I know that the Nitsche's method is a very attractive methods since it allows to take into account Dirichlet type boundary conditions or contact with friction boundary conditions in a weak way without the use of Lagrange multipliers. And its advantage, which is to transform a Dirichlet boundary condition into weak terms similarly as a Neumann boundary condition, is paid by the fact that the implementation is model dependent.

However, it seems to be too general for me. Can you give me more specific idea of this method? A simple example would be appreciated.

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    $\begingroup$ I don't think I quite understand your question. You correctly identify why the method was invented (to handle Dirichlet conditions in the weak form). What do you mean by "However, it seems to be too general for me. Can you give me more specific idea of this method? A simple example is costly."? $\endgroup$ Jun 11, 2015 at 23:15
  • $\begingroup$ @WolfgangBangerth : I need a (simple) example for this idea. It's so abstract for me. $\endgroup$ Jun 12, 2015 at 9:23
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    $\begingroup$ @Oliver: I'm assuming you mean "costly" as in "dear", "precious", that is, "appreciated"? I've taken the liberty of changing the word; if you disagree, feel free to roll back the edit. $\endgroup$ Jun 12, 2015 at 13:07

1 Answer 1


Nitsche's method is related to discontinuous Galerkin methods (indeed, as Wolfgang points out, it is a precursor to these methods), and can be derived in a similar fashion. Let's consider the simplest problem, Poisson's equation: $$ \left\{\begin{aligned} -\Delta u &= f \qquad\text{on }\Omega,\\ u &= g \qquad\text{on }\partial\Omega. \end{aligned}\right. \tag{1} $$ We are now looking for a variational formulation that

  1. is satisfied by the (weak) solution $u\in H^1(\Omega)$ (i.e., consistent),
  2. is symmetric in $u$ and $v$,
  3. admits a unique solution (which means that the bilinear form is coercive).

We start as usual by taking the strong form of the differential equation, multiplying by a test function $v\in H^1(\Omega)$ and integrating by parts. Starting with the right-hand side, we obtain $$ \begin{aligned} (f ,v) = (-\Delta u,v)&=(\nabla u,\nabla v) - \int_{\partial\Omega} \partial_\nu u v\,ds \\ &= (\nabla u,\nabla v) - \int_{\partial\Omega} \partial_\nu u v\,ds - \int_{\partial\Omega} (u-g)\partial_\nu v\,ds \end{aligned} $$ where in the last equation we have added the productive zero $0=u-g$ on the boundary. Rearranging the terms to separate linear and bilinear forms now gives a variational equation for a symmetric bilinear form that is satisfied for the solution $u\in H^1(\Omega)$ of $(1)$.

The bilinear form is however not coercive, since you cannot bound it from below for $u=v$ by $c\|v\|_{H^1}^2$ (as we don't have any boundary conditions for arbitrary $v\in H^1(\Omega)$, we cannot use Poincaré's inequality as usual -- this means we can make the $L^2$ part of the norm arbitrarily large without changing the bilinear form). So we need to add another (symmetric) term that vanishes for the true solution: $\eta\int_{\partial\Omega} (u-g)v\,ds$ for some $\eta>0$ large enough. This leads to the (symmetric, consistent, coercive) weak formulation: Find $u\in H^1(\Omega)$ such that $$ (\nabla u,\nabla v) - \int_{\partial\Omega} \partial_\nu u v\,ds - \int_{\partial\Omega} u\partial_\nu v\,ds +\eta\int_{\partial\Omega} u v\,ds = -\int_{\partial\Omega} g\partial_\nu v\,ds + \eta\int_{\partial\Omega} g v\,ds + \int_\Omega f v\,dx \qquad\text{for all }v\in H^1(\Omega). $$

Taking instead of $u,v\in H^1(\Omega)$ discrete approximations $u_h,v_h\in V_h\subset H^1(\Omega)$ yields the usual Galerkin approximation. Note that since it's non-conforming due to the boundary conditions (we are looking for the discrete solution in a space that is larger than the one we sought the continuous solution in), one cannot deduce well-posedness of the discrete problem from that of the continuous problem. Nitsche now showed that if $\eta$ is chosen as $ch^{-1}$ for $c>0$ sufficiently large, the discrete problem is in fact stable (with respect to a suitable mesh-dependent norm).

(This is not Nitsche's original derivation, which predates discontinuous Galerkin methods and starts from an equivalent minimization problem. In fact, his original paper does not mention the corresponding bilinear form at all, but you can find it in, e.g., Freund and Stenberg, On weakly imposed boundary conditions for second-order problems, Proceedings of the Ninth Int. Conf. Finite Elements in Fluids, Venice 1995. M. Morandi Cecchi et al., Eds. pp. 327-336.)

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    $\begingroup$ Your first sentence is not wrong, but historically inaccurate: Nitsche's idea came first and inspired the development of discontinuous Galerkin methods. That said, this doesn't take away from the otherwise excellent answer. $\endgroup$ Jun 12, 2015 at 17:24
  • $\begingroup$ @WolfgangBangerth You are of course correct; no causality was implied, only correlation. But it is important to give proper attribution, especially to people who otherwise get short-shifted. I'll edit to make that clear. $\endgroup$ Jun 12, 2015 at 18:05
  • $\begingroup$ Questions: 1. Could you elaborate more on the coercivity issue prior to adding the additional boundary term? 2. What does "non-conforming" here mean? 3. I thought I read that stability is an automatic result of coercivity of the bilinear form..? Though this explanation is quite good (the only explanation I've been able to find in fact), can anyone link to another overall explanation of the method (and/or its derivation) just for comparison? Even if I could locate the original paper, not sure it would be much help. The Freund and Stenberg paper only gives a short synopsis and a couple specific $\endgroup$
    – Nights
    Sep 22, 2015 at 8:10
  • $\begingroup$ Nonconformity: the discrete solution space $V_h$ is not a subspace of the continuous solution space $H_g^1(\Omega)$ - because the Dirichlet boundary conditions are enforced only in a weak sense. Here is a potentially useful link. $\endgroup$
    – GoHokies
    Sep 22, 2015 at 8:55
  • $\begingroup$ @Nights I have edited the answer to address your points (except that in your second paragraph, obviously). $\endgroup$ Sep 22, 2015 at 11:15

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