Solve the system of differential equations: $\partial_i E_j + \partial_j E_i = 2C_{ij} ~(1\le i,j\le n)$ Let $E_i: \mathbb R^n \to \mathbb R~(1\le i\le n)$ be differentiable functions, and let $C_{ij}=C_{ji}$ be constants. Denote by $\partial_i$ the partial derivative $\frac{\partial}{\partial x_i}$. My question is to find solutions of the following equations:
$$
\partial_iE_j + \partial_j E_i = 2C_{ij}
$$
I notice that there is a good particular solution:
$$
E_i(x_1,\dots,x_n)=\sum_k C_{ik}x_k + C_i 
$$
where $C_i$ are constants too. I was wondering if this is the only solution.
 A: Let $F_i = E_i - \sum_j C_{ij}x_j$. Then
$$ \partial_i F_j + \partial_j F_i = \partial_i E_j + \partial_j E_i - (C_{ji}+C_{ij}) = 0, $$
so given your particular solution, it suffices to find the solutions to the system
$$ \partial_i F_j + \partial_j F_i = 0. \tag{*} $$
Firstly, the diagonal equations give $\partial_i F_i = 0$, so $F_i$ does not depend on $x_i$. The trick now is to look at second derivatives, and use their commutativity and $(*)$:
$$ \partial_k \partial_j F_i = -\partial_k \partial_i F_j = -\partial_i \partial_k F_j = \partial_i \partial_j F_k = \partial_j \partial_i F_k = -\partial_j \partial_k F_i. $$
So $\partial_k \partial_j F_i = 0$. Since this is true for any $k$, we must certainly have $ \partial_j F_i = A_{ij} $ independent of all the variables, with $A_{ij}=-A_{ji}$ from $(*)$. And then $F_i = \sum_j A_{ij} x_j + B_i $ since we already know all the partial derivatives of $F_i$ (we could eliminate one at a time, but it should be clear what's going on here if you try the first couple).
Hence the general solution is
$$ E_i = \sum_{j} (C_{ij}+A_{ij})x_j + B_i, $$
where $A_{ij}+A_{ji}=0$.
