## Description

1. (0 points) Please sign the Academic Integrity Checklist. If you do not sign the Academic

Integrity Checklist you will receive a 0 for this assignment.

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)

3

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.