# Pde problem with Neumann BC's

Let $U \subset\mathbb{R}^n$ be a bounded open set with smooth boundary $\partial U$. Consider the Neumann boundary problem $$-\Delta u +u=f, \quad \left.\frac{\partial u}{\partial n}\right|_{\partial U}=0.$$

(a) Define the notion of weak solution.

(b) Prove existence of weak solutions.

(c) Assume $f \in C^{\infty}(\bar U)$ and the weak solution is smooth. Show that $u$ satisfies the Neumann boundary condition.

(a): Considering Green's identity and boundary condition i conclude that $u \in H^1(U)$ is weak solution of our problem if $\int_U (Dv Du + vu) dx = \int_U fv dx$ for any $v\in H^1(U)$.

(b): Using standard norm on $H^1(U)$ and defining bounded linear functional on $H^{1*}(U)$ $$F_f(v) = \int_U fv dx$$ i can show the existence and uniqueness of weak solution $u \in H^1(U)$ via Riesz representation theorem.

Part (c) is causing me problems. I don't even understand the question and i have no idea how to do it. Can someone help me?

• Have you tried to incorporate the boundary conditions in the definition of the notion of solution, as is usually done when dealing with dirichlet b.cs? (sorry if the comment is useless, I don't know the answer but this was my first idea...) – user67133 Apr 4 '13 at 17:59

You are right on (a) and (b). For (c), you assume that $u\in C^\infty(\bar U)$ satisfies the weak formulation. You have to show $\partial u/\partial n = 0$ on the boundary. This can be done as usual: testing with a smooth function and integrating by parts.
• Yes, you have integrated by parts and used the boundary condition. Now, you go the other way round. Then, you arrive at $\int_\Omega -\Delta u \, \phi \, dx = \int_\Gamma \partial u/\partial n \, \phi \, d s$ (maybe the signs are not correct). Using your PDE, the first term is zero... – gerw Apr 4 '13 at 19:07
• i just use Green's identity and i arrive at: $\int_U (D v D u + v \Delta u)dx = \int_{\partial U}v\frac{\partial u}{\partial n}dS$. How did you arrive at yours and i also don't see why the first term (left hand side) is zero :( – UrošSlovenija Apr 4 '13 at 19:29
• You need the smoothness of $u$ in order to show that $-\Delta u + u = f$ holds pointwise in $U$. Actually, the smothness of $f$ follows. – gerw Apr 5 '13 at 12:17
• You can assume, e.g., $v \in C^\infty(\bar U)$ or something like that. Then it looks fine. – gerw Apr 8 '13 at 7:27