CHAPTER II MATHEMATICAL BACKGROUND OF THE BOUNDARY ELEMENT METHOD

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1 CHAPTER II MATHEMATICAL BACKGROUND OF THE BOUNDARY ELEMENT METHOD For the second chapter in the thesis, we start with surveying the mathematical background which is used directly in the Boundary Element Method (BEM). In the first part we show the properties of Gaussian integral. In the second part we introduce the Potential problem. In the third and fourth part we give an overview of the form of theorems which are frequently mentioned in BEM; The divergence and, the second form of Green s Theorem. In the fifth part we state the Laplace s equation in the Polar coordinate. In the sixth part we give the properties of Dirac delta function. In the seventh part is concerned with fundamental of the Laplace s equation. Finally, in the last two parts we discuss about the basic integral equation of both boundary and internal points. 2.1 Gaussian Integration To approximate the definite integral linear transformation as shown in Figure 2.1 b a f(x)dx we can change variables using a f(x) f(x) f(t) f(t) transform a b x -1 1 t Figure 2.1 Transformation from the domain a x b to 1 t 1 Suppose x = At + B then a = ( 1)A + B and b = A + B

2 6 By solving these equations we obtain the transformation Hence I = b a x = ( b + a 2 ) + (b a )t (2.1) 2 f(x)dx = b a 2 Gaussian integration assumes an approximation of the form 1 1 ( 1 ) f 2 (b a) + (b a)t dt (2.2) 1 1 f(t)dt n w i f(t i ) (2.3) i=1 where n is the chosen number of Gauss integration points, t i are the zeros of the Legendre polynomial and w i are Gaussian weights given by the formula d n P n (t) = 1 2 n n! dt n (tn 1) n (2.4) w i = 2(1 t2 i ) n 2[ P n 1 (t i ) ] 2 (2.5) The proof of these formulas can be found in Buchann and Turner (1992). The proof confirms that the Gauss n-point integration will compute integrals of polynomials of degree no more than 2n 1 exactly. Programming formula (2.3) is straightforward once the points, t i, and the weights, w i, are known. Finally, we obtain the approximation to the integral (2.2). The four points and Gaussian weights are shown in Table 2.1.

3 7 Table 2.1 Point and Gaussian weights. Gauss point Point (t i ) Gaussian weight (w i ) Potential Problem A scalar function, φ, defined on a domain bounded by a closed curve, is called a harmonic function if it is a solution of the Laplace equation 2 φ = 0 (2.6) on the domain. Suppose that u is a harmonic function, i.e. 2 u = 0 in (2.7) and is subject to the Dirichlet condition u = u on 1 and the Neumann condition q = q on 2

4 8 where n is a unit outward normal vector to the boundary and = as shown in Figure u = u n 2 u = 0 q = q Figure 2.2 Laplace s equation posed on the domain with the boundary The combination of the partial differential equation with these boundary conditions is called a potential problem. It has become traditional in the boundary element literature to write the flux term,, as q. Problems governed by the Laplace equation appear in different fields of applied science and engineering such as fluid flow, electric and magnetic potential fields, stress and strain analysis of elastic solids, groundwater transport, etc. 2.3 The Divergence Theorem Let be a regular domain with boundary as shown in Figure 2.1 and let F be a vector function continuously differentiable at every point of. The divergence or the Gauss Theorem states.fda = F.nds (2.8) where n is the unit outward normal to the boundary. 2.4 The Second Form of Green s Theorem Suppose u and v are two scalar functions defined in bounded by the closed curve as shown in Figure 2.2. Assume that they are continuous and admit continuous

5 9 partial derivatives. From the basic property of the gradient, we have.(u v) = u 2 v + u. v (2.9) and.(v u) = v 2 u + v. u (2.10) Subtracting the two equations and integrating over we obtain (u v v u)da = (u 2 v v 2 u)da (2.11) Applying the Gauss Theorem (2.8) to the left hand side then gives (u v v u)da = (u v v u).nds (2.12) Using the property of the gradient on the right hand side gives ( (u v v u)da = u v v ) ds (2.13) Hence from (2.11) and (2.13) we obtain (u 2 v v 2 u)da = ( u v v ) ds (2.14) This is the well-known second form of Green s theorem. 2.5 Laplace Operator in Polar Coordinate We start with setting u = u(x, y), r = xi + yj and r = r = x 2 + y 2 and suppose x = r cos φ, y = r sin φ We finally get an important equation expressed as 2 u = 2 u x + 2 u 2 y = 2 u 2 r r r u (2.15) r 2 θ 2

6 10 we note that if u depends only on r, we obtain 2 u(r) = 2 u r r r (2.16) 2.6 The Dirac Delta Function In many applications in engineering, the concept of a source at a point is required. The Dirac delta function was originally introduced as a suitable mathematical representation of this concept. The definition of the Dirac delta δ function is given by 1. δ(x) = 0 for x 0 2. δ(x) has a singularity at x = 0 3. δ(x)dx = 1 A property which is usually used is f(x)δ(x x 0 )dx = f(x 0 ) (2.17) for every continuous function f. A complete discussion and more properties can be found in Humi and Miller (1992) [27]. 2.7 Fundamental Solution of the Laplace s Equation Suppose that r is the position vector of a point, Q, relative to the point, P, inside. Surround P by a small disc, center P radius ε, as shown in Figure 2.3. The point P will be called the source point and the point Q will be called the field point. Suppose that r 0 is the position vector of P and r 1 is the position vector of Q.

7 11 ε ε 2 1 r 0 P n q = q u = u r r 1 Q Figure 2.3 Neighborhood of the point P in the domain Let r = r 1 r 0 with r = r. Suppose that u (r) is a fundamental solution of the Laplace equation which satisfies 2 u (r) + δ(r) = 0 (2.18) where δ(r) is the Dirac delta function. We see that 2 u (r) = 0 (2.19) everywhere except where r 1 = r 0, i.e. when r = 0. To find the fundamental solution, we first find the solution of the homogeneous equation (2.19). In cylindrical co-ordinate, 2 u (r) = 2 u r r and since the solution depends only on the variable r, we obtain d 2 u dr r r + 1 r 2 2 u φ 2 (2.20) du dr = 0 (2.21) This ordinary differential equation can be solved analytically and then we get u = Cln(r) + D (2.22) where C and D are constant. It is conventional to set D = 0.

8 12 Let ε be the disc bounded by the curve ε with center P and radius ε as shown in Figure 2.3. Integrating on ε of equation (2.18) we obtain ε 2 u (r)da = Applying the property (2.17) of the Dirac delta function yields hence ε δ(r)da (2.23) ε δ(r)da = 1 (2.24) ε 2 u (r)da = 1 (2.25) Applying the divergence theorem on the left hand side of (2.23) we get ε ε 2 u (r)da = u.nds = Taking partial derivative in (2.22) we obtain ε ds (2.26) = C r (2.27) Substituting (2.27) in (2.26) and calculating last term yields ε ds = 0 C rdθ = C (2.28) r Substitution (2.28) in (2.26) and then in (2.25) we obtain C = 1 (2.29) Finally, from (2.22) and (2.29) we obtain u (r) = 1 lnr (2.30) which is the fundamental solution of Laplace s equation in two dimensions. In a similar way, in three dimensions the fundamental solution of Laplace s equation is u (r) = 1 4πr (2.31)

9 13 More details of both fundamental solutions can be seen in Gilbert and Howard (1990) or Paris and Gañas (1997) [13]. 2.8 Basic Integral Equation for Internal Points Historically, there have been several ways to obtain the integral representation of the potential at an internal point. We follow the approach of Davies and Crann (1996). From Figure 2.3, applying the second form of Green s theorem (2.14) with v = u in the region ε we have ε (u 2 u u 2 u)da = ε (u u )ds (2.32) Since, from (2.7) and (2.19), both u and u are harmonic in ε the integral on the left hand side is zero, therefore (u u )ds = ε (u u )ds (2.33) Substituting the fundamental solution (2.30) in (2.33) and use the fact that = q and then taking the limit as ε 0 we have 1 (u (lnr) qlnr)ds = lim ε 0 ε u ( 1 1 lnr)ds + lim ( ε 0 qlnr)ds ε In the limit as ε 0, since u is continuous partial derivative then u u P and q q P hence 1 (u (lnr) qlnr)ds = u P lim ε 0 ε ( 1 ε )ds + q 1 P lim ( lnε)ds (2.34) ε ε 0 Computing the integral to the right hand side and rearranging the equation we obtain u P = 1 (u (lnr) qlnr)ds (2.35) or u P = 1 [( lnr)q u( 1r 2 )r.n ] ds (2.36)

10 14 i.e. u P = (u q uq )ds, where q = Equation (2.37) is known as the basic formula for the internal solution. (2.37) obtained whenever the value of u and q are known everywhere on the boundary. However, for properly-posed problems we can specify only one of u and q at each boundary point (Weinberger, 1965) [28] and consequently equation (2.37) is of little help as it stands. In section 2.9 we show how to develop a boundary integral equation from which we are able to obtain u and q everywhere on the boundary and hence, via (2.37), everywhere inside the domain. It is 2.9 Basic Integral Equation for Boundary Points Suppose that P is a point on the boundary at which there is a kink with angle α as shown in Figure 2.4 (a). If the boundary is smooth at P then α = π. P α P θ 2 α θ 1 ε n ε ε ε ε (a) (b) Figure 2.4 Internal angle of at the boundary node From Figure 2.4 (b), applying the second form of Green s theorem with v = u to the region ε, where ε is the difference of ε and, then (u 2 u u 2 u)da = (u u )ds + (u u )ds (2.38) ε ε As in the previous section, the left hand side is zero therefore u )ds = (u u )ds (2.39) ε (u ε ε

11 15 Taking limits as ε 0 gives Therefore 1 (u (u (lnr) u )ds = lim ε 0 (u )ds + lim ε 0 ε qlnr)ds = u P lim θ 2 ε 0 θ 1 ε 1 ε εdθ + q P lim ( ε 0 θ 1 (u )ds (2.40) θ 2 1 lnε)εdθ Computing the integral to the right hand side and using the fact that α = θ 1 θ 2, we obtain αu P = or αu P = (u (lnr) qlnr)ds (2.41) [( lnr)q u( 1r 2 r.n) ] ds (2.42) i.e. cu P = (u q uq )ds, where c = α (2.43) Equation (2.43) is known as the basic formula for the boundary element method. It is clear that if P is an interior point then α = and hence c = 1. If P is a boundary point and the curve is smooth, then α = π and c = 1. 2