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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.

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up vote 5 down vote accepted

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.

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Right - this is Cartan–Kähler theory? (I'm annoyed with myself because despite having put it into this form I wasn't aware of this aspect of the theory.) Intuitively, then, this arises because one can apply $\partial_m$ to the original equations and then commute it past the $\partial_i \partial_j$ operators to generate new equations involving derivatives of $A_ij$ and $B^i_{jk}$ - makes sense. Typically, then, one uses this to obtain new equations involving $f$ and its derivatives, and then if one can eliminate them one obtains an integrability condition involving $A, B$? – Sharkos Aug 6 '12 at 14:28
@Sharkos: You don't need Cartan-Kähler theory for this, just ODE in the form of the Frobenius Theorem is enough. Your description of the practical consequence is correct; using the given equations to compare $\partial_if_{jk}$ with $\partial_jf_{ik}$ could, a priori lead to new relations among $f,\partial_1f,\ldots,\partial_nf$, but, if the $A$ and $B$ functions are such that these relations are identities, then the Frobenius Theorem says that there is a solution to the equations with $f,\partial_1f,\ldots,\partial_nf$ specified arbitrarily at any given point. – Robert Bryant Aug 6 '12 at 14:47

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