I was wondering if I can solve somehow the charge conservation equation on a specific domain and boundary conditions.
There is no time dependency, thus the equation reads:
$$\vec{\nabla}\cdot\vec{J}=0$$
Or in other words:
$$\frac{\partial J_x}{\partial x}+\frac{\partial J_y}{\partial y}=0$$
The domain is a rectangle with edges $(a,b)$.
I'll define the edges as follows
$t$ - Is the top edge
$b$ - Is the bottom edge
$l$ - Is the left edge
$r$ - Is the right edge
The $x$ axis is perpendicular to the left and right edges and the $y$ axis is perpendicular to the top and bottom edges.
The left and bottom edges are isolated boundaries.
The top edge is where all current should come out.
and the right edge is the current source.
So, the only boundary conditions that I know are:
$$J_y(b)=0$$
$$J_x(l)=0$$
$$J_x(t)=0$$
$$J_x(r)=f(y)$$
Where $f(y)$ is a known function.
So, is this kind of differential equation is solvable?
And, does these four boundary conditions are enough to produce a unique solution?
And if so, how can I solve it?
Thanks
Unfortunately not. The reason is that the problem is not well-posed. The reason for this is that you are trying to find two functions $J_x$ and $J_y$ (because $\bf J$ is a vector with two components) but you have only one equation that $J_x$ and $J_y$ should satisfy.
It is possible to demonstrate that you have infinite solutions. In particular, if ${\bf J}(x,y)=(J_x(x,y),J_y(x,y))^T$ is a solution of $\nabla \cdot {\bf J}=0$, then ${\bf J}(x,y)=(J_x(x,y),J_y(x,y)+h(x))^T$ is again a solution for any functions $h(x)$ with the only condition $h(b)=0$.