Existence of first order PDEs?

Question

It seems the follow theorem is classical, but I don't know how to proof it: For $x\in\Omega\subset R^n$, where $\Omega$ is a domain with smooth boundary, consider the system of PDEs: \begin{equation}\label{eq:1} \frac{\partial f(x)}{\partial x_i}=f_i(x),\quad i=1,2,\ldots,n, f_i\in C^\infty(\Omega),\tag{1} \end{equation} if the compitable condition $$ \frac{\partial f_i}{\partial x_j}=\frac{\partial f_j}{\partial x_i},\quad\forall i,j=1,2,\ldots,n,\forall x\in\Omega $$ is satisfied, then there exists $f\in C^\infty(\Omega)$, which solves \eqref{eq:1}.

So, can anyone give a detailed proof? OR, at least, provid some exactly references?

Answer 1

I assume $f_i$ is a function unrelated to $f$?

If we let $v$ be the one-form dual to the vector field $f_i$, you are asking, in other words, for a solution to $df = v$, where $dv = 0$. Under the additional assumption that $\Omega$ is contractible, existence of this $f$ is guaranteed since the first deRham cohomology group of the $n$-sphere is trivial. (For a counterexample for general $\Omega$, consider the vector field $(-y,x)/(x^2+y^2)$ in an annulus centered at the origin in $\mathbb{R}^2$.)

Answer 2

Think in $\Bbb R^3$ for a minute. The hypothesis

$\dfrac{\partial f}{\partial x_i}(x) = f_i(x) \tag{1}$

says that the vector field

$\vec F(x) = (f_1(x), f_2(x), f_3(x))^T \tag{2}$

is the gradient of the function $f(x)$:

$\vec F(x) = \nabla f(x); \tag{3}$

as such,

$\nabla \times \vec F = 0. \tag{4}$

As is well-known, this situation is reversible: for $\vec F(x)$ satisfying (4), there is a scalar function $f(x)$ such that (3) applies, at least locally. And if one carefully examines the relations

$\dfrac{\partial f_i}{\partial x_j} = \dfrac{\partial f_j}{\partial x_i}, \; \; 1 \le i, j \le 3, \tag{5}$

one sees that (4) is exactly what they say; so that resolves the situation for $n = 3$, again, locally at least. For $n > 3$, it is perhaps easiest to revert to the language of differential forms; setting

$\omega = \sum_1^n f_i dx_i, \tag{6}$

we have

$d\omega = \sum_{1 \le i < j \le n} (\dfrac{\partial f_i}{\partial x_j} - \dfrac{\partial f_j}{\partial x_i}) dx_i \wedge dx_j = 0 \tag{7}$

this last equality holding by our hypothesis on the derivatives of the $f_i(x)$. In the parlance of differential forms, $\omega$ is said to be closed. It is substance of the Poincare lemma, which may be found at http://en.m.wikipedia.org/wiki/Closed_and_exact_differential_forms, that such forms, of no matter degree $p$ they may be, are also exact, that is such $\omega = d\alpha$ for some $p-1$ form $\alpha$, provided the domain $\Omega$ is simple enough; "star-shaped" is the criterion oft' applied, though the simpler notion convex will do to ensure such $\alpha$ exists, locally at least. In the present case, this means $\alpha$ is a function such that

$\dfrac{\partial \alpha}{\partial x_i} = f_i \tag{8}$

on appropriate $\Omega$. The linked citing explains the conditions which must be satisfied by $\Omega$ for the Poincare lemma to bind.

The preceeding presents, in broad outline, how the existence of solutions to the given equation may be established. Such are "funny" problems as far as PDEs go; for example it is not at all clear to me how issues such as boundary/initial data fit into the differential forms-based view of these matters; but it is certain that the domain $\Omega$ must be somewhat carefully specified if a solution is to be had. But then again, that's cohomology for you!

Hope this helps. Cheers,

and as ever,

Fiat Lux!!!

Answer 3

If the domain is convex (WLOG suppose it contains the origin), it is then pretty easy to show that $\displaystyle f(x)=\sum_i\int_0^x f_i\,dx^i\,,$ where we are integrating over a straight line connecting the origin to $x$, solves the equations. It's really just the fundamental theorem of calculus.