Given the 2-D unsteady Eulerian fluid velocity field
$$\underline{u}(\underline{r}, t) = a\Big[-y\underline{\hat{\imath}} + (x – bt)\underline{\hat{\jmath}}\Big]\quad\text{with}\quad\underline{r}_0 = (x_0,y_0,z_0)$$
find(i) the equation of the pathline for the fluid particle at $\underline{r}_0$ at $t = 0$
(ii)the equation of the streamline at time $t$ which passes through $\underline{r}_0$
So I see that
$$\frac{dx}{dt} = -ay,\,\frac{dy}{dt} = ax – abt,\,\frac{dz}{dt} = 0\quad\text{with}\quad x=x_0,\, y=y_0,\, z=z_0\,\,\text{at}\,\, t=0$$
So then I start to solve
$$z=z_0,\,\frac{d^2x}{dt^2} = -a\frac{dy}{dt} = -a^2x + a^2bt$$
Then (after lots of working) this leads to a sol. of
$$x = x_0\cos(at) + B\sin(at) + bt,\,\, y = \Big(y_0 + \frac{b}{a}\Big)\cos(at) + D\sin(at) – \frac{b}{a}$$
Now, my problem is that I don't know what to do next to find the Pathline? How do I find these two unknown constants?
Best Answer
My solution is different than yours. The pathlines are curves $(x(t), y(t), z(t))$ satisfying the system of ODEs \begin{alignat*}{3} x'(t) & = -ay(t), && \qquad x(0) && = x_0 \tag{1}\\ y'(t) & = a(x(t) - bt), && \qquad y(0) && = y_0 \tag{2}\\ z'(t) & = 0, && \qquad z(0) && = z_0. \tag{3} \end{alignat*} Equation $(3)$ implies that $z(t) = z_0$. Differentiating $(1)$ with respect to $t$ and substituting $(2)$ we find that $$ x''(t) = -ay'(t) = -a^2(x(t) - bt) $$ and rearranging yields $$ x''(t) + a^2x(t) = a^2bt. $$ It can be shown that the general solution is $$ x(t) = A\cos(at) + B\sin(at) + bt. $$ The initial condition $x(0) = x_0$ gives $A = x_0$, and the initial condition $x'(0) = -ay(0) = -ay_0$ (This is essentially $(1)$ at time $t=0$) gives $$ -ay_0 = aB + b \implies B = \frac{-ay_0 - b}{a} = -y_0 - \frac{b}{a}. $$ Consequently, $$ x(t) = x_0\cos(at) - \left(y_0 + \frac{b}{a}\right)\sin(at) + bt $$ and the equation of the pathline for any time $t\ge 0$ is $$ \begin{cases} \, x(t) & = x_0\cos(at) - \left(y_0 + \frac{b}{a}\right)\sin(at) + bt \\ \, y(t) & = -\frac{1}{a}\Big[-ax_0\sin(at) - \left(ay_0 + b\right)\cos(at) + b\Big] \\ \, z(t) & = z_0. \end{cases} $$