logo

Mass Conservation Law in the Inviscid Burgers' Equation 📂Partial Differential Equations

Mass Conservation Law in the Inviscid Burgers' Equation

Theorem

{ut+uux=0,t>0u(t,x)=f(x),t=0 \begin{cases} u_{t} + u u_{x} = 0 & , t>0 \\ u(t,x) = f(x) & , t=0 \end{cases}

Considering the solution uu of the inviscid Burgers’ equation up to the interval [a,b][a,b], let’s define the mass MM in the following way.

Ma,b(t):=abu(t,x)dx M_{a,b}(t) := \int_{a}^{b} u(t,x) dx

And if we call the burst time tt_{\ast}, the following holds for t(0,t)t \in ( 0 , t_{\ast}).

ddtMa,b(t)=(12u2(t,b)12u2(t,a)) {{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 12u2(t,b)\displaystyle {{1} \over {2}} u^2 (t,b) is called the outflux, and 12u2(t,a)\displaystyle {{1} \over {2}} u^2 (t,a) is called the influx. If the change in Ma,b(t)M_{a,b}(t) over time is negative, it means that uu is diminishing on [a,b][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, F(u):=12u2\displaystyle F(u) := {{1} \over {2}} u^2 is referred to as the Flux Function.

ut+uxF(u)=0 {{\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 11 dimensions.

Derivation

Since uu is a continuous function, the following holds.

ddtabu(t,x)dx=abtu(t,x)dx {{d } \over {dt} } \int_{a}^{b} u(t,x) dx = \int_{a}^{b} {{\partial } \over { \partial t} } u(t,x) dx

Since ut=uuxu_{t} = - u u_{x}, the following holds.

abtu(t,x)dx=abuuxdx \int_{a}^{b} {{\partial } \over { \partial t} } u(t,x) dx = - \int_{a}^{b} u u_{x} dx

Given uux=x(12u2)\displaystyle u u_{x} = {{\partial} \over {\partial x}} \left( {{1 } \over {2}} u^2 \right) and ut=uuxu_{t} = - u u_{x}, the following holds.

abuuxdx=abx(12u2)dx=(12u2(t,b)12u2(t,a)) - \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)


Especially when f(x)dx<\displaystyle \int_{-\infty}^{\infty} f(x) dx < \infty, the following holds for t[0,t)t \in [0,t_{\ast}).

u(t,x)dx=f(x)dx \int_{-\infty}^{\infty} u(t,x) dx = \int_{-\infty}^{\infty} f(x) dx