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Harmonic Functions 6 HARMONIC FUNCTIONS This chapter is devoted to the Laplace equation. We introduce two of its important properties, the maximum principle and the rotational invariance. Then we solve the equation in series form in rectangles, circles, and related shapes. The case of a circle leads to the beautiful Poisson formula. 6.1 LAPLACE’S EQUATION If a diffusion or wave process is stationary (independent of time), then ut 0 ≡ and utt 0. Therefore, both the diffusion and the wave equations reduce to the Laplace≡ equation: uxx 0 in one dimension = u u uxx u yy 0 in two dimensions ∇·∇ = = + = u u uxx u yy uzz 0 in three dimensions ∇·∇ = = + + = A solution of the Laplace equation is called a harmonic function. In one dimension, we have simply uxx 0, so the only harmonic functions in one dimension are u(x) A Bx. But=this is so simple that it hardly gives us a clue to what happens =in higher+ dimensions. The inhomogeneous version of Laplace’s equation u f (1) = with f a given function, is called Poisson’s equation. Besides stationary diffusions and waves, some other instances of Laplace’s and Poisson’s equations include the following. 152 6.1 LAPLACE’S EQUATION 153 1. Electrostatics. From Maxwell’s equations, one has curl E 0 and div E 4πρ, where ρ is the charge density. The first equation implies= E grad= φ for a scalar function φ (called the electric potential). Therefore,=− φ div(grad φ) div E 4πρ , = =− =− which is Poisson’s equation (with f 4πρ ). =− 2. Steady fluid flow. Assume that the flow is irrotational (no eddies) so that curl v 0, where v v(x, y, z) is the velocity at the position (x, y, z), assumed= independent=of time. Assume that the fluid is incompressible (e.g., water) and that there are no sources or sinks. Then div v 0. Hence v grad φ for some φ (called the velocity potential) and φ= div v =−0, which is Laplace’s equation. = − = 3. Analytic functions of a complex variable. Write z x iy and = + f (z) u(z) iv(z) u(x iy) iv(x iy), = + = + + + where u and v are real-valued functions. An analytic function is one that is expressible as a power series in z. This means that the powers are not xmyn but zn (x iy)n. Thus = + ∞ n f (z) an z = n 0 = (an complex constants). That is, ∞ n u(x iy) iv(x iy) an(x iy) . + + + = n 0 + = Formal differentiation of this series shows that ∂u ∂v ∂u ∂v and ∂x = ∂y ∂y =−∂x (see Exercise 1). These are the Cauchy–Riemann equations. If we differ- entiate them, we find that uxx vyx vxy u yy, = = =− so that ∆u = 0. Similarly ∆v = 0, where is the two-dimensional laplacian. Thus the real and imaginary parts of an analytic function are harmonic. 4. Brownian motion. Imagine brownian motion in a container D. This means that particles inside D move completely randomly until they hit the bound- ary, when they stop. Divide the boundary arbitrarily into two pieces, C1 and C2 (see Figure 1). Let u(x, y, z) be the probability that a particle that begins at the point (x, y, z) stops at some point of C1. Then it can be deduced that ∆u = 0 in D u 1onC1 u 0onC2. = = Thus u is the solution of a Dirichlet problem. 154 CHAPTER 6 HARMONIC FUNCTIONS Figure 1 As we discussed in Section 1.4, the basic mathematical problem is to solve Laplace’s or Poisson’s equation in a given domain D with a condition on bdy D: u f in D ∂u = ∂u u h or h or au h on bdy D. = ∂n = ∂n + = In one dimension the only connected domain is an interval a x b .We will see that what is interesting about the two- and three-dimensional{ ≤ ≤cases} is the geometry. MAXIMUM PRINCIPLE We begin our analysis with the maximum principle, which is easier for Laplace’s equation than for the diffusion equation. By an open set we mean a set that includes none of its boundary points (see Section A.1). Maximum Principle. Let D be a connected bounded open set (in ei- ther two- or three-dimensional space). Let either u(x, y) or u(x, y, z)bea harmonic function in D that is continuous on D D (bdy D). Then the maximum and the minimum values of u are attained= on∪bdy D and nowhere inside (unless u constant). ≡ In other words, a harmonic function is its biggest somewhere on the boundary and its smallest somewhere else on the boundary. To understand the maximum principle, let us use the vector shorthand x (x, y) in two dimensions or x (x, y, z) in three dimensions. Also, the = = 1/2 1/2 radial coordinate is written as x (x2 y2) or x (x2 y2 z2) . The maximum principle asserts| that|=there+exist points| |=xM and+xm on+ bdy D such that u(xm) u(x) u(xM) (2) ≤ ≤ 6.1 LAPLACE’S EQUATION 155 Figure 2 for all x D (see Figure 2). Also, there are no points inside D with this property ∈(unless u constant). There could be several such points on the boundary. ≡ The idea of the maximum principle is as follows, in two dimen- sions, say. At a maximum point inside D, if there were one, we’d have uxx 0 and u yy 0. (This is the second derivative test of calculus.) So ≤ ≤ uxx u yy 0. At most maximum points, uxx < 0 and u yy < 0. So we’d get a contradiction+ ≤ to Laplace’s equation. However, since it is possible that uxx 0 u yy at a maximum point, we have to work a little harder to get a proof.= = Here we go. Let ǫ>0. Let v(x) u(x) ǫ x 2. Then, still in two dimen- sions, say, = + | | v u ǫ(x2 y2) 0 4ǫ>0inD. = + + = + But v vxx vyy 0 at an interior maximum point, by the second deriva- tive test =in calculus!+ Therefore,≤ v(x) has no interior maximum in D. Now v(x), being a continuous function, has to have a maximum some- where in the closure D D bdy D. Say that the maximum of v(x)is = ∪ attained at x0 bdy D. Then, for all x D, ∈ ∈ 2 2 u(x) v(x) v(x0) u(x0) ǫ x0 max u ǫl , ≤ ≤ = + | | ≤ bdy D + where l is the greatest distance from bdy D to the origin. Since this is true for any ǫ>0, we have u(x) max u for all x D. (3) ≤ bdy D ∈ Now this maximum is attained at some point xM bdy D.Sou(x) u(xM) for all x D, which is the desired conclusion. ∈ ≤ ∈ The existence of a minimum point xm is similarly demonstrated. (The absence of such points inside D will be proved by a different method in Section 6.3.) UNIQUENESS OF THE DIRICHLET PROBLEM To prove the uniqueness, suppose that u f in D v f in D = = u h on bdy D v h on bdy D. = = 156 CHAPTER 6 HARMONIC FUNCTIONS We want to show that u v in D. So we simply subtract equations and let w u v. Then w 0i≡nD and w 0 on bdy D. By the maximum principle = − = = 0 w(xm) w(x) w(xM) 0 for all x D. = ≤ ≤ = ∈ Therefore, both the maximum and minimum of w(x) are zero. This means that w 0 and u v. ≡ ≡ INVARIANCE IN TWO DIMENSIONS The Laplace equation is invariant under all rigid motions. A rigid motion in the plane consists of translations and rotations. A translation in the plane is a transformation x′ x ay′ y b. = + = + Invariance under translations means simply that uxx u yy ux′x′ u y′ y′ . A rotation in the plane through the angle α is giv+en by= + x x cos α y sin α ′ = + (4) y x sin α y cos α. ′ =− + By the chain rule we calculate ux ux cos α u y sin α = ′ − ′ u y ux sin α u y cos α = ′ + ′ uxx (ux cos α u y sin α)x cos α (ux cos α u y sin α)y sin α = ′ − ′ ′ − ′ − ′ ′ u yy (ux sin α u y cos α)x sin α (ux sin α u y cos α)y cos α. = ′ + ′ ′ + ′ + ′ ′ Adding, we have 2 2 uxx u yy (ux x u y y )(cos α sin α) ux y (0) + = ′ ′ + ′ ′ + + ′ ′ · ux x u y y . = ′ ′ + ′ ′ This proves the invariance of the Laplace operator. In engineering the laplacian is a model for isotropic physical situations, in which there is no preferred direction. The rotational invariance suggests that the two-dimensional laplacian ∂2 ∂2 2 = ∂x2 + ∂y2 should take a particularly simple form in polar coordinates. The transforma- tion x r cos θ y r sin θ = = 6.1 LAPLACE’S EQUATION 157 has the jacobian matrix ∂x ∂y ∂r ∂r cos θ sin θ j = ∂x ∂y = r sin θ r cos θ − ∂θ ∂θ with the inverse matrix ∂r ∂θ sin θ cos θ − 1 ∂x ∂x r j− . = ∂r ∂θ = cos θ θ sin ∂y ∂y r 1 (Beware, however, that ∂r/∂x (∂x/∂r)− .) So by the chain rule we have = ∂ ∂ sin θ ∂ cos θ , ∂x = ∂r − r ∂θ ∂ ∂ cos θ ∂ sin θ . ∂y = ∂r + r ∂θ These operators are squared to give ∂2 ∂ sin θ ∂ 2 cos θ ∂x2 = ∂r − r ∂θ ∂2 sin θ cos θ ∂2 cos2 θ 2 = ∂r 2 − r ∂r∂θ sin2 θ ∂2 2 sin θ cos θ ∂ sin2 θ ∂ + r 2 ∂θ2 + r 2 ∂θ + r ∂r ∂2 ∂ cos θ ∂ 2 sin θ ∂y2 = ∂r + r ∂θ ∂2 sin θ cos θ ∂2 sin2 θ 2 = ∂r 2 + r ∂r∂θ cos2 θ ∂2 2 sin θ cos θ ∂ cos2 θ ∂ .
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