18.311 — MIT (Spring 2015)
Rodolfo R. Rosales (MIT, Math. Dept., E17-410, Cambridge, MA 02139).
March 20, 2015.
Answers to Problem Set # 04.
Contents
1 DiAn27. Non-dimensional form
1.1 DiAn27 statement: Non-dimensional form (beam equation boundary value problem) . . . . . . .
1.2 DiAn27 answer: Non-dimensional form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
2
2
2 ExPD01. General solution of the 1D wave equation
2.1 ExPD01 statement: General solution of the 1D wave equation . . . . . . . . . . . . . . . . .
2.2 ExPD01 answer: General solution of the 1D wave equation . . . . . . . . . . . . . . . . . .
2
2
3
3 ExPD02. Initial value problem for the 1D wave equation
3.1 ExPD02 statement: Initial value problem for the 1D wave equation . . . . . . . . . . . . . .
3.2 ExPD02 answer: Initial value problem for the 1D wave equation . . . . . . . . . . . . . . .
3
3
4
4 Linear 1st order PDE #06
4.1 Statement: Linear 1st order PDE #06 (general sln. for ux + 2 x uy = y + x u) . . . . . . . . . .
4.2 Answer: Linear 1st order PDE #06 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
5
5
5 Haberman Book problem 63.07
5.1 Statement: Haberman problem 63.07 (flow density curve fitting Lincoln Tunnel data) . . . . . .
5.2 Answer: Haberman problem 63.07 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
6
6
6 Haberman Book problem 72.01
6.1 Statement: Haberman problem 72.01 (waiting time at light – braking distance theory) . . . . .
6.2 Answer: Haberman problem 72.01 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
7
8
7 Haberman Book problem 73.01
7.1 Statement: Haberman problem 73.01 (solve initial value problem) . . . . . . . . . . . . . . . .
7.2 Answer: Haberman problem 73.01 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
9
9
8 Haberman Book problem 74.02
10
8.1 Statement: Haberman problem 74.02 (solve initial value problem) . . . . . . . . . . . . . . . . 10
8.2 Answer: Haberman problem 74.02 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
9 Haberman Book problem 77.04
11
9.1 Statement: Haberman problem 77.04 (nonlinearity matters) . . . . . . . . . . . . . . . . . . . 11
9.2 Answer: Haberman problem 77.04 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1
18.311 MIT, (Rosales)
Dimensional analysis.
DiAn27.
2
List of Figures
6.1
7.1
8.1
1
Haberman’s problem 72.01: Red traffic light turns green . . . . . . . . . . . . . . . . . . .
Haberman’s problem 73.01: Solve by characteristics a traffic flow problem . . . . . . . . .
Haberman’s problem 74.02: Initial value problem with c quadratic in ρ . . . . . . . . . .
8
9
11
DiAn27. Non-dimensional form
1.1
DiAn27 statement: Non-dimensional form (beam equation boundary value problem)
Consider the problem (Beam equation Boundary Value Problem)
∂ 2 ũ
∂ 4 ũ
+
β
= 0,
∂ x̃4
∂ t̃2
0 < x̃ < d and t̃ > 0,
(1.1)
with some initial data and the boundary conditions: ũ and ũx̃ vanish at both x̃ = 0 and x̃ = d. Here (a)
β > 0, and d > 0 are dimensional constants; (b) ũ measures the deviation of the beam from straight; and
(c) tildes (i.e.: ũ, x̃, and t̃) denote dimensional variables.
1. What are the dimensions of ũ, β, and d?
2. Introduce a-dimensional variables 1 u, x,
and t, so that the equation takes the form
where 0 < x < 1.
1.2
utt + uxxxx = 0,
(1.2)
DiAn27 answer: Non-dimensional form
1. The units are:
[ũ] = [d] = length,
and
[β] =
(length)4
.
time2
2. There is a single length, d, provided. Thus it should be . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . x =
Similarly . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . u =
x̃
d
ũ
.
.
d
t̃ p
A unit of time can be constructed using β and d. Thus take . . . . . . . . . . . . . . . . . . . . . . . . . t =
d2
Then the equation takes the form in (1.2), with 0 < x < 1.
2
β.
ExPD01. General solution of the 1D wave equation
2.1
ExPD01 statement: General solution of the 1D wave equation
Consider the wave equation for u = u(x, t), where c > 0 is a constant,
utt − c2 uxx = 0.
1
That is, x = x̃/L, t = t̃/T , and u = ũ/U , for appropriate choices of a length L, a time T , and units of ũ, U .
(2.1)
18.311 MIT (Rosales).
Exercises in Partial Differentiation and Linear PDE.
ExPD02.
3
Introduce the new independent variables η = x − c t, and ξ = x + c t. Change variables to write the
equation for u as a function of these new variables: u = u(η, ξ). Using this transformed form of the
equation, integrate it twice to show that it must be
u = f (η) + g(ξ),
(2.2)
for some arbitrary functions f and g. This shows that any solution of the wave equation (2.1) must have the
form u = f (x − c t) + g(x + c t).
2.2
ExPD01 answer: General solution of the 1D wave equation
Write u = u(η, ξ), where η = x − c t and ξ = x + c t. Then the chain rule yields:
ut = c uξ − c uη ,
ut t = c2 uξ ξ − 2 c2 uξ η + c2 uη η ,
ux = uξ + uη ,
ux x = uξ ξ + 2 uξ η + uη η .
Thus, in these coordinates the wave equation reduces to uξη = 0. Integrating this equation with respect
to η, we find: uξ = F (ξ), where F is some arbitrary function. Integrating once again, we obtain equation
(2.2), where f is the indefinite integral of F .
3
3.1
ExPD02. Initial value problem for the 1D wave equation
ExPD02 statement: Initial value problem for the 1D wave equation
In problem ExPD01 we showed that the general solution of the 1-D wave equation ut t − c2 ux x = 0,
with c > 0 a constant, has the form u(x, t) = f (x − c t) + g(x + c t) — where f and g are arbitrary
functions. Using this result, show that if initial values
u(x, 0) = U (x)
and ut (x, 0) = V (x),
(3.1)
are given for the wave equation, then the solution is
1
1
u = (U (x − c t) + U (x + c t)) +
2
2c
Z
x+c t
V (s) ds.
(3.2)
x−c t
In particular, show that if U and V are periodic functions of x (of period P ) then:
u is periodic in x (of period P ) and periodic in t (of period T = P/c).
Use the GBNS GScheme script in the 18311 MatLab Toolkit and check how closely the numerical calculation
reproduces the periodicity in time (for initial conditions where, say, U is periodic of period 2 and V ≡ 0.)
Note that in the GBNS GScheme, c = 1.
18.311 MIT (Rosales).
3.2
Exercises in Partial Differentiation and Linear PDE.
ExPD02.
4
ExPD02 answer: Initial value problem for the 1D wave equation
As shown in problem ExPD01, the general solution to the wave equation has the form
u(x, t) = f (x − c t) + g(x + c t),
(3.3)
where f and g are arbitrary functions. Thus, to satisfy the initial conditions posed in equation (3.1), we
must have
U (x) = f (x) + g(x) and V (x) = −c f 0 (x) + c g 0 (x),
(3.4)
where the primes denote derivatives. Now, multiply the second equation here by c−1 , take the derivative
of the first equation, and add (or subtract) the resulting equations. This yields:
1
1
0
0
f (x) =
U (x) − V (x) ,
(3.5)
2
c
1
1
U 0 (x) + V (x) .
(3.6)
g 0 (x) =
2
c
Integrate these equations to obtain:
f (x) =
g(x) =
Z 0
1
U (x) +
V (s) ds + c1 ,
c x
Z x
1
1
U (x) +
V (s) ds + c2 ,
2
c 0
1
2
(3.7)
(3.8)
where c1 and c2 are integration constants. Substituting this into (3.4) we see that c1 + c2 = 0 is needed
for (3.4) to apply. Finally, substituting this answer into (3.3), we obtain (as requested)
Z x+c t
1
1
u = (U (x − c t) + U (x + c t)) +
V (s) ds.
(3.9)
2
2 c x−c t
Assume now that both U and V are periodic functions, of period P > 0. Then
u(x + P , t) =
=
Z x+P +c t
1
1
(U (x + P − c t) + U (x + P + c t)) +
V (s) ds
2
2 c x+P −c t
Z x+c t
1
1
(U (x − c t) + U (x + c t)) +
V (s) ds, = u(x, t),
2
2 c x−c t
where we have used that (for a periodic function of period P )
Z b+P
Z b
V (s) ds =
V (s) ds.
a+P
a
This shows that u is periodic in x, of period P .
Let us now investigate the periodicity in time of the solution. We have
Z x+c t+P
1
1
(U (x − c t − P ) + U (x + c t + P )) +
V (s) ds
2
2 c x−c t−P
Z x+c t+2 P
1
1
=
(U (x − c t) + U (x + c t)) +
V (s) ds,
2
2 c x−c t
Z
1 P
= u(x, t) +
V (s) ds,
c 0
P
u(x, t + ) =
c
(3.10)
18.311 MIT, (Rosales)
Linear 1st order PDE #06.
5
where we have used (3.10) in the second and third line. This shows that
The solution u is periodic in time, of period P/c, if and
only if the average of V over one period vanishes.
(3.11)
As a simple example where U and V are periodic, but u fails to be periodic in time, consider the case:
U = 0 and V = 1. The solution is then u = t, which is (clearly) not periodic in time — though it is periodic
in space, of any and all periods.
Remark 3.1 The observant reader will notice, at this point, that
The result in (3.11) is not what the problem statement requests!
This is an intentional “typo”, put here for the following reason: There is a “trap” into which some students
fall when asked to show that something applies. Given the desired result, they produce some “argument”
leading to it, without checking that the steps in the argument are logical and consistent. If, when answering
this exercise, you managed to “prove” what the problem statement requests you to do, then BEWARE:
you just fell into the trap mentioned above! This is not good, and it shows that you are not being sufficiently
careful with your arguments and reasoning.
4
4.1
Linear 1st order PDE #06
Statement: Linear 1st order PDE #06 (general sln. for ux + 2 x uy = y + x u)
Consider the p.d.e.
ux + 2 x uy = y + x u.
(4.1)
Part 1. Write the characteristic form for this equation, and use it to write the general solution u = u(x, y)
to the p.d.e. — the general solution should involve an arbitrary function f = f (·).
Part 2 (challenge). Find u = u(x, y) if u = u(0, y) = 1 + y 2 for 1 < y < 2. In which region is the solution
u uniquely defined by this? That is: What is the region Ω of the x-y plane with the following property: if
u1 and u2 are both solutions of (4.1) such that uj (0, y) = 1 + y 2 for 1 < y < 2 and j = 1, 2, then u1 = u2
in Ω. Give examples of solutions such that u1 6= u2 outside Ω.
4.2
Answer: Linear 1st order PDE #06
Part 1. The characteristic equations for (4.1) are
dy/dx = 2 x
and du/dx = y + x u.
(4.2)
These have the general solution
y =µ+x
2
and u =
Z
ν+
x
2
−s2 /2
(µ + s ) e
2
ds ex /2 ,
(4.3)
0
where µ and ν are arbitrary constants. If we now use µ to label each characteristic curve, then we can
take ν as some arbitrary function of µ (i.e. ν = f (µ)). Then, since we can write µ = y − x2 , we obtain
Z x
2
2
2
2 −s2 /2
u = f (y − x ) +
(y − x + s ) e
ds ex /2 .
(4.4)
0
MIT, (Rosales)
Haberman Book problem 63.07.
6
This is the general solution to the p.d.e. in (4.1).
Part 2. In order to satisfy the condition u(0, y) = 1 + y 2 for 1 < y < 2, we must take f (µ) = 1 + µ2 for
1 < µ < 2. Thus
Z x
2
2
2 −s2 /2
2 2
(y − x + s ) e
ds ex /2 .
(4.5)
u = 1 + (y − x ) +
0
Clearly, this solution is uniquely defined in the region 1 < y − x2 < 2. There are infinitely many solutions
that satisfy the problem in this region, but differ outside. Just take f (µ) = 1 + µ2 for 1 < µ < 2, but
arbitrary for µ ≥ 2 and µ ≤ 1.
5
5.1
Haberman Book problem 63.07
Statement: Haberman problem 63.07 (flow density curve fitting Lincoln Tunnel data)
Consider exercise 61.3. Suppose that the drivers accelerate in such a fashion that
a2
α = ut + u ux = − ρx , where a > 0 is a constant.
ρ
(a) Physically interpret this situation.
(b) If u only depends on ρ, and the equation for conservation of cars is valid, show that
(5.1)
a
du
=− .
(5.2)
dρ
ρ
(c) Solve the differential equation in part (b), subject to the condition that u(ρmax ) = 0. The resulting
flow–density curve fits quite well to the Lincoln Tunnel data.
(d) Show that a is the velocity that corresponds to the road’s capacity.
(e) Discuss objections to the theory for small densities.
5.2
Answer: Haberman problem 63.07
Notice that, from exercise 61.3, we know that α above in equation (5.1) is the car acceleration. Now we
have:
(a) We can motivate (5.1) as follows:
• Drivers will accelerate if they see the density going down ahead of them and (conversely) slow down if the
opposite is true. The simplest way to model this behavior is to make the car acceleration α proportional
to the gradient of the density ρx , with a negative proportionality factor. That is, take α = −κρx , for
some κ > 0.
• We also expect the response of the drivers to the gradient of the density to be less intense as the density
goes up (for “large” densities the velocities, hence the accelerations, will be small.) A simple way to
model this is to take κ above inversely proportional to the density. That is, take: κ = −a2 /ρ, which then
leads to (5.1).
(b) If u only depends on ρ (i.e.: u = u(ρ)), then we can use the result in exercise 63.6, namely:
du
α = −ρ
ux . Thus:
dρ
2
2
a2
a2
du
du
du
− ρx = α = −ρ
ux = −ρ
ρx , =⇒
= 2.
ρ
dρ
dρ
dρ
ρ
But
du
< 0, thus equation (5.2) follows from the last equality here.
dρ
MIT, (Rosales)
Haberman Book problem 72.01.
7
(c) Equation (5.2) is an O.D.E. for u = u(ρ), with general solution
u = −a ln(ρ) + constant.
(5.3)
Since u(ρmax ) = 0, we can rewrite this in the form:
ρ
u = −a ln
.
ρmax
(5.4)
This yields the flow–density relationship
q = ρ u = −a ρ ln
ρ
ρmax
.
(5.5)
(d) From equation (5.5) it follows that
dq
= −a
dρ
ln
ρ
ρmax
+1 .
(5.6)
1
Thus q achieves its maximum (road capacity = qM ) at ρM = ρmax , with qM = aρM . In particular
e
uM = a,
(5.7)
which gives an interpretation of a as the car velocity at the maximum flow rate (road’s capacity.)
(e) Note that this theory gives unbounded car velocities as the density vanishes, so it has to be taken
with some care then. This problem arises from the second assumption in (a): while generally true
that the drivers response to ρx will be more intense as the density goes down, it will certainly not go
to infinity as ρ vanishes!
On the other hand, since one should not be using a continuum approximation for small densities
anyway, this problem is not too troublesome.
6
Haberman Book problem 72.01
6.1
Statement: Haberman problem 72.01 (waiting time at light – braking distance theory)
Show that if u = u(ρ) is determined by “braking distance theory” (see exercise 61.2), then the waiting time
per car after a traffic light turns green is the same as the human reaction time for braking.
Recall that (below the speed limit, which certainly applies in the circumstances of this problem) “braking
distance theory” gives the following density-flow velocity relationship
1 u
=
+ 1 L,
ρ
V
(6.1)
where V is the trigger velocity in the law,2 and L is the car length. The trigger velocity V is related to
the reaction time τ by the equation V τ = L.
2
V is defined by: For each amount V in car velocity, the separation between cars must augment by one car length.
MIT, (Rosales)
6.2
Haberman Book problem 72.01.
8
Answer: Haberman problem 72.01
The relationship (6.1) yields, for car densities large enough to ensure that the cars velocity is below the
speed limit,
1
q=
− ρ V = (ρj − ρ) V ,
(6.2)
L
where ρj =
1
L
is the jamming density. Thus
L
c(ρj ) = −V = − .
τ
(6.3)
From the solution to the red light turns green problem (see figure 6.1), a car at a distance d from the traffic
c(!) = x/t
x=
)t
c(! j
c(0
)
x=
t
t
! = !j
!=0
0
x
Red Light Turns Green.
Figure 6.1: Haberman’s problem 72.01. Characteristics in space-time for the “red traffic light turns green”
problem. The critical characteristics are x = c(0) t and x = c(ρj ) t, and the solution is
(a) For x ≤ c(ρj ) t, ρ = ρj .
(b) For c(ρj ) t ≤ x ≤ c(0) t, ρ is given by the implicit equation c(ρ) = x/t.
(c) For x > c(0) t, ρ = 0.
light must wait a time t = −d/c(ρj ) before it starts moving. Thus
τW = waiting time per car = −
where L is the car length and we used (6.3) above.
L
= τ = reaction time,
c(ρj )
(6.4)
MIT, (Rosales)
7
7.1
Haberman Book problem 73.01.
9
Haberman Book problem 73.01
Statement: Haberman problem 73.01 (solve initial value problem)

 ρmax
1
ρ(x, 0) =
ρ
 2 max
0
Assume that the traffic density is
initially given by
for x < 0,
for 0 < x < a,
for a < x,
(7.1)
where a > 0 is some fixed length, and that the car flow velocity is related to the car density by
ρ
ρ
ρ
u = umax 1 −
=⇒ q = umax 1 −
ρ =⇒ c = umax 1 − 2
.
ρmax
ρmax
ρmax
(7.2)
Sketch the initial density. Determine and sketch the density at all later times.
7.2
Answer: Haberman problem 73.01
Using (7.2) we solve for all the characteristics in the problem (see the left frame in figure 7.1).
t
(4)
!(x, t)
(5)
Solution
!
max
(1)
(2)
(3)
x
0
x = - t u max
a
x
x = a + t umax
- t u max
0
a
a + t umax
Characteristics in space time.
Figure 7.1: Haberman’s problem 73.01. The left frame shows the characteristics for the problem posed in
equations (7.1–7.2). The right frame shows a plot of the density (as a function of x) at an arbitrary time
t ≥ 0 (for t = 0, ρ has discontinuities at x = 0 and x = a).
Region 1: The characteristics starting on x = ζ < 0 at t = 0, have speed c = −umax , and carry the
value ρ = ρmax . They are given by x = −umax t + ζ and ρ = ρmax . We conclude that:
ρ = ρmax , for x < −umax t and t ≥ 0.
Region 2: The characteristics starting on 0 < x = ζ < a at t = 0, have speed c = 0, and carry the
1
1
value ρ = ρmax . They are given by x = ζ and ρ = ρmax . We conclude that:
2
2
1
ρ = ρmax , for 0 < x < a and t ≥ 0.
2
MIT, (Rosales)
Haberman Book problem 74.02.
10
Region 3: The characteristics starting on a < x = ζ at t = 0, have speed c = umax , and carry the
value ρ = 0. They are given by x = umax t + ζ and ρ = 0. We conclude that:
ρ = 0, for a + umax t < x and t ≥ 0.
Region 4: The discontinuity in the initial conditions at x = 0 generates an expansion fan of characteristics, centered at x = 0 and t = 0. The characteristics in this expansion fan are given by x = c(ρ0 ) t
1
x
and ρ = ρ0 , where ρmax ≤ ρ0 ≤ ρmax . This yields c(ρ) = for ρ. We conclude:
2
t
x
1
, for − umax t ≤ x ≤ 0 and t ≥ 0.
ρ = ρmax 1 −
2
t umax
Region 5: The discontinuity in the initial conditions at x = a generates an expansion fan of characteristics, centered at x = a and t = 0. The characteristics in this expansion fan are given by
1
x−a
x = c(ρ0 ) t + a and ρ = ρ0 , where 0 ≤ ρ0 ≤ ρmax . This yields c(ρ) =
for ρ. Hence:
2
t
x−a
1
, for a ≤ x ≤ umax t + a and t ≥ 0.
ρ = ρmax 1 −
2
t umax
Putting all the regions (1–5) together, gives the solution displayed on the right in figure 7.1.
8
8.1
Haberman Book problem 74.02
Statement: Haberman problem 74.02 (solve initial value problem)
Assume that u(ρ) = um (1 − ρ2 /ρ2j ), where um is the speed limit and ρj is the jamming density. For the
initial conditions:

for x < 0,
 ρ0
ρ(x, 0) =
(8.1)
ρ0 (L − x)/L for 0 < x < L,

0
for L < x,
where 0 < ρ0 < ρj and 0 < L, determine and sketch ρ(x, t).
8.2
Answer: Haberman problem 74.02
d(ρ u)
Note: c = c(ρ) =
= um
dρ
3 ρ2
1− 2
ρj
!
is a decreasing function of ρ. Solve using characteristics:
Region (1) x < 0 at t = 0. Here ρ = ρ0 on x = c0 t + ζ, where ζ < 0
and c0 = c(ρ0 ) — with c0 < um . Eliminating ζ, it follows that . . . . . . . . . . . . . . ρ = ρ0 for x < c0 t.
ρ0 (L − ζ)
ρ0 (L − ζ)
Region (2) 0 ≤ x ≤ L at t = 0 . Here ρ =
on x = c
t + ζ,
L
L
where 0 ≤ ζ ≤ L.


s
−L ρ0
4 t (um − c0 ) λ
1 − 1 +
,
Eliminating ζ, it follows that . . . . . . . . . . . . . . . . ρ =
2 t (um − c0 )
L2
where λ = um t + L − x, and c0 t ≤ x ≤ um t + L.
MIT, (Rosales)
Haberman Book problem 77.04.
11
Region (3) L < x at t = 0. Here ρ = 0 on x = um t + ζ,
where L < ζ. Eliminating ζ, it follows that . . . . . . . . . . . . . . . . . . . . . . . . . . . ρ = 0 for um t + L < x.
Summarizing, we have (see figure 8.1)

ρ0


!
r


−L ρ0
4 t (um − c0 ) λ
ρ(x, t) =
1− 1+

2 t (um − c0 )
L2



0
for x < c0 t,
for c0 t ≤ x ≤ um t + L,
(8.2)
for um t + L < x,
!(x, t)
!0
Solution,
case c < 0.
0
x
0
c t
0
u t+L
m
Figure 8.1: Haberman’s problem 74.02. Solution the the initial value problem posed in equation (8.1), for
c(ρ0 ) < 0, plotted for some arbitrary t > 0. The case c(ρ0 ) > 0 is similar.
9
9.1
Haberman Book problem 77.04
Statement: Haberman problem 77.04 (nonlinearity matters)
Show [ ρ2 ] 6= [ ρ ]2 .
9.2
Answer: Haberman problem 77.04
[ ρ2 ] − [ ρ ]2 = (ρ21 − ρ20 ) − (ρ1 − ρ0 )2
= (ρ21 − ρ20 ) − (ρ21 − 2 ρ0 ρ1 + ρ21 )
= 2 ρ0 ρ1 .
This is non-zero, unless either ρ0 = 0 or ρ1 = 0.
THE END.