I am working through some practice problems for my PDE class and I came across the following:

Let $U\in \mathbb{R}^n$ be a smooth, bounded, connected open set. Let $\Gamma_1$, $\Gamma_2$, be two disjoint subsets of $\partial U$ of positive $(n-1)$-dimensional measure such that $\Gamma_1\cup\Gamma_2=\partial U$. Define the set

$$ H=\{\phi\in C^{\infty}(\overline{U}) : \text{dist}(\text{spt}\phi,\Gamma_1)>0 $$ and define the Hilbert space $\check{H}^1(U)$ as the closure of H in the standard $H^1(U)$ norm.

a) Prove the following Poincare inequality for functions in $\check{H}^1(U)$: there exists a constant $C>0$ such that

$$ \int_U u^2dx\le C\int_U|Du|^2dx $$

for all $u\in \check{H}^1(U).

b) Consider the following problem: Given $f\in L^2(u)$, find $u\in\check{H}^1(U)$ such that

$$ \int_U Du\cdotp D\phi dx=\int_U f\phi dx $$

for all $\phi\in\check{H}^1(U)$. Prove the existence of a unique solution of this problems.

c) Explain what boundary value problem you solved in the weak sense in part (b).

So I have proven the above for the Poisson Equation and this seems extremely similar but I am getting confused with $\check{H}^1(U)$. Any help or insight would be great.


  • $\begingroup$ What confuses you? Can you be more specific and also write down what you have done so far. $\endgroup$ – Peter Feb 29 '16 at 10:50

Part (a) is the proof that we only need the trace of a Sobolev function to be zero on a set of positive measure (with respect to the boundary measure) in order to apply the Poincare inequality.

Part (b) Now equipped with the Poincare inequality, we can show that the bilinear form defined by the left hand side is bilinear, bounded and coercive, and that the linear form defined by the right hand side is a continuous linear functional, and thus the Lax-Milgram Theorem gives us existence and uniqueness.

Part (c) The BVP solved is $$-\Delta u=f\quad x\in\Omega,$$ $$u=0\quad x\in\Gamma_1,$$ $$\frac{\partial u}{\partial n}=0\quad x\in\Gamma_2.$$ To see this, multiply the equation by $v\in \check{H}^1(\Omega)$, and integrate by parts, and you will obtain the given weak formulation.

  • $\begingroup$ For part c, when you integrate it by parts you get: $$ \int_{U}Du\cdot Dv dx - \int_{\partial U}\frac{\partial u}{\partial \nu}\cdot v dS=\int_U f\cdot vdx $$ My thought was then to use that $\Gamma_1\cup\Gamma_2=\partial U$ to break up $\int_{\partial U} \frac{\partial u}{\partial \nu}\cdot v$ as $$ \int_{\partial U} \frac{\partial u}{\partial \nu}\cdot v= \int_{\Gamma_1}\frac{\partial u}{\partial \nu}\cdot v+\int_{\Gamma_2}Du\cdot v $$ but I don't understand why the conditions for $u$ on $\Gamma_1$ and $\Gamma_2$ come from in this? $\endgroup$ – user133541 Feb 29 '16 at 17:44
  • $\begingroup$ You know that the sum is zero by looking at the weak formulation, and the second integral is zero because $v$ is zero on $\Gamma_2$. $\endgroup$ – Ellya Feb 29 '16 at 18:01
  • $\begingroup$ That is, we know that $v|_{\Gamma_2}=0$, so $\int_{\partial U}\frac{\partial u}{\partial\nu}v=\int_{\Gamma_1}\frac{\partial u}{\partial\nu}v+\int_{\Gamma_2}\frac{\partial u}{\partial\nu}v=\int_{\Gamma_1}\frac{\partial u}{\partial\nu}v$. Then since $u$ satisfies the weak formulation, we deduce that what is left is also zero. $\endgroup$ – Ellya Mar 1 '16 at 15:02

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