# Constructing an $C^{\infty}$ function, radius $\delta$, centered at $(a,b)$

I'm wondering how to construct a $$C^{\infty}$$ function which is positive on an open disk of radius $$\delta$$, centered at the point $$(a,b)$$ and vanishing outside of the disk. I know that in 1D, a function like $$f(x)=e^{-(b-x)^{-1}} \cdot e^{-(a-x)^{-1}}$$ would work, but what would the analog be in 2D?

I'm thinking that the function would have the form $$u(x,y)=e^{x}e^{y}$$. Then, using Euler's Formula, we would have $$u(x,y)=u(?, ?) = r_1\cdot(cos(\theta_1)+isin(\theta_1)) \cdot r_2(cos(\theta_2)+isin(\theta_2))$$. But then, we have a function in terms of $$r$$'s and $$\theta$$'s. Am I taking the right approach here?

For the circle of radius one centered in $$(0,0)$$ a function that satisfies your requirements is:

$$f(x,y)=\begin{cases}e^{-\frac{1}{(x^2+y^2-1)^2}}\ &\text{if}\ x^2+y^2<1\\ 0\ &\text{otherwise}\end{cases}$$

The general case for the circle of radius $$\delta$$ and center$$(a,b)$$ follows by translation and dilatation

$$\hat{f}(x,y):=f\left(\frac{(x,y)-(a,b)}{\delta}\right)$$

In general, given a $$C^\infty(\mathbb{R})$$ function $$h$$ with compact support $$[0,1]$$, the function

$$g((x,y)):=h\left(\sqrt{x^2+y^2}\right)$$ works for the unitary circle

• So does that mean, for radius $\delta$ and center $(a,b)$, we would have $f(x,y)=e^{-1/(((x-a)/\delta)^2+((y-b)/\delta)^2-1)^2}$ Mar 3, 2020 at 16:39
• @natn2323 yes, that is one of the possible functions Mar 3, 2020 at 17:13

If $$f(t)>0$$ for $$0\leq t <\delta^{2}$$ and $$f(t)=0$$ for $$t >\delta^{2}$$ then $$f(\|(x,y)-(a,b)\|^{2})$$ satisfies your requirements.