Streaklines compared with streamlines

Velocity field: $u = Bx(1+At) = x(1+0.5t)$, $v = Cy = y$.

Streamlines through (1, 1) at fixed t

$$\frac{dy}{dx} = \frac{y}{x(1+At)}\;\Rightarrow\; \ln y = \frac{1}{1+At}\ln x\;\Rightarrow\; y = x^{1/(1+0.5t)}$$

Points: for $x = 1, 2, 3, 4$: $t=0$: $y = 1, 2, 3, 4$; $t=1$: $y = 1, 1.587, 2.080, 2.520$; $t=2$: $y = 1, 1.414, 1.732, 2.0$.

Streakline

Take the streakline observed at $t = 3$ s, formed by all the particles that passed through (1, 1) at release times $\tau$ from 0 to 3 s. Integrate the path of a particle released at $\tau$:

$$\frac{dy}{dt} = y \Rightarrow y = e^{(t-\tau)},\qquad \frac{dx}{dt} = x(1+0.5t)\Rightarrow x = \exp\left[(t-\tau) + 0.25(t^2-\tau^2)\right]$$

At $t = 3$:

Eliminating $\tau$ (with $s = \ln y = t-\tau$): $x = y^{1+0.5t}e^{-0.25(\ln y)^2}$, and at $t=3$, $x = y^{2.5}e^{-0.25(\ln y)^2}$.

 y (m)
 20 |                                        * (190.6, 20.1)  tau = 0
    |
 12 |                     * (108.6, 12.2)  tau = 0.5
    |
  7 |           * (54.6, 7.4)  tau = 1
  4 |     * (24.2, 4.5)  tau = 1.5
  1 |*(1,1) tau = 3    [(3.3, 1.65) and (9.5, 2.7) lie close to the start]
    +------------------------------------------------- x (m)
     0         50        100       150       200

Comparison: a streamline is the instantaneous curve tangent to the velocity. Through (1, 1) it is the straight line $y = x$ at $t = 0$, and it becomes flatter with time ($y = x^{1/(1+0.5t)}$). The streakline at $t = 3$ s is much longer and lies well below the line $y = x$ (at $y = 20$ m, $x$ is already 190 m), because particles released earlier have been carried far in the $x$-direction by the growing $u$. It does not lie on any one streamline. The flow is unsteady, so streaklines, pathlines and streamlines are different curves; they would coincide only in steady flow.

Answer: streamlines $y = x,\ x^{2/3},\ x^{1/2}$ at $t = 0, 1, 2$ s; the streakline at $t = 3$ s runs from $(1,1)$ to $(190.6, 20.1)$, as given by the table above.