= 2 x y 2. (1)

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1 COMPLEX ANALYSIS PART 5: HARMONIC FUNCTIONS A Let me start by asking you a question. Suppose that f is an analytic function so that the CR-equation f/ z = 0 is satisfied. Let us write u and v for the real and the imaginary parts of f respectively: f = u + iv. Now the CR-equation becomes (u/ z) + i(v/ z) = 0. Thus u/ z = 0 and v/ z = 0. This shows both u and v are analytic. Is my argument correct? Q No. The conclusion is wrong. It is known that real-valued analytic functions must be constants. This would imply that u and v are constants and hence so is f = u+iv. So there must be something wrong with the argument. I have to go over this argument once again to find where the mistake is... Ah! I ve found it: the functions u/ z and v/ z in general are not real-valued, even though u and v are. So we cannot go from (u/ z) + i(v/ z) = 0 to u/ z = 0 and v/ z = 0. A Good. Now I try to remedy the situation. I apply /z to both sides of the CRequation to get (/z)(/ z)f = 0. Now z z = 1 2 ( x i ) y where is the Laplacian operator given by 1 2 ( x + i ) = 1 y 4. = 2 x y 2. (1) I rewrite (/z)(/ x)f = 0 as f = 0, or (u + iv) = 0. Thus u + i v = 0 from which I conclude that u = 0 and v = 0. Am I correct now? Q I think so. The Laplacian operator is real in the sense that, if u is real-valued, so is u. This is clear from (1). Thus the previous mistake is avoided. A We call a function u harmonic if it satisfies the Laplace equation u = 0. Thus we have proved that the real part u and the imaginary part v of an analytic function f = u + iv are harmonic. Actually, f itself is harmonic. So is its complex conjugate 1

2 f = u iv, because we can safely put f = f = 0 in view of the fact that is a real operator. Q OK. So there is a link between harmonic functions and analytic functions: the real part of an analytic function is a harmonic function. A Actually, this link is rather complete: locally, every (real-valued) harmonic function is the real part of an analytic function. Q Why do you use the word locally? A This is because the statement in general is not true globally. For example, take the principal branch of the logarithm function Log z. We know that it is an analytic function. So its real part is a harmonic function. What is the real part of Log z? Q Let me see... Put z in the polar form, say z = re iθ with π < θ < π. Then Log z = Log re iθ = log r + iθ. So the real part of Log z is log r. Since r = z, the real part of Log z is log z. A Did you notice something unusual here, concerning the domain of log z? Q The domain of log z consists of all z with z 0. But the domain of Log z is smaller: it is the slit plane C \ {x R : x 0}. A Why is there a discrepancy of domains between Log z and log z? Q I can explain. Again, write z = re iθ with r > 0 and π < θ < π. We have r = z and θ = Arg z. Then Log z = log r + iθ = log z + i Arg z. So, when z goes along a circle around 0 once, log z remains the same, but Arg z is changed by 2π. A Good. Let us start with the harmonic function u(z) = log z, which is defined for all z 0. If we consider u(z) = u(x + iy) as a function of x and y, we may write u(x, y) = log x 2 + y 2 = 1 2 log(x2 + y 2 ), which is defined for all (x, y) (0, 0). I claim that there is no analytic function f(z) which is defined for all z 0 such that u(z) is the real part of f(z). Q Let me try to find out why this is so. If, instead of z = 0, we consider a small domain excluding the negative real axis, then such f exists: just let f(z) = log z +i Arg z = Log z. But as we have mentioned above, Log z cannot be extended to an analytic function defined for all z 0. 2

3 A You ve got the right idea, but your argument does not prove what I have claimed. If you write this in an exam, I m afraid that you won t receive any mark. Q Why? A Because it did not answer the question. The correct proof should be like this. Assume the contrary that there is an analytic function f(z) defined for z 0 such that Re f(z) = log z. Write f(z) f(x + iy) = u(x, y) + iv(x, y) (where u(x, y) = log z = (1/2) log(x 2 + y 2 )). Write v 0 (x, y) = Arg z, which is defined on D Log = {z : z is not a negative real number, nor zero }. On D Log, according to the Cauchy-Riemann equations, we have Put w = v v 0... v x = u y = v 0 x and v y = u x = v 0 y. Q Stop. Let me finish this. Putting w = v v 0, we have w/x = 0 and w/y = 0. So w must be a constant, say C. Hence we have v(x, y) = v 0 (x, y)+c = Arg(x+iy)+C for x + iy in D log. Now we can finish the proof by following my previous argument. A Good. At this point let us introduce a definition: given a (real-valued) harmonic function u(x, y), by a harmonic conjugate of u we mean another harmonic function v(x, y) having the same domain such that f(z) = f(x + iy) = u(x, y) + iv(x, y) is analytic. We have seen that a function v(x, y) defined on D Log is a harmonic conjugate of u(x, y) = (1/2) log(x 2 + y 2 ) if and only if v is Arg (x + iy) plus a constant. In fact, our above argument shows that, on a connected domain, harmonic conjugates of a given harmonic function are determined up to constants. Q Is it hard to find harmonic conjugates? A It depends on the given harmonic function. u(x, y) = x 2 y 2 we need to do is to solve For example, it is easy to check that is a harmonic function. To find its harmonic conjugate v(x, y), all You know how to solve these equations. v x = u y = 2y and v y = u x = 2x. 3

4 Q Yes. Here is my answer: v(x, y) = 2xy + C. So f(z) = f(x + iy) u(x, y) + iv(x, y) = x 2 y 2 + i(2xy + C) = (x 2 2ixy y 2 ) + C 0 = (x + iy) 2 + C 0 = z 2 + C 0, where C 0 = ic. Now let us return to our original investigation: tell me why, locally, harmonic conjugates exist. A OK. Let s get back to a real-valued harmonic function, say h(x, y) h(x+iy) = h(z). Take any point z 0 in its domain and a disk D(z 0 ; r) = {z : z z 0 < r} in the domain. Since we only consider local issues, we restrict our attention to functions on D(z 0 ; r). Now z z h = 1 ( 2 x + i ) y 1 ( 2 x i ) h = 1 h = 0. y 4 Thus g (/z)h satisfies the Cauchy-Riemann equation and hence is analytic. Let f be a primitive of 2g, i.e. f is an analytic function defined on D(z 0 ; r) such that f = 2g. Write f = u + iv. Then f = f x = u x + i v x and 2g = 2 h z = h x ih y Thus 2g = f gives h x = u x and h y = v x = u y. This shows h = u + C, where C is a constant. Hence h is the real part of the analytic function u + C + iv f + C. Q Why does 2g has a primitive on D(z 0 ; r)? A An easy way to see this is to use power series expansions. Since 2g is analytic on D(z 0 ; r), it has a power series expansion 2g(z) = n=0 a n(z z 0 ) n, with a radius of convergence r. Now the series n=0 (n + 1) 1 a n (z z 0 ) n+1 has the same radius of convergence and hence it converges to f on D(z 0, r). Using term by term differentiation, it is easy to see that f = 2g. Of course we have to justify the procedure of term by term differentiation... Q Could you skip this justification? It is rather technical, but I can handle it. 4

5 A Are you sure? Could you briefly describe it? Q OK. Suppose that f n converges to f uniformly, where f n are analytic. Take an appropriate region D. Then f n(z) = (1/2πi) D (w z) 2 f n (w) dw for all z in D. Let n 0. The right hand side converges to (1/2πi) D (w z) 2 f(w) dw, which is equal to f (z). A Yes, this is enough to handle our situation. Q I have another question: why is f equal to f/x? A This is because the derivative f can be viewed as the limit of a quotient along the horizontal direction. There is another way to see this. We have already seen f = f/z them out, we have and we also have the Cauchy Riemann equation 0 = f/z. Spelling f = 1 2 ( ) f x if y Adding these identities, we have f = f/x. and 0 = 1 2 ( ) f x + if. y Q What property of harmonic functions do you consider to be the most important or most outstanding one? A Harmonic functions have many very nice properties. If I m allowed to pick one only, I would pick the mean value property, which says that the average of a harmonic function over a circle is equal to its value at the center of the circle. That is, if h is a harmonic function and if the closed disk D(z 0 ; r) {z : z z 0 r} is contained in its domain, then h(z 0 ) = 1 2π h(z 0 + re it ) dt. (2) 2π 0 or h(x 0, y 0 ) = 1 2π 2π h(x 0 0 +r cos t, y 0 +r sin t) dt. This roughly says that the average of a harmonic function over a circle is equal to its value at the center of the circle. Actually, the mean value property characterizes harmonic functions: if a continuous function h on an open set in the complex plane has the mean value property, then h is harmonic. Q Why do harmonic functions possess the mean-value property? A There are many ways to see this. One may use Green s formula. We follow a method making use of the fact that we have just proved. In the present situation, we may 5

6 assume that h is the real part of an analytic function, say f. The circle D(z 0 ; r) is described by the parametric equation z = z 0 + re it (0 t 2π). Applying Cauchy s formula, we get f(z 0 ) = 1 2πi = 1 2πi D(z 0 ;r) 2π 0 f(z) z z 0 dz f(z 0 + re it ) re it d(re it ) = 1 2π 2π 0 f(z 0 + re it ) dt in view of d(re it ) = re it dt. Taking the real parts on both sides, we obtain (2). Q Show me an application of the mean value property. A A consequence of the mean-value property is the maximum principle, which says that a harmonic function defined on a connected open set cannot attain is maximum unless it is a constant function. Q How do you deduce this from the mean value property? A Well, let us assume the contrary that h attains its maximum at some point z 0 in its domain U, which is a connected open set. Let D(z 0 ; r) be a closed disk centered at z 0 contained in U. Then (2) above holds. By assumption, h(z 0 + re it ) h(z 0 ). If the strict inequality h(z 0 + re it ) < h(z 0 ) actually holds for some t, then we have (1/2π) 2π 0 h(z 0 + re it ) dt < h(z 0 ), contradicting (2). So h(z 0 + re it ) = h(z 0 ) for all t. By varying r, we see that h(z) = h(z 0 ) for z in a neighborhood of z 0. By following the usual connectedness argument (which was used in part 3 to prove that if f is an analytic function on a connected open set having a vanishing power series expansion at a point, then f = 0), we can show that h(z) = h(z 0 ) for all z in U. Q I have heard a theorem called the maximum modulus principle for analytic functions. A This theorem says that The modulus f(z) of an analytic function f(z) on a connected open set never attains its maximum unless f Q How do you prove this? is a constant. A In many ways. But let us prove this by using what we ve just learned, namely, the maximum principle for harmonic functions. Assume the contrary that f(z) attains its maximum at z 0. We can choose some real number α such that f(z 0 ) = e αi f(z 0 ). 6

7 Let h(z) be the real part of e αi f(z). Then h is a harmonic function attaining its maximum at z 0 and hence h is a constant. Thus the real part of an analytic function e iα f is constant. Therefore f is constant as well. Q What is the maximum principle for harmonic functions good for? A This can be used for proving the uniqueness of solution to Dirichlet s problem: u = 0 on D and u = g, D where g is a given continuous function defined on D, the boundary of a bounded connected open set D. Indeed, suppose that both u 1 and u 2 are solutions to this problem. We can show u 1 = u 2 as follows. Note that u 2 u 1 vanishes on D. Since u 2 u 1 is a continuous function on the compact set D = D D, it attains its maximum in D. Suppose that this maximum is attainable in D. Then the maximum Principe tells us that u 2 u 1 is a constant. This constant is zero because u 2 u 1 vanishes at boundary points. Consequently u 1 = u 2. So suppose that the maximum of u 2 u 1 is attained at a boundary point. Then we have u 2 u 1 0. Replacing u 2 u 1 by u 1 u 2, in the same way we can show u 1 u 2 0. Hence u 1 = u 2. Q How do you solve Dirichlet s problem? A This question belongs to another subject: boundary value problems for partial differential equations. Let me just give a few words here: a solution can be written down explicitly, if we know the Green s function of the domain. In the 2-dimensional case, this Green s function can be found if we know a conformal map from this domain onto the unit disk. Practically we often solve it by numerical methods, such as finite difference method. We will solve Dirichlet s problem for the unit disk by writing down an explicit solution. We leave it to the next time. 7