EqWorld http://eqworld.ipmnet.ru

Exact Solutions > Linear Partial Differential Equations >Second-Order Elliptic Partial Differential Equations > Poisson Equation

3.2. Poisson Equation ∆w + Φ(x) = 0

The two-dimensional Poisson equation has the following form:

∂2w

∂x2 +∂2w

∂y2 + Φ(x,y) = 0 in the Cartesian coordinate system,

1r

∂

∂r

(r∂w

∂r

)+

1r2

∂2w

∂ϕ2 + Φ(r,ϕ) = 0 in the polar coordinate system.

3.2-1. Domain: –∞ < x < ∞, –∞ < y < ∞.

Solution:

w(x,y) =1

2π

∫ ∞

−∞

∫ ∞

−∞Φ(ξ,η) ln

1√(x − ξ)2 + (y − η)2

dξ dη.

3.2-2. Domain: –∞ < x < ∞, 0 ≤ y < ∞. First boundary value problem.

A half-plane is considered. A boundary condition is prescribed:

w = f (x) at y = 0.

Solution:

w(x,y) =1π

∫ ∞

−∞

yf (ξ) dξ

(x − ξ)2 + y2 +1

2π

∫ ∞

0

∫ ∞

−∞Φ(ξ,η) ln

√(x − ξ)2 + (y + η)2

√(x − ξ)2 + (y − η)2

dξ dη.

3.2-3. Domain: 0≤ x < ∞, 0≤ y < ∞. First boundary value problem for the Poisson equation.

A quadrant of the plane is considered. Boundary conditions are prescribed:

w = f1(y) at x = 0, w = f2(x) at y = 0.

Solution:

w(x,y) =4π

xy

∫ ∞

0

f1(η)η dη

[x2 + (y −η)2][x2 + (y +η)2]+

4π

xy

∫ ∞

0

f2(ξ)ξ dξ

[(x−ξ)2 +y2][(x+ξ)2 +y2]

+1

2π

∫ ∞

0

∫ ∞

0Φ(ξ,η) ln

√(x−ξ)2 + (y +η)2

√(x+ξ)2 + (y −η)2

√(x−ξ)2 + (y −η)2

√(x+ξ)2 + (y +η)2

dξ dη.

3.2-4. Domain: 0≤ x ≤ a, 0 ≤ y ≤ b. First boundary value problem for the Poisson equation.

A rectangle is considered. Boundary conditions are prescribed:

w = f1(y) at x = 0, w = f2(y) at x = a,

w = f3(x) at y = 0, w = f4(x) at y = b.

Solution:

w(x,y) =∫ a

0

∫ b

0Φ(ξ,η)G(x,y, ξ,η) dη dξ

+∫ b

0f1(η)

[∂

∂ξG(x,y, ξ,η)

]

ξ=0dη −

∫ b

0f2(η)

[∂

∂ξG(x,y, ξ,η)

]

ξ=a

dη

+∫ a

0f3(ξ)

[∂

∂ηG(x,y, ξ,η)

]

η=0dξ −

∫ a

0f4(ξ)

[∂

∂ηG(x,y, ξ,η)

]

η=b

dξ.

1

2 POISSONEQUATION

Two forms of representation of the Green’s function:

G(x,y, ξ,η) =2a

∞∑

n=1

sin(pnx) sin(pnξ)pn sinh(pnb)

Hn(y,η) =2b

∞∑

m=1

sin(qmy) sin(qmη)qm sinh(qma)

Qm(x, ξ),

where

pn =πn

a, Hn(y,η) =

sinh(pnη) sinh[pn(b − y)] for b ≥ y > η ≥ 0,sinh(pny) sinh[pn(b − η)] for b ≥ η > y ≥ 0,

qm =πm

b, Qm(x, ξ) =

sinh(qmξ) sinh[qm(a − x)] for a ≥ x > ξ ≥ 0,sinh(qmx) sinh[qm(a − ξ)] for a ≥ ξ > x ≥ 0.

3.2-5. Domain: 0≤ r ≤ R. First boundary value problem for the Poisson equation.

A circle is considered. A boundary condition is prescribed:

w = f (ϕ) at r = R.

Solution in the polar coordinates:

w(r,ϕ) =1

2π

∫ 2π

0f (η)

R2 − r2

r2 − 2Rr cos(ϕ − η) + R2 dη +∫ 2π

0

∫ R

0Φ(ξ,η)G(r,ϕ, ξ,η)ξ dξ dη,

where

G(r,ϕ, ξ,η) =1

4πln

r2ξ2 − 2R2rξ cos(ϕ − η) + R4

R2[r2 − 2rξ cos(ϕ − η) + ξ2].

ReferencesBabich, V. M., Kapilevich, M. B., Mikhlin, S. G., et al., Linear Equations of Mathematical Physics[in Russian], Nauka,

Moscow, 1964.Butkovskiy, A. G., Green’s Functions and Transfer Functions Handbook, Halstead Press–John Wiley & Sons, New York,

1982.Polyanin, A. D., Handbook of Linear Partial Differential Equations for Engineers and Scientists, Chapman & Hall/CRC,

2002.

Poisson Equation

Copyright c© 2004 Andrei D. Polyanin http://eqworld.ipmnet.ru/en/solutions/lpde/lpde302.pdf