Special Applied Mathematics Course Boundary Integral Methods for Fluid Flow

Special Applied Mathematics Course Boundary Integral Methods For Fluid Flow

Contents 1. Note 2. Introduction and revision 3. Boundary integral methods for potential flow 4. Boundary integral methods for Stokes flow 5. Examples of Green’s functions for Stokes flow 6. The Boundary Integral Equation for Stokes flow 7. The Boundary Integral Method for Stokes flow 8. Applications 2

1 Note

These lecture notes draw heavily on the contents of the following book: Pozrikidis, C.: Boundary Integral & Singularity Methods for Linearised Viscous Flow, Cam- bridge. 3

2 Introduction and revision

2.1 Index notation In standard index notation, we represent the components of a vector by a subscript, so

ui for i = 1,..., 3 represents the vector u = (u1, u2, u3). We will make extensive use in this course of the Kronecker delta, which is deﬁned as

0 if i 6= j δ = ij 1 if i = j So δ12 = 0 and δ22 =1 and so on. We will also need to use the alternating tensor deﬁned as

0 if any of i, j, k are equal ǫ = 1 if i, j, k are in cyclic order ijk   −1 otherwise, so, for example, ǫ123 = ǫ231 = 1, ǫ132 = −1, ǫ133 = 0. The alternating tensor provides a convenient way of expressing a vector (cross) product in index notation.

Exercise: Show that ǫijkajbk is equal to the ith component of a × b by writing out the components.

Einstein’s summation convention: This states that whenever we see an index repeated, we sum over that index. The range of the sum will be obvious from the context. So, in three dimensions, i = 1, 2, 3 and 3 aii means aii. i X=1 2.2 The delta function The delta function is an example of a generalised function or distribution. Informally, we express the delta function as

0 if x 6= 0 δ(x)= ∞ if x = 0 We will not need the formal theory behind the delta function here (see any book on Func- tional Analysis). However, we will make use of the following property. Sifting property: The action of a delta function on another function inside an integral is to pick out the value of that function at a particular point. In fact, ∞ δ(x)f(x)dx = f(0). −∞ Z 4

More generally, ∞ δ(x − a)f(x)dx = f(a). −∞ Z The delta function sifts out all values of f except that at x = a.

2.3 Principal value integrals Recall that a principal value integral is a particular way of interpreting an improper (i.e. divergent) integral. For example, consider the integral,

1 1 dx. − x Z 1 There is an obvious problem with the integrand at x = 0 – it blows up! Working na¨ıvely, we integrate to get

1 [ ln |x| ]−1 = ln |1|− ln |− 1| = ln 1 − ln 1 = 0.

But to get this result, we have fudged over the difficulty at x = 0. So let’s do the first integral again but more carefully this time. First, anticipating a problem at x = 0, we re-write it as − ǫ 1 1 1 lim dx + dx . → ǫ 0 −1 x ǫ x h Z Z i Integrating, we have lim ln |ǫ|− ln |− 1| + ln |1|− ln |ǫ| ǫ→0 h = 0 i as before. That’s fine, but a little thought shows that we could equally have written the integral as − 2ǫ 1 1 1 lim dx + dx . → ǫ 0 −1 x ǫ x h Z Z i Now if we integrate, we get

lim ln |2ǫ|− ln |− 1| + ln |1|− ln |ǫ| ǫ→0 h i = ln 2 6= 0. So we get a completely different result. This is no real surprise as the original integral we are working with is divergent! However, at least we see that it is possible to view the integral in such a way that we get a finite result. At this stage, since there are different ways to interpret the improper integral, we find it useful to define a principal way of making that interpretation. We define the principal value of the integral, writing 1 1 P V dx = 0. − x Z 1 5

That is, we choose the first interpretation, that − 1 1 ǫ 1 1 1 P V dx = lim dx + dx = 0. → −1 x ǫ 0 −1 x ǫ x Z h Z Z i An analogy The question arises: why choose the principal value of the integral? Why not define some different value? We will see that, in the context of zero Reynolds number fluid motion, by taking the principal value we are able to produce a continuous velocity field over the flow domain. By way of a simpler analogy, consider the function, sin x f(x)= . x Clearly, f(0) is undefined. We know that

lim f(x) = 1, x→0 by using the Taylor series expansion. This suggests that if we are interested in producing a continuous function, we should make the deﬁnition f(0) = 1 (a little like deﬁning the principal value of the integral above). In this way, the function

1 if x = 0 f(x)= sinx if x 6= 0 x is made continuous and diﬀerentiable everywhere.

2.4 Boundary Integral Methods Over the last 30 or so years, a very popular method for solving Stokes flows in intricate geometries has developed and matured. This nifty method calculates a Stokes flow solely by reference to what happens at the flow boundaries. As we will see, we can write the velocity at any point in the flow field purely in terms of the flow quantites on the boundaries. The method is called the boundary integral method. Its advantage lies in the following observation:

• Since reference is only made to boundary values, the dimension of the problem is eﬀectively reduced by one. So a three dimensional problem eﬀectively becomes a two dimensional problem. This brings about a tremendous saving in computing time.

Its power lies in the following observation • The boundary integral method can cope with any geometry at all. Since many practical applications involve very complex geometries, this constitutes a major advantage over other methods such as ﬁnite diﬀerences which are very clumsy if the geometry is not simple. 6

The boundary integral method can be applied to both potential flows and Stokes flows. Since the implementation of the former is slightly easier, it is these which we shall discuss first.

3 Boundary integral methods for potential ﬂow

We start with a discussion of boundary integral methods for potential ﬂow. The fundamental ideas are the same as for Stokes ﬂow.

3.0.1 Revision of potential ﬂow

Recall from Hydrodynamics I that a potential ﬂow refers to a ﬂow which is both inviscid and irrotational. The latter means that

∇× u = 0.

So we can deﬁne a scalar potential, φ, such that

u = ∇φ.

In addition, conservation of mass requires that

∇ · u = 0.

Substituting u = ∇φ, we ﬁnd

∇2φ = 0. (3.1)

So a potential flow satisfies Laplace’s equation. Laplace’s equation arises in a wide variety of applied topics including elasticity, electrodynamics, magnetism, and so on. For a potential flow, the boundary condition at a solid wall is

u · n = 0, where n is the normal to the wall. This is called the normal flow condition. It states that fluid cannot flow through the solid wall. Since u = ∇φ, ∂φ ∇φ · n =0 or = 0. (3.2) ∂n

So, a typical problem might require us to solve (3.1) subject to (3.2). Under what circum- stances might we want to use the boundary integral method? Well, if we are asked to solve Laplace’s equation in a nice simple geometry, we can do that fairly easily. For example, we can solve the problem ∇2φ = 0 7 in the circle 0 ≤ r ≤ 1 with φ =1on r = 1. In polars, assuming no θ dependence, Laplace’s equation becomes 1 d dφ r = 0. r dr dr Integrating, we ﬁnd φ = A log r + B. We set A = 0 to prevent blow-up at r = 0, leaving φ = 1 as the solution. But what if the domain is not a circle but is instead some complicated geometry that is not easy to deﬁne mathematically? Then we cannot work out the solution so easily and the boundary integral method comes in very useful.

3.1 The boundary integral equation for potential ﬂow in 2D To motivate the method in two-dimensions, we start with Green’s second identity,

ψ∇2φ − φ∇2ψ = ∇ · (ψ∇φ − φ∇ψ). (3.3)

Suppose we want to solve Laplace’s equation over a domain D, with n the unit normal at the boundary pointing into the solution domain. Suppose now we choose ψ = G, where G satisﬁes the following equation

2 ∇ G + δ(x − x0) = 0, (3.4) and may (or may not) be required to satisfy a given condition on a given boundary. The second term is the Dirac delta function, deﬁned so that

0 if x 6= x δ(x − x )= 0 0 ∞ if x = x 0 We call G a fundamental solution of Laplace’s equation or Green’s function1 (we will discuss these in more detail later). We saw above that a solution to the problem for G is given by

G = λ log r, (3.5) where λ is some constant. The form (3.5) is called the free-space Green’s function. The value of λ will be chosen in a moment.

We now wish to integrate (3.3) over D. To avoid the singularity at x = x0 we delete from ′ ′ D a small disk, Dǫ, of radius ǫ centred at x0. Let D = D − Dǫ. Integrating over D , we have (G∇2φ − φ∇2G)dS = ∇ · (G∇φ − φ∇G)dS = 0 ′ ′ ZD ZD 1Strictly speaking one should only refer to this as a Green’s function when it satisﬁes a boundary value problem, that is G adopts prescribed values on a given boundary 8 since both ∇2G = 0 and ∇2φ = 0 within D′. Now, since the integrand in the middle integral is regular throughout D′, we may apply the divergence theorem to obtain

(G∇φ − φ∇G) · n dl = 0, ZC+Cǫ where C is the boundary to D and Cǫ is the boundary to Dǫ, and l is arc length along a contour.

Consider now the integral around Cǫ,

Iǫ = (G∇φ − φ∇G) · n dl = 0. ZCǫ In the limit ǫ → 0, the radius of the disk shrinks to zero. It is known that sufficiently close to the singular point, the Green’s function G may be approximated by its free-space form given above, G(x, x0)= λ log r. Accordingly, ∂φ φ I = λ log r − λ dl. ǫ ∂r r ZCǫ Since r = ǫ on Cǫ, letting ǫ → 0, we find ∂φ φ(x ) I = λ log ǫ (x ) dl − λ 0 dl ǫ ∂r 0 ǫ ZCǫ ZCǫ ∂φ φ(x ) = 2πλǫ log ǫ (x ) − 2πλǫ 0 = −2πλφ(x ) ∂r 0 ǫ 0 in the limit ǫ → 0. So, 2 2 (G∇ φ − φ∇ G)dS = 2πλφ(x0). ZC Rearranging, 1 φ(x )= − (φ∇2G − G∇2φ)dS. 0 2πλ ZC If we choose λ = −1/2π, we obtain the boundary integral equation for potential flow

φ(x0)= (φ∇G − G∇φ) · n dl. (3.6) ZC This means that the free-space Green’s function (3.5) is chosen to adopt the form 1 G(x, x )= − log r. 0 2π

Notes 9

1. Equation (3.6) applies for any Green’s function satisfying (3.4) and any chosen boundary conditions. In the particular case where we choose G to be the free-space Green’s function, we find 1 φ(x )= − (φ∇ log r − log r∇φ) · n dl. 0 2π ZC Or, 1 φ(x )= log r n ·∇φ dl − φn ·∇ log r dl . (3.7) 0 2π ZC ZC This is called Green’s third identity. 2. In some problems, it may prove expedient to make another choice for the Green’s function. We’ll see examples as we go along. Remember that the Green’s function G solves the equation 2 ∇ G + δ(x − x0) = 0. By taking different boundary conditions for G we can derive different Green’s functions. 3. Equation (3.7) is an example of an integral equation. Although φ appears as the subject on the left hand side, the function φ and its derivatives also appear on the right hand side inside the integrals.

How do we solve for φ? On the RHS of (3.9) φ appears inside integrals which are deﬁned over the boundary C. So the right hand side is only concerned with boundary values of φ. This suggests the following way to proceed:

(i) Take the ﬁeld point x0 to lie on the boundary C. Then the integral equation involves only values of φ on the boundary. (ii) Solve the integral equation for the boundary values of φ (see later discussion). (iii) Once the boundary values of φ are known we can compute the right hand side of (3.7) for any x0. Hence we can ﬁnd φ everywhere in the domain D.

Points (i-iii) form the basic steps of the boundary integral method. The same basic proce- dure also applies to Stokes ﬂows, as we shall see later.

The ﬁrst step (i) above requires that we take the ﬁeld point x0 to lie on C. Let’s work with a general Green’s function and keep the boundary integral equation in the form,

φ(x0)= φ(x) n ·∇G(x, x0)dl(x) − G(x, x0) n ·∇φ(x)dl(x). (3.8) ZC ZC We’ve explicitly written out the dependencies of each function for clarity. On the right hand side of (3.8), there are two integrals,

G(x, x0) n ·∇φ(x)dl(x) and φ(x) n ·∇G(x, x0)dl(x). ZC ZC 10

We call the ﬁrst integral the single layer potential (SLP) We call the second integral the double layer potential (DLP)

How does each behave as x0 approaches C, as is required by step (i)? It can be shown that the SLP is continuous as the point x0 approaches and then crosses C. In contrast, as x0 crosses C, the double-layer potential undergoes a jump discontinuity so that

P V 1 lim { φ(x) [n ·∇G(x, x0)] dl(x)} = φ(x)[n ·∇G(x, x0)] dl(x) ± φ(x0), x0→C 2 ZC ZC where the +/− sign applies if x0 approaches the contour C from the inside/outside re- spectively2 (see the Problem Sheet). The superscript P V means that we are taking the principal value of the integral. To complete step (i) of the boundary integral method, we let the singular point x0 approach C from the inside in (3.8) and ﬁnd,

P V 1 φ(x )= φ(x) [n ·∇G(x, x )] dl(x)+ φ(x ) − G(x, x ) n ·∇φ(x)dl(x). 0 0 2 0 0 ZC ZC So, 1 P V φ(x )= φ(x) [n ·∇G(x, x )] dl(x) − G(x, x ) n ·∇φ(x)dl(x). 2 0 0 0 ZC ZC We now introduce the ﬂux q, which is deﬁned3 so that q ≡ n ·∇φ. Then we have

1 P V φ(x )= φ(x) [n ·∇G(x, x )] dl(x) − G(x, x ) q(x)dl(x). (3.9) 2 0 0 0 ZC ZC If φ is prescribed on the boundary, then we have a Fredholm integral equation of the ﬁrst kind for the ﬂux q,

G(x, x0) q(x)dl(x)= F (x0), ZC where P V 1 F (x )= φ(x) [n ·∇G(x, x )] dl(x) − φ(x ). 0 0 2 0 ZC It is called an integral equation of the ﬁrst kind since the subject (in this case q) appears inside the integral but not on the right hand side. If instead q is prescribed on the boundary C, we have a Fredholm integral equation of the second kind for the boundary distribution of φ,

1 P V φ(x )= φ(x) [n ·∇G(x, x )] dl(x)+Φ(x ), 2 0 0 0 ZC 2For multiply-connected regions, inside is taken to mean the area into which the normal vector points 3Why do we call this the flux? Fick’s law states that a substance flows down its concentration gradient (like heat, for example). If φ represents the concentration of some substance, then −∇φ is its flow rate according to Fick’s law. Then the flux out of a closed region with inward pointing normal n will be n ·∇φ. 11 where

Φ(x0)= − G(x, x0) q(x)dl(x). ZC It is called an integral equation of the second kind because the subject (in this case φ) appears both inside the integral and outside of it.

3.2 The boundary integral equation at a corner The arguments in the previous section relied on the use of the divergence theorem, which is applicable provided that the boundary C is piecewise smooth4. However, there is a difficulty with allowing the point x0 to approach a point on a curve, such as a sharp corner, where the normal vector is not continuous. In this case, the argument needs some modification. In practice, we are often interested in computing solutions in domains which involve corners or kinks. In this section, we address how to adapt the boundary integral method for potential flow to such regions. To proceed, we go back to the beginning of the argument: Green’s second identity. This states that ψ∇2φ − φ∇2ψ = ∇ · (ψ∇φ − φ∇ψ). Integrating over the domain, D, we obtain

ψ∇2φ − φ∇2ψ dS = ∇ · (ψ∇φ − φ∇ψ)dS ZD ZD = (ψ n ·∇φ − φ n ·∇ψ)dS ZC by the divergence theorem, which we note is still valid for a boundary with a corner. As before, φ satisﬁes Laplace’s equation, ∇2φ = 0. For the sake of argument, we take ψ to be the free-space Green’s function, 1 ψ = G = − log r, 2π satisfying 2 ∇ G + δ(x − x0) = 0. So, − φ∇2G dS = (G n ·∇φ − φ n ·∇G)dS ZD ZC 2 Consider now a point x0 which lies outside of the domain D. It follows that ∇ G = 0 leaving 1 0= (G n ·∇φ − φ n ·∇G)dS. 2π ZC 4A curve is smooth if its normal vector is a continuous function of position. Such a curve or surface is also called Lyapunov. A curve is piecewise smooth if its normal vector is continuous at ﬁnitely many points. One such example is a square (e.g., Kreysig §10.5). 12

Substituting for G, 1 0= (log r n ·∇φ − φ n ·∇ log r)dS. 2π ZC Since r = |x − x0|, ∂ ∂ x − x y − y x − x ∇ log r = , log r = 0 , 0 = 0 . ∂x ∂y r2 r2 r2 Thus, 1 x − x 0= (log r n ·∇φ − φ n · 0 )dl (3.10) 2π r2 ZC

(a) (b)

x 0

n θ n ε x 0

Figure 1: A domain with a sharp corner of angle α.

Now consider a domain including a sharp corner of local angle α as shown in figure 1. A close-up of the corner is shown to the right of the figure. To proceed with the formulation of the method, we deform the contour locally to skirt round the inside of the corner through a small arc of radius ǫ as shown in figure 1(b). We will call this circular arc Γ. Note that the point x0 lies outside of the deformed contour, so the most recent equation is applicable. Over Γ, the contribution from the right hand side is 1 x − x (log r n ·∇φ − φ n · 0 )dl. 2π r2 ZΓ On the arc, r = ǫ, x − x0 = ǫ cos θ and y − y0 = ǫ sin θ, where θ is the angle shown, and dl = ǫ dθ giving, 1 α x − x ǫ (log ǫ n ·∇φ − φ n · 0 )dθ. 2π ǫ2 Z0 The unit normal to the arc x − x n = 0 , r 13 leaving 1 α |x − x |2 ǫ (log ǫ n ·∇φ − φ 0 )dθ 2π ǫ3 Z0 1 α φ = ǫ (log ǫ n ·∇φ − )dθ. 2π ǫ Z0 Also, x − x x − x ∂φ y − y ∂φ n ·∇φ = 0 ·∇φ = 0 + 0 , r r ∂x r ∂y giving 1 α ∂φ ∂φ φ ǫ log ǫ cos θ + sin θ − dθ 2π ∂x ∂y ǫ Z0 (since x − x0 = ǫ cos θ and y − y0 = ǫ sin θ)

1 α ∂φ ∂φ = ǫ log ǫ cos θ + sin θ − φ dθ. 2π ∂x ∂y Z0 Now we let the arc Γ shrink to the corner point by taking the limit ǫ → 0, i.e. we consider

1 α ∂φ ∂φ lim ǫ log ǫ cos θ + sin θ − φ dθ ǫ→0 2π ∂x ∂y Z0

Note that limǫ→0 ǫ log ǫ = 0. So, 1 α x − x ∂φ y − y ∂φ lim ǫ log ǫ 0 + 0 − φ(x) dθ ǫ→0 2π r ∂x r ∂y Z0 1 α α = − φ(x ) dθ = − φ(x ). 2π 0 2π 0 Z0 In summary, 1 x − x α (log r n ·∇φ − φ n · 0 )dl = − φ(x ). 2π r2 2π 0 ZΓ So, 1 x − x (log r n ·∇φ − φ n · 0 )dl 2π r2 ZC 1 x − x 1 x − x = P V (log r n ·∇φ − φ n · 0 )dl + lim (log r n ·∇φ − φ n · 0 )dl, 2π r2 ǫ→0 2π r2 ZC ZΓ where the principal value P V refers to the fact that the small arc of radius ǫ has been excluded from the integration. Thus, (3.10) becomes 1 x − x (log r n ·∇φ − φ n · 0 )dl 2π r2 ZC 1 x − x α = P V (log r n ·∇φ − φ n · 0 )dl − φ(x ) = 0. 2π r2 2π 0 ZC 14

Rearranging, α 1 x − x φ(x )= P V (log r n ·∇φ − φ n · 0 )dl 2π 0 2π r2 ZC The previous equation was derived working with the free-space Green’s function. But it is equally applicable to any Green’s function. So we may write in general,

α P V φ(x )= − G(x, x ) n ·∇φ(x)dl(x)+ φ(x) n ·∇G(x, x )dl(x). 2π 0 0 0 ZC ZC Note that the P V is only required on the 2nd integral as the ﬁrst is continuous as x0 approaches the contour.

Note: If α = π, then there is no corner, and we recover our original boundary integral equation (3.9),

1 P V φ(x )= − G(x, x ) n ·∇φ(x)dl(x)+ φ(x) n ·∇G(x, x )dl(x), 2 0 0 0 ZC ZC 3.3 Summary: Potential Flow In this section we have:

• starting from Green’s second identity, derived the boundary integral equation (BIE) for potential ﬂow valid for a point x0 inside the domain of interest.

• introduced the notion of a Green’s function

• using the notion of a principal value, obtained the form of the BIE valid for a point x0 lying on the contour enclosing the domain of interest.

• derived the form of the BIE valid for a contour with a sharp kink 15

4 Boundary integral methods for Stokes ﬂow

The equations describing creeping fluid motion are called the Stokes equations 0 = −∇P + ρF + µ∇2u, ∇ · u = 0, (4.1) where ρ is the fluid density, µ is the viscosity, and F is a body force acting per unit mass (e.g., gravity). The first equation is the momentum equation for Stokes flow and is an expression of Newton’s second law of motion. The second equation is the continuity equation expressing conservation of mass. You will note that equation (4.1) is linear. This is a crucial property for the theory which we shall develop in the following lectures.

4.1 Derivation of the governing equations 4.1.1 Internal stress in a fluid What is the stress at any point in a fluid due to the surrounding fluid? Consider a parcel of fluid of volume V and surface S inside a larger fluid. The force acting on an element of the surface dS with unit outward normal n is defined to be

(σij nj)dS or (σ · n)dS. So the total force acting on the surface S is

σij nj dS. ZS 4.1.2 Conservation of mass Consider a closed volume V which is fixed in space inside a moving fluid. As the fluid moves past, the rate of change of the mass of fluid within V must equal the net amount of fluid coming into (or out of) it, assuming there are no sources of fluid within V . Mathematically, d ρ dV + ρu · n dS = 0, dt ZV ZS where ρ is the fluid density, S is the volume surface, and n is the normal to V pointing outwards. Using the fact that V is fixed in space and applying the divergence theorem, this becomes ∂ρ + ∇ · (ρu) dV = 0. ∂t ZV Since our choice of volume V wasn arbitrary, it musto be true that ∂ρ + ∇ · (ρu) = 0. (4.2) ∂t If ρ is constant, this reduces to ∇ · u = 0. (4.3) Equation (4.3) therefore expresses the principle of mass conservation. It is usually referred to as the continuity equation. 16

4.1.3 Newton’s second law

We now derive the equation of motion of a slow moving viscous fluid. Let F represent some body force per unit mass acting on the fluid (e.g., gravity). We apply Newton’s second law of motion to a volume of fluid V (t), with surface area S(t) and unit outward normal n, moving with the flow. Thus, d ρu dV = ρ F dV + σ n dS. dt i i ij j ZV (t) ZV (t) ZS(t) The left hand side expresses the rate of change of momentum of the fluid; the right hand side the forces acting on the fluid. The last term represents the stresses acting over the surface of the chosen volume by the surrounding fluid. For a fluid moving under conditions of Stokes flow, the momentum term is neglected, leaving

0 = ρ Fi dV + σijnj dS. ZV (t) ZS(t)

Applying the divergence theorem to the last term on the right hand side, we ﬁnd ∂σ 0 = ρ F dV + ij dV. (4.4) i ∂x ZV (t) ZV (t) j

Assuming that the ﬂuid is Newtonian, we may use the constitutive relation relating the stress in a ﬂuid to the rate of strain

σij = −P δij + 2µeij, where 1 ∂u ∂u e = i + j ij 2 ∂x ∂x j i is called the rate of strain tensor. This is a measure of how rapidly ﬂuid parcels are being stretched. Accordingly, ∂σij ∂P ∂ = − + 2 (µeij). ∂xj ∂xi ∂xj

Substituting this relation into the above, we obtain ∂P ∂ −ρ Fi + − 2 (µeij)dV = 0. (4.5) ∂x ∂x ZV (t) i j

Assuming the viscosity µ is constant, (4.5) therefore reduces to

2 2 ∂P ∂ ui ∂ uj − ρ Fi + − µ 2 + dV = 0. (4.6) V (t) ∂xi ∂xj ∂xi∂xj ! Z n o 17

But ∂uj/∂xj ≡ ∇ · u = 0. Also, since our original volume was chosen arbitrarily, the integrand in (4.6) must vanish, as the statement must be true for all possible volume choices. Finally, we have 2 ∂ui ∂P ∂ ui = 0, 0 = ρ Fi − + µ 2 , ∂xi ∂xi ∂xj or

∇ · u = 0, 0 = ρ F −∇P + µ∇2u. (4.7)

These are the Stokes equations for an incompressible Newtonian ﬂuid moving under the conditions of Stokes ﬂow.

4.2 The Lorentz reciprocal relation The boundary integral method for potential flow discussed above took as its point of de- parture Green’s second identity (3.3). To develop the boundary integral method for Stokes flow, we use a similar identity valid for a velocity and pressure field which satisfy the Stokes equations of motion (4.1). It is called the Lorentz reciprocal relation. It was derived by H. A. Lorentz in 1907 (his paper is reproduced in the Journal of Engineering Mathematics 1996, volume 30). Before introducing the relation, we take a step back and recast (4.1) in terms of the stress tensor for viscous flow. The stress tensor σ, or in index notation, σij, was defined in section (4.1.3) for a Newtonian fluid as

σij = −P δij + 2µeij, (4.8) where P is the fluid pressure, δij is the Kronecker delta, and eij is the rate of strain tensor, defined as 1 ∂u ∂u e = i + j . ij 2 ∂x ∂x j i We only discuss Newtonian fluids in this course. Taking the divergence of (4.8), we obtain

∂σij ∂ ∇ · σ = = (−P δij + 2µeij ) ∂xj ∂xj ∂P ∂ ∂u ∂u = − + µ i + j , ∂x ∂x ∂x ∂x i j j i 2 ∂P ∂ ui ∂ ∂uj = − + µ 2 + µ . ∂xi ∂xj ∂xi ∂xj But, from the continuity equation, ∂u i = 0, (4.9) ∂xi 18 and so we are left with 2 ∂σij ∂P ∂ ui = − + µ 2 . ∂xj ∂xi ∂xj In vector form,

∇ · σ = −∇P + µ∇2u.

We now see that an alternative way to writing the Stokes ﬂow momentum equation with no body force (F = 0), 0 = −∇P + µ∇2u, is to write

∇ · σ = 0. (4.10)

In index notation, ∂σ ij = 0. ∂xj

We are now in a position to derive the reciprocal relation for zero Reynolds number flow. 5 ′ Consider two different, independent regular Stokes flows with velocity fields ui and ui, ′ and respective stress tensors, σij and σij. In other words we have two flows satisfying the momentum equations ′ ∇ · σ = 0 and ∇ · σ = 0 We now compute ′ ′ ∂σij ∂ ′ ∂ui ui = (uiσij) − σij , ∂xj ∂xj ∂xj by the product rule of differentiation. Continuing, ′ ∂ ′ ∂u ∂u ∂u = (u σ ) − −P δ + µ i + j i , ∂x i ij ij ∂x ∂x ∂x j j i j ′ ′ ∂ ′ ∂ui ∂ui ∂uj ∂ui = (u σij)+ P − µ + . ∂x i ∂x ∂x ∂x ∂x j i j i j So ′ ′ ∂σij ∂ ′ ∂ui ∂uj ∂ui u = (u σij) − µ + , (4.11) i ∂x ∂x i ∂x ∂x ∂x j j j i j by continuity. By swapping primed and unprimed variables we may also write down

′ ′ ′ ∂σij ∂ ′ ∂ui ∂uj ∂ui ui = (uiσ ) − µ + . (4.12) ∂x ∂x ij ∂x ∂x ∂x j j j i j 5By regular we mean there are no singular points in the domain of interest, that is neither of the velocity ﬁelds or the stress ﬁelds blow up in the domain. 19

Now (4.11)-(4.12) yields

′ ′ ∂σij ∂σij ∂ ′ ′ ui − ui = (uiσij − uiσij). ∂xj ∂xj ∂xj

But, by the momentum equation for each ﬂow, ′ ∂σ ∂σ ij = 0 ij = 0, ∂xj ∂xj and we are left with

∂ ′ ′ (uiσij − uiσij) = 0. (4.13) ∂xj

This is the Lorentz reciprocal identity for Stokes ﬂow. In vector form it is written as

′ ′ ∇ · (u · σ − u · σ ) = 0.

Note: To derive the above we assumed the two flows were for fluids with the same viscosity. To generalize to flows with different viscosities, see the Problem Sheet.

4.3 Green’s functions Before proceeding to the boundary integral formulation, we must first discuss the notion of a Green’s function for Stokes flow. These are analagous to Green’s functions that you may have encountered previously. A Green’s function results from a singularly forced differential equation. For example, consider simple harmonic motion of a mass on a spring. If x(t) is the displacement from equilibrium at time t, the equation of motion is by Newton’s second law d2x m + kx = 0, dt2 where k is the stiffness of the spring. If we apply an additional force, F (t), to the spring, the equation becomes d2x m + kx = F (t). dt2 If F (t) is a nice smooth function, we say this is a regularly forced equation. But if F is a delta function, which is a singular function, we have

d2x m + kx = δ(t), dt2 and we say the system is singularly forced. A practical example would be hitting the mass suddenly with a hammer. The sudden blow of the hammer can be modelled with a delta function. We call the solution to the singularly forced equation a fundamental solution or Green’s function. 20

Equally, we can add a forcing term to the equation for Stokes ﬂow. For example, including the eﬀect of gravity, the momentum equation becomes

0 = −∇P + ρg + µ∇2u, where ρ is the ﬂuid density. If instead we singularly force the Stokes ﬂow with a delta function, we have

2 0 = −∇P + b δ(x − x0)+ µ∇ u. (4.14)

We have included the forcing term b δ(x − x0).

This represents a point force applied at the point x0 in space pointing in the b direction, where b = bxi + byj + bzk is a constant vector. b

We will write the solution to (4.14) in the general form 1 u = G (x, x )b , (4.15) i 8πµ ij 0 j where the function Gij is as yet to be determined. The factor of 1/8πµ is purely for convenience. We call (4.15) a Green’s function. Note that the Green’s function

G(x, x0) depends on the current position in space, x, and the position of the point force x0.

We call x the ﬁeld point. We call x0 the pole. Note that summation over j is implied in (4.15), so it really means

3 1 ui = Gijbj, 8πµ j X=1 We write a similar expression for the pressure solution, 1 P = p b (4.16) 8π j j where pj is yet to be determined. The factor of 1/8π is purely for convenience. 21

In index notation, the momentum equation is

2 ∂P ∂ ui 0= − + bi δ(xi − xO,i)+ µ 2 . ∂xi ∂xk Substituting in the Green’s function solution (4.15) and (4.16), we ﬁnd

2 1 ∂pj 1 ∂ Gij 0= − bj + bi δ(xi − xO,i)+ 2 bj. 8π ∂xi 8π ∂xk This can be re-written as

2 1 ∂pj 1 ∂ Gij 0= − bj + δijbj δ(xi − xO,i)+ 2 bj, 8π ∂xi 8π ∂xk in which case the arbitrary common factor bj may be removed to leave

2 ∂pj ∂ Gij −8πδ(xi − x0,i)= − + 2 . (4.17) ∂xi ∂xk

So we must ensure that the Green’s function Gij and the pressure vector pj satisfy this equation. In due course we will consider speciﬁc solutions of (4.17), that is speciﬁc examples of Green’s functions. First, we discuss generic properties held by all possible Green’s functions.

4.3.1 Properties of Green’s functions for Stokes ﬂow 1. Conservation of mass

We have shown that a Stokes ﬂow Green’s function, Gij is a solution of equation (4.17), together with a suitable choice for the pressure vector pj. The velocity ﬁeld associated with the Green’s function was given above,

ui = Gijbj.

Taking the divergence, we have

∂ui ∂ = (Gijbj) = 0, ∂xi ∂xi since ∂u i = 0, ∂xi by continuity. Suppose we integrate over a closed volume V with unit outward normal ni, then ∂ui ∂ dV = (Gijbj)dV = 0. ∂x ∂x ZV i ZV i 22

Applying the divergence theorem6, we obtain ∂ (Gijbj)dV = niGijbj dS = 0. ∂x ZV i ZS Since b is a constant vector, this shows that

Gij(x, x0)ni dS = 0. (4.18) ZS This integral identity is satisfied by all Green’s functions and effectively expresses conservation of mass of the flow.

2. Symmetry of the Green’s function A Green’s function is symmetric in the following sense:

Gij(x, x0)= Gji(x0, x). i.e. swapping i and j and also x and x0 leaves the Green’s function unchanged. Proof: See Appendix C.

5 Examples of Green’s functions for Stokes ﬂow

As we will see, Green’s functions will form a vital component of the boundary integral formulation for Stokes ﬂow. To consolidate our understanding of Green’s functions, we now compute some examples.

5.1 The free space Green’s function or Stokeslet Consider a point force in free space. We call the associated Green’s function the free space Green’s function or Stokeslet. The analysis depends on whether we are in two or three dimensions.

5.1.1 Stokeslet in 3D As discussed above, the Green’s function will be a solution of the equation

2 0 = −∇P + b δ(x − x0)+ µ∇ u, with 1 ui = Gij (x, x )bj. 8πµ 0 Here the forcing term b δ(x − x0)

6For a more formal argument, see Appendix D 23 becomes inﬁnite as x → x0 only and so corresponds to an isolated singularity in free space. We wish to ﬁnd G. To proceed, we re-express the delta function in the more convenient form, (see Appendix A) 1 δ(xˆ)= −∇2 . 4πr We take the singularly forced Stokes equation, namely,

2 0 = −∇P + b δ(x − x0)+ µ∇ u, and write it in the equivalent form b 1 ∇2 = −∇P + µ∇2u, (5.1) 4π r How do we deal with the pressure term? The Stokes pressure is a harmonic function, i.e. it satisﬁes Laplace’s equation, ∇2P = 0 (this is left as an exercise). If we write

1 ∂ ∂ ∂ 1 1 1 P = − b · , , = − b ·∇ 4π ∂x ∂y ∂z r 4π r we can conﬁrm that P is a harmonic function (see Problem Sheet 1). For the next step, we introduce a scalar function H, and write the velocity ﬁeld in the form

1 ∂ ∂ 2 uj = bi · − δij∇ H, µ ∂xˆ ∂xˆ i j where δij is the Kronecker delta. Using a more compact vector notation, we write this as 1 u = b · (∇∇ − I∇2)H, µ where I is the identity matrix,

1 0 0 0 1 0 ,   0 0 1   and ∂ ∂ ∇∇ ≡ ∂xˆi ∂xˆj is an example of a dyadic product. For any two vectors a and b, the dyadic product

ab = aibj.

Note that ab is a matrix with ijth entry aibj. 24

Summary: The singularly forced Stokes momentum equation to be satisﬁed is b 1 ∇2 = −∇P + µ∇2u, (5.2) 4π r and thus far we have written 1 1 P = − b ·∇ , (5.3) 4π r and 1 u = b · (∇∇ − I∇2)H. µ

Now, note that ∂2u 1 ∂2H = b · (∇∇ − I∇2) . ∂xˆ2 µ ∂xˆ2 Consequently, 1 ∇2u = b · (∇∇ − I∇2)∇2H. (5.4) µ Substituting (5.3) and (5.4) into (5.2), we have

1 1 1 1 b · I∇2 = b · ∇∇ + b · (∇∇ − I∇2)∇2H. 4π r 4π r Rearranging, 1 1 1 1 0 = b · ∇∇ − b · I∇2 + b · (∇∇ − I∇2)∇2H, 4π r 4π r or 1 0 = b · ∇∇− I∇2 ∇2H + . 4πr We want this to be true for any choice of b, so 1 0= ∇∇ − I∇2 ∇2H + . (5.5) 4πr We can satisfy then (5.5) if we choose H such that 1 ∇2H = − . (5.6) 4πr To ﬁnd the solution, recall from above that we may express the delta function in the form 1 δ(xˆ)= −∇2 . 4πr Taking ∇2 of both sides of (5.6), 1 ∇4H = −∇2 = δ(xˆ). 4πr 25

So now we know that H satisiﬁes the forced biharmonic equation

∇4H = δ(xˆ).

Recalling that in spherical polars (independent of ψ and ψ),

1 ∂ ∂f ∇2f = r2 = 0, r2 ∂r ∂r we see that, away from x = x0, 1 ∂ ∂ 1 ∂ ∂H ∇4H = r2 r2 = 0. r2 ∂r ∂r r2 ∂r ∂r Integrating, ∂ 1 ∂ ∂H r2 r2 = A, ∂r r2 ∂r ∂r for constant A. Integrating again,

1 ∂ ∂H A r2 = − + B, r2 ∂r ∂r r for constant B. Integrating once more, ∂H A B r2 = − r2 + r3 + C, ∂r 2 3 for constant C. Integrating for a fourth time, A B C H = − r + r2 − + D, 2 6 r for constant D. As usual, we integrate the singularly forced equation over a sphere of radius a centred at the singularity, obtaining

∇4H dV = δ(xˆ)dV = 1, ZV ZV by the sifting property of the delta function. Now,

∇4H = ∇ · (∇∇2H), so ∇ · (∇∇2H)dV = 1, ZV and on applying the divergence theorem,

(∇∇2H) · n dS = 1. ZS 26

But for the sphere, n = ˆr, and so ∂ ∇2H dS = 1. ∂r ZS Now, A B C H = − r + r2 − + D, 2 6 r as found above, and thus 1 ∂ ∂H ∇2H = r2 . r2 ∂r ∂r But 2 1 ∇2r = , ∇2r2 = 6, ∇2 = 0, r r and so A ∇2H = − + B. r Accordingly, ∂ A A A ∇2H dS = dS = dS = 4πa2 = 4πA = 1, ∂r r2 a2 a2 ZS ZS ZS implying that A = 1/4π, and the values of B, C and D are immaterial (since we have just seen that we only need to ﬁx A to be 1/4π to get a solution to the equation). Therefore, we can set B = C = D = 0, leaving r H = − . 8π Originally, we wrote 1 u = b · (∇∇ − I∇2)H. µ So now we know that 1 r u = b · (∇∇ − I∇2) − . µ 8π 1 r 1 r = b · (∇∇) − + b · (I∇2) µ 8π µ 8π Since ∇2r = 2/r, this gives 1 r 1 1 u = b · (∇∇) − + b · I , µ 8π µ 4πr i.e., 1 2 u = b · ∇∇(−r)+ I , 8πµ r In index notation, 1 ∂ ∂ 2 uj = bi (−r)+ δij , 8πµ ∂xˆi ∂xˆj r 27

Noting that ∂r xˆ = j , ∂xˆj r we have 1 ∂ xˆj 2 uj = bi − + δij , 8πµ ∂xˆi r r 1 rδ − xˆ (ˆx /r) 2 = b − ij j i + δ , 8πµ i r2 ij r using the quotient rule of diﬀerentiation and using the fact that

∂xˆj = δij. ∂xî Simplifying, 1 δij xîxˆj 2δij uj = bi − + + , 8πµ r r3 r leaving, 1 xˆ xˆ δ u = b i j + ij . j 8πµ i r3 r We are now in a position to write the velocity field in the form of (4.15), i.e. 1 ui = Gij (x, x )bj, 8πµ 0 where the Stokeslet Green’s function xˆ xˆ δ G = i j + ij . ij r3 r

Note that the Stokeslet is sometimes called the Oseen-Burger’s tensor. We write the pressure associated with the Stokeslet as 1 P = p b , 8π j j (see equation 4.16). Above, we wrote

1 1 1 ˆx 1 2ˆx P = − b ·∇ = − b · − = b · 4π r 4π r3 8π r3 1 2ˆx = b j 8π j r3 in index notation. So the Stokeslet pressure is xˆ p = 2 j . j r3 28

5.1.2 Stokeslet in 2D

In two dimensions, we can follow an analysis similar to that presented in the last section to show that the Stokeslet is given by

xîxˆj Gij = −δij log r + , r2 and the corresponding velocity field is 1 ui = Gij (x, x )bj, 4πµ 0 where the 4π is included purely for convenience. The pressure field is 1 P = p (x, x )b , 4π j 0 j where xˆj pj = 2 . r2

5.2 Green’s function for flow above a plane wall Above, we looked at the free-space Green’s functions in two and three-dimensional space. Sometimes it is more convenient to work with a Green’s function which vanishes over a boundary of the flow. This will become more clear later. For the moment, suppose we are interested in computing a Stokes flow which involves a flat wall at x = −1. We would like to compute a Green’s function G such that

G(x, x0) = 0, whenever the ﬁeld point x lies on the wall. The correct form of the Green’s function is given by Blake (1971) to be

IM D IM SD IM Gij(x, x0)= Sij(x − x0) − Sij(x − x0 ) + 2Gij (x − x0 ) − 2Gij (x − x0 ) (5.7)

IM where x0 is the singular point, and x0 = (−x0 − 2,y0, z0) is its image on reﬂection in the plane wall. Furthermore, δ x x S (x) = ij + i j , ij |x| |x|3

δ x x GD(x) = ± ij − 3 i j , ij |x|3 |x|5 δ x − δ x GSD(x) = x GD(x) ± j1 i i1 j . ij 1 ij |x|3 The plus sign applies for j = 2, 3 and the minus sign for j = 1. Sij is the Stokeslet discussed above. GD is called a potential dipole and GSD a Stokeslet doublet. 29

The construction of (5.7) is complicated and we will not go through it here. Let’s instead just check that it does indeed have the desired property,

Gij(x, x0) = 0, when x = (−1,x,y) i.e. the Green’s function always vanishes on the wall at x = −1.

We will check by considering components of Gij. For simplicity, we consider the case when IM IM x0 = (0, 0, 0). In this case x0 = (−2, 0, 0). Also, note that |x−x0| = |x−x0 |. Henceforth denote |x − x0| = R. We our choice of x0, we have simply

x − x0 = (x,y,z)= x and IM x − x0 = (x + 2,y,z). Since we only look at points on the wall at x = −1, we thus have

x − x0 = (−1,y,z) and IM x − x0 = (1,y,z).

Look at Gyy:

1 y2 S (x − x )= S (x)= + yy 0 yy R R3

1 y2 S (x − xIM )= + = S (x − x ) yy 0 R R3 yy 0

δ x − δ x GSD(x − xIM ) = 1 × GD (x − xIM )+ 21 2 21 2 , yy 0 yy 0 R3 so SD IM D IM Gyy (x − x0 )= Gyy(x − x0 ).

So the full Green’s function gives

IM D IM SD IM Gyy(x − x0)= Syy(x − x0) − Syy(x − x0 ) + 2Gyy(x − x0 ) − 2Gyy (x − x0 )

D IM D IM = Syy(x − x0) − Syy(x − x0) + 2Gyy(x − x0 ) − 2Gyy(x − x0 ) = 0 as required.

Next, look at Gxy:

1 y2 Sxy(x − x )= Sxy(x)= − 0 R R3 30 since x − x0 = (−1,y,z).

IM 1 y Syy(x − x )= + 0 R R3 IM since x − x0 = (1,y,z). δ x − δ x GSD(x − xIM ) = 1 × GD(x − xIM )+ 21 1 11 2 , xy 0 ij 0 R3 so y GSD(x − xIM )= GD (x − xIM ) − , xy 0 xy 0 R3 So the full Green’s function gives

IM D IM SD IM Gxy(x − x0)= Sxy(x − x0) − Sxy(x − x0 ) + 2Gxy(x − x0 ) − 2Gxy (x − x0 )

1 y2 1 y y = − − + + 2GD − 2GD + 2 . R R3 R R3 xy xy R3 = 0 as required.

By the symmetry property of the Green’s function, Gyx will also be zero. See the Problem sheet for the remaining components.

5.3 The stress tensor Thus far we have discussed the velocity field and associated pressure field for flow due to a point force in two or three dimensional space. In 3D, we wrote 1 u = G (x, x )b , (5.8) i 8πµ ij 0 j and 1 P = p (x, x )b . (5.9) 8π j 0 j We can write the stress tensor associated with the flow in a similar way, 1 σ (x, x )= T (x, x )b . (5.10) ik 0 8π ijk 0 j

For the Stokeslet in 3D, Tijk takes the form,

∂Gij ∂Gkj Tijk(x, x0)= −δik pj(x, x0)+ + . ∂xk ∂xi Substituting this into (5.10) and the result into (5.8), we can conﬁrm by diﬀerentiation that

∂u ∂u σ = −P δ + µ i + k , (5.11) ik ik ∂x ∂x k i 31 as in §4.1.3.

At this stage we note that the stress tensor, σik is symmetric (we can conﬁrm this by swapping i and k in (5.11). Since 1 σ (x, x )= T (x, x )bj, ik 0 8π ijk 0 and we expect σik = σki, it follows that

Tijk = Tkji.

Using all the information at our disposal so far, we can obtain the following identity

8π Tijk(x, x )ni(x)dS(x)= − 4π δjk, (5.12) 0   ZD 0   where D is a closed surface, and the unit normal, n, points out of D. Note that

• 8π applies when x0 is located inside D.

• 4π applies when x0 is located on D.

• 0 applies when x0 is located outside D.

Derivation First, note that the hydrodynamic force acting on a blob of fluid of volume V , surface S, with unit outward normal n, due to the surrounding fluid is, by definition (see §4.1.1),

∂σ F = σ n dS = ik dV, ik i ∂x ZS ZV i by the divergence theorem. But the singularly-forced Stokes momentum equation states that ∇ · σ + bδ(x − x0)= 0, or ∂σik + bkδ(x − x0) = 0. ∂xi So, ∂σ F = ik dV = −b δ(x − x )dV. ∂x k 0 ZV i ZV Using (5.10),

1 F = σ n dS = b T n dS = −b δ(x − x )dV. ik i 8π j ijk i k 0 ZS ZS ZV 32

Writing bk = δjkbj, we have

1 b T n dS = −b δ δ(x − x )dV. 8π j ijk i j jk 0 ZS ZV But since b is an arbitrary vector, it must be true that 1 T n dS = −δ δ(x − x )dV, 8π ijk i jk 0 ZS ZV and the result follows on using the properties of the delta function,

1 if x0 lies inside V 1 δ(x − x0)dV = 2 if x0 lies exactly on S V  Z  0 if x0 lies outside V . The middle property is explained in Appendix B.

5.4 Summary Thus far, with regard to Stokes ﬂow, we have

• Discussed the governing equations

• Discussed the notion of Green’s functions, Gij(x, x0)

• Computed the Green’s function in free space

• Discussed the Green’s function with a plane wall boundary

• Introduced the stress tensor Tijk(x, x0) 33

6 The Boundary Integral Equation for Stokes ﬂow

At long last we are in a position to derive the boundary integral equation for Stokes ﬂow. Our starting point is the Lorentz reciprocal identity derived in section 4.2,

∂ ′ ′ (uiσik − uiσik) = 0. (6.1) ∂xk

Consider two different flows with velocity and stress fields (u, σ) and (u′, σ′) respectively. ′ We choose u to be the velocity field induced by a point force of strength b at the point x0, i.e. ′ 1 ′ 1 u = G b , σ (x)= T b . i 8πµ ij j ik 8π ijk j Substituting into (6.1), we obtain,

∂ (Gijσik − µ uiTijk) bj = 0. ∂xk

Dropping the irrelevant bj,

∂ (Gij(x, x0)σik(x) − µ ui(x)Tijk(x, x0)) = 0. (6.2) ∂xk

We now divide the argument into two streams.

1. x0 lies outside V. We integrate equation (6.2) over a closed volume V with (possibly disjoint) boundary D, as shown in the ﬁgure.

D V

D n

∂ [G (x, x )σ (x) − µ u (x)T (x, x )] dV (x) = 0. ∂x ij 0 ik i ijk 0 ZV k Applying the divergence theorem,

[Gij(x, x0)σik(x) − µ ui(x)Tijk(x, x0)] nk dS(x) = 0. ZD 34 where the unit vector nj points out of the volume V in accordance with the stipulation of the divergence theorem.

For later convenience we redeﬁne it here to point into the volume V and write the equation for convenience as 1 1 − G (x, x )f (x)dS(x)+ u (x)T (x, x )n dS(x) = 0. (6.3) 8πµ ij 0 i 8π i ijk 0 k ZD ZD where fi = σiknk as usual.

Important: We were able to apply the divergence theorem because the integrand is analytic everywhere inside the domain of integration. This is a requirement of the theorem. The integrand is analytic because x0 is outside V and so neither Gij nor Tijk become singular inside V .

2. x0 lies inside V . It is a requirement of the divergence theorem that the functions inside the integrals remain regular (i.e. do not become infinite) over the whole domain in question. Once x0 lies inside V , this is no longer true of, for example, Gij(x, x0), which becomes infinite at the singular point x0. To circumvent this problem, we integrate (6.2) over the volume enclosed by V but excluding a a small sphere of radius ǫ ≪ 1 with surface Sǫ and volume Vǫ and centred ′ at the point x0 as shown in the figure. Call this region V = V − Vǫ.

D V

Sε D x n ε 0

n Vε

Accordingly, ∂ (Gij(x, x0)σik(x) − µ ui(x)Tijk(x, x0)) dV (x) = 0. ′ ∂x ZV k Now V ′ is a multiply-connected domain. Fortunately, the divergence theorem is easily extended to such domains (see the Problem Sheet). Applying the divergence theorem,

[Gij(x, x0)σik(x) − µ ui(x)Tijk(x, x0)] nk dS(x) = 0 (6.4) ZSǫ+D where the unit normal is chosen to point into the volume of integration. The idea now is to take the limit ǫ → 0. 35

Let’s concentrate on the integral over Sǫ,

[Gij (x, x0)σik(x) − µ ui(x)Tijk(x, x0)] nk dS(x). (6.5) ZSǫ The integrand involves terms Gij(x, x0) and Tijk(x, x0). The ﬁeld point, x, lies somewhere on the surface Sǫ, and as ǫ gets smaller, the possibile values of x become closer and closer to x0. So it’s worth examining the behaviour of Gij and Tijk as x → x0. In fact we ﬁnd that regardless of the type of Green’s function and stress tensor, both behave like the Stokeslet as x → x0, and so δ xˆ xˆ xˆ xˆ xˆ G ∼ ij + i j , T ∼−6 i j k . (6.6) ij ǫ ǫ3 ijk ǫ5

The radius of the sphere Sǫ is ǫ = |x − x0|. The normal to the surface is n = x − x0, so the unit normal is (x − x ) x − x n = 0 = 0 . |x − x0| ǫ

With our usual notation, x − x0 = xˆ, so, xˆ xˆ n = , n = i ǫ i ǫ So the integral (6.5) becomes, as ǫ → 0, δ xˆ xˆ xˆ xˆ xˆ ij + i j σ (x) − µ u (x) −6 i j k n dS(x). ǫ ǫ3 ik i ǫ5 k ZSǫ Taking spherical polars (r,ψ,θ), so that dS = r2 sin θ dθ dψ, we have δ xˆ xˆ xˆ xˆ xˆ ij + i j σ (x) − µ u (x) −6 i j k n ǫ2 sin θ dθdψ. ǫ ǫ3 ik i ǫ5 k ZSǫ So,

xîxˆj xîxˆjxˆk ǫδij + σ (x) + 6µ ui(x) n sin θ dθdψ. (6.7) ǫ ik ǫ3 k ZSǫ Look at the first term: xˆ xˆ ǫδ + i j σ (x)n sin θ dθdψ ij ǫ ik k ZSǫ As ǫ → 0, x → x0 and σik(x) → σik(x0), so we can write,

xˆixˆj σ (x ) ǫδij + n sin θ dθdψ ik 0 ǫ k ZSǫ Butx ˆi ∼ ǫ andx ˆj ∼ ǫ, so the whole term is of the order

σik(x0) [ǫδij + ǫ] nk sin θ dθdψ ZSǫ 36 which equals zero in the limit ǫ → 0. So the ﬁrst term of (6.7) can be dropped, leaving, xˆ xˆ xˆ 6µ u (x) i j k n sin θ dθdψ. i ǫ3 k ZSǫ So, xˆ xˆ 6µ u (x) i j xˆ n sin θ dθdψ. i ǫ3 k k ZSǫ But (x − x ) ǫ2 xˆ n = (x − x ) k 0,k ∼ = ǫ. k k k 0,k ǫ ǫ Therefore, xˆ xˆ 6µ u (x) i j sin θ dθdψ. i ǫ2 ZSǫ In the limit ǫ → 0, ui(x) → ui(x0), so xˆ xˆ 6 6µ u (x ) i j sin θ dθdψ = µ u (x ) xˆ xˆ dS, i 0 ǫ2 ǫ4 i 0 i j ZSǫ ZSǫ since dS = ǫ2 sin θdθdψ. Now, from the Problem Sheet,

4 4 xˆixˆj dS = δijπǫ , 3 ZSǫ and so the integral is 8µπ uj(x0).

Summary: We have just shown that

lim [Gij(x, x0)σik(x) − µ ui(x)Tijk(x, x0)] nk dS(x) = 8µπ uj(x0). ǫ→0 ZSǫ So (6.4) becomes

[Gij(x, x0)σik(x) − µ ui(x)Tijk(x, x0)] nk dS(x)= −8µπ uj(x0). ZD Rearranging, we have 1 1 u (x )= − G (x, x )σ (x)n dS(x)+ u (x)T (x, x )n dS(x). j 0 8πµ ij 0 ik i 8π i ijk 0 k ZD ZD

We deﬁne f = σ · n to be the stress (also called traction), and instead write 1 1 uj(x )= − fi(x) Gij(x, x )dS(x)+ ui(x)T (x, x )n dS(x). (6.8) 0 8πµ 0 8π ijk 0 k ZD ZD 37

This is the boundary integral equation for Stokes ﬂow. We will abbreviate this to BIE.

Notes: 1. Equation (6.8) is the culmination of all of our eﬀorts so far.

2. It is quite remarkable. It says that the velocity at any point in the ﬂow ﬁeld, x0, can be expressed purely in terms of boundary values of the velocity and stress.

3. The choices for Gij and Tijk are up to us. The only requirement is that they satisfy the singularly forced momentum equation (4.14). 4. Equation (6.8) is tremendously powerful as it allows us to deal with any boundary geometry we like.

5. It is vital to remember that in equation (6.8), ni is the unit inward normal.

6. Equation (6.8) applies for a ﬁeld point x0 located inside but not actually on the boundary S. We will say more on this below.

Terminology: Equation (6.8) involves the two terms

fi(x) Gij(x, x0)dS(x) ZD and

ui(x)Tijk(x, x0)nk dS(x). ZD The ﬁrst is called the single-layer potential and the second is called the double-layer potential.

Summary: When the point x0 lies outside a domain D, the boundary integral equation is 1 1 0= − G (x, x )f (x)dS(x)+ u (x)T (x, x )n dS(x) 8πµ ij 0 i 8π i ijk 0 k ZD ZD

When the point x0 lies inside a domain D, the boundary integral equation is 1 1 u (x )= − f (x) G (x, x )dS(x)+ u (x)T (x, x )n dS(x). j 0 8πµ i ij 0 8π i ijk 0 k ZD ZD 38

7 The Boundary Integral Method for Stokes ﬂow

The BIE (6.8) applies when x0 lies somewhere in the flow field. Thus, given knowledge of the boundary values of u and f, we can compute the flow anywhere. But how do we acquire knowledge of the boundary values in the first place? That is, how do we complete our original goal and solve for the fluid velocity and traction across the flow field?

To proceed, we need to obtain the form of (6.8) when the point x0 lies on the boundary. To determine this, we will need to examine carefully the behaviour of the single and double layer potential contributions to the BIE in turn.

7.1 The single-layer potential The single-layer potential is given by

SL Ij (x0)= fi(x) Gij(x, x0)dS(x). ZS We wish to ask the following question: What happens as the point x0 approaches and crosses over the boundary S? Does ISL vary continuously or does it jump across the boundary?

Since we expect to compute a fluid flow with continuous velocity and stress fields, fi(x) will be continuous across S. Given this fact, it can be shown that ISL is continuous across S provided that S does not have any kinks (i.e. it has a continuously varying normal vector).

7.2 The double-layer potential The double-layer potential is given by

DL Ij (x0)= ui(x)Tijk(x, x0)nk dS(x). (7.1) ZS

We ask the same question as before: What happens as the point x0 approaches and crosses over the boundary S? Does IDL vary continuously or does it jump across the boundary? Denote by S+ the side of S lying in the flow into which the normal vector points (see the diagram below). Denote by S− the side of S exterior to the flow. Recall that in (7.1) the normal vector points into the fluid consistent with the figure. To proceed, we re-write (7.1) in the equivalent form

DL Ij (x0)= ui(x) − ui(x0) Tijk(x, x0)nk dS(x)+ ui(x0) Tijk(x, x0)nk dS(x). S S Z h i Z Let

Yj(x0)= ui(x) − ui(x0) Tijk(x, x0)nk dS(x), S Z h i and let

Qj(x0)= ui(x0) Tijk(x, x0)nk dS(x). ZS 39

S− S+ n

S+ Fluid S− n

Then,

DL Ij (x0)= Yj(x0)+ Qj(x0). (7.2)

The reason for doing this is that the integral

Yj(x0)= ui(x) − ui(x0) Tijk(x, x0)nk dS(x) S Z h i is certainly continuous as x0 moves across S, and we can instead focus on Qj. From section 5.3, we have the identity:

8π T (x, x )ni(x)dS(x)= 4π δ , ijk 0   jk ZS 0   where S is a closed surface, and the unit normal, n, points into7 S as shown in the ﬁgure above. Note that

• 8π applies when x0 is located inside S.

• 4π applies when x0 is located on S.

• 0 applies when x0 is located outside S.

Recalling that Tijk = Tkji, the relation can also be written

8π T (x, x )n (x)dS(x)= 4π δji, ijk 0 k   ZS 0  

So, when x0 is inside S,

+ Qj(x0) = 8πδij ui(x0) = 8πuj (x0)= Qj , say.

7Note that for the identity quoted in §5.3 the normal points out of S. Thus there is a minus sign diﬀerence on the right hand side between the identities. 40

When x0 is outside S − Qj(x0)=0= Qj , say.

When x0 is precisely on S,

S Qj(x0) = 4πuj(x0)= Qj , say.

In summary,

+ S − Qj = 8πuj(x0), Qj = 4πuj(x0), Qj = 0. (7.3)

From (7.2), DL Ij (x0)= Yj(x0)+ Qj(x0). + So as x0 → S from inside the ﬂow,

DL + + Ij (x0) → Yj + Qj ,

+ + − where Yj is the limiting value of Yj as x0 → S . Also, as x0 → S from outside of the ﬂow, DL − − Ij (x0) → Yj + Qj , − − where Yj is the limiting value of Yj as x0 → S . + − S But since Yj is continuous as x0 crosses S, we have Yj = Yj = Yj , say, and so

lim IDL(x ) = Y S + Q+, (7.4) + j 0 j j x0→S

DL x S − lim − Ij ( 0) = Yj + Qj . (7.5) x0→S

DL−P V Deﬁne the principal value of the double-layer potential, Ij as its value when x0 lies on S, namely P V DL−P V Ij = ui(x)Tijk(x, x0)nk dS(x). ZS The superscript P V on the integral reminds us that x0 lies precisely on S.

Taking x0 to lie on S in (7.2) we have

DL−P V S S Ij (x0)= Yj + Qj and so

S DL−P V S DL−P V Yj = Ij (x0) − Qj = Ij (x0) − 4πuj(x0) (7.6) using (7.3). 41

Therefore, putting (7.3), (7.4), (7.5), and (7.6) together, we have

P V DL S + lim Ij (x0)= Yj + Qj = 4πuj (x0)+ ui(x)Tijk(x, x0)nk dS(x), (7.7) x →S+ 0 ZS and

P V DL S − lim Ij (x0)= Yj + Qj = −4πuj(x0)+ ui(x)Tijk(x, x0)nk dS(x). (7.8) x →S− 0 ZS

So across S, the double layer potential IDLP experiences a jump of size

+ − Qj − Qj = 8πuj(x0).

7.3 The BIE with x0 on the boundary The BIE (6.8) is 1 1 u (x )= − ISL(x )+ IDL(x ), j 0 8πµ j 0 8π j 0

+ where x0 lies inside the ﬂow. Taking the limit x0 → S and using (7.7), we obtain

1 1 P V u (x )= − ISL(x )+ 4πu (x )+ u (x)T (x, x )n dS(x) . j 0 8πµ j 0 8π j 0 i ijk 0 k ZS Rearranging, we get

1 1 SL 1 DL−P V uj(x )= − I (x )+ I (x ). 2 0 8πµ j 0 8π j 0

Or, in full,

1 1 1 P V u (x )= − f (x) G (x, x )dS(x)+ u (x)T (x, x )n dS(x). (7.9) 2 j 0 8πµ i ij 0 8π i ijk 0 k ZS ZS This is the form of the BIE when x0 is positioned precisely on the boundary S.

Notes: 1. Equation (7.9) is almost identical to the BIE, except for the half factor on the LHS. 2. Equation (7.9) could equally we have been derived by starting with (6.3) and then taking − x0 → S .

We have now completed the formulation of the Boundary Integral Method. Specifically, we have obtained an integral equation to be solved for the values of the fluid velocity and traction on the boundary of the domain. Once this equation is solved, we may use equation (6.8) to find the fluid velocity at any point within the flow domain. 42

Summary: When the point x0 lies outside a domain D, the boundary integral equation is 1 1 − G (x, x )f (x)dS(x)+ u (x)T (x, x )n dS(x) 8πµ ij 0 i 8π i ijk 0 k ZD ZD

When the point x0 lies inside a domain D, the boundary integral equation is 1 1 uj(x )= − fi(x) Gij(x, x )dS(x)+ ui(x)T (x, x )n dS(x). 0 8πµ 0 8π ijk 0 k ZD ZD

When the point x0 lies on the boundary of D, the boundary integral equation is

1 1 1 P V u (x )= − f (x) G (x, x )dS(x)+ u (x)T (x, x )n dS(x). 2 j 0 8πµ i ij 0 8π i ijk 0 k ZD ZD 43

8 Applications

The boundary integral method as we have derived it has two very powerful features

• It provides the ﬂow solution with reference only to the values on the boundary

• It produces solutions for ﬂow over complex geometries or topography

In this section, we discuss some applications. We will in each case derive the requisite boundary integral equation. In the last section, we will see how to implement a numerical method to ﬁnd the solution.

8.1 Shear ﬂow over an obstacle Simple shear ﬂow over a wall at y = 0 is described by the solution

u = λ y.

The velocity profile is seen in the figure below. Suppose instead we are interested in computing flow over a wall with an obstacle attached to it. Practical examples of this type of flow include coating flows in manufacturing micro- electronic circuitry or blood flow over a thrombus. The figure below shows the geometry of the flow. The wall is at y = 0. The obstacle sits on the wall as shown.

A C x

Figure 2: Shear ﬂow over an object attached to a plane wall

Let P represent the volume occupied by the obstacle. Let C represent the wall area covered by the obstacle. Let A represent the wall area left uncovered by the obstacle.

Aim: To derive a boundary integral equation for the velocity ﬁeld at a general point x0.

Method: We start by splitting the desired velocity ﬁeld into two parts like so

∞ u = u + uD where 44

• u∞ represents the background ﬂow obtaining in the absence of the obstacle, i.e. the undisturbed shear ﬂow. So ∞ u = λy i.

• uD represents the disturbance due to the obstacle. A long way from the obstacle this disturbance ﬂow decays to zero. If there is no obstacle, uD ≡ 0 everywhere.

Together, the background and disturbance ﬂows combine to make up the total ﬂow u. Similarly, for the traction, the total traction

∞ f = f + f D, where f ∞ and f D are the background and disturbance parts of the traction respectively.

From equation (6.8), we have the boundary integral equation for Stokes flow 1 1 u (x )= − f (x) G (x, x )dS(x)+ u (x)T (x, x )n dS(x), (8.1) j 0 8πµ i ij 0 8π i ijk 0 k ZS ZS where x0 is a point within the flow field and not on the boundary. Now, apply the BIE (8.1) to the disturbance flow, uD: 1 1 uD(x )= − f D(x) G (x, x )dS(x)+ uD(x)T (x, x )n dS(x), (8.2) j 0 8πµ i ij 0 8π i ijk 0 k ZS ZS where S includes the whole of A, P and a surface extending from the wall to infinity and back (indicated by the dotted line in the figure). Since the Green’s function Gij and stress tensor Tijk decay to zero at infinity the integrals over the surface extending to infinity equal zero. Now we can make some simplifications. The total flow satisfies no-slip, u = 0, on the wall A and on the obstacle P . Therefore uD + u∞ = 0 or uD = −u∞ on A and on P . Substituting into (8.2),

D 1 D 1 ∞ u (x )= − f (x) Gij(x, x )dS(x) − u (x)T (x, x )n dS(x). j 0 8πµ i 0 8π i ijk 0 k ZA,P ZA,P ∞ But, ui =0 on A so

D 1 D 1 ∞ u (x )= − f (x) Gij(x, x )dS(x) − u (x)T (x, x )n dS(x). (8.3) j 0 8πµ i 0 8π i ijk 0 k ZA,P ZP Note that the second integral is now taken only over the obstacle P . Equation (8.3) is valid for a point x0 lying inside the ﬂow.

For the next step, we recall the integral equation we derived for a point x0 lying outside the integration domain, equation (6.3): 1 1 − Gij(x, x )fi(x)dS(x)+ ui(x)T (x, x )n dS(x) = 0. (8.4) 8πµ 0 8π ijk 0 k ZS ZS 45

If we choose S to be the union of P and C and x0 to lie in the ﬂow, then x0 is certainly outside the domain of integration. Applying (8.4) for the background ﬂow,

1 ∞ 1 ∞ − Gij(x, x0)fi (x)dS(x)+ ui (x)Tijk(x, x0)nk dS(x) = 0, (8.5) 8πµ P,C 8π P,C ∞ Z ∞ ∞ Z where fi = σik nk. But ui = 0 on the wall C so 1 ∞ 1 ∞ − G (x, x )f (x)dS(x)+ u (x)T (x, x )n dS(x) = 0. (8.6) 8πµ ij 0 i 8π i ijk 0 k ZP,C ZP Rearranging,

∞ 1 ∞ u (x)T (x, x )n dS(x)= Gij(x, x )f (x)dS(x). (8.7) i ijk 0 k µ 0 i ZP ZP,C which is valid for a point x0 lying within the ﬂow.

Since both (8.3) and (8.7) are valid for a point x0 within the ﬂow, we can combine them. Substituting (8.7) into (8.3) we get

D 1 D 1 ∞ u (x ) = − f (x) Gij(x, x )dS(x) − u (x)T (x, x )n (x)dS(x) j 0 8πµ i 0 8π i ijk 0 k ZA,P ZP 1 1 ∞ = − f D G dS(x) − f G dS(x) 8πµ i ij 8πµ i ij ZA,P ZP,C 1 D 1 1 ∞ = − f Gij dS(x) − fi Gij dS(x) − f Gij dS(x). 8πµ i 8πµ 8πµ i ZA ZP ZC Note that in the last step the disturbance and background tractions in the integral over P have combined to give the total traction. ∞ Adding ui to both sides, we obtain ∞ 1 1 u (x )= u (x ) − f D(x) G (x, x )dS(x) − f (x) G (x, x )dS(x) j 0 j 0 8πµ i ij 0 8πµ i ij 0 ZA ZP 1 ∞ − f (x)Gij (x, x )dS(x). (8.8) 8πµ i 0 ZC This is a boundary integral equation for the velocity ﬁeld. If we choose a Green’s function 8 such that Gij(x, x0)= 0 when x is on the wall , it simpliﬁes to

∞ 1 u (x )= u (x ) − f (x) G (x, x )dS(x). (8.9) j 0 j 0 8πµ i ij 0 ZP Thus it can sometimes be a good idea to use a particular Green’s function in order to simplify the formulation. Notes: 1. Equation (8.9) is a boundary integral equation for the velocity involving the unknown traction fi(x) and velocity ui(x). To obtain the solution, we must solve for these simulta- neously. 8The form of the required Green’s function was given in §5.2 46

8.2 Flow past a rotating object Consider the slow flow of a viscous fluid around a rigid object rotating about an axis passing through O perpendicular to the page as shown in the figure. Let x be the position vector of a point P relative to O. n

x P

O Ω solid

fluid S

If the rigid body is rotating with angular velocity Ω, where the vector Ω points along the axis of rotation (out of the page), the velocity of the point P is9

u(x)= Ω × x.

Making use of the alternating tensor deﬁned in §2.1, we can write this in index notation as

ui(x)= ǫilmΩlxm. (8.10)

To find the velocity of the fluid, we use the boundary integral formulation. The BIE equation (6.8) states that 1 1 u (x )= − f (x) G (x, x )dS(x)+ u (x)T (x, x )n dS(x) (8.11) j 0 8πµ i ij 0 8π i ijk 0 k ZS ZS when x0 lies in the fluid. Using (8.10), this becomes 1 Ω u (x )= − f (x) G (x, x )dS(x)+ ǫ l x T (x, x )n dS(x). j 0 8πµ i ij 0 ilm 8π m ijk 0 k ZS ZS An identity states that

8π ǫ xmT (x, x )n (x)dS(x)= −ǫ x ,m 4π (8.12) ilm ijk 0 k jlm 0   ZS 0   when x0 is inside (8π), right on (4π) or outside S (0), and where the normal vector points outside S as shown in the figure. Note that x0,m means the mth component of x0. Hence, since x0 is in the fluid outside S, the double layer potential vanishes leaving 1 u (x )= − f (x) G (x, x )dS(x), (8.13) j 0 8πµ i ij 0 ZS 9See any book on classical mechanics 47 for a point x0 lying inside the fluid.

In fact, (8.13) also applies when x0 lies on the surface S. To see this, we can either appeal to §7.1 and note that the single layer potential is a continuous function as x0 approaches S and hence (8.13) is equally valid on the surface, or we may simply apply the BIE for a point x0 lying on S, namely equation (7.9), 1 1 1 P V u (x )= − f (x) G (x, x )dS(x)+ u (x)T (x, x )n dS(x). 2 j 0 8πµ i ij 0 8π i ijk 0 k ZS ZS Recall that the P V on the second integral means that the integral is evaluated with x0 on S. Using (8.10) and then identity (8.12) we ﬁnd 1 1 1 u (x )= − f (x) G (x, x )dS(x) − ǫ Ω x . 2 j 0 8πµ i ij 0 2 jlm l 0,m ZS But when x0 lies on S, uj(x0)= ǫjlmΩlx0,m, and so we have 1 uj(x )= − fi(x) Gij(x, x )dS(x) 0 8πµ 0 ZS again. Note: Problem 12 on the Problem Sheet asks you to perform a similar analysis to the above for an object moving at constant speed through a viscous liquid.

8.3 Flow past a liquid drop attached to a wall Consider the flow of a viscous liquid past a drop of another liquid of the same viscosity attached to a plane wall. We assume that the drop does not move along the wall, but can be deformed by the flow. We also expect the passing flow to generate a fluid motion inside the drop. Label the outer fluid as fluid 1, and the fluid in the drop as fluid 2. Let n be the unit normal vector pointing into the outer flow.

A C x

Let P represent the surface of the drop exposed to the ﬂuid. Let C represent the wall area covered by the drop. Let A represent the wall area left uncovered by the drop.

We wish to derive a boundary integral equation for the velocity ﬁeld at a general point x0 both in the outer ﬂow and inside the drop. 48

To begin, we label the velocity and traction ﬁelds in ﬂuid j = 1, 2 as

(j) (j) u (x0), f (x0), at a point x0. We split the velocity and traction ﬁelds in ﬂuid 1 into a background and a disturbance part,

∞ ∞ u(1) = u + u(1)D, f (1) = f + f (1)D.

Here,

• u∞ represents the background ﬂow found in the absence of the drop, given by

∞ u = λy i.

• u(1)D represents the disturbance outer flow in fluid 1 due to the drop. A long way from the drop this disturbance flow decays to zero. If there is no drop, u(1)D ≡ 0 everywhere.

Now, apply the BIE (8.1) for a point inside fluid 1 to the disturbance flow, u(1)D: 1 1 u(1)D(x )= − f (1)D(x) G dS(x)+ u(1)D(x)T n dS(x), (8.14) j 0 8πµ i ij 8π i ijk k ZS ZS where S includes the whole of A, P and a surface extending from the wall to infinity and back (indicated by the dotted line in the figure). Since the Green’s function Gij and stress tensor Tijk decay to zero at infinity the integrals over the surface extending to infinity equal zero. The total flow in fluid 1 satisfies no-slip, u(1) = 0, on the wall A. Therefore u(1)D + u∞ = 0 or u(1)D = −u∞ on A. Substituting into (8.14),

1 1 ∞ u(1)D(x ) = − f (1)D G dS(x) − u T n dS(x) j 0 8πµ i ij 8π i ijk k ZA,P ZA 1 + u(1)DT n dS(x). 8π i ijk k ZP ∞ But ui =0 on A so 1 1 u(1)D(x )= − f (1)D G dS(x)+ u(1)DT n dS(x). j 0 8πµ i ij 8π i ijk k ZA,P ZP ∞ Adding ui to both sides, we have

∞ 1 1 u(1)(x )= u (x ) − f (1)D G dS(x)+ u(1)DT n dS(x) (8.15) j 0 j 0 8πµ i ij 8π i ijk k ZA,P ZP for a point x0 inside ﬂuid 1. 49

For the next step, recall the integral equation we derived for a point x0 lying outside the domain of integration, equation (6.3): 1 1 − G (x, x )f (x)dS(x)+ u (x)T (x, x )n dS(x) = 0. (8.16) 8πµ ij 0 i 8π i ijk 0 k ZS ZS First, we take S to be the union of P and C and apply this equation for the background ∞ flow ui for a point x0 lying inside fluid 1 and therefore outside the domain of integration. We obtain, 1 ∞ 1 ∞ 0= − fi Gij dS(x)+ ui Tijknk dS(x). 8πµ P,C 8π P,C ∞ Z Z Since ui = 0 on the wall C, this simplifies to

1 ∞ 1 ∞ 0= − f Gij dS(x)+ u T n dS(x). (8.17) 8πµ i 8π i ijk k ZP,C ZP which is valid for a point x0 lying inside ﬂuid 1.

Since both (8.15) and (8.17) apply for a point x0 inside ﬂuid 1 we can combine them. Adding (8.17) to (8.15) we ﬁnd

(1) ∞ 1 (1)D 1 (1)D u (x ) = u (x ) − f Gij dS(x)+ u T n dS(x) j 0 j 0 8πµ i 8π i ijk k ZA,P ZP 1 ∞ 1 ∞ − f G dS(x)+ u T n dS(x) 8πµ i ij 8π i ijk k ZP,C ZP ∞ 1 1 ∞ 1 = u (x ) − f (1)D G dS(x) − f G dS(x) − f (1) G dS(x) j 0 8πµ i ij 8πµ i ij 8πµ i ij ZA ZC ZP 1 + u(1)T n dS(x). (8.18) 8π i ijk k ZP At this point, we decide to simplify matters by choosing a Green’s function which vanishes on the wall, so that Gij(x, x0)= 0 when x is on the wall. Then (8.18) reduces to

(1) ∞ 1 (1) 1 (1) u (x )= u (x ) − f Gij dS(x)+ u T n dS(x). (8.19) j 0 j 0 8πµ i 8π i ijk k ZP ZP This equation is valid for a point x0 inside ﬂuid 1.

Next we apply equation (8.16) with S as the union of P and C but this time for the flow in the drop, fluid 2, and for a point x0 lying in fluid 1. We obtain, 1 1 0= − f (2)G + u(2)T n dS(x). 8πµ i ij 8π i ijk k ZP,C ZP,C (2) But ui = 0 on the wall C by the no-slip condition. Also Gij has now been chosen to vanish on the wall. So,

1 (2) 1 (2) 0= − f Gij dS(x)+ u T n dS(x) (8.20) 8πµ i 8π i ijk k ZP ZP 50 for a point x0 lying inside ﬂuid 1.

Since both (8.19) and (8.20) apply for a point x0 inside ﬂuid 1 we can combine them. Subtracting (8.20) from (8.19), we ﬁnd

(1) ∞ 1 (1) 1 (1) u (x ) = u (x ) − f Gij dS(x)+ u T n dS(x) j 0 j 0 8πµ i 8π i ijk k ZP ZP 1 1 + f (2)G dS(x) − u(2)T n dS(x) (8.21) 8πµ i ij 8π i ijk k ZP ZP ∞ 1 1 (1) (2) = uj (x0) − ∆fi Gij dS(x)+ ui − ui Tijknk dS(x), 8πµ P 8π P Z Z h i where (1) (2) ∆fi = fi − fi is the jump in traction at the interface between the drop and the outer liquid. Physically, we expect the velocity ﬁeld to be continuous at the interface, so that

(1) (2) ui (x)= ui (x) when x lies on the interface. Hence (8.21) reduces to

(1) ∞ 1 u (x )= u (x ) − ∆fi(x) Gij(x, x )dS(x), j 0 j 0 8πµ 0 ZP and is valid for a point x0 inside ﬂuid 1. This formula should be compared with (8.9) in §8.1 for ﬂow over a solid object attached to a wall.

Appendix A: Alternative form of the delta function In section §5.1 we re-wrote the delta function as follows 1 δ(xˆ)= −∇2 . 4πr To see why this is true, note that the free-space Green’s function G for Laplace’s equation in three-dimensions satsiﬁes

∇2G = δ(xˆ), (8.22) where xˆ = x − x0. Away from x = x0 the delta function is zero, leaving

∇2G = 0.

Deﬁne spherical polar coordinates (r, θ, ψ) centred on x0, so that

r = |x − x0|, 51 i.e. r is the radial distance from the ﬁeld point to the pole. Then, working in spherical polars and noting that we expect the solution to be radially symmetric we have 1 ∂ ∂G ∇2G = r2 = 0, r2 ∂r ∂r away from x = x0. Integrating, we see that one solution is A G = − . r for constant A. To ﬁx A we integrate () over a small sphere, volume V , radius ǫ, and surface S, which is centred at x0. So,

2 ∇ G dV = δ(x − x0)dV = 1 ZV ZV on using the sifting property of the delta function. Applying the divergence theorem10 in fact, we ﬁnd ∂G dS = δ(x − x )dV = 1 ∂r 0 ZS ZV Thus, since the surface area of the sphere is 4πǫ2, we ﬁnd

−4πA = 1 and so A = −1/4π. Thus, 1 G = − . (8.23) 4πr is a solution of ∇2G = δ(xˆ). Accordingly, we can write

1 δ(xˆ)= −∇2 . 4πr Note: we can alternatively derive (8.23) using a Fourier transform method.

Appendix B: Properties of the delta function We demonstrate the properties of the delta function given in §5.3, namely

1 if x0 lies inside V 1 δ(x − x0)dV = 2 if x0 lies exactly on S V  Z  0 if x0 lies outside V . 10Note that it is a little dicey using the divergence theorem over a domain containing a singularity! This calculation should more properly be performed by ﬁrst isolating the singularity inside a small ball. See Stakgold “Green’s Functions and Boundary Value Problems”, Chapter 4 52

We will work in two dimensions, in which case we have

1 if x0 lies inside V 1 δ(x − x0)dS = 2 if x0 lies exactly on S S  Z  0 if x0 lies outside V . where S is a two-dimensional surface enclosed by a contour C with unit outward normal n. The argument is readily extended to three dimensions. To proceed we re-write the two-dimensional delta function in a manner analogous to that introduced11in §5.1.1, 1 δ = ∇2 log r. 2π Then, we integrate over S, 1 1 δ(x − x )dS = ∇2 log r dS = n ·∇ log r dl, 0 2π 2π ZS ZS ZC by the divergence theorem.

Consider the case when the singular point, x0, lies on the contour C. We indent C inwards in the form of a circular arc, radius ǫ, so that x0 now lies outside S as shown. Now, since Γ

ε θ

x0 x0 is outside S, we have 1 δ(x − x )dS =0= n ·∇ log r dl 0 2π ZS ZC 1 P V 1 = lim n ·∇ log r dl + n ·∇ log r dl , ǫ→0 2π 2π ZC ZΓ where the principal value P V means integration around C excluding the indented portion Γ. Integrating over the indented arc, Γ, we ﬁnd, 1 1 π 1 1 n ·∇ log r dl, = − ǫdθ = − , 2π 2π ǫ 2 ZΓ Z0 since the outward normal n points in the inward radial direction normal to the arc. Thus,

1 1 P V 0= − + n ·∇ log r dl 2 2π ZC So, 1 P V 1 n ·∇ log r dl = . 2π 2 ZC 11 2 1 Recall that in §5.1.1, we wrote δ(xˆ)= −∇ ( 4πr ). 53

Hence, when x0 lies precisely on the contour C, 1 P V 1 δ(x − x )dS = n ·∇ log r dl = . 0 2π 2 ZS ZC The other results follow from the standard sifting property of the delta function.

Appendix C: Symmetry of the Green’s function for Stokes ﬂow It was stated in §4.3.1 that the Green’s function is symmetric, that is

Gij(x1, x2)= Gji(x2, x1). To show this, we apply a similar argument used to demonstrate the symmetry of the Green’s function for potential ﬂow12. We start with the Lorentz reciprocal relation (4.13) valid for ′ ′ two ﬂows (ui,σik) and (ui,σik) which contain no singular points within the domain of interest,

∂ ′ ′ (uiσik − uiσik) = 0. (C.1) ∂xk Suppose that the domain in question is enclosed by a boundary S and that the Green’s function in question vanishes on the boundary, that is

Gij(x, x0) = 0 when x lies on S and where x0 is the location of the singularity. To apply the Lorentz reciprocal identity, we consider the volume which consists of the inside of S but with two small spheres of radius ǫ centred around two points x1 and x2 removed as shown. Call this ′ ′ volume V and its bounding surfaces S = S S1 S2. S S

S1 S2 x n 1 n x 2

n S

Note that the unit normals are taken to point out of V ′. We take 1 1 ui(x)= Gim(x, x )am, σ = T (x, x )aj, (C.2) 8πµ 1 ik 8π ijk 1

12See, for example, Garabedian, Partial Diﬀerential Equations, p.244. 54 and

′ 1 ′ 1 u (x)= G (x, x )b , σ = T (x, x )b , (C.3) i 8πµ im 2 m ik 8π ijk 2 j for arbitrary constant vectors a and b. Integrating (C.1) over the volume V ′ within S′,

∂ ′ ′ (uiσik − uiσik)dV = 0, ′ ∂x ZV k or, using the divergence theorem13,

′ ′ (uiσiknk − uiσiknk)dS = 0. ′ ZS Accordingly,

′ ′ ′ ′ (uiσiknk − uiσiknk)dS + (uiσiknk − uiσiknk)dS = 0. ZS ZS1,S2 ′ The ﬁrst integral on the left hand side vanishes since both ui and ui vanish on S. So,

′ ′ (uiσiknk − uiσiknk)dS = 0. ZS1,S2 Consider now the integral ′ I11 = uiσiknkdS. ZS1 Using (C.2) and (C.3) this becomes 1 ajbm Gim(x, x ) T (x, x )n dS 64π2µ 2 ijk 1 k ZS1 Letting the radius ǫ of S1 tend to zero, this becomes 1 xˆ xˆ xˆ x x i j k 2 ajbmGim( 1, 2) − 6 5 nk dS. 64π µ S1 ǫ Z Note that we have approximated Tijk by its Stokeslet form (compare §6). From the working in §6, we have the result that xˆ xˆ xˆ x x i j k x x Gim( 1, 2) − 6 5 nk dS = 8π Gjm( 1, 2). S1 ǫ Z So, 1 I = ajbmGjm(x , x ). 11 8πµ 1 2 Next consider

′ 1 I = uiσ n dS = ambj Gim(x, x )T (x, x )n dS. 12 ik k 16π2µ 1 ijk 2 k ZS1 ZS1 ′ 13Note that this is permitted since the integrand is regular throughout S 55

Letting ǫ → 0, 1 δ xˆ xˆ x x im i m I12 ∼ 2 ambj Tijk( 1, 2) + 3 nkdS. 16π µ S1 ǫ ǫ Z Note that we have approximated Gim by its Stokeslet form (compare §6). From the working in §6, we have the result that

δim xˆixˆm + 3 nk dS → 0 S1 ǫ ǫ Z as ǫ → 0. So I12 → 0. Following similar reasoning we may determine the values of

′ ′ I21 = uiσiknk dS, I22 = uiσiknkdS. ZS2 ZS2 In fact, we ﬁnd 1 1 I = 0, I = ambjGjm(x , x )= ajbmGmj (x , x ). 21 22 8πµ 2 1 8πµ 2 1 Note that in the last step we switched the roles of m and j. This is permissible since they are both dummy indices. Finally, we ﬁnd that

′ ′ (uiσiknk − uiσiknk)dS = I11 + I12 − I21 − I22 ZS1,S2 1 1 = a b G (x , x ) − a b G (x , x ) = 0. 8πµ j m jm 1 2 8πµ j m mj 2 1 It therefore follows that Gij(x1, x2)= Gmj(x2, x1), which is the required symmetry property.

Appendix D: Conservation of mass for a point force in Stokes ﬂow In section (4.3.1), we argued using the divergence theorem that

Gij(x, x0)ni(x)dS = 0. (D.1) ZS We applied the divergence theorem for an integral containing a singularity at the pole x0. Here we provide a more formal, classical argument.

The velocity ﬁeld in response to a point force at x0 is

ui = Gij(x, x0) bj, (D.2) 56 where the Green’s function satisﬁes the singularly forced Stokes equation (4.14). We take the divergence of (D.2) and integrate over a volume V containing the singular point x0 but excluding a small sphere, Vǫ, of radius ǫ centred at x0. Call this punctured volume ′ V = V − Vǫ. Let S and Sǫ represent the surfaces of V and Vǫ respectively. We ﬁnd that

∂ui dV = uini dS, ′ ∂x ZV i ZS,Sǫ on applying the divergence theorem (note that the integrand is regular over the punctured region). Now,

xîxîxˆj δij xˆj δij uini dS = bj Gijni dS = + dS = + dS ǫ4 ǫ ǫ2 ǫ ZSǫ ZSǫ ZSǫ ZSǫ 2 2 on noting that ni =x ˆj/ǫ on Sǫ. Noting further that dS ∼ ǫ andx ˆj ∼ ǫ and taking the limit as ǫ → 0, we find that this integral vanishes and we are left with

∂ui dV = uini dS. ′ ∂x ZV i ZS Since ∇ · u = 0 over V ′ we deduce that

uini dS = 0 ZS and so (D.1) holds.