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I have given the Poisson equation with Dirichlet boundary conditions \begin{cases} -\Delta u & = f & \text{in} &\Omega \\ \quad u & = g & \text{on} & \partial\Omega \end{cases} For $g=0$ the weak variational formulation is $$\int_{\Omega} \nabla u \nabla v = \int_{\Omega} fv$$ Now I want to know: How could a weak variational formulation be defined if $g\neq0$?

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  • $\begingroup$ Do you know how to derive the weak formulation when $g=0$? $\endgroup$
    – Jeff
    Commented Apr 7, 2021 at 17:59
  • $\begingroup$ yes it is also linked in the text. Using Greens formula you get to the point $\int_{\Omega} \nabla u \nabla v - \int_{\partial \Omega} (v \nabla u) \cdot \mathbf{n} \, dS = \int_{\Omega} fv$ and now you have to apply the boundary condition I guess. $\endgroup$
    – holly
    Commented Apr 7, 2021 at 18:29
  • $\begingroup$ Maybe you can find answer in Page79 of this PDF $\endgroup$
    – robothead
    Commented Jun 21, 2021 at 14:46

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Usually one reduces to the case of $g=0$, which is equivalent to interpreting $u=g$ on the boundary in the trace sense. For example, if you can find an $H^2$ function $w$ with $w=g$ on $\partial \Omega$ in the trace sense, then you set $v=u-w$ and find that $v$ satisfies $-\Delta v = -\Delta u + \Delta w = f + \Delta w$ and $v=0$ on $\partial \Omega$. So you obtain an equation for $v$ with zero boundary condition with a modified source term $f + \Delta w$.

You can also just keep the same weak variational form, but instead look for solutions $u\in H^1(\Omega)$ with $u=g$ in the trace sense on $\partial \Omega$ (you still take test functions $v\in H^1_0(\Omega)$). This is a linear closed subspace of $H^1$ (it is the space $w + H^1_0(\Omega)$, where $w$ is given above).

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  • $\begingroup$ Sorry for asking on an old post. But does this not mean we need to know if a function $w$ such that $w=g$ in trace sense exists already before solving the poisson problem itself? Basically it is amounting to saying that if a $w\in H^2$ has trace equal to $g$ then the problem of nonzero boundary is same as problem of zero boundary. Are they not be the same without this assumption? $\endgroup$
    – Manraj
    Commented Feb 4, 2022 at 21:14
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    $\begingroup$ Yes, you need the existence of $w$ with trace equal to $g$ on the boundary. If no such function existed, then the Poisson equation would have no solution, so one has to make such an assumption. $\endgroup$
    – Jeff
    Commented Feb 5, 2022 at 0:17
  • $\begingroup$ Thank you for your answer. Are you familiar with the line of questioning which explores the existence of said w? I found a similar line of thought in Gilbarg and Trudinger. There in chapter 2 they have classified the boundary points where the classical Dirichlet problem for harmonic functions can be solved via Perrons method. The points where it can be solved are called regular points. Is there a similar characterization for weak solutions? $\endgroup$
    – Manraj
    Commented Feb 5, 2022 at 5:50
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    $\begingroup$ I'm not overly familiar with this for weak solutions. For the classical Perron method, one needs an exterior sphere condition at each boundary point. For weak solutions, probably less boundary regularity is needed, but you would have to consult another reference (Evans generally assumes strong boundary regularity, like $C^1$). If you look at a good reference book on Sobolev spaces, they will prove trace theorems in the weakest setting possible. It's been a while since I looked at this, but perhaps a Lipschitz boundary is sufficient. $\endgroup$
    – Jeff
    Commented Feb 5, 2022 at 10:33
  • $\begingroup$ Ah, ok. I have found one such characterization in Adams. At least for $\mathbb{R}^n$ it was simply the corresponding Besov space. In particular passing from functions in $W^{m,p}(\mathbb{R}^{n+1})$ to their traces on surfaces of codimension 1 results in a loss of smoothness corresponding to 1/p of a derivative which is the corresponding Besov space $B^{m-\frac{1}{p};p,p}(\mathbb{R}^n)$. Probably, for looking in bounded domains, one can use extension and approximation results with required regularity as you suggested. Thanks! $\endgroup$
    – Manraj
    Commented Feb 5, 2022 at 14:42

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