Ex2-9

Solution to Chapter 2, Exercise 9

Question: Let $$C$$be a curve with parametric equations $$x=\phi(t)$$and $$y=\psi(t)$$, $$a \leq t \leq b$$.

(a) If $$\phi$$and $$\psi$$ are of bounded variation and continuous, show that $$L=\lim_{|\Gamma|\to 0}l(\Gamma)$$.

(b) If $$\phi$$and $$\psi$$ are continuously differentiable, show that $$L=\int_a^b([\phi'(t)]^2+[\psi'(t)]^2)^{1/2}dt$$.

Proof:

(a) If $$\phi$$and $$\psi$$are of bounded variation, we know that $$C$$is rectifiable. We want to prove that for any $$M0$$such that if $$|\Gamma|<\delta$$, then $$l(\Gamma)>M$$. Choose $$\mu>0$$so that $$M+\muM+\mu$$ and let $$\bar{\Gamma}=\{\bar{t_0},\bar{t_1},...,\bar{t_k}\}$$.

$$\phi$$and $$\psi$$ are uniformly continuous since they are continuous on a compact set. Then there exists $$\delta'>0$$such that if $$|s-t|<\delta'$$, then

(i) $$|\phi(s)-\phi(t)|<\frac{\mu}{\sqrt{8}(k+1)}$$

(ii) $$|\psi(s)-\psi(t)|<\frac{\mu}{\sqrt{8}(k+1)}$$

for all $$s,t\in[a,b]$$.

Let $$\Gamma=\{t_0,t_1,...,t_m\}$$be any partition that satisfies

(iii) $$|\Gamma|<\delta'$$

(iv) $$|\Gamma|<\min\{\overline{t_i}-\overline{t_{i-1}}\vert1\leq i \leq m\}$$

We have

$$ \begin{align} l(\Gamma)&=\sum_{i=1}^m ([\phi(t_i)-\phi (t_{i-1})]^2+[\psi(t_i)-\psi (t_{i-1})]^2)^{1/2}\\ &=\sum' + \sum ''\text{(1)} \end{align} $$

where $$\sum'$$ is extended over all $$i$$such that $$(t_{i-1},t_i)$$does not contain some $$\bar{t_j}$$while $$\sum ''$$is extended over all $$i$$such that $$(t_{i-1},t_i)$$contains some $$\bar{t_j}$$. Notice that any $$(t_{i-1},t_i)$$ can contain at most one $$\bar{t_j}$$ by (iv), and the number of terms of $$\sum ''$$is at most $$k+1$$.

Now we consider the partition $$\Gamma \cup \bar{\Gamma}$$. $$l(\Gamma \cup \bar{\Gamma})$$can be written as

$$ l(\Gamma \cup \bar{\Gamma})=\sum' + \sum ''' $$

where $$\sum '$$is the same as before while $$\sum '$$is obtained from $$\sum $$by replacing each $$([\phi(t_i)-\phi (t_{i-1})]^2+[\psi(t_i)-\psi (t_{i-1})]^2)^{1/2}$$term by $$([\phi(t_i)-\phi (\bar{t_{j}})]^2+[\psi(t_i)-\psi (\bar{t_{j}})]^2)^{1/2} + ([\phi(\bar{t_j})-\phi (t_{i-1})]^2+[\psi(\bar{t_j})-\psi (t_{i-1})]^2)^{1/2}$$.

Then each term in $$\sum $$will be smaller than or equal to those in $$\sum $$and hence $$\sum  \geq \sum $$. Thus, $$l(\Gamma \cup \bar{\Gamma}) \geq l(\Gamma)$$. Similarly, we can show that $$l(\Gamma \cup \bar{\Gamma}) \geq l(\bar{\Gamma})$$.

$$\sum '''$$has at most $$2(k+1)$$ terms and by $$(i),(ii),(iii)$$, we have

$$ \begin{align} \sum '''&< \left( \frac{\mu^2}{8(k+1)^2}+\frac{\mu^2}{8(k+1)^2} \right)^{1/2}2(k+1)\\ &=\left( \frac{\mu^2}{4(k+1)^2} \right)^{1/2}2(k+1)\\ &=\mu. \end{align} $$

Since $$l(\Gamma \cup \bar{\Gamma}) = \sum ' + \sum '''$$, we have

$$ \begin{align} \sum ' &= l(\Gamma \cup \bar{\Gamma}) - \sum '''\\ &> l(\Gamma \cup \bar{\Gamma}) - \mu. \end{align} $$

Recall that $$\bar{\Gamma}$$ has been chosen so that $$l(\bar{\Gamma})>M+\mu$$. Then

$$ \begin{align} l(\Gamma) &= \sum ' + \sum ''\\ &\geq \sum '\\ &> l(\Gamma \cup \bar{\Gamma}) - \mu\\ &\geq l(\bar{\Gamma}) - \mu\\ &> M+\mu-\mu\\ &=M \end{align} $$

Take $$\delta=\min\{ \delta',\min\{\overline{t_i}-\overline{t_{i-1}}\}\}$$, then for any $$\Gamma$$ with $$|\Gamma|<\delta$$, we will have $$l(\Gamma)>M$$.

(b) Let $$\Gamma=\{t_0,t_1,...,t_m\}$$. By Mean Value Theorem,

$$ \begin{align} \phi(t_i)-\phi(t_{i-1})&=\phi'(c_i)(t_i-t_{i-1})\\ \psi(t_i)-\psi(t_{i-1})&=\psi'(d_i)(t_i-t_{i-1}) \end{align} $$

for some $$c_i$$and $$d_i$$ in $$(t_{i-1},t_i)$$. Then

$$ \begin{align} l(\Gamma)&=\sum_{i=1}^m ([\phi(t_i)-\phi (t_{i-1})]^2+[\psi(t_i)-\psi (t_{i-1})]^2)^{1/2}\\ &=\sum_{i=1}^m ([\phi'(c_i)(t_i-t_{i-1})]^2+[\psi'(d_i)(t_i-t_{i-1})]^2)^{1/2}\\ &=\sum_{i=1}^m ([\phi'(c_i)]^2+[\psi'(d_i)]^2)^{1/2}(t_i-t_{i-1}) \end{align} $$

If we take the limit as $$|\Gamma| \to 0$$on both sides, the left hand side converges to $$L$$ by part (a) while the limit of the right hand side is the integral

$$ \int_a^b ([\phi'(t)]^2+[\psi'(t)]^2)^{1/2}~dt $$