91Ó°ÊÓ

A velocity field describes how the velocity of a fluid changes in space over time. It is essential in understanding how fluid particles move through a medium. In this particular exercise, the velocity field is given by the vector equation \( \mathbf{v} = 5x\mathbf{i} + 2y\mathbf{j} \), where \( x \) and \( y \) are spatial coordinates, and \( \mathbf{i} \) and \( \mathbf{j} \) are unit vectors in the x and y directions, respectively.
This velocity field indicates that the velocity of a particle increases with its x-coordinate and y-coordinate, which means the further you are in the positive x or y direction, the faster the particle moves.

By breaking it into components, \( u(x,y) = 5x \) and \( v(x,y) = 2y \), we can analyze how velocity is distributed in each directional axis. This breakdown is crucial because it tells us that the motion in the x-direction depends directly on the x-position, and similarly, the y-motion depends directly on the y-position. Understanding this dependence is a vital part of velocity field analysis.

Differential equations are used to model changes and dynamics in fluid systems. In this context, they describe the rate of change of a particle's position in terms of time and position-related factors.
For the given problem, the differential equations are:

• \( \frac{dx}{dt} = u(x,y) = 5x \)
• \( \frac{dy}{dt} = v(x,y) = 2y \)

These equations are first-order, meaning they relate first derivatives to the function itself. By solving these equations, we can determine how a fluid particle traces a path over time, called a pathline.
The solution involves separating variables and integrating, which generally involves some constants of integration that depend on initial conditions. Solving these differential equations helps to derive the functional form of the pathlines \( x(t) \) and \( y(t) \), which describe how positions in x and y evolve over time.

Initial conditions specify the state of a fluid system at the beginning of analysis. They are necessary to uniquely determine the solution to a differential equation.
In fluid dynamics, initial conditions typically involve the initial positions and velocities of fluid particles. For the given exercise, we know the particle originates at the point (1, 2). Using this information, we can find the specific values of the constants from our integrated equations.

Applying these initial conditions:

• For \( x(t) = C_2 e^{5t} \): Set \( t = 0 \) to find \( C_2 \). Since \( x(0) = 1 \), \( C_2 = 1 \).
• For \( y(t) = C_4 e^{2t} \): Set \( t = 0 \) to find \( C_4 \). Since \( y(0) = 2 \), \( C_4 = 2 \).

These initial condition values are crucial as they tailor the general solutions from the differential equations to the specific conditions of the problem.

Boundary conditions are constraints necessary to define the feasible domain of the fluid flow. They dictate how the fluid can behave at the edges or boundaries of its environment. In this exercise, boundary conditions are given as constraints on the x and y domain: \( 0 \leq x \leq 100 \) m and \( 0 \leq y \leq 100 \) m.
To determine the behavior of the pathlines within these boundaries, we check when the pathline equations \( x(t) = e^{5t} \) and \( y(t) = 2e^{2t} \) exceed these limits:

• For \( x(t): e^{5t} \leq 100 \), solving gives \( t \leq \frac{\ln(100)}{5} \approx 0.92 \).
• For \( y(t): 2e^{2t} \leq 100 \), solving gives \( t \leq \frac{\ln(50)}{2} \approx 1.92 \).

The minimum of these values \( t \leq 0.92 \) defines how long the particle remains within this domain. Analyzing and applying boundary conditions ensure that the pathline calculations reflect realistic physical constraints of the system.