Mass Conservation Law in the Inviscid Burgers' Equation
Theorem
$$ \begin{cases} u_{t} + u u_{x} = 0 & , t>0 \\ u(t,x) = f(x) & , t=0 \end{cases} $$
Considering the solution $u$ of the inviscid Burgers’ equation up to the interval $[a,b]$, let’s define the mass $M$ in the following way.
$$ M_{a,b}(t) := \int_{a}^{b} u(t,x) dx $$
And if we call the burst time $t_{\ast}$, the following holds for $t \in ( 0 , t_{\ast})$.
$$ {{d} \over {dt}} M_{a,b}(t) = - \left( {{1} \over {2}} u^2 (t,b) - {{1} \over {2}} u^2 (t,a) \right) $$
Description
The burst time is not mathematically a function and physically signifies a moment when multiple states overlap simultaneously.
To simply put it in terms of formulas, the change in mass equates to the net sum of what goes in and out.
The right-hand side $\displaystyle {{1} \over {2}} u^2 (t,b)$ is called the outflux, and $\displaystyle {{1} \over {2}} u^2 (t,a)$ is called the influx. If the change in $M_{a,b}(t)$ over time is negative, it means that $u$ is diminishing on $[a,b]$, indicating that outflux is greater than influx. Conversely, a positive means the mass is increasing, which implies influx is greater than outflux, aligning the expression meaningfully.
Putting this together, $\displaystyle F(u) := {{1} \over {2}} u^2$ is referred to as the Flux Function.
$$ {{\partial u} \over {\partial t}} + {{\partial u} \over {\partial x}} F(u) =0 $$
In this sense, the Burgers’ equation is also referred to as the mass conservation law in $1$ dimensions.
Derivation
Since $u$ is a continuous function, the following holds.
$$ {{d } \over {dt} } \int_{a}^{b} u(t,x) dx = \int_{a}^{b} {{\partial } \over { \partial t} } u(t,x) dx $$
Since $u_{t} = - u u_{x}$, the following holds.
$$ \int_{a}^{b} {{\partial } \over { \partial t} } u(t,x) dx = - \int_{a}^{b} u u_{x} dx $$
Given $\displaystyle u u_{x} = {{\partial} \over {\partial x}} \left( {{1 } \over {2}} u^2 \right)$ and $u_{t} = - u u_{x}$, the following holds.
$$ - \int_{a}^{b} u u_{x} dx = - \int_{a}^{b} {{\partial} \over {\partial x}} \left( {{1} \over {2}} u^2 \right) dx = - \left( {{1} \over {2}} u^2 (t,b) - {{1} \over {2}} u^2 (t,a) \right) $$
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Especially when $\displaystyle \int_{-\infty}^{\infty} f(x) dx < \infty$, the following holds for $t \in [0,t_{\ast})$.
$$ \int_{-\infty}^{\infty} u(t,x) dx = \int_{-\infty}^{\infty} f(x) dx $$