# Delta function of two variables.

How can we transfer equation

$$\iint \delta\left(f\left(x,y\right)-t\right)\, \mathrm{d}x\,\mathrm{d}y,$$

into line integral? Where $$t$$ is a parameter and a constant value of $$t$$ denotes a closed curve in XY-plane.

## 1 Answer

$$\delta(u) = \lim_{n \to \infty} 2n 1_{|u|< 1/n}$$ (limit in the sense of distributions)

$$\iint \delta(f(x,y) - a)\phi(x,y) dxdy \overset{def}= \lim_{n \to \infty}\iint 2n 1_{|f(x,y) - a|<1/n}\phi(x,y) dxdy$$

If there is a $$C^1$$ curve $$\gamma(t),t \in (a,b)$$ such that $$|f(x,y)-a| < 1/n \implies |dist((x,y),\gamma)| and $$f$$ is $$C^1$$ on $$\gamma\circ (a,b)$$ with non-vanishing gradient then $$1_{|f(x,y) - a|<1/n} \approx 1_{dist((x,y),\gamma)< \frac1{n \|\nabla f(x,y)\|}}$$ which as a distribution is close to $$\frac{1_{dist((x,y),\gamma)< 1/n} }{\|\nabla f(x,y)\|}$$ and hence for $$\phi \in C^0_c(\Bbb{R}^2)$$ $$\lim_{n \to \infty}\iint 2n 1_{|f(x,y) - a|<1/n}\phi(x,y) dxdy = \lim_{n \to \infty}\iint 2n \frac{1_{dist((x,y),\gamma)< 1/n} }{\|\nabla f(x,y)\|} \phi(x,y)dxdy$$ $$=\int_a^b \frac{\phi(\gamma(t))}{\|\nabla f(\gamma(t))\|} dt$$

When $$f$$ vanishes on several $$C^1$$ curves, assuming there are finitely many in each compact, it becomes

$$\lim_{n \to \infty}\iint 2n 1_{|f(x,y) - a|<1/n}\phi(x,y) dxdy =\sum_j\int_a^b \frac{\phi(\gamma_j(t))}{\|\nabla f(\gamma_j(t))\|} dt$$