2
$\begingroup$

I'm trying to create a function for a research project, but I fear my math knowledge is insufficient to derive it from the attached diagram I've created showing its desired behavior. I'm hoping someone will recognize the behavior and provide some leads on similar functions to lead me in the right direction.

$\hskip 1.1in$ pic

For the axes we have $X \in [0,1]$ and $Y \in (0, \infty)$. The dotted lines are isometric lines of the Z axis. The Z values and exact shape of the isolines are not as important as the monotonicity of the graph w.r.t. both X and Y. I'm not against having a piecewise function for $Y>1$, $Y=1$, and $Y<1$ if necessary but a single function would be a bonus.

-Jeff

  • 0
    Oh, contour lines, of course.2012-04-13

4 Answers 4

0

The simplest thing I can think of is $z=\frac{1−y}{1+4x^2}+1.$ Its contour lines are parabolae.

  • 0
    No fair, you said "the exact shape of the isolines are not as important"! :) I'm glad you found what you were looking for, in any case.2012-04-15
1

It looks like it ought to be something like $ z = y^{-g(x)}$ where $g$ is some strictly decreasing function $[0,1] \to (0,\infty)$.

The rough sketch does not give much clue to the exact choice of $g$; you could try something like $g(x) = 10 - 9x$ as a starting point.

  • 0
    When y<1 rather.2012-04-14
1

Try something like $(x,y) \mapsto 1+e^{-(x^2+y^2)}-e^{-(x^2+(y-2)^2)}$? May need a little scaling...

wxMaxima contour plot

  • 0
    Thank you so much. This captures exactly what I was looking for!2012-04-14
0

After some thought, to me this looks like it could be an implicit relationship between $x$, $y$, and $z$:

$y = 1+(1-z)\textrm{e}^{(1-z)^2x}$

Solving this equation for $z$ explicitly in terms of $x$ and $y$ is not possible with elementary functions though.

In case it helps, I found this by subtracting $1$ from your contour values, so that they are balanced around $0$ at the horizontal contour line. Then I negated them, so that positives would naturally correspond to contours above the horizontal line. The actual shape of your contour lines reminds me of $y=1+k\mathrm{e}^{k^2x}$ where the $k$ in the exponent is really only squared to make it positive. $k$ ranges from $-1$ to $\infty$. $k$ plays the role of the transformed contour values, and then we have my guess at the relationship.

(And maybe replacing $\mathrm{e}$ by some other base would make contour lines more precisely in line with yours.)

  • 0
    No, that wouldn't simplify this particular stab at your problem. With $z$ both "down below" and up in the exponent, it will not be possible to solve for $z$ explicitly using usual functions. However, if you are willing to work with the [product log function](http://www.wolframalpha.com/input/?i=W%28x%29), then breaking up piecewise does help and you have the two functions defined [here](http://www.wolframalpha.com/input/?i=solve+for+z%3A+y+%3D+%281-z%29*exp%28%281-z%29^2*x%29%2B1). I'd be curious if this works - let me know.2012-04-14