Continuity equation

Remarks

By the divergence theorem, the integral law says $\int_V \left( \partial_t \rho + \nabla \cdot (\rho \mathbf{u}) \right) dV = 0$ for every small ball $V$. The integrand is continuous, so if it were nonzero at a point it would keep its sign on a small ball around that point, and the integral would not vanish. The second form follows from $\nabla \cdot (\rho \mathbf{u}) = \mathbf{u} \cdot \nabla \rho + \rho \, \nabla \cdot \mathbf{u}$. Example: for constant density, the equation reduces to $\nabla \cdot \mathbf{u} = 0$, the condition of incompressible flow. Example: the one-dimensional gas $\rho(x, t) = 1/(1 + t)$, $u(x, t) = x/(1 + t)$ satisfies it, since $\partial_t \rho = -1/(1 + t)^2$ and $\partial_x (\rho u) = 1/(1 + t)^2$.

Going deeper

For a student

The continuity equation is the first of the equations of motion of a fluid, next to the momentum equation and the energy equation. It has the form of a conservation law $\partial_t \rho + \nabla \cdot \mathbf{j} = 0$ with flux $\mathbf{j} = \rho \mathbf{u}$, and the same form governs electric charge, probability in quantum mechanics and heat. In the divergence form it couples the density to the velocity; for an incompressible fluid it becomes the constraint $\nabla \cdot \mathbf{u} = 0$ in the Navier–Stokes equations. The passage from the integral law to the pointwise law needs the fields to be continuously differentiable; for flows with shocks only the integral form survives.