6.1 Smooth curves in the plane  (Page 3/4)

The proof is now complete.

REMARK It should be evident that the preceding theorem can easily be generalized to a piecewise smooth function $f,$ i.e., a function that is continuous on $[a,b],$ continuously differentiable on each subinterval $(t_{i-1},t_i)$ of a partition $\{t_0<t_1<...<t_n\},$ and whose derivative $f^{\text{'}}$ is absolutely integrable on $(a,b).$ Indeed, just apply the theorem to each of the subintervals $(t_{i-1},t_i),$ and then carefully piece together the piecewise linear functions on those subintervals.

Now we are ready to define what a smooth curve is.

By a smooth curve from a point $z_1$ to a different point $z_2$ in the plane, we mean a set $C\subseteq C$ that is the range of a 1-1, smooth, function $\phi :[a,b]\to C,$ where $[a,b]$ is a bounded closed interval in $R,$ where $z_1=\phi \left(a\right)$ and $z_2=\phi \left(b\right),$ and satisfying ${\phi }^{\text{'}}\left(t\right)\ne 0$ for all $t\in (a,b).$

More generally, if $\phi :[a,b]\to R^2$ is 1-1 and piecewise smooth on $[a,b],$ and if $\{t_0<t_1<...<t_n\}$ is a partition of $[a,b]$ such that ${\phi }^{\text{'}}\left(t\right)\ne 0$ for all $t\in (t_{i-1},t_i),$ then the range $C$ of $\phi$ is called a piecewise smooth curve from $z_1=\phi \left(a\right)$ to $z_2=\phi \left(b\right).$

In either of these cases, $\phi$ is called a parameterization of the curve $C.$

Note that we do not assume that $|{\phi }^{\text{'}}|$ is improperly-integrable, though the preceding theorem might have made you think we would.

REMARK Throughout this chapter we will be continually faced with the fact that a given curve can have many different parameterizations.Indeed, if ${\phi }_1:[a,b]\to C$ is a parameterization, and if $g:[c,d]\to [a,b]$ is a smooth function having a nonzero derivative, then ${\phi }_2\left(s\right)={\phi }_1\left(g\left(s\right)\right)$ is another parameterization of $C.$ Since our definitions and proofs about curves often involve a parametrization, we will frequently need to prove that the results we obtain are independent of the parameterization.The next theorem will help; it shows that any two parameterizations of $C$ are connected exactly as above, i.e., there always is such a function $g$ relating ${\phi }_1$ and ${\phi }_2.$

Let ${\phi }_1:[a,b]\to C$ and ${\phi }_2:[c,d]\to C$ be two parameterizations of a piecewise smooth curve $C$ joining $z_1$ to $z_2.$ Then there exists a piecewise smooth function $g:[c,d]\to [a,b]$ such that ${\phi }_2\left(s\right)={\phi }_1\left(g\left(s\right)\right)$ for all $s\in [c,d].$ Moreover, the derivative $g^{\text{'}}$ of $g$ is nonzero for all but a finite number of points in $[c,d].$

Because both ${\phi }_1$ and ${\phi }_2$ are continuous and 1-1, it follows from [link] that the function $g={\phi }_1^{-1}\circ {\phi }_2$ is continuous and 1-1 from $[c,d]$ onto $[a,b].$ Moreover, from [link] , it must also be that $g$ is strictly increasing or strictly decreasing. Write ${\phi }_1\left(t\right)=u_1\left(t\right)+i{v}_1\left(t\right)\equiv (u_1\left(t\right),v_1\left(t\right)),$ and ${\phi }_2\left(s\right)=u_2\left(s\right)+i{v}_2\left(s\right)\equiv (u_2\left(s\right),v_2\left(s\right)).$ Let $\{x_0<x_1<...<x_p\}$ be a partition of $[a,b]$ for which ${\phi }_1^{\text{'}}$ is continuous and nonzeroon the subintervals $(x_{j-1},x_j),$ and let $\{y_0<y_1<...<y_q\}$ be a partition of $[c,d]$ for which ${\phi }_2^{\text{'}}$ is continuous and nonzero on the subintervals $(y_{k-1},y_k).$ Then let $\{s_0<s_1<...<s_n\}$ be the partition of $[c,d]$ determined by the finitely many points $\left\{y_k\right\}\cup \left\{g^{-1}\left(x_j\right)\right\}.$ We will show that $g$ is continuously differentiable at each point $s$ in the subintervals $(s_{i-1},s_i).$

Fix an $s$ in one of the intervals $(s_{i-1},s_i),$ and let $t={\phi }_1^{-1}\left({\phi }_2\left(s\right)\right)=g\left(s\right).$ Of course this means that ${\phi }_1\left(t\right)={\phi }_2\left(s\right),$ or $u_1\left(t\right)=u_2\left(s\right)$ and $v_1\left(t\right)=v_2\left(s\right).$ Then $t$ is in some one of the intervals $(x_{j-1},x_j),$ so that we know that ${\phi }_1^{\text{'}}\left(t\right)\ne 0.$ Therefore, we must have that at least one of $u_1^{\text{'}}\left(t\right)$ or $v_1^{\text{'}}\left(t\right)$ is nonzero. Suppose it is $v_1^{\text{'}}\left(t\right)$ that is nonzero. The argument, in case it is $u_1^{\text{'}}\left(t\right)$ that is nonzero, is completely analogous. Now, because $v_1^{\text{'}}$ is continuous at $t$ and $v_1^{\text{'}}\left(t\right)\ne 0,$ it follows that $v_1$ is strictly monotonic in some neighborhood $(t-\delta ,t+\delta )$ of $t$ and therefore is 1-1 on that interval. Then $v_1^{-1}$ is continuous by [link] , and is differentiable at the point $v_1\left(t\right)$ by the Inverse Function Theorem. We will show that on this small interval $g=v_1^{-1}\circ v_2,$ and this will prove that $g$ is continuously differentiable at $s.$