# proof of closed differential forms on a simple connected domain

###### lemma 1.

Let ${\gamma}_{0}$ and ${\gamma}_{1}$ be two regular homotopic curves in $D$ with the 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 could regularize $\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

$${F}^{\prime}(s)={\int}_{0}^{1}\frac{d}{dt}\left[a{x}_{s}+b{y}_{s}\right]\mathit{d}t={\left[a{x}_{s}+b{y}_{s}\right]}_{0}^{1}.$$ |

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 $F:D\to \mathbb{R}$ 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 $h\in \mathbb{R}$ 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

$$F(x+h,y)-F(x,y)={\int}_{0}^{h}a(x+t,y)\mathit{d}t.$$ |

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. ∎

Title | proof of closed differential forms on a simple connected domain |
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Canonical name | ProofOfClosedDifferentialFormsOnASimpleConnectedDomain |

Date of creation | 2013-03-22 13:32:49 |

Last modified on | 2013-03-22 13:32:49 |

Owner | paolini (1187) |

Last modified by | paolini (1187) |

Numerical id | 9 |

Author | paolini (1187) |

Entry type | Proof |

Classification | msc 53-00 |

Related topic | SubstitutionNotation |