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I am wondering about the definition of weak solution to the nonhomogeneous problem $$-\Delta u = 0 \text{ in }\Omega$$ $$u = g \text{ in }\partial\Omega$$ given $g \in H^{\frac 12}(\partial\Omega)$.

It should be something like: $u \in H^1(\Omega)$ satisfies $$\int_\Omega \nabla u \nabla v - \int_{\partial\Omega} v \partial_\nu u = 0\quad\text{for all $v \in H^1(\Omega)$}\tag{1}$$ and $\gamma(u) = g$ where $\gamma$ is the trace operator. Edit: as suggested by Tomas, the normal derivative is not defined for $H^1$ functions. So I am not sure what the natural weak formulation should be for this problem. I could test with $H^1_0$ functions instead, but this seems unnatural and we obtain a mixed bilinear form.


Let $G$ extend $g$, and consider the homogeneous problem $$-\Delta w = \Delta G \text{ in }\Omega$$ $$w = 0 \text{ in }\partial\Omega$$ This is well-posed via Lax-Milgram and we have $w \in H^1_0(\Omega)$ such that $$\int_\Omega \nabla w \nabla \varphi = -\int_\Omega \nabla G \nabla \varphi$$ for all $\varphi \in H^1_0(\Omega)$. So then set $u=w+G \in H^1(\Omega)$, then $\gamma(u) = g$ and $u$ satisfies $$\int_\Omega \nabla u\nabla \varphi = 0\quad\text{for all $\varphi \in H^1_0(\Omega)$.}$$


How do I reconcile this with what I think should be the weak form $\text{(1)}$? Notice the different spaces the test functions lie in.

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  • $\begingroup$ When $u\in H^1(\Omega)$, what do you mean by $\partial_\nu u$? $\endgroup$ – Tomás Jan 2 '15 at 20:31
  • $\begingroup$ @Tomás Hmm. You're right that the normal derivative is not defined. OK then, what should my weak formulation be? I could of course test with $H^1_0(\Omega)$ functions to remove the problem term, and this would reconcile what I got at the end. But this seems unsatisfactory to me. $\endgroup$ – weasd Jan 2 '15 at 21:11
  • $\begingroup$ I've always wondered why when people say weak formulation it is assumed $u\in H^!$ and not $u\in H^2\,.$ I guess the former is more natural because you can more easily exploit the Hilbert space structure, but still... $\endgroup$ – JLA Jan 2 '15 at 21:28
  • $\begingroup$ I will write an answer. You should change $H^{-1/2}$ by $H^{1/2}$. $\endgroup$ – Tomás Jan 2 '15 at 21:41
  • $\begingroup$ Actually the test function is only $v\in H_0^1$ so you never consider the trace value, hence no integration over the boundary. I will write more details later. $\endgroup$ – spatially Jan 2 '15 at 22:52
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First we look for a distributional solution. Remember that, as an distribution, $\Delta u$ is defined by $$\langle\Delta u,v \rangle=-\langle\nabla u,\nabla v\rangle,\ \forall\ v\in C_0^\infty(\Omega). \tag{1}$$

From $(1)$, we can say that a solution in the distributional sense, is a function $u\in H^1(\Omega)$ with $Tu=g$ satisfying $$\int_\Omega \nabla u\nabla v=0,\ \forall\ v\in C_0^\infty(\Omega). \tag{2}$$

By density we may conclude from $(2)$ that $$\int_\Omega \nabla u\nabla v=0,\ \forall v\in H_0^1(\Omega). \tag{3}$$

This is not the only weak formulation for this problem, however, it is the one which comes from a variational problem, to wit, let $F:\{u\in H^1(\Omega):\ Tu=g \}\to \mathbb{R}$ be defined by $$Fu=\frac{1}{2}\int_\Omega |\nabla u|^2.$$

Note that $(3)$ can be rewritten as $$\langle F'(u),v\rangle=0,\ \forall\ v\in H_0^1(\Omega). \tag{4}$$

Also note that, once $\{u\in H^1(\Omega):\ Tu=g \}$ is a closed convex set of $H^1(\Omega)$ and $F$ is a coercive, weakly lower semi continuous function, we have that $F$ has a unique global minimum which satisfies $(4)$.

By not considering any kind of derivative of $u$, you can also use another weak formulation: let $C_0^{1,\Delta}(\overline{\Omega})=\{u\in C_0^1(\overline{\Omega}):\ \Delta u \in L^\infty(\Omega)\}$. A "very" weak solution, is a function $u\in L^1(\Omega)$ satisfying $$\int_\Omega u\Delta v=-\int_{\partial\Omega}g\frac{\partial v}{\partial \nu },\ \forall\ v\in C_0^{1,\Delta}(\overline{\Omega}).$$

In your setting, I mean, when $g\in H^{1/2}(\Omega)$, it can be proved that both definitions are equivalent. For references, take a look in the paper Elliptic Equations Involving Measures from Veron. It has a PDF version here. Take a look in page 8.

To conclude, I would like to adress @JLA, which gave a comment in OP; in the end, what we really want is a $H^2$ function (or more regular), because we are working with the Laplacean and it is natural to have two derivatives.

It can be proved, by using regularity theory, that $u$ is in fact in $H^2$, however, there is a huge difference between proving that $u$ is in $H^2$ after finding it in $H^1$ by the above methods and finding directly $u\in H^2(\Omega)$ by another method. Note, for example, that none of the methods above, does apply if we change $H_0^1(\Omega)$ by $H^2(\Omega)$.

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  • $\begingroup$ Hmm, it seems if we prove the existence via the direct method of calculus (like you wrote with the functional $F$) we don't a Poincare inequality. This approach seems useful on bad domains, unless I'm missing something. $\endgroup$ – riem Jan 7 '15 at 9:43
  • $\begingroup$ @riem, there is no Poincare inequality in $H^1(\Omega)$. $\endgroup$ – Tomás Jan 7 '15 at 10:42
  • $\begingroup$ I meant instead of making the Dirichlet condition zero like in the OP, which would need Poincare inequality for Lax-Milgram on the space $H^1_0$. $\endgroup$ – riem Jan 7 '15 at 11:09
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    $\begingroup$ Ok, now I understood. Yes, it seems you are right. At least for those domains, where the trace operator is well defined. $\endgroup$ – Tomás Jan 7 '15 at 11:14
  • $\begingroup$ Tomas, do you have a reference for the fact that the constrained minimisation problem satisfies the variational problem? I know if $J$ is a functional on a Banach space $X$, and if $J:U \to \mathbb{R}$ satisfies certain assumptions and if $U = X$ the whole Banach space then $J'(u,v) = 0$ for all $v \in X$. But if $U$ is a closed convex subset of $X$ I cannot find a reference. All I can find is that it satisfies a variational inequality. $\endgroup$ – riem Feb 5 '15 at 21:03
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From my experience, there are three types of Laplace equation PDEs, namely, (D) for Dirichlet problem \begin{cases}\tag D -\Delta u=f\\ u=0 \end{cases} (N) for Neumann problem \begin{cases}\tag N -\Delta u=f\\ \partial_\nu u=0 \end{cases} and (R) for Robin problem \begin{cases}\tag R -\Delta u=f\\ \partial_\nu u+\alpha u=0 \end{cases}

We use test function $v\in H_0^1$ for problem $(D)$ and $v\in H^1$ for problem $(N)$ and $(R)$.

Let's first look at problem $(D)$. You already discovered that the problem \begin{cases} -\Delta u=f\\ u=0 \end{cases} and problem \begin{cases} -\Delta u=0\\ u=g \end{cases} are more or less the same thing. So let's focus on the first one. To get a sense what should be the proper test space, we assume we have a smooth solution $u\in H^1(\Omega)\cap C^\infty(\bar \Omega)$. Then by trace theorem, we have $u\in H_0^1(\Omega)$. Hence, if we ask to use Lax-Milgram, we need to fit test space with the desired properties of $u$, and hence the test space must be $H_0^1(\Omega)$. Finally the weak formulation $$ \int_\Omega \nabla u\cdot\nabla v=\int_\Omega fv\,\, \text{ for all }v\in H_0^1(\Omega)$$ will let us use Lax-Milgram.

Use the same way we say $H^1$ is the proper choice of $(N)$.

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