Graduate How are the equations of continuity derived in 1D?

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SUMMARY

The derivation of the equations of continuity in one-dimensional fluid mechanics is established through the transformation between Lagrangian and Eulerian descriptions. The equation of continuity is expressed as $$\partial_t \rho + \vec{\nabla} \cdot \vec{j} = 0$$ where $$\vec{j} = \rho \vec{v}$$ represents the mass-current density. The relationship between the Lagrangian velocity $$\vec{v}_L$$ and the Eulerian velocity $$\vec{v}$$ is crucial, as it allows for the expression of mass conservation in terms of the Eulerian framework. The derivation also incorporates the Jacobian of the transformation and the substantial time derivative, leading to the conclusion that the mass density and entropy are conserved in ideal fluid motion.

PREREQUISITES
  • Understanding of Lagrangian and Eulerian fluid dynamics
  • Familiarity with the concept of mass-current density in fluid mechanics
  • Knowledge of Jacobians in transformations
  • Basic principles of conservation laws in physics
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  • Study the derivation of the Euler equations in fluid dynamics
  • Learn about the substantial derivative and its applications in fluid mechanics
  • Explore the implications of the continuity equation in compressible and incompressible flows
  • Investigate the relationship between entropy and fluid dynamics in thermodynamic systems
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aliens123
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In Landau-Lifshitz Volume 6 Fluid-Mechanics the following problem is given
landau1.PNG
Where the equation of continuity is given earlier:
Landau2.PNG
As is Euler's equation:
landau3.PNG
And the equation of continuity for entropy:
landau4.PNG
I don't understand how this conclusion was reached. I can understand the derivation for the equation of continuity , but I have no idea how you could derive it from Euler's equation:
$$ \frac{\partial \rho}{\partial t} + \nablda \cdot ( \rho} \vec{v} )=0 $$
In 1d:
$$ \frac{\partial \rho}{\partial t} + \frac{\partial }{\partial a} ( \rho} v )=0 $$
I tried integrating this with
$$\int_{t_0}^{t} dt$$: $$\rho - \rho_0 + \frac{\partial }{\partial a} \int_{t_0}^{t} \rho \frac{\partial x}{\partial t}dt = 0$$
But I can't see any way to simplify this...

I also do not see how the next two equations were derived. Any help would be appreciated.
 
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This is simply the transformation between Lagrange and Euler descriptions of the fluid.

In the Lagrange description you use some standard configuration of a continuum mechanical system. For fluids you can choose the configuration at the initial time ##t##. Then one defines ##\vec{x}(t,\vec{y})## as the position of the fluid element at time ##t##, which has been located at ##\vec{y}## at time ##t=0##. We assume that the map ##\vec{y} \mapsto \vec{x}(t,\vec{y})## is smooth and invertible (a diffeomorphism).

Then the velocity of the fluid element is simply given by
$$\vec{v}_L(t,\vec{y})=\partial_t \vec{x}(t,\vec{y}).$$
That's the Lagrangian description of the fluid motion (and that's why I put an ##L## at the velocity).

The more familiar Eulerian description is found when thinking in terms of an observer looking at the fluid at position ##\vec{x}## and just describes the properties of the fluid element at time ##t## which is just at this position. Then you get the usual velocity field of fluid dynamics
$$\vec{v}(t,\vec{x})=\vec{v}_L[t,\vec{y}(t,\vec{x})],$$
where now we simply write ##\vec{y}(t,\vec{x})## for the inverse function of ##\vec{x}(t,\vec{y})##, i.e., ##\vec{y}(t,\vec{x})## gives the initial ##\vec{y}## of the fluid element which is at ##\vec{x}## at time ##t##.

From this you get the acceleration of a given fluid element as follows: In Lagrangian description it's simply
$$\vec{a}_L(t,\vec{y})=\partial_t \vec{v}_L(t,\vec{y}),$$
but now we want to express this in the Eulerian description. To that end we note that
$$\vec{v}_L(t,\vec{y})=\vec{v}[t,\vec{x}(t,\vec{y})].$$
And then we get
$$\vec{a}_L(t,\vec{y})=\vec{a}(t,\vec{x})=\partial_t \vec{v}[t,\vec{x}(t,\vec{y})]=\partial t \vec{v}(t,\vec{x}) + [\partial_t \vec{x}(t,\vec{y}) \cdot \vec{\nabla} \vec{v}(t,\vec{x}) = \partial_t \vec{v}(t,\vec{x}) + (\vec{v} \cdot\vec{\nabla}) \vec{v}(t,\vec{x}).$$
This you can generalized to the "substantial time derivative" of any quantity, and define
$$\mathrm{D}_t f(t,\vec{x})=\partial_t f(t,\vec{x}) + (\vec{v} \cdot \vec{\nabla}) f(t,\vec{x}).$$
Now again consider a fixed fluid element at the initial position ##\vec{y}##. Then the mass density in Lagrangian description is given by ##\rho_L(t,\vec{y})##.

Now for these given fluid particles the mass is conserved, i.e.,
$$\mathrm{d} m = \mathrm{d}^3 x \rho_L=\text{const}. \; \Rightarrow \; \partial_t \mathrm{d} m=0.$$
The partial time derivative has to be taken with ##\vec{y}## fixed.

In terms of the Eulerian density we have
$$\mathrm{d} m=\mathrm{d}^3 x \rho=\mathrm{d}^3 \vec{y} \mathrm{det} \frac{\partial \vec{x}}{\partial \vec{y}} \rho=\mathrm{d}^3 \vec{y} J(t,\vec{y}) \rho_L(t,\vec{y}).$$
Here ##J## is the Jacobian of the diffeomorphism ##\vec{y} \mapsto \vec{x}(t,\vec{y})##. The time derivative is given by
$$\partial_t J=(\vec{\nabla} \cdot \vec{v}) J.$$
From this you get
$$\partial_t \mathrm{d}m =\mathrm{d}^3 y J (\partial_t \rho_L(t,\vec{y}) + \rho_L \vec{\nabla} \cdot \vec{v}_L)=0.$$
Now we can rewrite this again in terms of the Eulerian description, using ##\partial_t \rho_L=\mathrm{D}_t \rho##:
$$\mathrm{D}_t \rho + \rho \vec{\nabla} \cdot \vec{v}=0$$
or writing the material time derivative out
$$\partial_t \rho + (\vec{v} \cdot \vec{\nabla}) \rho + \rho \vec{\nabla} \cdot \vec{v} = \partial_t \rho +\vec{\nabla} \cdot (\rho \vec{v})=0.$$
This is the equation of continuity for the mass,
$$\partial_t \rho + \vec{\nabla} \cdot \vec{j},$$
where ##\vec{j}=\rho \vec{v}## is the mass-current density.

That the ideal fluid motion is adiabatic characterizes the ideal fluid which by definition does not dissipate energy into heat, i.e., the entropy per unit mass ##s## is constant for the corresponding fluid element, which means
$$\partial_t s_L(t,\vec{y})=\mathrm{D}_t s(t,\vec{x})=\partial_t s + (\vec{v} \cdot \vec{\nabla})s=0.$$
This only means that the entropy of each fluid element is conserved, indeed we have
$$\partial_t (\rho s) + \vec{\nabla} \cdot (\vec{j} s)=(\partial_t \rho + \vec{\nabla} \cdot \vec{j}) + \rho \partial_t s +(\vec{j} \cdot \vec{\nabla}) s=\rho (\partial_t s + \vec{v} \cdot \vec{\nabla} s)=0.$$
 
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