I am looking for a 2D function, $f(x, y)$ which increases monotonically over the range $(0,0)$ to $(1,1)$. In other words, it will be $0$ at $(0,0)$ and $1$ at $(1,1)$. It will also evaluate to $0$ whenever $x = 0$ or $y = 0$. It's monotonic in the sense that the value at $(0.5,0.6)$ will be greater than at $(0.5, 0.5)$. The closest guess that I have to the function that I am looking for is:
Plot3D[1/((1+e^(-x*y))),{x,0,1},{y,0,1}]
But I would also like to be able to "bias" or make the function rise faster with $x$ relative to $y$ or vice-versa. Ideally, I would like to specify the surface with 2 variables. One that controls / scales how $f(x,y)$ rises relative to $x$, and another that controls / scales how $f(x,y)$ rises relative to $y$.
Edit:
I am looking for a sigmoid-shaped curve. The function does not need to be exactly equal to 1 at $z(1,1)$. However, $z(1,1)$ needs to be higher than any other $z(x,y)$ across the space $x[0:1], y[0:1]$. A closer function than the one above is:
$$z=1/\left(1+e^{-(x^a*y^b)}\right),$$
where $a$ and $b$ are used to "bias" or change the slope of the curve with respect to $x$ vs. $y$. However, that curve is close to 1 at $z(2,2)$ instead of $z(1,1)$. So $x$ and $y$ need to be scaled. Those scaling factors will be a function of $a$ and $b$. They cannot be computed directly since the exponent needs to be 0 to satisfy the equation exactly. However, if the exponent evaluates to a sufficiently small number, such as $0.01$, that should be close enough for my purposes.
Best Answer
Here's a simple example: $f(x,y)=xy$. You can change the powers of $x$ and $y$ to 'bias' one of $x$ or $y$.
Here's another function that is "sigmoid-shaped" and is close to your required values, as per your comments:
$$f(x,y)=\left( \frac{1}{1+e^{-100x+50}}\right)\left( \frac{1}{1+e^{-100y+50}}\right)$$
This is approximately $0$ at $(0,0)$ and $1$ and $(1,1)$, and is sigmoid-shaped in the sense that when either $x=1$ or $y=1$ the resulting slice of the 3D-plot is a sigmoid, as seen here:
Theoretically, you could also make this exactly equal to $1$ at $(1,1)$ and $0$ at $(0,0)$ by multiplying by a constant and changing the values of $50$ in the exponents appropriately to ensure that the function went through the points $(0,0,0)$ and $(1,1,1)$.