Existence theorem for planar curves (II)

We follow with the above entry taking an 'easy' function $k$: find a planar curve as a   $y=f(x)$ where $k(x)=2x$.

By the given method in the last entry, we integrate $k$ between $a$ and $x$. Take $a=0$, obtaining  $g(x)=x^2$ (here $m=0$). Then the solution is
$$y(x)=\int_0^x \frac{g(t)}{\sqrt{1-g(t)^2}}dt.$$
In this case,
$$\int_0^x\frac{t^2}{\sqrt{1-t^4}}dt.$$
This integral is not possible to compute!

Other example. Take $k(x)=e^x$. Now
$$\int_0^x e^t dt=e^x-1.$$
If we take $m=-1$, then $g(x)=e^x$ and finally $$f(x)=\int:0^x\frac{e^t}{\sqrt{1-e^{2t}}}dt=\mbox{arc}\sin (e^x).$$In order to make sense in the integrand, we are assuming that  $x<0$. The solution is $y(x)=\mbox{arc}\sin (e^x)$ whose graphic is

Existence theorem for planar curves

We know how to calculate the curvature of a regular curve of  $\mathbb{R}^2$ even if it is not parametrized by the length-arc. We pose the next

Question: let $k:I\subset \mathbb{R}\rightarrow\mathbb{R}$ be a smooth function. Does exist a curve  $\alpha:I\rightarrow\mathbb{R}^2$ such that its curvature is $k$, that is, $\kappa(s)=k(s)$?

Answer: yes.

How do we find such a curve? We can do the next approach (and an answer) thanks to the curvature and using our knowledge from Calculus. We do an answer only for curves that are graphs of a function  $y=f(x)$. But this is enough because any curve is locally the graph of a function.

Thus the given function $k=k(s)$ is now a function $k=k(x)$ where $x\in (a,b)$ and $k$ is an easy function. The curve $\alpha$ that we are looking for writes then as  $\alpha(x)=(x,f(x))$ and its curvature is
$$\kappa(x)=\frac{f''(x)}{(1+f'(x)^2)^{3/2}}.$$
Then we have to solve the following

Problem: let $k:(a,b)  \rightarrow\mathbb{R}$ be a smooth function. Does exist a function $y=f(x)$ such that $$\frac{f''(x)}{(1+f'(x)^2)^{3/2}}=k(x)?$$

Let us use the notation from the high-school with $y=f(x)$. We have to find $y=y(x)$ such that
$$\frac{y''}{(1+y'^2)^{3/2}}=k(x).$$
For this, let us integrate. With the change of variable given by $z=y'$, we have
$$\frac{z'}{(1+z^2)^{3/2}}=k(x).$$
But the left-hand side is the derivative of $x/\sqrt{1+z^2}$. Thus, integrating from $a$ to $x$, and returning with $y'$, we have
$$\frac{y'}{\sqrt{1+y'^2}}-m=\int_{a}^x k(t)dt\Rightarrow \frac{y'}{\sqrt{1+y'^2}}:=g(x),$$
where $g(x)=m+\int_{a}^x k(t)dt$ and $m\in \mathbb{R}$. Let us observe that the integral do exist because the integrand is a continuous function, even more, it is differentiable. As a consequence: there are many solutions because in the integral of $k$ we can add a constant and the above formula holds again.

Now we have to find $y$. Then
$$y'=\frac{g}{\sqrt{1-g^2}},$$ where we suppose $g>0$: on the contrary, we change of sign. Now let us integrate again! $$y(x)=\int_a^x \frac{g(t)}{\sqrt{1-g(t)^2}}dt.$$
The integral exists answering positively to the initial question.

Remark: to find the curve, we have to integrate twice!

We give an explicit example. If $k=0$, then $f''=0$, so $f(x)=ax+b$, which it is a straight-line.

The following example is when $k$ is a constant function, and we have to get a circle. Recall that if the radius is $r>0$, then the curvature is $1/r$. Thus we pose the next

Question: find $y=y(x)$ when $k(x)=c>0$.

Hint: for the integral constants, take  $a=0$,  $f(0)= -1/c$ and $f'(0)=0$ (why?)