Velocity Potential Function Analysis
Problem Statement
For the velocity components given as:
v = ax sin(xy)
1. Understanding the Relationship
For a velocity potential φ, the velocity components are related by:
u = ∂φ/∂x
v = ∂φ/∂y
Given:
u = ay sin(xy)
v = ax sin(xy)
2. Integration with Respect to x
We have:
∂φ/∂x = ay sin(xy)
Integrating with respect to x:
φ = ay ∫sin(xy) dx
Using the substitution t = xy, dt = y dx, dx = dt/y:
φ = ay ∫sin(t) (dt/y) = a ∫sin(t) dt = -a cos(xy) + f(y)
Where f(y) is a function that depends only on y.
3. Determining the Complete Potential
Step 3.1: Differentiating with respect to y:
∂φ/∂y = ax sin(xy) + f'(y)
Step 3.2: From the second condition:
∂φ/∂y = ax sin(xy)
Step 3.3: Comparing the two equations:
ax sin(xy) + f'(y) = ax sin(xy)
Therefore:
f'(y) = 0
Step 3.4: Integrating to find f(y):
f(y) = C
Where C is a constant.
Step 3.5: The complete velocity potential function is:
Verification
Let’s verify our solution by checking if the velocity components derived from φ match the given ones:
u = ∂φ/∂x = -a·(-sin(xy))·y = ay sin(xy)
v = ∂φ/∂y = -a·(-sin(xy))·x = ax sin(xy)
Given velocity components:
u = ay sin(xy)
v = ax sin(xy)
Physical Interpretation
This potential flow has the following characteristics:
- The flow is irrotational (curl V = 0) as expected for a potential flow
- The velocity components form a symmetric pattern where both x and y directions have similar influence
- The equipotential lines (φ = constant) represent lines where cos(xy) = constant
- The flow pattern has periodic behavior due to the trigonometric functions involved
- The constant ‘a’ serves as a scaling factor for the velocity field
