1. The problem statement, all variables and given/known data A time-dependent two-dimensional fluid flow is given, in a Cartesian coordinates system (x, y), by the velocity field: u = (y, t-x) Show that, at time t = 0, the streamlines of this flow are circles centred on the origin. Find equation of the streamline that passes through the point (0, 1) Find the equation of the path of a particle released at the point (0,1) at t=0. 2. Relevant equations For streamlines in u = (u, v): dx(t)/u = dy(t)/w Particle path: dx/dt = u dy/dt = v 3. The attempt at a solution I think I'm okay with the streamline part but here it is anyway: Streamline: Using the given equation for finding streamlines, plug in the values and rearrange to get: (t-x) dx = y dy y2/2 = tx - x2/2 + C Rearrange to show x2+y2 = 2tx + C For t = 0 , it reduces to x2+y2 = + C, which is a circle. Streamline passing though (0, 1) is given in: x2+y2 = C = 02+12 + C C = 1 Streamline is x2+y2 = 1 _______________________________________________________________ Now, onto the particle path. This is the bit I'm stumped on. I've plotted the stream flow in mathematica and intuitively it seems the path should be: (x, y) = (sin[ t ] + t, cos[ t ]) But my answer is coming out slightly different. Here goes... Velocity u = (y, t-x) Particle path: dx/dt = u dy/dt = v So dx/dt = y dy/dt = t-x Taking the 2nd derivative of the first part, we get d2x/dt2 = dy/dt = t - x This gives a differential equation for x(t): d2x/dt2 + x = t Solving the left hand side: d2x/dt2 + x = 0 Using characteristic method to find particular solution, setting xc(t) = Aeλt Thus: λ = ±i Giving: xc(t) = Aeit + Be-it Then to solve for the full differential equation, d2x/dt2 + x = t Try particular solution xp(t) = t This appears to work, giving finally: x(t) = Aeit + Be-it + t Then, from the information above, we have: dx/dt = y giving y(t) = Aieit - Bie-it + 1 So, final solution for positions x(t) and y(t) are: x(t) = Aeit + Be-it + t y(t) = Aieit - Bie-it + 1 The information states that the particle is released at (x, y) = (0, 1) at time t=0 So this can be used to find constants A and B: x(0) = Ae0 + Be0 + 0 = 0 A + B = 0 A = -B y(0) = Aie0 - Bie0 + 1 = 1 Substitute A = -B -Bi - Bi + 1 = 1 -2Bi = 0 B = 0 Therefore: A = -B = 0 So both functions x(t) and y(t) are reduced to zero for the given boundary conditions. And this is where I am lost. Thanks for any help anyone can give, and thanks for taking the time to look at this because I know it's a large post!