Math 752 Spring 2015 Class 3/23 1 Harmonic Functions Definition 1. G open subset of C. A function u : G ! C is harmonic if u has continuous second partial derivatives and @2u @2u + = 0: @x2 @y2 The left-hand side is called the Laplacian of u, denoted by r2u. Let 0 2= G and rewrite the differential equation above in polar coordi- nates: @2u 1 @u 1 @2u + + = 0: @r2 r @r r2 @θ2 Sketch of derivation: Start with ur = ux cos θ + uy sin θ uθ = uxr sin θ + uyr cos θ and differentiate again and simplify which eventually leads to 1 1 u + u = u + u − u rr r2 θθ xx yy r r Examples of harmonic functions: iθ n U1(re ) = r cos(nθ); (n ≥ 0) iθ n U2(re ) = r sin(nθ); (n ≥ 1); We note that r2u = 0 is linear in u, hence a complex valued function is harmonic if and only if its real and imaginary part are harmonic. Hence, u(reiθ) = rjnjeinθ is harmonic for n 2 Z. A very important example of real harmonic functions comes from an- alytic functions. Let F be analytic and let u = <F and v = =F . The Cauchy-Riemann equations are ux = vy; uy = −vx: Theorem 1. f on a region G is analytic iff <f = u and =f = v are har- monic functions which satisfy the Cauchy-Riemann differential equations. Proof. u(x; y) = <f(x + iy), similarly v = =f. Let h be real. Divide the identity f(z + h) − f(z) = u(x + h; y) − u(xy) + ifv(x + h; y) − v(x; y)g by h and let h ! 0 through real values of h. This leads to 0 f = ux + ivx; and letting h ! 0 through imaginary values gives with a similar calculation 0 f = −iuy + vy: Equating real and imaginary parts gives the forward implication of the theo- rem. It is an elementary but lengthy exercise to show the reverse implication: Show first (using continuity of the partials of u and v) u(x + s; y + t) − u(x; y) = ux(x; y)s + uy(x; y)t + '(s; t); v(x + s; y + t) − v(x; y) = vx(x; y)s + vy(x; y)t + (s; t); where '(s; t) and (s; t) are o(s; t) as (s; t) ! (0; 0), and note now that f(z + s + it) − f(z) u s + u t + i(v s + v t) '(s; t) + i (s; t) = x y x y + s + it s + it s + it The C-R equations simplify this to f(z + s + it) − f(z) '(s; t) + i (s; t) = u (z) + iv (z) + ; s + it x x s + it which gives the claim by letting (s; t) ! (0; 0). Theorem 2. Let u be real valued and harmonic on some disk jzj < r. Then there is a harmonic function v on the disk (called the harmonic conjugate) such that f = u + iv is analytic on G. Proof. Wanted: v, infinitely differentiable, so that vx = −uy; vy = ux; because then vxx + vyy = −uyx + uxy = 0. Consider the form P dx + Qdy := uydx − uxdy: We have Py = uyy = −uxx = Qx, hence the form is exact, and since any disk is simply connected, Calculus III gives: there exists a potential function v with vx = P = uy and vy = Q = −ux. Done! Corollary 1. If u : G ! C is harmonic, then u is infinitely differentiable. Remark. One can show that the existence of the harmonic conjugate on an open set G is equivalent to G being simply connected..