Description
2. Consider the advection equation:
∂tu + a∂xu = 0, −∞ < x < ∞, t > 0 and u(x, 0) = u0(x), (1)
with a > 0 a constant.
Discretize the x-t plane by choosing a mesh width ∆x and time step
∆t and define the discrete mesh points (xj , tn) by
xj = j∆x, j = …, −1, 0, 1, 2, … and tn = n∆t, n = 0, 1, 2, …
(a) (2 points) Show that the following difference scheme is not stable:
U
n+1
j − U
n
j
∆t
+ a
U
n
j+1 − U
n
j−1
2∆x
= 0. (2)
We will now derive a stable scheme.
(b) (1 point) From (1) it is clear that ∂tu = −a∂xu. Show that ∂ttu = a
2∂xxu.
(c) (2 points) Using the Taylor series expansion of u(x, t + ∆t) around t and question (b),
show that
u(x, t + ∆t) = u(x, t) − ∆ta∂xu(x, t) + 1
2
(∆t)
2
a
2
∂xxu(x, t) + h.o.t. (3)
(h.o.t = higher order terms).
(d) (2 points) Approximating the x-derivatives in (3) by central differences, show that we
obtain the following finite difference scheme:
U
n+1
j = U
n
j −
∆t
2∆x
a
U
n
j+1 − U
n
j−1
+
∆t
2
2∆x
2
a
2
U
n
j+1 − 2U
n
j + U
n
j−1
. (4)
(e) (3 points) Define ν = a∆t/∆x. Using Fourier analysis, find the amplification factor.
Under what condition on ν is the finite difference scheme (4) stable?
(over)
3. Given is the following problem:
∂tu = ∂x (a(x)∂xu), u(0, t) = 0, u(1, t) = 0, u(x, 0) = u0(x), where a(x) > 0. (5)
(a) (2 points) Find the finite difference scheme for (5) using the θ-method in time. You
may use that central differences in space for the ∂x (a(x)∂xu) term is given by
∂x (a(x)∂xu) ≈
aj+1/2
(Uj+1 − Uj ) − aj−1/2
(Uj − Uj−1)
(∆x)
2
. (6)
where aj+1/2 = a(xj+1/2
) and xj+1/2 =
1
2
(xj + xj+1).
(b) (2 points) Show that the maximum principle is satisfied if
2∆t(1 − θ) max
x∈[0,1]
a(x) ≤ (∆x)
2
.
(c) (5 points) Show that (6) is a consistent discretization of ∂x (a(x)∂xu).
4. (4 points) Let 0 < x < 1 and 0 < t < TF and consider the following mixed initial-boundary
value problem:
∂tu = 2∂xxu, u(0, t) = u(1, t) = 1, u(x, 0) = f(x), (7)
where f(x) = 2 if x = 0.5 and f(x) = 1 otherwise.
Suppose the mesh points are chosen to
satisfy
0 = x0 < x1 < x2 < … < xJ−1 < xJ = 1
where xj − xj−1 = ∆x for all j = 1, 2, …, J. Implement the θ-method for problem (7). To
solve the implicit system you need to solve a matrix system of the form AUn+1 = F. Write
down the matrix A and vector F. Now compute a solution using θ =
1
2
and J = 10. Make two
plots: in the first plot plot the solution at all computed time levels between 0 and TF = 0.07
using ∆t = ∆x
2
(so plot the solution at t = 0, t = ∆t, t = 2∆t, …, t = TF ); in the second plot,
do the same but use ∆t = 0.5∆x
2
.
Explain the difference in solution.
(over)
5. Being able to capture boundary layer effects in fluid dynamics is very important and the grid
plays an important role in being able to capture these effects.
Let 0 < x < 1 and 0 < t < 10 and consider the following advection-diffusion problem:
∂tu + ∂xu −
1
Re
∂xxu = 0, (8)
where Re is the Reynolds number. Consider the following boundary conditions u(0, t) = 1
and u(1, t) = 0 and initial condition u(x, 0) = 1 − x. The exact steady-state solution to this
problem is given by
u(x) = exp(Re) − exp(Re x)
exp(Re) − 1
. (9)
To discretize (8), we first use a uniform grid, where
xi = (i − 1)∆x, i = 1, 2, …, N + 1, (10)
where ∆x = 1/N. As finite difference method, we use, for i = 2, 3, …, N,
U
n+1
i = U
n
i −
∆t
∆x
U
n
i − U
n
i−1
+
1
Re
∆t
∆x
2
U
n
i+1 − 2U
n
i + U
n
i−1
, (11)
(a) (5 points) Take Re = 40. Implement the finite difference discretization (11). For this,
take N = 64 and ∆t small enough such that your scheme is stable. In one figure, plot
the exact steady solution given by (9) and the numerical approximation at time t = 10.
At t = 10, compute the error E = maxi=1,…,N |Ui − u(xi)|. Compute this error also
when N = 128. Based on these two errors, what would you say is the order of the finite
difference method given by (11)?
We will now compute the solution on the following non-uniform grid:
xi =
(
2(1 − c)(i − 1)/N, for i = 1, 2, …, N/2 + 1,
1 − c +
2c(i−1−N/2)
N
, for i = N/2 + 1, N/2 + 2, …, N + 1,
(12)
where c = (2/Re) ln(N).
The reason to use this non-uniform grid is that for high Reynolds
numbers Re the solution rapidly changes in a small neighbourhood of the point x = 1. This
non-uniform grid captures these changes better than a uniform grid.
A finite difference approximation to (8), on the non-uniform grid (12), is given by
U
n+1
i = U
n
i −
∆t
∆xi
U
n
i − U
n
i−1
+
1
Re
∆t
∆xi+1
U
n
i+1 − U
n
i
∆xi+1
−
U
n
i − U
n
i−1
∆xi
, (13)
for i = 2, 3, …, N, and where ∆xi = xi − xi−1.
(b) (3 points) Take Re = 40. Implement the finite difference discretization (13). For this,
take N = 64 and ∆t small enough such that your scheme is stable. In one figure, plot
the exact steady solution given in part (a) and the numerical approximation at time
t = 10. At t = 10, compute the error E = maxi=1,…,N |Ui − u(xi)|. Compute this error
also when N = 128. Based on these two errors, what would you say is the order of the
finite difference method given by (13)?
(c) (1 point) Now take Re = 1000. In the same figure, plot the solution computed on
the uniform grid and the solution on the non-uniform grid. Use N = 64. Explain the
difference.
