# Derivation of weak form for variational problem

My question is about understanding the derivation of the weak form of a variational problem (to be used for the solution via the finite element method).

The problem is as follows (it is an image processing problem, $\mathbf{u}$ is a 2D vector field of displacements I is an image function (see Horn-Schunk optical flow algorithm)):

Minimize the following functional w.r.t. the vector field $\mathbf{u}$ $$E(\mathbf{u}) = \int_\Omega \left\|\mathbf{u}\cdot\nabla I + \frac{\partial I}{\partial t}\right\|^2dx + \lambda\int_\Omega \nabla \mathbf{u} : \nabla \mathbf{u} dx$$

(: denotes the Frobenius inner product http://en.wikipedia.org/wiki/Frobenius_inner_product#Frobenius_product)

The apparent weak form of the problem is given in the following form (http://code.google.com/p/debiosee/wiki/DemosOptiocFlowHornSchunck):

$$E(\mathbf{u}) = \int_\Omega \left(\mathbf{u}\cdot\nabla I + \frac{\partial I}{\partial t}\right) \left(\mathbf{v}\cdot\nabla I\right) dx + \lambda\int_\Omega\nabla\mathbf{u}:\nabla\mathbf{v}dx$$

Where $\mathbf{v}$ is the test function.

What I would like to know is if there is a direct way to come to this conclusion from the original problem or the way to go is to formulate the weak form of the Euler-Lagrange equation of the original functional?

Additionally I would like to know if any of you knew about a vector calculus textbook you could recommend for mastering these kinds of operations in their vector form without working out the derivation componentwise?

• I don't know any component-wise derivation, but the standard derivation from an objective functional ($E(\boldsymbol{u})$ you gave) to the weak form is using the fact that for the minimizer $\boldsymbol{u}$: $\displaystyle\lim_{\epsilon\to 0}\frac{d}{d\epsilon}E(\boldsymbol{u}+\epsilon\boldsymbol{v}) = 0$ for any $\boldsymbol{v}$ in the test space. The weak form should be a bilinear form instead of a functional, i.e., it should have two inputs, namely the weak form should be $B(\boldsymbol{u},\boldsymbol{v})$ in your case. Commented Feb 25, 2013 at 23:19
To answer your first equation, there is no direct way by looking at the functional to derive the variational problem. The only way I know is to use the method same as deriving Euler-Lagrange equation like you said. $\newcommand{\b}{\mathbf}$
Let $\b{v}\in V$. $V$ is the test space. Normally the difference between $V$ and the space $\b{u}$ lies in is the boundary condition. Since $\b{u} = \mathrm{argmin}_{\b{v}} E(\b{v})$, adding any perturbation would result the functional being bigger, we have for any $\b{v}\in V$: $$\lim_{\epsilon\to 0}\frac{d}{d\epsilon}E(\b{u}+\epsilon\b{v}) = 0$$ First compute: \begin{aligned} E(\b{u}+\epsilon\b{v}) &= \int_\Omega \left|(\b{u}+\epsilon\b{v})\cdot\nabla I + \frac{\partial I}{\partial t}\right|^2 + \lambda\int_\Omega \nabla (\b{u}+\epsilon\b{v}): \nabla (\b{u}+\epsilon\b{v}) \\ &= \int_\Omega \left\{(\b{u}+\epsilon\b{v})\cdot\nabla I + \frac{\partial I}{\partial t}\right\}^2 + \lambda\int_\Omega \nabla \b{u}: \nabla \b{u} + \lambda \epsilon^2\int_\Omega \nabla \b{v}: \nabla \b{v} + 2\lambda \epsilon\int_\Omega \nabla \b{u}: \nabla \b{v} \end{aligned} I am replace your notation of $\|\cdot\|$ by absolute value since it is a scalar inside. Taking derivative with respect to $\epsilon$: \begin{aligned} \frac{d}{d\epsilon}E(\b{u}+\epsilon\b{v}) &= 2\int_\Omega \left((\b{u}+\epsilon\b{v})\cdot\nabla I + \frac{\partial I}{\partial t}\right) \frac{d}{d\epsilon}\left((\b{u}+\epsilon\b{v})\cdot\nabla I + \frac{\partial I}{\partial t}\right) \\ &\quad + 2\lambda \epsilon\int_\Omega \nabla \b{v}: \nabla \b{v}+2\lambda\int_\Omega \nabla \b{u}: \nabla \b{v} \\ &= 2\int_\Omega \left((\b{u}+\epsilon\b{v})\cdot\nabla I + \frac{\partial I}{\partial t}\right) \left(\mathbf{v}\cdot\nabla I\right) + 2\lambda \epsilon\int_\Omega \nabla \b{v}: \nabla \b{v}+2\lambda\int_\Omega \nabla \b{u}: \nabla \b{v} \end{aligned} Letting $\epsilon \to 0$: $$0=\lim_{\epsilon \to 0}\frac{d}{d\epsilon}E(\b{u}+\epsilon\b{v}) = 2\int_\Omega \left(\b{u}\cdot\nabla I + \frac{\partial I}{\partial t}\right) \left(\mathbf{v}\cdot\nabla I\right) +2\lambda\int_\Omega \nabla \b{u}: \nabla \b{v}$$ And this gives us the variational problem: For any $\b{v}\in V$: $$\int_\Omega \left(\b{u}\cdot\nabla I \right) \left(\mathbf{v}\cdot\nabla I\right) +\lambda\int_\Omega \nabla \b{u}: \nabla \b{v} = \int_\Omega \frac{\partial I}{\partial t} \left(\mathbf{v}\cdot\nabla I\right)$$ This is in the form of $B(\b{u},\b{v}) = F(\b{v})$, which can be the bilinear form for finite element method.