Consider a degree- d {\displaystyle d} univariate polynomial p ( x ) := ∑ k = 0 d a k x k {\displaystyle p(x):=\textstyle \sum _{k=0}^{d}a_{k}x^{k}} . In order to be harmonic, this polynomial must satisfy
0 = ∂ 2 ∂ x 2 p ( x ) = ∑ k = 2 d k ( k − 1 ) a k x k − 2 {\displaystyle 0={\tfrac {\partial ^{2}}{\partial x^{2}}}p(x)=\sum _{k=2}^{d}k(k-1)a_{k}x^{k-2}} at all points x ∈ R {\displaystyle x\in \mathbb {R} } . In particular, when d = 2 {\displaystyle d=2} , we have a polynomial p ( x ) = a 0 + a 1 x + a 2 x 2 {\displaystyle p(x)=a_{0}+a_{1}x+a_{2}x^{2}} , which must satisfy the condition a 2 = 0 {\displaystyle a_{2}=0} . Hence, the only harmonic polynomials of one (real) variable are affine functions x ↦ a 0 + a 1 x {\displaystyle x\mapsto a_{0}+a_{1}x} .
In the multivariable case, one finds nontrivial spaces of harmonic polynomials. Consider for instance the bivariate quadratic polynomial p ( x , y ) := a 0 , 0 + a 1 , 0 x + a 0 , 1 y + a 1 , 1 x y + a 2 , 0 x 2 + a 0 , 2 y 2 , {\displaystyle p(x,y):=a_{0,0}+a_{1,0}x+a_{0,1}y+a_{1,1}xy+a_{2,0}x^{2}+a_{0,2}y^{2},} where a 0 , 0 , a 1 , 0 , a 0 , 1 , a 1 , 1 , a 2 , 0 , a 0 , 2 {\displaystyle a_{0,0},a_{1,0},a_{0,1},a_{1,1},a_{2,0},a_{0,2}} are real coefficients. The Laplacian of this polynomial is given by Δ p ( x , y ) = ∂ 2 ∂ x 2 p ( x , y ) + ∂ 2 ∂ y 2 p ( x , y ) = 2 ( a 2 , 0 + a 0 , 2 ) . {\displaystyle \Delta p(x,y)={\tfrac {\partial ^{2}}{\partial x^{2}}}p(x,y)+{\tfrac {\partial ^{2}}{\partial y^{2}}}p(x,y)=2(a_{2,0}+a_{0,2}).}
Hence, in order for p ( x , y ) {\displaystyle p(x,y)} to be harmonic, its coefficients need only satisfy the relationship a 2 , 0 = − a 0 , 2 {\displaystyle a_{2,0}=-a_{0,2}} . Equivalently, all (real) quadratic bivariate harmonic polynomials are linear combinations of the polynomials 1 , x , y , x y , x 2 − y 2 . {\displaystyle 1,\quad x,\quad y,\quad xy,\quad x^{2}-y^{2}.}
Note that, as in any vector space, there are other choices of basis for this same space of polynomials.
A basis for real bivariate harmonic polynomials up to degree 6 is given as follows: ϕ 0 ( x , y ) = 1 ϕ 1 , 1 ( x , y ) = x ϕ 1 , 2 ( x , y ) = y ϕ 2 , 1 ( x , y ) = x y ϕ 2 , 2 ( x , y ) = x 2 − y 2 ϕ 3 , 1 ( x , y ) = y 3 − 3 x 2 y ϕ 3 , 2 ( x , y ) = x 3 − 3 x y 2 ϕ 4 , 1 ( x , y ) = x 3 y − x y 3 ϕ 4 , 2 ( x , y ) = − x 4 + 6 x 2 y 2 − y 4 ϕ 5 , 1 ( x , y ) = 5 x 4 y − 10 x 2 y 3 + y 5 ϕ 5 , 2 ( x , y ) = x 5 − 10 x 3 y 2 + 5 x y 4 ϕ 6 , 1 ( x , y ) = 3 x 5 y − 10 x 3 y 3 + 3 x y 5 ϕ 6 , 2 ( x , y ) = − x 6 + 15 x 4 y 2 − 15 x 2 y 4 + y 6 {\displaystyle {\begin{aligned}\phi _{0}(x,y)&=1\\\phi _{1,1}(x,y)&=x&\phi _{1,2}(x,y)&=y\\\phi _{2,1}(x,y)&=xy&\phi _{2,2}(x,y)&=x^{2}-y^{2}\\\phi _{3,1}(x,y)&=y^{3}-3x^{2}y&\phi _{3,2}(x,y)&=x^{3}-3xy^{2}\\\phi _{4,1}(x,y)&=x^{3}y-xy^{3}&\phi _{4,2}(x,y)&=-x^{4}+6x^{2}y^{2}-y^{4}\\\phi _{5,1}(x,y)&=5x^{4}y-10x^{2}y^{3}+y^{5}&\phi _{5,2}(x,y)&=x^{5}-10x^{3}y^{2}+5xy^{4}\\\phi _{6,1}(x,y)&=3x^{5}y-10x^{3}y^{3}+3xy^{5}&\phi _{6,2}(x,y)&=-x^{6}+15x^{4}y^{2}-15x^{2}y^{4}+y^{6}\end{aligned}}}
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Cf. Corollary 1.8 of Axler, Sheldon; Ramey, Wade (1995), Harmonic Polynomials and Dirichlet-Type Problems ↩