Deformation of integration paths

Question

First of all: I'm more of an algebraic person. So happily differentials/integrals are not what I deal with a lot. However, I got this exercise to solve:

Let $\Omega \subset \mathbb{R}^n$ open, $K: \Omega \rightarrow \mathbb{R}^n$ a $C^1$ vector field with $\partial_iK_j = \partial_j K_i$ for $i, j= 1, ..., n$.

Let $\gamma, \gamma ' \in C^1([0,1]; \Omega)$ be curves with $\gamma(0) = \gamma'(0)$ and $\gamma(1) = \gamma' (1)$. Let $\Phi \in C^2 ( [0,1] \times [0,1]; \mathbb{R}^n)$ be a homotopy between $\gamma$ and $\gamma'$.

Then it is to show that $\int_{\gamma} K d x= \int_{\gamma '} K d x$.

I know this holds for continous homotopies already, but I have to present the solution to a group of first years and the proof for general case seems a bit lengthy. But, here I have a $C^2$ homotopy, and I'd like to use it by showing

$\frac{\partial}{\partial s} \int_{\Phi(s, -)} K dx = 0$

How can I do this?

Answer 1

Pull back the vector field $K$ to the domain $Q=[0,1]^2$ of $\Phi$ as follows: For a given point $u\in Q$ the function $$\psi(U):=K\bigl(\Phi(u)\bigr)\cdot \bigl(d\Phi(u).U\bigr)$$ is a linear functional on the tangent space $T_u$. Therefore there is a unique vector $a\in T_u$ such that $\psi(U)=a\cdot U$. This vector $a$ is called the pullback of $K$ at $u$ and will be denoted by $K^*(u)$. From the defining equality $$K^*(u)\cdot U=K\bigl(\Phi(u)\bigr)\cdot \bigl(d\Phi(u).U\bigr)$$ one easily deduces that (a) for any curve $\gamma^*$ in $Q$ with image curve $\Phi(\gamma^*)=:\gamma$ one has $$\int\nolimits_\gamma K\cdot dx=\int\nolimits_{\gamma^*} K^*\cdot du\ ,$$ and that (b) one has $${\partial K_2^*\over\partial u_1}-{\partial K_1^*\over\partial u_2}\equiv0$$ when $\partial_iK_j-\partial_jK_i\equiv0$ in $\Omega$.

Now apply Green's theorem to $K^*$, $Q$ and $\partial Q$.

Answer 2

Let $\gamma_s(t)$ be the path $\Phi(s,\cdot)$, and let its derivatives be $\gamma'$ (w.r.t. $t$) and $\dot\gamma$ (w.r.t. $s$). We know that $\dot\gamma(0)=\dot\gamma(1)=0$. We also know that $((a\nabla)K,b) = ((b\nabla)K,a)$.

Now, $$\begin{aligned} \frac{d}{ds}\int_{\gamma_s}K &= \int_0^1dt\,\frac{d}{ds}(K(\gamma_s(t)), \gamma_s'(t)) \\&= \int_0^1 ((\dot \gamma\nabla)K, \gamma') + (K, \dot\gamma') \\&= \int_0^1 ((\gamma'\nabla)K, \dot\gamma) + (K, \dot\gamma') \\&= \int_0^1 -(K, \dot\gamma') + (K,\dot\gamma') \\& = 0, \end{aligned}$$ where $(\gamma'\nabla)K = \frac{d}{dt}K(\gamma_s(t))$, and the second last line is obtained from integration by parts ($\dot\gamma$ vanishes at $t=0,1$).

Answer 3

The proof can be done as follows: $$\frac{d}{d s}\int_{\Phi(s,-)}K(x)\cdot dx = \frac{d}{d s}\int_0^1K(\Phi(s,t))\cdot \frac{\partial\Phi}{\partial t}(s,t)dt$$ We use one of the classical results of measure theory to exchange integral and derivative: $$\int_0^1\frac{d}{d s}\left(K(\Phi(s,t))\cdot \frac{\partial\Phi}{\partial t}(s,t)\right)dt =$$ $$=\int_0^1\left(DK(\Phi(s,t))\frac{\partial\Phi}{\partial s}\right)\cdot \frac{\partial\Phi}{\partial t}(s,t) + K(\Phi(s,t))\cdot \frac{\partial^2\Phi}{\partial s\partial t}(s,t)dt$$ Now we use partial integration on the second term, notice that the boundary terms vanish since $\frac{\partial \Phi}{\partial s}(s,t)$ vanishes for $t=0$ and $t=1$: $$\int_0^1K(\Phi(s,t))\cdot \frac{\partial^2\Phi}{\partial s\partial t}(s,t)dt =$$ $$=\int_0^1\left(\frac{d}{dt}K(\Phi(s,t))\right)\cdot \frac{\partial\Phi}{\partial s}(s,t)dt =$$ $$=\int_0^1\left(DK(\Phi(s,t))\frac{\partial\Phi}{\partial t}(s,t)\right)\cdot \frac{\partial\Phi}{\partial s}(s,t)dt=$$ $$=\int_0^1\left(DK(\Phi(s,t))\frac{\partial\Phi}{\partial s}(s,t)\right)\cdot \frac{\partial\Phi}{\partial t}(s,t)dt$$ Where in the last line we use that $DK$ is symmetric (i.e. the assumption that $\frac{\partial K^i}{\partial x^j} = \frac{\partial K^j}{\partial x^i}$) and that $(Au)\cdot v = (A^tv)\cdot u$. In conclusion we obtain that $$\frac{d}{d s}\int_{\Phi(s,-)}K(x)\cdot dx =0$$ As you wanted.