Potential theory for initial-boundary value problems ofunsteady Stokes flow in two dimensions

Shidong Jiang

1 Introduction

Integral equations have been of great theoretical importance for analyzingboundary value problems. There is a large amount of literature devoted tothe classical potential theory and its applications on solving the boundaryvalue problems of elliptic partial differential equations (see, for example,[5, 20, 23, 25, 26, 31, 32, 36, 37, 40]). For elliptic problems, integral equationshave been coupled with finite element methods in numerical computationand the resulting boundary element methods have been very popular inengineering science (see, for example, [2, 3, 24]).

For time-dependent problems, integral equation methods are less suc-cessful in numerical computation. The primary reason is that the directimplementation of integral equation methods for time-dependent problemsis computationally expensive as compared with finite difference or finite ele-ment methods. Indeed, the discretizations of integral equations usually leadto dense linear systems. And for time dependent problems, the layer poten-tials involve integration in both space and time, which makes the evaluationof layer potentials and time marching extremely expensive.

The invention of the Fast Multipole Method (FMM) (see, for example,[16, 17, 4]) has dramatically changed the landscape of the field of scien-tific computing. Tremendous progress has been made in designing fast andaccurate numerical algorithms using FMM and its descendents to solve inte-gral equations for various problems in electromagnetics, elasticity, and fluidmechanics (see, for example, [6, 14, 12, 13, 33, 45, 46, 47]). The numer-ical tools for solving the heat equation using integral equations have alsobeen developed recently (see, for example, [18, 19, 28, 15, 43, 44]). Hyper-bolic potentials have also been applied to study time-dependent problems forscattering problems in electromagnetics (see, for example, [30]). When thehurdle of computational cost has been overcome, integral equation methodsoffer several advantages as compared with standard finite difference and/orfinite element methods. First, problems of complex geometry can be dealtwith more easily. Second, the artificial boundary conditions are avoided forexterior problems. Third, the dimension of the problem is reduced by onefor certain problems. Fourth, the influences of the initial data, nonhomo-

1

geneous term, and boundary data are clearly separated in integral equationformulations, and this is physically more attractive for many problems.

In this paper, we study the integral equation formulations for the time-dependent linearized viscous incompressible flow (also called the unsteadyStokes flow or linearized Navier-Stokes flow):

ρ∂u

∂t= −∇p+ µ∆u, (1)

∇ · u = 0. (2)

We first derive Green’s formula for the unsteady Stokes flow with nonho-mogeneous terms in a bounded domain. We then show that the pure initialvalue problem can be solved using the initial potential; the nonhomoge-neous problem can be solved using the volume potential. Next, we derivethe jump relation of the double layer potential and apply it to derive aboundary integral equation for the Dirichlet problem. Finally, we derive thejump relation of the single layer potential and apply it to derive a boundaryintegral equation for the Neumann problem. Since the problem is linear,the integral equation formulations for general initial-boundary value prob-lems of the unsteady Stokes flow are readily available. Here we observethat many numerical tools for solving integral equations for the heat equa-tion [18, 19, 28, 15, 43, 44] can be applied to solve integral equations forthe unsteady Stokes flow. Thus, fast numerical algorithms can be readilydeveloped to solve integral equations associated with the unsteady Stokesflow.

Remark 1.1. In [38], single and double layer potentials have been usedto show the existence and uniqueness of the boundary value problems forthe unsteady Stokes flow in Lipschitz domains. Our definition of the dou-ble layer potential is different from that of [38]. We have symmetrized thedouble layer potential to make it physically meaningful and more parallel tothe steady Stokes case (see, for example, [37]). As compared with [38], weoffer elementary but more comprehensive and systematic study of integralequation methods for the unsteady Stokes flow, which hopefully will be moreeasily accessible to physicists and engineers and more useful for numericalcomputation.

Remark 1.2. As pointed out by one of referees to this paper, Odquist [34]has proposed the layer potential formulation for unsteady Stokes flow in 1932where he aimed at the proof of the solvability of the Stokes problem. Subse-quently, Leray [27] constructed the potential theory for 2-dimensional convex

2

domains based on another kernel and proved the above-mentioned existencetheorem in 1934.

The paper is organized as follows. First, we present some estimatesabout the singularities of the Green’s function. We then derive Green’sformula for the unsteady Stokes flow in Section 3. In Section 4, we studythe pure initial value problem of the unsteady Stokes flow. In Section 5, westudy the nonhomogeneous problem using the volume potential. In Section6, we study the Dirichlet problem using double layer potentials. In Section7, we study the Neumann problem using single layer potentials. Finally, weconclude this paper with a short discussion.

2 Analytical preliminary

2.1 Notation

We use x, y to denote points in R2, i.e., x = (x1, x2), y = (y1, y2). We use tand τ to denote time variables. A bounded domain in R2 is denoted by D,its boundary is denoted by S. S is assumed to be a Lyapunov curve. Theoutward unit normal vector on S is denoted by n = (n1, n2). Vectors aregenerally boldfaced. The Einstein summation convention will be used.

2.2 Fundamental solutions for the unsteady Stokes flow - theunsteady Stokeslet

The fundamental solution for the unsteady Stokes flow has been derived byseveral researchers (see, for example, [35, 39, 8, 9, 10, 22, 42, 21]). Theunsteady Stokeslet Gij (i, j = 1, 2) for the velocity field (also called Oseen’stensor in literature since Oseen [35] was the first one to define it) and theassociated pressurelet pi (i = 1, 2) satisfy the following six equations

ρ∂Gij∂t

= −∂pj∂xi

+ µ∇2Gij + δijδ(x)δ(t), i, j = 1, 2, (3)

∂Gij∂xi

= 0, j = 1, 2. (4)

The explicit expressions of Gij and pi are given by the following formulae(see, for example, [22, 42, 21])

Gij(x, t) = − 1

2πρ(δij−

2xixjr2

)1− e−r2/4νt

r2+

1

4πρ(δij−

xixjr2

)e−r

2/4νt

νt, (5)

3

pi(x, t) =1

2π

xir2δ(t), (6)

where r = |x| =√x21 + x22 and ν = µ/ρ is the kinematic viscosity.

2.3 Properties of the unsteady Stokeslet

We now consider the case where the source is at a general point y and timeτ . We write Gij(x, t; y, τ) to denote the unsteady Stokeslet in this case. Dueto translation invariance, we actually have Gij(x, t; y, τ) = Gij(x− y, t− τ)and similarly pi(x, t; y, τ) = pi(x− y, t− τ).

We first point out some connections between the unsteady Stokeslet andfundamental solutions of other related PDEs. The Laplace kernel Gl satisfiesthe equation −∆Gl = δ(x − y) and is given by the formula Gl(x, y) =12π ln 1

|x−y| . The heat kernel Gh satisfies the equation ρ∂Gh∂t = µ∆Gh + δ(x−

y)δ(t − τ) and is given by the formula Gh(x, t; y, τ) = 14πρ

1ν(t−τ)e

− |x−y|2

4ν(t−τ) ,

where ν = µ/ρ is the kinematic viscosity.Straightforward computation shows that

G11 +G22 = Gh,∂Gij∂t

= δij∂Gh∂t− ν ∂2Gh

∂xi∂xj, (7)

and

limτ→t−

Gij(x, t; y, τ) = − 1

2πρ(δij −

2(xi − yi)(xj − yj)|x− y|2

)1

|x− y|2

=1

ρ

∂2Gl(x, y)

∂xi∂xj=

1

ρ

∂2Gl(x, y)

∂yi∂yj.

(8)

For the fundamental solutions of pressure, we have

pi(x, t; y, τ) = −∂Gl(x, y)

∂xiδ(t− τ), i = 1, 2;

∂pj∂yi

=∂pi∂yj

. (9)

For x 6= y and τ < t, it is easy to see that Gij = Gij(x, t; y, τ) satisfies theforward heat equation in x and t and backward heat equation in y and τ :

−ρ∂Gij∂τ− µ∇2

yGij = 0, (y, τ) ∈ (D −Bε(x))× [0, t). (10)

From (4),∂Gij∂xi

= −∂Gij∂yi

, and Gij = Gji, we have

∂Gij∂yi

=∂Gji∂yi

= 0. (11)

4

We now provide some estimates ofGij = Gij(x, t; y, τ), whose proof is similarto that of the heat kernel in [11] (see also [8]).

Lemma 2.1. (Estimates of unsteady Stokeslet)

1. For any fixed t > τ and x ∈ R2, Gij is bounded and thus absolutelyintegrable as a function of y in any bounded domain D ∈ R2.

2. For any 0 < α < 1,

|Gij(x, t; y, τ)| ≤ const.

|t− τ |α|x− y|2−2α. (12)

Thus, for a fixed x ∈ D, Gij is absolutely integrable on D × [0, t] as afunction of y and τ ; and for a fixed x ∈ S (S is a curve in R2), Gij isabsolutely integrable on S × [0, t] as a function of y and τ .

3. For any 1/2 < α < 1,

| ∂∂yk

Gij(x, t; y, τ)| ≤ const.

|t− τ |α|x− y|3−2α. (13)

Thus, for a fixed x ∈ D, ∂∂yk

Gij is absolutely integrable on D × [0, t]as function of y and τ .

The following integrals will be used subsequently, all of which can be calcu-lated using polar coordinates.

Lemma 2.2. Let Sa be a circle of radius a centered at the origin and Ba bea ball of radius a centered at the origin. Then∫

Sa

µ∂Gij(x, t)

∂xknkdsx = −δij

a2e−a2/4νt

8νt2, (14)

∫Ba

ρGij(x, t)dx =1

2δij(1− e−a

2/4νt), (15)

and ∫Sa

∂Gl(x, 0)

∂xinjdsx = −1

2δij . (16)

5

3 Green’s formula for the unsteady Stokes flow

Consider the unsteady Stokes flow with a nonhomogeneous term in a boundeddomain D with boundary S:(1) the governing equation:

ρ∂ui∂t

= − ∂p

∂xi+ µ∆ui + fi(x, t), (x, t) ∈ D × [0, T ], i = 1, 2; (17)

(2) the incompressibility condition:

∂ui∂xi

= 0, (x, t) ∈ D × [0, T ]; (18)

(3) initial condition:

ui(x, 0) = u0i(x), x ∈ D, t = 0. (19)

Multiplying Gij with the governing equation with the variables changed toy and τ in (17), we obtain

Gij(x, t; y, τ)

{ρ∂uj∂τ

+∂p

∂yj− µ∆yuj − fj(y, τ)

}= 0. (20)

Consider first the case that x ∈ D. Integrating the above equation overthe domain D−Bε(x) with Bε(x) a small ball of radius ε centered at x, weobtain∫ t

0

∫D−Bε(x)

Gij(x, t; y, τ)

{ρ∂uj∂τ

+∂p

∂yj− µ∆yuj − fj(y, τ)

}dydτ = 0.

(21)

Applying integration by parts on the first term, Green’s theorem on thesecond term, and Green’s second identity on the third term, we obtain∫ t

0

∫D−Bε(x)

uj

(−ρ∂Gij(x, t; y, τ)

∂τ− µ∆yGij

)dydτ −

∫ t

0

∫D−Bε(x)

p∂Gij∂yj

dydτ

= µ

∫ t

0

∫S−Sε

(Gij∂uj∂yk− uj

∂Gij∂yk

)nkdsydτ −∫ t

0

∫S−Sε

pGijnjdsydτ

−∫D−Bε(x)

ρGij(x, t; y, τ)uj(y, τ)|t0dy +

∫ t

0

∫D−Bε(x)

Gij(x, t; y, τ)fj(y, τ)dydτ.

(22)

6

Since the Stokeslet satisfies the backward heat equation in y and τ anddivergence free condition (see (10), (11)), the left hand side of (22) is zero.Thus

0 =

∫ t

0

∫S−Sε

Gij

(µ∂uj∂yk

nk − pnj)dsydτ −

∫ t

0

∫S−Sε

µ∂Gij∂yk

nkujdsydτ

−∫D−Bε

ρGij(x, t; y, τ)uj(y, τ)|t0dy +

∫ t

0

∫D−Bε

Gij(x, t; y, τ)fj(y, τ)dydτ.

(23)

Furthermore,

−∫D−Bε

ρGij(x, t; y, τ)uj(y, τ)|t0dy

=

∫D−Bε

ρGij(x, t; y, 0)u0j(y)dy −∫D−Bε

∂2Gl(x, y)

∂yi∂yjuj(y, t)dy

=

∫D−Bε

ρGij(x, t; y, 0)u0j(y)dy −∫D−Bε

∂

∂yj

[∂Gl(x, y)

∂yiuj(y, t)

]dy

=

∫D−Bε

ρGij(x, t; y, 0)u0j(y)dy −∫S−Sε

∂Gl(x, y)

∂yiuj(y, t)njdsy.

(24)

where (8) is used to obtain the first equality, the incompressibility of uj isused to derive the second equality, and in the third equality we have appliedGreen’s theorem.Combining (23), (24), we obtain

0 =

∫ t

0

∫S−Sε

Gij

(µ∂uj∂yk

nk − pnj)dsydτ

−∫ t

0

∫S−Sε

µ∂Gij∂yk

nkujdsydτ −∫S−Sε

∂Gl(x, y)

∂yiuj(y, t)njdsy

+

∫D−Bε

ρGij(x, t; y, 0)u0j(y)dy

+

∫ t

0

∫D−Bε

Gij(x, t; y, τ)fj(y, τ)dydτ.

(25)

7

(1) When |x−y| = ε is very small, since the velocity, its first derivative, andthe pressure are all continuous, we have∣∣∣∣∫ t

0

∫Sε

Gij

(µ∂uj∂yk

nk − pnj)dsydτ

∣∣∣∣≤M

∫ t

0

∫ 2π

0

1

(t− τ)αε2−2αεdθdτ

≤Mε2α−1 → 0, as ε→ 0 for 1/2 < α < 1,

(26)

where (12) is used in the first inequality.(2) Using (14), we have

limε→0

∫ t

0

∫Sε

µ∂Gij∂yk

nkuj(y, τ)dsydτ

= limε→0

∫ t

0uj(x, τ)

∫Sε

µ∂Gij∂yk

nkdsydτ

= limε→0

∫ t

0uj(x, τ)

(−δij

ε2e−ε2/4ν(t−τ)

8ν(t− τ)2

)dτ

= −1

2limε→0

∫ ∞ε2/4νt

ui(x, t− ε2/4νλ)e−λdλ

= −1

2ui(x, t).

(27)

(3) Using (16), we have∫Sε

∂Gl(x, y)

∂yiuj(y, t)njdsy → uj(x, t)

∫Sε

∂Gl(x, y)

∂yinjdsy

= uj(x, t)(−1

2δij) = −1

2ui(x, t), ε→ 0.

(28)

Taking the limit of ε→ 0 in (25) and then using (26), (27), (28), we obtain

ui(x, t) =

∫ t

0

∫SGij

(µ∂uj∂yk

nk − pnj)dsydτ

−∫ t

0

∫Sµ∂Gij∂yk

nkujdsydτ −∫S

∂Gl(x, y)

∂yiuj(y, t)njdsy

+

∫DρGij(x, t; y, 0)u0j(y)dy

+

∫ t

0

∫DGij(x, t; y, τ)fj(y, τ)dydτ.

(29)

8

To symmetrize the first two terms on the right hand side, we use the incom-pressibility conditions for both Gij and ui. We have∫ t

0

∫SGij

∂uk∂yj

nkdsydτ =

∫ t

0

∫D

∂

∂yk

(Gij

∂uk∂yj

)dydτ

=

∫ t

0

∫D

∂Gij∂yk

∂uk∂yj

dydτ =

∫ t

0

∫D

∂Gik∂yj

∂uj∂yk

dydτ

=

∫ t

0

∫D

∂

∂yk

(uj∂Gik∂yj

)dydτ =

∫ t

0

∫Suj∂Gik∂yj

nkdsydτ,

(30)

where the first and the last equalities follow from Green’s theorem, thesecond and the fourth equalities follow from the incompressibility of uk andGik, respectively, and the third equality follows from the interchange of thedummy summation variables j and k.Furthermore, since pi(x, t; y, τ) = ∂Gl(x,y)

∂yiδ(t− τ), we may write∫

S

∂Gl(x, y)

∂yiuj(y, t)njdsy =

∫ t

0

∫Spi(x, t; y, τ)uj(y, τ)njdsydτ, (31)

with the understanding of∫ t0 uj(y, τ)δ(t− τ)dτ = uj(y, t).

Combining (29), (30), and (31), we obtain

ui(x, t) =

∫ t

0

∫SGij(x, t; y, τ)σj(y, τ)dsydτ

−∫ t

0

∫STij(x, t; y, τ)uj(y, τ)dsydτ

+

∫DρGij(x, t; y, 0)u0j(y)dy

+

∫ t

0

∫DGij(x, t; y, τ)fj(y, τ)dydτ,

(32)

where

σj = µ

(∂uj∂yk

+∂uk∂yj

)nk − pnj

=

[µ

(∂uj∂yk

+∂uk∂yj

)− pδjk

]nk

= σjknk(y)

(33)

9

is the jth component of the surface force and

Tij(x, t; y, τ) = µ

(∂Gij∂yk

+∂Gik∂yj

)nk + pinj

= −[µ

(∂Gij∂xk

+∂Gik∂xj

)− pi(x, t; y, τ)δjk

]nk(y)

= Tijknk

(34)

is the stress tensor associated with the Stokeslet.When x /∈ D, it is easy to see that the left side of (32) should be zero.

Moreover, the jump relation of the double layer potential (69) shows thatthe left side of (32) should be 1

2ui(x, t) for x ∈ S. Combining all these facts,we obtain∫ t

0

∫SGij(x, t; y, τ)σj(y, τ)dsydτ −

∫ t

0

∫STij(x, t; y, τ)uj(y, τ)dsydτ

+

∫DρGij(x, t; y, 0)u0j(y)dy +

∫ t

0

∫DGij(x, t; y, τ)fj(y, τ)dydτ

=

ui(x, t), x ∈ D,12ui(x, t), x ∈ S,0, x /∈ D.

(35)

We call (35) the Green’s formula for the unsteady Stokes flow (there isa short, nonrigorous and incorrect derivation for this formula in [22]). Itactually provides the so called “direct” boundary integral equation formu-lation for the unsteady Stokes flow. The first term of the left side of (35)is called the single layer potential of the unsteady Stokes flow, the secondterm is called the double layer potential, the third term is called the initialpotential, the last term is called the volume potential which deals with thenonhomogeneous term of the governing equation.

We now analyze these four terms and study various problems associatedwith the unsteady Stokes flow.

4 Pure initial-value problem

Theorem 4.1. (Solution of initial-value problem). Suppose that f ∈ C2(R2)has compact support in R2 and satisfies the condition

∇ · f =∂fi∂xi

= 0. (36)

10

Define

ui(x, t) =

∫R2

ρGij(x, t; y, 0)fj(y)dy. (37)

Then

1. u ∈ C2∞(R2 × (0,∞)),

2. ρ∂u∂t = −∇p+ µ∆u, where p is an arbitrary function of t only,

3. ∇ · u = 0, t > 0,

4.

limt→0+

ui(x, t) = limt→0+

∫R2

ρGij(x, t; y, 0)fj(y)dy = fi(x). (38)

Proof. 1. For t > 0, Gij is infinitely differentiable with respect to t, so isui(x, t). To show that ui is twice differentiable with respect to x, wesimply make change of variable and write

ui(x, t) =

∫R2

ρGij(y, t)fj(x− y)dy. (39)

2. To show that u and p = g(t) satisfy the governing equation, we merelynote that (a). Gij satisfies the heat equation and ∇p = 0; (b). thesecond derivatives of Gij are actually bounded at y = x for t > 0.

3. The fact that u is divergence free can also be easily proved by notingthat Gij is divergence free and its first derivatives are bounded aty = x for t > 0.

4. Fix x ∈ R2, ε > 0. Choose δ > 0 such that

|fj(y)− fj(x)| < ε if |y − x| < δ. (40)

Let Bδ(x) be the ball of radius δ centered at x. We split the integralinto two parts

ui(x, t) =

∫R2−Bδ(x)

+

∫Bδ(x)

:= I1 + I2, (41)

and split I1 further into two parts with one part being time-independent

I1 = J1 + J2, (42)

11

where

J1 =

∫R2−Bδ(x)

∂2Gl(x, y)

∂yi∂yjfj(y)dy (43)

and

J2 =

∫R2−Bδ(x)

(1

2π

[δij −

2(xi − yi)(xj − yj)|x− y|2

]e−|x−y|

2/4νt

|x− y|2

+1

4π

[δij −

(xi − yi)(xj − yj)|x− y|2

]e−|x−y|

2/4νt

νt

)fj(y)dy.

(44)

(a) For J1, we have

J1 =

∫R2−Bδ(x)

∂

∂yj

(∂Gl(x, y)

∂yifj(y)

)dy

= −∫Sδ(x)

∂Gl(x, y)

∂yifj(y)njdsy

→ −fj(x)

∫Sδ(x)

∂Gl(x, y)

∂yinjdsy

=1

2fi(x),

(45)

where we used the divergence free condition of fj in the firstequality, Green’s theorem in the second equality, and (16) to ob-tain the final result.

(b) Since δ is fixed, the integrand of J2 tends to 0 as t → 0. HenceJ2 → 0 as t→ 0.

(c) For I2, we have

I2 → fj(x)

∫Bδ(x)

ρGij(x, t; y, 0)dy = fj(x)1

2δij(1−e−δ

2/4νt), (46)

where the second equality follows from (15).Again since δ is fixed, we have

I2 →1

2fi(x) as t→ 0. (47)

And (38) follows by combining (41)–(47).

12

5 Nonhomogeneous problem and the volume po-tential

Theorem 5.1. (Solution of nonhomogeneous problem). Suppose that f ∈C21 (R2 × [0,∞)) and f has compact support. Define u by the formula

ui(x, t) =

∫ t

0

∫R2

Gij(x, t; y, τ)fj(y, τ)dydτ (48)

and p by the formula

p(x, t) =

∫ t

0

∫R2

pj(x, t; y, τ)fj(y, τ)dydτ

=

∫R2

∂Gl(x, y)

∂yjfj(y, t)dy

=1

2π

∫R2

xj − yj|x− y|2

fj(y, t)dy =1

2π

∫R2

yj|y|2

fj(x− y, t)dy.

(49)

Then

1. u ∈ C21 (R2 × (0,∞)), p ∈ C2

1 (R2 × (0,∞)),

2. ρ∂u∂t = −∇p+ µ∆u + f(x, t),

3. ∇ · u = 0,

4. limt→0 ui(x, t) = 0 for each point x ∈ R2.

Proof. 1. Since Gij has a singularity at (0, 0), we cannot directly justifydifferentiating under the integral sign.

First we change variables, to write

ui(x, t) =

∫ t

0

∫R2

Gij(y, τ)fj(x− y, t− τ)dydτ. (50)

As f ∈ C21 (R2 × [0,∞)) has compact support and Gij = Gij(y, τ) is

bounded near τ = t > 0, we compute

∂ui(x, t)

∂t=

∫ t

0

∫R2

Gij(y, τ)∂fj(x− y, t− τ)

∂tdydτ

+

∫R2

Gij(y, t)fj(x− y, 0)dydτ,

(51)

13

∂ui(x, t)

∂xk=

∫ t

0

∫R2

Gij(y, τ)∂fj(x− y, t− τ)

∂xkdydτ, (52)

and

∂2ui(x, t)

∂xk∂xl=

∫ t

0

∫R2

Gij(y, τ)∂2fj(x− y, t− τ)

∂xk∂xldydτ. (53)

Since f ∈ C21 (R2 × [0,∞)) has compact support D, Gij(y, τ) is abso-

lutely integrable on D× (0, t), and Gij(y, t) is absolutely integrable on

D for t > 0, we see that ui,∂ui(x,t)∂t , ∂ui(x,t)∂xk

, ∂2ui(x,t)∂xk∂xl

are all continuous

on R2 × (0,∞), i.e., u ∈ C21 (R2 × (0,∞)).

The proof for p is similar as we note that (a) time derivative can bedirectly passed into integration; (b) ∂Gl

∂yj=

xj−yj2π|x−y|2 is absolutely inte-

grable on a bounded domain and thus after change of variable spatialderivatives can also be directly passed into the integral sign to fj .

2. We then calculate

ρ∂ui(x, t)

∂t+

∂

∂xip(x, t)− µ∆xui(x, t)

=

∫ t

0

∫R2

Gij(y, τ)[(ρ∂

∂t− µ∆x)fj(x− y, t− τ)]dydτ

+1

2π

∫R2

yj|y|2

∂fj(x− y, t)∂xi

dy

+ ρ

∫R2

Gij(y, t)fj(x− y, 0)dydτ

:= T1 + T2 + T3.

(54)

We now show that T1 = −T2 − T3 + fi(x, t) and thus the governingequation is satisfied.

T1 =

∫ t

0

∫BR(0)

Gij(y, τ)[(−ρ ∂∂τ− µ∆y)fj(x− y, t− τ)]dydτ

=

∫ δ

0

∫BR(0)

+

∫ t

δ

∫BR(0)−Bε(0)

+

∫ t

δ

∫Bε(0)

:= J1 + J2 + J3,

(55)

14

where BR(0) is a ball of fixed radius R centered at the origin whichcontains the compact support of f , Bε(0) is a ball of radius ε.

Using the estimate (12), we have

|J1| ≤ (‖ft‖L∞ + ‖D2f‖L∞)

∫ δ

0

∫BR

const.

|τ |α|y|2−2αdydτ

≤ Cδ1−α, 0 < α < 1.

(56)

Similarly,

|J3| ≤ (‖ft‖L∞ + ‖D2f‖L∞)

∫ t

δ

∫Bε

const.

|τ |α|y|2−2αdydτ

≤ Cε2α, 0 < α < 1.

(57)

Integrating by parts, we have

J2 =

∫ t

δ

∫BR−Bε

[(ρ∂

∂τ− µ∆y)Gij(y, τ)]fj(x− y, t− τ)dydτ

− ρ∫BR−Bε

Gij(y, t)fj(x− y, 0)dy

+ ρ

∫BR−Bε

Gij(y, δ)fj(x− y, t− δ)dy

− µ∫ t

δ

∫Sε

Gij(y, τ)∂fj(x− y, t− τ)

∂yknkdsydτ

+ µ

∫ t

δ

∫Sε

∂Gij(y, τ)

∂ykfj(x− y, t− τ)nkdsydτ

:= 0− T3 +K1 +K2 +K3,

(58)

where the first term on the right side vanishes since Gij satisfies theheat equation.

15

we have

limδ→0

K1 = ρ

∫BR−Bε

Gij(y, 0)fj(x− y, t)dy

=

∫BR−Bε

∂2Gl(y)

∂yi∂yjfj(x− y, t)dy

=

∫BR−Bε

[∂

∂yi

(∂Gl(y)

∂yjfj(x− y, t)

)− ∂Gl(y)

∂yj

∂fj(x− y, t)∂yi

]dy

= −∫Sε

∂Gl(y)

∂yjfj(x− y, t)nidsy +

∫BR−Bε

∂Gl(y)

∂yj

∂fj(x− y, t)∂xi

dy

=1

2π

∫Sε

yj|y|2

fj(x− y, t)nidsy −1

2π

∫BR−Bε

yj|y|2

∂fj(x− y, t)∂xi

dy

→ 1

2fi(x, t)− T2, as ε→ 0.

(59)

Using the estimate (12), we have

|K2| ≤ C‖Dft‖L∞∫ t

δ

∫Sε

const.

|τ |α|ε|2−2α|dsy|dτ

≤ Cε2α−1, 1/2 < α < 1.

(60)

Using (14), we have

K3 → δij

∫ t

δfj(x, t− τ)

ε2e−ε2/4ντ

8ντ2dτ

→∫ t

0fi(x, t− τ)

ε2e−ε2/4ντ

8ντ2dτ as δ → 0

=1

2

∫ ∞ε2/4νt

fi(x, t− ε2/4νλ)e−λdλ

→ 1

2fi(x, t), as ε→ 0.

(61)

Combining (54)–(61), we obtain

ρ∂ui(x, t)

∂t+

∂

∂xip(x, t)− µ∆xui(x, t) = fi(x, t). (62)

3. The proof of divergence free condition is entirely similar.

16

4. Finally, we note

‖u(·, t)‖L∞ ≤ ‖f‖L∞∫ t

0

∫BR

const.

|τ |α|y|2−2αdydτ, 0 < α < 1

≤ C‖f‖L∞t1−α → 0 as t→ 0.

(63)

Remark 5.2. We can certainly combine Theorems 4.1 and 5.1 to solve aninitial value problem with nonhomogeneous term.

6 Dirichlet problem and properties of the doublelayer potential

Theorem 6.1. (Solution of the Dirichlet problem). Suppose that f is acontinuous function on S × [0, T ] satisfying the following compatibility con-ditions

f(x, 0) = 0, x ∈ S,∫Sf(y, t) · n(y)ds(y) = 0, t ∈ [0, T ].

(64)

Suppose further that φ is a continuously differentiable function on S× [0, T ].Define the double layer potential for the velocity by the formula:

ui(x, t) =

∫ t

0

∫STij(x, t; y, τ)φj(y, τ)dsydτ, (65)

where the kernel Tij is defined in (34).Define the associated pressure field by the formula:

p(x, t) = ρ

∫SGl(x, y)nj(y)

∂φj(y, t)

∂tdy

+ 2µ

∫S

∂2Gl(x, y)

∂yj∂yknk(y)φj(y, t)dy.

(66)

Then

1.

ρ∂u(x, t)

∂t= −∇p(x, t) + µ∆u(x, t), (x, t) ∈ D × (0, T ]; (67)

17

2.

∇ · u(x, t) = 0, (x, t) ∈ D × (0, T ]; (68)

3. for x ∈ S, the integrals in the definition of u exist in the usual sense;

4. u satisfies the following jump relation:

limε→0+

ui(x± εn(x), t) = ui(x, t)±1

2φi(x, t), x ∈ S; (69)

5. u satisfies the following boundary condition

u(x, t) = f(x, t), (x, t) ∈ S × (0, T ] (70)

if the density φ satisfies the equations:

−1

2φi(x, t) +

∫ t

0

∫STij(x, t; y, τ)φj(y, τ)dsydτ = fi(x, t), i = 1, 2.

(71)

Proof. 1. We first show the velocity defined by (72) and the pressuredefined by (66) satisfy the governing unsteady Stokes equation.

Substituting (34) into (65), we obtain

ui(x, t) =

∫S

xi − yi2π|x− y|2

nj(y)φj(y, t)dsy

− µ∫S

∫ t

0

(∂Gij∂xk

+∂Gik∂xj

)nk(y)φj(y, τ)dτdsy

(72)

For x ∈ D, differentiation can be passed into integration and we cal-culate

ρ∂ui(x, t)

∂t= ρ

∫S

∂Gl(x, y)

∂yinj(y)

∂φj(y, t)

∂tdsy

− ρµ∫S

(∂Gij(x, t; y, t)

∂xk+∂Gik(x, t; y, t)

∂xj

)nk(y)φj(y, t)dsy

− ρµ∫S

∫ t

0

(∂2Gij(x, t; y, τ)

∂xk∂t+∂2Gik(x, t; y, τ)

∂xj∂t

)nk(y)φj(y, τ)dτdsy

= ρ

∫S

∂Gl(x, y)

∂yinj(y)

∂φj(y, t)

∂tdsy

− 2µ

∫S

∂3Gl(x, y)

∂xi∂xj∂xknk(y)φj(y, t)dsy

− ρµ∫S

∫ t

0

(∂2Gij(x, t; y, τ)

∂xk∂t+∂2Gik(x, t; y, τ)

∂xj∂t

)nk(y)φj(y, τ)dτdsy,

18

(73)

where the second equality follows (8). Similarly,

−µ∆ui(x, t) = −µ∫S

∂

∂xi∆xGl(x, y)nj(y)φj(y, t)dsy

+ µ2∫S

∫ t

0

(∂∆xGij∂xk

+∂∆xGik∂xj

)nk(y)φj(y, τ)dτdsy

= ρµ

∫S

∫ t

0

(∂2Gij∂xk∂t

+∂2Gik∂xj∂t

)nk(y)φj(y, τ)dτdsy,

(74)

where the second equality follows from the fact that Gl is harmonicfor x 6= y and Gij satisfy the heat equation for x 6= y and t 6= τ . Also,

∂p(x, t)

∂xi= ρ

∫S

∂Gl(x, y)

∂xinj(y)

∂φj(y, t)

∂tdsy

+ 2µ

∫S

∂3Gl(x, y)

∂xi∂yj∂yknk(y)φj(y, t)dsy

= −ρ∫S

∂Gl(x, y)

∂yinj(y)

∂φj(y, t)

∂tdsy

+ 2µ

∫S

∂3Gl(x, y)

∂xi∂xj∂xknk(y)φj(y, t)dsy.

(75)

Adding the above three equations (73)-(75), we obtain

ρ∂ui(x, t)

∂t+∂p(x, t)

∂xi− µ∆ui(x, t) = 0. (76)

2. The fact that the velocity is divergence free can be easily proved by asimilar computation.

3. We now show that even for x ∈ S, the integrals in (72) exist in theusual sense.We divide the kernel in the second integral in (72) into two parts:

−µ(∂Gij∂xk

+∂Gik∂xj

)nk(y) = Kij(x, t; y, τ) + Kij(x, t; y, τ), (77)

19

where Kij and Kij are defined by the formulae

Kij(x, t; y, τ) =ni(y)(xj − yj)

2π|x− y|2|x− y|2e−

|x−y|24ν(t−τ)

4ν(t− τ)2

− ni(y)(xj − yj) + nj(y)(xi − yi)2π|x− y|2

4ν

|x− y|2

(1− e−

|x−y|24ν(t−τ) − |x− y|

2

4ν(t− τ)e− |x−y|

2

4ν(t−τ)

),

(78)

Kij(x, t; y, τ) =(xk − yk)nk(y)

2π|x− y|2

(δij − 2(xi − yi)(xj − yj)|x− y|2

)|x− y|2e−

|x−y|24ν(t−τ)

4ν(t− τ)2

−(δij −

4(xi − yi)(xj − yj)|x− y|2

)4ν

|x− y|2

(1− e−

|x−y|24ν(t−τ) − |x− y|

2

4ν(t− τ)e− |x−y|

2

4ν(t−τ)

)].

(79)

Combining (72), (77), (79) with a further splitting, we obtain

ui(x, t) =

∫SBij(x, y)φ(y, t)dsy

+

∫S

∫ t

0Kij(x, t; y, τ)[φj(y, τ)− φj(y, t)]dτdsy

+

∫S

∫ t

0Kij(x, t; y, τ)φj(y, τ)dτdsy,

(80)

where Bij is defined by the formula

Bij(x, y) =(xi − yi)nj(y)

2π|x− y|2+

∫ t

0Kij(x, t; y, τ)dτ. (81)

(a) For fixed x ∈ S, we claim that Bij is absolutely integrable on Sas a function of y.

First, applying change of variable λ = |x−y|24ν(t−τ) , we have dλ =

|x−y|24ν(t−τ)2dτ ,

∫ t

0

|x− y|2e−|x−y|24ν(t−τ)

4ν(t− τ)2dτ =

∫ ∞|x−y|2/4νt

e−λdλ

= e−|x−y|2

4νt ,

(82)

20

and ∫ t

0

4ν

|x− y|2

(1− e−

|x−y|24ν(t−τ) − |x− y|

2

4ν(t− τ)e− |x−y|

2

4ν(t−τ)

)dτ

=

∫ ∞|x−y|2/4νt

1− e−λ − λe−λ

λ2dλ

=1− e−

|x−y|24νt

|x− y|2/4νt.

(83)

Combining (79), (81), (82), and (83), we obtain

Bij(x, y) =(xi − yi)nj(y)

2π|x− y|2+

∫ t

0Kij(x, t; y, τ)dτ

=1

2π|x− y|2

((xi − yi)nj(y) + (xj − yj)ni(y)e−

|x−y|24νt

−[(xj − yj)ni(y) + (xi − yi)nj(y)]1− e−

|x−y|24νt

|x− y|2/4νt

.

(84)

Since e−λ → 1 and (1 − e−λ)/λ → 1 as λ → 0, we see theBij(x, y) ≤M |x− y| and thus is absolutely integrable on S.

(b) Next, for fixed x ∈ S and t > 0, we show that Kij is absolutelyintegrable on S × (0, t) and thus the associated integral operatoris compact. First, a well known result in classical potential theory(see, for example, page 79 of [25]) states that

|(xk − yk)nk(y)| ≤M |x− y|2. (85)

Second, it is obvious that

|(xi − yi)(xj − yj)| ≤M |x− y|2. (86)

Third, note that for any λ > 0, λφe−λ ≤ M for any φ ≥ 0and |(1 − e−λ − λe−λ)/λφ| ≤ M for any 0 < φ ≤ 2. Withλ = |x− y|2/4ν(t− τ), we easily see that∣∣∣∣∣∣ |x− y|

2e− |x−y|

2

4ν(t−τ)

4ν(t− τ)2

∣∣∣∣∣∣ ≤ M

|t− τ |α|x− y|2−2α, 1/2 < α < 1 (87)

21

and ∣∣∣∣ 4ν

|x− y|2

(1− e−

|x−y|24ν(t−τ) − |x− y|

2

4ν(t− τ)e− |x−y|

2

4ν(t−τ)

)∣∣∣∣≤ M

|t− τ |α|x− y|2−2α, 1/2 < α < 1.

(88)

Combining (79), (85)–(88), we obtain

|Kij(x, t; y, τ)| ≤ M

|t− τ |α|x− y|2−2α, 1/2 < α < 1, (89)

and thus Kij is absolutely integrable on S × (0, t).

(c) Since φ is continuously differentiable, it is Holder continuous withexponent β (0 < β < 1). Then a similar derivation shows that

|Kij(x, t; y, τ)[φj(y, τ)−φj(y, t)]| ≤M

|t− τ |α|x− y|3−2α−β, 1−β/2 < α < 1,

(90)

and we see that Kij(x, t; y, τ)[φj(y, τ)− φj(y, t)] is absolutely in-tegrable on S × (0, t).

4. We now derive the jump relation of the double layer potential. Wewrite

ui(x, t) =

∫SBij(x, y)φj(y, t)dsy

+

∫S

∫ t

0Kij(x, t; y, τ)[φj(y, τ)− φj(y, t)]dτdsy

+

∫S

∫ t

0Kij(x, t; y, τ)[φj(y, τ)− φj(y, t)]dτdsy

+

∫S

∫ t

0Kij(x, t; y, τ)φj(y, t)dτdsy.

(91)

From the above estimates about Bij , Kij , and Kij , it is easy to see thatthe first three integrals in the above equation are continuous across theboundary S. We will denote the last integral by Ji(x, t), i.e.,

Ji(x, t) =

∫S

∫ t

0Kij(x, t; y, τ)φj(y, t)dτdsy. (92)

22

Integrating over τ first, we find

Ji(x, t) =

∫S

(xk − yk)nk(y)

2π|x− y|2

[(δij −

2(xi − yi)(xj − yj)|x− y|2

)e−|x−y|2

4νt

−(δij −

4(xi − yi)(xj − yj)|x− y|2

)1− e−

|x−y|24νt

|x− y|2/4νt

φj(y, t)dsy=

∫S

(xk − yk)nk(y)

2π|x− y|2φi(y, t)

e− |x−y|24νt − 1− e−|x−y|2

4νt

|x− y|2/4νt

dsy+

∫S

(xi − yi)(xj − yj)(xk − yk)nk(y)

π|x− y|4

21− e−

|x−y|24νt

|x− y|2/4νt− e−

|x−y|24νt

φj(y, t)dsy:= J1

i (x, t) + J2i (x, t).

(93)

Since (xk−yk)nk(y)2π|x−y|2 = ∂Gl(x,y)

∂n(y) and the jump relation of the double layer

potential of the Laplace equation (see, for example, page 80 of [25])states that

limε→0+

∫S

∂Gl(x± εn(x), y)

∂n(y)φ(y)dsy =

∫S

∂Gl(x, y)

∂n(y)φ(y)dsy±

1

2φ(x), x ∈ S,

(94)

we find that

limε→0+

J1i (x± εn(x), t) = J1

i (x, t), x ∈ S, (95)

as

limy→x

e− |x−y|24νt − 1− e−|x−y|2

4νt

|x− y|2/4νt

= 0. (96)

Similarly, we find that (see, for example, page 61 of [37])

TSijk = 4(xi − yi)(xj − yj)(xk − yk)nk(y)

|x− y|4, (97)

where TSijk is the stress tensor for 2D steady Stokes flow. We also find(see, for example, page 61 of [37]) that jump relation for the stress

23

tensor TSijk is as follows:

limε→0+

∫STSijk(x±εn(x), y)nk(y)φj(y)dsy = ±2πφi(x)+

∫STSijk(x, y)nk(y)φj(y)dsy.

(98)

Note that we have changed notations in [37] so that (97) and (98)match our notation. Note further that

limy→x

21− e−

|x−y|24νt

|x− y|2/4νt− e−

|x−y|24νt

= 1. (99)

Combining (88), (92)-(94), we obtain

limε→0+

J2i (x± εn(x), t) = J2

i (x, t)± 1

2φi(x, t), x ∈ S. (100)

Combining (81)-(83), (95), (100), we obtain the jump relation of thedouble layer potential for the unsteady Stokes flow as follows:

limε→0+

ui(x± εn(x), t) = ui(x, t)±1

2φi(x, t), x ∈ S. (101)

5. This simply follows from the jump relation of the double layer poten-tial.

Remark 6.2. For exterior problems, the boundary integral equations havethe form (as compared with (71) for interior problem)

1

2φi(x, t) +

∫ t

0

∫STij(x, t; y, τ)φj(y, τ)dsydτ = fi(x, t), i = 1, 2. (102)

7 Neumann problem and properties of the singlelayer potential

Define the single layer potential for the velocity by the formula

ui(x, t) =

∫ t

0

∫SGij(x, t; y, τ)φj(y, τ)dsydτ, (103)

24

and the associated pressure by the formula

p(x, t) =

∫ t

0

∫Spj(x, t; y, τ)φj(y, τ)dsydτ

=

∫S

∂Gl(x, y)

∂yjφj(y, t)dsy.

(104)

We first have the following theorem regarding the single layer potential itself.

Theorem 7.1. The kernel Gij is absolutely integrable as a function of yand τ for x ∈ R2. Thus the single layer potential operator is compact fromC(S × [0, T ]) to C(S × [0, T ]) and the single layer potential is continuousacross the boundary.

Proof. The absolute integrability simply follows from the estimate (12) andthe subsequent statement in Lemma 2.1. Since the kernel Gij is absoluteintegrable, the single layer potential operator is compact. The fact that thepotential is continuous across the boundary S can also easily be shown usingthe fact that the estimate (12) is true for both x ∈ D and x ∈ S.

It is also easy to see that the following theorem holds by a similar calculationas in Section 6.

Theorem 7.2. Suppose that φ is a continuous function on S × [0, T ] andthat u and p are defined by (103) and (104), respectively. Then

1.

ρ∂u(x, t)

∂t= −∇p(x, t) + µ∆u(x, t), x /∈ S, t ∈ (0, T ], (105)

2.

∇ · u(x, t) = 0, x /∈ S, t ∈ (0, T ], (106)

3.

limt→0

ui(x, t) = 0, x ∈ R2, i = 1, 2. (107)

In the Neumann problem of fluid dynamics, the surface force is specified onthe boundary, that is,[µ

(∂ui(x, t)

∂xk+∂uk(x, t)

∂xi

)− p(x, t)δik

]nk(x) = fi(x, t), (x, t) ∈ S × (0, T ].

25

(108)

Thus we will not consider the jump relation of the normal derivative of thesingle layer potential. Instead, we consider the stress tensor

σik(x, t) = µ

(∂ui(x, t)

∂xk+∂uk(x, t)

∂xi

)− p(x, t)δik

=

∫ t

0

∫ST ∗ijk(x, t; y, τ)φj(y, τ)dsydτ,

(109)

where T ∗ijk is defined by the formula

T ∗ijk(x, t; y, τ) = µ

(∂Gij(x, t; y, τ)

∂xk+∂Gkj(x, t; y, τ)

∂xi

)− pj(x, t; y, τ)δik.

(110)

We also introduce T ∗ij given by the formula

T ∗ij(x, t; y, τ) = T ∗ijk(x, t; y, τ)nk(x)

=

[µ

(∂Gij(x, t; y, τ)

∂xk+∂Gkj(x, t; y, τ)

∂xi

)− pj(x, t; y, τ)δik

]nk(x)

= Tji(y, t;x, τ),

(111)

where Tij is defined by (34).We have the following theorem regarding the jump relation of the singlelayer potential in a form that is suitable for handling the boundary condition(108). Its proof is entirely similar to that of Theorem 6.1.

Theorem 7.3. (Jump relation of the single layer potential operator) Sup-pose that φ is a continuously differentiable function on S × [0, T ]. Then

1. the integrals∫S

∫ t0 T∗ij(x, t; y, τ)φj(y, τ)dsydτ (i = 1, 2) exist in the

usual sense for x ∈ S,

2. for x ∈ S, the following jump relation holds:

limε→0+

σik(x± εnk(x), t)nk(x) = σik(x, t)nk(x)∓ 1

2φi(x, t), (112)

26

3. u and p are solutions of initial-boundary value problem specified byequations (105)–(108) if the density φ satifies the following equations:

1

2φi(x, t) +

∫ t

0

∫ST ∗ij(x, t; y, τ)φj(y, τ)dsydτ = fi(x, t), i = 1, 2.

(113)

Remark 7.4. For exterior problems, the boundary integral equations havethe form (as compared with (113) for interior problems)

−1

2φi(x, t)+

∫ t

0

∫ST ∗ij(x, t; y, τ)φj(y, τ)dsydτ = fi(x, t), i = 1, 2. (114)

Remark 7.5. The existence and uniqueness of the solution to the bound-ary integral equation (113) associated with interior Neumann problem wasproved in [38] for fi ∈ L2(S) (i = 1, 2) when S is a Lipschitz curve. For exte-rior Neumann problems, simple integration by parts shows that the nullspaceof (114) contains all functions of the form g(t)n(x) with g(t) an arbitraryfunction in L2([0, T ]) and n(x) the unit outward normal function on S. Sincethe kernel T ∗ij is adjoint to Tij in spatial variable, it is easy to see that theboundary integral equation (102) associated with exterior Dirichlet problemhas a unique solution; and that the boundary integral equation (71) asso-ciated with interior Dirichlet problem has a nullspace containing functionsof the form g(t)η(x) with g(t) an arbitrary function in L2([0, T ]) and η aparticular function in L2(S). Finally, the existence of the solution to (71)can be proved in a similar fasion as in [38].

8 Conclusions and discussions

We have analyzed the properties of the unsteady Stokeslet and developedan analytical machinery for unsteady Stokes flows in two dimensions. It isshown that the unsteady Stokeslet is a linear combination of the Laplacekernel and the heat kernel (and their derivatives and integrals). The initialvalue problem, the nonhomogeneous problem, the Dirichlet problem, andthe Neumann problem have been analyzed and each of them has a simpleintegral equation formulation via the initial potential, the volume potential,double layer potential, and single layer potential, respectively. Since theproblem is linear, the more general initial-boundary value problem withnonhomogeneous problem can be solved via integral equations by simplesuperpositions.

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The analysis here can be easily carried out for three dimensional unsteadyStokes flows and will be discussed in a subsequent paper. The numericalapplications of these integral equation formulations are also currently underinvestigation and will be reported on a later date. We would like to pointout that a full numerical machinery for solving the heat equation usingintegral equations has been developed (see, for example, [19, 28, 15, 43, 44]).The resulting algorithm is accurate, robust, and nearly optimal in termsof computational speed and storage requirement. Though details will bedifferent, we expect that the numerical algorithm for the unsteady Stokesflow based on the integral equation formulations developed in this paper willalso be accurate, robust, and nearly optimal in computational complexity.Finally, we expect that the integral equation formulations developed in thispaper can be used to solve the fully nonlinear Navier-Stokes equation in asimlilar fasion as in [12].

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