MATH319 WINTER 2013 ASSIGNMENT 3 SOLUTIONS
Jan Feys
Question 1 is worth six points, questions 2 and 3 are worth four points each, questions 4
and 5 and 6 are worth three points each and question 7 is worth four points.
Extra, Burgers’ equation. Solve Burgers’ equation
ut + u ux = 0
with u = u(x, t) and u(x, 0) = x. Find the explicit solution u(x, t) and make a sketch of the
characteristic lines. Also graph the evolution in time of u(x, t) from the initial condition.
Solution: One partial mark for finding that the characteristics have the form
x = x0 t + x0
where x0 is a chosen point on the x-axis. One partial mark for making a picture and drawing the
characteristics as a fan emanating from the x-axis. One partial mark for implementing a change of
variables
w=
x
,
t+1
z = t.
One more partial mark for finding the reduced partial differential equation to be uz = 0. One
partial mark for finding that after implementing the side condition, the solution is
u(x, t) =
x
.
t+1
One partial mark for making a graph of this solution with time; it is a straight line, initially
diagonal, with slope decreasing in time.
Chapter 3.1, exercise 3c. Solve the problem
ut = 2uxx ,
0 ≤ x ≤ 3,
u(0, t) = 0,
u(3, t) = 0
1
t≥0
for the initial condition
u(x, 0) = f (x) = sin3
πx 3
.
Write out the complete solution including the separation of variables and the analysis of
the three cases for c. You can use expressions from the book or from class to verify your
work.
Solution: One partial mark for finding the eigenvalues c = −λ2 = −n2 π 2 /9 and writing
Xn (x) = sin (nπx/3). At some point it should be noted that n = 1, 2, . . . . An additional
partial mark for finding the spatial eigenfunction Tn (t) = exp (−2n2 π 2 t/9). One partial mark
for writing the general solution
∞
X
u(x, t) =
bn sin
nπx 3
n=1
2n2 π 2
t
exp −
9
where bn are constants to be determined. One partial mark for finding b1 = 3/4 and b3 = −1/4
and all other bn to be zero.
Chapter 3.1, exercise 6c. Solve the problem
ut = uxx ,
−π ≤ x ≤ π,
u(−π, t) = u(π, t),
t≥0
ux (−π, t) = ux (π, t)
for the initial condition
u(x, 0) = f (x) = 4 + cos2 (3x).
Write out the complete solution including the separation of variables and the analysis of
the three cases for c. You can use expressions from the book or from class to verify your
work.
Solution: One partial mark for finding the eigenvalues c = −λ2 = −n2 and writing
Xn (x) = an cos (nx) + bn sin (nx).
Here n = 0, 1, 2, . . . . An additional partial mark for finding the spatial eigenfunction Tn (t) =
exp (−n2 t). One partial mark for writing the general solution
u(x, t) =
∞
X
(an cos (nx) + bn sin (nx)) exp −n2 t
n=0
2
where an and bn are constants to be determined. One partial mark for finding a0 = 1/2 and
a6 = 9/2 and all other an and bn to be zero.
Chapter 3.2, exercise 4. Use the Maximum/Minimum principles to find the strictest constant C and D such that the solution u(x, t) of the problem
ut = kuxx ,
0 ≤ x ≤ π,
u(0, t) = 0,
u(π, t) = 0
t≥0
u(x, 0) = f (x) = 5 sin (3x) − 3 sin (5x)
satisfies C ≤ u(x, t) ≤ D for all 0 ≤ x ≤ π and t ≥ 0.
Solution: The heat equation does not have to be solved to find the constants C and D. One partial
mark for noting that the minimum and maximum on the boundaries is 0 and therefore a non zero
minimum or maximum may only be reached at t = 0. One partial mark for finding the extreme
√
of f (x) to be 4 2 at x = π/4, 3π/4 and −8 at x = π/2. One partial mark for concluding that
√
C = −8 and D = 4 2.
Chapter 3.2, exercise 6. Consider the problem, where α > 0,
ut = 2uxx ,
0 ≤ x ≤ 2,
t≥0
ux (0, t) = −α,
ux (2, t) = 0
1
u(x, 0) = f (x) = −αx 1 − x
4
(a) By trying a function of the form u(x, t) = Ax2 + Bx + Ct deduce that a solution to this
system is
u(x, t) = αx
1
x − 1 + αt.
4
In fact, it is the unique solution (you do not have to show this).
(b) Suppose α1 and α2 are two constants and u1 (x, t) and u2 (x, t) are the associated solutions. Show that
|u1 (x, 0) − u2 (x, 0)| ≤ |α1 − α2 |.
3
This means that for α1 and α2 close to one another, the corresponding initial conditions
are also close.
(c) Using the solution found in part (a), show that
|u1 (x, t) − u2 (x, t)| → ∞
for t → ∞. Conclude that for these boundary conditions there is no continuous dependence on the initial condition.
Solution: One partial mark for doing all of the following in part (a): putting the model solution into the PDE and obtaining C = 2A, plugging it into the first boundary condition to obtain
B = −α and plugging it into the other boundary condition to get 4A + B = 0. One partial mark
for showing in part (b) that
1 1
|u1 (x, 0) − u2 (x, 0)| = −α1 x 1 − x + α2 x 1 − x 4
4
1 = |α1 − α2 | x 1 − x 4
≤ |α1 − α2 |
cause
1
x 1− x
4
has a maximum of 1 on 0 ≤ x ≤ 2. One partial mark for part (c) by writing
1
|u1 (x, t) − u2 (x, t)| = |α1 − α2 | x
x − 1 + t → ∞
4
when t → ∞.
Chapter 3.3, exercise 6. Solve
ut = kuxx ,
0 ≤ x ≤ L,
u(0, t) = a,
ux (L, t) = b
u(x, 0) = f (x) = bx + a.
4
t≥0
Assume that a, b, k and L are given constants.
Solution: One partial mark for finding the steady state solution to be us (x) = bx + a. One partial
mark for finding the transient solution to be ut (x, t) = 0. One partial mark for finding the final
solution to be
u(x, t) = us (x) + ut (x, t) = bx + a.
Alternatively the three marks may be earned by directly proposing u(x, t) = bx + a as a solution
but an explicit verification must be made in this case.
Chapter 3.3, exercise 7. Solve
ut = uxx ,
0 ≤ x ≤ π,
ux (0, t) = 2,
t≥0
u(π, t) = 4
u(x, 0) = f (x) = 4 − 2π + 2x + 7 cos
3x
.
2
Solution: One partial mark for finding the steady state solution to be
us (x) = 2x + 4 − 2π.
One partial mark for finding the general form of the transient solution to be
2 !
∞
X
1
1
t
u (x, t) =
an cos
+ n x exp −
+n t .
2
2
n=0
One partial for finding a1 = 7 and all the other an = 0, or writing out
9t
3x
t
u (x, t) = 7 cos
exp −
.
2
4
One partial mark for composing to the final answer.
5