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Find velocity component from continuity

Short answer

Continuity gives dv/dy=-2xy; integrating with v=0 at y=0 gives v=-xy^2. Vorticity zeta_z = dv/dx - du/dy = -(x^2+y^2), so the flow is rotational.

The x-component of velocity in a steady, incompressible two-dimensional flow is u=x2yu = x^2 y. Using the continuity equation, find the y-component vv, given that v=0v = 0 at y=0y = 0. Also find the z-component of vorticity.

Given: u=x2yu=x^2y, incompressible 2D flow, v=0v=0 at y=0y=0.

Continuity

∂u∂x+∂v∂y=0\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0 ∂u∂x=2xy ⇒ ∂v∂y=−2xy\frac{\partial u}{\partial x}=2xy\ \Rightarrow\ \frac{\partial v}{\partial y}=-2xy

Integrate with respect to yy:

v=−xy2+f(x)v=-xy^2+f(x)

The condition v=0v=0 at y=0y=0 for all xx gives f(x)=0f(x)=0. Hence

v=−xy2v=-xy^2

Vorticity

ζz=∂v∂x−∂u∂y=(−y2)−(x2)=−(x2+y2)\zeta_z=\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}=(-y^2)-(x^2)=-(x^2+y^2)

The vorticity is not zero, so the flow is rotational. Its magnitude increases with distance from the origin, and the minus sign means clockwise rotation.

Answer: v=−xy2v=-xy^2; ζz=−(x2+y2)\zeta_z=-(x^2+y^2) s−1^{-1} (rotational).

✓ Answer checked by Claude Opus 5.5 (high effort)

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