My book (Stewart's Essential Calculus) states Green's Theorem as follows:
Let $C$ be a positively oriented, piecewise-smooth, simple closed curve in the plane and let $D$ be the region bounded by $C$. If $P$ and $Q$ have continous partial derivatives on an open region that contains $D$, then $$\int_C P \,dx+Q\,dy =\iint_D \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \right)dA $$
My question is why are the partial derivatives required to be continous?
I have seen a few examples of Greens Theorem not working when the partial derivatives do not exist but cannot seem to find one when the partial derivatives have a discontinuity.
My first idea is that the requirement of continuity insures that no "kinks" in the vector field occur but then I found Darboux's theorem which seems to insure that discontinous derivatives will never have this issue.
I expect a technical understanding of the details of Green's Theorem to be out of my reach currently (my book doesn't even include a full proof) but is there any intuitive justification of the continuity requirement on the derivatives?
Best Answer
Continuity of the partial derivatives is strong sufficient condition.
Consider the simplest proof of Green's theorem for a rectangular region $D = [a,b] \times [c,d]$. One step is to show that (with $P_y := \frac{\partial P}{\partial y}$)
$$\tag{*}\oint_C P(x,y) \, dx = -\int_D P_y(x,y) \, dA,$$
which reduces to
$$\int_a^b P(x,c) \, dx - \int_a^b P(x,d) \, dx = -\int_a^b \int_c^d P_y(x,y) \, dx \, dy = -\int_a^b \left(\int_c^d P_y(x,y) \, dy\right) \, dx$$
We have to be sure that we can evaluate the double integral as an iterated single integral in any order and we have to apply the fundamental theorem of calculus to obtain
$$\int_c^d P_y(x,y) \, dy = P(x,d) - P(x,c),$$
in showing that (*) is true.
All of these steps can be justified if the partial derivatives are continuous.
However, there is a more general form of Green's theorem for Lebesgue integrals which requires only that $P$ and $Q$ be absolutely continuous. In this case, the partial derivatives exist almost everywhere, are integrable, and the fundamental theorem still holds even though the derivatives need not be continuous everywhere.
For example, observe that Green's theorem still holds for $D = [0,1] \times [0,1]$ and
$$P(x,y) = 0 \\ Q(x,y) = \begin{cases}yx^2\sin(1/x), \, \, x \neq 0 \\ 0, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, x = 0 \end{cases}$$
even though $\frac{\partial Q}{\partial x}$ is not continuous on $\{(x,y): x = 0, 0 < y \leqslant 1 \}$.