A variational problem with a lagrangian , what is the lagrangian? given the 2 PDE
$$ \Delta u-au_{tt}+u_{t}=0$$
and $$ \Delta u + Du*Df=0 $$
here $ \Delta u $ is the Laplacian $ Du= grad(u) $ is the gradient and * means scalar product $u_{t} = \frac{\partial u}{\partial t}$
my doubt is what term should i include to get the linear part of the equations $ u_{t} $ and $Du*Df $ from the Euler Lagrange equations in $ R^{n} $ thanks
 A: Well, for the second case, the Lagrangian you want is
$$
\Lambda_D(u) = \tfrac12 \int_D\  e^{f(x)}\ |\nabla u|^2\ \mathrm{d}x
$$
where $D$ is an appropriate domain in $\mathbb{R}^n$.  The first case is similar (if you are assuming that $a$ is a constant), for then, it will just be
$$
\Lambda_D(u) = \tfrac12 \int_{D'}\  e^{t}\ \bigl(|\nabla u|^2-a|u_t|^2\bigr)\ \mathrm{d}x\mathrm{d}t
$$
where $D'$ is an appropriate domain in $\mathbb{R}^n\times \mathbb{R}$.  If $a$ is not a constant, there are conditions.
A: Here we will assume that the definition of the Laplacian is $\Delta=\vec{\nabla}^2$, and that $a$ is a non-zero constant. 
I) The Lagrangian density

$$\tag{1a} {\cal L}_1~=~ \frac{e^{-t/a}}{2}\left(a\dot{u}^2 - (\vec{\nabla} u)^2\right)$$ 

has Euler-Lagrange equation 
$$0~\stackrel{(1a)}{=}~\frac{\partial {\cal L}_1}{\partial u}
~\stackrel{EL}{=}~\frac{d}{dt}\frac{\partial {\cal L}_1}{\partial \dot{u}}
+\vec{\nabla}\cdot\frac{\partial {\cal L}_1}{\partial \vec{\nabla}u}$$
$$\tag{1b} ~\stackrel{(1a)}{=}~ \frac{d}{dt}\left(e^{-t/a}a\dot{u}\right)  - \vec{\nabla}\cdot \left(e^{-t/a}\vec{\nabla}u\right)
~=~e^{-t/a}\left(a\ddot{u}-\dot{u}-\Delta u\right)  $$
equal to OP's first PDE

$$\tag{1c} a\ddot{u}-\dot{u}-\Delta u~\stackrel{(1b)}{=}~0.$$

II) The Lagrangian density 

$$\tag{2a} {\cal L}_2~=~ \frac{e^f}{2} (\vec{\nabla} u)^2$$ 

has Euler-Lagrange equation
$$ 0~\stackrel{(2a)}{=}~\frac{\partial {\cal L}_2}{\partial u}
~\stackrel{EL}{=}~\vec{\nabla}\cdot\frac{\partial {\cal L}_2}{\partial \vec{\nabla}u}$$
$$\tag{2b}~\stackrel{(2a)}{=}~\vec{\nabla}\cdot \left(e^f\vec{\nabla}u\right)
~=~e^f\left(\Delta u +\vec{\nabla}f \cdot \vec{\nabla}u\right) $$
equal to OP's second PDE

$$\tag{2c} \Delta u +\vec{\nabla}f \cdot \vec{\nabla}u~\stackrel{(2b)}{=}~0.$$

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[Note added: OP's question was originally cross-posted to mathoverflow and migrated back to Math.SE, so that OP's question appeared in duplicates on Math.SE. This answer (which was composed without knowing of the duplicate entry) originally appeared under the original Math.SE question, but is now moved here to collect all answers in one place.] 
