construction of a curve connecting two points

Question

Let $a,b,c$ be positive reals numbers. Assume $a<b$. I'm trying to construct a $C^1$ function (meaning a function with continuous derivative) $f$ with the following properties:

1. $f$ is increasing and supported on $[a,b]$;
2. $f(a)=0$ and $f(b)=c$;
3. $f'(a)=0$ and $f'(b)=1$;

Geometrically $f$ is a curve connecting $(a,0)$ and $(b,c)$ with additional derivative requirements at two end points.

I think such a function is very useful in application but I can't find any book containing methods of constructions. I guess the construction may be tedious. Do you know any reference providing constructions of lots of special functions?

Answer 1

You can look for a polynomial of degree 4 of the form $$ p(x)=(x-a)^2(A\,x+B) $$ where the constants $A$ and $B$ are chosen to satisfy the conditionsconditions at the point $b$.

The special case $a=0$, $b=c=1$ gives $p(x)=x^2(2-x)$, with $p'(x)=x(4-3\,x)\ge0$ if $0\le x\le1$.The general case is then $$ c\,p\Bigl(\frac{x-a}{b-a}\Bigr), $$ which is increasing on $[a,b]$.

Answer 2

Just write down the formula for a degree 4 polynomial or higher and you should be able to solve for the coefficients to satisfy the constraints.

Answer 3

Using linear interpolation one reduces to the same problem with $a=b=c=0$, replacing the condición $f'(b)=1$ by the more general $f'(1)=\lambda>0$. Now the conditions mean that the tangents to the graph are horizontal at $t=0$ and have slope $\lambda$ at $t=1$. There are different ways to construct such a smooth function. One is through the so-called bump functions, as found typically in many Differential Calculus texts. But one can also use splines. Try one of them $$ f(t)=n(1-\tfrac{1}{n}\lambda)t^{n-1}(1-t)+t^n $$ with $n>\lambda$. What is important here is that the function is strictly increasing, hence a difeo, which makes necessary a possibly very high degree $n$.

In general you can produce anything form $[a,b]$ to $[c,d]$ with any chosen slopes at $a$ and $b$.

Answer 4

Not very different from the other answers, but maybe easier to understand and use: $$ f(x) = \frac{(a-x)^2 \left( (a-b+2c)x -ab +ac -3bc + b^2 \right)}{(a-b)^3} $$