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While reading literature about boundary element and finite element methods I have repeatedly seen that some integral kernels are singular and others are hypersingular. Could you explain what is the difference between singular and hypersingular and how to tell where a certain kernel belongs?

"Weakly singular" also seems unclear to me, what is that one?

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Consider an integral of the sort $$ \int_{\mathbb R^n}K(x,y)f(y)\,dy $$ for some smooth function $f$ and a kernel $K$. To allow the integral to converge at infinity we'll assume that $f$ has compact support. Integrals of that sort appears in the integral equations to which, in particular, boundary value problems for PDE are reduced (not always in $\mathbb R^n$, but on surfaces too).

If $|K(x,y)|\sim |x-y|^{-\alpha}$ when $y\to x$, $\alpha>0$, the kernel is called singular, because it tends to infinity as $y\to x$. If $\alpha<n$ the kernel (and the integral) is called weakly singular, since the integrand is absolutely summable. For $\alpha>n$ a kernel is called hypersingular, because in this case the estimate on $|K|$ is not enough to guaranty convergence. Such integrals are often taken in the sense of the principal value. It can be done under some additional assumptions on $K$ such as $$ \int_{|x-y|=\varepsilon}K(x,y)\,ds_y=0, \quad \varepsilon>0. $$ For example, the values of the simple-layer potential for the Laplace equation (n=3) $$ \frac{1}{4\pi}\int_S\frac{f(y)}{|x-y|}ds_y, \quad x\in S, $$ is a weakly singular integral on a smooth bounded surface $S$ and the double layer potential $$ \frac{1}{4\pi}\int_Sf(y)\frac{\partial }{\partial n_y}\frac1{|x-y|}ds_y, \quad x\in S, \qquad (*) $$ is singular, but not weakly singular, because $\left|\frac{\partial }{\partial n_y}\frac1{|x-y|}\right|\sim|x-y|^{-2}$, $S$ is a 2-dimensional surface and $$ \frac{1}{4\pi}\int_S\left|\frac{\partial }{\partial n_y}\frac1{|x-y|}\right|ds_y=+\infty $$ for some points $x\in S$. Nevertheless for smooth $f$ the integral $(*)$ converges in the sense of the principal value.

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Also, sometimes other regularization devices, such as analytic continuation, are necessary. – paul garrett May 19 '13 at 16:41
Thank you very much, I had some feeling that it's related to the power, but the relation to the actual number varied and I just failed to guess that the power is compared to dimension. – Juris May 19 '13 at 20:43

Roughly speaking, the differentiation of certain Cauchy principal singular integrals gived rise to hypersingular integrals which are interpreted in the Hadamard finite-part sense.

In 2D, if the singularity is 1/(t-x) and the integral is over some interval of t containing x, then the differentiation of the integral wrt x gives a hypersingular integral with 1/(t-x)^2. Hypersingular integrals are not integrals in the ordinary Riemman sense.

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