5 Introduction to harmonic functions

1 Topic 5 NotesJeremy Orloff5 Introduction to harmonic IntroductionHarmonic functions appear regularly and play a fundamental role in math, physics andengineering. In this topic we ll learn the definition, some key properties and their tightconnection to complex analysis. The key connection to is that both the real andimaginary parts of analytic functions are harmonic . We will see that this is a simpleconsequence of the Cauchy-Riemann equations. In the next topic we will look at someapplications to harmonic functionsWe start by defining harmonic functions and looking at some of their functionu(x,y) is called harmonic if it is twice continuously differen-tiable and satisfies the following partial differential equation: 2u=uxx+uyy= 0.

2 (1)Equation 1 is called Laplace s equation. So a function is harmonic if it satisfies Laplace sequation. The operator 2is called the Laplacian and 2uis called the Laplacian Del notationHere s a quick reminder on the use of the notation . For a functionu(x,y) and a vectorfieldF(x,y) = (u,v), we have(i) =( x, y)(ii)gradu= u= (ux,uy)(iii)curlF= F= (vx uy)(iv)divF= F=ux+vy(v)div gradu= u= 2u=uxx+uyy(vi)curl gradu= u= 0(vii)div curlF= F= Introduction TO harmonic Analytic functions have harmonic piecesThe connection between analytic and harmonic functions is very strong. In many respectsit mirrors the connection between ezand sine and +iyand writef(z) =u(x,y) +iv(x,y).

3 Theorem (z) =u(x,y) +iv(x,y) is analytic on a regionAthen bothuandvareharmonic functions is a simple consequence of the Cauchy-Riemann equations. Sinceux=vywehaveuxx= ,uy= vximpliesuyy= haveuxx+uyy=vyx vxy= harmonic . We can handlevsimilarly. we know an analytic function is infinitely differentiable we knowuandvhavethe required two continuous partial derivatives. This also ensures that the mixed partialsagree, complete the tight connection between analytic and harmonic functions we show thatany harmonic function is the real part of an analytic (x,y) is harmonic on a simply connected regionA, thenuis the realpart of an analytic functionf(z) =u(x,y) +iv(x,y). is similar to our proof that an analytic function has an antiderivative.

4 First wecome up with a candidate forf(z) and then show it has the properties we need. Here arethe details broken down into steps Find a candidate, call itg(z), forf (z):If we had an analyticfwithf=u+iv, then Cauchy-Riemann says thatf =ux , let s defineg=ux is our candidate forf .2. Show thatg(z) is analytic:Writeg= +i , where =uxand = uy. Checking the Cauchy-Riemannequations we have[ x y x y]=[uxxuxy uyx uyy]Sinceuis harmonic we knowuxx= uyy, so x= y. It is clear that y= the Cauchy-Riemann equations, so it is Letfbe an antiderivative ofg:5 Introduction TO harmonic FUNCTIONS3 SinceAis simply connected our statement of Cauchy s theorem guarantees thatg(z)has an antiderivative inA.

5 We ll need to fuss a little to get the constant of integrationexactly right. So, pick a base pointz0inA. Define the antiderivative ofg(z) byf(z) = zz0g(z)dz+u(x0,y0).(Again, by Cauchy s theorem this integral can be along any path inAfromz0toz.)4. Show that the real part s writef=U+iV. So,f (z) =Ux iUy. By constructionf (z) =g(z) =ux means the first partials ofUanduare the same, soUandudiffer by at most aconstant. However, also by construction,f(z0) =u(x0,y0) =U(x0,y0) +iV(x0,y0),So,U(x0,y0) =u(x0,y0) (andV(x0,y0) = 0). Since they agree at one point we musthaveU=u, the real part offisuas we wanted to infinitely definition we only require a harmonic functionuto have continuous secondpartials.

6 Since the analyticfis infinitely differentiable, we have shown that so isu! harmonic the real and imaginary parts of an analytic function, then wesayuandvare harmonic (z) =u+ivis analytic then so isif(z) = v+iu. So, ifuandvare harmonicconjugates and so areuand A second proof thatuandvare harmonicThis fact is important enough that we will give a second proof using Cauchy s integralformula. One benefit of this proof is that it reminds us that Cauchy s integral formula cantransfer a general question on analytic functions to a question about the function 1/z. Westart with an easy to derive real and imaginary parts off(z) = 1/zare harmonic away from the forg(z) =f(z a) =1z aaway from the pointz= have1z=xx2+y2 iyx2+ Introduction TO harmonic FUNCTIONS4It is a simple matter to apply the Laplacian and see that you get 0.

7 We ll leave the algebrato you! The statement aboutg(z) follows in either exactly the same way, or by noting thatthe Laplacian is translation proof thatfanalytic impliesuandvare are proving that iff=u+ivis analytic thenuandvare harmonic . So, supposefis analytic at the means there is a disk of some radius, sayr, aroundz0wherefis analytic. Cauchy sformula saysf(z) =12 i Crf(w)w zdw,whereCris the circle|w z0|=randzis in the disk|z z0|< , since the real and imaginary parts of 1/(w z) are harmonic , the same must be trueof the integral, which is limit of linear combinations of such functions . Since the circle isfinite andfis continuous, interchanging the order of integration and differentiation is nota Maximum principle and mean value propertyThese are similar to the corresponding properties of analytic functions .

8 Indeed, we deducethem from those corresponding (Mean value property) Ifuis a harmonic function thenusatisfies the meanvalue property. That is, supposeuis harmonic on and inside a circle of radiusrcenteredatz0=x0+iy0thenu(x0,y0) =12 2 0u(z0+rei )d +ivbe an analytic function withuas its real part. The mean valueproperty forfsaysu(x0,y0) +iv(x0,y0) =f(z0) =12 2 0f(z0+rei )d =12 2 0u(z0+rei ) +iv(z0+rei )d Looking at the real parts of this equation proves the (Maximum principle) Supposeu(x,y) is harmonic on a open regionA.(i) Supposez0is inA. Ifuhas a relative maximum or minimum atz0thenuis constanton a disk centered atz0.(ii) IfAis bounded and connected anduis continuous on the boundary ofAthen theabsolute maximum and absolute minimum ofuoccur on the proof for maxima is identical to the one for the maximum modulus proof for minima comes by looking at the maxima of analytic functions we only talked about maxima because we had to use themodulus in order to have real values.

9 Since| f|=|f|we couldn t use the trick of turningminima into maxima by using a minus Introduction TO harmonic Orthogonality of curvesAn important property of harmonic conjugatesuandvis that their level curves are orthog-onal. We start by showing their gradients are +iyand suppose thatf(z) =u(x,y) +iv(x,y) is analytic. Thenthe dot product of their gradients is 0, u v= proof is an easy application of the Cauchy-Riemann equations. u v= (ux,uy) (vx,vy) =uxvx+uyvy=vyvx vxvy= 0In the last step we used the Cauchy-Riemann equations to substitutevyforuxand vxforuy. The lemma holds whether or not the gradients are 0. To guarantee that the level curvesare smooth the next theorem requires thatf (z)6= +iyand suppose thatf(z) =u(x,y) +iv(x,y)is analytic.

10 Iff (z)6= 0 then the level curve ofuthrough (x,y) is orthogonal to the levelcurvevthrough (x,y). technical requirement thatf (z)6= 0 is needed to be sure that the level curvesare smooth. We need smoothness so that it even makes sense to ask if the curves areorthogonal. We ll discuss this below. Assuming the curves are smooth the proof of thetheorem is trivial: We know from that the gradient uis orthogonal to the levelcurves ofuand the same is true for vand the level curves ofv. Since, by Lemma , thegradients are orthogonal this implies the curves are , we show thatf (z)6= 0 means the curves are smooth. First note thatf (z) =ux(x,y) iuy(x,y) =vy(x,y) +ivx(x,y).