1. Use the second-order accurate central difference approximation and the first-order forward difference approximation to evaluate ∂(cos 2x) at x = Π/6 to four decimal places. A step size of Δx = 0.2 is to be used.

Compare this to the actual value of the derivative and compute the percent error to three decimal places for each approximation, defined by

%Error = 100(Numerical value - Analytical value/Analytical value )

Your percent error should include the proper sign (positive or negative).

2. Consider the following two-dimensional convection-diffusion equation:

∂u/∂t = u∂u/dx + v(∂²u/∂x² + ∂²u/∂y²)

Obtain an explicit finite difference equation using first-order forward time, first-order backwards in space (for the convective term), and second-order central spatial differencing (for the diffusion terms). You do not have to derive the difference approximations; simply use them. Remember to include all super- and subscripts, as well as the "order" term including the lowest-order error term for each variable.

3. Consider a fluid bounded by two parallel plates extended to infinity such that no end effects are encountered. The walls and the fluid are initially at rest. The lower wall is suddenly accelerated in the x-direction. The coordinate system is such that the lower wall coincides with the xz-plane to which the y-axis is perpendicular. The spacing between the two plates is denoted by h.

The equation for this problem is

∂u/∂t = v∂²u/∂y²

where u is the kinematic viscosity of the fluid. It is required to compute the velocity profile u = u(y,t). The problem has the following initial and boundary conditions

Initial Condition: 

t = 0, u = 1.0 m/s for y=0

  u = 0.0 m/s for 0 < y ≤ h

Boundary Condition:

t ≥ 0, u = 1.0 m/s for y=0            

u = 0.0 m/s for 0 < y ≤ h          

The fluid has a kinematic viscosity of 0.000217 m²/s and h = 40 mm. Your solution should be run until t = 1.08 seconds. Various time steps are to be used to investigate the numerical schemes and the effect of the time step on stability and accuracy.

Write a two code to solve this problem for the following schemes and conditions.

a) FTCS Explicit with time steps 0.002 and 0.00233 sec

b) FTCS Implicit with time steps 0.002 and 0.01 sec

If j = 1 at the lower surface and a step size of Dy = 0.001 m is used, then j at the upper surface will be 41. Note that n = 1 corresponds to t = 0.

For each case of each method, plot the velocity profile for four difference times on the same graph: 0.27 sec, 0.54 sec, 0.81sec and 1.08 sec. (You will have 4 plots in all, each with four curves.) Remember to use proper plotting techniques, including title, axis labels, units, legend, and legibility. Submit your plots in hardcopy, and a copy of your codes both in hardcopy and electronically on Canvas.

The file names for your codes should be:

firstname_lastname.ftcsexp.sp15.ext

firstname_lastname.ftcsimp.sp15.ext

whereext is the proper extension for the computer language you are using. MATLAB and other similar software is not permitted for this assignment.