Math 575-Lecture Fundamental solutions for point force: Stokeslet, stresslet

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1 Math 575-Lecture 2 Fundamental solutions for point force: tokeslet, stresslet Consider putting a point singular force with strength F in a tokes flow of infinite extent. We then need to solve P µ U = δ(x)f, U = (.) The equations should be understood in the sense of distributions. By linearity, the solutions can be written as P = 8π Π F, U = 8πµ G F and the stress is σ = P I +µ( U + U T ) = 8π T F. The tensors G, T, Π then give basic information of the system. To derive the expression of G formally, we use the fundamental solution of the Laplace equation: ϕ = 4π x (.2) The Dirac delta can be replaced by ϕ. Taking the divergence, we have P = δ F = ( ϕ F ). Hence P = ϕ F because both P and ϕ F decay at infinity, from which Π is derived. Plugging this back into the equation, µu = F ϕ ϕf (.3) where ϕ = ζ is defined to be the distribution that satisfies ζ, g = ϕ, g for all g Cc (R 3 ). It s found that ζ = x /8π. Hence, G and T can be derived correspondingly: Π(x) = 2x x 3, (.4) G(x) = I x + xx x 3, (.5) T (x) = 6 xxx x 5. (.6) G is called the tokeslet and T is called the stresslet. One can check that these really solve the equations in the sense of distribution. Even though T is for stress in the definition, it s found that u(x) = T (x) M for any constant vector M is also a good velocity field for the tokes equations. Hence, both G and T are the fundamental solutions.

2 We first introduce two useful formulas. Fix any piecewise smooth, closed surface with outer normal N. Using the formulas of the tokeslet and stresslet, we have that G(x y) N(y)d y =, (.7) P V T (x y) N(y)d y = {8π, 4π, }I, (.8) where P V means Cauchy s principal value. In the second formula, it s 8π if x is inside, 4π if x is on and otherwise. I is the unit second order tensor. An approach for solutions of tokes equations From the derivative above, if we let χ play the role of µ ϕ, then we find that will solve the tokes equations if u = ( χ χi) F, p = µ ( χ) F 2 χ =. The bi-harmonic equation is natural for tokes equations. 2 Boundary Integral formulation Let x Ω. We construct Ω ɛ = Ω \ B(x, ɛ). Take p = 8π Π(x x ) F, u = 8πµ G(x x ) F and σ = p I + µ( u + u T ) = 8π T (x x ) F. Applying the reciprocal theorem on Ω ɛ, Ω ɛ (n 8π T (x x ) F ) ( 8πµ G(x x ) F ) (n σ)d =. Ω ɛ = + B(x, ɛ). n is the normal of the fluid domain. Hence, n is the inner normal of B(x, ɛ). Let N be the outer normal of B(x, ɛ), we find (n 8π T (x x ) F ) 8πµ G(x x ) F (n σ)d = (N 8π T (x x ) F ) 8πµ G(x x ) F (N σ)d B(x,ɛ) 2

3 By the arbitrariness of F, we obtain 8π T (x x ) n 8πµ G(x x ) f d = 8π T (x x ) N 8πµ G(x x ) (N σ)d B(x,ɛ) where f = t = n σ = σ n is the force per unit area acting on the fluid. Note that as x x, G is much bigger than T. However, as x x, G decays like ɛ and the surface area is like ɛ2. The second term tends to zero as ɛ. Now, let us check the first term. N = x x x x, then we find B(x,ɛ) 8π T (x x ) Nd = B(x,ɛ) The next step is to apply Taylor expansion on u: u(x) = u(x ) + B(x; x )ɛ 8π ( 6(x x )(x x ) ɛ 4 )d where the norm of B is uniformly bounded. Then, we find that as ɛ, we have lim ɛ B(x,ɛ) 8π ( 6(x x )(x x ) ɛ 4 )d = u(x ) 8π lim ( 6 (x x )(x x ) ɛ ɛ 4 )d B(x,ɛ) = u(x ) ( 6xx)d 8π B(,) Using spherical coordinates, π 2π xxd = sin θ cos φ, sin θ sin φ, cos θ sin θ cos φ, sin θ sin φ, cos θ sin θdφdθ B(,) The offdiagonal terms either contains cos φ or sin φ and the integral is zero. The the diagonal terms. we compute for example π 2π Hence, we obtain u(x ) = sin 3 θ cos 2 φdφdθ = π π sin 3 θdθ = 4π 3 8π T (x x ) n 8πµ G(x x ) f d, 3

4 or u(x ) = G(x x) f 8πµ d x + u(x) T (x x) n(x)d x 8π This is the boundary integral formulation. This form shows that the field at x is the superposition of fields generated by the singularities on the surface. 3 Point source and source dipole We now consider another singularity: point source. This means we consider This has a potential flow solution By linearity, U = Qδ(x) U = φ. U = 4π Q We find φ = Q 4πr. Consequently, the point source is = x x 3. Now, we consider the source dipole: one is placed at x + d h/2 with density Q and the other is placed at x d h/2 with strength Q. Then, as h, the leading order field is given by u = 4π Qhd x (x x ). Now, if we denote the dipole strength as p = Qhd (we assume Qh is O()), we have where u = 4π p d d = x (x x ) = x (x x ) = I x xx x 5 This is related to G through the relation 2 G = I x 3 3 xx x 5. (3.) 4

5 4 Force dipole solutions: rotlets and stresslets We now place a point force F at x + d h/2 with force F and the other one is placed at x + d h/2 with force F. The field is then given by u = 8πµ d x G(x x ) F h = 8πµ d xg(x x ) F h This field is called the tokeslet doublet. Let P = d F h which is O(). The filed is written as We compute that We can show that u i = 8πµ jg ik P jk j G ik = ( x 3 (δ ikx j δ ij x k δ kj x i ) + 3 x ix j x k x 5 ) A ij B ij = A ijb ij + A A ijb A ij In other words, we can take the symmetric part at the same time. Hence, we find We find that Let D jk = P jk 3 P mmδ jk, we find 8πµu = 2 ( jg ik + k G ij )P jk + 2 ( jg ik k G ij )P A jk. 2 ( jg ik + k G ij ) = ( δ kj x 3 x i + 3 x ix j x k x 5 ) 2 ( jg ik + k G ij )P jk = 2 ( jg ik + k G ij )D jk = 2 T jikd jk The symmetric part is the stresslet. Consider the antisymmetric part, one find easily that 2 ( jg ik k G ij ) = x 3 (δ ikx j δ ij x k ) = jik rot This is called the rotlet. The second field can be written as 8πµ rot jik DA jk. 5

6 Then, Further, there exists a vector L such that rot D A jk = 2 ɛ jkml m. jik DA jk = 2 x 3 (δ ikx j δ ij x k )ɛ jkm L m = 2 x 3 (ɛ jimx j ɛ ijm x j )L m = ɛ ijmx j L m x 3 ome people also call this second order tensor as the rotlet G c (x) ij = ɛ ijkx k x 3. (4.) It can be shown that L is the torque. The force dipole can be decomposed as stresslet and rotlet. The stresslet on one side gives the Cauchy stress in the point force singularity and gives the velocity field in this force dipole singularity. 6