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2.1. Variational Formulation 47

such a solution, the validity of the differential equation (1.1) is no longer re-
quired pointwise but in the sense of some integral average with “arbitrary”
weighting functions ϕ. In the same way, the boundary condition (1.2) will
be weakened by the renunciation of its pointwise validity.
For the present, we want to confine the considerations to the case of
homogeneous boundary conditions (i.e., g ≡ 0), and so we consider the
following homogeneous Dirichlet problem for the Poisson equation: Given
a function f :Ω → R, find a function u : Ω → R such that

−∆u = f in Ω , (2.1)
u = 0 on ∂Ω . (2.2)

In the following let Ω be a domain such that the integral theorem of
d
Gauss is valid, i.e. for any vector field q :Ω → R with components in
1
C(Ω) ∩ C (Ω) it holds

∇· q(x) dx = ν(x) · q(x) dσ . (2.3)
Ω ∂Ω
Let the function u : Ω → R be a classical solution of (2.1), (2.2) in the
1
sense of Definition 1.1, which additionally satisfies u ∈ C (Ω) to facili-
tate the reasoning. Next we consider arbitrary v ∈ C (Ω) as so-called test
∞
0
functions. The smoothness of these functions allows all operations of differ-
entiation, and furthermore, all derivatives of a function v ∈ C (Ω) vanish
∞
0
on the boundary ∂Ω. We multiply equation (2.1) by v, integrate the result
over Ω, and obtain

f, v = f(x)v(x) dx = − ∇· (∇u)(x) v(x) dx
0
Ω Ω

= ∇u(x) ·∇v(x) dx − ∇u(x) · ν(x) v(x) dσ (2.4)
Ω ∂Ω

= ∇u(x) ·∇v(x) dx .
Ω
The equality sign at the beginning of the second line of (2.4) is obtained
by integration by parts using the integral theorem of Gauss with q = v∇u.
The boundary integral vanishes because v =0 holds on ∂Ω.
1
If we define, for u ∈ C (Ω), v ∈ C (Ω), a real-valued mapping a by
∞
0

a(u, v):= ∇u(x) ·∇v(x) dx ,
Ω
then the classical solution of the boundary value problem satisfies the
identity
∞
a(u, v)= f, v for all v ∈ C (Ω) . (2.5)
0 0