The standard weak formulation of the Neumann problem for the Poisson equation is to find $u \in H^1 ( \Omega)$ such that for every $v \in H^1 ( \Omega)$:

$$ \int_{\Omega} \nabla u \nabla v d x = \int_{\Omega} fv d x + \int_{\partial \Omega} gv d s $$

for given $f \in L^2 ( \Omega)$ and $g \in H^{- 1 / 2} ( \Omega)$.

My goal is to introduce an arbitrary coordinate change and arrive at an analogous weak formulation.

I start with the classical formulation $- \Delta u ( x) = f ( x)$ and $\partial_n u ( x) = g (x)$ with everything fine and smooth. Let now $\phi : \Omega \rightarrow \tilde{\Omega} : x \mapsto \phi ( x) = : \tilde{x}$ be $C^{\infty}$ and bijective and define $\tilde{u} \circ \phi = u$. Applying the chain rule to $ \Delta_x ( \tilde{u} \circ \phi)$ we find the differential operator $L$:

$$ L \tilde{u} ( \tilde{x}) = a_{i j} ( x) \partial_{i j} \tilde{u} ( \tilde{x}) + b_i ( x) \tilde{u} ( \tilde{x}), $$

where $a_{i j} ( x) = \partial_k \phi_i ( x) \partial_k \phi_j ( x)$, $b_i ( x) = \Delta \phi_i ( x)$ and we used the usual summation convention over repeated indices. Yes the coordinates are all mixed up: I left out a composition with $\phi^{- 1}$ because it's messy.

It now holds that $\Delta_x u ( x) = L \tilde{u} (\phi ( x))$ for all $x \in \Omega$.

Question 1: My operator is not in divergence form, so deriving the weak formulation is going to be a real mess. What would be a better approach?

Question 2: How do I transform the boundary conditions? I'd like something like

$$ \partial_{\nu} \tilde{u} ( \tilde{x}) = \tilde{g}( \tilde{x}) . $$

The chain rule applied to $\partial_n ( \tilde{u} \circ \phi)$ results in $\partial_{\nu} \tilde{u} ( x) = \nabla \tilde{u} ( x) \nu ( x) = \nabla \tilde{u} ( x) D \phi ( x) n ( x)$, which is partly natural since the normal vector field is transformed with the differential, but is far from what I'd need during the partial integration deriving the weak formulation, which would rather be (assuming $L$ were in divergence form)

$$ \partial_{\nu} \tilde{u} = a_{i j} \partial_i \tilde{u} \nu_j .$$

This must be pretty basic, but I'm a bit confused here...

  • $\begingroup$ To my knowledge only Piola transformation(see Mathematical Elasticity by PG Ciarlet in section 1.7, original transformation normalized by $|D\phi(x)|$) would preserve the surface integration. $\endgroup$
    – Shuhao Cao
    Nov 9, 2012 at 22:20
  • $\begingroup$ Thanks, that seems reasonable, but I still don't understand how the boundary conditions change. See however this proposed solution: mathoverflow.net/questions/111935/… $\endgroup$
    – AnCo
    Nov 10, 2012 at 16:19

1 Answer 1


I suggest you to apply the coordinate transform to the weak formulation, rather than to the classical formulation.


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