1MTH302 Chapter 5Some Applications of Harmonic FunctionsRemark: Correction to Schwarz’s Lemma in Chapter 3Schwarz’s Lemma (Ref Th 2.5, Unit 2 of Study Guide)Let f be an analytic function from the unit disc ?(0,1) into itself and ?(0) = 0. Theneither (i) ?(?) = ?? for all ? in ?(0,1) for some |?| ≤ 1,or (ii) |?(?)| < ? for all ? in ?(0,1).Laplace’s EquationLaplace’s equation under conformal transformationMany physical problems of interest in heat conduction, fluid flow, electrostatic potential,gravitation, etc, are governed by Laplace’s equation in a given domain Ω, that is,φ xx (x, y) + φ yy (x, y) = 0 or 2 φ (x, y) = 0for all x, y in Ω, subject to appropriate boundary conditions on the boundary ∂Ω of thedomain Ω.If the boundary of Ω has an irregular shape then the Laplace’s equation can be very difficultto solve.Now suppose we are able to map the given domain Ω in the z-plane to a regular domain, likethe unit disc or a half space, in the w-plane, via a conformal mapping w = f (z).We know that φ must be the real part or imaginary part of an analytic function.We may assume that φ is the real part of an analytic function g(z). We can then define ananalytic function G(w) in f (Ω) via2G(w) = g(f -1(w)) = (g ∘ f -1)( z).If U (u, v) = Re(G(w)), then U is a harmonic function in f (Ω) and we haveU (u, v) = Re(G(w)) = Re(g(f -1(w)) = Re(g(z)) = φ (x, y),where (u, v) = f (x+iy).Thus, if we are able to find U (u, v) by solving Laplace’s equation in f (Ω), we will be able tosolve the corresponding Laplace’s equationφ xx (x, y) +φ yy (x, y) = 0 in Ω.FactIt is know that the temperature distribution φ (x, y) in a domain Ω maintained by heat sources(or sinks) and insulation around the boundary ∂Ω of Ω obeys Laplace’s equation in its steadystate, that is,φ xx (x, y) + φ yy (x, y) = 0for all x, y in Ω.Laplace’s Equation in Polar FormInstead of using the Cartesian form of the Laplace’s equation , we can also use the followingpolar form ( , ) r of Laplace’s equation given by2 22 2 21 10r r r r ExampleFind a harmonic function ( , ) x y in the upper half plane y > 0 such that100 0( ,0)50 0xxx Solution3Since the boundary condition is constant on the rays x ≥ 0 and x ≤ 0, it is reasonable toexpect the harmonic function have to take constant value on each ray in the upper half plane.Thus we try( , ) Arg x y a z b ,where a and b are constants to be determined,Thus we have100 Arg 050a b ba b Hence,50b a 100, Thus ( , ) Arg 100 arctan( ) 100 x y z 50 50 yx ExampleFind a harmonic function u(x,y) in the upper half plane y > 0 such that0( ,0)0x au x T a x bx b Solution We can try, u x y z a z b ( , ) Arg ( ) Arg ( ) where , , are constants to be determined.4Then we have( ,0)x au x a x bx b Hence0 T 0 Thus 0, , T Tand u x y z a z b ( , ) Arg( ) Arg( ) T T ExampleFind the steady state temperature distribution φ (x, y) in the closed unit discΩ ={(x, y) : x2 + y2 ≤ 1}such that the arc Γ+ = {(x, y) : x2 + y2 = 1, y > 0} is maintained at 1∘C andthe arc Γ- = {(x, y) : x2 + y2 = 1, y < 0} is maintained at -1∘C.SolutionWe note that the Mobius map w = f (z) = (1+z)/ (1- z) maps Ω onto H,and Γ+ is mapped onto U+ = {w = it : t > 0}while Γ- is mapped onto U- = {w = it : t < 0}.5Ω HNow observe that the functionU(w) = 2/π Arg w is harmonic in H(since it is the real part of the analytic function 2/π Log w)and is +1 on {w =(0, v) : v > 0}and -1 on {w =(0, v) : v < 0}.To solve the original problem, we now only have to map this function back to Ω as follows:φ (x, y) = U (u, v) = 2/π Arg w = 2 π Arg [(1+z)/ (1- z)] = 2/ π arctan (2y/(1 – x2 – y2 ))Recall that φ xx (x, y) + φ yy (x, y) = 0 is also denoted by 2 φ (x, y) = 0ExampleSolve the following Laplace’s equation.6SolutionLet w = f (z) = z2 Then U w w w ( ) [Arg ( 1) Arg ( 1)] 100 satisfies the Laplace’s equation in the upper half w-plane.Hence the corresponding solution φ to the Laplace’s equation in the first quadrant of thez-plane is given by2 22 2 2 2100( , ) ( , ) [Arg( 1) Arg( 1)]100[Arg( 1) Arg( 1)]100[Arg( 1 2 ) Arg( 1 2 )]x y U u v w wz zx y ixy x y ixy
