440 Mixed Boundary Value Problems or u(r, θ, z)= 1 π 2  a 0  2π 0  R ξ +arctan  ξ R  z R 3 f(ρ, ϑ) dϑ ρ dρ, (6.3.19) where R 2 = r 2 + ρ 2 − 2rρcos(θ −ϑ)+z 2 ,  1,2 (x)= 1 2  (x + r) 2 + z 2 ∓  (x − r) 2 + z 2 , (6.3.20) the operator L(k)σ(r, θ)isgivenby L(k)σ(r, θ)= 1 2π  2π 0 λ(k, θ − ϑ)σ(r, ϑ) dϑ (6.3.21) = ∞  n=−∞ k |n| e inθ  1 2π  2π 0 e −inϑ σ(r, ϑ) dϑ  , (6.3.22) λ(k, ϑ)= 1 − k 2 1+k 2 − 2k cos(ϑ) = ∞  n=−∞ k |n| e inϑ , (6.3.23) and ξ = z  a 2 − ρ 2   2 2 (a) − a 2 =   2 2 (a) −  2 1 (ρ)   2 2 (a) −  2 2 (ρ)  2 (a) (6.3.24) =  a 2 − ρ 2  a 2 −  2 1 (a) a =  a 2 − ρ 2   2 2 (a) − r 2  2 (a) . (6.3.25) Equation 6.3.18 is recommended in those cases when the integrals can be evaluated exactly while Equation 6.3.19 is more convenient when the integrals must be computed numerically. To illustrate Fabrikant’s results, consider the case when f(r, θ)=w 0 ,a constant. In this case, L(ρ)f(ρ, θ)= ∞  n=−∞ ρ |n| e inθ  1 2π  2π 0 e −inϑ w 0 dϑ  = w 0 (6.3.26) because all of the terms in the summation vanish except n =0. Therefore, d dη   η 0 L(ρ)f(ρ, θ) ρ  η 2 − ρ 2 dρ  = w 0 , (6.3.27) and L   2 1 (η) rη 2  d dη   η 0 L(ρ)f(ρ, θ) ρ  η 2 − ρ 2 dρ  (6.3.28) = ∞  n=−∞   2 1 (η) rη 2  |n| e inθ  1 2π  2π 0 e −inϑ w 0 dϑ  = w 0 . © 2008 by Taylor & Francis Group, LLC Green’s Function 441 From Equation 6.3.18 u(r, θ, z)= 2w 0 π  a 0 d 1 (η)  r 2 −  2 1 (η) = 2w 0 π arcsin   2 1 (η) r      a 0 (6.3.29) = 2w 0 π arcsin   2 1 (a) r  . (6.3.30) Fabrikant also considered the case when the mixed boundary condition reads  u z (r, θ, 0) = σ(r, θ), 0 ≤ r<a, 0 ≤ θ<2π, u(r, θ, 0) = 0,a<r<∞, 0 ≤ θ<2π. (6.3.31) In this case he showed that the solution is u(r, θ, z)=4C   2 (a)  2 (0)   g(x) 0 ρdρ  g 2 (x) − ρ 2 L  ρr x 2  σ(ρ, θ)  dx √ x 2 − r 2 , (6.3.32) u(r, θ, z)=4C  a 0   η 0 ρdρ  η 2 − ρ 2 L  ρr  2 2 (η)  σ(ρ, θ)  d 2 (η)   2 2 (η) − r 2 , (6.3.33) or u(r, θ, z)= 2C π  a 0  2π 0 arctan  ξ R  σ(ρ, ϑ) R dϑ ρ dρ, (6.3.34) where g(x)=x  1+z 2 /(r 2 − x 2 )andR and ξ have been defined earlier. The constant coefficient C equals −1/(2π)inclassical potential problems and different values in other applications. Equation 6.3.32 and Equation 6.3.33 are best when the integrals can be evaluated exactly while Equation 6.3.34 should be used otherwise. To illustrate Equation 6.3.32 through Equation 6.3.34, consider the case when σ(r, θ)=σ 0 ,aconstant. Then, L  ρr x 2  σ(ρ, θ)= ∞  n=−∞  ρr x 2  |n| e inθ  1 2π  2π 0 e −inϑ σ 0 dϑ  = σ 0 , (6.3.35) and  g(x) 0 ρdρ  g 2 (x) − ρ 2 L  ρr x 2  σ(ρ, θ)=σ 0 g(x). (6.3.36) Therefore, using Equation 6.3.32, u(r, θ, z)=4Cσ 0   2 (a)  2 (0) g(x) dx √ x 2 − r 2 (6.3.37) =4Cσ 0   2 (a)  2 (0) √ x 2 − r 2 − z 2 x 2 − r 2 xdx (6.3.38) © 2008 by Taylor & Francis Group, LLC 442 Mixed Boundary Value Problems =2Cσ 0   2 (a)  2 (0) d(x 2 − r 2 ) √ x 2 − r 2 − z 2 − 2Cσ 0 z 2   2 (a)  2 (0) d(x 2 − r 2 ) (x 2 − r 2 ) √ x 2 − r 2 − z 2 (6.3.39) =4Cσ 0  x 2 − r 2 − z 2     2 (a)  2 (0) − 4Cσ 0 z arctan  √ x 2 − r 2 − z 2 z        2 (a)  2 (0) (6.3.40) =4Cσ 0  a 2 −  2 1 (a) − 4Cσ 0 z arctan   a 2 −  2 1 (a) z  , (6.3 .41) where  2 2 (0) = r 2 +z 2 and  2 2 (a)+ 2 1 (a)=a 2 +r 2 +z 2 .Fabrikanthasextended his work to spherical coordinates 12 and crack problems. 13 12 Fabrikant, V. I., 1987: Mixed problems of potential theory in spherical coordinates. Z. Angew. Math. Mech., 67, 507–518. 13 Fabrikant, V. I., and E. N. Karapetian, 1994: Elementary exact method for solving mixed boundary value problems of potential theory, with applications to half-plane contact and crack problems. Quart. J. Mech.Appl.Math., 47, 159–174. © 2008 by Taylor & Francis Group, LLC Chapter 7 Conformal Mapping Conformal mapping is a method from classical mathematical physics for solv- ing Laplace’s equation. It is readily shown that an analytic function w = ξ + iη = f(z), where z = x + iy,transforms Laplace’s equation in the xy- plane into Laplace’s equation in the ξη-plane. The objective here is to choose amapping so that the solution is easier to obtain in the new domain. This method has been very popular in fields such as electrostatics and hydrodynamics. In the case of mixed boundary value problems, this technique has enjoyed limited success because the transformed boundary conditions are very complicated and the corresponding solution to Laplace’s equation is dif- ficult to find. In this chapter we illustrate some of the successful transforma- tions. 7.1 THE MAPPING z = w + a log(w) During their study of fringing fields in disc capacitors, Sloggett et al. 1 used this mapping to find the potential in the upper half-plane y>0wherethe potential equals V along the line y = πa when −∞ <x<aln(a) − a.Along 1 Sloggett, G. J., N. G. Barton, and S. J. Spencer, 1986: Fringing fields in disc capaci- tors. J. Phys., Ser. A, 19, 2725–2736. 443 © 2008 by Taylor & Francis Group, LLC 444 Mixed Boundary Value Problems −4 −3 −2 −1 0 1 2 3 4 0 0.5 1 1.5 2 2.5 3 3.5 4 − 0 0 0.5 1 1.5 0.005 0.025 0.1 0.3 1 1.5 2 3 x y a = 2/π Figure 7.1.1:Theconformal mapping z = w + a log(w)witha =2/π.Ifw = ξ + iη,the dark (solid) lines are lines of constant ξ while the lighter (dashed) lines give η.Theheavy dark line corresponds to the line −∞ <ξ<0 − and η =0. y =0,u(x, 0) =0.Figure 7.1.1 illustrates this mapping. In particular, the following line segments are mapped along the real axis in the w-plane: z-plane w-plane −∞ <x<aln(a) − ay=(aπ) + −∞ <ξ<−aη=0 −∞ <x<aln(a) − ay=(aπ) − −a<ξ<0 − η =0 −∞ <x<1 y =0 0 + <ξ<1 η =0 1 <x<∞ y =0 1<ξ<∞ η =0 Here the (·) + and (·) − denote points just above or below (·), respectively. From Poisson’s integral, u(ξ, η)= V π  ∞ 0 η (ξ −s) 2 + η 2 ds = V π − V π arctan  η ξ  . (7.1.1) Therefore, for a given value of ξ and η we can use Equation 7.1.1 to compute the potential. Then the conformal mapping provides the solution for the corresponding x and y.Figure 7.1.2 illustrates this solution. For a given z, Newton’s method wasusedtosolveforw.Thenthe potential follows from Equation 7.1.1. © 2008 by Taylor & Francis Group, LLC 446 Mixed Boundary Value Problems 0.2 0.4 0.6 0.8 1 1.25 1.5 0.2 0.4 0.6 0.8 1 1.25 1.5 π x/(2b) π y/(2b) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Figure 7.2.1:Theconformal mapping tanh[πz/(2b)] = sn(w,k)whena/b = 1 2 and k = tanh[πa/(2b)]. The solid, dark line gives values of (w)while the dashed, horizontal lines denote (w). can be used to solve Equation 7.2.1 through Equation 7.2.4, where z = x+ iy, w = ξ+iη, k =tanh[πa/(2b)], and sn(·, ·)isone of the Jacobi elliptic functions. Figure 7.2.1 illustrates lines of constant ξ and η within a portion of the original xy-plane. It shows that the original semi-infinite strip has been mapped into a rectangular region with 0 <ξ<Kand 0 <η<K  ,where K and K  arecomplete elliptic integrals of the first kind for moduli k and k  = √ 1 − k 2 . Applying the conformal mapping Equation 7.2.5, the problem becomes ∂ 2 u ∂ξ 2 + ∂ 2 u ∂η 2 =0, 0 <ξ<K, 0 <η<K  , (7.2.6) subject to the boundary conditions u(ξ, 0) = 0,u(ξ,K  )=1, 0 <ξ<K, (7.2.7) and u ξ (0,η)=u ξ (K, η)=0, 0 <η<K  . (7.2.8) The solution to Equation 7.2.6 through Equation 7.2.8 is simply u(ξ, η)= η/K  .Therefore, lines of constant η/K  give u(x, y)viaEquation7.2.5. Figure 1.1.4 illustrates the solution. © 2008 by Taylor & Francis Group, LLC Conformal Mapping 447 −1 +1 +1−1 S S d d S mm S S m m S z−plane w−plane y η ξ x λ y= Figure 7.3.1:The conformal mapping used to map the half-plane above the boundary conditions S m ∪ S d into the half-plane η>0inthew-plane. 7.3 THE MAPPING z = w + λ √ w 2 − 1 This mapping is useful in converting an elliptic shaped domain in the xy-plane into a rectangular one in the ξη-plane where w = ξ + iη.Toillustrate this conformal mapping, let us solve Laplace’s equation in the half-plane ∂ 2 u ∂x 2 + ∂ 2 u ∂y 2 =0, −∞ <x<∞,η(x, 0) <y<∞, (7.3.1) subject to the boundary conditions lim y→∞ u(x, y) → 0, −∞ <x<∞, (7.3.2) and ∂u ∂n     S d =0, and u| S m =0. (7.3.3) We begin by solving theproblem ∂ 2 u ∂ξ 2 + ∂ 2 u ∂η 2 =0, −∞ <ξ<∞, 0 <η<∞, (7.3.4) with the boundary conditions lim |ξ|→∞ |u(ξ, η)| < ∞, −∞ <η<∞, (7.3.5) © 2008 by Taylor & Francis Group, LLC 448 Mixed Boundary Value Problems −2 −1 0 1 2 0 0.5 1 1.5 2 0.6 0.7 0.7 0.7 0.7 0.7 7 0.8 0.8 0.8 0.8 0.9 0.9 0.9 0.9 0.99 099 0.99 099 x y C/V = 0.25 −2 −1 0 1 2 0 0.5 1 1.5 2 0. 3 0.3 0.3 0.3 0.3 0.3 0.4 0.4 0.4 0.4 0.4 0 .4 0.5 0.5 0.5 0.5 0.5 0.5 0. 7 0.7 0.7 0.7 099 099 99 099 x y C/V = 0.50 −2 −1 0 1 2 0 0.5 1 1.5 2 0.001 0.001 0.001 0.001 0.001 0 01 0 . 0.1 0.1 0.1 0.1 0.1 0.2 5 0.25 0.25 0.25 0.25 0.25 0.5 0.5 0.5 0.5 0.7 5 0.75 . 75 0.75 09 099 099 099 x y C/V = 0.75 −2 −1 0 1 2 0 0.5 1 1.5 2 1e−2 0 1e−20 1e−20 e−20 1e−20 1e−20 0.25 0.25 0.25 0.25 0.5 0.5 0.5 0.5 0.7 5 0.75 0 .75 0.75 09 099 099 x y C/V = 1.00 Figure 7.3.2:Plotsofthe potential u(x, y)/V when λ =0.5forvarious values of C/V . lim η→∞ u(ξ, η) → 0, −∞ <ξ<∞, (7.3.6) and  u η (ξ,0) = 0, |ξ| < 1, u(ξ, 0) = V, |ξ| > 1. (7.3.7) The solution to this problem is u(ξ, η)=V + C   i  w 2 − 1  , (7.3.8) because u(ξ, 0) = V + C   i  ξ 2 − 1  = V (7.3.9) if |ξ| > 1, and ∂u(ξ,0) ∂η = C   − ξ  ξ 2 − 1  =0 (7.3.10) if |ξ| < 1. Therefore, u(x, y)=V + C   i −λz + √ z 2 + λ 2 − 1 1 − λ 2  , (7.3.11) where z = x + iy and C is a free parameter. Figure 7.3.2 illustrates u(x, y)whenλ =0.5. In the construction of the conformal mapping and solution, it is important to take the branch cut of √ w 2 − 1sothatitliesalong the real axis in the complex w-plane. © 2008 by Taylor & Francis Group, LLC Conformal Mapping 449 7.4 THE MAPPING w = ai(z − a)/(z + a) Let us solve 3 Laplace’s equation in a domain exterior to an infinitely long cylinder of radius a ∂ 2 u ∂r 2 + 1 r ∂u ∂r + 1 r 2 ∂ 2 u ∂θ 2 =0,a≤ r<∞, 0 < |θ| <π, (7.4.1) subject to the boundary conditions lim r→∞ u(r, θ) → 0, 0 ≤|θ|≤π, (7.4.2) and  u r (a, θ)=1, 0 ≤|θ| <α, u(a, θ)=0,α<|θ| <π. (7.4.3) We begin by introducing the conformal mapping: w = ˜ ξ + i˜η = ia z −a z + a , (7.4.4) where z = x + iy.Equation7.4.1 then becomes ∂ 2 u ∂ ˜ ξ 2 + ∂ 2 u ∂ ˜η 2 =0, −∞ < ˜ ξ<∞, 0 < ˜η<∞, (7.4.5) with the boundary conditions lim ˜η→∞ u( ˜ ξ, ˜η) → 0, −∞ < ˜ ξ<∞, (7.4.6) and  u ˜η ( ˜ ξ,0) = 2a 2 /( ˜ ξ 2 + a 2 ), | ˜ ξ| <atan(α/2), u( ˜ ξ,0) = 0, | ˜ ξ| >atan(α/2). (7.4.7) We now nondimensionalize ˜ ξ and ˜η as follows: ξ = ˜ ξ a tan(α/2) = − sin(θ) tan(α/2) 2ar r 2 + a 2 +2ar cos(θ) , (7.4.8) and η = ˜η a tan(α/2) =cot(α/2) r 2 − a 2 r 2 + a 2 +2ar cos(θ) . (7.4.9) Figure 7.4.1 illustrates this conformal mapping. Equation 7.4.5 then becomes ∂ 2 u ∂ξ 2 + ∂ 2 u ∂η 2 =0, −∞ <ξ<∞, 0 <η<∞, (7.4.10) 3 See Iossel’, Yu. Ya., 1971: A mixed two-dimensional stationary heat-conduction prob- lem for a cylinder. J. Engng. Phys., 21, 1145–1147. © 2008 by Taylor & Francis Group, LLC Conformal Mapping 451 −2 −1 0 1 2 −2 −1 0 1 2 0 0.5 1 1.5 x/a y/a −u(x,y)/a Figure 7.4.2:The solution of Equation 7.4.1 subject to the mixed boundary conditions Equation 7.4.2 and Equation 7.4.3 when α = π/4. To solve these dual integral equations, we define A(k)by A(k)=  1 0 g(t)J 0 (kt) dt. (7.4.16) Aquickcheck shows that Equation 7.4.16 satisfies Equation 7.4.15 identically. On the other hand, integrating Equation 7.4.14 with respect to ξ,wefind that  ∞ 0 A(k)sin(kξ) dk = −2a arctan[ξ tan(α/2)], |ξ| < 1. (7.4.17) Next, we substitute Equation 7.4.16 into Equation 7.4.17, interchange the order of integration,thenapply Equation 1.4.13 and obtain  ξ 0 g(t)  ξ 2 − t 2 dt = −2a arctan[ξ tan(α/2)], |ξ| < 1. (7.4.18) Applying the results from Equation 1.2.13 and Equation 1.2.14, g(t)=− 4a π d dt   t 0 ζ arctan[ζ tan(α/2)]  t 2 − ζ 2 dζ  (7.4.19) = − 2at tan(α/2)  1+t 2 tan 2 (α/2) . (7.4.20) Finally, if we substitute Equation 7.4.16 and Equation 7.4.20 into Equation 7.4.13, © 2008 by Taylor & Francis Group, LLC [...]... difficulties that arise in a mixed boundary value problem where one of the boundary conditions is a Robin condition Here he suggests a method for solving this problem 4 See Strakhov, I A., 1969: One steady-state heat-conduction problem for a polygonal region with mixed boundary conditions J Engng Phys., 17, 990–994 © 2008 by Taylor & Francis Group, LLC 454 Mixed Boundary Value Problems Figure 7.5.2: The... UU(icount,jcount) = zeta r; VV(icount,jcount) = zeta i; distance = abs(F); © 2008 by Taylor & Francis Group, LLC 462 Mixed Boundary Value Problems 1 0.8 0.6 0.4 0.2 0 0 0 0.2 0.5 0.4 1 0.6 1.5 0.8 y/b 1 2 x/b Figure 7.6.3: Solution to Laplace’s equation with the mixed boundary value problems given by Equation 7.6.2 through Equation 7.6.4 when a = 2, b = 1, and = 1 end; end; end % now use Newton’s method... Phys., 23, 1191–1193 © 2008 by Taylor & Francis Group, LLC 458 Mixed Boundary Value Problems y z−plane 3 2 1 −a x a η 1 0 w−plane ai 2 3 _ 0 ξ + Conformal Mapping w = √ z 2 − a2 0 u(x,y)/A −0.2 −0.4 −0.6 −0.8 −1 −2 2 1.5 −1 1 0 x y 0.5 1 2 0 Problem 1 solve Laplace’s equation:8 ∂2u ∂2u + 2 = 0, ∂x2 ∂y 0 < x < a, 0 < y < b, (7.6.1) subject to the boundary conditions u(x, b) = 0, 0 < x < a, (7.6.2) 8 See... and w-planes, © 2008 by Taylor & Francis Group, LLC 460 Mixed Boundary Value Problems respectively In the z-plane, 0 < x/a < Kz and 0 < y/b < Kz , where Kz and iKz are the quarter-periods of the elliptic function Therefore, we must choose kz so that Kz /Kz = a/b Using MATLAB R , this is done as follows: % By guessing kz , called k, find the closest value % of Kz /Kz to a/b diff = 10000; for n = 1:19999... y < ∞, (7.5.1) subject to the boundary conditions lim |u(x, y)| < ∞, y→∞ lim |u(x, y)| < ∞, y→−∞ and   u(−b, y) = T0 , u(0, y) = T0 ,  −uy (x, 0) + hu(x, 0) = 0, −b < x < ∞, (7.5.2) −b < x < 0, (7.5.3) −∞ < y < ∞, −∞ < y < 0, 0 < y < ∞ (7.5.4) In the previous problem we used conformal mapping to transform a mixed boundary value problem with Dirichlet and/or Neumann boundary conditions into a simple... π/2 ei dζ , +1 ζ dζ ζ (7.5.20) ζ2 where we have introduced B=− 2i π ∞ 0 ζ +i ζ −i © 2008 by Taylor & Francis Group, LLC /2 − ei π/2 ei ζ dζ +α ζ ∞ 0 ζ +i ζ −i /2 ei ζ dζ ζ2 + 1 (7.5.21) 456 Mixed Boundary Value Problems In the case of Equation 7.5.21, the integration can occur along any curve in the first quadrant of the w-plane that does not pass through the point ζ = i Let us now evaluate Equation...452 Mixed Boundary Value Problems u(ξ, η) = −2a tan(α/2) 1 0 ∞ t 1 + t2 tan2 (α/2) 0 J0 (kt)e−kη cos(kξ) dk dt (7.4.21) √ = − 2 a tan(α/2) × 1 t 0 t2 + η 2 + ξ 2 + (t2 + η 2 − ξ 2 )2 + 4η 2 ξ 2 dt [(t2 + η 2 − ξ 2 )2... a given w, we can compute the corresponding z/b from Equation 7.5.5 The same w is then used to find Ψ(w) and the value of u(x, y) Figure 7.5.3 illustrates the solution when = 2 e−i w ei ζ − Problems 1 Solve Laplace’s equation7 ∂2u ∂2u + 2 = 0, ∂x2 ∂y −∞ < x < ∞, 0 < y < ∞, subject to the boundary conditions lim u(x, y) → 0, |x|→∞ 0 < y < ∞, 7 See Karush, W., and G Young, 1952: Temperature rise in a... code is % Find the corresponding values of kw , Kw and Kw [sn,cn,dn] = ellipj(ell*K z/a,k z); k w = k z*sn; K w = ellipke(k w); k prime = sqrt(1-k w*k w); K prime w = ellipke(k prime); Having found kz , Kz , Kz and kw , we are ready to find the values of ξ and η corresponding to a given x and y This is done in two steps First we find for a given x and y an approximate value of ξ and η where 0 < ξ < Kw... the w-plane Therefore, the boundary conditions become 0 < η < ∞, [Ψ(iη)] = T0 , (7.5.12) and −i(w2 + 1) dΨ(w) + w2 Ψ(w) dw = 0, 0 < ξ < ∞ (7.5.13) w=ξ At this point we introduce the function χ(w) defined by Ψ(w) = χ(w) − 5 2iT0 log(w) π (7.5.14) Stoker, J J., 1947: Surface waves in water of variable depth Quart Appl Math., 5, 1–54 6 Chester, C R., 1961: Reduction of a boundary value problem of the third . Group, LLC 462 Mixed Boundary Value Problems 0 0.5 1 1.5 2 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 x/b y/b Figure 7.6.3:Solution to Laplace’s equation with the mixed boundary value problems given. problem for a polygonal region with mixed boundary conditions. J. Engng. Phys., 17, 990–994. © 2008 by Taylor & Francis Group, LLC 454 Mixed Boundary Value Problems Figure 7.5.2:Theconformal. Karapetian, 1994: Elementary exact method for solving mixed boundary value problems of potential theory, with applications to half-plane contact and crack problems. Quart. J. Mech.Appl.Math., 47, 159–174. ©