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Home , Harmonic function

Lecture 1

Harmonic Functions

1.1 The Definition

Definition 1.1. Let Ω denote an in R3. A real valued u(x, y, z) on Ω with continuous second partials is said to be harmonic if and only if the Laplacian ∆u = 0 identically on Ω. Note that the Laplacian ∆u is defined by ∂2u ∂2u ∂2u ∆u = + + . ∂x2 ∂y2 ∂z2

We can make a similar definition for an open set Ω in R2.Inthatcase, u is harmonic if and only if

∂2u ∂2u ∆u = + =0 ∂x2 ∂y2 on Ω. Some basic examples of harmonic functions are

2 2 2 3 u = x + y 2z , Ω=R , − 1 3 u = , Ω=R (0, 0, 0), r − where r = x2 + y2 + z2. Moreover, by a theorem on complex variables, the real part of an on an open set Ω in 2 is always harmonic. p R Thus a function such as u = rn cos nθ is a harmonic function on R2 since u is the real part of zn.

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1.2 The Maximum Principle

The basic result about harmonic functions is called the maximum principle. What the maximum principle says is this: if u is a harmonic function on Ω, and B is a closed and bounded region contained in Ω, then the max (and min) of u on B is always assumed on the boundary of B. Recall that since u is necessarily continuous on Ω, an absolute max and min on B are assumed. The max and min can also be assumed inside B, but a harmonic function cannot have any local extrema inside B. In particular, all isolated critical points of a harmonic function correspond to saddle points. They are neither maximas nor minimas. Note that a saddle point, the second derivative test may or may not apply. We are not using it in any way. Theorem 1.1 (The weak Maximum Principle). Suppose u(x, y, z) is a harmonic function on Ω. Then an isolated critical point of u in Ω cannot give a relative max or min of u. In particular, if a harmonic function on Ω has only isolated critical points, then if E is a closed solid region E in Ω, the absolute max and min of u on E occur on the boundary of E.

Proof. Let p be an isolated critical point, and suppose p gives a local max for u.Thusu(q) u(p) for all q Ω sufficiently near p.Let>0. By the continuity of u, there≤ exists a closed∈ ball B about p such that u(q) u(p)  and u(q) =0forallq B such that q = p. Consider the level≥ surface− S = ∇q B6 u(q)=u(p)∈  .ThenS is a6 closed surface contained in B, and { u(∈q) =| 0 at any point−q }of S.ThusS is a smooth surface (the implicit function∇ theorem6 guarantees this), and the outer unit normal to S is given by u n = ∇ − u |∇ | at each point of S. Note the minus sign is neessary to obtain the outer normal since u points in the direction of greatest increase of u.Now ∇ u n dS < 0, ∇ · ZZS since u n < 0onS. But the divergence theorem says ∇ · u n dS = ∆udV ∇ · ZZS ZZZE where E is the region S bounds. Since u is harmonic, ∆u =0onE,so we have a contradiction. Thus p cannot give a relative max. Similarly, it cannot give a relative min. Hence p corresponds to a saddle point. 3

Finally, let E be a closed solid region in Ω. By a result from Calculus 3, the absolute max and min of u on E are assumed and occur either on the boundary of E or at a critical point of u in the interior of E. But no critical point of u is a local max or min, so the max and min of u in E occur on the boundary. What is essentially the same proof goes through in the two dimensional case. We will leave this to your imagination. By a theorem in , the critical points of an analytic function are always isolated, so the theorem applies to any harmonic function which is the real part of an analytic function. For example, u = rn cos nθ is the real part of the analytic function zn. Hence, if D is a closed disk in R2, such a function takes its max and min values on D at a point on the boundary of D.

1.3 The

The celebrated heat equation of gives an interest- ing application of the maximum principle. The heat equation says that if T (x, y, z, t) is the temperature function on the domain Ω, then there exists a constant c such that ∂T = c∆ T, ∂t x,y,z where ∂2T ∂2T ∂2T ∆ T = + + . x,y,z ∂2x ∂2y ∂2z The heat equation is derived from Gauss’s theorem. We will carry this out below. To see how the maximum principle applies to the heat equation, suppose that the temperature function T (x, y, z, t) has reached a steady state. That ∂T is, ∂t = 0. This occurs exactly when T is independent of time t.Whena steady state prevails, the heat equation implies that the usual Laplacian of T satisfies ∆T = 0. The main consequence of the maximum principle for a steady state temperature function T on Ω is that the hottest and coldest points on any closed bounded subset B of Ω have to occur an the boundary of B. In particular, there are no hot spots in the interior of B. This result may be of some practical interest. When you pick up a hot pan that has been sitting for awhile, avoid contact with the edge of the skillet, since that is where the hottest point is. The coldest point is likely at the end of the handle, as far away from the flat part as possible. 4

The heat equation is derived in most books on vector analysis, Stewart’s being an unfortunate exception. A brief derivation goes as follows. Heat flows with the vector field F = k T where k is a real constant. Therefore the rate of change of heat flowing∇ across a closed surface S in Ω (i.e. the flux across S)is

F n dS = div(F) dV. − · − ZZS ZZZE Here, E is the region S bounds. The amount of heat in E is proportional to

TdV ZZZE sotherateofchangeofheatinE is ∂T ∂T TdV= dV. ∂t ∂t ZZZE ZZZE By conservation of energy, we can therefore say ∂T dV = c ∆ TdV, ∂t x,y,z ZZZE ZZZE for an appropriate constant c. Since the region E is arbitrary, it follows that ∂T = c∆ T. ∂t x,y,z This completes the derivation of the heat equation.

1.4 The Average Value Property

The average value property says that if u is a harmonic function defined on an open set Ω (in R2 or R3), then the value of u at the center p of a disk (or ball) in Ω is the average value of u on the boundary of the disk (or ball). More precisely, the average value property says that if u is harmonic 2 on Ω R , p Ω, and Br is a closed disk in Ω centred at p,then ⊂ ∈ 1 u(p)= uds, 2πr ZSr where Sr is the sphere which bounds Br. In the 3-dimensional case, the result is that 1 u(p)= udS. 4πr2 ZZSr 5

Notice that in both formulas, the integral is the average value of u on the boundary of Br. The proof of the average value property is another application of the divergence theorem. The proof isn’t hard, but time doesn’t permit us to give all the details here. In the 3-dimensional case, one considers the function 1 I(r)= udS. 4πr2 ZZSr

We have to show I(r)=u(p) for all r. The derivative I0(r)satisfies

I0(r)=4π u n dS, ∇ · ZZSr which is shown by using spherical coordinates. Thus, by Gauss’s theorem and the fact that u is harmonic, I0(r) = 0 for all r,soI is constant. It then follows from an elementary argument that I(r)=u(p) for all r,whichgives the formula. The proof in the 2-dimensional case is similar. The average value property implies a stronger maximum principle. Theorem 1.2 (The strong maximum principle). Suppose u is a non- constant harmonic function on an open set Ω in R2 or R3 and B is a closed ball in Ω. Then the absolute max and min of u on B are assumed on the boundary of B.

Proof. The max and min are assumed, since u is continuous. Suppose p B is a point not on the boundary of B at which the max of a harmonic function∈ u is assumed. We will show u is constant on B.Nowu(p) is the average value of u on any sphere Sr of radius r centered at p and contained in B, so it follows that u is constant on every such Sr.Inthiscase,u is constant on the largest ball B0 centred at p and contained in β. By applying the same argument to points of B0, we eventually get that the max occurs at the center of B, in which case it follows from the same argument that u is constant on B.