Using these relations, one gets: ∫ ∫ Ω Ω ρ ρ ∂ ∂ u ∂ ∂ t ⋅ ⋅ v + ∫ ∫ Ω Ω μ μ ∇ ∇ u ⋅ ⋅ ∇ ∇ v + ∫ ∫ Ω Ω ρ ρ ( u ⋅ ⋅ ∇ ∇ ) u ⋅ ⋅ v − − ∫ ∫ Ω Ω p ∇ ∇ ⋅ ⋅ v = ∫ ∫ Ω Ω f ⋅ ⋅ v + ∫ ∫ ∂ ∂ Ω Ω ( μ μ ∂ ∂ u ∂ ∂ n ^ ^ − − p n ^ ^ ) ⋅ ⋅ v ∀ ∀ v ∈ ∈ V . In the same fashion, the continuity equation is multiplied for a test function q belonging to a space Q and integrated in the domain Ω Ω : ∫ ∫ Ω Ω q ∇ ∇ ⋅ ⋅ u = 0. ∀ ∀ q ∈ ∈ Q . The space functions are chosen as follows: V = [ H 0 1 ( Ω Ω ) ] d = { v ∈ ∈ [ H 1 ( Ω Ω ) ] d : v = 0 on Γ Γ D } , Q = L 2 ( Ω Ω ) Considering that the test function v vanishes on the Dirichlet boundary and considering the Neumann condition, the integral on the boundary can be rearranged as: ∫ ∫ ∂ ∂ Ω Ω ( μ μ ∂ ∂ u ∂ ∂ n ^ ^ − − p n ^ ^ ) ⋅ ⋅ v = ∫ ∫ Γ Γ D ( μ μ ∂ ∂ u ∂ ∂ n ^ ^ − − p n ^ ^ ) ⋅ ⋅ v ⏟ ⏟ v = 0 on Γ Γ D + ∫ ∫ Γ Γ N ∫ ∫ Γ Γ N ( μ μ ∂ ∂ u ∂ ∂ n ^ ^ − − p n ^ ^ ) ⏟ ⏟ = h on Γ Γ N ⋅ ⋅ v = ∫ ∫ Γ Γ N h ⋅ ⋅ v . Having this in mind, the weak formulation of the Navier–Stokes equations is expressed as: find u ∈ ∈ L 2 ( R + [ H 1 ( Ω Ω ) ] d ) ∩ ∩ C 0 ( R + [ L 2 ( Ω Ω ) ] d ) such that: { ∫ ∫ Ω Ω ρ ρ ∂ ∂ u ∂ ∂ t ⋅ ⋅ v + ∫ ∫ Ω Ω μ μ ∇ ∇ u ⋅ ⋅ ∇ ∇ v + ∫ ∫ Ω Ω ρ ρ ( u ⋅ ⋅ ∇ ∇ ) u ⋅ ⋅ v − − ∫ ∫ Ω Ω p ∇ ∇ ⋅ ⋅ v = ∫ ∫ Ω Ω f ⋅ ⋅ v + ∫ ∫ Γ Γ N h ⋅ ⋅ v ∀ ∀ v ∈ ∈ V , ∫ ∫ Ω Ω q ∇ ∇ ⋅ ⋅ u = 0 ∀ ∀ q ∈ ∈ Q .