Most of the questions are more measure theory and integration related but I need to give some context, so I will now.

Consider the quasilinear 2nd-order partial differential equation $$-\text{div}(a(x,u,\nabla u)) + c(x,u,\nabla u) = g$$

Assume that we have a boundary $\Gamma$ which is divided (up to a zero-measure set) into disjoint open parts $\Gamma_{D}$ and $\Gamma_{N}$ such that $\text{meas}_{n-1}(\Gamma \setminus \Gamma_{D} \cup \Gamma_{N}) = 0$, and then consider the so-called mixed boundary conditions(Dirichlet and Neumann conditions) $$u|_{\Gamma} = u_{D} \text{ }\text{ on }\Gamma_{D}$$ $$\nu\cdot a(x,u,\nabla ) + b(x,u) = h \text{ } \text{ on }\Gamma_{N}$$

Assume we have the weak formulation identity $$\int_{\Omega}a(x,u,\nabla u)\cdot \nabla v + c(x,u,\nabla u)v dx + \int_{\Gamma_{N}}b(x,u)vdS = \int_{\Omega}gv dx + \int_{\Gamma_{N}}hvdS$$

We call $u \in W^{1,p}(\Omega)$ a weak solution to the 2nd-order PDE and boundary conditions if $u|_{\Gamma_{D}} = u_{D}$ and if the weak formulation identity holds for any $v \in W^{1,p}_{o}(\Omega) = \{ v \in W^{1,p}(\Omega): v|_{\Gamma} = 0\}$.

If we are then given that $u$ is a weak solution and $u \in C^{2}(\bar{\Omega})$ and we are also given that $$\int_{\Omega}(\text{div}a(x,u,\nabla u)-c(x,u,\nabla u) + g)vdx + \int_{\Gamma_{N}}(h-b(x,u)-\nu\cdot a(x,u,\nabla u))vdS = 0$$ where $a \in C^{1}(\bar{\Omega}\times\mathbb{R}\times \mathbb{R}^{n};\mathbb{R}^{n})$, $c \in C^{0}(\bar{\Omega}\times \mathbb{R}\times \mathbb{R}^{n})$, $b \in C^{0}(\bar{\Gamma} \times \mathbb{R})$, $g \in C(\bar{\Omega})$, $h \in C(\Gamma_{N})$.


  1. Considering $v|_{\Gamma} = 0$, the boundary integral vanishes, with $v$ arbitrarily chosen in $W^{1,p}_{o}(\Omega)$. How do you conclude that $\text{div}a(x,u,\nabla u) - c(x,u,\nabla u) + g = 0$ almost everywhere?

  2. Why does $\text{div}a(x,u,\nabla u) - c(x,u,\nabla u) + g = 0$ even hold everywhere due to the smoothness of $a$ and $c$?

  3. Noting that the first integral vanishes. How does putting a arbitrary $v \in W^{1,p}_{o}$ into the given equation show that $$\nu \cdot a(x,u,\nabla u) + b(x,u) = h \text{ on } \Gamma_{N}$$ which is the second boundary condition.

Thanks let me know if something is unclear.

  • $\begingroup$ I think we use the fact that $W^{1,p}_{o}(\Omega)$ is dense in $L^{1}(\Omega)$ and $C^{\infty}_{o}(\Omega)$ is dense in $L^{1}(\Omega)$ for question 1 and 2. $\endgroup$ – user100431 May 15 '14 at 13:18

I think noting that $C^{\infty}_{o}(\Omega)$ is dense in $W^{k,p}_{o}(\Omega)$ and then using The du Bois-Reymond lemma(Check also Fundamental Lemma of Calculus of variations, see Wikipedia entry 'This is a link to the Lemma') would resolve the questions.

Noting that $\text{div}a(x,u,\nabla u) - c(x,u,\nabla u) + g$ is a continuous function(which therefore is also locally integrable) and then taking $v \in C^{\infty}_{o}(\Omega) \subset W^{k,p}_{o}(\Omega)$. We can do this since the integral equation in question must hold for all $v \in W^{k,p}_{o}(\Omega)$. We can now apply du Bois-Reymond lemma to get the desired result of question 1., so we have shown $$\text{ div}a(x,u,\nabla u) - c(x,u,\nabla u) + g = 0 \text{ a.e. } x \in \Omega$$ It then follows that $$\text{ div}a(x,u,\nabla u) - c(x,u,\nabla u) + g = 0 \text{ everywhere }$$ since $(\text{ div}a(x,u,\nabla u) - c(x,u,\nabla u) + g)^{-1}(\mathbb{R} \setminus \{0\})$ is open in $\mathbb{R}^{n}$, and the Lebesgue measure of an open set is zero iff the open set is the emptyset. I'm still working on question 3. Does this proof seem fine?


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