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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?

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1 Answer 1

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In general, not every differential equation is coming from a Lagrangian, see Douglas' theorem on that matter.

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.

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