Degenerate scale problem for plane elasticity in a multiply connected region with outer elliptic boundary

Transcription

1 Arch Appl Mech 00) 80: DOI 0.007/s ORIGINAL Y. Z. Chen X. Y. Lin Z. X. Wang Degenerate scale problem for plane elasticity in a multiply connected region with outer elliptic boundary Received: 4 April 009 / Accepted: 7 August 009 / Published online: September 009 Springer-Verlag 009 Abstract This paper investigates the degenerate scale problem for plane elasticity in a multiply connected region with an outer elliptic boundary. Inside the elliptic boundary, there are many voids with arbitrary configurations. The problem is studied on the relevant homogenous boundary integral equation. The suggested solution is derived from a solution of a relevant problem. It is found that the degenerate scale and the non-trivial solution along the elliptic boundary in the problem are same as in the case of a single elliptic contour without voids. The present study mainly depends on integrations of several integrals, which can be integrated in a closed form. Keywords oundary integral equation Degenerate scale Plane elasticity Multiply connected region Introduction Many researchers were attracted by the boundary integral equation IE). Some pioneering works for IE were initiated in [,]. The basic theory for IE could be found from [3,4] The development of the boundary element method was summarized in [5]. In earlier times, Christiansen pointed out that some integral equations of the first kind with logarithmic kernel are known to have a non-unique solution in some exceptional cases [6]. This means that the relevant homogenous equation has a non-trivial solution. Alternatively speaking, the well-known IE formulation, particularly, for the exterior Dirichlet problem, does not work when some particular scales are encountered. The most typical example is as follows. One considers the D-Laplace equation ux, y) = 0 for an exterior circular boundary, and gets a solution u = ln r. If the radius of the inner circle takes a =, the boundary value at the inner circle will be vanishing, or equals zero. In this exceptional case, a zero boundary value or ux, y) = 0 on the boundary) corresponds to a non-vanishing solution or u = ln r in the exterior problem bounded by a circular hole). This is the degenerate scale problem. This result is universal. However, if the configuration of the void is arbitrary, it is more difficult to get: ) the degenerate scale and ) the relevant non-trivial solution. The degenerate scale problem in IE is a particular boundary value problem in plane elasticity. The problem typically arises from the exterior Dirichlet problem in plane elasticity. In fact, once the degenerate scale is reached, the relevant homogenous equation for the boundary tractions has a non-trivial solution. Alternatively speaking, the non-homogeneous equation has non-unique solution, or multiple solutions. Clearly, the degenerate scale represents an illness condition. Therefore, one must avoid meeting illogical solution caused by occurrence of the degenerate scale. Y. Z. Chen ) X. Y. Lin Z. X. Wang Division of Engineering Mechanics, Jiangsu University, 03 Zhenjiang, Jiangsu, People s Republic of China chens@ujs.edu.cn Tel.: Fax.:

2 056 Y. Z. Chen et al. The degenerate scale problems were studied by many researchers using a variety of methods. The degenerate scale can be evaluated by solving some IE in a normal scale [7 9]. Numerical procedure was developed to evaluate the degenerate scale directly from the zero value of a determinant [0]. Degenerate scale for multiply connected Laplace problems was solved []. This paper investigates degenerate scale problem for plane elasticity in a multiply connected region with an outer elliptic boundary. Inside the elliptic boundary, there are many voids with arbitrary configurations. The problem is studied on the relevant homogenous IE. The merit of this study is to evaluate a non-trivial solution for the homogenous IE. It is assumed that the tractions on all the inner void boundaries are equal to zero and tractions on the outer elliptic boundary are equal to some functions. Therefore, all the integrations are performed on the outer elliptic boundary only. The degenerate scale for the problem is found. It is found that the degenerate scale and the non-trivial solution along the elliptic boundary in the problem are same as in the case of a single elliptic contour without voids. The present study mainly depends on integrations of several integrals, which can be integrated in the closed forms. Degenerate scale problem for plane elasticity with a single elliptic boundary The degenerate scale problem in the multiply connected region has a close relation with the same problem in a single connected boundary. Therefore, the degenerate scale problem for plane elasticity with a single elliptic boundary is introduced. Without losing generality, we introduce the IE for plane elasticity for a single elliptic boundary Fig. ). The source point is denoted by ξξ,ξ ), and the field point is denoted by xx, x ). In the plane strain case, the IE is as follows [4,] u iξ) + P ij ξ, x)u jx)dsx) = and the integral kernels Pij ξ, x) and U ij ξ, x) are defined by Pij ξ, x) = 4π ν) U ij ξ, x)p jx)dsx), i =,, for ξ Ɣ) ) { r, n + r, n ) ν)δ ij + r,i r, j ) + ν)n i r, j n j r,i ) } ) r Uij ξ, x) = { } κlnr)δij + r,i r, j 0.5δ ij 8π ν)g 3) where Kronecker deltas δ ij is defined as, δ ij = fori = j, δ ij = 0fori = j,κ = 3 4ν, G denotes the shear modulus of elasticity, ν is the Poisson s ratio, and the normal nn, n ) at the boundary point always directs at outward side. In addition, we have r = x ξ ) + x ξ ), r, = x ξ r = cos α, r, = x ξ r = sin α 4) where the angles α are indicated in Fig.. Note that the IE can be used for both exterior and interior boundary value problems. Substituting u i ξ) = 0andu j x) = 0 into the left-hand side of Eq. ), we can obtain the following homogeneous equation U ij ξ, x)p jx)dsx) = 0 i =,, for ξ Ɣ) 5) Equation 5) reveals that regardless of the exterior and the interior problems same homogeneous equation is obtained. ased on Eq. 5), the degenerate scale problem can be formulated as follows. One wants to find a particular scale such that the integral equation 5) has a non-trivial solution p j x) = 0 for j =,, x ). Here, p j x) = 0 j =, ) is a trivial solution.

3 Degenerate scale problem for plane elasticity 057 a) z-plane x y) b) plane Exterior Region b z θ a x x) * n α ξ ξ, ξ ) x,x ) x Fig. a An exterior region to the elliptic boundary, b a mapping on plane The following analysis depends on the complex variable function method in plane elasticity [3]. In this method, the stresses σ x,σ y,σ xy ), the resultant forces X, Y ) and the displacements u,v) are expressed in terms of two complex potentials φ z) and ψ z) such that σ x + σ y = 4Reφ z) σ y σ x + iσ xy = [ zφ z) + ψ z)] 6) f = Y + ix = φ z) + zφ z) + ψ z) 7) Gu + iv) = κφ z) zφ z) ψ z) 8) where z = x + iy denotes complex variable, G is the shear modulus of elasticity, κ = 3 ν)/ + ν) is for the plane stress problems, κ = 3 4ν is for the plane strain problems, and ν is the Poisson s ratio. In the present study, the plane strain condition is assumed thoroughly. In the following, we occasionally rewrite the displacements u, v asu, u,σ x,σ y,σ xy as σ,σ,σ, and x, y asx, x, respectively. Assume that there is an ellipse with two half-axis a and b. The following mapping function is introduced Fig. ) z = ω) = R + m ), with 0 m < ) 9) which maps the elliptical contour and its exterior region in z-plane) into a unit circle and its exterior region in -plane) [3]. From Eq. 9), we have R + m), b = R m), R = a + b)/andm = a b)/a + b). After using the conformal mapping, the following functions are introduced φ) = φ z) z=ωζ), ψ) = ψ z) z=ωζ), φ z) = φ )/ω ) φ z) = φ )ω ) φ )ω ) ω )) 3, ψ z) = ψ )/ω ) 0) Therefore, the stresses, resultant forces and displacements can be expressed as φ ) ) σ x + σ y = 4Re ω ) σ y σ x + iσ xy = ) ω)[φ )ω ) φ )ω )] ω )) 3 + ψ ) ω ) ) f = Y + ix = φ) + ω)φ ) ω ) + ψ) )

4 058 Y. Z. Chen et al. Gu + iv) = κφ) ω)φ ) ψ) 3) ω ) After using the conformal mapping technique, two solutions for the homogeneous equation 5) have been obtained previously. In the first solution, the relevant complex potentials and the degenerate scale are as follows [] φ) = ln + ln R), ψ) = κln + ln R) + m m 4) R cr = expm/κ) 5) In Eq. 5), the subscript cr means that it is the first critical value for R, or R cr is the first degenerate scale. The merit of Eqs. 4)and5) is as follows. If one substitutes: ) the p j x) derived from the complex potentials shown by Eq. 4), and ) the R value shown by Eq. 5), or R = R cr = expm/κ) into Eq. 5), the homogenous equation 5) is satisfied. Similarly, in the second solution, the relevant complex potentials and the degenerate scale are as follows: φ) = iln + ln R), ψ) = i κln + ln R) + ) m m 6) R cr = exp m/κ) 7) The merit of Eqs. 6)and7) is similar to the first case. The degenerate scales shown by Eqs. 5)and7) were obtained by Chen in an earlier time [4]. In the literature, instead of Eq. 3), the following kernel was suggested [4] Uij ξ, x) = { } κ lnr)δij + r,i r, j 8) 8π v)g which is different from Eq. 3) by a constant. It was proved that the kernel could be used to the interior problem, since the tractions applied on the boundary of finite region must be in equilibrium. However, this kernel is derived from a fundamental solution that is not expressed in a pure deformable form. Thus, this kernel cannot be used to the exterior boundary value problem with applied loadings not in equilibrium [5]. 3 Formulation and solution of the degenerate scale problem for a multiply connected region with outer elliptic boundary 3. Formulation Without losing generality, we introduce the IE for plane elasticity for a multiple connected region with outer elliptic boundary. Inside the elliptic boundary, there are many voids k k =, 3, N) with arbitrary configurations Fig. ). The source point is denoted by ξξ,ξ ), and the field point is denoted by xx, x ). In this case, the IE is as follows u iξ) + Pij ξ, x)u jx)dsx) = U ij ξ, x)p jx)dsx), i =,, for ξ, = N ) 9) where the integral kernels Pij ξ, x) and U ij ξ, x) have been defined by Eqs. )and3). As before, after substituting u i ξ) = 0andu j x) = 0 into left-hand side of Eq. 9), the following homogeneous equation is obtainable Uij ξ, x)p jx)dsx) = 0, i =,, for ξ, = N ) 0)

5 Degenerate scale problem for plane elasticity 059 x y) 3 b N n a x x) ξ ξ, ξ ) n α ξ ξ, ξ ) x,x ) Fig. A multiply connected region with an outer elliptic boundary x In the degenerate scale problem, one needs to find same particular size such that Eq. 0)has a non-trivial solution for p j x) j =, ), orp j x) = 0 j =, ) for x, = N ).Thisis the highest demand for the formulation. However, we can propose a lower demand for the formulation. For example, the lower demand is as follows: p j x) = 0 j =, ) for x and p j x) = 0 j =, ) for x k, k =, 3,...,N. Clearly, it is not easy to obtain a non-trivial solution from Eq. 0) directly. However, it can be obtained from the previous result for a single elliptic contour case []. 3. Evaluation of some integrals in the closed forms In order to solve the degenerate scale problem addressed, one must evaluate some integrals in advance. In the derivation, the mapping function shown by Eq. 9) is still used Fig. ). For a point = e iθ on the unit circle, we have dθ = d ) i In the meantime, we can let z t = e iα ) In the first group of evaluation, three integrals K, K and K 3 are defined as follows Fig. 3) d K = ln z t dθ = Re lnz t)dθ = Im lnz t), for z Ɣ, t Ɣ) 3) K ik 3 = K 3 = z t z t dθ = e iα dθ or K = cos αdθ, sin αdθ, for z Ɣ, t Ɣ) 4) where Ɣ denotes the elliptic contour Fig. 3). Note that, in Eqs. 3)and4), the integration dθ is performed for argument θ in = e iθ. In addition, the subscript denotes the unit circle. After some manipulations, from Eqs. a)anda) in Appendix A, we will find the following results: K = ln z t dθ = Im lnz t) d = π ln R, for z Ɣ, t Ɣ) 5)

6 060 Y. Z. Chen et al. z-plane x z or x,x ) x * plane α θ b o t: or ξ ξ, ξ ) a x θo + S Γ Fig. 3 Mapping relation for z to and t to o with both z and t on the elliptic contour Ɣ K = cos αdθ = πm, K 3 = sin αdθ = 0, for z Ɣ, t Ɣ) 6) In the second group of evaluation for three integrals, the points z for integration) is on the elliptic contour Ɣ, or z Ɣ. However, the point t now is an inner point to the elliptic contour, or t S +,wheres + denotes the finite elliptic region Fig. 4). In this case, the three integrals L, L and L 3 are defined as follows Fig. 4): d L = ln z t dθ = Re lnz t)dθ = Im lnz t), for z Ɣ, t S+ ) 7) L il 3 = L 3 = z t z t dθ = e iα dθ or L = cos αdθ, sin αdθ, for z Ɣ, t S + ) 8) After some manipulations, from Eqs. a)anda3) in Appendix A, we will find the following results L = ln z t dθ = π ln R, for z Ɣ, t S + ) 9) L = cos αdθ = πm, L 3 = sin αdθ = 0, for z Ɣ, t S + ) 30) It is interesting to point out that the integrals K, K and K 3 shown by Eqs. 5) and6) and the integrals L, L and L 3 shownbyeqs.9)and30) take the same value. However, they have different definitions. 3.3 Solution As claimed above, the non-trivial solution cannot be obtained from Eq. 0) directly. However, it can be obtained from previous result for a single elliptic contour case []. We can prove that, the homogeneous equation 0) can be satisfied if the following conditions are fulfilled: ) p j x) = 0 j =, ) for x k, k =, 3,...,N. ) p j x) = 0 j =, ) for x, p j x) are derived from the complex potentials shown by Eq. 4), or by Eq. 6). 3) the R value is equal to R cr = expm/κ) shownbyeq.5), or R cr = exp m/κ) shown by Eq. 7).

7 Degenerate scale problem for plane elasticity 06 z-plane x plane * + S Γ z or b xx,x ) t: or ξ ξ, ξ ) α a x θ Fig. 4 Mapping relation for z to with z on the elliptic contour z Ɣ) and t in the elliptic contour t S + ) Clearly, after using the first condition, or p j x) = 0 j =, ) for x k, k =, 3,...,N,Eq.0) can be reduced to Uij ξ, x)p jx)dsx) = 0, i =,, for ξ, = N ) 3) Equation 3) can be rewritten as Uij jx)dsx) = 0, i =,, for ξ ) 3) U ij ξ, x)p jx)dsx) = 0, i =,, for ξ k, k =, 3,...,N) 33) First, the homogenous equation 3) is studied. For convenience in derivation, Eq. 3) is rewritten in the following form I = 0, with I = U ξ, x)p x)dsx) + U ξ, x)p x)dsx) ) for ξ ) 34) I = 0, with I = U ξ, x)p x)dsx) + U ξ, x)p x)dsx) ) for ξ ) 35) Clearly,inEqs.34)and35), p x)dsx) corresponds to dx and p x)dsx) corresponds to dy in Eq. ). Thus, we have { } p x)dsx) = Imd φ) + ω)φ ) + ψ) 36) ω ) { } p x)dsx) = Red φ) + ω)φ ) + ψ) ω ) 37) For a point = e iθ on the unit circle, we have = /, dθ = d for = e iθ on the unit circle) 38) i Substituting Eq. 4) into Eqs. 36)and37) and using Eq. 38) yields p x)dsx) = κ + )dθ 39) p x)dsx) = 0 40)

8 06 Y. Z. Chen et al. Note that in the integral kernel U ij ξ, x), there are r, = cos α, r, = sin α, r, r, = cos α, r, r, = sin α Fig. 3). Therefore, after using Eqs. 39)and40), the integral I in Eq. 34) can be rewritten as where I = 6π v)g κ I + I ) for ξ ) 4) I = ln rx,ξ)dθ, I = cos αdθ 4) Since both points xx, x ) and ξξ,ξ ) are located on the elliptic contour Fig. 3) I is equal to K shown by Eq. 5), and I is equal to K shownbyeq.6). Therefore, we have I = ln rx,ξ)dθ = π ln R, I = cos αdθ = πm 43) Substituting Eq. 43) into Eq. 4), we will find I = κ ln R + m) 44) 8 v)g Finally, substituting the third condition, or R = R cr = expm/κ) into Eq. 44), we will find Similarly, the integral I in Eq. 35) can be rewritten as I = I R=Rcr = 0 45) sin αdθ for ξ ) 46) In fact, the integral I is equal to K 3 shown by Eq. 6). Therefore, we have I = sin αdθ = 0 for ξ ) 47) Finally, the above-mentioned assertion is proved. Alternatively speaking, under the above-mentioned three conditions, the homogeneous equation 3) is satisfied. As shown by Eqs. 5), 6), 9)and30), if t orξξ,ξ )) is located in the elliptic contour, or t S +, the three integrals ln rx,ξ)dθ, cos αdθ and sin αdθ take the same values as in the case of t Ɣ Figs. 3, 4). Therefore, under the above-mentioned three conditions, the homogeneous equation 33) is satisfied. Alternatively, we can conclude the obtained result as follows. When the degenerate scale is reached, or R = R cr = expm/κ), the homogeneous equations 0) has a non-trivial solution which is shown by first and second conditions mentioned above, or ) p j x) = 0 j =, ) for x k, k =, 3,...,N, ) p j x) = 0 j =,, for x ) are derived from the complex potentials shown by Eq. 4). Similarly, fromthenon-trivialsolutionandthedegeneratescaleshownbyeqs. 6) and7), similar result can be found. From above-mentioned analysis, we see that under above-mentioned three conditions, the final solution is obtained. After using the first condition, or p j x) = 0 j =, ) for x k, k =, 3,...,N, the integral equation can be simplified in the form of Eq. 3). In this case, the unknown functions are reduced to two, or p j x) = 0 j =, ) for x.ineq.3), or Eqs. 3) and33), all integrations are performed on the elliptic boundary x. However, the point ξ in Eq. 3) is now defined by ξ, = N. The Eq. 3) is satisfied for ξ under conditions ) and 3), and this result is from a previous study []. In this paper, we prove that the Eq. 3) is also satisfied for ξ k,k =, 3,...,N) under conditions ) and 3). In fact, the kernel Uij ξ, x) defined in Eq. 3) only depends on the position of ξ and x. Thus, if the Eq. 3) is satisfied for one point of ξ, the equation is satisfied for ξ k k =, 3,...,N) under conditions ) and 3). After some manipulations, the assertion is proved.

9 Degenerate scale problem for plane elasticity 063 Table Computed degenerate scales for a doubly connected region a d, = f b/a, d/b) and a d, = f b/a, d/b) see Fig. 5; Eq. 48)) d/b b/a a d, = f b/a, d/b) Exact a a d, = f b/a, d/b) Exact b a Exact from Eq. 5) b Exact from Eq. 7) y b r=0.4b a x d Fig. 5 Doubly connected region with the outer ellipse and inner circle 4 Numerical illustration To examine the theoretical result mentioned above, a numerical example is presented below. It is found that coordinate transform method for finding the degenerate scale is effective [7,9]. This method has now been used to the case of a doubly connected infinite region [6]. Using this method, a numerical examination is presented below. The merit of the coordinate transform method is as follows [7,9,6]. After using the coordinate transform, the original homogenous IE in a degenerate scale can be reduced to a non-homogenous IE in the normal scale. The reduced non-homogenous IE in the normal scale can be solved with unique solution. This is the basic idea in the method. However, the detailed derivation is a little bit complicated than the mentioned basic idea. In the example, an ellipse with two half-axes a, b) contains a circular hole with a radius r = 0.4b. The center of the circular hole is shifted by a distance d Fig.5). In computation, 80 divisions are used for the discretization of the ellipse, and 30 divisions for the circle. ν = 0.3 is used in computation. Thus, an algebraic equation with 0 unknowns was formulated. For a/b = 0., 0.,...,.0 andd/b = 0.4, 0.0 and 0.4, the computed degenerate scales for size a can be expressed as a d, = f b/a, d/b) and a d, = f b/a, d/b) 48) The computed results for a d, and a d, are listed in Table.InTable, the exact results for a single ellipse are also attached, which is derived from Eqs. 5) and7). It is seen from tabulated results that; ) there is no influence from the position of the inner circular hole, ) the deviations from the computed results to the exact solutions for the simple elliptic contour case are very small.

10 064 Y. Z. Chen et al. 5 Conclusions In this paper, it is proved that degenerate scale for the outer elliptic contour does not depend on the voids involved in the contour. This result is proved exactly. As claimed previously, the degenerate scale will cause illness solution for IE. It is assumed that the ellipse has a ratio b/a = /3 with m = 0.5 andκ =.8. In this case, from Eq. 5) the first solution is R = R cr = expm/κ) =.4900, a = R + m) =.7349 and b = R m) = From Eq. 7), the second solution is R = R cr = exp m/κ) = , a = R + m) = and b = R m) = Therefore, one should avoid using the mentioned sizes in computation, particularly, in the Dirichlet boundary value problem. Previously, a numerical examination for degenerate scale problem for ellipse-shaped ring region was carried out [0]. The degenerate scale for half-axis is denoted by a cr. Computed result proved that if one takes the real scale a = 0.9a cr or a =.a cr, the illness solution disappears. Acknowledgments This project was supported by National Natural Science Foundation of China. Appendix A: Evaluation of some integrals in the closed forms In order to solve the degenerate scale problem, one must evaluate some integrals in advance. In the derivation, the mapping function shown by Eq. 9) is still used here Fig. ). For a point = e iθ on the unit circle, we have dθ = d i a) In the meantime, we can let z t = e iα a) In the first group of evaluation, three integrals K, K and K 3 are defined as follows Fig. 3) d K = ln z t dθ = Re lnz t)dθ = Im lnz t), for z Ɣ, t Ɣ) a3) z t K ik 3 = z t dθ = e iα dθ or K = cos αdθ, K 3 = sin αdθ, for z Ɣ, t Ɣ) where Ɣ denotes the elliptic contour. Note that, in Eqs. a3) anda4), the integration dθ is performed for argument θ in = e iθ. In addition, the subscript denotes the unit circle. Since both z and t are located on the ellipse, or z Ɣ, t Ɣ, wecanlet z = R + m ) ), t = R o + mo, z t = R o ) m ) a5) o Therefore, from Eqs. a3)anda5) wehave a4) K = K + K + K 3 a6) d K = Im ln R = π ln R a7)

11 Degenerate scale problem for plane elasticity 065 d K = Im ln o ) a8) K 3 = Im ln m ) d o = 0 a9) Clearly, in Eq. a9), the function ln m o ) can be considered as an analytic function in the region outside of the unit circle, since m/ o ) <, and the result is obtained. For the integral K, we have see Fig. 3) d K = Im ln o ) = π+θ o θ o = Re ln sin θ θ π o dθ = ln ln o )dθ = 0 sin u π 0 ln sin θ θ o dθ ) π/ du = 4 ln sinv) dv = 0 Finally, from Eqs. a6)toa0), we have d K = ln z t dθ = Im lnz t) = π ln R, for z Ɣ, t Ɣ) a) For the points and o on the unit circle, there are = / and o = / o. Therefore, the integral K ik 3 can be rewritten and evaluated immediately K ik 3 = e iα z t dθ = z t dθ = m o o m dθ = m o d = πm, for z Ɣ, t Ɣ) o m i a) In Eq. a), function m o o m can be considered as an analytic function with the principal portion G ) = m at infinity [3]. Clearly, Eq. a) can be rewritten as K = cos αdθ = πm, K 3 = sin αdθ = 0, for z Ɣ, t Ɣ) a3) In the second group of evaluation for three integrals, the points z for integration) is on the elliptic contour Ɣ,orz Ɣ. However, the point t now is an inner point to the elliptic contour, or t S +, where S + denotes the finite elliptic region Fig. 4). In this case, the three integrals L, L and L 3 are defined as follows Fig. 4) d L = ln z t dθ = Re lnz t)dθ = Im lnz t), for z Ɣ, t S+ ) a4) z t L il 3 = z t dθ = e iα dθ, or L = cos αdθ,l 3 = sin αdθ, for z Ɣ, t S + ) a5) Clearly, ln z t is a single-valued function. However, lnz t) is a multiple-valued function. In this case, one may define a branch cut line from point t toz = a to infinity Fig. 4). However, from the integrand lnz t)dθ in Eq. a4) we see that, the final result for L does not depend on the taken branch. In Eq. a4), the term lnz t) can be written in the form lnz t) = ln z + ln t ) = ln R + ln + ln + m ) ) t z + ln a6) R + m/) 0 a0)

12 066 Y. Z. Chen et al. Note that the function ln z t ) = lnz t) ln z here t S+ ) is an analytic single-valued function defined in the region outside the elliptic contour. This property is simply because both functions lnz t) and ln z have same increment πi when z goes around the elliptic contour. This property does not change after conformal t mapping. Thus, ln R+m/) ) = ln z t ) is also an analytic single-valued function defined in the region outside the unit circle. Clearly, ln + m ) is also a single-valued function in the region outside the unit circle. From Eqs. a4)anda6), we have L = L + L + L 3 + L 4 a7) d L = Im ln R = π ln R a8) d L = Im ln = Im ln ) ln dln ) = Im ln =πi ln =0 = Im π ) = 0 a9) Ɣ L 3 = Im ln + m ) d = 0 a0) ) t d L 4 = Im ln R + m) = 0 a) In Eqs. a0) anda), two functions ln + m ) and ln ) can be considered as a single-valued R +m) function defined in the region outside of unit circle, as mentioned above. Finally, we have L = t ln z t dθ = π ln R, for z Ɣ, t S + ) a) It is interesting to point out that the integral K shown by Eq. a) and the integral L shown by Eq. )take the same value. However, they have different definitions. Similar to Eq. a), the integral for L il 3 is defined and evaluated as L il 3 = e iα dθ = = Re Rm + ) t R + m) t z t z t dθ = d i = Im Rm + ) t R + m) t dθ Rm + ) t R + m) t = Imπim) = πm, for z Ɣ, t S + ) a3) Note that, the function R + m) t has no zero for at the region outside of the unit circle. Second, the Rm +) t R +m) t has its principal portion at infinity G ) = m. Thus, the result shown by Eq. a3) is obtained. Note that in Eq. a3), the relation = / is used for the point on the unit circle. Clearly, Eq. a3) can be rewritten as L = cos αdθ = πm, L 3 = sin αdθ = 0, for z Ɣ, t S + ) a4) It is interesting to point out that the integral K ik 3 shownbyeq.a) and the integral L il 3 shown by Eq. a3) take the same value. However, they have different definitions. d

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