I consider the domain $$ A:=\left\{(x,y)\in\mathbb{R}^2:\lvert x-x_0\rvert < d_1, \lvert y-y_0\rvert < d_2\right\} $$ with $d_1,d_2>0, (x_0,y_0)\in\mathbb{R}^2$ and a function $a\in C(A)$. My question is, if $a$ has an antidervative.

What I know is: $u$ is continious on $A$ and therefore differentiable on $A$. And $A$ is simple connected (because $A$ is a rectangle.)

Does this already imply, that $a$ has an antiderviative?

  • 3
    $\begingroup$ What would an "antiderivative" mean in that situation? $\frac{\partial F}{\partial x} = a$, or $\frac{\partial F}{\partial y} = a$? $\endgroup$ – Daniel Fischer Oct 17 '13 at 16:02
  • $\begingroup$ I do not know, i refer to math.stackexchange.com/questions/528615/… where I do not know if it's right in general and if the antidervative A exists in particular. $\endgroup$ – math12 Oct 17 '13 at 16:04
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    $\begingroup$ You have that (such a function exists; $$F(x,y) = \int_{x_0}^x a(\xi,y)\,d\xi$$ does the job). But that's not usually called an "antiderivative". $\endgroup$ – Daniel Fischer Oct 17 '13 at 16:14

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