Tamtam Proceedings - lamsin

A theoretical study of free surface flow 2513. Solution of the free surface problemWe transform the equation (6) inγ(x) = − F 2 [|∇ψ| 2 (x, 1 + γ(x)) + 2 ∂ψ]2∂y (x, 1 + γ(x))(7)and we put :T (b, γ) = − F 22[|∇ψ| 2 (x, 1 + γ(x)) + 2 ∂ψ∂y (x, 1 + γ(x)) ].The problem can be formulated as : given a function y = b(x) which represents anobstacle, find a function γ : R → R (free surface) such that T (b, γ) = γ with ψ verifyingthe equations (1)-(4). This is equivalent to solve the equation :T 1 (b, γ) = γ − T (b, γ) = 0 for a fixed b.For b = γ = 0, ψ = 0 verifies equations (1)-(6). So T (0, 0) = 0 and T 1 (0, 0) = 0.To solve T 1 (b, γ) = 0, we use the implicit function theorem at a neighbourhood of(b, γ) = (0, 0). Consider the change of variables :{∼x = x∼y =y−b(x)1+γ(x)−b(x) , (8)we transform the domain Ω γ bin the following infinite strip Q:Q = { (x, y) ∈ R 2 / − ∞ < x < +∞, 0 < y < 1 } .We put ψ(x, y) = ∼ ψ( ∼ x, ∼ y) then ∼ ψ verifies :∼∆ψ ∼ + P γ bψ = 0 in Q (9)∼ψ( ∼ x, 0) = −b( ∼ x), ∼ x ∈ R (10)∼ψ( ∼ x, 1) = −γ( ∼ x),∼x ∈ R (11)P γ bis an operator defined by :∂ 2∂ 2P γ b = a 1∂ ∼ x∂ ∼ y + a 2∂ ∼ y 2 + a ∂3∂ ∼ ywhere∼y(b ′ − γ ′ ) − b ′a 1 = 2, a 2 = ( a 111 + γ − b2 )2 − 1 +(1 + γ − b) 2TAMTAM –Tunis– 2005