Chapter6 - Chapter 6 Harmonic Functions The shortest route...

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Unformatted text preview: Chapter 6 Harmonic Functions The shortest route between two truths in the real domain passes through the complex domain. J. Hadamard 6.1 Definition and Basic Properties We will now spend a chapter on certain functions defined on subsets of the complex plane which are real valued. The main motivation for studying them is that the partial differential equation they satisfy is very common in the physical sciences. Recall from Section ?? the definition of a harmonic function: Definition 6.1. Let G ⊆ C be a region. A function u : G → R is harmonic in G if it has continuous second partials in G and satisfies the Laplace 1 equation u xx + u yy = 0 in G . There are (at least) two reasons why harmonic functions are part of the study of complex analysis, and they can be found in the next two theorems. Proposition 6.2. Suppose f = u + iv is holomorphic in the region G . Then u and v are harmonic in G . Proof. First, by Corollary 5.2, f is infinitely differentiable, and hence so are u and v . In particular, u and v have continuous second partials. By Theorem 2.13, u and v satisfy the Cauchy–Riemann equations u x = v y and u y =- v x in G . Hence u xx + u yy = ( u x ) x + ( u y ) y = ( v y ) x + (- v x ) y = v yx- v xy = 0 in G . Note that in the last step we used the fact that v has continuous second partials. The proof that v satisfies the Laplace equation is completely analogous. 1 For more information about Pierre-Simon Laplace (1749–1827), see http://www-groups.dcs.st-and.ac.uk/ ∼ history/Biographies/Laplace.html . 63 CHAPTER 6. HARMONIC FUNCTIONS 64 Proposition 6.2 shouts for a converse theorem. There are, however, functions which are harmonic in a region G but not the real part (say) of an holomorphic function in G (Exercise 3). We do obtain a converse of Proposition 6.2 if we restrict ourselves to simply connected regions. Theorem 6.3. Suppose u is harmonic on the simply connected region G . Then there exists a harmonic function v such that f = u + iv is holomorphic in G . Remark. The function v is called a harmonic conjugate of u . Proof. We will explicitly construct the holomorphic function f (and thus v = Im f ). First, let g = u x- iu y . The plan is to prove that g is holomorphic, and then to construct an antiderivative of g , which will be almost the function f that we’re after. To prove that g is holomorphic, we use Theorem 2.13: first because u is harmonic, Re g = u x and Im g =- u y have continuous partials. Moreover, again because u is harmonic, they satisfy the Cauchy–Riemann equations:...
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This note was uploaded on 06/18/2011 for the course MATH 375 taught by Professor Marchesi during the Spring '10 term at Binghamton.

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Chapter6 - Chapter 6 Harmonic Functions The shortest route...

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