Ideal Fluids in Motion — Question 4

Question 4

Streamline crossing has an exception at a stagnation point

Consider the steady field $\mathbf v=(ax,-ay)$ with $a>0$. Away from the axes, a streamline satisfies $dy/dx=v_y/v_x$. At the origin, both velocity components vanish.

1. Derive the streamline family away from the axes. Explain why two distinct streamlines cannot cross at a point where velocity is nonzero and single-valued.

2. Describe motion on the positive $y$ axis. Does a parcel starting at $y_0>0$ reach the origin in finite time and then choose one of the $x$-axis branches?

Question 4 - Solution

Streamline crossing has an exception at a stagnation point

(a) Derivation and calculation

Away from the axes,$$\frac{dy}{dx}=-\frac yx\Rightarrow\frac{dy}{y}=-\frac{dx}{x}.$$$$\ln|y|+\ln|x|=C\Rightarrow\boxed{xy=C_1}.$$A smooth single-valued nonzero velocity field determines a unique local trajectory; distinct streamlines cannot cross transversely there.

(b) Result and interpretation

On the positive $y$ axis,$$x=0,\qquad\dot y=-ay\Rightarrow\boxed{y=y_0e^{-at}}.$$$$t=\frac1a\ln\frac{y_0}{y}\to\infty\quad(y\to0^+).$$The parcel never reaches the origin in finite time. The geometrical meeting of axis branches at the stagnation point does not permit arbitrary parcel splitting.

Physical check

At the origin $\mathbf v=0$, so the usual nonzero-vector tangent-direction argument does not apply.