Variational formulation-exercise

Question

I have the following exercise: let $\Omega$ an open to $\mathbb{R}^n$, and let $v \in H^1(\Omega)$ and $u \in H^1_0(\Omega)$.

1. Prove that $$ -\langle \Delta u, v \rangle_{H^{-1},H^1_0}= \displaystyle\int_{\Omega} \nabla u(x) \cdot \nabla v(x) dx $$ I don't understand how we can answer this question. I'm lost. Help me please.

Answer 1

Since $u$ does not have second order derivatives, you can define $\Delta u$ only in the sense of distributions, that is, you first consider $$T_u(\phi):=\int_\Omega u\phi\,dx$$ for $\phi\in C^\infty_c(\Omega)$, and then take its distributional derivatives $$\frac{\partial T_u}{\partial x_i}(\phi)=-\int_\Omega u\frac{\partial \phi}{\partial x_i}\,dx$$ and $$\frac{\partial^2 T_u}{\partial x_i^2}(\phi)=\int_\Omega u\frac{\partial^2 \phi}{\partial x_i^2}\,dx.$$ Then $$\Delta T_u(\phi)=\int_\Omega u\Delta \phi\,dx.$$ If you now integrate by parts you get $$\Delta T_u(\phi)=-\int_\Omega \nabla u\cdot\nabla \phi\,dx$$ and by Holder's inequality $$|\Delta T_u(\phi)|\le ||\nabla u||_{L^2}||\nabla \phi||_{L^2}\le ||\nabla u||_{L^2}||\phi||_{H^1}$$ This is telling you that $\Delta T_u$ is a linear continuous operator from $H^1_0(\Omega)$ (you can extend it by density of $C^\infty_c(\Omega)$ to $H^1_0(\Omega)$), and so it belongs to $H^{-1}$.

Answer 2

The expression you want to prove relates to the inner product between functions in the Sobolev spaces H^(-1)(Ω) and H^1(Ω). To prove this, you'll need to use the properties of Sobolev spaces and integration by parts.

First, let's break down the notation:

• Ω is an open subset of R^n, representing the spatial domain.
• v ∈ H^1(Ω) means that v is a function in the Sobolev space H^1(Ω), which implies that v is a square-integrable function in Ω, and its weak gradient (∇v) is also square-integrable.
• u ∈ H^1₀(Ω) means that u is a function in the Sobolev space H^1₀(Ω), which means it's a square-integrable function in Ω with zero Dirichlet boundary conditions.

You want to prove the identity:

-⟨Δu, v⟩_(H^(-1), H^1₀) = ∫_Ω ∇u(x)⋅∇v(x) dx

Here's a step-by-step proof:

1. Start with the left-hand side of the equation:

-⟨Δu, v⟩_(H^(-1), H^1₀)

This is the duality pairing between H^(-1) and H^1₀, where Δu is in H^(-1) and v is in H^1₀.

2. Now, let's express Δu as a distribution. Since u ∈ H^1₀(Ω), you can use the weak Laplacian definition:

⟨Δu, φ⟩ = ∫_Ω ∇u ⋅ ∇φ dx for all φ ∈ C₀^(∞)(Ω)

Here, φ is a smooth test function with compact support in Ω.

3. Replace φ with v in the above equation since v is in H^1 and has compact support:

⟨Δu, v⟩ = ∫_Ω ∇u ⋅ ∇v dx

4. Now, you have shown that -⟨Δu, v⟩_(H^(-1), H^1₀) = ∫_Ω ∇u ⋅ ∇v dx, which is the right-hand side of the equation.

So, you have proven that -⟨Δu, v⟩_(H^(-1), H^1₀) = ∫_Ω ∇u ⋅ ∇v dx, as desired. This demonstrates the relationship between the weak Laplacian (-Δu) and the gradient (∇u) in Sobolev spaces H^(-1) and H^1₀.