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PoissonPlusHydro

-              v14
+              v15
 = Update Formulas =
+To solve the first source term ODE,
+[[latex($\frac{d}{dt}(\rho u) = - \rho \triangledown \phi$)]]
+replace with discrete derivatives,
+[[latex($\frac{(\rho u)_i^{n+1} - (\rho u)_i ^{n+1/2}}{dt} = -\frac{\rho_i^{n+1/2}(\phi_{i+1}^{n+1/2}- \phi_i^{n+1/2})}{dx}$)]]
+which becomes,
+[[latex($\frac{\rho_i^{n+1}(u_i^{n+1} - u_i^{n+1/2})}{dt} = -\frac{\rho_i^{n+1/2}(\phi_{i+1}^{n+1/2}- \phi_i^{n+1/2})}{dx}$)]]
+since
+[[latex($\rho_i^{n+1} = \rho_i^{n+1/2}$)]]
+we can just cancel the rho's and solve for the updated u,
+[[latex($u_i^{n+1}=-\frac{dt}{dx}[\phi_{i+1}^{n+1/2}-\phi_i^{n+1/2} ] + u_i^{n+1/2}$)]]
+This requires a dt (which is the same for the hydro step, solved using the CFL condition), and a gravitational potential phi. Phi is solved using the Jacobi solver, with the source function given by
+[[latex($f(i)=4 \pi G (\rho_i^{n+1/2} - \bar{\rho})$)]]
+and boundary conditions
+[[latex($\phi(0) = \phi(mx), ~\phi(mx+1) = \phi(1)$)]]
+To solve the 2nd source term ODE,