# variational formulation

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 ?

Thanks!

• 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. Commented Dec 14, 2013 at 21:22

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 $$something = 0.$$
• thanks ; would that $something$ be $\Delta(u-g) + v.\nabla(\nabla u . v) + \frac{\partial^2 u}{\partial t} + 2 \frac{\partial}{\partial t}(\nabla u . v)$ ? Commented Dec 13, 2013 at 23:47
• 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 ? Commented Dec 13, 2013 at 23:49
• (also, I forgot the $\lambda$ term of course, in the second part) Commented Dec 13, 2013 at 23:55
• 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. Commented Dec 14, 2013 at 21:42