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[18] in stics g Γ ieo h afore- the of time ε stegradient, the is nb reflecting a by in ⊂ ε ∂M particle odnt the denote to oc field. force odenote to g . hnon- th X t hits F 2 MEDETNURSULTANOV,WILLIAMTRAD,ANDLEOTZOU modelled as simply connected two-dimensional domains with small absorbing win- dows on the boundary representing ion channels or target binding sites; the quantity sought is then the mean time for a diffusing ion or receptor to exit through an ion channel or reach a binding site [22, 7, 17]. All of this lead to the narrow escape theory in applied mathematics and computational biology; see [21, 26, 24]. There has been much progress for this problem in the setting of planar domains, and we refer the readers to [7, 17, 26, 1] and references therein for a complete bibliography. An important contribution was made in the planar case by [1] to introduce rigor into the computation of [17]. The use of layered potential in [1] also cast this problem in the mainstream language of elliptic PDE and facilitates some of the approach we use in this article. Few results exists for three dimensional domains in Rn or Riemannian manifolds; see [4, 21, 25, 5, 16] and references therein. The additional difficulties introduced by higher dimension are highlighted in the introduction of [1] and the challenges in geometry are outlined in [25]. In the case when M is a domain in R3 with Euclidean metric and Γε,a is a single small disk absorbing window, [21, 25] gave an expansion for the average of the expected first arrival time, averaged over M, up to an unspecified O(1) term. These results were improved upon in [16], by using geometric microlocal analysis. Namely, the authors derived the bounded term and estimated the remaining term, moreover, they obtain these results for general Reimaniann manifolds. The case when Γε,a is a small elliptic window was also addressed in [21, 25, 16]. We also mention [20], where the author gave a short review of related works (up to 2012). When M is a three dimensional ball with multiple circular absorbing windows on the boundary, an expansion capturing the explicit form of the O(1) correction in terms of the Neumann Green’s function and its regular part was done in [4]. The method of matched asymptotic used there required the explicit computation of the Neumann Green’s function, which is only possible in special geometries with high degrees of symmetry/homogeneity. In these results one does not see the full effects of local geometry. This result was also rigorously proved in [3] but with a better estimate for the error term. Much less has been done for the case of non-zero force field. For instance, all works we metioned above, except [1], deal with the diffusion without a force field, that is F =0. We could find two works concerning this case: [23] and [1]. Both these works consider M being a domain in R2 or R3. In [23], the authors generalize the method of [21, 26, 24] to obtain the leading-order term of the average of the expected first arrival time for two and three dimensional cases. For planar domains, the authors in [1], by using layer potential techniques, derive assymptotic expansion up to O(ε) term. In this paper, we derive all the main terms of the expected value of the first arrival time for Riemaniann manifolds of dimension three in the presence of a force field. The window or target, Γε, is considered to be a small geodesic ellipse of eccentricity 2 + E √1 a and size ε 0 (to be made precise later). To investigate [τΓε X0 = x], we needed− to know→ a singularity structure of the Neumann Green’s function,| which is given by the equation
∆ G(x, z) div (F (z)G(x, z)) = δ (z); g,z − g,z − x (1.1) ∂ G(x, z) F (z) ν G(x, z) = 1 ; νz − · z |∂M − |∂M| −φ(z) ∂M e G(x, z)dh(z)=0. R NARROW ESCAPE PROBLEM 3 is required. For the case F = 0, the authors in [25] highlighted the difficulty in obtaining a comprehensive singularity expansion of G(x, z) x,z∈∂M,x=6 z in a neigh- bourhood of the diagonal x = z when M is a bounded domain| in Rn, but it turns out that even when M is a{ general} Riemannian manifold this question can be treated via pseudo-differential techniques, which is done in [16]. Here, we generalize result of [16] for the case F =0. Knowledge of the singularity structure allows us to derive the mean first arrival6 time of a Brownian particle on a Riemannian manifold with a single absorbing window which is a small geodesic ellipse. Our method extends to multiple windows but we present the single window case to simplify notations. The paper is organized as follows. In Section 2, we introduce the notations. In Section 3, we formulate the problem, state and discuss the main results of this pa- per. Section 4 deals with computing the singular structure of the Neumann Green’s function. Finally, in Section 5 we carry out the asymptotic calculation using the tools we have developed. The appendix characterizes the expected first arrival time E [τΓε,a X0 = x] as the solution of an elliptic mixed boundary value problem. This is classical| in the Euclidean case (see [19]) but we could not find a reference for the general case of a Riemannian manifold with boundary.
2. Preliminaries 2.1. Notation. Throughout this paper, (M,g,∂M) be a compact connected ori- entable Riemannian manifold with non-empty smooth boundary. The corresponding volume and geodesic distance are denoted by dg( ) and dg( , ), respectively. By M we denote the volume of M. · · · | | Let ι∂M : ∂M ֒ M is the trivial embedding of the boundary ∂M into M. This → ∗ allows us to define the following metric h := ι∂M g be the metric on the boundary ∂M. Set d ( ) and d ( , ) to be the volume and the geodesic distance on the boundary h · h · · given by metric h. By ∂M we denote the volume of ∂M with respect to dh. For x ∂M, let E| (x),| E (x) T ∂M be the unit eigenvectors of the shape ∈ 1 2 ∈ x operator at x corresponding respectively to the principal curvatures λ1(x), λ2(x). We will drop the dependence in x from our notation when there is no ambiguity. We ♭ ♭ ♭ choose E1 and E2 such that E1 E2 ν is a positive multiple of the volume form ∧ ∧ ♭ ♯ dg (see p.26 of [12] for the “musical isomorphism” notation of and ). Here we use ν to denote the outward pointing normal vector field. By H(x), we denote the mean curvature of ∂M at x. We also set II (V ) := II (V,V ), V T ∂M x x ∈ x to be the scalar second fundamental quadratic form (see pages 235 and 381 of [12] for definitions). Note that, in defining II and the shape operator, we will follow geometry literature (e.g.[12]) and use the inward pointing normal so that the sphere embedded in R3 would have positive mean curvature in our convention.
2.2. Boundary normal coordinates. In this work, we will often use the boundary normal coordinates. Therefore, we briefly recall its construction. For a fixed x ∂M, 0 ∈ we will denote by Bh(ρ; x0) ∂M the geodesic disk of radius ρ > 0 (with respect ⊂ 2 to the metric h) centered at x0 and Dρ to be the Euclidean disk in R of radius ρ centered at the origin. In what follows ρ will always be smaller than the injectivity radius of (∂M,h). Letting t =(t , t , t ) R3, we will construct a coordinate system 1 2 3 ∈ x(t; x0) by the following procedure: 4 MEDETNURSULTANOV,WILLIAMTRAD,ANDLEOTZOU
Write t R3 near the origin as t =(t′, t ) for t′ =(t , t ) D . Define first ∈ 3 1 2 ∈ ρ ′ x((t , 0); x0) := expx0;h(t1E1 + t2E2), where expx0;h(V ) denotes the time 1 map of h-geodesics with initial point x0 and initial velocity V T ∂M. The coordinate t′ D x((t′, 0); x ) is then an h- ∈ x0 ∈ ρ 7→ 0 geodesic coordinate system for a neighborhood of x0 on the boundary surface ∂M. We can extend this to become a coordinate system for points in M near x0 so that t x(t; x0) is a boundary normal coordinate system with t3 > 0 in M as the boundary7→ defining function. Readers wishing to know more about boundary normal coordinates can refer to [11] for a brief recollection of the basic properties we use here and Prop 5.26 of [12] for a detailed construction. ′ ′ For convenience we will write x(t ; x0) in place of x((t , 0); x0). The boundary coordinate system t x(t; x0) has the advantage that the metric tensor g can be expressed as 7→
3 2 ′ 2 (2.1) gj,k(t)dtjdtk = hα,β(t , t3)dtαdtβ + dt3, j,kX=1 α,βX=1 ′ ′ where hα,β(t , 0) = hα,β(t ) is the expression of the boundary metric h in the h-geodesic ′ coordinate system x(t ; x0). We will also use the rescaled version of this coordinate system. For ε> 0 sufficiently small we define the (rescaled) h-geodesic coordinate by the following map (2.2) xε( ; x ): t′ =(t , t ) D x(εt′; x ) B (ε; x ), · 0 1 2 ∈ 7→ 0 ∈ h 0 where D is the unit disk in R2.
3. The main results Here we state and disscus the main results of this paper. We begin with formulating the problem. Let (Xt, Px) be the Brownian motion on M starting at x, generated by the differential operator u ∆ u + g(F, u), → g ∇g ∗ where F is a force field given by potential φ, that is F = gφ. For x ∂M and ε> 0, let Γ ∂M be a small geodesic ellipse define as ∇ ∈ ε,a ⊂ ∗ ∗ 2 −2 2 (3.1) Γ := exp ∗ (εt E (x )+ εt E (x )) t + a t 1 . ε,a { x ;h 1 1 2 2 | 1 2 ≤ }
Denote by τΓε,a the first time the Brownian motion Xt hits Γε,a, that is τ := inf t 0 : X Γ . Γε,a { ≥ t ∈ ε,a} We aim to investigate the mean sojourn time, that is the expected value u (x) := E[τ X = x], ε,a Γε,a | 0 and its avarege expected value over M
M −1 E[τ X = x]d (x). | | Γε,a | 0 g ZM Namely, we want to derive asymptotic expansion for these quantities as ε 0. → NARROW ESCAPE PROBLEM 5
In the Appendix, we show that the mean sojourn time, uε,a, satisfies the following elliptic mix boundary value problem ∆ u + g(F, u )= 1; g ε,a ∇g ε,a − (3.2) uε,a = 0; |Γε,a ∂ν uε,a =0. |∂M\Γε,a ′ o Let G( , ) (M M) solve (1.1). For x M , Greens formula used in con- junction with· · ∈ (1.1) D and× (3.2) yields the following∈ integral representation for the mean sojourn time uε,a
(3.3) u (x)= (x)+ G(x, z)∂ u (z)d (z)+ C ε,a G νz ε,a h ε,a Z∂M where 1 C := u (z)d (z), (x) := G(x, z)d (z). ε,a ∂M ε,a h G h | | Z∂M ZM Where satisfies the following boundary value problem G ∆ (x)+ g(F (x), (x)) = 1; gG ∇gG − (3.4) ∂ (x) = Φ(x) ; ν G |x∈∂M − |∂M| (x)dh(x)=0. ∂M G Where Φ is a weighted volume defined in Theorem 3.2. R In order to derive an asymptotic expansion for uε,a, we need to derive asymptotics for Cε,a. The first step within said program is to exploit the vanishing Dirichlet boundary condition of Γ . In doing so, we restrict x M o to x Γ which yields ε,a ∈ ∈ ε,a
0= (x) + G(x, z)∂ν uε,a(z)dh(z) + Cε,a. G |Γε,a z Z∂M Γε,a
What follows, is the definition of the restricted Neumann Gre ens function, defined as the Schwartz kernel to the operator
G : f G(x, y)f(y)d (y) . ∂M 7→ h Z∂M ∂M ∞ ∞ k Here G∂M : C (∂M) C (∂M) can be extended to H ( ∂M). Using a parametrix construction, in conjunction→ with Fourier techniques and h omogeneous distributions, we can show that the kernel G∂M attains the following form for x, y ∂M near the diagonal. ∈ Proposition 3.1. There exists an open neighbourhood of the diagonal Diag := (x, y) ∂M ∂M x = y { ∈ × | } such that in this neighbourhood, the singularity structure of G∂M (x, y) is given by: 1 1 G (x, y)= d (x, y)−1 (H(x)+ ∂ φ(x)) log d (y, x) ∂M 2π g − 4π ν h 1 exp−1 (y) exp−1 (y) (3.5) + II x;h II ∗ x;h 16π x exp−1 (y) − x exp−1 (y) | x;h |h ! | x;h |h !! 1 exp (y) h F k(x), x;h + R(x, y), − 4π x exp (y) | x;h |h 6 MEDETNURSULTANOV,WILLIAMTRAD,ANDLEOTZOU where F k is the tangential part of the force field F and R( , ) C0,µ(∂M ∂M), for all µ < 1, is called the regular part of the Green’s function· · and∈ is the× Hodge-star operator (i.e. rotation by π/2 on the surface ∂M). ∗
We will use the formula in Proposition 3.1 to derive the mean first arrival time of a Brownian particle on a Riemannian manifold with a single absorbing window which is a small geodesic ellipse. As mentioned earlier, our method extends to multiple windows but we present the single window case to simplify notations. We first state the result when the window is a geodesic disk of the boundary ∂M around a fixed point since the statement is cleaner:
Theorem 3.2. Let (M,g,∂M) be a smooth compact Riemannian manifold of dimen- ∗ sion three with boundary. Fix x ∂M and let Γε be a boundary geodesic ball centered at x∗ of geodesic radius ε> 0. ∈ i) For each x / Γ , ∈ ε E[τ X = x]= C + (x) Φ(x∗)G(x∗, x)+ r (x), Γε | 0 ε G − ε with rε Ck(K) Ck,Kε for any integer k and compact set K M which does not containk xk∗. The≤ function is the solution of (3.4). The constant⊂ C is given by G ε Φ(x∗) (H(x∗)+ ∂ φ(x∗))Φ(x∗) C = ν log ε + R(x∗, x∗)Φ(x∗) (x∗) ε 4ε − 4π − G (H(x∗)+ ∂ φ(x∗))Φ(x∗) 3 ν 2 log 2 + O(ε log ε), − 4π − 2 where R(x∗, x∗) is the evaluation at (x, y)=(x∗, x∗) of the kernel R(x, y) in Propo- sition 3.1 and
φ(z)−φ(x) Φ(x) := e dg(z). ZM ii) One has that the integral of E[τ X = x] over M satisfies Γε,a | 0
E[τ X = x]d (x)= C M + (x)d (x) Φ(x∗) G(x, x∗)d (x). Γε,a | 0 g ε| | G g − g ZM ZM ZM Theorem 3.2 does not realize the full power of Proposition 3.1 as it does not see the non-homogeneity of the local geometry at x∗ (only the mean curvature H(x∗) shows up). This is due to the fact that we are looking at windows which are geodesic balls. If we replace geodesic balls with geodesic ellipses, we see that the second fundamental E form term in (3.5) contributes to a term in [τΓε,a X0 = x] which is the difference of principal curvatures. |
Theorem 3.3. Let (M,g,∂M) be a smooth Riemannian manifold of dimension three ∗ with boundary. Fix x ∂M and let Γε,a be a boundary geodesic ellipse given by (3.1) with ε> 0. ∈ i) For each x M Γ , ∈ \ ε,a E[τ X = x]= C + (x) Φ(x∗)G(x∗, x)+ r (x), Γε | 0 ε G − ε NARROW ESCAPE PROBLEM 7 with rε Ck(K) Ck,Kε for any integer k and compact set K M which does not containk xk∗. The≤ function is the solution of (3.4). The constant⊂ C is given by G ε K Φ(x∗) (H(x∗)+ ∂ φ(x∗))Φ(x∗) C = a ν log ε ε,a 4π2aε − 4π + R(x∗, x∗)Φ(x∗) (x∗) − G ∗ ∗ ∗ 2 2 2 1/2 (H(x )+ ∂ν φ(x ))Φ(x ) 1 log ((t1 s1) + a (t2 s2) ) ′ ′ 3 ′ 2 1/2 − ′ 2 1/2− dt ds − 16π D (1 s ) D (1 t ) ∗ ∗ ∗ Z −| | Z 2 −|2 | 2 (λ1(x ) λ2(x ))Φ(x ) 1 (t1 s1) a (t2 s2) 1 ′ ′ + − 3 ′ 2 1/2 − 2 − 2 − 2 ′ 2 1/2 dt ds 64π D (1 s ) D (t s ) + a (t s ) (1 t ) Z −| | Z 1 − 1 2 − 2 −| | + O(ε log ε), where R(x∗, x∗) is the evaluation at (x, y)=(x∗, x∗) of the kernel R(x, y) in Propo- sition 3.1, D is the two dimensional unit disk centered at the origin, and π 2π sin2 θ −1/2 Φ(x) := eφ(z)−φ(x)d (z), K := cos2 θ + dθ g a 2 a2 ZM Z0 ii) One has that the integral of E[τ X = x] over M satisfies Γε,a | 0 E[τ X = x]d (x)= C M + (x)d (x) Φ(x∗) G(x, x∗)d (x). Γε,a | 0 g ε| | G g − g ZM ZM ZM 4. The Neumann Green’s Function Here, we investigate the Neumann Green’s function. Namely, we derive its singular structure on the boundary near the diagonal. By Neumann Green’s function, G( , ) ′(M M), we mean the solution to the following equation · · ∈ D × ∆ G(x, z) div (F (z)G(x, z)) = δ (z); g,z − g,z − x (4.1) ∂ G(x, z) F (z) ν G(x, z) = 1 ; νz − · z |∂M − |∂M| −φ(z) ∂M e G(x, z)dh(z)=0. By using Green’s identityR to e−φ(y)G(z,y)= e−φ(x)G(z, x) (∆ G(y, x) div (F (x)G(y, x))) d (x), − g − g g ZM we obtain 1 e−φ(y)G(z,y) e−φ(z)G(y, z)= e−φ(x)G(y, x)d (x) e−φ(x)G(z, x)d (x) . − ∂M h − h | | Z∂M Z∂M Therefore, by the last condition in (4.1), we obtain (4.2) G(z,y)= eφ(y)−φ(z)G(y, z). Therefore, we can check
φ(y)−φ(z) G(z,y)dh(z)= e G(y, z)dh(z)=0, Z∂M Z∂M eφ(y)−φ(z) ∂ G(z,y)= eφ(y)−φ(z) ∂ G(y, z) F (z) ν G(y, z) = νz νz − · z |∂M − ∂M | | 8 MEDETNURSULTANOV,WILLIAMTRAD,ANDLEOTZOU for z ∂M, and finally ∈ ∆ G(z,y)+g(F (z), G(z,y)) = eφ(y)−φ(x) g( φ(z), φ(z))G(y, z) g,z ∇g,z ∇g,z ∇g,z ∆ φ(z)G(y, z) 2g( φ(z), G(y, z))+∆ G(y, z) − g,z − ∇g,z ∇g,z g,z g(F (z), φ(z))G(y, z)+ g(F (z), G(y, z)) . − ∇g,z ∇g,z Therefore, we conclude that G(z,y) satisfies the differential equation (with respect to the first variable)
∆g,zG(z,y)+ g(F (z), g,zG(z,y)) = δy(z); ∇φ(y)−φ(z) − (4.3) ∂ G(z,y) = e ; νz |z∈∂M − |∂M| ∂M G(z, x)dh(z)=0. This indicates thatR solves the problem (3.4). Now, we let f C∞(∂M). Using this G ∈ f, we introduce uf , the solution to the following auxiliary problem (4.4) ∆ u (z)+ g (F (z), u (z))=0, u = f C∞(∂M). g f z ∇g f f |z∈∂M ∈ By using Green’s identity and the Divergence form theorem to
(4.5) u (x)= u (z) (∆ G(x, z) div (F (z)G(x, z))) d (z), f − f g,z − g,z g ZM we compute
u (x)= ∆ u (z)G(x, z)d (z) (u (z)∂ G(x, z) ∂ u (z)G(x, z)) d (z) f − g f g − f νz − νz f h ZM Z∂M G(x, z)g (F (z), u (z))d (z)+ u (z)G(x, z)F (z) ν d (z). − z ∇g f g f · z h ZM Z∂M Since functions uf , G(x, z) satisfy (4.4), (4.1) respectively, we conclude that 1 u (x)= G(x, z)∂ u (z)d (z)+ f(z)d (z). f νz f h ∂M h Z∂M | | Z∂M Restricting x to ∂M, we have that 1 (4.6) f(x) = G(x, z)∂ u (z)d (z) + f(z)d (z) |∂M νz f h ∂M h Z∂M ∂M Z∂M | | Let Λ Ψ1 (∂M) be the Dirichlet-to-Neumann map associated to the boundary F ∈ cl value problem (4.4) and G∂M (x, y) be the Schwartz kernel of the operator
f f(y)G(x, y)d (y) → h Z∂M ∂M which takes C∞(∂M) C∞(∂M). Then we can rewrite (4.6) in the following way → f(x)= G∂M ΛF f + P f where P is a smoothing operator, that is P Ψ−∞(∂M). In operator form this is ∈ (4.7) I = G∂M ΛF + P.
Since ΛF is an elliptic pseudo-differential operator, we can construct G∂M via a stan- dard left parametrix construction. NARROW ESCAPE PROBLEM 9
4.1. Symbolic Expansion for the symbol of the Dirichlet-to-Neumann map. We compute here the first two terms of the asymptotic expansion for the symbol of the Dirichlet-to-Neumann map. We will use this to obtain the corresponding terms for the symbol of G∂M . We will follow [13] and adapt some of their results for the drift case. In boundary normal coordinates, we decompose our differential operator in the following way 2 (4.8) ∆ g (F, )= D 3 + iE(x)D 3 + Q(x, D ′ ) − g − x ∇g· x x x where e e 1 αβ n E(x) := h (x)∂ n h (x) F (x) −2 x αβ − Xα,β e αβ Q(x, Dx′ ) := h (x)Dxα Dxβ Xα,β e 1 αβ αβ β α i h (x)∂ α log δ(x)+ ∂ α h (x) D β + iF (x)h (x)D α . − 2 x x x β x Xα,β We will need the following modification of Proposition 1.1 in [13]
′ 1 Proposition 4.1. There exists a pseudo-differential operator AF (x, Dx ) Ψcl(∂M) which depends smoothly on x3 such that ∈
∆ g (F, )= D 3 + iE(x) iA (x, D ′ ) (D 3 + iA (x, D ′ )) , − g − x ∇g· x − F x x F x modulo a smoothing operator. e
Proof. We construct an asymptotic series for the symbol of AF (x, Dx′ ) using a ho- mogeneity argument. The proposition can be re-stated as the construction of some −∞ pseudo-differential operator AF (x, Dx′ ) modulo Ψ which satisfies the following statement
0 = ∆ + g (F, )+ D 3 + iE(x) iA (x, D ′ ) (D 3 + iA (x, D ′ )) g x ∇g· x − F x x F x Due to decomposition (4.8), the probleme becomes the construction of a classical, first order pseudo-differential operator AF (x, Dx′ ) which satisfies the following operator equation 2 A Q + i[D 3 , A ] EA =0 F − x F − F modulo a smoothing operator. Reduction of the above operatore equation to thee pseudo-differential symbol calculus yields the following equation (modulo S−∞).
1 µ µ (4.9) ∂ aD a q + ∂ 3 a Ea =0 µ! ξ x − x − |Xµ|≥0 e e where a is the full symbol of AX and q˜ is the full symbol of Q˜ given by
′ αβ 1 αβ αβ α β q(x, ξ )= h (x)ξ ξ i h (x)∂ α log δ(x)+ ∂ α h (x) F (x)h (x) ξ α β − 2 x x − α β α,β α,β X ′ ′ X e =: q2(x, ξ )+ q1(x, ξ )
e e 10 MEDETNURSULTANOV,WILLIAMTRAD,ANDLEOTZOU
Let us write a(x, ξ′) a (x, ξ′) ∼ j j≤1 X j ∗ ′ where aj S1,0(T ∂M), and is homogeneous of degree j in ξ . Collecting terms which are homogeneous∈ of degree 2 in (4.9) yields the following a2 q =0 = a = q 1 − 2 ⇒ 1 ± 2 For consistency, we will choose a1 := q2. Next,p we will collect the terms which are homogeneous of degree 1 in (4.9)e as− follows e p 2a a + ∂ a D α a e q + ∂ 3 a Ea =0 0 1 ξα 1 x 1 − 1 x 1 − 1 α X Solving for a0, we have that e e 1 a = ∂ a D α a q + ∂ 3 a Ea 0 −2a ξα 1 x 1 − 1 x 1 − 1 1 α ! X 1 e e = ∂ q D α q q ∂ 3 q + E q ξα 2 x 2 − 1 − x 2 2 2 q2 α ! X p p p p We can apply the samep recursive argumente indefinitelye e for alle degreese e of homogeneity 1 j in order to obtaine a for every j 1. For our purposes, the construction of − 1−j ≥ a0 is sufficient. This completes the proof. Proposition 1.2 of [13] yields the required calculation for the Dirichlet to Neumann operator, given by Λ f = δ1/2A fdx1 dx2 mod Ω2(∂M) F F ∧ ∂M Via a one-to-one correspondence, we can associate to this n 1 differential form a symbol, denoted by b(x, ξ′) S1 (T ∗∂M) given by − ∈ cl (4.10) b(x, ξ′)= δ1/2a(x, ξ′). 4.2. Explicit Calculation for the Neumann Green’s function. In this subsec- tion, we will extract the singular part of G∂M on the diagonal. Since ΛF is elliptic and (4.7), we construct G∂M as a standard left parametrix ′ −1 of order 1. Let p(x, ξ ) Scl (∂M) be its symbol with the following asymptotic expansion− ∈ p(x, ξ′) p (x, ξ′), ∼ −j j≥1 X where p S−j(T ∗∂M) for each j 1. From (4.7), we deduce that −j ∈ ≥ 1=(p#b)(x, ξ′)+ S−∞(T ∗∂M)
1 µ µ 1/2 −∞ ∗ = ∂ ′ pD δ a + S (T ∂M), µ! ξ x |Xµ|≥0 where a is the symbol constructed in Section 4.1. By matching the terms with the same orders, we obtain ′ ′ χ(ξ ) p−1(x, ξ )= 1/2 ′ δ q2(x, ξ ) p e NARROW ESCAPE PROBLEM 11
′ ′ ′ ′ ′ χ(ξ ) χ(ξ )a0(x, ξ ) µ χ(ξ ) µ 1/2 ′ p−2(x, ξ )= ∂ ′ D δ q2(x, ξ ) . 1/2 ′ ′ − ξ ′ x δ q2(x, ξ ) q2(x, ξ ) |µ|=1 q2(x, ξ ) X p Here, χ is a smoothp cutoff function,p non-zero outside ofp some sufficiently largee neigh- e e e bourhood of the origin. The choice of terms p for j 3 can be done via standard, −j ≥ iterative parametrix arguments which were used to obtain p−1 and p−2. For the sake of brevity, such computations shall be omitted and are unnecessary for our purposes. It should be noticed that for x = x∗, the origin of our geodesic disk, as a result of boundary normal co-ordinates, the symbol terms are reduce to χ(ξ′) p (x∗,ξ′)= . −1 ξ′ | | In addition, we have that
′ ′ ∗ ′ ′ ∗ ′ χ(ξ ) χ(ξ )a0(x ,ξ ) µ χ(ξ ) µ ′ p−2(x ,ξ )= ∂ ′ D q2(x, ξ ) . ′ ′ ξ ′ x ξ ξ − q2(x, ξ ) |µ|=1 ∗ | | | | X p x=x 1/2 p e Notice that in p−2, the δ term is peeled off by thee product rule and vanishes as a result of ∂xk hij =0 at the origin. The calculation of the principle symbol for G∂M as well as the the next highest order term yield the following asymptotic for the kernel centred at x∗, evaluated at x B (ρ; x∗) ∂M ∈ h ⊂ ∗ ∗ ′ G∂M (x , x) = Φ G∂M (0, t )
1 −iξ′·t′ ∗ ′ ′ 1 −iξ′·t′ ∗ ′ ′ −3 = 2 e p−1(x ,ξ )dξ + 2 e p−2(x ,ξ )dξ +Ψcl (∂M). 4π R2 4π R2 Z Z The first term evaluates to ′ 1 −iξ′·t′ ∗ ′ ′ 1 −iξ′·t′ χ(ξ ) ′ 2 e p−1(x ,ξ )dξ = 2 1/2 e ′ dξ . 4π R2 4π δ R2 ξ Z Z | | Furthermore, we can split the above integral into a singular and regular part as follows
1 −iξ′·t′ 1 ′ 2 e ′ dξ + Ireg. 4π R2 ξ Z | | The regular part is of no interest to us and will be lost in the error as this computation is done in order to isolate the most prevalent singularities occurring in G∂M . This idea can be carried forth in the computation for the second integral. We have that
1 −iξ′·t′ 1 ′ ′ −1 ′ 1 ′ −1 2 e ′ dξ = ( ξ )(t )= t . 4π R2 ξ F | | 2π | | Z | | Next, we use the previous arguments and proceed to split the second term up as follows ′ 1 −iξ′·t′ a0(0,ξ ) ′ 2 e ′ 2 dξ 4π R2 ξ Z | | ′ ′ 1 −iξ ·t 1 µ ′ −1/2 µ ′ ′ e ∂ ′ (q (x, ξ )) D q (x, ξ ) dξ . 2 ′ ξ 2 x=x∗ x 2 ∗ − 4π R2 ξ x=x |µ|=1 Z | | X p e e 12 MEDETNURSULTANOV,WILLIAMTRAD,ANDLEOTZOU
′ αβ We recall that we have that q2(x, ξ ) = h (x)ξαξβ. Thus, we have the following identity e µ αβ µ αβ Dx h (x) Dx h (x)ξαξβ = ′ . 2q1(x, ξ ) q In boundary normal co-ordinates, centred at x = x∗, this identity evaluates to 0. Thus, we have that the second integral is vanishing. We are now left to compute the following integral
′ 1 −iξ′·t′ 1 ′ α 3 I(t ) := 2 e ′ 3 ∂ξα q2Dx q2 q1 ∂x q2 + E q2 dξ . R2 8π ξ − − ! ∗ Z | | α x=x X p p p p ′ e e ′ e α e e e Furthermore, since Dxα q2(x, ξ )=0 and q1(x, ξ )= iX (x)ξα in boundary normal co-ordinates, we have that I is given by p e e ′ 1 −iξ′·t′ 1 ′ α ′ 3 αβ I(t )= 2 e ′ 3 E ξ ∂x h (x)ξαξβ iF (x)ξα dξ 8π R2 ξ | | − − ∗ Z | | q x=x ∗ αβ E(x ) ′ ′ 1 e 1 ′ ′ ∂ 3 h (x)ξ ξ −iξ ·t ′ −iξ ·t x α β ′ = 2 e ′ 2 dξ 2 e ′ 3 dξ 8π R2 ξ − 8π R2 ξ p ∗ e Z | | Z | | x=x ′ ′ i α ∗ −iξ ·t ξα ′ 2 F (x ) e ′ 3 dξ . − 8π R2 ξ Z | | We calculate each term in the integral in order of increasing difficulty. Using propo- sition 8.17 [Lee] we can ascertain the following
∗ 1 αβ 3 ∗ ∗ 3 ∗ E(x )= h (x)∂x3 hαβ(x) F (x )=2H(x ) F (x ). −2 − − α,β x=x∗ X e ∗ ∗ Where H(x ) denotes the mean curvature of (M,g,∂M) at x . Thus, we have that ∗ ∗ 3 ∗ ∗ 3 ∗ E(x ) −iξ′·t′ 1 ′ 2H(x ) F (x ) ′ −2 2H(x ) F (x ) ′ 2 e ′ 2 dξ = − ( ξ )= − log t . 8π R2 ξ 2 F | | − 4π | | e Z | | Next, we calculate the third term by making the following observation ξ ∂ ξ′ −1 = α . ξα | | − ξ′ 3 | | We can re-write the third term as follows
i α ∗ −iξ′·t′ ξα ′ i α ∗ −iξ′·t′ ′ −1 ′ 2 F (x ) e ′ 3 dξ = 2 F (x ) e ∂ξα ξ dξ −8π R2 ξ 8π R2 | | α α X Z | | X Z 1 α ∗ −iξ′·t′ ′ −1 ′ = 2 tαF (x ) e ξ dξ 8π R2 | | α X Z 1 = t F α(x∗) ( ξ′ −1)(t′) 2 α F | | α X t = F α(x∗) α . 4π t′ α X | | NARROW ESCAPE PROBLEM 13
Lastly, we compute the second term. It should be noted that via rotation with basis ♭ ♭ ♭ E1 E2 ν , we find that the scalar second fundamental form is diagonalised in our co-ordinate∧ ∧ system. Thus, we have the following
′ 1 −iξ′·t′ ∂x3 ξ h ′ 2 e |′ 3| dξ − 8π R2 ξ Z | | ′ ′ 11 12 22 1 −iξ ·t (∂x3 h (x)+2∂x3 h (x)+ ∂x3 h (x)) x=x∗ ′ = 2 e ′ 4 | dξ −8π R2 ξ Z ∗ 2 ∗ 2| | 1 −iξ′·t′ λ1(x )ξ1 + λ2(x )ξ2 ′ = 2 e ′ 4 dξ −8π R2 ξ Z | | 1 λ (x∗)+ λ (x∗) π λ (x∗)t2 + λ (x∗)t2 λ (x∗)t2 + λ (x∗)t2 = 1 2 π log t′ 2 1 1 2 1 1 2 2 4π2 2 | | − 4 t′ 2 − t′ 2 | | | |
′ Here λ1,λ2 denote the associated principle curvatures. Therefore, we have that I(t ) is given by
2H(x∗) F 3(x∗) I(t′)= − log t′ − 4π | | 1 π λ (x∗)t2 + λ (x∗)t2 λ (x∗)t2 + λ (x∗)t2 + H(x∗)π log t′ 2 1 1 2 1 1 2 2 4π2 | | − 4 t′ 2 − t′ 2 | | | | 1 F α(x∗)t + α 4π t′ α X | |
Therefore, we have that
∗ 3 ∗ −1 H(x ) ∗ F (x ) ∗ I(exp ∗ (x)) = log d (x , x)+ log d (x , x) x − 4π h 4π h −1 −1 1 exp ∗ (x) ⋆ exp ∗ (x) ∗ x ∗ x + IIx −1 IIx −1 16π exp ∗ (x) − exp ∗ (x) x h x h | −|1 | | 1 exp ∗ (x) ∗ k ∗ x + hx F (x ), −1 . 4π exp ∗ (x) | x |h
This yields the following asymptotic for G∂M
∗ 3 1 H(x ) X ∗ G (x∗, x)= d (x∗, x)−1 log d (x∗, x)+ x log d (x∗, x) ∂M 2π h − 4π h 4π h −1 −1 1 exp ∗ (x) ⋆ exp ∗ (x) ∗ x ∗ x + IIx −1 IIx −1 16π exp ∗ (x) − exp ∗ (x) x h x h | −|1 | | 1 exp ∗ (x) ∗ k ∗ x −3 + hx F (x ), −1 +Ψ (∂M). 4π exp ∗ (x) | x |h 14 MEDETNURSULTANOV,WILLIAMTRAD,ANDLEOTZOU
We can make a further refinement on the above series by invoking Corollary 2.5 from [16] in order to write the following 1 H(x∗) F 3(x∗) G (x∗, x)= d (x∗, x)−1 log d (x∗, x)+ log d (x∗, x) ∂M 2π g − 4π h 4π h −1 −1 1 exp ∗ (x) ⋆ exp ∗ (x) ∗ x ∗ x + IIx −1 IIx −1 16π exp ∗ (x) − exp ∗ (x) x h x h | −|1 | | 1 exp ∗ (x) ∗ k ∗ x −3 + hx F (x ), −1 +Ψ (∂M). 4π exp ∗ (x) | x |h The above work yields an expression for G∂M centred at the origin of the window ∗ x Γε,a. We however need an expansion for G∂M (x, y) for x, y close in Γε,a. In particular,∈ we need an expression given by (2.2). In order to do so, we simply invoke the following proposition, which was proven in [16].
Proposition 4.2. Let x0 ∂M and λ1(x0) and λ2(x0) be the eigenvalues of the shape ∈ ε ′ ε ′ ′ ′ ′ operator at x0. Assume that x = x (s ; x0), y = x (t ; x0), r = s t and t = s+rω. Then, for ε> 0 sufficiently small, we have that | − | −1 −1 −1 −1 1 ′ dg(x, y) = ε r + εr A (ε,s ,r,ω),
−1 −1 −1 −1 2 ′ dh(x, y) = ε r + εr A (ε,s ,r,ω),
exp−1 y exp−1 y (s t )2 (s t )2 II x;h , x;h = λ (x ) 1 − 1 + λ (x ) 2 − 2 + εR1(t′,ω,r), x exp−1 y exp−1 y 1 0 r2 2 0 r2 ε | x;h |h | x,h |h ! exp−1 y exp−1 y (s t )2 (s t )2 II x;h , x;h = λ (x ) 1 − 1 + λ (x ) 2 − 2 +εR2(t′,ω,r), x ∗ exp−1 y ∗ exp−1 y 2 0 r2 1 0 r2 ε | x;h |h | x,h |h ! exp (y) F (t′)(s t )+ F (t′)(s t ) h F k(x), x;h = 1 1 − 1 2 2 − 2 + εR3(t′,ω,r), x exp (y) r ε | x;h |h 1 2 3 D R 1 1 2 3 where A , A , and A are smooth in [0,ε0] S and Rε, Rε, Rε are smooth in D S2 [0, ρ] with derivatives of all orders× × uniformly× bounded in ε. × × Proof. The first two statements were proved in Corollaries 2.6 and 2.3 of [16], respec- tively. The next two results were proved in Corollary 2.9 of [16]. The last one follows from Lemma 2.8.
5. Proof of Theorems 3.2 and 3.3 In this section we give a proof for Theorems 3.2 and 3.3. We recall that the mean sojourn time, uε,a, satisfies the elliptic mixed boundary value problem (3.2), this is proved in the Appendix. Therefore, by using Green’s identity, we show that uε,a time satisfies the integral equation
(5.1) u (x)= (x)+ G(x, z)∂ u (z)d (z)+ C , ε,a G νz ε,a h ε,a Z∂M where x M 0, G is the Neumann Green function we disscused in Section 4 and ∈ 1 C := u (z)d (z), (x) := G(x, z)d (z). ε,a ∂M ε,a h G h | | Z∂M ZM NARROW ESCAPE PROBLEM 15
From (4.3), it follows that sutisfies G ∆ (x)+ g(F (x), (x)) = 1; gG ∇gG − (5.2) ∂ (x) = Φ(x) ; ν G |x∈∂M − |∂M| (x)dh(x)=0. ∂M G By taking the trace of theR integral equation (5.1) to x Γε,a, we obtain ∈ 0= (x)+ G (x, z)∂ u (z)d (z)+ C . G ∂M νz ε,a h ε,a Z∂M Therefore, Proposition 3.1 gives 1 (x) C = d (x, y)−1∂ u (y)d (y) −G − ε,a 2π g ν ε,a h ZΓε,a H(x) ∂ φ(x) − ν log d (x, y)∂ u (y)d (y) − 4π h ν ε,a h ZΓε,a 1 exp−1(y) ⋆ exp−1(y) (5.3) + II x II x ∂ u (y)d (y) 16π x exp−1(y) − x exp−1(y) ν ε,a h ZΓε,a | x |h | x |h 1 exp (y) + h F k(x), x;h ∂ u (y)d (y) 4π x exp (y) ν ε,a h | x;h |h
+ R(x, y)∂νuε,a(y)dh(y). ZΓε,a Since F = φ, the fact that u satisfies (3.2) implies ∇g ε,a div (eφ u )= eφ(∆ u + F u )= eφ. g ∇g ε,a g ε,a ·∇g ε,a − By the divergence form theorem, we know that
div (eφ u )(z)d (z)= eφ∂ u (z)d (z). g ∇g ε,a g ν ε,a h ZM Z∂M Thereofore, by integrating the penultimate equation, we derive the following compat- ibility condition
(5.4) eφ∂ u (z)d (z)= eφ(z)d (z). ν ε,a h − g Z∂M ZM We will use the coordinate system given by (5.5) D (s ,s ) xε(s , as ; x∗) Γ , ∋ 1 2 7→ 1 2 ∈ ε,a ε ∗ where x ( ; x ): a Γε,a is the coordinate defined in Section (2). To simplify notation we· will dropE → the x∗ in the notation and denote xε( ; x∗) by simply xε( ). Note that in these coordinates the volume form for ∂M is· given by · (5.6) d = aε2(1 + ε2Q (s′))ds ds , s′ D h ε 1 ∧ 2 ∈ ′ for some smooth function Qε(s ) whose derivatives of all orders are bounded uniformly in ε. We denote ′ ε (5.7) ψε(s ) := ∂ν uε(x (s1, as2)). Then, in this coordinate system, the compatibility condition become
φ(xε(s′)) ′ 2 ′ ′ 1 φ(z) (5.8) e ψε(s )(1 + ε Qε(s ))ds = 2 e dg(z). D −aε Z ZM 16 MEDETNURSULTANOV,WILLIAMTRAD,ANDLEOTZOU
We can re-write this as follow
φ(x∗) φ(xε(s′)) φ(x∗) 2 φ(x∗) ′ ′ ′ 1 φ(z) e ψ(x)dx+ (e e )+ ε e Qε(s ) ψε(s )ds = 2 e dg(z). D D − −aε M Z Z h i Z Since φ is smooth, we conclude that ∗ Φ(x ) ˜ ′ ′ ′ (5.9) ψ(x)dx = 2 + ε Qε(s )ψε(s )ds , D − aε D Z Z ′ where Φ is the function defined in Theorem (3.2) and Q˜ε(s ) is some smooth function whose derivatives of all orders are bounded uniformly in ε. Next, we re-write (5.3) in this coordinate system given by (5.5). To do this, let us first introduce the following operators. Consider ′ f(s ) ′ (5.10) Laf = a 1/2 ds D ((t s )2 + a2(t s )2) Z 1 − 1 2 − 2 acting on functions of the disk D. By [21] we have that (5.11) L K −1(1 t′ 2)−1/2 =1, a a −| | on D where π 2π 1 Ka = 1/2 dθ. 2 0 2 sin2 θ Z cos θ + a2
1/2 ∗ By (4.4) in [16], this is the unique solution in H (D) to Lau =1. Next we denote
′ 2 2 2 1/2 ′ ′ Rlog,af(t ) := a log (t1 s1) + a (t2 s2) f(s )ds , D − − Z 2 2 2 ′ (t1 s1) a (t2 s2) ′ ′ R∞,af(t ) := a − 2 − 2 − 2 f(s )ds , D (t s ) + a (t s ) Z 1 − 1 2 − 2 1 2 ′ F (0)(t1 s1)+ aF (0)(t2 s2) ′ ′ RF,af(t ) := a − 2 2 2 −1/2 f(s )ds . D ((t s ) + a (t s ) ) Z 1 − 1 2 − 2 Remark 5.1. In [16], it was showed that the operators Rlog,a and Rlog,a are bounded maps from H1/2(D)∗ to H3/2(D). By repeating the arguments, one can show that this is also true for RF,a. We unwrap the right hand side of (5.3) term by term in the following five lemmas. Note that the first four of them are proved in [16]. We repeat them here for the convenience of the readers. Lemma 5.2. We have the following identity
d (x, y)−1∂ u (y)d (y)= εL ψ (t′)+ ε3 ψ (t′) g ν ε,a h a ε Aε ε ZΓε,a ε ′ 1/2 ′ ∗ 1/2 ′ with x = x (t ) for some ε : H (D; ds ) H (D; ds ) with operator norm bounded uniformly in ε. A → NARROW ESCAPE PROBLEM 17
Proof. By using Proposition 4.2 and (5.6), we see that in the coordinate system (5.5) it follows
−1 1 ′ ′ dg(x, y) ∂νuε,a(y)dh(y)=aε 2 2 1/2 ψε(s )ds D ((s t ) + a(s t ) ) ZΓε,a Z 1 − 1 2 − 2 1 ′ ′ 2 ′ ′ 3 A (s , t )(1 + ε Qε(s )) + Qε(s ) ′ ′ + aε 2 2 1/2 ψε(s )ds . D ((s t ) + a(s t ) ) Z 1 − 1 2 − 2
Therefore, by Lemma 2.11 of [16], the second term of the right-hand side can be 3 written as ε εψ, for operator ε which satisfies the requirement of the statement. A A
1/2 ′ ∗ From now on, we will denote by ε any operator which takes H (D; ds ) H1/2(D; ds′) whose operator norm is boundedA uniformly in ε. → For the second term of the right-hand side of (5.3), the following lemma holds. Lemma 5.3. We have the following identity
(H(x) ∂ φ(x)) log d (x, y)∂ u (y)d (y)= (H(x∗) ∂ φ(x∗))Φ(x∗) log ε − ν h ν ε,a h − − ν ZΓε,a 2 ∗ ∗ ′ 3 + ε (H(x ) ∂ φ(x ))R ψ (t )+ O 1/2 D (ε log ε)+ ε log ε ψ . − ν log,a ε H ( ) Aε ε where x = xε(t′) and Φ is the function defined in Theorem (3.2). Proof. By using Proposition (4.2) and (5.6), we obtain
(H(x) ∂ φ(x)) log d (x, y)∂ u (y)d (y) − ν h ν ε,a h ZΓε,a 2 ε ′ ε ′ ′ ′ = aε log ε(H(x (t )) ∂ν φ(x (t ))) ψε(s )ds − D Z 2 ε ′ ε ′ 2 2 1/2 ′ ′ + aε (H(x (t )) ∂νφ(x (t ))) log ((t1 s1) + a(t2 s2) ) ψε(s )ds − D − − Z 2 ε ′ ε ′ 2 ′ ′ ′ ′ aε (H(x (t )) ∂ν φ(x (t ))) log 1+ ε A(s , t ) ψε(s )ds − − D Z 4 ε ′ ε ′ 2 2 1/2 ′ ′ ′ aε (H(x (t )) ∂ν φ(x (t ))) log ε((t1 s1) + a(t2 s2) ) Qε(s )ψε(s )ds − − D − − Z 4 ε ′ ε ′ 2 ′ ′ ′ ′ ′ aε (H(x (t )) ∂ν φ(x (t ))) log 1+ ε A(s , t ) Qε(s )ψε(s )ds . − − D Z Apart the first and second terms, all terms of the right-hand side can be written as ε3 ψ . The first term, by (5.9), is equals to Aε ε ∗ ∗ ∗ 3 ε ′ ε ′ ′ ′ ′ (H(x ) ∂ν φ(x ))Φ(x ) log ε + ε log ε(H(x (t )) ∂νφ(x (t ))) Q˜ε(s )ψε(s )ds − − − D Z +(H(x∗) ∂ φ(x∗) H(xε(t′)+ ∂ φ(xε(t′)))) log εΦ(x∗), − ν − ν consequently, by Lemma 2.11 of [16], it is equal to
∗ ∗ ∗ 3 (H(x ) ∂ φ(x ))Φ(x ) log ε + O 1/2 D (ε log ε)+ ε log ε ψ . − − ν H ( ) Aε ε 18 MEDETNURSULTANOV,WILLIAMTRAD,ANDLEOTZOU
While the second term is ε2(H(x∗) ∂ φ(x∗))R ψ (t′) − ν log,a ε + ε2(H(xε(t′)) ∂ φ(xε(t′)) H(x∗)+ ∂ φ(x∗))R ψ (t′). − ν − ν log,a ε Since H and φ are smooth functions, we derive that the last term of the above expression is ε3 ψ . Aε ε For the forth term of (5.3), we have the following lemma. Lemma 5.4. We have the following identity exp−1(y) ⋆ exp−1(y) II x II x ∂ u (y)d (y) x exp−1(y) − x exp−1(y) ν ε,a h ZΓε,a | x |h | x |h = ε2(λ (x∗) λ (x∗))R ψ (t′)+ ε3 ψ . 1 − 2 ∞,a ε Aε ε Proof. By Proposition 4.2 and (5.6), we see that exp−1(y) ⋆ exp−1(y) II x II x ∂ u (y)d (y) x exp−1(y) − x exp−1(y) ν ε,a h ZΓε,a | x |h | x |h 2 2 2 ∗ ∗ 2 (s1 t1) a (s2 t2) ′ ′ = a(λ1(x ) λ2(x ))ε − 2 − 2 − 2 ψε(s )ds − D (s t ) + a (s t ) Z 1 − 1 2 − 2 3 1 ′ ′ 2 ′ ′ 2 ′ ′ ′ + aε R (t ,s ) R (t ,s ) (1 + ε Q(s ))ψε(s )ds D − Z 2 2 2 ∗ ∗ 4 (s1 t1) a (s2 t2) ′ ′ ′ + a(λ1(x ) λ2(x ))ε − 2 − 2 − 2 Q(s )ψε(s )ds . − D (s t ) + a (s t ) Z 1 − 1 2 − 2 We use Lemma 2.11 of [16] to complete the proof. Next, we stady the forth term of (5.3). Lemma 5.5. The folloing is true exp (y) h F k(x), x;h ∂ u (y)d (y)= ε2R ψ + ε3 ψ . x exp (y) ν ε,a h F,a ε Aε ε ZΓε,a | x;h |h Proof. By Proposition 4.2 and (5.6), we get exp (y) h F k(x), x;h ∂ u (y)d (y) x exp (y) ν ε,a h ZΓε,a | x;h |h ′ ′ 2 F1(t )(s1 t1)+ F2(t )(s2 t2) ′ ′ = aε −2 2 2−1/2 ψε(s )ds D ((s t ) + a (s t ) ) Z 1 − 1 2 − 2 3 3 ′ ′ 2 ′ ′ ′ + aε R (t ,s )(1 + ε Q(s ))ψε(s )ds D Z ′ ′ 4 F1(t )(s1 t1)+ F2(t )(s2 t2) ′ ′ ′ + aε −2 2 2−1/2 Q(s )ψε(s )ds D ((s t ) + a (s t ) ) Z 1 − 1 2 − 2 Finally, Lemma 2.11 of [16] implies that the last two terms are ε3 ψ . Aε ε Finally, let us look to the last term of (5.3). By Lemma 5.1 in [16], we know that for operator T : C∞(D) ′(D) defined by the integral kernel ε c →D R(xε(t′; x∗), xε(s′; x∗)) R(x∗, x∗), − NARROW ESCAPE PROBLEM 19 we have T 1/2 D ∗ 1/2 D = O(ε log ε). Therefore, by using Lemmas 5.2-5.5, we k εkH ( ) →H ( ) re-write (5.3) in the following way ε 1 (x∗) C = L ψ + (H(x∗) ∂ φ(x∗))Φ(x∗) log ε −G − ε,a 2π a ε 4π − ν ε2 λ (x∗) λ (x∗) (H(x∗) ∂ φ(x∗))R 1 − 2 R + R ψ − 4π − ν log,a − 4π ∞,a F,a ε ∗ ∗ ∗ 2 3 R(x , x )Φ(x )+ ε aT ψ + O 1/2 D (ε log ε)+ ε log ε ψ . − ε ε H ( ) Aε ε This is equivalent to 2π (H(x∗) ∂ φ(x∗))Φ(x∗) log ε R(x∗, x∗)Φ(x∗) (x∗) C − ν =(L ε )ψ ε − G − ε,a − 4π a − R ε 2 +εaT ψ + O 1/2 D (log ε)+ ε log ε ψ , ε ε H ( ) Aε ε where H(x∗) ∂ φ(x∗) λ λ 1 := − ν R 1 − 2 R + R . R 2 log,a − 8 ∞,a 2 F,a −1 Applying La to both sides, we obtain 2π (H(x∗) ∂ φ(x∗))Φ(x∗) log ε R(x∗, x∗)Φ(x∗) (x∗) C − ν L−11 ε − G − ε,a − 4π a −1 −1 ′ (5.12) =(I εL + εL T )ψ + O 1/2 D ∗ (log ε), − a R a ε ε H ( ) ′ 1/2 D ∗ 1/2 D ∗ for some Tε : H ( ) H ( ) with operator norm O(ε log ε). As we mentioned → 1/2 ∗ 3/2 in Remark 5.1, Rlog,a, R∞,a, and RF,a are bounded maps from H (D) to H (D). Therefore, the right side can be inverted by Neumann series to deduce
2πCε,a −1 −1 (5.13) ψ = L 1+ C O 1/2 D ∗ (1) + O 1/2 D ∗ (ε log ε). ε − ε ε,a H ( ) H ( ) Let us integrate this over D and use (5.9), then we derive ∗ 2πCε,a −1 ′ ′ −1 Φ(x ) La 1(s )ds + Cε,aO(1) + O(ε log ε)= 2 . − ε D − aε Z Note that
−1 ′ ′ 1 1 ′ 2π (5.14) La 1(s )ds = ′ 2 1/2 ds = , D K D (1 s ) K Z a Z −| | a and hence, the previous equation gives K Φ(x∗) (5.15) C = a + C′ , ε,a 4π2aε ε,a ′ with Cε,a = O(log ε). We put this into (5.13), to obtain K Φ(x∗) (5.16) ψ = a L−11+ ψ′ , ε − 2πaε2 a ε ′ −1 where ψ 1/2 D ′ ∗ = O(ε log ε). Let us insert this into (5.9), then we obtain k εkH ( ;ds ) ∗ ′ ′ ′ KaΦ(x ) ˜ ′ −1 ′ ′ ˜ ′ ′ ′ ′ (5.17) ψε(s )ds = Q(s )La 1(s )ds + ε Q(s )ψε(s )ds D − 2πaε D D Z Z Z where 1 ε ′ ∗ ∗ Q˜(s′)= (eφ(x (s )) eφ(x ))+ ε2eφ(x )Q (s′) ε − ε 20 MEDETNURSULTANOV,WILLIAMTRAD,ANDLEOTZOU and Q is the function involved into volume form (5.6). If we use Taylor expansion to eφ at x = x∗, we obtain ϕ ϕ(x∗) ϕ(x∗) 2 e = e + εg ∗ (e ϕ(x) ∗ , t E + t E )+ O(ε ). x ∇g |x=x 1 1 2 2 Therefore
φ(xε(t)) −1 φ(x∗) −1 e [La 1](t)dt =e [La ](t)dt D D Z Z ∗ ∗ ϕ(x ) ∗ −1 + εgx e gϕ(x) x=x∗ , E1(x ) t1[La 1](t)dt ∇ | D Z ∗ ∗ ϕ(x ) ∗ −1 + εgx e gϕ(x) x=x∗ , E2(x ) t2[La 1](t)dt ∇ | D Z + O(ε2 ). Noting that −1 −1 t1[La ](t)dt = t2[La 1](t)dt =0, D D Z Z we obtain
φ(xε(t)) −1 φ(x∗) −1 2 (5.18) e [La 1](t)dt = e [La 1](t)dt + O(ε ). D D Z Z Therefore, from (5.17) it follows that
′ ′ ′ (5.19) ψε(s )ds = O(1) + O(log ε)= O(log ε). D Z Next, we put (5.15) and (5.16) into (5.12) to obtain
∗ ∗ ∗ 2π ∗ ∗ ∗ ∗ ′ (H(x ) ∂ν φ(x ))Φ(x ) log ε −1 R(x , x )Φ(x ) (x ) Cε,a − La 1 ε − G − − 4π ! K Φ(x∗) K Φ(x∗) = a L−1 L−11 a L−1T ′L−11 2πaε a R a − 2πaε a ε a ′ −1 −1 ′ ′ + ψ ε L L T ψ + O 1/2 D ∗ (log ε). − a R − a ε H ( ) Therefore, recalling that ′ T 1/2 D ∗ 1/2 D ∗ = O(ε log ε), 1/2 D ∗ 3/2 D ∗ = O(1), k εkH ( ) →H ( ) kRkH ( ) →H ( ) ′ −1 ψ 1/2 D ′ ∗ = O(ε log ε), k εkH ( ;ds ) we derive 2π (H(x∗) ∂ φ(x∗))Φ(x∗) log ε R(x∗, x∗)Φ(x∗) (x∗) C − ν L−11 ε − G − ε,a − 4π a ∗ KaΦ(x ) −1 −1 ′ = L L 1+ ψ + O 1/2 D ∗ (log ε). 2πaε a R a ε H ( ) Let us integrate this over D and take into account (5.19), (5.14), then (H(x∗) ∂ φ(x∗))Φ(x∗) C′ = − ν log ε ε,a − 4π 2 ∗ ∗ ∗ ∗ ∗ Ka Φ(x ) −1 −1 + R(x , x )Φ(x ) (x ) 3 La La 1(t)dt + O(ε log ε). − G − 8π a D R Z NARROW ESCAPE PROBLEM 21
−1 Since La is self-adjoint, we can express the last integral more explicitly:
−1 −1 ′ ′ −2 ′ 2 −1/2 ′ 2 −1/2 La La 1(s )ds = Ka (1 s ) , (1 s ) . D R h −| | R −| | i Z Moreover,
s1 1 1 ′ ′ ′ 2 1/2 2 2 2 1/2 ′ 2 1/2 dt ds =0. D (1 s ) D ((s t ) + a (s t ) ) (1 t ) Z −| | Z 1 − 1 2 − 2 −| | Indeed, consider the following two changes of variables for the left-hand side
(s1,s2, t1, t2)=(r cos α,r sin α, ρ cos β, ρ sin β), (s ,s , t , t )=( r cos α,r sin α, ρ cos β, ρ sin β). 1 2 1 2 − − The results differ by multiplying by 1, which means that the left-hand side is 0. Therefore, we know that −
−1 −1 ′ ′ La RF,aLa 1(s )ds =0. D Z Finally, recalling the definition of and the relation between C′ and C , we obtain R ε,a ε,a K Φ(x∗) (H(x∗) ∂ φ(x∗))Φ(x∗) C = a − ν log ε ε,a 4π2aε − 4π + R(x∗, x∗)Φ(x∗) (x∗) − G ∗ ∗ ∗ 2 2 2 1/2 (H(x ) ∂ν φ(x ))Φ(x ) 1 log ((t1 s1) + a (t2 s2) ) ′ ′ − 3 ′ 2 1/2 − ′ 2 1/2− dt ds − 16π D (1 s ) D (1 t ) ∗ ∗ ∗ Z −| | Z 2 2−| | 2 (λ1(x ) λ2(x ))Φ(x ) 1 (t1 s1) a (t2 s2) 1 ′ ′ + − 3 ′ 2 1/2 − 2 − 2 − 2 ′ 2 1/2 dt ds 64π D (1 s ) D (t s ) + a (t s ) (1 t ) Z −| | Z 1 − 1 2 − 2 −| | + O(ε log ε). In case of the disc, that is a =1, the last two terms explicitly calculated in Lemmas 4.5 and 4.6 in [16]. Therefore, we have (5.20) Φ(x∗) (H(x∗) ∂ φ(x∗))Φ(x∗) C := C = − ν log ε + R(x∗, x∗)Φ(x∗) (x∗) ε ε,1 4aε − 4π − G (H(x∗) ∂ φ(x∗))Φ(x∗) 3 (5.21) − ν 2 log 2 + O(ε log ε). − 4π − 2 Next, let us recall that u (x)= E[τ X = x]= (x)+ C Φ(x∗)G(x∗, x)+ r (x) ε,a Γǫ,a | 0 G ǫ,a − ǫ for each x M Γ . Here the remainder r is given by ∈ \ ǫ,a ǫ (5.22) r (x)= (G(x, y) G(x, x∗))∂ u (y)d (y). ǫ − ν ǫ h Z∂M Let K M be a compact subset of M which has positive distance from x∗ and consider⊂x K. Writing out this integral in the xǫ( ; x∗) coordinate system and ∈ · using (5.7), (5.6), and the expression of ψǫ derived in (5.16), we get
M ′ 2 ′ ′ (5.23) rǫ(x) = ǫ −| ′| 2 1/2 L(x, ǫs )(1 + ǫ Qǫ(s ))ds D 2π(1 s ) Z −| | 3 ′ ′ ′ 2 ′ ′ + aǫ ψǫ(s )L(x, ǫs )(1 + ǫ Qǫ(s ))ds D Z 22 MEDETNURSULTANOV,WILLIAMTRAD,ANDLEOTZOU for some function L(x, s′) jointly smooth in (x, s′) K D. The second integral 1/2 D ∗ ∈ 1/2×D ′ formally denotes the duality between H ( ) and H ( ). The estimate for ψǫ derived in (5.16) now gives for any integer k and any compact set K not containing ∗ x , rǫ Ck(K) Ck,Kǫ. This gives us the first parts of Theorems 3.2 and 3.3. Finally,k k we≤ compute the average expected value over M. Let us writte
v(x)= G(x, y)∂νuε,a(y)dh(y). ZΓǫ,a Then v(x)= u (x) (x) C , ε,a − G − ε,a so that ∆ v(x)+ g(F (x), v(x)) = 0; g ∇g v(x) = G∂M (x, y)∂νuε,a(y)dh(y). ( |∂M Γǫ,a Sicne G Ψ−1(∂M) we knowR that v(x) H1/2(∂M). ∂M ∈ cl |∂M ∈ Let f ∞ be a sequence of smooth functions such that f ∂ u in H−1/2(∂M). { j}j=1 j → ν ε,a Let v ∞ be functions which sutisfy { j}j=1 ∆ v (x)+ g(F (x), v (x)) = 0; g j ∇g j vj(x) = G∂M (x, y)fj(y)dh(y). ( |∂M Γǫ,a Then v v in H1(M). Therefore,R we compute j →
G(x, y)∂νuǫ(y)dh(y)dg(x) = lim G(x, y)fj(y)dh(y)dg(x) j→∞ ZM ZΓǫ,a ZM Z∂M
= lim fj(y) G(x, y)dg(x)dh(y) j→∞ Z∂M ZM
= ∂ν uε,a(y) G(x, y)dg(x)dh(y). ZΓǫ,a ZM Recalling (4.2), we derive
φ(y)−φ(x) ∂ν uε,a(y) G(x, y)dg(x)dh(y)= ∂νuε,a(y) e G(y, x)dg(x)dh(y) ZΓǫ,a ZM ZΓǫ,a ZM = eφ(z)d (z) e−φ(x)G(x∗, x)d (x)+ O(ε) − g g ZM ZM ∗ = Φ(x∗) eφ(x )−φ(x)G(x∗, x)d (x)+ O(ε) − g ZM = Φ(x∗) G(x, x∗)d (x)+ O(ε) − g ZM Therefore, (5.1) implies
u (x)= (x) Φ(x∗) G(x, x∗)d (x)+ M C . ε,a G − g | | ε,a ZM ZM ZM This gives us the second parts of Theorems 3.2 and 3.3. NARROW ESCAPE PROBLEM 23
6. Appendix A -Elliptic Equation for E[τ X = x] Γ| 0 In this appendix we show that u(x) := E[τΓ X0 = x] satisfies the boundary value problem (3.2). This is standard material but we| could not find a suitable reference which precisely addresses our setting. As such we are including this appendix for the convenience of the reader. Let (M,g,∂M) be an orientable compact connected Riemannian manifold with non-empty smooth boundary oriented by dg. Let us consider the operator u ∆ u + g(F, u), → g ∇g where F is a force field, which is given by F = gφ for a smooth up to the boundary potential φ. We can re-write this operator in the∇ following way 1 ∆F := ∆ +g(F, )= div (eφ ). g · g · ∇g· eφ g ∇g· Note that eφ is a smooth positive function on M. Moreover, there exist constants such that (A.1) 0
M˜ M M =Γ. Note that, the Brownian motion, starting at any point M Γ, hits \ ∩ \ M˜ M if and only if it hits Γ. Therefore, the parabolicity condition of (M,µ) gives \ (A.4) lim Γ(t, x)=1, x M. t→∞ P ∈ Further, let us define the mean first arrival time u, as ∞ (A.5) u(x) := E[τ X = x] := td (t, x), Γ| 0 PΓ Z0 where the integral is a Riemann-Stieltjes integral. To investigate u, let us recall some properties of . By Remmark 2.1 in [6], it follows that PΓ t∆F 1 (t, x)= e mix 1 (x), − PΓ F t∆mix F F where e is the semigroup with infinitesimal generator ∆mix, and ∆mix is the F weighted Laplace operator ∆g corresponding to the Dirichlet boundary condition on Γ and Neumann boundary condition on ∂M Γ, which is defined as follows \ F 1 F 2 (A.6) D(∆ ) := u H (M) : ∆ u L (M,µ) u =0, ∂ u c =0 mix { ∈ g ∈ |Γ ν |Γ } (A.7) ∆F u = ∆F u u D(∆F ). mix g ∈ mix Since eφ is smooth and satisfies (A.1), we conclude that, as a set, L2(M)= L2(M,µ). 1 F 2 2 Moreover, for u H (M), the conditions ∆g u L (M,µ) and ∆gu L (M) are ∈ F ∈ ∈ equivalent. Therefore, D(∆mix) = D(∆mix), where ∆mix is the classical Laplace operator ∆g corresponding to the Dirichlet boundary condition on Γ and Neumann boundary condition on ∂M Γ: \ 1 2 (A.8) D(∆ ) := u H (M) : ∆ u L (M) u =0, ∂ u c =0 mix { ∈ g ∈ |Γ ν |Γ } (A.9) ∆ u = ∆ u u D(∆ ). mix g ∈ mix −1/2 In (A.6) and (A.8), we define ∂νu H (∂M) using the same method for defining the Dirichlet to Neumann map. That∈ is, for u H1(M) such that ∆ u L2(M), the ∈ g ∈ distribution ∂ u H−1/2(∂M) acts on f H1/2(∂M) via ν |∂M ∈ ∈ (A.10) ∂ u , f := ∆ u(z)v (z)d (z)+ g( u(z), v (z))d (z), h ν |∂M i g f g ∇g ∇g f g ZM ZM 1 where vf H (M) is the harmonic extension of f. We say that ∂ν u ω = 0, for ∈ 1/2| non-empty open set ω ∂M, if ∂νu ∂M , f ∂M = 0 for all f H (∂M) such that f = 0. Note⊂ that if u sufficientlyh | regular,| i for instance ∈u H2(M), then |∂M\ω ∈ ∂νu ∂M , f is equal to the boundary integral of ∂ν u ∂M and f. h Next,| wei note that |
F φ(z) (∆ u,u) 2 = e g( u(z), u(z))d (z), mix L (M,µ) − ∇g ∇g g ZM and therefore, (A.1) implies that F F F c1(∆mixu,u)L2(M) < (∆mixu,u)L2(M,µ) F that the spectrum of ∆mix consists eigenvalues with finite multiplicity accumulating at . Hence, ∆F satisfies the quadratic estimate −∞ mix ∞ F 2 F 2 −1 2 dt 2 t∆ (1 + t (∆ ) ) u 2 C u 2 , k mix mix kL (M,µ) t ≤ k kL (M,µ) Z0 2 for some C > 0 and all u L (M,µ); see for instance [14, p. 221]. Therefore, ∆mix admits the functional calculus∈ defined in [15]. Remark 6.1. The functional calculus in [15] is defined for a concrete operator, F which is denoted by T in the notation used in that article. However, ∆mix satisfy all necessary conditions to admit this functional calculus. t∆F Therefore, the semigroup e mix , which is contracting by Hille-Yosida theorem [10, Theorem 8.2.3], can be defined as follows t∆F 1 tζ F −1 2 e mix u = e (ζ ∆ ) udζ, u L (M), 2πi − mix ∈ Zγa,α where a (τ, 0), α (0, π ), and γ is the anti-clockwise oriented curve: ∈ ∈ 2 a,α γ := ζ C : Reζ a, Imζ = Reζ a tan α . a,α { ∈ ≤ | | | − | } F Let ε> 0 such that a + ε< 0. Then ∆mix + ε is also a negative self-adjoint operator, t(∆F +ε) and hence generates contracting semigroup, e mix , as above. By definition, we obtain, for u L2(M,µ), ∈ t∆F 1 tζ F −1 (A.11) e mix u = e (ζ ∆ ) udζ 2πi − mix Zγa,α e−tε = et(ζ+ε)(ζ + ε (∆F + ε))−1udζ 2πi − mix Zγa,α −tε e tξ F −1 −tε t(∆F +ε) = e (ξ (∆ + ε)) udξ = e e mix u, 2πi − mix Zγa+ε,α where γa+ε,α = γa,α + ε Reξ < 0 . Let f1 the constant function on M equals 1. By Theorem 8.2.2 in [10],⊂ we { know, for} λ> 0, ∞ F −1 −λt t(∆F +ε) (λ (∆ + ε)) f = e e mix f dt. − mix 1 1 Z0 Let us choose λ = ε, then, by using (A.11), we obtain ∞ ∞ F −1 −εt t(∆F +ε) t∆F (∆ ) f = e e mix f dt = e mix f dt − mix 1 1 1 Z0 Z0 and hence, ∞ (A.12) 1 (t, x)dt = ((∆F )−1f )(x) < . − PΓ − mix 1 ∞ Z0 Therefore, the dominated convergence theorem implies b ∞ lim ( Γ(b, x) Γ(t, x)) dt = 1 Γ(t, x)dt < . b→∞ P − P − P ∞ Z0 Z0 26 MEDETNURSULTANOV,WILLIAMTRAD,ANDLEOTZOU Hence, by using (A.5) and integration by parts, we obtain b b u(x) = lim Γ(b, x)b Γ(t, x)dt = lim ( Γ(b, x) Γ(t, x)) dt < b→∞ P − 0 P b→∞ 0 P − P ∞ ∞ Z Z = 1 (t, x)dt. − PΓ Z0 Therefore, by (A.12), we obtain ∆F u = f = 1. mix − 1 − In particular, u D(∆ ), and hence, ∈ mix u =0, ∂ u =0. |Γ ν |∂M\Γ We see that (3.2) is satisfied. References [1] Habib Ammari, Kostis Kalimeris, Hyeonbae Kang, and Hyundae Lee. Layer potential techniques for the narrow escape problem. J. Math. Pures Appl. (9), 97(1):66–84, 2012. 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[26] Amit Singer, Zeev Schuss, and David Holcman. Narrow escape, part II: The circular disk. Journal of statistical physics, 122(3):465–489, 2006. School of Mathematics and Statistics, University of Sydney Email address: [email protected] School of Mathematics and Statistics, University of Sydney Email address: [email protected] School of Mathematics and Statistics, University of Sydney Email address: [email protected]