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Poisson kernel

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Mathematical concept
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In mathematics, and specifically inpotential theory, thePoisson kernel is anintegral kernel, used for solving the two-dimensionalLaplace equation, givenDirichlet boundary conditions on theunit disk. The kernel can be understood as thederivative of theGreen's function for the Laplace equation. It is named forSiméon Poisson.

Poisson kernels commonly find applications incontrol theory and two-dimensional problems inelectrostatics.In practice, the definition of Poisson kernels are often extended ton-dimensional problems.

Two-dimensional Poisson kernels

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On the unit disc

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In thecomplex plane, the Poisson kernel for the unit disc[1] is given byPr(θ)=n=r|n|einθ=1r212rcosθ+r2=Re(1+reiθ1reiθ),   0r<1.{\displaystyle P_{r}(\theta )=\sum _{n=-\infty }^{\infty }r^{|n|}e^{in\theta }={\frac {1-r^{2}}{1-2r\cos \theta +r^{2}}}=\operatorname {Re} \left({\frac {1+re^{i\theta }}{1-re^{i\theta }}}\right),\ \ \ 0\leq r<1.}

This can be thought of in two ways: either as a function ofr andθ, or as a family of functions ofθ indexed byr.

IfD={z:|z|<1}{\displaystyle D=\{z:|z|<1\}} is the openunit disc inC,T is the boundary of the disc, andf a function onT that lies inL1(T), then the functionu given byu(reiθ)=12πππPr(θt)f(eit)dt,0r<1{\displaystyle u(re^{i\theta })={\frac {1}{2\pi }}\int _{-\pi }^{\pi }P_{r}(\theta -t)f(e^{it})\,\mathrm {d} t,\quad 0\leq r<1}isharmonic inD and has a radial limit that agrees withfalmost everywhere on the boundaryT of the disc.

That the boundary value ofu isf can be argued using the fact that asr → 1, the functionsPr(θ) form anapproximate unit in theconvolution algebraL1(T). As linear operators, they tend to theDirac delta function pointwise onLp(T). By themaximum principle,u is the only such harmonic function onD.

Convolutions with this approximate unit gives an example of asummability kernel for theFourier series of a function inL1(T) (Katznelson 1976). LetfL1(T) have Fourier series {fk}. After theFourier transform, convolution withPr(θ) becomes multiplication by the sequence {r|k|} ∈1(Z).[further explanation needed] Taking the inverse Fourier transform of the resulting product {r|k|fk} gives theAbel meansArf off:Arf(e2πix)=kZfkr|k|e2πikx.{\displaystyle A_{r}f(e^{2\pi ix})=\sum _{k\in \mathbb {Z} }f_{k}r^{|k|}e^{2\pi ikx}.}

Rearranging thisabsolutely convergent series shows thatf is the boundary value ofg +h, whereg (resp.h) is aholomorphic (resp.antiholomorphic) function onD.

When one also asks for the harmonic extension to be holomorphic, then the solutions are elements of aHardy space. This is true when the negative Fourier coefficients off all vanish. In particular, the Poisson kernel is commonly used to demonstrate the equivalence of the Hardy spaces on the unit disk, and the unit circle.

The space of functions that are the limits onT of functions inHp(z) may be calledHp(T). It is a closed subspace ofLp(T) (at least forp ≥ 1). SinceLp(T) is aBanach space (for 1 ≤ p ≤ ∞), so isHp(T).

On the upper half-plane

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Theunit disk may beconformally mapped to theupper half-plane by means of certainMöbius transformations. Since the conformal map of a harmonic function is also harmonic, the Poisson kernel carries over to the upper half-plane. In this case, the Poisson integral equation takes the formu(x+iy)=Py(xt)f(t)dt,y>0.{\displaystyle u(x+iy)=\int _{-\infty }^{\infty }P_{y}(x-t)f(t)\,dt,\qquad y>0.}

The kernel itself is given byPy(x)=1πyx2+y2.{\displaystyle P_{y}(x)={\frac {1}{\pi }}{\frac {y}{x^{2}+y^{2}}}.}

Given a functionfLp(R){\displaystyle f\in L^{p}(\mathbb {R} )}, theLp space of integrable functions on the real line,u can be understood as a harmonic extension off into the upper half-plane. In analogy to the situation for the disk, whenu is holomorphic in the upper half-plane, thenu is an element of the Hardy space,Hp,{\displaystyle H^{p},} and in particular,uHp=fLp{\displaystyle \|u\|_{H^{p}}=\|f\|_{L^{p}}}

Thus, again, the Hardy spaceHp on the upper half-plane is aBanach space, and, in particular, its restriction to the real axis is a closed subspace ofLp(R).{\displaystyle L^{p}(\mathbb {R} ).} The situation is only analogous to the case for the unit disk; theLebesgue measure for the unit circle is finite, whereas that for the real line is not.

On the ball

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For the ball of radiusr,BrRn,{\displaystyle r,B_{r}\subset \mathbb {R} ^{n},} the Poisson kernel takes the formP(x,ζ)=r2|x|2rωn1|xζ|n{\displaystyle P(x,\zeta )={\frac {r^{2}-|x|^{2}}{r\omega _{n-1}|x-\zeta |^{n}}}}wherexBr,ζS{\displaystyle x\in B_{r},\zeta \in S} (the surface ofBr{\displaystyle B_{r}}), andωn1{\displaystyle \omega _{n-1}} is thesurface area of the unit (n − 1)-sphere.

Then, ifu(x) is a continuous function defined onS, the corresponding Poisson integral is the functionP[u](x) defined byP[u](x)=Su(ζ)P(x,ζ)dσ(ζ).{\displaystyle P[u](x)=\int _{S}u(\zeta )P(x,\zeta )\,d\sigma (\zeta ).}

It can be shown thatP[u](x) is harmonic on the ballBr{\displaystyle B_{r}} and thatP[u](x) extends to a continuous function on the closed ball of radiusr, and the boundary function coincides with the original function u.

On the upper half-space

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An expression for the Poisson kernel of anupper half-space can also be obtained. Denote the standard Cartesian coordinates ofRn+1{\displaystyle \mathbb {R} ^{n+1}} by(t,x)=(t,x1,,xn).{\displaystyle (t,x)=(t,x_{1},\dots ,x_{n}).} The upper half-space is the set defined byHn+1={(t;x)Rn+1t>0}.{\displaystyle H^{n+1}=\left\{(t;x)\in \mathbb {R} ^{n+1}\mid t>0\right\}.}The Poisson kernel forHn+1 is given byP(t,x)=cnt(t2+x2)(n+1)/2{\displaystyle P(t,x)=c_{n}{\frac {t}{\left(t^{2}+\left\|x\right\|^{2}\right)^{(n+1)/2}}}}wherecn=Γ[(n+1)/2]π(n+1)/2.{\displaystyle c_{n}={\frac {\Gamma [(n+1)/2]}{\pi ^{(n+1)/2}}}.}

The Poisson kernel for the upper half-space appears naturally as theFourier transform of theAbel transformin whicht assumes the role of an auxiliary parameter. To wit,P(t,x)=F(K(t,))(x)=Rne2πt|ξ|e2πiξxdξ.{\displaystyle P(t,x)={\mathcal {F}}(K(t,\cdot ))(x)=\int _{\mathbb {R} ^{n}}e^{-2\pi t|\xi |}e^{-2\pi i\xi \cdot x}\,d\xi .}In particular, it is clear from the properties of the Fourier transform that, at least formally, the convolutionP[u](t,x)=[P(t,)u](x){\displaystyle P[u](t,x)=[P(t,\cdot )*u](x)}is a solution of Laplace's equation in the upper half-plane. One can also show that ast → 0,P[u](t,x) →u(x) in a suitable sense.

See also

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References

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  1. ^"complex analysis - Deriving the Poisson Integral Formula from the Cauchy Integral Formula".Mathematics Stack Exchange. Retrieved2022-08-21.
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