I would like to minimize: $$J = \int_{\Omega} \|\nabla u - \nabla g\|^2 + \lambda \|\frac{\partial u}{\partial t} + \nabla u.v||^2 ~\text{dx dy dt}$$ where $u(x,y,t)$ is the unknown function, the integral is over a fixed spatio-temporal domain $\Omega$, $\nabla$ is the spatial gradient, $v(x,y,t)$ is a velocity (vector) field. $g(x,y,t)$ and $v(x,y,t)$ are known. $\lambda$ is a constant.

Computing and cancelling the variation of $J$ using a neighborhood $u+\epsilon h$ of $u$, I obtain:

$$ ``\frac{\partial J}{\partial u}" = \int_{xyt} -\Delta (u-g).h + \lambda (\frac{\partial u}{\partial t} + \nabla u.v)(\frac{\partial h}{\partial t} + \nabla h.v) = 0$$

If only the term $\Delta (u-g).h$ was present, I could directly say that if the integral is zero for all $h$, then $\Delta (u-g)$ is zero locally. I'd obtain the standard Poisson equation.

However, the addition of the other term makes me wonder how I can turn this expression into a local PDE to solve via finite differences. Any idea ?


  • $\begingroup$ Your derivative dJ/du is missing a factor of two. It doesn't matter if you set the result to zero, but it needs to be there anyway to be correct. $\endgroup$ Dec 14, 2013 at 21:22

1 Answer 1


You already have a variational formulation of your problem. You can get the strong form if you integrate back by parts so that you get a formula of the form $$ \int (something) h \; dx \; dt = 0. $$ (In fact, you will probably get multiple integrals to take into account boundary and initial terms coming out of the integration by parts.) The strong form is then $$ something = 0. $$

  • $\begingroup$ thanks ; would that $something$ be $\Delta(u-g) + v.\nabla(\nabla u . v) + \frac{\partial^2 u}{\partial t} + 2 \frac{\partial}{\partial t}(\nabla u . v)$ ? $\endgroup$
    – WhitAngl
    Dec 13, 2013 at 23:47
  • $\begingroup$ and for the boundary terms, when deriving the standard Poisson equation, it always reads like "we use a function space for $h$ that is zero at infinity". Can't something similar be done in this case to avoid the spurious terms in the integration by parts ? $\endgroup$
    – WhitAngl
    Dec 13, 2013 at 23:49
  • $\begingroup$ (also, I forgot the $\lambda$ term of course, in the second part) $\endgroup$
    – WhitAngl
    Dec 13, 2013 at 23:55
  • $\begingroup$ The "something" looks about right. As for boundary conditions: If you know the boundary values of $u$ (Dirichlet values), then $h$ is zero there. That's because $h$ is a variation of $u$. But if you don't have boundary values for $u$ on a finite domain, then you can't omit these terms. Likewise, from the time integration you will get a final time term that you can't get rid of. $\endgroup$ Dec 14, 2013 at 21:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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