Math Problem Statement

Consider a viscous fluid flowing with steady flow: \rho u \frac{du}{dx} = \mu \frac{d^2 u}{dx^2} - \frac{dP}{dx}

Given that \frac{dP}{dx} is constant, find u(x)

Solution

To find the velocity profile u(x)u(x) for the viscous fluid under the given conditions, we start with the equation:

ρududx=μd2udx2dPdx\rho u \frac{du}{dx} = \mu \frac{d^2 u}{dx^2} - \frac{dP}{dx}

Step 1: Analyze the given equation

We are given that dPdx\frac{dP}{dx} is constant, so let dPdx=C\frac{dP}{dx} = C where CC is a constant. The equation becomes:

ρududx=μd2udx2C\rho u \frac{du}{dx} = \mu \frac{d^2 u}{dx^2} - C

This is a second-order, nonlinear ordinary differential equation (ODE).

Step 2: Simplify and rearrange the equation

Rearrange the equation to isolate the second derivative of uu:

μd2udx2=ρududx+C\mu \frac{d^2 u}{dx^2} = \rho u \frac{du}{dx} + C

To simplify further, divide by μ\mu:

d2udx2=ρμududx+Cμ\frac{d^2 u}{dx^2} = \frac{\rho}{\mu} u \frac{du}{dx} + \frac{C}{\mu}

Step 3: Integrate the equation

This ODE can be challenging to solve directly, but we can try integrating with respect to xx by using an intermediate variable. Define:

v=dudxv = \frac{du}{dx}

This transforms the equation to:

dvdx=ρμuv+Cμ\frac{dv}{dx} = \frac{\rho}{\mu} u v + \frac{C}{\mu}

This is still a nonlinear differential equation, so a direct integration might not be straightforward. Let's multiply both sides by vv to see if the equation simplifies:

vdvdx=ρμuv2+Cμvv \frac{dv}{dx} = \frac{\rho}{\mu} u v^2 + \frac{C}{\mu} v

Now, notice that vdvdx=12ddx(v2)v \frac{dv}{dx} = \frac{1}{2} \frac{d}{dx} (v^2), so the equation becomes:

12ddx(v2)=ρμuv2+Cμv\frac{1}{2} \frac{d}{dx} (v^2) = \frac{\rho}{\mu} u v^2 + \frac{C}{\mu} v

This equation is still complex, but we can attempt a simpler approach by considering the boundary conditions.

Step 4: Assume a particular solution

Given the complexity, one approach is to assume a solution form based on typical velocity profiles. A common approach is to assume:

u(x)=Ax2+Bx+Du(x) = Ax^2 + Bx + D

where AA, BB, and DD are constants to be determined by substituting this into the original equation.

First, compute the first and second derivatives:

dudx=2Ax+B\frac{du}{dx} = 2Ax + B d2udx2=2A\frac{d^2 u}{dx^2} = 2A

Substitute into the original equation:

ρ(Ax2+Bx+D)(2Ax+B)=μ(2A)C\rho (Ax^2 + Bx + D)(2Ax + B) = \mu(2A) - C

This leads to a quadratic equation in xx that can be solved for the constants AA, BB, and DD. However, because the original equation is complex, it's often simpler to solve numerically or use specific boundary conditions to simplify further.

Final Velocity Profile u(x)u(x)

The general form of u(x)u(x) is quadratic:

u(x)=Ax2+Bx+Du(x) = Ax^2 + Bx + D

The constants AA, BB, and DD are determined by boundary conditions or further simplification depending on the specific problem setup (e.g., no-slip conditions at boundaries, symmetry, etc.).

Would you like to go through the integration in detail or discuss specific boundary conditions to determine the constants? Here are some related questions:

  1. What boundary conditions are typical for this type of fluid flow problem?
  2. How does viscosity affect the velocity profile compared to an inviscid fluid?
  3. What are common numerical methods for solving nonlinear differential equations?
  4. How would the solution change if the pressure gradient dPdx\frac{dP}{dx} was not constant?
  5. What physical scenarios might justify a quadratic velocity profile?

Tip: In fluid dynamics, simplifying assumptions and boundary conditions often play a critical role in solving complex differential equations. Be mindful of the physical context when choosing a method.

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Math Problem Analysis

Mathematical Concepts

Fluid Dynamics
Viscous Flow
Nonlinear Differential Equations

Formulas

Navier-Stokes equation
Boundary conditions for fluid flow

Theorems

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Suitable Grade Level

Advanced Undergraduate or Graduate Level