# Deriving Lagrangian for common a class of PDEs

I am interested in constructing a Lagrangian for a PDE of type

$$u_t(t,x) - F(u,u_t,u_x)=0$$

such that for some functional $$\quad I[u] = \int D[u]\mathcal{L}[u,u_t,u_x]$$ its associated Euler Lagrange Equations satisfy the above PDE

$$\frac{\partial \mathcal{L}}{\partial u} - \partial_t\frac{\partial \mathcal{L}}{\partial u_t} - \partial_x\frac{\partial \mathcal{L}}{\partial u_x} = u_t(t,x) - F(u,u_t,u_x)=0.$$

Really I am trying to understand which class of PDE's are the kind that are the Euler Lagrange Equations associtated with some extremal functional $$I[u]$$.

Is there a way to work backwards from the PDE to $$\mathcal{L}[u,u_t,u_x]$$? For example, if $$F(u,u_t,u_x) = u_x$$, how would you do it, if possible?

So even if we assume $$F(u,u_t,u_x)=F(u,u_t)$$ you might not be able to find a suitable Lagrangian for your problem. On that note, note that $$\tilde{F}(u,u_t,u_x)=\partial_t u(x,t)-F(u,u_t,u_x)$$ is again a function of the same form as $$F$$ and therefore you are essentially asking if every first-order PDE in $$(x,t)$$ $$\tilde{F}(u,u_t,u_x)=0$$ is coming from a Lagrangian. And the answer is no. Giving some conditions or having a more explicit form (or even estimates, for that matter) of $$F$$ might help.