3
$\begingroup$

A lot of texts derive the variational form of a PDE as follows. First, life begins with a conservation law for the field $q$:

$$\partial_t \int_\omega G(q)\;dx + \int_{\partial\omega} F(q, \nabla q, \ldots)\cdot\nu\;ds = \int_\omega f\;dx$$

for all control volumes $\omega$, where $\nu$ is the unit outward normal vector to $\partial\omega$, $F$ is the flux function, $G$ takes $q$ to the density of some extensive quantity, and $f$ is a source term. We then apply the divergence theorem (assuming things are nice enough) to arrive at a strong form

$$\partial_t G(q) + \nabla\cdot F(q, \nabla q, \ldots) = f.$$

Then we multiply everything by a test function $v$ and push the divergence over onto $v$ using the usual tricks. So the argument goes conservation law $\rightarrow$ strong form $\rightarrow$ weak or variational form. I find the indirectness kind of annoying. Can we go directly from a conservation law to a variational form instead?

You could imagine that, for any $u$,

$$\int_\omega u\;dx = \int_\Omega u\cdot \mathbf 1_\omega\;dx$$

where $\mathbf 1_\omega$ is the indicator function of the set $\omega$. Likewise, assuming that the boundary of $\omega$ is nice enough, for some reasonable class of vector fields $F$, we can also say that

$$\int_{\partial\omega} F\cdot\nu\;ds = -\int_\Omega F\cdot\nabla\mathbf 1_\omega\;dx$$

in a distributional sense. Substituting these two relations into our original conservation law, we can then take arbitrary scalar multiples and sums to arrive at a variational form

$$\int_\Omega\left\{\partial_tG(q)\cdot v - F(q, \nabla q, \ldots)\cdot\nabla v - f\cdot v\right\}\,dx = \ldots$$

in a distributional sense for any simple function $v$, i.e. $v = \sum_i\alpha_i\mathbf{1}_{\omega_i}$ for some finite collection of control volumes $\{\omega_i\}$. (I've left off some boundary terms for brevity.) We can then take a limit as $v$ approaches a function in, say, $H^1(\Omega)$ or whatever space of test functions is most appropriate.

It's been a few years since I had measure theory or distribution theory. Can this argument be made rigorous, without an intermediate assumption that the PDE has a strong solution? I'd be perfectly happy if it worked for some common problem or class of problems; for example, the generalized Poisson equation where the conductivity coefficient has a jump discontinuity across some smooth hypersurface. I don't know that doing so has much practical use, I just find it more satisfying. Virtually every book I have on Galerkin methods or PDE theory either goes the conservation / strong / variational route or just starts with variational forms and ignores their derivation entirely.

$\endgroup$

1 Answer 1

2
$\begingroup$

Alright guess I'll have to try and do it myself then. Assumptions: $\Omega$ is a nice enough compact domain that $C^\infty(\Omega)$ is dense in $H^1(\Omega)$, and $u$ is a solution of the conservation form of the diffusion equation -- for all smooth control volumes $\omega$,

$$-\int_{\partial\omega} k\nabla u\cdot\nu\;ds = \int_\omega f\; dx$$

where $\nu$ is the unit outward normal vector, $k$ is the (strictly positive) diffusion coefficient, and $f$ is in $L^2(\Omega)$. We'll take the boundary conditions to be $u|_{\partial\Omega} = 0$ for simplicity; adding an extra term takes care of the Neumann or Robin cases. Assume additionally that $u$ is in $H_0^1(\Omega)$. Let $v$ be an arbitrary smooth function that vanishes on $\partial\Omega$. Using the smooth coarea formula, we can write

$$\int_\Omega k\nabla u\cdot\nabla v\;dx = \int_\Omega k\nabla u\cdot \frac{\nabla v}{|\nabla v|}|\nabla v|\; dx = \int_{-\infty}^\infty\int_{\partial\{x: v(x) \ge t\}}k\nabla u\cdot\frac{\nabla v}{|\nabla v|}\;ds\;dt\ldots $$

but we also know that, if $t$ is a regular value of $v$, the unit outward normal to the hypersurface $v^{-1}(t)$ is equal to $-\nabla v/|\nabla v|$, hence

$$\ldots = -\int_{-\infty}^\infty\int_{\partial\{x: v(x) \ge t\}}k\nabla u\cdot\nu\;ds\;dt = \int_{-\infty}^\infty\int_{\{x: v(x) \ge t\}}f\;dx\;dt\ldots $$

The latter equality is because of our assumption that $u$ is a solution of the conservation form of the equations and that $v$ is smooth + Sard's theorem. Finally the latter integral is just a rearrangement of the integral

$$\ldots = \int_\Omega f\,v\;dx$$

We can then pass to the limit for $v$ an arbitrary element of $H_0^1(\Omega)$ to say that $u$ is a solution of the variational problem

$$\int_\Omega k\nabla u\cdot\nabla v\;dx = \int_\Omega fv\; dx$$

for all $v$, QED.


I thought this would directly involve some distribution theory in order to be able to make statements like the gradient of an indicator function being a surface measure. But the coarea formula lets you shove all that under the hood. The same idea (+ some extra terms for initial conditions) seems like it should work for more problems as long as they can be expressed as a divergence in spacetime.

Please let me know if there are any mistakes or holes in the argument.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.