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Let $U\subset \mathbb{R}^2$ be an open set containing the unit circle $S^1$. For each $k\in \mathbb{Z}$, consider the path $\lambda_k:[0, 2\pi]\to U$ given by $\lambda_k(t) = (\cos kt, \sin kt)$. If $w:U\to(\mathbb{R^2})$ is any continuous $1$-form, prove that $\int_{\lambda_k}w = k\int_{\lambda_1}w$

By the definition of Stieltjes integration along a path:

$$\int_{\lambda_k}w = \lim_{|P|\to 0} \sum_{i}^k w(\lambda_k(\zeta_j))\cdot[\lambda_k(t_j)-\lambda_k(t_{j-1})]$$

I don't think it will work. I'm thinking in using change of variables for Stieltjes intgegrals:

$$\int_c^d f(t) d\alpha = \int_a^b f\circ\phi(s)d(\alpha\circ \phi)$$

for $f$ continuous, $\alpha$ of bounded variation and $\phi$ being a crescent homeomorphism. The integral is negative if $\phi$ is decrescent. But does such homeomorphism exists? If so, I need to end up with:

$$\int_{\lambda_k} w = k\int_{\lambda_1}w$$

but to arrive at $\int_{\lambda_1}w$ I would have to arrive at this integral, I'd have to think about integrals along paths, but the change of variables theorem is for normal Stieltjes integrals. I know that there is this theorem:

When the $1$-form is continuous and the path is rectifiable and continuous, and each $\int_a^b a_i(\lambda(t)) dx_i$ exists, where $w = \sum_1^n a_i dx_i$, then

$$\int_{\lambda} w = \sum_{i=1}^n \int_a^b a_1(\lambda(t))dx_i$$

and for each $k$, my path is rectifiable, so I think this theorem can be used, right? However, I'll deviate from path integration and move to simple Stieltjes integration.

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  • $\begingroup$ Seems to me we have $$\int_{\lambda_k} w = \int_0^{2\pi} w(\lambda_k(t))\cdot \lambda_k'(t)\, dt.$$ Does that look familiar? $\endgroup$
    – zhw.
    Aug 26 '17 at 17:42
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A continuous $1$-form $w$ can be written as $w = \sum_{i=1}^n a_i dx_i$, where $a_i:U\to\mathbb{R}$ are continuous functions. The integral of a continuous $1$-form is given by $$\int_{\lambda_k}w= \sum_{i=1}^n\int_0^{2\pi}a_i(\lambda_k(t))\lambda_{k,i}^\prime (t)\,dt.$$ In your case $n=2$ and $\lambda_k^\prime(t)=(-k\sin (kt), k\cos (kt))$ so you have $$\int_{\lambda_k}w=k \int_0^{2\pi}-a_1(\cos (kt), \sin (kt))\sin (kt)+a_2(\cos(kt), \sin(kt))\cos (kt)\,dt.$$ If you consider the change of variables $s=kt$ so $ds=kdt$ you get $$\int_{\lambda_k}w= \int_0^{2k\pi}-a_1(\cos s, \sin s)\sin s+a_2(\cos s, \sin s)\cos s\,ds\\=\sum_{j=1}^k \int_{(2j-1)\pi}^{2j\pi}-a_1(\cos s, \sin s)\sin s+a_2(\cos s, \sin s)\cos s\,ds.$$ Making the change of variables $s=(2j-1)\pi+r$ and using the periodicity of $\cos$ and $\sin$ you have that the right-hand side becomes $$\sum_{j=1}^k \int_{0}^{2\pi}-a_1(\cos r, \sin r)\sin r+a_2(\cos r, \sin r)\cos r\,dr=k\int_{\lambda_1}w.$$

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