# Question for divergence theorem

In general, the Divergence Theorem:

for a bounded domain $\Omega \subset \mathbb{R}^n$ with $C^1$ boundary $\partial \Omega$ and for a vector field $F \in C^1 (\overline \Omega),$ $$\int_\Omega \nabla \cdot F dx= \int_{\partial\Omega} F \cdot n dS$$ where $n$ is the unit outward normal to $\partial \Omega$.

I was wondering if the theorem holds for vector fields $F \in C^1(\Omega)\cap C(\overline \Omega).$ If this is true, for $u\in C^2(\Omega)\cap C^1(\overline \Omega),$ Green’s identities hold and Green’s representation formula for such $u$ (not in $C^2(\overline \Omega)$.

Please let me know if it is true and/or any reference for this question. Thanks in advance!

The conditions under which the divergence theorem holds can be much weaker than what you propose. We do not even need that the partial derivatives of the vector field be continuous on $\Omega$ -- bounded and Lebesgue integrable is sufficient.

There are even weaker forms. This paper by Bochner is a good starting point as a reference.

To see that continuity of the partial derivatives on the boundary is not necessary, consider the example (for the analogous Green's theorem) where $\Omega = [0,1]^2$ and

$$P(x,y) = 0 \\ Q(x,y) = \begin{cases}yx^2\sin(1/x), \, \, x \neq 0 \\ 0, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, x = 0 \end{cases}$$

In this case, we have

$$\iint_{\Omega} \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \right)dA = \int_{\partial \Omega} P \,dx+Q\,dy$$

even though $\frac{\partial Q}{\partial x}$ is not continuous at points in $\{(x,y): x = 0, 0 < y \leqslant 1 \}$.

• Thanks for your answer. Wold you give any specific book which state the desirable theorem? I cannot find suitable one. Jul 15 '18 at 0:59
• 0706: You're welcome. Let me see what I can find. Most books don't even prove the theorem or sketch the proof for special regions with the strongest assumptions about the continuity of the partial derivatives. Books like Spivak and Munkres on manifolds are more concerned with more general surfaces and higher dimensions and again assume the strongest conditions for the vector field. I believe Mathematical Analysis by Apostol, but only the first edition, may have what you are looking for.
– RRL
Jul 15 '18 at 1:08
• I have one more question. You said "The conditions under which the divergence theorem holds can be much weaker than what you propose". My condition is $F \in C^1(\Omega)\cap C(\overline \Omega).$ It seems not true that the fact the partial derivatives of the vector field be continuous in $\Omega$ implies they are bounded and Lebesgue integrable on $\Omega.$ Thus we need $F \in C^1(\Omega)\cap C(\overline \Omega)$ with the condition partial derivatives are bounded and Lebesgue integrable on $\Omega.$ Is it true? Jul 30 '18 at 3:41
• I guess the condition you mentioned is $F$ is bounded and Lebesgue integrable. Would you give the condition of $F$ more precisely under which the divergence theorem holds? Thanks in advance! Jul 30 '18 at 4:04
• Yes partial derivatives continuous in $\Omega$ is not enough to ensure they are bounded and integrable. $f(x) = 1/x$ is continuous in $(0,1)$ but not bounded and integrable. We certainly want the partial derivatives (or at least the combination as the divergence) to be integrable so that $\int_\Omega \nabla \cdot F$ exists -- note this does not require that they be bounded (e.g. $1/\sqrt{x}$ is integrable on $(0,1)$
– RRL
Jul 30 '18 at 5:21