I find myself staring blankly at a system of PDEs in $n$ dimensions which has "one equation per component" of the Hessian of the unknown function - that is, it specifies the Hessian in terms of the derivative and the value of the function:


$\qquad \partial\_i \partial\_j f + B\_{ij}^{\ \ \ k} \partial\_k f + A\_{ij} f = 0$


Here, both $A\_{ij}(x)$ and $B\_{ij}^{\ \ \ k}(x)$ are symmetric in $i,j$ and there is an implicit $k$ summation - my question is simply this:


* Are there further integrability
conditions, or is there generically a
solution for $f(x)$ given e.g. its value
and gradient at the origin?


Apologies if this is trivial, but I've got a complete mental block.



Integrability conditions for 'componentwise' systems of linear PDEs
The answer is, in general, yes. There are further integrability conditions. What you should be considering is the system of $1$-forms on $\mathbb{R}^n\times \mathbb{R}\times \mathbb{R}^n$ (with coordinates $(x^i,u,p\_i)$ given by
$$
\begin{align}
\theta &= du - p\_i\ dx^i\\\
\theta\_i &= dp\_i + (B^k\_{ij}p\_k + A\_{ij}u)\ dx^j
\end{align}
$$
If this system is Frobenius, i.e., if all of the $d\theta\_i$ are in the algebraic ideal generated by $\{\theta,\theta\_1,\ldots,\theta\_n\}$, then you will have what you want. Otherwise, the coefficients of these 2-forms modulo $\{\theta,\theta\_1,\ldots,\theta\_n\}$ will give you more equations on $u,p\_1,\ldots,p\_n$ that have to be satisfied.


