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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?

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2 Answers 2

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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

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

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