A confusion about null Lagrangian in Evans‘s PDE textbook. Assume the smooth Lagrangian function：
$$ L: M^{m\times n}\times \mathbb{R}^m\times \overline{U}\to\mathbb{R}$$
is given.
The function L is called a null Lagrangian if the system of Euler-Lagrange equations
$$ -\sum_{i=1}^{n}(L_{p_i}^k(Du,u,x))_{x_i}+L_{z^k}(Du,u,x)=0\qquad in\quad U，k=1,\cdots,m.$$
is automatically solved by all smooth functions $u: U\to\mathbb{R}^m$.
And the author says that “In the scalar case that $m=1$ the only null Lagrangians are the boring examples where $L$ is linear in the variable $p$.”
However, I don’t know how to prove it. It seems not obvious at least for me. I do hope someone can help me.
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Maybe I should write the problem in a more simple form. (If you don’t want to read the above which seems to bring some notation nightmare, you can just read the following problem)
Suppose that $f=f(p,z,x): \mathbb{R}\times \mathbb{R}\times I\to\mathbb{R}$ is a smooth function, where $I$ is an interval. If
$$\frac{d}{dx}D_1f(u',u,x)=D_2f(u',u,x)$$
for any smooth function $u=u(x)\in C^{\infty}(I;\mathbb{R})$, then $f=kp$ for some constant
$k$.
 A: You can construct (For example, let $u$ be a polynomial multiplied by a cutoff function centered at $x_0$) a smooth function $u$ with
$$ u(x_0) = A \qquad \partial_{x_i}u(x_0)= B_i \qquad \partial^2_{x_{ij}}\,u(x_0) = C_{ij}$$
at an interior point $x_0$ of $U$. Now plug this $u$ into your Euler-Lagrangian and evaluate it at $x_0$,
$$ - \sum^n_{i,j=1}(\partial_{p_i p_j} L(B,A,x_0))(\partial^2_{x_{ij}}\,u(x_0)) - \sum^n_{i=1}(\partial_{p_i z} L(B,A,x_0))(\partial_{x_{i}}\,u(x_0)) - \sum^n_{i=1}\partial_{p_i x_i} L(B,A,x_0) \\ + \partial_{z}L(B,A,x_0) = 0$$
Since $A, B, C$ are arbitrary we see $$\partial_{p_i p_j}L(B,A,x_0) = \partial_{p_iz}L(B,A,x_0) = 0$$
So far we have for some smooth function $f_i$ and $g$
$$L(p,z,x) = g(z,x) + \sum^n_{i=1}f_i(x)p_i$$
Now plug this form into your Euler-Lagrangian again and find $\partial_z g(z,x) = \partial_{x_i}f_i(x) = 0$.
So In the end you have $L$ of this form
$$L(p,z,x) = h(x) + z(\text{div} f_i)(x) +\sum^n_{i=1} f_i(x) p_i$$
I think it cannot be simplified further. Probably Evans means $L(p,z,x)$ is affine in $p$.
