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# 数学代写|偏微分方程作业代写Partial Differential Equations代考|The Euler Equations: A Derivation

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## 数学代写|偏微分方程作业代写Partial Differential Equations代考|Conservation of Mass and the Continuity Equation

Consider the fluid inside a set fixed piece $W \subseteq \Omega$.
The mass of the fluid in $W$ at time $t$ is
$$m_{W}(t):=\iiint_{W} \rho(x, y, z, t) d x d y d z .$$
Note that $m_{W}(t)$ has dimensions of mass. We use differentiation under the integral sign (cf. Section A.9) to find that
$$\frac{d m_{W}(t)}{d t}=\iiint_{W} \rho_{t}(x, y, z, t) d x d y d z .$$
Now we ask the fundamental question: How can $m_{W}$, the total mass of fluid in $W$, change with time? The simple answer is that fluid can enter or leave $W$ across the boundary $\partial W$. At any point $(x, y, z)$ on the boundary and time $t$, the fluid flux across the boundary is instantaneously $\rho \mathbf{u} \cdot \mathbf{n}$, where $\mathbf{n}$ denotes the outer unit normal to $W$. Note that the fluid flux $\rho \mathbf{u} \cdot \mathbf{n}$ has dimensions of mass per unit area per unit time. The total (net) amount of fluid leaving $W$ at time $t$ is then
$$\iint_{\partial W} \rho \mathbf{u} \cdot \mathbf{n} d S .$$
Note that the above has dimensions of mass per unit time. Balancing (2.54) with (2.55), we find
$$\iiint_{W} \rho_{t}(x, y, z, t) d x d y d z=-\iint_{\partial W} \rho \mathbf{u} \cdot \mathbf{n} d S .$$

## 数学代写|偏微分方程作业代写Partial Differential Equations代考|Conservation of Linear Momentum and Pressure

We now turn to the conservation of linear momentum (mass $\times$ velocity). This conservation principle is dictated by Newton’s Second Law which states that the rate of change of linear momentum must be balanced by the net forces. Let us first motivate why computing the time rate of change of linear momentum is far more complicated than simply taking the partial derivative with respect to $t$. We already implicitly addressed this in deriving the continuity equation but it is useful to focus on this point for spatial descriptions of velocity and acceleration. If you are familiar with the notion of a material derivative (cf. Section 2.8.6), or comfortable with the derivation of the continuity equation, you may skip the following comment (digression).

## 数学代写|偏微分方程作业代写PARTIAL DIFFERENTIAL EQUATIONS代考|Gas Dynamics: The Compressible Euler Equations

We consider an ideal gas, for example air. A gas is an example of a compressible fluid; as a set region of the fluid (a set collection of fluid particles) moves with time, its volume can change. Alternatively, the density of the fluid can vary with space and time. So what is pressure $p$ in a gas? Is it an additional state variable which we need the PDEs to solve for? No, for a compressible fluid, the pressure at any point and time is a function of the density; that is,
$$p(x, y, z, t)=f(\rho(x, y, z, t)),$$
for some function $f$. As you might expect, $f$ is an increasing function as, intuitively, the greater the density the greater the pressure. The particular choice of function will depend on the exact nature of the gas. For many gases, the function $f$ is simply a power law; i.e., $f(\rho)=c_{0} \rho^{\gamma}$, for some constants $c_{0}>0$ and $\gamma>0$.

Substituting (2.64) into (2.63), we arrive at the compressible Euler equations:
$$\left{\begin{array}{l} \frac{\partial \mathbf{u}}{\partial t}+\mathbf{u} \cdot \nabla \mathbf{u}=-\frac{1}{\rho} \nabla f(\rho) \ \frac{\partial \rho}{\partial t}+\operatorname{div}(\rho \mathbf{u})=0 \end{array}\right.$$
These are equations for gas dynamics and consist of a system of four partial differential equations in four unknown state variables $\left(\rho, u_{1}, u_{2}, u_{3}\right)$. These Euler equations are quasilinear first order. The reader should note the parallel to the inviscid Burgers equation in $1 \mathrm{D}\left(u u_{x}\right.$ vs. $\left.\mathbf{u} \cdot \nabla \mathbf{u}\right)$. Indeed, discontinuities and shock waves are signature solutions to these equations of gas dynamics.

## 数学代写|偏微分方程作业代写PARTIAL DIFFERENTIAL EQUATIONS代考|CONSERVATION OF MASS AND THE CONTINUITY EQUATION

d米在(吨)d吨=∭在ρ吨(X,是,和,吨)dXd是d和.

∬∂在ρ在⋅nd小号.

∭在ρ吨(X,是,和,吨)dXd是d和=−∬∂在ρ在⋅nd小号.

## 数学代写|偏微分方程作业代写PARTIAL DIFFERENTIAL EQUATIONS代考|GAS DYNAMICS: THE COMPRESSIBLE EULER EQUATIONS

p(X,是,和,吨)=F(ρ(X,是,和,吨)),

$$\left{∂在∂吨+在⋅∇在=−1ρ∇F(ρ) ∂ρ∂吨+div⁡(ρ在)=0\对。$$