While solving a fluid dynamical problem, the following integral equation arised: $$ \iint_\mathscr{D} \frac{(x-x')^2 f(x',y') \, \mathrm{d}x' \mathrm{d}y'}{\left( (x-x')^2 + (y-y')^2 \right)^{3/2}} = 1 $$ for the unknown function $f$ on the domain $\mathscr{D}$ defined as a square of unit length centered at the origin, i.e. $x,y \in [-1/2,1/2]$
What I tried is to expand $f$ in the form of multivariate Taylor series and evaluate the resulting integrals term by term. This approach does not seem to be of help since identification with the right-hand side of the above equation is not possible.
It would be great if someone here could provide useful hints that could help solve this integral equation.
Thank you!
I know that I'm late, but this might help (though I'm not sure about the limits of integration.).
First of all, due to the polar symmetry in the denominator, we can do the following substitution: $$ \left\{ \begin{aligned} &x'=x+r\cos\theta\\ &y'=y+r\sin\theta \end{aligned} \right. $$
Thus, the integral becomes $$ \begin{aligned} \iint_{\mathscr{D}}\frac{(x'-x)^2f(x',y')\text dx'\text dy'}{((x'-x)^2+(y'-y)^2)^{\frac{3}{2}}}&=\iint_{\mathscr{D}}\frac{r^2\cos^2\theta f(x+r\cos\theta,y+r\sin\theta)r\text dr\text d\theta}{r^3}\\ &=\iint_{\mathscr{D}}\cos^2\theta f(x+r\cos\theta,y+r\sin\theta)\text dr\text d\theta \end{aligned} $$ Since a square region in polar coordinates is discontinuous, I'll do the same which is 8 times the right triangle delimited by the center of the square, a corner and the midpoint of a contiguous side. Therefore, $r(\theta)$. Namely, $$ \left\{ \begin{aligned} &1/2=x+r\cos\theta\iff r\in(0,(1/2-x)\sec\theta)\\ &\theta\in(0,\pi/4) \end{aligned} \right. $$ Finally, what's left is doing what you suggested, that is, approximating the function via Taylor series up to whatever order you want, solve the integrals and solve for the coefficients of the power series. The resulting integral should look like this: $$8\int_0^{\pi/4}\int_0^{(1/2-x)\sec\theta}\cos^2\theta(a_0+a_1(x+r\cos\theta)+a_2(y+r\sin\theta)+...)\text dr\text d\theta\overset{!}{=}1$$ I'd suggest developing the series around $(x',y')=(0,0)$.