Fluid Mechanics Questions A steady two-dimensional flow (pure straining) is given by $u = k x$, $v = -k y$, for $k$ constant. 


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*Find the equation for a general streamline of the flow.

*At $t = 0$,the fluid on the curve $x^2 + y^2 = a^2$ is marked. Find the equation for this material fluid curve for $t > 0$.

*Does the area within the curve change in time, and why?


My Attempt:
I have solved a differential equation to find the streamlines by eliminating the parameter $s$; I get $xy  = A$ where $A$ is some constant. 
I am stuck on the second part; I know that the initial position lies on the circle given, but how do I find the equation of the material curve? Will it be a function of $x$, $y$ and $t$?
Any help would be greatly appreciated!
 A: Part (2)
In the Lagrangian description of the flow, a fluid particle at a point $\mathbf{x}_0=(x_0,y_0)$ at time $t = 0$ is located at a point $\mathbf{x} =(x,y)$ at time $t > 0$ with coordinates given by
$$x= X(t;\mathbf{x}_0), \,\, y = Y(t;\mathbf{x}_0),$$
where the Eulerian velocity field is related to the Lagrangian mapping by
$$\tag{*}\frac{d}{dt}X(t;\mathbf{x}_0) = u[X(t;\mathbf{x}_0),Y(t;\mathbf{x}_0),t],\\ \frac{d}{dt}Y(t;\mathbf{x}_0) = v[X(t;\mathbf{x}_0),Y(t;\mathbf{x}_0),t] $$
In steady, 2D straining flow the velocity field has the form 
$$u(x,y,t) = kx,\,\,  v(x,y,t) = -ky$$
 and (*) simplifies to 
$$\tag{**}\frac{dX}{dt} = kX,\\ \frac{dY}{dt} =- kY$$
Solving (**) subject to initial conditions $X(0,\mathbf{x_0})= x_0, \,Y(0,\mathbf{x}_0) = y_0$ we get
$$X(t;\mathbf{x}_0) = x_0e^{kt}, \,\,\, Y(t;\mathbf{x}_0) = y_0 e^{-kt},$$
and for points $(x_0,y_0)$ on the circle of radius $a$ centered at the origin we have the evolution of the material curve given by 
$$X(t;\theta) = a\cos(\theta)e^{kt}, \,\,\, Y(t;\theta) = a \sin (\theta) e^{-kt}\quad (0 \leqslant \theta < 2\pi)$$
Part (3)
Intuitively you can see that the circle of marked fluid particles is  deformed into an eccentric elliptical shape as it is elongated in the positive and negative $x-$directions and squashed in the $y-$ directions.  The enclosed area remains the same for incompressible flow with $\nabla \cdot \mathbf{u} = 0$ -- which holds in this case.
In incompressible flow the density $\rho$ remains constant (within any fluid parcel that moves with the flow).  The total mass of fluid particles at $t= 0$ within a cylindrical region with cross-section $\Omega(0)$ in the plane and bounded by the circle $C(0)$ of radius $a$ must remain constant through time as the region $\Omega(t)$ is convected and deformed with the flow. Since the flow is two-dimensional, there is no motion in the direction normal to the plan so we can say that the mass per unit length in the $z-$direction remains constant. Thus,
$$\frac{d}{dt}\int_{\Omega(t)} \rho \, dA = 0,$$
and since the density does not vary with time the area $\int_{\Omega(t)} \, dA$ is constant.
This can be justified rigorously by the Reynolds transport theorem.
