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Lectures 22 and 23 Harmonic Functions As Cauchy -Riemann Equations Tell Us, the Real and the Imag

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Lectures 22 and 23 Harmonic Functions As Cauchy -Riemann Equations Tell Us, the Real and the Imag Lectures 22 and 23 Harmonic Functions As Cauchy -Riemann equations tell us, the real and the imag- inary parts of a complex analytic function have some special prop- erties. Apart from being inter-related they have the special prop- erty of possessing partial derivatives of all order. Similarly, the maximum modulus theorem tells us about certain distinct fea- tures of the modulus function of a complex analytic function. Such properties of real valued functions can be studied on their own and such a study can either be carried out using the knowl- edge of complex functions or independently. Also, an independent enquiry can lead to better understanding of the theory of com- plex functions themselves. The class of harmonic functions and the wider class of subharmonic functions substantiate this view with many such instances. From the application point of view, few ideas surpass the notion of harmonic functions. Definition 1 A real valued function u = u(x, y) defined on a domain D in C, is called harmonic with respect to the variables x, y, if it possesses continuous second order partial derivatives and satisfies the Laplace’s equation: ∂2u ∂2u 52u := + = 0. (1) ∂x2 ∂y2 Polar coordinate form of Laplace: x = r cos θ, y = r sin θ, the Laplace’s equation takes the form 1 ∂ ∂u ∂2u r r + = 0. (2) ∂r ∂r ∂θ2 Verify this. Observe that this form of Laplace’s equation is not applicable at the origin, since the polar coordinate transformation is singular there. However, the polar coordinate form has many advantages of its own and is loved by physicists and engineers. As an application, from (2), it follows easily that ln r = ln |z| is well defined and harmonic throughout C\{0}. In fact as an easy exercise, prove that any function of r alone, i.e., independent of θ, is harmonic iff it is of the form a ln r +b, where a, b are constants. Remark 1 (i) Harmonic functions arise in the study of graviational fields, electrostatic fields, steady-state heat conduction, incompressible fluid flows etc. etc.. (ii) Mathematically, they are very close to holomorphic functions. Given a holomorphic function f = u + ıv, as i a straight forward consequence of Cauchy-Riemann equations and the property of an analytic function possessing continuous derivatives of all order, it follows that u and v are both harmonic. In this case we call v the harmonic conjugate of u. Observe that, by considering the function ıf(x, y), it then follows that u is the harmonic conjugate of −v. We shall see the converse of this as a theorem. iii) There is nothing very special about considering only real val- ued functions for the definition of harmonic functions. We could even allow complex valued functions in the above definition. Then it follows that a complex function is harmonic iff its real and imag- inary parts are harmonic. Thus, it suffices to treat only the real 2 valued functions, in the study of harmonic functions. iv) From the linearity of the differential operator, 52, it follows that the set of all harmonic maps on a domain forms a vector space. In particular all linear functions ax + by are harmonic. However, it is not true that product of two harmonic functions is harmonic. For example, xy is hamonic but x2y2 is not. iv) Harmonicity is quite a delicate property. If φ is a smooth real valued function of a real variable and u is harmonic, then, in gen- eral, φ ◦ u need not be harmonic. Indeed, φ ◦ u is harmonic for all harmonic u iff φ is linear(exercise). Likewise, if f : D1 −→ D2 is a smooth complex valued function of two real variables then u ◦ f need not be harmonic. However, under conformal mapping we have some positive result which we shall see below. Theorem 1 Let u(x, y) be a harmonic function on a simply connected domain. Then u is the real part of an analytic function in D, i.e., u has a harmonic conjugate v throughout D. ∂u ∂u Proof: We first consider the function g(z) = − ı . From ∂x ∂y Cauchy-Riemann equations it follows that g is holomorphic. (Thus ∂u ∂u is also harmonic and has an harmonic conjugate − .) If v ∂x ∂y were a conjugate of u, then the differential dv is given by, ∂v ∂v ∂u ∂u dv := dx + dy = − dx + dy =: ?du ∂x ∂y ∂y ∂x Thus we must try to show that ?du is an exact differential. Now the differential g(z)dz can be written as, ∂u ∂u ? g(z)dz = − ı (dx + ıdy) = du + ı du ∂x ∂y 3 and by Cauchy’s theorem we have, Z g(z)dz = 0 γ for all cycles γ in D, since D is simply connected. On the other hand, since du is the exact differential, its integral on any cycle is zero(see primitive existence theorem). Hence it follows that Z ?du = 0 (3) γ for all cycles γ in D. This in turn implies that ?du is an exact differential, i.e., ?du = dv for some v defined throughout D. ♠ Remark 2 The statement of the theorem for convex domains can be proved just by integration. Observe that the simply con- nectivity assumption on the domain is important. A typical ex- ample is the ln r which is harmonic throughout C \{0} but there is no holomorphic function on C \{0} whose real part is equal to ln r. For any function which may fit this bill has to be a branch of ln z none of which is avalaible all over C \{0}. 4 Theorem 2 Suppose w = f(z) is a one-to-one, conformal mapping of a domain D1 in the xy-plane onto a domain D2 uv-plane. Let C1 be a smooth curve in D1 and C2 = f(C1). Let φ(u, v) be a real valued function with continuous partial derivatives of second order on D2 and let ψ be the composite function φ ◦ f on D1. Then (i) ψ is harmonic if and only if φ is harmonic (ii) φ is con- stant on C2 if and only if ψ is constant on C1 (or in other words, f takes the level curves of ψ to those of φ. (iii) The normal derivative of φ along C2 vanishes iff the nor- mal derivative of ψ along C1 vanishes. Proof: (i) Take z1 ∈ D1 and let z2 = f(z1). We can then choose open discs Bj around these points contained in D1 and D2 re- spectively and such that f(B1) ⊂ B2. Assuming φ is harmonic, ˆ ˆ we can choose a conjugate φ to it in B2 so that g = φ + ıφ is holomorphic. But then g ◦ f is holomorphic in B1. Its real part is nothing but φ ◦ f = ψ. Therefore, ψ is harmonic at z1. Since z1 is arbitrary, ψ is harmonic in D1. The converse follows by taking −1 f : D2 → D1 and interchanging the role of φ and ψ. (ii) This is just set theory. (iii) Let z1 be a point on C1,T,N be tangent and normal vectors to C1 at z1. Then the derivative Df of f at z1 maps T onto a tangent vector to C2 at z2. By conformality of f it also maps N onto a nonzero normal vector, λn, where n is the unit normal vector to C2, λ =6 0. to C2 at z2. We also have, by chain rule (grad φ) ◦ Df = grad ψ. Now the normal derivative of ψ along C1 vanishes iff grad ψ · N = 0 iff [(grad φ) ◦ Df] · N = 0 iff 5 grad φ · Df(N) = 0 iff grad φ · n = 0 iff the normal derivative of φ along C2 vanishes. ♠ Example 1 It is not true, in general that the normal derivative of ψ along C2 is a constant if normal derivative of φ along C1 is a constant. An an example take D1 = {(x, y): x > 0, }, f(z) = z2 and C1 = {(x, 0) : x0} and φ(u, v) = v. Then C2 is the ∂v positive u-axis, the normal derivative of u along C2 is ∂v = 1. We have ψ(x, y) = 2xy and the normal derivative of along C1 is ∂ψ ∂y = 2x which is not a constant along C1. Mean Value Theorem and Maximum Principle Theorem 3 Mean Value Property : Let u be harmonic in a domain D and Br(z0) ⊂ D. Then 1 Z 1 Z 2π ıθ u(z0) = u(z) d(arg(z − z0)) = u(z0 + re ) dθ.(4) 2π |z−z0|=r 2π 0 Proof: Let v be a harmonic conjugate of u on Br(z0) so that f = u + ıv is holomorphic. By Cauchy’s integral formula we have 1 Z f(z) 1 Z 2π f(z ) = dz = f(z + reıθ)dθ. 0 |z−z |=r 0 0 2πı 0 z − z0 2π Upon comparing real parts we get (4). ♠ Theorem 4 Let u be a harmonic function on a domain D. If u is constant on a non empty open subset, then it is a constant on the whole of D. Proof: First assume that D is simply connected. Let f be a holomorphic function f = u + ıv. If U is a non empty open connected set on which u is a constant, by CR equations it follows 6 that f is also a constant on U. By Identity Theorem, this implies that f is a constant on D and hence u is also so. In the general case, fix z0 ∈ U.
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