# Space of solutions to a system of first-order PDEs

I would like to know what is known (both explanations and references) about the spaces of smooth solutions to linear systems of PDEs of the following form:

Let $g_{1},...,g_{n}$ be smooth functions on $\mathbb{R}^{n}$ with the integrability condition $\partial{g_{i}}/\partial{x^{j}}=\partial{g_{j}}/\partial{x^{i}}$ and consider the space of smooth functions $f$ on $\mathbb{R}^{n}$ satisfying $\partial{f}/\partial{x^{i}}=fg_{i}$ for all $i$.

Similarly for the $g_{i}$ and $f$ being holomorphic on $\mathbb{C}^{n}$, and replacing $\mathbb{R}^{n}, \mathbb{C}^{n}$ with open contractible subsets.

My hope is that the answer is there is a unique solution, up to scaling.

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You can rewrite your system as $\vec{\nabla} \ln f = \vec{g}$... – anon Jul 29 '11 at 17:20
Does that help? – Pascal Jul 29 '11 at 17:22
Also, what if $f$ has zeroes and what if I'm interested in complex values? I would be happy to say that there exists a contractible open cover of my space over which there is a unique solution on each covering set, if that would help solve the problem about branches of the logarithm. – Pascal Jul 29 '11 at 17:31
Note that $f$ can only have zeros where $\vec{g}$ has singularities. Remember that $\vec{g}$ is the given and $f$ the unknown, so that when looking at the latter in terms of the former we are working with exponentials instead of logarithms and therefore don't have to worry about branches. – anon Jul 29 '11 at 17:38

You can rewrite the system of PDEs as $\vec{\nabla} \ln f = \vec{g}$. If the vector field $\vec{g}$ has a potential function (which I believe is true if and only if its Jacobian matrix is symmetric, but I don't recall the source of this fact off-hand), then we may denote the potential as $G$ and solve the system as $f=\exp G$, which is unique up to rescalings. (There may be a negative sign in the exponential depending on what definition of potential you're using.) On the other hand, if $\vec{g}$ doesn't have a potential then it isn't a gradient field and hence no $f$ exists.
As anon pointed your system is $\frac{\partial}{\partial x_i} \ln(f)= g_i$ then by Froebenius theorem, the PDE system has a solution if and only if the $\nabla g$ is a symmetric matrix, which is true given your assumption that $\frac{\partial}{\partial x_i} g_j= \frac{\partial}{\partial x_j} g_j$.