Question

Assume a,f∈C(Ω). Let u be a smooth solution to { −Δu+a(x)u=f(x),∂ ν​ u=0,​ x∈Ω,x∈∂Ω.

​ (i) Assume that a≡0. Show that one has necessarily ∫ Ω​ f(x)dx=0 and for any constant c∈R,u−c is a solution to

Answer 1

When a ≡ 0, ∫ Ω f(x)dx = 0. For any constant c ∈ ℝ, the function u−c is a solution to -∆u + a(x)u = f(x).

Step-by-step explanation:

To show that ∫ Ω f(x)dx = 0 when a ≡ 0, we need to use the divergence theorem. The equation -∆u + a(x)u = f(x) can be written as ∇ · (∇u - au) = f(x), where ∇ is the gradient operator. We can rewrite this equation as ∇ · ∇u - ∇ · (au) = f(x). Since the second term on the left-hand side is zero (due to the boundary condition), we have ∇ · ∇u = f(x). Applying the divergence theorem to this equation gives us ∫ Ω ∇ · ∇u dx = ∫ ∂Ω ∇u · dS = ∫ Ω f(x)dx, where ∂Ω is the boundary of Ω and dS is the outward normal vector on the boundary.

Since ∇ · ∇u is equal to the Laplacian of u, we have ∆u = f(x). But since a ≡ 0, the equation becomes -∆u = f(x). This is a standard Poisson equation which has a unique solution if and only if ∫ Ω f(x)dx = 0. Therefore, when a ≡ 0, we necessarily have ∫ Ω f(x)dx = 0.

Now, let's prove that u - c is a solution to -∆u + a(x)u = f(x) for any constant c ∈ ℝ. Since a ≡ 0, the equation becomes -∆u = f(x). Taking the Laplacian of u - c, we have ∆(u - c) = ∆u - ∆c = -∆u = f(x). Therefore, u - c is a solution to -∆u + a(x)u = f(x) for any constant c ∈ ℝ.