Let Vn be the space of homogeneous harmonic polynomials of degree n

1. Space of Harmonic Polynomials

Let R[x, y] be the space of polynomials in x, y over R. The differential operator

∆ = ∂²

∂x² + ∂²

∂y²

defines a linear operator on R[x, y]. Elements of V = ker ∆ are called harmonic polynomials, i.e. P ∈ V if and only if ∆P = 0.

A polynomial P (x, y) is called homogenuous of degree n if P (λx, λy) = λⁿP (x, y)

for all λ > 0. Let Vn be the space of homogeneous harmonic polynomials of degree n. Then it forms a vector subspace of V. Let P be a homogeneous polynomial of degree n. Then

P (r cos θ, r sin θ) = rⁿP (cos θ, sin θ).

Denote P (cos θ, sin θ) = Φ(θ). Using polar coordinate for

∆ = 1 r

∂

∂r

 r ∂

∂r

 + 1

r²

∂²

∂θ², we see that ∆P = 0 if and only if

Φ⁰⁰(θ) + n²Φ(θ) = 0.

By solving this O.D.E, we obtain Φ(θ) = a cos nθ + b sin nθ. Hence

P (x, y) = rⁿ(a cos nθ + b sin nθ) = aC_n(x, y) + bS_n(x, y).

Here Cn(x, y) = Re(x + iy)ⁿand Sn(x, y) = Im(x + iy)ⁿ. This shows that Vnis spanned by {C_n(x, y), S_n(x, y)}. Moreover, it is easy to see {C_n(x, y), S_n(x, y)} is linearly independent (you can use Polar coordinate again. Hence we simply use the fact that {cos nθ, sin nθ} is linearly independent.) In other words, {C_n(x, y), S_n(x, y)} forms a basis for V_n.

Notice that every polynomial can be written as a sum of homogeneous polynomials. In fact, given a polynomial f (x, y) =P

i,jaijxⁱy^j of degree N, we set fn(x, y) =P

i+j=naijxⁱy^j, then f (x, y) = f0(x, y)+f1(x, y)+· · ·+fN(x, y). Moreover fn(x, y) is homogeneous of degree n. If ∆f = 0, then

∆f0+ ∆f1+ · · · + ∆fN = 0.

Since constants and degree one polynomials are already harmonic, we see that

∆f2+ ∆f3+ · · · + ∆fN = 0.

Since ∆fi are homogeneous of different degree, {∆fi : 2 ≤ i ≤ N } is linearly independent.

Hence ∆f_i(x, y) = 0 for 0 ≤ i ≤ N. In other words, for any f ∈ V, f ∈ L∞

n=0V_n. It is obvious that L∞

n=0Vn is a subspace of V. Hence we obtain V =

∞

M

n=0

V_n.

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