proof of closed differential forms on a simple connected domain

lemma 1.

Let ${\gamma }_0$ and ${\gamma }_1$ be two regular homotopic curves in $D$ withthe same end-points.Let $\sigma :[0,1]\times [0,1]\to D$ be the homotopy between ${\gamma }_0$ and ${\gamma }_1$ i.e.

$$\sigma (0,t)={\gamma }_0(t),\sigma (1,t)={\gamma }_1(t).$$

Notice that we may (and shall) suppose that $\sigma$ is regular too. In fact $\sigma ([0,1]\times [0,1])$ is a compact subset of $D$. Being $D$ open this compact set has positive distance from the boundary $\partial D$. So we couldregularize $\sigma$ by mollification leaving its image in $D$.

Let $\omega (x,y)=a(x,y)dx+b(x,y)dy$ be our closed differential form and let $\sigma (s,t)=(x(s,t),y(s,t))$.Define

$$F(s)={\int }_0^{1}a(x(s,t),y(s,t))x_t(s,t)+b(x(s,t),y(s,t))y_t(s,t)dt;$$

we only have to prove that $F(1)=F(0)$.

We have

$$F^{\prime }(s)=\frac{d}{ds}{\int }_0^{1}a{x}_t+b{y}_{t}dt$$

$$={\int }_0^1{a}_x{x}_s{x}_t+a_y{y}_s{x}_t+a{x}_{ts}+b_x{x}_s{y}_t+b_y{y}_s{y}_t+b{y}_{ts}dt.$$

Notice now that being $a_y=b_x$ we have

$$\frac{d}{dt}\left[a{x}_s+b{y}_s\right]=a_x{x}_t{x}_s+a_y{y}_t{x}_s+a{x}_{st}+b_x{x}_t{y}_s+b_y{y}_t{y}_s+b{y}_{st}$$

$$=a_x{x}_s{x}_t+b_x{x}_s{y}_t+a{x}_{ts}+a_y{y}_s{x}_t+b_y{y}_s{y}_t+b{y}_{ts}$$

hence

$$F^{\prime }(s)={\int }_0^1\frac{d}{dt}\left[a{x}_s+b{y}_s\right]𝑑t={\left[a{x}_s+b{y}_s\right]}_0^1.$$

Notice, however, that $\sigma (s,0)$ and $\sigma (s,1)$ are constant hence $x_s=0$ and $y_s=0$ for $t=0,1$. So $F^{\prime }(s)=0$ for all $s$ and $F(1)=F(0)$.∎

Lemma 2.

Let us fix a point $(x_0,y_0)\in D$ and define a function $F:D\to ℝ$ by letting $F(x,y)$ be the integral of $\omega$ on any curve joining $(x_0,y_0)$ with $(x,y)$. The hypothesis assures that $F$ is well defined.Let $\omega =a(x,y)dx+b(x,y)dy$. We only have to prove that $\partial F/\partial x=a$ and $\partial F/\partial y=b$.

Let $(x,y)\in D$ and suppose that $h\in ℝ$ is so small that for all $t\in [0,h]$ also $(x+t,y)\in D$. Consider the increment $F(x+h,y)-F(x,y)$.From the definition of $F$ we know that $F(x+h,y)$ is equal to the integral of $\omega$ on a curve which starts from $(x_0,y_0)$ goes to $(x,y)$ and then goes to $(x+h,y)$ along the straight segment $(x+t,y)$ with $t\in [0,h]$.So we understand that

$$F(x+h,y)-F(x,y)={\int }_0^{h}a(x+t,y)𝑑t.$$

For the integral mean value theorem we know that the last integral is equal to $ha(x+\xi ,y)$ for some $\xi \in [0,h]$ and hence letting $h\to 0$ we have

$$\frac{F(x+h,y)-F(x,y)}{h}=a(x+\xi ,y)\to a(x,y)\mathit{\hspace{1em}}h\to 0$$

that is $\partial F(x,y)/\partial x=a(x,y)$. With a similar argument (exchange $x$ with $y$) we prove that also $\partial F/\partial y=b(x,y)$.∎

Theorem.

Just notice that if $D$ is simply connected, then any two curves in $D$ with the same end points are homotopic. Hence we can apply Lemma 1 and then Lemma 2 to obtain the desired result.∎