Let $U$ be connected. A function $u \in H^1(U)$ is a weak solution of Neumann's problem \begin{equation} (*)\qquad\left\{ \begin{array}{rl} -\Delta = f & \text{in } U \\ \frac{\partial u}{\partial \nu} = 0 & \text{on } \partial U \end{array} \right. \end{equation} if $$ \int_U Du \cdot Dv \; dx = \int_Ufv \; dx $$ for all $v \in H^1(U)$. Let $f\in L^2(U)$. Prove $(*)$ has a weak solution if and only if $$ \int_U f \; dx =0. $$

For the only if part I set $v=1$. However I do not see where to start the if part. I was thinking of the Lax-Milgram theorem. That's where I'm now.

  • $\begingroup$ $-\Delta$ is an elliptic operator on $H^1(U)$, so Lax Milgram is the correct approach. $\endgroup$ – AlexR Apr 14 '14 at 9:16
  • $\begingroup$ But I cannot see where I need it. I mean $(f,v)_{L^2(U)}$ is a bounded linear functional on $H^1(U)$. $\endgroup$ – simon Apr 14 '14 at 9:26
  • $\begingroup$ Never mind i got it. Thx $\endgroup$ – simon Apr 14 '14 at 10:49
  • $\begingroup$ Would instead of a direct Lax-Milgram approach, an (alternative) version of Theorem 4 (iii) (p.321) using the Fredholm Alternative also be something which can solve the `if ' part? $\endgroup$ – xpnerd Feb 1 '17 at 11:02

For the Neumann problem $\,(\ast)\,$ in a bounded domain $U\subset\mathbb{R}^n$, $n\geqslant 2$, satisfying the cone condition, to prove that assumption $$ f\in \{ L^2(U)\,\colon\;\int\limits_{U}f\,dx=0\}\tag{1} $$ implies the existence of a weak solution $u\in H^1(U)$, it is convenient to introduce the space $$ \widetilde{H}^1(U)=\{w\in H^1(U)\colon\,\int\limits_{U}\!w\,dx=0\}. $$ Notice that $\widetilde{H}^1(U)$ is a Hilbert space with inner product $$ (u,v)\overset{\rm def}{=}\int\limits_{U}\nabla u\cdot\nabla v\,dx $$ satisfying the condition $$ (u,u)=0\;\;\Longrightarrow\;\;u=0 $$ by virtue of the Poincaré inequality $$ \|u\|^2_{L^2(U)}\leqslant C\int\limits_{U}|\nabla u|^2\,dx \quad \forall\,u\in \widetilde{H}^1(U)\tag{2} $$ which requires certain regularity of the boundary $\partial U$. Note that the cone condition is not precisely the regularity of $\partial U$ for $(2)$ to be valid — it just proves to be the least complicated suitable general restriction on $\partial U$. Denote $$ \bar{u}\overset{\rm def}{=}\frac{1}{|U|}\int\limits_{U}u\,dx, $$ with notation $|U|$ standing for the $n$-dimensional Lebesgue measure of domain $U\subset \mathbb{R}^n$. Since $u-\bar{u}\in \widetilde{H}^1(U)$ for any $u\in H^1(U)$, the Poincaré inequality can be as well rewritten in the form $$ \|u-\bar{u}\|^2_{L^2(U)}\leqslant C\int\limits_{U}|\nabla (u-\bar{u})|^2\,dx =C\int\limits_{U}|\nabla u|^2\,dx \quad \forall\,u\in H^1(U). $$         The rest of the proof is easy. Consider a linear functional $$ \Lambda(v)=\int\limits_{U}fv\,dx $$ on $\widetilde{H}^1(U)$. Due to $(2)$, the linear functional $\Lambda$ is bounded on the Hilbert space $\widetilde{H}^1(U)$. Hence, by the Riesz representation theorem, there is a unique $u\in\widetilde{H}^1(U)$ such that $$ \Lambda(v)=(u,v)\quad \forall\,v\in \widetilde{H}^1(U),\tag{3} $$ which immediately implies the integral identity $$ \int\limits_{U}\nabla u\cdot\nabla v\,dx=\int\limits_{U}fv\,dx \quad \forall\,v\in \widetilde{H}^1(U).\tag{4} $$ To complete the proof, notice that, in fact, $(4)$ is valid as well for all $u\in H^1(U)$. Indeed, due to the assumption $(1)$, for any $v\in H^1(U)$ we have $$ \int\limits_{U}fv\,dx=\int\limits_{U}f(v-\bar{v})\,dx= \int\limits_{U}\nabla u\cdot\nabla (v-\bar{v})\,dx= \int\limits_{U}\nabla u\cdot\nabla v\,dx $$ by virtue of $(3)$ since $v-\bar{v}\in \widetilde{H}^1(U)$. Thus, there is a unique $u\in\widetilde{H}^1(U)\subset H^1(U)$ such that $$ \int\limits_{U}\nabla u\cdot\nabla v\,dx=\int\limits_{U}fv\,dx \quad \forall\,v\in H^1(U). $$ Q.E.D

Remark.  Being valid for general real bilinear forms, not necessarily symmetric, the Lax-Milgram theorem looks too much advanced for this rather trivial case when all the inner product axioms are met by the symmetric bilinear form $\,(\cdot,\cdot)$. Generally, the Lax-Milgram theorem is to be applied in cases where the Riesz representation theorem is inapplicable, e.g., in case of a Dirichlet problem for the equation $-\Delta u+\partial_{x_m}u=f$.

  • $\begingroup$ In the second-to-last centered equation, I understand that $\nabla (v-\bar{v})=\nabla v$ since $\bar{v}$ is a constant value. But how does $v-\bar{v} \in \widetilde{H}^1(U)$ explain the $v=v-\bar{v}$ in $\int_U fv \, dx = \int_U f(v-\bar{v}) \, dx$? $\endgroup$ – Cookie Feb 7 '15 at 6:50
  • $\begingroup$ @dragon: $v-\bar{v} \in \widetilde{H}^1(U)$ explains why $$\int\limits_{U}f(v-\bar{v})\,dx= \int\limits_{U}\nabla u\cdot\nabla (v-\bar{v})\,dx,$$ while $\int_U fv \, dx = \int_U f(v-\bar{v}) \, dx$ is explained by the fact that $\int\limits_{U}f\,dx=0$. $\endgroup$ – mkl314 Feb 7 '15 at 22:48
  • $\begingroup$ It seems like you mean in $(3),(4)$ instead $\forall v\in\tilde{H}^1$? Also both immediately preceding the second-to-last equation and in the last equation? $\endgroup$ – charlestoncrabb Nov 20 '15 at 23:13
  • $\begingroup$ @charlestoncrabb: Of course, "there is a unique u".Typo corrected. Thanx. $\endgroup$ – mkl314 Nov 26 '15 at 13:57
  • $\begingroup$ What is cone condition ? $\endgroup$ – lanse7pty Jan 10 '16 at 11:45

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