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When putting the Poisson Equation into weak form we usually get to solve this: Find $u\in H^1$ s.t. $$ \int_\Omega\textrm{grad}u \cdot \textrm{grad} v=\int_\Omega fv\quad \forall v \in H^1(\Omega) $$ This is called the primal formulation. We can put boundary conditions to the problem, since due to the trace theorem in a sense we can set $$ u|_{\partial\Omega}=u_0$$ Here these are essential BCs, i.e. they have to be forced upon our solution apart from the formulation. Here this can be done because $u$ comes from a space that allows for imposing values on the boundary.

Now suppose we reformulate the Poisson problem to the dual-mixed formulation: Find $\sigma \in H(div),u \in L^2$ so that

$$ \int_\Omega\sigma\cdot \tau+\int_\Omega u \thinspace div\thinspace \tau %% =-\int_{\partial\Omega} \tau \cdot n \ u \ {\rm d} s \qquad \forall \tau\in \mathbf{H}(div)$$ $$ \int_\Omega \thinspace div \thinspace \sigma v=- \int_\Omega fv \qquad\forall v \in L^2 $$ How can we still impose Dirichlet BCs, since u is now Element $L^2$ and no trace theorem allows for setting u on the boundary (unless that it is not just a meaningless change on a 0-measure)? Is the question clear?

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In dual-mixed formulation the roles of Dirichlet and Neumann conditions are flipped when compared to the primal formulation. Thus, in order to impose condition $u|_{\partial \Omega} = u_0$ you add a loading term $\int_{\partial \Omega}u_0 v\,dx$ like you would do in the case of inhomogeneous Neumann condition and primal formulation.

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