Let's solve the problem step by step as you requested: --- **Step 1: Weak Formulation and Necessary Condition** We are given Neumann's problem: \[ \begin{cases} -\Delta u = f & \text{in } U \\ \frac{\partial u}{\partial \nu} = 0 & \text{on } \partial U \end{cases} \] The weak form is: \( u \in H^1(U) \) is a weak solution if \[ \int_U \nabla u \cdot \nabla v \, dx = \int_U f v \, dx, \quad \forall v \in H^1(U) \] Let's test with \( v = 1 \), which is admissible since \( 1 \in H^1(U) \): \[ \int_U \nabla u \cdot \nabla 1 \, dx = \int_U f \cdot 1 \, dx \] But \( \nabla 1 = 0 \), so the left side is 0, which gives: \[ 0 = \int_U f \, dx \] **Explanation:** For the weak solution to exist, it is **necessary** that \( \int_U f \, dx = 0 \). --- **Step 2: Sufficiency (Existence of Solution if \( \int_U f = 0 \))** Suppose \( \int_U f \, dx = 0 \). The bilinear form \( a(u, v) = \int_U \nabla u \cdot \nabla v \, dx \) is continuous and coercive on the subspace \[ V = \left\{ v \in H^1(U) : \int_U v \, dx = 0 \right\} \] because constants are the kernel of the Laplacian with Neumann BCs. For any \( f \) with zero mean, the linear functional \( L(v) = \int_U f v \, dx \) is continuous and well defined on \( V \). By the **Lax-Milgram theorem**, there exists a unique \( u \in V \) such that \[ \int_U \nabla u \cdot \nabla v \, dx = \int_U f v \, dx, \quad \forall v \in V \] Since the solution is unique up to an additive constant, a weak solution exists. **Explanation:** If \( \int_U f \, dx = 0 \), a weak solution **does exist** (unique up to constants) via the Lax-Milgram theorem. --- ## Final Answer **The Neumann problem (3) has a weak solution if and only if** \[ \int_U f \, dx = 0. \]

**Step 1: Deriving the necessary condition for existence** *Explanation:* Starting from the weak formulation of the Neumann problem: \[ \int_U \nabla u \cdot \nabla v \, dx = \int_U f v \, dx, \quad \forall v \in H^1(U). \] Choose \( v = 1 \), which is admissible because constant functions are in \( H^1(U) \). Then: \[ \int_U \nabla u \cdot \nabla 1 \, dx = \int_U f \cdot 1 \, dx, \] but \(\nabla 1 = 0\), so the left side is zero: \[ 0 = \int_U f \, dx. \] *Therefore,* **a necessary condition** for the existence of a weak solution is: \[ \boxed{\int_U f \, dx = 0.} \] --- **Step 2: Showing sufficiency and existence of solutions** *Explanation:* Assuming \(\int_U f \, dx = 0\), consider the subspace: \[ V = \left\{ v \in H^1(U) : \int_U v \, dx = 0 \right\}. \] On \( V \), the bilinear form \[ a(u,v) = \int_U \nabla u \cdot \nabla v \, dx \] is continuous, coercive, and symmetric. The linear functional \[ L(v) = \int_U f v \, dx \] is continuous on \( V \) because \( f \) has zero mean and is in \( L^2(U) \). By the **Lax-Milgram theorem**, there exists a unique \( u \in V \) such that: \[ a(u,v) = L(v), \quad \forall v \in V. \] This \( u \) is a weak solution of the Neumann problem, unique up to an additive constant (since the problem's kernel consists of constants). *Hence,* **a weak solution exists if and only if**: \[ \boxed{\int_U f \, dx = 0.} \] --- ## **Final Answer:** **The Neumann problem (3) has a weak solution if and only if** \[ \boxed{\int_U f \, dx = 0.} \]