Amath-Math 586/Atm S 581 Due by 5 p.m. Friday, 10 June 2005
Final Project (Please do by yourself without consulting with anyone)
1. Consider the 1D advection equation:
qt + qx = 0, 0 < x < 1, t > 0,
with initial condition q(x, 0) = 0 and boundary condition q(0, t) = sin(50t).
(a) Discretize this problem with a finite element method like in class, using piecewise linear
chapeau basis functions and trapezoidal time differencing, a grid spacing ∆x = 0.01, and a
timestep ∆t = 0.005. Note that the left boundary condition specifies the coefficient of the
basis function centered on the left boundary, so the unknowns are the coefficients aj(t) of
the basis functions centered at the gridpoints xj = j∆x , j = 1,…,100, corresponding to the
interior gridpoints and the right boundary. Using a sparse solver, solve the resulting
tridiagonal system and overplot the numerical solution q(x, 0.9) on the exact solution at
this time. How well do they compare?
(b) Now discretize the problem using exactly the same kind of finite element method, but on
a nonuniform grid xj = 1 - (1 –
ξ
j)1/2,
ξ
j = 0.02j, j = 1,…,50. This grid starts with the
same grid spacing as in part (a) near x = 0, but the grid stretches greatly as we move close
to the right boundary x = 1. Using trapezoidal time differencing and the same timestep
as before, again calculate the numerical solution q(x, 0.9) and overplot it on the result of
part (a). Does the stretched grid affect the numerical solution? Why or why not?
2. Use the Matlab PDE Toolbox to solve the following initial-boundary value problem:
222
222
0
uuu
txy
∂∂∂
−+=
∂∂∂
in the right-triangular domain Ω: {x, y > 0, 0 < x + y < 1} over the time interval 0 < t < 2 with
IC u(x, y, 0) = 0 and BCs u(x+y=1, t) = u(x, 0, t) = 0, u(0, y, t) = sin(
π
y)sin(
π
t). Use the
default grid generated by the toolbox. To show me you got this to work, please print out a 3D
surface plot of the solution at t = 2 with the grid shown.