Consider a wire carrying a current $I$, I need to find the current density distribution in the wire of a cylinder shape. Let the density function be $j(x,y)$, in the circle $D:x^2+y^2
We have the simple relation that \begin{align} \iint_{D}j(x,y)dxdy=I \end{align}
Considering the magnetic field generated by the current, the density has a trend getting in the center because we know that when two wires enjoying the same direction of current parrallel to each other, they will have ampere force pointing to each other. However, the Coulomb force keep them apart. That is to say, the ampere force will equal the Coulomb force when they have the distribution. Thus the following relation is established: \begin{aligned} \forall(x,y)\in D,\iint_{D}\frac{\mu_0 j(u,v)}{2\pi \rho}e_{\rho}dudv=\iint_{D}-\frac{ j(u,v)}{2\pi\epsilon_0 \rho w^2}e_{\rho}dudv \end{aligned} this relation is set up, where $e_\rho$ is the unit vector pointing from $(x,y)$ to $(u,v)$, $w$ is the speed, $\rho=\sqrt{(u-x)^2+(v-y)^2}$ is the distance, So \begin{aligned} \forall(x,y)\in D,\iint_{D}\frac{j(u,v)}{\rho}e_{\rho} dudv=0 \end{aligned} But I have no idea how to evaluate $j(x,y)$, thanks for your attention!