# Coupled pdes of the first order

May question is about possible approaches to solve the following system $$\begin{array}{rcl} \nabla{n}&=&n\,\mathbf{E},\\ \nabla\cdot\mathbf{E}&=&1-n, \end{array}$$ in general with some boundary conditions (BC). In particular, I'm interested to solve it in the polar coordinates $$\begin{array}{rcl} \partial_rn&=&nE_r,\\ \partial_{\varphi}n&=&rnE_{\varphi},\\ \partial_r{(rE_r)}+\partial_{\varphi}{(E_{\varphi})}&=&r(1-n), \end{array}$$ plus some BC.

Any suggestions?

Update 1

According to request of @nicoguarolet let's say that boundary conditions are as follows $$\begin{array}{c} \left.n(r,\varphi)\right|_{r=r_L}=n_0\approx1,~ \left.E_{\varphi}\equiv E_r(r,\varphi)\right|_{r=r_L}=E_0\approx0,\\ \left.\partial_rE(r,\varphi)\right|_{r=r_R}=0,\\ \end{array}$$ where $$[r,\,\varphi]\in[r_L,\,r_R]\times[\varphi_L,\,\varphi_R].$$ Hope, that I haven't missed somewhat.

• Boundary conditions are important for the solution of PDE, can you add them? – nicoguaro Apr 30 at 19:24
• @nicoguaro Original post was updated. – Oleg Kravchenko Apr 30 at 19:39

That's not a valid system of PDEs. In particular, the equation $$\nabla n = n\mathbf E$$ implies that you can find a function $$n$$ whose gradient equals a given vector field (n\mathbf E), but that's not possible for arbitrary vector fields. To see this, apply the curl to both sides: $$\nabla \times \nabla u = \nabla\times (n\mathbf E)$$ and remember that $$\nabla \times \nabla = 0$$. In other words, the equation will only have a solution if $$\nabla\times (n\mathbf E) = 0,$$ but your other equation does not guarantee that that is the case.