Prove that the Lagrangian Density $\mathscr{L}$, which generates a given set of Euler-Lagrange equations, is not unique [duplicate]

Prove that the Lagrangian Density $\mathscr{L}$, which generates a given set of Euler-Lagrange equations, is not unique.

Hint 1: Adding a divergence to $\mathscr{L}$ does not alter the Euler-Lagrange equation.

Attempt: Let $\mathscr{L´}=\mathscr{L}+\sum_{k}\frac{\partial f_k}{\partial x_k}$

Where $$\mathscr{L}=\mathscr{L}\big(x_k, \varphi_j, \frac{\partial \varphi_k}{\partial x_k}\big)$$ $$f_k=f_k(\varphi_j)$$ $j=1,...,m$ indexes the dependent field variables.

$k=1,...,n$ indexes the independent variables.

Hint 2: Then prove that: $$\frac{\delta \mathscr{L´}}{\delta \varphi_j}=\frac{\delta \mathscr{L}}{\delta \varphi_j}$$

Attempt: Now we define $$\frac{\delta \mathscr{L}}{\delta \varphi_j}=\frac{\partial \mathscr{L}}{\partial \varphi_j}-\sum_{l} \frac{\partial}{\partial x_l} \frac{\partial \mathscr{L}}{\partial (\partial \varphi_j/\partial x_l)}$$

But i can´t keep following the hints to get the proof done:

marked as duplicate by ZeroTheHero, Kyle Kanos, Qmechanic♦ lagrangian-formalism StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 9 '17 at 2:09

Recall where the Euler Lagrange equations come from: extremizing the action. So let's look at what adding a divergence will do to the action, which, if you recall, is defined as an integral over our spacetime manifold, $\mathcal M$ as $$S[\varphi] =\int_{\mathcal M }\! \mathrm d^dr \ \mathcal L[\varphi,\nabla\varphi]$$ We'll add a divergence to the Lagrange density by letting $\mathcal L\rightarrow\mathcal L + \nabla\cdot f$, so our new action is: \begin{align} S'[\varphi] &=\int_\mathcal M\! \mathrm d^dr \ (\mathcal L+\nabla\cdot f)\\ &=S[\varphi]+\oint_{\partial\mathcal M}\mathrm da\cdot f \end{align} where we have used the amazing power of Gauss' theorem to change our $4$-dimensional volume integral over spacetime, to a $3$-dimensional integral over the boundary of spacetime, $\partial \mathcal M$. Now we demand that $f$ be well-behaved, in the sense that it vanishes on the boundary. We see then that the action is unchanged, so it will have the same extrema as before, and hence the same Euler-Lagrange equations.