solve the boundary value problem ∆ g = − , in Ω h ,∂g∂ν = − | Ω h || ∂ Ω h | , on ∂ Ω h , (cid:90) ∂ Ω h gdσ = 0 . (8)4pplying the Green’s formula and using (3) and (8), we obtain u ε ( x ) = g ( x ) + (cid:90) Γ ε N ∂ Ω h ( x, z ) ∂u ε ( z ) ∂ν z dσ ( z ) + C ε , (9)where C ε = 1 | ∂ Ω h | (cid:90) ∂ Ω h u ε ( z ) dσ ( z ) . Let x ∈ Γ ε . Substitute the Robin boundary condition and the structure of Neumann function(5) into (9), we obtain βα − α φ ε ( x ) = g ( x ) + 12 π (cid:90) Γ ε | x − z | φ ε ( z ) dσ ( z ) + (cid:90) Γ ε v ∂ Ω h ( x, z ) φ ε ( z ) dσ ( z ) + C ε , x ∈ Γ ε , where φ ε ( x ) = ∂u ε ( x ) /∂ν x . Note that the mean curvature H ( z ) = 0 for z ∈ Γ ε since Γ ε is assumedto be flat.By a simple change of variables, the above equation can be written as βα − αε ˜ φ ε ( x ) = g ( εx ) + 12 π (cid:90) Γ | x − z | ˜ φ ε ( z ) dσ ( z ) + ε (cid:90) Γ v ∂ Ω h ( εx, εz ) ˜ φ ε ( z ) dσ ( z ) + C ε , (10)where Γ = { x/ε : x ∈ Γ ε } , and ˜ φ ε ( x ) = εφ ε ( εx ), x ∈ Γ .Define two integral operators L, L : L ∞ (Γ ) → L ∞ (Γ ) by L [ ϕ ]( x ) = 12 π (cid:90) Γ | x − z | ϕ ( z ) dz,L [ ϕ ]( x ) = (cid:90) Γ v ∂ Ω h ( εx, εz ) ϕ ( z ) dz. Since v ∂ Ω is bounded, one can easily see that L is bounded independently of ε . The integraloperator L is a also bounded (see the proof in Appendix A).So we can write (10) as βα − αε ˜ φ ε ( x ) = g ( εx ) + ( L + εL ) ˜ φ ε ( x ) + C ε . Collecting ˜ φ ε terms, we have (cid:2) I + αε ( L + εL ) (cid:3) ˜ φ ε ( x ) = εα (cid:18) βα − g ( εx ) − C ε (cid:19) . (11)Assume here α < α and αε (cid:28)
1. It is easy to see that( I + αε ( L + εL )) − = I − αε ( L + εL ) + O ( α ε ) . Noting that g is C in ¯Ω h , and g ( εx ) = g ( x ∗ )(1 + O ( ε )) on Γ , we have from (11) that˜ φ ε ( x ) = εα (cid:2) I − αε ( L + εL ) + O ( α ε ) (cid:3) ( ˜ C ε + O ( ε )) . where ˜ C ε := βα − g ( x ∗ ) − C ε . (12)5y the compatibility condition (7), we can see that ˜ C ε = O (( αε ) − ). Then collecting terms wehave ˜ φ ε ( x ) = εα ˜ C ε − ( εα ) ˜ C ε ( L [1] + εL [1]) + O ( αε ) . (13)Plug (13) into the compatibility condition (7), we obtain παε ˜ C ε − α ε ˜ C ε (cid:90) Γ ( L [1] + εL [1])( x ) = −| Ω h | + O ( αε ) , which implies ˜ C ε = (cid:18) I + αεπ (cid:90) Γ ( L [1] + εL [1])( x ) + O ( α ε ) (cid:19) (cid:18) − | Ω h | παε + O ( ε ) (cid:19) = − | Ω h | παε − | Ω h | π ε M − | Ω h | v ∂ Ω ( x ∗ , x ∗ ) + O ( α ) , where M = (cid:82) Γ L [1] dx , and (cid:82) Γ L [1]( x ) = π v ∂ Ω ( x ∗ , x ∗ ) + O ( ε ). Hence from (12), we have C ε = | Ω h | παε + | Ω h | π ε M + | Ω h | v ∂ Ω ( x ∗ , x ∗ ) + βα − g ( x ∗ ) + O ( α ) . (14)Substitute ˜ C ε into (13), we have˜ φ ε ( x ) = − | Ω h | πε − α | Ω h | π ( M − πL [1]) + O ( α ε ) , and hence ∂u ε ( x ) ∂ν x = φ ε ( x ) = 1 ε ˜ φ ε (cid:16) xε (cid:17) = − | Ω h | πε − α | Ω h | π ε (cid:104) M − πL [1] (cid:16) xε (cid:17)(cid:105) + O ( α ) . (15)In order to obtain the solution to (3), it remains to calculate the second term in (9). Combining(4), (5) and (15) yields (cid:90) Γ ε N ∂ Ω h ( x, z ) ∂u ε ( z ) ∂ν z dσ ( z )= (cid:90) Γ ε N ∂ Ω h ( x, z ) (cid:20) − | Ω h | πε − α | Ω h | π ε (cid:16) M − πL [1] (cid:16) xε (cid:17) + O ( α ) (cid:17)(cid:21) dσ ( z )= −| Ω h | N ∂ Ω h ( x, x ∗ ) + O ( αε )= − | Ω h | π | x − x ∗ | + | Ω h | v ∂ Ω h ( x, x ∗ ) . (16)provided that dist( x, Γ ε ) ≥ c for some constant c > u ε ( x ) = | Ω h | παε + | Ω h | Mπ ε + βα − | Ω h | π | x − x ∗ | + Φ( x ) + O ( α ) , where Φ( x ) = g ( x ) − g ( x ∗ ) + | Ω h | [ v ∂ Ω h ( x ∗ , x ∗ ) − v ∂ Ω h ( x, x ∗ )]is a bounded function depending only on Ω h . 6 Application to the narrow escape problem in dendritic spine
In this section we use the asymptotic solution to the Robin-Neumann model to approximate thecalcium ion diffusion time in a dendritic spine domain. Without loss of generality, let Ω n be placedalong the x -axis with Γ ε at x = 0 and ∂ Ω a at x = L . Since ε (cid:28)
1, we assume u ( x ) is constantin each cross section of Ω n and thus is a function of x only. The three-dimensional problem (1)restricted in Ω n is then approximated by the one-dimensional problem d udx = − , < x < L,u = 0 , x = L. By direct calculation we obtain the solution to the above problem as u ( x ) = − x + Cx + 12 L − CL, < x < L, where C is a constant. Solution u satisfies the Robin boundary condition at x = 0. Evaluating u and du/dx at x = 0 yields the Robin condition dudx (0) + 1 L u (0) = L . By the continuity of u and ∂u/∂ν on Γ ε , we obtain the Robin-Neumann boundary value problemin Ω h : ∆ u ε = − , in Ω h ,∂u ε ∂ν = 0 , on ∂ Ω r ,∂u ε ∂ν + αu ε = β, on Γ ε . where α = 1 /L and β = L/
2. Applying Theorem 3.1 we obtain the asymptotic expansion of u ε as u ε ( x ) ≈ | Ω h | Lπε + | Ω h | Mπ ε + L − | Ω h | π | x − x ∗ | , (17)Note that the leading term coincides with (2), which is obtained in [15, 4] using numerical simula-tion. In the rest of this section we shall conduct numerical experiments to verify the asymptotic expansion(17). We shall compare the asymptotic solution (17) with the solution u to the original problem(1) obtained numerically with the finite element method. We shall also confirm the coefficients inthe first two terms in (17).For simplicity we confine ourself to the case when the connection part Γ ε is a disk of radius ε .In this case the constant M in (6) has an explicit and elegant value M = 8 / n is chosen to be a cylinder with radius ε and length L , and whose axis is parallelto the normal of Γ ε . To nondimensionalize our problem, the numerical results are regarded as usingconsistent units throughout this section.For the first experiment, we select the spine head as an unit ball and set ε = 0 . , L = 1 . u on ∂ Ω h , i.e. the whole surface of the spine.7 a) (b) (c) Figure 2: (a) numerical solution u on the surface of the spine; (b) asymptotic solution u ε on thesurface of the spine head; (c) relative error between u ε and u on the surface of the spine head. ε u u r u ε ( u ε − u ) /u ε .Observe that the MFPT u is relatively large in the spine head and decreases monotonically tozero towards the end of the spine neck. This is consistent with our intuition about the underlyingphysical process. In Figure 2(b) we plot the asymptotic solution u ε according to (17) on ∂ Ω h , i.e.the surface of the spine head. Note that u is relatively constant but is smaller for points closer tothe connection part, which is also consistent with the physical intuition. In Figure 2(c) we plot therelative error between u and u ε , computed as ( u ε − u ) /u , on the surface of the spine head. Themaximal relative error is seen to be about 0 . L = 1 . ε decreasesfrom 0 . .
01 in a step size of 0 .
01. For each value of ε , we compare the value of the numericalsolution u to the original problem (1), the numerical solution of the Robin-Neumann model (3),which is denoted by u r , and the asymptotic solution u ε given in (17). In Table 1 we list the valueof u, u r , u ε , as well as the relative error ( u ε − u ) /u , at the center of the spine head. Clearly we havea good match of the solutions and small relative error for all values of ε .The asymptotic solution u ε in (17) at a fixed x can be considered as a quadratic polynomialof 1 /ε with leading coefficients | Ω h | L/π = 4 / | Ω h | M/π ≈ .
13. We now confirm thesecoefficients by fitting the value of u by a quadratic polynomial of 1 /ε in Table 1. The result isplotted in Figure 3. Clearly the coefficients of the fitting polynomial matches well with those of8igure 3: Quadratic fitting of the data in the first two columns of Table 1. L u u r u ε ( u ε − u ) /u L .the asymptotic solution.Another important parameter in the asymptotic solution (17) is L , the length of the spine neck.For the next experiment, we fix the neck radius ε = 0 .
05 and let the neck radius L increases from1 . . .
0. In Table 2 we list the value of u, u r , u ε , as well as the relative error( u ε − u ) /u , at the center of the spine head. Clearly we have a good match of the solutions andsmall relative error for each value of L .Finally we conduct numerical experiments on three different shapes of the spine head, whichmay correspond to different types of spine. The numerical solution u , the asymptotic solution u ε and the relative error between them are shown in Figure 4. The first row shows the results when thespine head is a relatively flat three dimensional domain. The second one is a relatively thin domainand the third one is a non-convex domain. From the relative error ( u ε − u ) /u , which is describedin the third column, we can see that the relative error is small enough to show that asymptoticformula u ε is a good approximation to the MFPT. In this study, we used the Robin-Neumann model to compute the NEP in a three-dimensionaldomain with a long neck, which is often referred as a dendritic spine domain. This is a following uppaper of [14], where the Robin-Neumann model is presented to solve the NEP in a two-dimensional9 a) (b) (c)(d) (e) (f)(g) (h) (i)
Figure 4: The numerical solution u , the asymptotic solution u ε and the relative error ( u ε − u ) /u for three different types of spine. 10nalogue of a dendritic spine domain. In this paper, we derived asymptotic expansion formula forthree-dimensional Robin-Neumann model. Our results demonstrate that this asymptotic expansionformula could approximate the MFPT up to at least second leading order using this model, whichhas not been reported previously.The work of Hyundae Lee was supported by the National Research Foundation of Korea grant(NRF-2015R1D1A1A01059357). The work of Yuliang Wang was supported by the Hong KongRGC grant (No. 12328516) and the NSF of China (No. 11601459). Lemma 6.1. (Single layer potential on an surface) Let Γ ⊂ R be a bounded domain with sup x,z ∈ Γ | x − z | ≤ . The integral operator L : L ∞ (Γ ) (cid:55)→ L ∞ (Γ ) defined by L [ φ ]( x ) = (cid:90) Γ | x − z | φ ( z ) dz is bounded, i.e. (cid:107) S [ φ ] (cid:107) L ∞ (Γ ) ≤ C (cid:107) φ (cid:107) L ∞ (Γ ) , where C is a constant independent of x .Proof. Let B x, ⊂ R denote the disk centered at x with radius 2. For given x ∈ Γ we have (cid:90) Γ | x − z | d z ≤ (cid:90) B x, | x − z | d z = (cid:90) π (cid:90) ρ ρ d ρ d θ = 4 π. Hence (cid:12)(cid:12)(cid:12)(cid:12)(cid:90) Γ | x − z | φ ( z ) d z (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:107) φ (cid:107) L ∞ (Γ ) (cid:90) Γ | x − z | d z ≤ π (cid:107) φ (cid:107) L ∞ (Γ ) , x ∈ Γ . Therefore L is bounded. The value of (cid:90) Γ (cid:90) Γ | x − y | dxdy is π , where Γ is unit disk. Proof.
First, fix point y , draw a circle centered at y with radius r , where the distance between x and y is r . Let s be the distance between 0 and y .Then we have (cid:90) Γ (cid:90) Γ | x − y | dxdy = (cid:90) πs (cid:90) − s πdrds + (cid:90) πs (cid:90) s − s s arccos s + r − sr drds. (18)When r ≤ − s , by simple calculation, we can calculate the first term of (18) as (cid:90) πs (cid:90) − s πdrds = 23 π . ds and dr , the second term of (18) becomes4 π (cid:90) (cid:90) − r s arccos s + r − sr dsdr + 4 π (cid:90) (cid:90) r − s arccos s + r − sr dsdr. (19)By calculations, we get the value of the first term of (19) as + π − √
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