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5.The mappingf1(z) =z4takes the shaded region in Figure 7.56 onto the upper semi-disk ofradius 1. The upper semi-disk is mapped onto the first quadrant byf2(w1) =i1-w11 +w1.Thus the mappingw=φ(z) =i1-z41 +z4takes the shaded region in Figure 7.56 onto the first quadrant. It is also a conformal mapping, beingthe composition of such mappings. We now determine the image of the boundary. The rays at angle0 andπ/4 are mapped onto the interval [-1,1] byf1. The quarter of a circle is mapped to the uppersemi-circle byf1. The mappingf2takes the interval [-1,1] onto the upper half of the imaginaryaxis, and the semi-circle onto the right half of the real axis.Thus the problem in the first quadrant of thew-plane becomes:ΔU= 0,U(α) = 100 ifα >0, U(iα) = 0 ifα >0.The solution in thew-plane isU(w) =aArg (w) +b, whereaandbmust be chosen in according tothe boundary conditions:aArg (w)+b= 100, when Arg = 0 andaArg (w)+b= 0, when Arg =π2.Thusb= 100 anda=-200π. HenceU(w) =-200πArgw+ 100 and sou(z) =U(φ(z)) =-200πArgi1-z41 +z4+ 100.9.To solve the given problem, we can proceed as in Example 7.3.8 and make the necessary changes.A much quicker way is based is to use the solution in Example 7.3.8 and superposition, as follows.Letu1(z) denote the solution in Example 7.3.8. Letu2(z) = 100. It is clear thatu2is harmonicfor allz.Thusu2-u1is harmonic in the shaded region of Fig.18.On the real axis, we haveu2(z)-u1(z) = 100-0 = 100. On the upper semi-circle, we haveu2(z)-u1(z) = 100-100 = 0.Thus 100-u1is the solution, whereu1is the solution in Example 18.104.22.168.This problem is very similar to that in Example 7.3.3. The first step is to map the region toan annular region bounded by concentric circles.This can be done by using the linear fractionaltransformationφ(z) =4z-14-z.
Section 7.3Solving Dirichlet Problems with Conformal Mappings169Thenφ(z) =4z-14-z=z-141-z4=φ14(z).Usingφ, we map the outer circle to the unit circle and the inner circle to the circle of radius14andcenter at 0. As in Example 7.3.3, the solution isU(w) = 100 + 100ln|w|+ ln 4ln(1/4)= 100-100ln|w|+ ln 4ln 4=-100ln|w|ln 4.Thus the solution isu(z) =-100ln 4ln4z-14-z.17.(a) The solution of the Dirichlet Problem in Figure 7.6.7 is obtained by applying the Poissonformula (7.3.5), withf(s) =sif-1< s <1,f(s) =-1 ifs <-1 andf(s) = 1 ifs >1:u(x+iy)=yπZ1-1s(x-s)2+y2ds+yπZ-1-∞-1(x-s)2+y2ds+yπZ∞1ds(x-s)2+y2=I1+I2+I3.We compute each integral separately:I1=-yπZ1-1x-s(x-s)2+y2ds+yπZ1-1x(x-s)2+y2ds=yπ12ln((x-s)2+y2)1-1+xyπZ1-1ds(s-x)2+y2=-y2πln((x+ 1)2+y2)-ln((x-1)2+y2)+xyπZ1-1ds(s-xy)2+ 1=-y2πln(x+ 1)2+y2(x-1)2+y2+xπZ1s=-1duu2+ 1(Letu=s-xy.)=-y2πln(x+ 1)2+y2(x-1)2+y2+xπtan-1s-xy1s=-1=-y2πln(x+ 1)2+y2(x-1)2+y2+xπhtan-11 +xy+ tan-11-xyi.