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i'm trying to solve the Poisson equation: $$ \begin{split} -\Delta u &= f \quad \text{in } \Omega\\ u &= g \quad \text{on } \partial \Omega, \end{split} $$

where $ \Omega$ is bounded a Lipschitz-continues domain and $u \in H^2(\Omega), f,g \in L^2(\Omega)$. The least squares finite element method minimizes the following energy functional:

$$ J(u;f,g) := \frac{1}{2} \int_{\Omega} (-\Delta u-f)^2 + \int_{\partial\Omega} (u -g)^2 $$

this method leads to the following variational problem: $$ a(u,v) = l(v), \quad \forall v \in V \subset H^2(\Omega) $$ where

$$ \begin{split} a(u,v) &= \int_{\Omega} \Delta u\Delta v + \int_{\partial\Omega} uv \\ l(v) &= \int_{\Omega} -f\Delta v + \int_{\partial\Omega} gv \end{split} $$

It's clear that this bilinear operator is continues. Now i'm trying to proof that $a(v,v)$ is coercive, e.q.

$$ a(v,v) = \int_{\Omega} (\Delta v)^2 + \int_{\partial\Omega} v^2 \geq C\| v\|^2_{H^2(\Omega)} $$

It seems to be hard to prove that, so I tried to show this weaker inequality: $$a(v,v) = \int_{\Omega} (\Delta v)^2 + \int_{\partial\Omega} v^2 \geq C\| v\|^2_{L^2(\Omega)} $$ but with no success.

I don't have the opportunity to choose $u \in H^2(\Omega) \cap H^1_0(\partial \Omega)$, so i had to add the boundary residual to the energy functional $J$. I guess i'm working with the wrong energy.

This is my first question here and i'm thankful for every hint.

skymath

skymath
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  • Please continue writing the question body. – daw Jul 18 '19 at 14:24
  • i'm finished. Please don't hesitate to ask if i forgot some information – skymath Jul 18 '19 at 15:11
  • I think there is a small issue with the question: if $g$ is only in $L^2(\Omega)$, there is no natural interpretation for its value at the boundary. I believe you need at least $g \in H^{1/2}(\Omega)$ to make sense of it using the trace. – Roberto Rastapopoulos Jul 18 '19 at 15:43
  • could you please explain in more details? – skymath Jul 18 '19 at 15:48
  • You mean: the trace operator i guess only maps $H^{k+2}(\Omega) \rightarrow H^{k+2-1/2}(\partial \Omega)?$ so $g$ has to be from $H^{3/2}$? – skymath Jul 18 '19 at 15:51
  • But then i have to use the following energy functional: $$ J(u;f,g) := \frac{1}{2} |-\Delta u-f|^2_{L^2(\Omega)} + \frac{1}{2} |u -g|^2_{H^{3/2}(\partial\Omega)}$$

    which is pretty ugly and not practical :(

     – skymath Jul 18 '19 at 15:59
  • Sorry, I'm a bit rusty and made lots of typos. Why do you think you would need that functional? – Roberto Rastapopoulos Jul 18 '19 at 16:02
  • Ah ok, I found a reference on the LSFEM. I wasn't aware of it before, and you're right that norm that appears in the energy functional is related to the space of the data. – Roberto Rastapopoulos Jul 18 '19 at 16:05
  • Do you know a way to overcome this impractical functional? I would like to have all the residual-term measured in the $L^2$ norm. Is there some inequality like: $$ | u-g|{H^{3/2}(\partial \Omega)} \leq \tilde{C}| u-g|{L^2(\partial \Omega)} $$ – skymath Jul 18 '19 at 16:11
  • No, that certainly does not hold. Think of a more and more oscilatory $(u - g)$, for example. – Roberto Rastapopoulos Jul 18 '19 at 16:14
  • Where did $\Delta v$ even come from in the first place? The way that $\nabla v$ enters into the conventional weak formulation of the Poisson equation is by integrating by parts once, which converts $\Delta u$ into $\nabla u$. You can get $\Delta v$ as well by integrating by parts twice instead, but then $\nabla u$ becomes $u$. Either way you don't have $\Delta u \Delta v$. – Ian Jul 18 '19 at 16:46
  • It seems to me that if the goal is to minimize that $J$ then you would set $\int_\Omega (-\Delta u - f) v$ and $\int_{\partial \Omega} (u-g) v$ both equal to zero for all $v$ in some suitable class. – Ian Jul 18 '19 at 16:50
  • The energy functional i defined on the beginning was: $$ J(u) = \frac{1}{2}|-\Delta u - f |^2_{L^2(\Omega)} + \frac{1}{2} |u - g|^2_{L^2(\partial \Omega)}$$ If I calculate $\lim_{t \rightarrow 0} \frac{d}{dt}J(u+tv)$ i got: $$ \lim_{t \rightarrow 0} \frac{d}{dt}J(u+tv) = \left<-\Delta u - f, -\Delta v\right>{L^2(\Omega)} + \left<u-g, v\right>{L^2(\partial \Omega)}$$ after setting it equal to zero, i got: $$ \left<\Delta u, \Delta v\right>{L^2(\Omega)} + \left<u, v\right>{L^2(\partial \Omega)} = \left<f, -\Delta v\right>{L^2(\Omega)} + \left<g, v\right>{L^2(\partial \Omega)}$$ – skymath Jul 18 '19 at 17:42
  • Beware! The Least Squares Finite Element Method may be very tricky. E.g. Any employment for the Varignon parallelogram? and What is the difference between Finite Difference Methods, Finite Element Methods and Finite Volume Methods for solving PDEs? – Han de Bruijn Jul 25 '19 at 10:06

1 Answers1

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The first coercivity inequality does not hold. Take, for example, the disk-shape domain $\Omega = \{x, y: x^2 + y^2 \leq 1\}$ and consider a sequence of boundary data $g_n$, $n \in \mathbb R$, defined by $g_n = \cos(n\theta(x, y))$, where $\theta(\cdot, \cdot)$ denotes the angle of the argument in $[0, 2\pi]$. Take $v_n$ as being the solution of \begin{align} - \Delta v_n = 0, \qquad & \text{in $\Omega$}, \\ v_n = g_n, \qquad & \text{in $\partial \Omega$}. \end{align} Clearly, $$ \int_{\Omega} (\Delta v_n)^2 + \int_{\partial\Omega} v_n^2 = \int_{\partial \Omega} g_n^2 \leq C. $$ If the first coercivity inequality you wrote was true, it would imply: $$ \|v_n\|_{H^2(\Omega)} \leq C, $$ and thus, by the trace inequality $$ \|v_n\|_{H^{m}(\partial \Omega)} = \|g_n\|_{H^m(\partial\Omega)}\leq C, \qquad 0 \leq m \leq 3/2. $$ Given the form of $g_n$, this is clearly false for $m = 1$.

Roberto Rastapopoulos
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  • Thank you for that counter example. I also guessed, i chose the wrong energy functional. I'm trying to find other energy, so that i got the coercivity proven. – skymath Jul 18 '19 at 18:00
  • I think you do still get continuity with the $3/2$ Sobolev space, but coercivity seems a bit trickier... – Roberto Rastapopoulos Jul 19 '19 at 10:26
  • last time i read about this inverse inequality, i dont know exactly where: $$ |u|{H^{3/2}(\partial \Omega)} \leq C h^{-3/2} |u|{L^{2}(\partial \Omega)}$$

    would you say with the estimate obove, i would be able to prove this weaker coercivity: $$a(v,v) \geq C(h) |v|_{H^2(\Omega)}$$

     – skymath Jul 19 '19 at 12:38
  • You need to specify the spaces in which you think these inequalities are valid; what is $h$ here? – Roberto Rastapopoulos Jul 24 '19 at 14:11