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I am wanting to show existence of solutions to $$u_t +L(u) = f \;\;\text{on}\;\; \Omega$$ with initial condition $u|_{t=0} = u_0$ and Neumann boundary condition $\nabla u\cdot \nu = 0$ on ${\partial\Omega}$.

How do I do this with the Galerkin method? My problem is with the boundary condition. Recall that the Galerkin method requires a triple $V \subset H$ with continuous dense inclusion. In this case, $$V = \{u \in H^1 : \nabla u \cdot \nu = 0 \;\;\text{on}\;\; \partial\Omega\}$$ and $H = L^2$. But showing $V$ is dense in $H$ is a problem.

So I guess this is not how we do it. How else can I tackle this problem without putting the BC in the Hilbert space?

Thanks

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  • $\begingroup$ How do you define the trace of $\nabla u \cdot \nu$ on the boundary? Usually, this is not defined. $\endgroup$
    – gerw
    Apr 10 '13 at 17:45
  • $\begingroup$ @gerw Not sure.. but Neumann conditions are standard, so they must make sense right? $\endgroup$
    – maximumtag
    Apr 10 '13 at 18:22
  • $\begingroup$ Have a look for the difference of "natural" and "essential" boundary conditions in weak formulations of PDEs. $\endgroup$
    – gerw
    Apr 10 '13 at 18:25
  • $\begingroup$ $V$ is just the vanilla $H^1$, you don't modify anything for homogeneous Neumann is the natural boundary condition. $\endgroup$
    – Shuhao Cao
    Apr 11 '13 at 19:48
  • $\begingroup$ @ShuhaoCao Thanks but how do we know that the solution we get at the end actually satisfies the Neumann BC? Without assuming $u$ is smooth enough to do IBP. $\endgroup$
    – maximumtag
    Apr 11 '13 at 21:02
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Let us assume the equation is second order parabolic and $L = -\Delta$ for a moment. Multiplying an arbitrary test function $v\in H^1(\Omega)$, the integration by parts of $\displaystyle -\int_{\Omega} \Delta u v\, dx$ reads: $$ -\int_{\Omega} \Delta u v\, dx = \int_{\Omega} \nabla u \cdot \nabla v\, dx - \int_{\partial \Omega} (\nabla u \cdot \nu) v\, dS $$ Notice that the integration on boundary does not appear in the Galerkin weak formulation: $$ \int_{\Omega} u_t v\, dx + \int_{\Omega} \nabla u \cdot \nabla v\, dx = \int_{\Omega} f v\, dx \tag{1} $$ i.e., for any test function the boundary integral simply vanishes. For we didn't impose any boundary conditions on the test function $v$, $\nabla u \cdot \nu$ must vanish a.e. on the boundary. If it were not the case, we could always find a $v$ such that (1) does not hold.


EDIT: Now suppose we solved (1) for all $v\in H^1$, the question is why $\nabla u \cdot \nu$ vanishes on boundary. If we assume $u\in H^2$ and perform integration by parts on the $\displaystyle\int_{\Omega} \nabla u \cdot \nabla v\, dx$ backwards we have $$ \int_{\Omega} u_t v\, dx - \int_{\Omega} \Delta u \, v\, dx + \int_{\partial \Omega} (\nabla u \cdot \nu) v\, dS= \int_{\Omega} f v\, dx \tag{2} $$ Since $v$ is an arbitrary test function in $H^1$ (here is the trick, we just manipulating $v$ to get the results we want), we could let $v=0$ on $\partial\Omega$, then the boundary integral vanishes, and $$ \int_{\Omega} (u_t- \Delta u)v\, dx= \int_{\Omega} f v\, dx $$ holds for any $v\in H^1_0$. We could argue in $\Omega$: $u_t- \Delta u = f$. Now back to (2) we have $\displaystyle \int_{\partial \Omega} (\nabla u \cdot \nu) v\, dS = 0$ for any $v\in H^1$ when $v$ is not zero on boundary. Hence $\nabla u \cdot \nu = 0$.

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  • $\begingroup$ Thank you. You write "For we didn't impose any boundary conditions on the test function v, ∇u⋅ν must vanish a.e. on the boundary. If it were not the case, we could always find a v such that (*) does not hold." Why isn't this the argument people use to show that BC is satisfied? Instead, they say: assume $u \in H^2$, then IBP to get something similar to your first displayed equation (with additional LHS and RHS) and then deduce that the BC holds. $\endgroup$
    – maximumtag
    Apr 12 '13 at 8:43
  • $\begingroup$ @maximumtag Assuming $H^2$-regularity is handy for dealing with $\nabla u \cdot \nu$ term on boundary since this normal derivative on boundary is at least in $L^2(\partial \Omega)$. $\endgroup$
    – Shuhao Cao
    Apr 12 '13 at 16:52
  • $\begingroup$ @ShuhaoCao: may i ask you a question, you said that "the integration on boundary does not appear in the Galerkin weak formulation". IS it true for any cases? For example, we have a problem without zero condition on the boundary, can we take the test function on $H^1_0$? or we can only take on $H^1$? Sorry, my background is not good. $\endgroup$ Sep 15 '15 at 8:42
  • $\begingroup$ @Oliver No. For Dirichlet boundary condition of $u$ having certain values on boundary, you have to take test function $H^1_0$ as both test/trial functions so that it does change $u$'s value (plus a function extended to interior bearing $u$'s boundary value). Here in this case it is Neumann, test function uses $H^1$. The OP here was concerning a different question, that is, how to state the Hilbert space we need w/o specify the boundary condition in it. The answer is that the Neumann type boundary condition can be encoded into the variational form. $\endgroup$
    – Shuhao Cao
    Sep 15 '15 at 14:55

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