- #1
sara_87
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Homework Statement
If we want to show whether a kernel is weakly singular or not, what do we do?
eg. consider:
a) [tex]\int_0^x sin(x-s)y(s)ds[/tex]
b) [tex]\int_{-3}^3 \frac{y(s)}{x-s}ds[/tex]
Homework Equations
A discontinuous kernel k(x; s) is weakly singular (at x = s) if k is continu-
ous when x [tex]\neq[/tex] s and if [tex]\exists[/tex] constants v[tex]\in[/tex] (0; 1) and c > 0 such that [tex]\left|k(x,s)\right|\leq c \left|x-s\right|^{-v}[/tex] for x [tex]\neq[/tex] s on its set of defnition.
The Attempt at a Solution
a) I don't think this is weakly singular because when x=s, the kernel is continuous.
b) when x=s, the kernel is discontinuous because the equation tends to infinity. and we can say that v=1/2 and c=1.
I feel like this is wrong. but even if it's right, i think it lacks explanation