0
$\begingroup$

Let $\phi$ and $\psi$ be two smooth functions on $\Omega$ open subset of $\mathbb{R}^2$. There exists a function $u=u(x,y)$ such that: $$\left\{\begin{array}{lll} \frac{\partial u}{\partial x}=\phi(x,y)\\ \frac{\partial u}{\partial y}=\psi(x,y) \end{array}\right.$$ if and only if $\frac{\partial \phi}{\partial y}=\frac{\partial \psi}{\partial x}$.

Where I can find the proof of this statement? Is it a obvious consecuence of some famous theorem (like implicit funtion)?

Many thank!

$\endgroup$
  • $\begingroup$ Yes, by the symmetry of second derivatives aka. Schwarz' theorem. $\endgroup$ – flawr May 30 '16 at 16:18
  • $\begingroup$ Thanks! It is clear that if $u$ is a solution, then $\partial_y \phi$ must be equal to $\partial_x\psi$. How can I prove the converse? $\endgroup$ – FUUNK1000 May 30 '16 at 17:05
  • $\begingroup$ Integrate those terms again, and do some detail work with the integration constants. $\endgroup$ – flawr May 30 '16 at 17:11
  • $\begingroup$ @flawr weird..... I always cite that as clairaut's $\endgroup$ – qbert May 30 '16 at 18:20
  • $\begingroup$ @qbert I probably depends on where you are / what language it is used. Among german speakers it is widely known as Satz von Schwarz. $\endgroup$ – flawr May 30 '16 at 20:21
0
$\begingroup$

Your statement is false in general, unless $\Omega$ is simply connected.

On the other hand, although you can give a more or less straightforward proof in this situation, it is a very particular case of the fact that the de Rham cohomology (which is a homotopy invariant) vanishes for contractible spaces together with Stokes' theorem.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.