Poisson integral for a disk
For continuous real 2π-periodic boundary data g, u(r,θ)=(2π)⁻¹∫₋π^π[(1−r²)/(1−2r cos(θ−α)+r²)]g(α)dα is harmonic on 0≤r<1 and extends continuously to u(1,θ)=g(θ).
Key knowledge
For continuous real 2π-periodic boundary data g, u(r,θ)=(2π)⁻¹∫₋π^π[(1−r²)/(1−2r cos(θ−α)+r²)]g(α)dα is harmonic on 0≤r<1 and extends continuously to u(1,θ)=g(θ). It is the unique such harmonic solution on the closed unit disk.
Values and equations
u(r,θ)=(1/(2π))∫₋π^π ((1−r²)/(1−2r cos(θ−α)+r²)) g(α) dα
- g
- Continuous real 2π-periodic boundary data
- r,θ
- Polar radius 0≤r<1 and angle
- α
- Boundary integration angle
- u
- Harmonic interior solution with the stated boundary extension
Conditions and scope
- Boundary data are continuous; discontinuous data require a separately qualified boundary-limit statement.
- The formula uses the unit disk and the polar-angle convention.
Sources · 3
References
Sources retain their recorded scope. A source link alone does not establish verification of every statement.
- Notes on Diffy Qs: Dirichlet problem in the circle and the Poisson integralJiří LeblTextbook chapter
- Math 120B (Spring 2020), Lecture 19: Harmonic functions and the Poisson kernelJames McKernan / University of California, San DiegoCourse notes · PDF
- Complex Analysis (Spring 2026), Lecture 7: The Poisson integral formulaUniversity of California, BerkeleyCourse notes
Evidence and review
Lebl section 4.10.3 exact Poisson integral; UCSD Lecture19 Theorem19.1 and Lemma19.2 kernel; Berkeley Lecture7 section1 continuous-boundary existence theorem (proof omitted, displayed angular-exponent typo excluded).
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