The general form of Poisson's equation is

2 is the laplacian (also known as "del squared" or "div grad"). It is very important for modelling stuff like electrostatic or gravitational fields (and occasionally for irritating maths students).

Like many types of equations Poisson's equation can be solved by finding a particular solution and adding it to the general solution of the homogeneous equation2φ = 0 (known as Laplace's equation. This works because the laplacian is a linear operator. If φp is our particular solution and φg is our general solution of the homegenous equation then

2p + φg)=∇2φp + ∇2φg=ρ(x) + 0
Likewise if φ is a solution then clearly φ - φp solves the homogeneous equation.

As is often the case when dealing with this sort of problem we also want to know what kind of conditions we need to impose on the solution in order to make it unique. With Poisson's equation we are usually interested in solving it in a certain volume V bounded by a smooth surface S and the condition is normally one of the following:

  • φ(x) =g(x) on the boundary S (Dirichlet condition)
  • The derivative in the direction of the outward normal, ∂φ/∂n is given on the boundary S (Neumann condition)
  • α(x)(∂φ/∂n) + β(x)φ(x) = d(x) on the boundary S, with α everywhere non-zero, α and β positive functions.
We will now prove that these 3 conditions are indeed uniqueness conditions. We will be using roughly the same idea each time : consider 2 solutions φ1 and φ2 and show that they are the same. We will set Φ to be φ1 - φ2 and try to show that Φ is everywhere zero. Our main tool will be Green's first formula, which states that, for φ and ψ "nice" scalar fields and for a volume V bounded by a smooth surface S
Vφ∇2ψ + ∇ψ.∇φdv=∫Sφ(∂ψ/∂n).dS
(this is easily prooved by applying the divergence theorem to φ∇ψ). We will be using this with φ=ψ so we have
Vφ∇2φ + |∇φ|2dv=∫Sφ(∂φ/∂n).dS
We have Φ =φ1 - φ2 and φ1 and φ2 solve the equation ∇2 = ρ(x). hence ∇2Φ =0 (by linearity of the laplacian), Φ is harmonic.

Dirichlet condition : Applying Green's formula to Φ gives

But on S, φ1 = φ2 = g, hence Φ=0 on S and thus ∫V|∇2Φ|dv=0
The integral of a positive function is zero if and only if the function is identically null, hence |∇Φ|2=0 in V, ie Φ is constant. Φ is 0 on the boundary, therefore Φ is zero everywhere : φ1 = φ2. The solution is unique.

Neumann condition : proof is identical except that it is (∂Φ/∂n) which is 0 on S.

Mixed condition:
α(x)(∂Φ/∂n) + β(x)Φ(x) = 0 on S
Hence on S
(∂Φ/∂n) = - β(x)Φ(x)/α(x)
Sticking this in Green's first formula gives :
V|∇2Φ|dv = -∫Sβ(x)Φ2/α(x).dS
α and β are positive, therefore both the integrands are positive. We therefore have that the RHS (right hand side) is of the opposite sign of the LHS (left hand side). Hence the LHS and the RHS are both zero. Thus |∇Φ|2=0 as before, and from this it follows that Φ=0 and that the solution is unique.

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