1 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 practi- cal 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.