HARMONIC FUNCTIONS, GREEN's FUNCTIONS and POTENTIALS. 1. the Poisson Kernel of the Disk. the Dirichlet Problem Consists Of

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HARMONIC FUNCTIONS, GREEN’S FUNCTIONS and POTENTIALS. 1. The Poisson kernel of the disk. The Dirichlet problem consists of finding a harmonic function (∆u = 0) in a bounded domain in Rn with given boundary values. Physically, one seeks to find the electrostatic potential in a conductor (no charges in the interior), with its boundary held at a given potential. 2 Consider this problem for the disk Da = {x ∈ R ; |x| ≤ a}, with given boundary value h. To begin, using the Laplacian in polar coordinates (r, θ) in the plane: 1 1 u∆u(r, θ) = u + u + u rr r r r2 θθ we find that, for each n ≥ 1, the functions rn cos nθ, rn sin nθ are harmonic. Thus we may seek u(r, θ) of the form:

∞ X n u(r, θ) = A0 + r (An cos nθ + Bn sin nθ), n=1

1 and substituting u(a, θ) = h(θ) we find that An,Bn equal an times the periodic Fourier coefficients of h(θ), θ ∈ [0, 2π]. Using the definition if Fourier coefficients of h, we find: ∞ 1 Z 2π X rn u(r, θ) = [1 + 2 cos n(θ − φ)]h(φ)dφ. 2π an 0 n=1 Surprisingly, the infinite series can be summed explicitly: ∞ X rn a2 − r2 1 + 2 cos n(θ − φ) = . an r2 − 2ar cos(θ − φ) + a2 n=1 Using the Law of Cosines, we find the denominator equals the distance squared between the points x = (r, θ) (in the interior of the disk) and y = (a, φ) (on the boundary). Thus the solution may also be written (more geometrically) as an integral over the boundary, with respect to arc length dsy = adφ: 1 Z a2 − |x|2 u(x) = 2 h(y)dsy. 2πa ∂Da |x − y| This leads to the definition of the Poisson kernel for the disk: 1 a2 − |x|2 P (x, y) = , (|x| < a, |y| = a). 2πa |x − y|2

1 The Poisson kernel encodes the geometry of the domain (and the Laplacian), giving the solution of the Dirichlet problem as an integral (weighted average) of the boundary values, with ‘weight function’ given by P (x, y): Z u(x) = P (x, y)h(y)dsy. ∂Da

Exercise 1. (Poisson kernel for the exterior of the disk.) Consider the exterior Dirichlet problem:

2 ∆u = 0 in R \ Da = {r > a}, u(x) = h(x) for |x| = a, where h is a given continuous function on the boundary of the disk. This can be solved by a similar method: (i). Show that r−n cos nθ, r−n sin nθ (n ≥ 1) are harmonic functions (for r 6= 0). (ii). Show that the solution of the exterior problem is given in polar coordinates by:

∞ 1 Z 2π X an u(r, θ) = [1 + 2 cos n(θ − φ)]h(φ)dφ. 2π rn 0 n=1 (iii) (see below.) To sum the series, all we have to do is exchange r and a in the calculation done earlier for the interior. We obtain the expression (in geometric form):

1 Z |x|2 − a2 u(x) = 2 h(y)dsy. 2πa ∂Da |x − y| Thus the Poisson kernel for the exterior of the disk is: 1 |x|2 − a2 P (x, y) = , (|x| > a, |y| = a). 2πa |x − y|2

Remark. Note that if h ≡ 1, this procedure would yield (in both cases) the solution u ≡ 1. Thus we have: 1 Z a2 − |x|2 1 Z |x|2 − a2 2 dsy = 1 for |x| < a; 2 dsy = 1 for |x| > a. 2πa ∂Da |x − y| 2πa ∂Da |x − y| That is: in both cases the Poisson kernel deﬁnes a probabilty distribution on the boundary of the region, depending on the point x (an interior or exterior

2 point.) There is a probabilistic interpretation: for any arc A ⊂ ∂Da of the circle, the integral of P (x, y)dsy over y ∈ A is the probability that Brownian motion started at the interior point x will ﬁrst hit the boundary at some point of A. In particular we ﬁnd that this solution of the exterior problem is bounded:

|u(x)| ≤ ||h|| for |x| > a, where ||h|| = sup |h(y)|. y∈∂Da

The solution is not unique, however: (iii). Show that if u(x) is a solution of the exterior Dirichlet problem (with |x| boundary value h on ∂Dr), then v(x) = log a + u(x) is also a harmonic function in the exterior of the disk, with the same boundary values. (Note that v is unbounded.)

Applications. A. Mean-value property. If u is harmonic in a disk, and continuous in the closed disk, the value at the center equals the average of the boundary values. This follows directly from the fact that the value of the Poisson kernel at the origin equals (2πa)−1. If u is harmonic in the exterior of a disk, continuous up to the boundary and bounded, then it is given by integration of the boundary values with respect to the Poisson kernel, and then we have:

lim u(x) = average of u over ∂Da. |x|→∞

This follows directly from the fact that, for the exterior Poisson kernel P (x, y): 1 lim P (x, y) = , |x|→∞ 2πa uniformly over y ∈ ∂Da, as can be seen directly from the expression above for P (x, y). Remark. The mean-value property can be used to deﬁne ‘harmonic function’ in a way that makes sense for continuous functions: u is harmonic if its average value over any circle equals the value at the center. B. Strong maximum principle. Let u be harmonic in a bounded open region D in the plane, and continuous on the closed region D¯. Let M be the maximum value of u over D¯. Then u cannot take the value M at any interior point x ∈ D, unless u is constant in D.

3 This can be proved starting from the mean-value property (seen in class.) Note this is diﬀerent from the weak maximum principle, which concludes only that the maximum value M must be attained at a boundary point (leaving open the possibility that it is also attained at an interior point.) C. Diﬀerentiability. Let u be a continuous function in the closed disk D¯a, with the property that its value at any point is given by integration over the boundary, with respect to the Poisson kernel:

Z 1 a2 − |x|2 u(x) = P (x, y)u(y)dsy,P (x, y) = 2 , (|x| < a, |y| = a). ∂Da 2πa |x − y|

Then u is in fact smooth and harmonic (∆u = 0) in the open disk Da. This folows from the fact that P (x, y) is smooth in x ∈ Da, for each ﬁxed y ∈ ∂Da. Changing coordinates to z = x − y, we have: 1 |a|2 − |z + y|2 1 z · y P (z + y, y) = = − (1 + 2 ), z 6= 0. 2πa |z|2 2πa |z|2

2 z·y Exercise 2. Show that, for each fixed y ∈ R , the function w(z) = |z|2 is 2 harmonic in {z ∈ R , z 6= 0}. 2. Green’s indentities and applications. 2 n Let u, v be C functions in a bounded open set D ⊂ R . Green’s iden- tities are obtained by applying the divergence theorem to the vector field v∇u, noting that: div(v∇u) = ∇u · ∇v + v∆u. Denote by n the outward unit normal at points of the boundary ∂D, and let ∂nu = ∇u · n be the normal derivative. The divergence theorem directly implies Green’s first indentity: Z Z I v∆udV = − ∇u · ∇vdV + v∂nudA. D D ∂D Exchanging u and v and taking the difference, we obtain Green’s second identity: Z I (v∆u − u∆v)dV = (v∂nu − u∂nv)dA. D ∂D (Here dV and dA denote volume integration in D, resp. integration with respect to area on ∂D.) An immediate consequence of the second identity is that the Laplacian is a symmetric differential operator on functions of

4 compact support (i.e., vanishing outside of some large ball) with respect to the L2 inner product: applying the identity to a large enough ball, the boundary terms vanishes and we have: h∆u, vi = hu, ∆vi.

n Example 1. Consider the Poisson problem in a bounded domain D ⊂ R , with non-homogeneous Neumann boundary conditions:

∆u = f in D; ∂nu = h on ∂D. This problem has physical meaning: we seek the electrostatic potential in a region with given charge density (represented by f), and given normal component of the electric field (En = −∂nu) on the boundary. And yet both existence and uniqueness fail! Given any solution, adding an arbitrary constant to it gives another solution (physically, the potential function is only defined up to a constant.) And Green’s first identity (taking v ≡ 1) implies we must also have (if a solution exists): Z I fdV = hdA. D ∂D Physically this says that the total charge in D must be proportional to the average value of the normal component of the electric field on the boundary.

Mean-Value property. One may use Green’s ﬁrst identity to prove the mean value property for harmonic function in any dimension. Let u be n harmonic in a disk Ba = {x ∈ R ; |x| < a}, continuous in the closed ball B¯a. Let 0 < R ≤ a, and consider the ball BR = {r < R} ⊂ Ba. Applying the identity to this region with v ≡ 1, we ﬁnd: I 0 = ∂nudA ∂BR Consider what this says in terms of polar coordinates: dA = Rn−1dω (where n−1 dω is the element of area on the unit sphere S = S ) and ∂nu is the radial derivative u : r I ur(R, ω)dω = 0, ∀0 < R ≤ a. S This implies: d I d 1 I 0 = u(r, ω)dω = ωn−1 ( udA), dr S dr A(Sr) Sr

5 where the last integral is the average value of u over the sphere Sr. Thus this average is constant in r, and has limit u(0) as r → 0. We conclude the mean value property: 1 I u(0) = udA. A(Sa) Sa In words: the value of a harmonic function at a point equals its average value on any sphere centered at that point. As a consequence, the strong maximum principle holds for harmonic functions in all dimensions. Remark: Note that the same argument shows that if ∆u ≤ 0, the average value of u over the sphere of radius r is decreasing in r (more precisely, non-increasing), so the value at the origin is greater than or equal to the average on Sr, for any r > 0. Twice-diﬀerentiable functions u satisfying ∆u ≤ 0 are called superharmonic (since they are “superaveraging”; the value at a point is greater than the average value on spheres centered at that point). Functions satisfying ∆u ≥ 0 are called subharmonic, and of course subharmonic functions are “subaveraging” in the same sense.

Uniqueness for the Poisson problem. Consider the non-homogeneous n problem with Dirichlet boundary condition in a bounded domain D ⊂ R : ∆u = f in D, u = h on ∂D.

To see that there is at most one solution, consider the problem solved by the diﬀerence of two solutions:

∆w = 0 in D, w = 0 on ∂D.

Then Green’s first identity (setting both functions equal to w) implies: Z Z I Z 2 2 0 = w∆wdV = − |∇w| dV + w∂nwdA = − |∇w| dV, D D ∂D D hence ∇w ≡ 0 in D, so w is constant in D, necessarily 0. Exercise 3. Consider the Poisson equation with Neumann boundary conditions (Example 1 above). Use Green’s first identity (as above) to show that any two solutions differ by a constant. Dirichlet’s principle. (Minimizing property of harmonic functions.) It is customary to define the energy of a smooth function u in a bounded region n D ⊂ R as: 1 Z E[u] = |∇u|dV. 2 D

6 (One reason for this terminology is that, in electrostatics, the energy of the electric ﬁeld is proportional to |E|2 = |∇u|2, where u is the potential.) It turns out that harmonic functions have the least possible energy for given boundary values, a property known as Dirichlet’s principle. A little more precisely, if u solves the Dirichlet problem in D with boundary values h, any other function v in D with the same boundary values has energy greater than that of u. To see this, consider w = v − u, which has zero boundary values. Since v = u + w, Z Z 1 2 E[v] = |∇u + ∇w| dV = E[u] + ∇u · ∇wdV + E[w] 2 D D Z I = E[u] + E[w] − (∆u)wdV + w∂nudA, D ∂D by Green’s ﬁrst identity. The last two terms are zero, since u is harmonic and w has zero boundary values. Since E[w] ≥ 0, this establishes Dirichlet’s principle (and also shows u is the unique energy minimizer for the given boundary values.) Exercise 4. [Strauss] There is also a “Dirichlet principle” for the Neu- n mann problem. Let D ⊂ R , h a given function on the boundary ∂D with H zero average value ( ∂D hdA = 0). Consider the “energy-type functional”: Z I 1 2 F[u] = |∇u| dV − hudA. 2 D ∂D

Let u be a harmonic function in D with ∂nu = h on ∂D. (Note that u is unique only up to a constant, but adding a constant to u doesn’t change the value of F[u], since h has zero average.) Show that, for any function v on D (without restriction on boundary values or normal derivatives) we have:

F[v] ≥ F[u].

(Hint: Consider w = v − u, and compute F[v] = F[w + u] using Green’s ﬁrst identity.)

n 3. The Poisson problem and Green’s function in R . Recall the Laplacian in polar coordinates (r, ω) is given by: n − 1 1 ∆u = u + u + ∆ u, rr r r r2 S

7 where the differential operator ∆S acts only on the angular coordinates ω. n−1 n Denote by ωn−1 the area of the unit sphere S in R (so ω1 = 2π, ω2 = 4π.) n Define the function G0(x) in R (except for the origin): 1 1 G0(x) = n−2 (n ≥ 3),G0(x) = − log r(n = 2), r = |x|. (n − 2)ωn−1r 2π n Exercise 5. Show that ∆G0 = 0 on R \{0}. n Now let D ⊂ R be a bounded domain with the origin in its interior, and (for R > 0 small) consider the domain DR = D \{r < R}. Applying Green’s second identity to G0 and an arbitrary function u : D → R we find, since G0 is harmonic in DR: I Z (u∂nG0 − G0∂nu)dA = − G0∆udV. ∂DR DR

Since ∂DR is the union of ∂D and the sphere (or circle) SR = ∂BR, we may also write this as: I Z Z (u∂rG0 − G0ur)dA − (u∂nG0 − G0∂nu)dA = G0∆udV. SR ∂D DR Our next goal is to take limits as R → 0. Consider ﬁrst the integral over n−1 SR, and recall dA = R dω (in polar coordinates) on SR. Also, on SR we have (if n ≥ 3): 1 1 G0 = n−2 , ∂rG0 = − n−1 . (n − 2)ωn−1R ωn−1R

Thus the integral over SR equals: 1 I R I − u(R, ω)dω + ur(R, ω)dω. (n − 2)ωn−1 S (n − 2)ωn−1 S Since |∇u| is bounded in D, the limit of this term as R → 0 is −(n − 2)u(0). Now consider the integral over DR, which is the diﬀerence of the integrals of G0∆u over D and over the ball BR. This last one, written in polar coordinates, equals: Z R Z 1 1 n−1 n−2 ∆u(r, ω)r dωdr. ωn−1 0 S r In absolute value, this is bounded above by R R R R |∆u|(r, ω)dω, which ωn−1 0 S clearly has limit zero as R → 0. We conclude: Z I u(0) = − G0∆udV − (u∂nG0 − G0∂nu)dA. D ∂D

8 Exercise 6. Show that the same formula is valid for n = 2 (using the above 2 definition for G0 in R .) n Definition. Green’s function in R with pole x0 is defined as: 1 1 Gx0 (x) = n−2 (n ≥ 3),Gx0 (x) = − log |x−x0|(n = 2). (n − 2)ωn−1|x − x0| 2π

n 2 From the above, for any bounded domain D in R , any C function u in D and any point x0 ∈ D, we have the representation formula: Z I

u(x0) = − Gx0 (x)∆u(x)dVx − (u∂nGx0 − Gx0 ∂nu)dA. D ∂D

Remark: Physically (at least for n = 3) Gx0 is the potential of a positive point charge located at x0. (Recall the electric ﬁeld equals minus the gradient of the potential, in electrostatics.) n Application: Poisson equation in R . Let f be a function of compact n support in R (meaning f ≡ 0 outside of a ball BR of suﬃciently large radius.)Suppose u is the solution of the whole-space Poisson equation:

∆u = f.

Assume the solution u to this problem has the property that the integrals over the sphere SR of u∂rGx and of urGx tend to 0 as R → ∞. (This is plausible on physical grounds: if the charge distribution is concentrated in a ﬁnite region of space, we expect both the potential and the electric ﬁeld to decay at large distances from the charge distribution.) Applying the representation formula to D = BR, and letting R → ∞, this assumption implies that the boundary term vanishes in the limit, and we have, for any n x ∈ R : Z Z u(x) = − Gx(y)f(y)dVy = − Gx(y)∆u(y)dVy. Rn Rn This gives a formula for the solution of the whole-space Poisson equation (in terms of Green’s function), although in a technical mathematical sense we haven’t actually proved that this formula does indeed solve the problem, or that the solution does have the decay properties we assumed to derive the formula.

9 Remark.(Physics) Using the fact that, formally, the Laplacian is a symmetric (self-adjoint) operator, this representation formula can (informally) be written in the “equivalent” form: Z u(x) = − u(y)∆Gx(y)dVy. Rn It is in this sense that “the Laplacian of Green’s function with pole at x is minus the delta function at x” (if n = 3):

∆Gx(y) = −δx(y)(n = 3). Maxwell’s equations div(E~ ) = κρ (where κ > 0 is a constant depending on the units, and ρ is the charge density) and curl(E~ ) = 0 (so E~ = −∇u) lead to −∆u = κρ, so for Green’s function with pole at x we have ρ = κδx, a 3 positive point charge at x ∈ R . (Though, as we saw earlier, ∆Gx = 0 on n R \{x}!)

Note that in some sense Gx0 is ‘superharmonic’. First, its Laplacian is minus the delta function, so it is “negative”. And it is certainly ‘superaveraging’, since it is inﬁnite at the pole x0. Quick remark on delta functions. In the Physics literature on often sees the expression: Z f(x) = δ(y − x)f(y)dVy, Rn n where δ(x − y) is supposed to be a “function vanishing on R \{x}” whose integral has this property for any f (in particular for f ≡ 1). Well, there simply is no function with this property. What we in fact have is a linear functional (a linear map from the space of continuous functions of compact support to the reals), denoted by δx: n δx : Cc(R ) → R, δx[f] = f(x).

One reason to use the integration symbol (not a very good one) is to consider “appproximations to the delta function”, for instance the heat kernel h(x, t). The equality: Z n lim h(y − x, t)f(y)dVy = f(x), f ∈ Cc( ), t→0 R + Rn can be interpreted as saying:

lim h(· − x, t)dV = δx, t→0+

10 in the sense both sides ‘have the same eﬀect’ on continuous functions f with compact support.

n 4. Green’s function for a domain D ⊂ R . n Deﬁnition. Let D ⊂ R be a bounded open set with smooth boundary, let x0 ∈ D. Green’s function for D with pole at x0 is the function

D Gx0 : D \{x0} → R with the properties: D 1. Gx0 is harmonic in D \{x0}; D 2. Gx0 (y) = 0 for y ∈ ∂D; D n 3. Gx0 (y) = Gx0 (y) + h(y), where Gx0 is Green’s function for R with pole at x0 and h is a harmonic function in D, continuous in D¯. The existence of such a function is easy to establish, provided we can show Dirichlet’s problem (ﬁnding a harmonic function with given boundary values) can be solved for D. All we have to do is let h be the harmonic D function in D with boundary values −Gx0 (y) on ∂D, then set Gx0 (y) =

Gx0 (y) + h(y). In the following we’ll assume the Dirichlet problem can always be solved and ﬁnd a formula for the solution (which can then be proven to actually yield a solution.) Let u : D → R be a smooth function, continuous on D¯. Adding the representation formula: Z I

u(x0) = − Gx0 (x)∆u(x)dVx − (u∂nGx0 − Gx0 ∂nu)dA. D ∂D to Green’s second identity applied to the arbitrary function u and the harmonic function h: Z I 0 = − h(x)∆u(x)dVx − (h∂nu − h∂nu)dA, D ∂D

D we ﬁnd (taking into account the fact that Gx0 = Gx0 + h = 0 on the boundary: Z I D D u(x0) = − Gx0 (x)∆u(x)dVx − u(∂nGx0 )dA, D ∂D the representation formula for the Domain D: it ‘represents’ a function u in D in terms of its Laplacian and its boundary values.

11 Application 1. The representation formula gives an expression for the solution of the Dirichlet problem: if ∆u = 0 in D and u = h on ∂D, we have: I D u(x) = P (x, y)h(y)dAy, where P (x, y) = −∂nGx (y) ∂D is the Poisson kernel for D. Application 2. We also obtain a formula for the solution of the Poisson equation: ∆u = f in D, u = h on ∂D: Z I D u(x) = − Gx (y)f(y)dVy + P (x, y)h(y)dAy, D ∂D where P (x, y) is the Poisson kernel (deﬁned in Application 1.)

We now turn to two important examples of domains with explicitly com- putable Green’s functions: half-spaces and balls. Green’s function for a half-space. (“Method of image charges.”) Let D ⊂ n n R be the upper-half space {x ∈ R ; xn > 0}, bounded by the hyperplane H = {xn = 0}. Green’s function for D with pole at x ∈ D is (for n ≥ 3):

D 1 1 1 Gx (y) = [ n−2 − n−2 ], y ∈ D \{x}, (n − 2)ωn−1 |y − x| |y − x¯| wherex ¯ = x − 2xnen is the reﬂection of x on H. The Poisson kernel can then be computed (note the outward unit normal to D is −en). In the following computation, we use the standard Calculus fact: ∂ y − x ∂ y − x¯ y + x |y − x| = n n , |y − x¯| = n n = n n , y ∈ H, ∂yn |y − x| ∂yn |y − x¯| |y − x| since |y − x| = |y − x¯| for y ∈ H.

D D ∂Gx 2 xn P (x, y) = −∂nGx (y) = (y) = n , x ∈ D, y ∈ H. ∂yn ωn−1 |x − y| So we have for the solution of Dirichlet’s problem on the upper-half space, with boundary values h on H: I 2xn h(y) u(x) = n dAy. ωn−1 H |x − y|

12 Remark. Note that, although we have a formula for the solution in terms of the boundary data h(x), the solution is not unique. (This is typical for Dirichlet or Neumann problems in unbounded regions.) To see this, just note that w(x) = xn is harmonic in the upper half-space, with zero boundary values; hence can be added to any solution to produce a new solution. On the other hand, it is true that bounded solutions of the Dirichlet problem are unique. Exercise. Use the same method to compute Green’s function and the Poisson kernel for the upper half-plane (n = 2).

n Green’s function for the ball. Let D = {x ∈ R ; |x| < a} be the ball of n n radius a in R . The map from R \{0} to itself given by: a2 x 7→ x¯ = x, (x 6= 0) |x|2 leaves the points on the sphere Sa = ∂D ﬁxed and is ‘idempotent’ (meaning (¯x¯) = x; verify!) So it is a kind of ‘reﬂection’ across the sphere (it sends D \{0} to the exterior of the sphere.) Green’s function for D is given by (for x ∈ D):

n−2 D 1 1 a 1 Gx (y) = [ n−2 − n−2 n−2 ], x 6= 0, y ∈ D \{x}; (n − 2)ωn−1 |y − x| |x| |y − x¯|

D 1 1 1 G0 (y) = [ n−2 − n−2 ], y ∈ D \{0}. (n − 2)ωn−1 |y| a D −1 (To verify that Gx (y) = 0 if |y| = a, we check that |y − x¯| = a|x| |y − x| if |y| = a.) D To compute the Poisson kernel, we ﬁnd the gradient of Gx (using the y−x ‘Calculus fact’: ∇y|y − x| = |y−x| ):

2 D 1 |x| ∇yGx (y) = − n (1 − 2 )y, x 6= 0, y 6= x; ωn−1|y − x| a

D 1 y ∇yG0 (y) = − n , y 6= 0. ωn−1 |y| This yields, for the Poisson kernel (since n = y/a on ∂D):

D y D 1 2 2 P (x, y) = −∂nGx (y) = − ·∇yGx (y) = n (a −|x| ), x ∈ D, y ∈ ∂D. a ωn−1a|x − y|

13 Exercise. Verify this, including the fact this expression also holds for x = 0. For the solution of Dirichlet’s problem with boundary values h, we ﬁnd, if n ≥ 3: a2 − |x|2 I h(y) u(x) = n dAy. ωn−1a Sa |y − x|

Exercise: Use the same reflection to find Green’s function and the Pois- son kernel for the disk of radius a, Da (confirming the result obtained via Fourier series in section 1!)

Exercise. For any n ≥ 2, ﬁnd a harmonic function on the exterior of the ball Ba, vanishing on the boundary of the ball. Use this function to show that solutions of the exterior Dirichlet problem for the ball are not unique.