$\omega - \lambda dx = dg$
by Hidenori
Proposition
If $\omega$ is a 1-form $fdx$ on $[0, 1]$ with $f(0) = f(1)$, show that there is a unique number $\lambda$ such that $\omega - \lambda dx = dg$ for some function $g$ with $g(0) = g(1)$.
Solution
Uniqueness
Suppose $\lambda$ and $g$ exist.
Let $c: [0, 1] \rightarrow [0, 1]$ such that $c(x) = x$. Then $c’ = [1]$. (The $1 \times 1$ identity matrix)
\[\begin{align*} \int_{c} dg &= \int_{\partial c} g \\ &= g(1) - g(0) = 0. \end{align*}\] \[\begin{align*} \int_{c} (\omega - \lambda dx) &= \int_{c} \omega - \int_{c} \lambda dx \\ &= \int_{c} fdx - \lambda \int_{c} dx \\ &= \int_{c} fdx - \lambda \int_{\partial c} x \\ &= \int_{c} fdx - \lambda (1 - 0) \\ &= \int_{c} fdx - \lambda \\ &= \int_{[0, 1]} c^*(fdx) - \lambda \\ &= \int_{[0, 1]} (f \circ c) (\det c') dx - \lambda \\ &= \int_{[0, 1]} f dx - \lambda. \end{align*}\]Therefore, $\lambda = \int_{[0, 1]} fdx$, so $\lambda$ must be unique, if it exists.
Existence
Let $\lambda = \int_{[0, 1]} fdx$ and $g(x) = \int_{0}^{x} (f(t) - \lambda) dt$. Then $g(0) = 0 = g(1)$. Moreover, $g(x) = D_1g \cdot dx = (f(x) - \lambda)dx = f(x)dx - \lambda dx = \omega - \lambda dx$.
Subscribe via RSS