Integral form of balance law

Question

Let $\rho: \mathbb{R} \times [0, \infty) \to \mathbb{R}$ be the density, $f: \mathbb{R} \to \mathbb{R}$ the flux of the density and $g: \mathbb{R} \to \mathbb{R}$ the source (or loss) term. The equation

$$ \partial_t\rho(x,t) + \partial_x f(\rho(x,t)) + g(\rho(x,t)) = 0 $$

is called one-dimensional balance law.

For $g=0$ the balance law reduces to a conservation law, for which the integral-form is given by

$$ \int_{x_1}^{x_2} \rho(x,t_2) \,\mathrm{d}x - \int_{x_1}^{x_2} \rho(x,t_1) \,\mathrm{d}x = \int_{t_1}^{t_2} f(\rho(x_1,t)) \,\mathrm{d}t - \int_{t_1}^{t_2} f(\rho(x_2,t)) \,\mathrm{d}t. $$

which implies that the change of mass inside an interval $[x_1,x_2]$ during a time interval $[t_1,t_2]$ is defined by the difference of the inflow and the outflow at $x_1$ and $x_2$ during this time interval.

Is there any literature for a derivation of the integral form for the balance law with $g \neq 0$ inclusive physical explanation? Any hints or suggestions are highly appreciated.

Answer 1

I would say that you just need to proceed the same way to derive the balance law. For interpretations, assume that $f_\rho$, $g$ are both positive. If we integrate the local form of the balance law $$\rho_t + f(\rho)_x = -g(\rho)$$ over the domain $x\in [x_1, x_2]$ and times $t \in [t_1, t_2]$, we find $$ \left[\int_{x_1}^{x_2} \rho\, \text d x \right]^{t_2}_{t_1} = \left[\int_{t_1}^{t_2} f(\rho)\, \text d t \right]^{x_1}_{x_2} - \underset{[x_1,x_2] \times [t_1,t_2]}{\iint} g(\rho)\, \text{d}x \text d t \, . $$ The left-hand side is the variation of mass $\int \rho \, \text d x$ between the times $t_1$ and $t_2$. The right hand side includes a first term which represents the variation of mass flux between both boundaries of the spatial domain (the mass entering $[x_1,x_2]$ at $x=x_1$ minus the mass which has left at $x=x_2$). The last term represents the loss of mass over the domain considered here, or if you prefer, $\iint -g \, \text d x \text d t$ represents the mass production.