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Chapter 9: Problem 6

Zeigen Sie, dass f眉r ein 3-mal stetig differenzierbares divergenzfreies Geschwindigkeitsfeld \(\mathbf{u}=(u, v)\) aus den instation盲ren STOKES- Gleichungen $$ \frac{\partial \mathbf{u}}{\partial t}=-\operatorname{grad} p+\frac{1}{R e} \Delta \mathbf{u} $$ f眉r den Druck die Gleichung \(\Delta p=0\) folgt.

Short Answer

Expert verified
For a divergence-free velocity field, the pressure must satisfy \( \Delta p = 0 \) under the Stokes equations.

Step by step solution

01

Understand the Stokes Equations

The Stokes equations describe the motion of an incompressible fluid and are given for the velocity field \( \mathbf{u} = (u, v) \) and pressure \( p \). In this exercise, we need to show that if the velocity field satisfies these equations, the pressure must satisfy \( \Delta p = 0 \). The Stokes equations in this context are \( \frac{\partial \mathbf{u}}{\partial t} = -\operatorname{grad} p + \frac{1}{Re} \Delta \mathbf{u} \).
02

Express Divergence-Free Condition

Since \( \mathbf{u} \) is divergence-free, we have \( \operatorname{div} \mathbf{u} = 0 \). This condition implies that the flow is incompressible, meaning the volume of any fluid element does not change as it moves.
03

Apply Divergence to the Stokes Equations

Take the divergence of both sides of the Stokes equations: \( \operatorname{div} \left( \frac{\partial \mathbf{u}}{\partial t} \right) = \operatorname{div} \left( -\operatorname{grad} p + \frac{1}{Re} \Delta \mathbf{u} \right) \). This simplifies to \( 0 = -\Delta p + \frac{1}{Re} \operatorname{div} (\Delta \mathbf{u}) \).
04

Simplify Using Divergence-Free Condition

Using the divergence-free condition \( \operatorname{div} \mathbf{u} = 0 \) repeatedly for smooth functions, \( \operatorname{div}(\Delta \mathbf{u}) = 0 \) because the Laplacian \( \Delta \) is a linear differential operator. This further simplifies our equation to \( -\Delta p = 0 \).
05

Conclude the Pressure Condition

Since we have \( -\Delta p = 0 \), it follows that \( \Delta p = 0 \). This shows that the pressure \( p \) satisfies the Laplace equation, given a divergence-free velocity field from the Stokes equations.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Divergence-Free Condition
The concept of being divergence-free is crucial in understanding fluid dynamics, especially for incompressible fluids. When we say that a velocity field \( \mathbf{u} = (u, v) \) is divergence-free, it implies that \( \text{div} \mathbf{u} = 0 \). This condition can be thought of as a mathematical way to express that the fluid density remains constant as it flows. To put it simply:
  • No new fluid is created or lost within a given volume.
  • The flow conserves mass as it moves.
For this reason, the divergence-free condition is often synonymous with incompressible flow in fluid dynamics. As we explored in the step-by-step solution, this condition simplifies the equations we work with, leading us to conclude properties such as the pressure being governed by a Laplace equation.
Incompressible Fluid
An incompressible fluid is one whose density remains unchanged regardless of the pressure applied. This concept is commonly represented by the condition \( \text{div} \mathbf{u} = 0 \) in fluid dynamics equations. In practical terms:
  • The volume of fluid elements remains constant during flow.
  • The mass conservation aligns with the divergence-free condition mentioned earlier.
The Stokes equations assume fluid incompressibility, simplifying the analysis by removing changes in density as a variable. As such, the incompressible fluid assumption allows analysts to focus directly on factors influencing the motion and pressure within the fluid.
Laplacian Operator
The Laplacian operator, denoted by \( \Delta \), is a crucial element in mathematical physics, particularly within the realm of fluid dynamics. It applies to scalar or vector fields, helping to define fields鈥� diffusions and rotations. The basic idea is:
  • It measures the rate at which a function diverges from its average value in the surrounding region.
  • In the context of a pressure field, it indicates how pressure changes spread out in a fluid.
In the given exercise, we take the Laplacian of both the pressure \( p \) and velocity \( \mathbf{u} \) fields. This step is important to simplify and understand Stokes equations. The use of Laplacian directly aids in demonstrating that under certain conditions, the pressure \( p \) satisfies the equation \( \Delta p = 0 \). This condition essentially points to a stable pressure distribution in the absence of other forces.
Velocity Field
A velocity field, \( \mathbf{u} = (u,v) \) in fluid dynamics, describes the velocity of every point in a fluid. This field gives a comprehensive view of the fluid鈥檚 movement throughout its domain.
  • It assigns a velocity vector to each point in space and time, illustrating how fluid particles move.
  • Understanding the velocity field is key to solving fluid flows and understanding the behavior of the fluid.
In the Stokes equations, the velocity field is assumed to be smooth and divergence-free. This assumption allows mathematicians and engineers to derive critical flow properties, including the incompressibility and Laplacian behaviour inherent in the system. Ultimately, the velocity field gives insights into how other flow characteristics, such as pressure, can be efficiently managed and predicted.

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Most popular questions from this chapter

Bestimmen Sie eine divergenzfreie L枚sung \(\mathbf{u}(x, y)=(u(x, y), v(x, y))\), d.h. div \(\mathbf{u}=0\), des Differentialgleichungssystems $$ \begin{aligned} u \frac{\partial u}{\partial x}+v \frac{\partial u}{\partial y} &=-\frac{\partial p}{\partial x}+\frac{1}{R e} \Delta u \\ u \frac{\partial v}{\partial x}+v \frac{\partial v}{\partial y} &=-\frac{\partial p}{\partial y}+\frac{1}{R e} \Delta v \end{aligned} $$ in einem Rechteckgebiet \(\Omega=\\{(x, y) \mid 0

Bestimmen Sie mit Hilfe des Separationsansatzes L枚sungen \(u(x, y)\) bzw. \(u(x, t)\) f眉r die partiellen Differentialgleichungen (a) \(u_{x x}=4 u_{y}, u(0, y)=u(\pi, y)=0\) (b) \(a^{2} u_{x x}=u_{t t}, a>0\)

Transformieren Sie die Differentialgleichung $$ \begin{array}{r} u_{x x}-u_{y y}=0 \text { f眉r }\left|x^{2}-y^{2}\right| \leq 1 \\ u(x, y)=x^{2}+y^{2} \text { f眉r }\left|x^{2}-y^{2}\right|=1 \end{array} $$ auf Hyperbelkoordinaten. Hinweis: Hyperbelkoordinaten sind durch die Transformation $$ \mathbf{x}(\rho, \phi)=\left(\begin{array}{l} x(\rho, \phi) \\ y(r, \phi) \end{array}\right):=\left(\begin{array}{l} \rho \cosh \phi \\ \rho \sinh \phi \end{array}\right) $$ gegeben.

Zeigen Sie, dass f眉r eine zweimal stetig differenzierbare L枚sung \(u(x, y)\) des Randwertproblems $$ -\Delta u=f \quad \text { in } \Omega, \quad u=0 \quad \text { auf } \Gamma_{d}, \quad \frac{\partial u}{\partial \mathbf{n}}=q \quad \text { auf } \Gamma_{n} $$ die Integralgleichung $$ \int_{\Omega}[\nabla u \cdot \nabla h-f h] d F-\int_{\Gamma_{n}} q h d s=0 $$ gilt, wobei \(f, q\) gegebene, integrierbare Funktionen sind und \(\Gamma=\Gamma_{d} \cup \Gamma_{n}, \Gamma_{d} \cap\) \(\Gamma_{n}=\emptyset\), der Rand von \(\Omega\) ist. \(h\) sei stetig differenzierbar und auf \(\Gamma_{d}\) gleich Null. \(\frac{\partial}{\partial \mathbf{n}}\) bezeichnet die Ableitung in Richtung der 盲u脽eren Normalen auf dem Rand \(\operatorname{von} \Omega\)
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