Home > Functional Analysis, Partial Differential Equations > Weak formulation for Laplace Equation with Robin boundary conditions

Weak formulation for Laplace Equation with Robin boundary conditions

Consider {\Omega \subset \Bbb{R}^N} an open set with Lipschitz boundary and consider on {\Omega} the following problem

\displaystyle \begin{cases} -\Delta u =f &\text{ in }\Omega \\ \frac{\partial u}{\partial n}+\beta u =0 & \text{ on }\partial \Omega. \end{cases}

where {\beta>0} is a constant. This is the Laplace equation with Robin boundary conditions. I will prove that the problem is well posed and for each {f \in L^2(\Omega)} there exists a solution {u \in H^2(\Omega)}.

First let’s find the weak (or variational) formulation of problem by multiplying with {v \in H^1(\Omega)} and by integrating by parts. We obtain

\displaystyle \int_\Omega \nabla u \cdot \nabla v +\beta \int_{\partial \Omega} uv=\int_\Omega fv,\ \forall v \in H^1(\Omega).

Note that the integral of {uv} on {\partial \Omega} is well defined, since {uv \in H^1(\Omega)} and each {H^1} function has a well defined trace on {\partial \Omega}.

Define {a(u,v)=LHS} of the previous inequality and {\varphi(v)=\int_\Omega fv}. First observe that {|\varphi(v)|\leq \|f\|_{L^2} \|v\|_{L^2}\leq \|f\|_{L^2}\|v\|_{H^1}}, therefore {\varphi} is a linear functional on {H^1(\Omega)}.

Secondly, we see that {a : H^1(\Omega) \times H^1(\Omega) \rightarrow \Bbb{R}} is bilinear and we want to prove that it is continuous and coercive. Note that

\displaystyle \left|\int_\Omega \nabla u \cdot \nabla v \right| \leq \|u\|_{H^1(\Omega)} \|v\|_{H^1(\Omega)}

and by the continuity of the trace application from {H^1(\Omega)} to {L^1(\partial \Omega)} we have

\displaystyle |\beta|\int_{\partial \Omega} | uv| \leq |\beta| \|u\|_{L^2(\partial \Omega)} \|v\|_{L^2(\partial \Omega)} \leq C^2 \|u\|_{H^1(\Omega)} \|v\|_{H^1(\Omega)}.

Using the above two inequalities the continuity of {a} can be proved.

Suppose that {a} is not coercive. Then there exists a sequence {(u_n) \subset H^1(\Omega)} such that {\|u_n\|_{H^1(\Omega)}=1} and {a(u_n,u_n) \rightarrow 0}. Since {u_n} is bounded in {H^1(\Omega)} and {H^1(\Omega)} is a Hilbert space, it contains a subsequence which converges weakly to {u \in H^1(\Omega)}. For simplicity assume that {u_n \rightharpoonup u} and since {\partial \Omega} is Lipschitz we know that {H^1(\Omega) \hookrightarrow L^2(\Omega)} we have {u_n \rightarrow u} in {L^2(\Omega)} and {\nabla u_n \rightharpoonup \nabla u} in {L^2(\Omega; \Bbb{R}^N)}.

We have

\displaystyle a(u_n,u_n)= \int_\Omega |\nabla u_n|^2 +\beta \int_{\partial \Omega} u_n^2

which implies that

\displaystyle \int_\Omega |\nabla u|^2 \leq \liminf \int_\Omega |\nabla u_n|^2=0

and therefore {u} is a constant. Moreover, we have that

\displaystyle \lim_{n \rightarrow \infty} \int_\Omega u_n^2=\int_\Omega u^2 =1.\ (1)

But we also have {u_n \rightarrow u} in {L^2(\partial \Omega)} which implies that

\displaystyle \int_{\partial \Omega}u^2 =\lim_{n \rightarrow \infty} \int_{\partial \Omega}u_n^2=0. (2)

We have thus reached a contradiction, since {u} is a constant, and this constant must be zero by {(2)} and strictly positive by {(1)}. Therefore {a} is coercive.

We can now apply Lax Milgram theorem to conclude that the weak formulation always has a solution for any {f \in L^2(\Omega)}.

Now we can ask ourselves if the solution of the variational problem is regular enough for the initial PDE to hold. If the solution {u} is in {H^2(\Omega)} then we can apply Green’s formula and recover the PDE and the boundary conditions as follows:

\displaystyle \int_\Omega (\Delta u +f)v= \int_{\partial \Omega}\left( \frac{ \partial u}{\partial n}+\beta u \right)v,\ \forall v \in H^1(\Omega),\ (3).

If we take {v} smooth with compact support the RHS vanishes and we obtain that

\displaystyle \int_\Omega (\Delta u +f)\varphi =0 ,\ \forall \varphi \in C_0^\infty(\Omega),

fact which guarantees that {-\Delta u=f} in {\Omega}. The boundary condition is a simple consequence of {(3)}.


  1. Fei
    April 1, 2013 at 4:48 am

    Hello, Beni, I have some questions in mind. May I ask you?

    In the integral of uv over the boundary, do you mean the u and v here is the trace of u and v instead of the original u and v? Evans tells us that only when u is continuous to the boundary, then Tu equals the value of u on the boundary. Otherwise, we are not sure they are the same.

    Another question is that to prove a is coercive, can we use poincare inequality if we require the domain is bounded?

    The last question is if the domain is not lipschitz continuous, can you get the same conclusion? I mean the existence of unique weak solution. In this case, the trace theorem can not be used. So how to prove a is bounded?

    Modern pde is new to me, and the above questions maybe easy for you. I am looking forward to your help. Thanks.

  2. April 1, 2013 at 10:08 am


    Yes, of course, everywhere where the integral is over the boundary, u,v are replaced by their traces.

    If Poincare’s inequality is applicable, you can surley use it.

    I’m not able to answer your question about the domains for which the boundary is not Lipschitz continuous. In general, regularity issues can be pretty tricky. In this case it is almost sure that if the boundary is not Lipschitz continuous, the trace may not exist, at least not in this form (the proof I’ve seen for the trace theorem uses extensively the fact that the boundary is Lipschitz continuous). You can search James B. Kennedy’s phd thesis. He treats problems with Robin boundary conditions, and maybe you can find more references there. (Here is his site: http://www.ciul.ul.pt/~jkennedy/)

    Thanks for reading my blog.

  3. Andrea
    May 3, 2013 at 7:01 pm

    Hello Beni,

    I have a quick question.
    Why do you use Lax-Milgram theorem, and not the Riesz’s representation theorem? Isn’t the bilinear form symmetric?

    Thank you

    • May 12, 2013 at 11:19 pm

      Lax-Milgram is a way to do this; maybe not the only one. I do not understand how can I use Riesz’s representation theorem here. Can you please give a few details?

  4. Thore
    January 26, 2016 at 1:28 pm

    Interesting, would these ideas extend to domains which have a varying Robin boundary condition? I.e., a $\beta$ varying along the boundary of $\Omega$? Also, what about the Dirac boundary condition u = 0. Does this follow from your thoughts as a limit?
    Do you know (or by yourself have) a publication about this that is citable?

    I realize, my comment is just a bunch of questions, so I would like to give a physically motivated example why the questions are relevant. The oscillations of a membrane shaped like $\Omega$ that is clamped, partially clamped or let lose at certain parts of the boundary would probably be described by such a differential equation. Therefore, it is very likely that the above thoughts extend to this regime.
    Anyways, you probably knew this.
    Good work.

    • January 27, 2016 at 11:39 pm

      If you impose some conditions, you may vary \beta along the boundary, while keeping roughly the same proof. Note that if \beta is variable, then it cannot be factored out of the integral. You need some kind of boundedness for \beta in order to prove the continuity of the bilinear form. For the coercivity you need that \beta is positive and that it is not zero everywhere on the boundary (to have the desired contradiction in the end). (note that if \beta is zero on all the boundary, we are in the Neumann case and the problem has solutions modulo a constant only in the case f has zero mean on the boundary)

  5. wolfi
    December 2, 2017 at 11:55 pm

    Hello Beni,
    if $\Omega$ is not connected, then $\nabla u = 0$ implies that $u$ is constant on each of its connected component. We also know that $Tu = 0$, where $T$ is the trace operator. How can we conclude from this that $u = 0$?

    • December 3, 2017 at 1:01 am

      If you only have a finite number of connected components then it is just like you applied the result for a connected set multiple times. If, however, the situation is more complex I don’t have a precise answer.

      If you have a function which is constant on each connected component and the trace is zero on the boundary, then each constant should be zero, since for each of the connected components you are in contact with a boundary where the trace is zero.

      • wolfi
        December 3, 2017 at 9:22 am

        Thank you for your answer.

        Your argument in the second paragraph seems to work also in the case of infinitely many connected components. But why is it true that “for each of the connected components you are in contact with a boundary”?

  6. December 3, 2017 at 3:45 pm

    An open connected component has non-void boundary (since sets without boundaries are either the whole space or the empty set). The boundary of a connected component is a subset of the boundary of the whole set.

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