NHSM Midterm Exam 1 Partial Differential Equations

November 2024 - Review - Semester 5, 2024/2025

Please, pay attention to writing quality, clarity, logical presentation, and appropriate use
of mathematical notations.

Exercise 1 (First order PDEs)

Find the integral curves of the following system of ODEs:

dx dy dz
(6) = = .
yz −xz xy (x 2 + y 2 )

Exercise 2 (First order PDEs)

Using first integrals, solve the following PDEs:

(6) u (x + y )ux + u (x − y )uy = x 2 + y 2

Exercise 3 (Method of characteristics)

Using the method of characteristics, solve the following problems:

 
(6) ut + ux − xuy = 0, u (x, y , 0) = x 2 + 2y e −x , (7) ux + uy + u = e x +2y , u (x, 0) = 0.

Exercise 4 (Modeling)

Consider the traffic flow in a highway (see figure below). Let x be distance along a road.
Traffic density ρ(x, t ) is defined as the number of vehicles per unit distance (length). To
experimentally measure ρ at x = x0 , for t = t0 , one selects a small spatial interval I =
(x0 − ∆, x0 + ∆x ) on the highway and
number of cars in I at t = t0
ρ ( x0 , t 0 ) ≈ .
2∆x
Let φ(x, t ) be the flux, i.e., the rate at which vehicles are crossing the point x and u (x, t )
their speed.

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 NHSM Midterm Exam 1 Partial Differential Equations
November 2024 - Review - Semester 5, 2024/2025

The flux φ(x, t ) has the dimension of cars per unit time. To measure φ(x0 , t0 ), one selects a
small time interval J = (t0 − ∆t, t0 + ∆t ) and

net number of cars that pass x0 for t ∈ J

φ ( x0 , t0 ) ≈
2∆t
(i) How can you express Q(t ), the quantity of vehicles between a and b?
(ii) Express the difference between the traffic inflow (at x = a) and outflow (x = b) in
terms of ρ and u then deduce the equation ρt + (uρ)x = 0.
(iii) Assume that u = 1 − ρ. Motivate this choice then deduce a nonlinear convection
equation of ρ.
(iv) Assume that u = c − ϵ(ρ′x /ρ). Motivate this choice then deduce a convection-diffusion
equation of ρ.
(v) Assume that u (x, t ) = f (ρ(x, t )), i.e., the cars speed only depends only on the density
of the traffic. Deduce a PDE in the dependent variable ρ. What is its type?

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 NHSM Midterm Exam 1 Partial Differential Equations
November 2024 - Review - Semester 5, 2024/2025

Exercise 1 (Solution)

(6) Let us find the integral curves of the system:

dx dy dz
= =
yz −xz xy (x 2 + y 2 )

We consider the first two terms and eliminate the common term z. We get dx y =− x .
dy

Thus, we obtain the solution x 2 + y 2 = c1 , where c1 is a real constant. We deduce that

v1 (x, y , z ) = x 2 + y 2 is a first integral.
Then, we use the first and third terms to get dx z = x (x 2 +y 2 ) = c1 x . Solving this
dz dz

equation gives z 2 = c1 x 2 + c2 , with c2 ∈ R. Consequently, z 2 − (x 2 + y 2 )x 2 = c2 .

Thus, v2 (x, y , z ) = z 2 − (x 2 + y 2 )x 2 is another first integral. It is easy to see that v1
and v2 are functionally independent as v2 depends on z whereas v1 does not.
We deduce that the general solution is a curve implicitly defined by:

x 2 + y 2 = c1 and z 2 − (x 2 + y 2 )x 2 = c2 , c1 , c2 ∈ R.

Exercise 2 (Solution)

(6) The characteristic system is:

dx dy dz
= = 2 .
z (x + y ) z (x − y ) (x + y 2 )
From the first two fractions:
dx dy dx + dy
= = , i.e., (x + y )d (x + y ) = 2xdx.
(x + y ) (x − y ) 2x

Then, v1 (x, y , z ) = (x + y )2 − 2x 2 = c1 , where c1 ∈ R, is a first integral.

On the other hand,
dx dy y dx xdy d (xy )
= , =⇒ = = 2 .
(x + y ) (x − y ) y (x + y ) x (x − y ) (x + y 2 )
Thus,
d (xy ) dz
2 2
= 2
z (x + y ) (x + y 2 )
Then, v2 (x, y , z ) = z 2 − 2xy = c2 is another first integral that is independent from
the first one (because the variable z appears in v2 and not in v1 ).
Consequently, the general solution can be written as
 
u (x, y )2 = 2xy + f (x + y )2 − 2x 2 ,

where f is an arbitrary C 1 function.

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 NHSM Midterm Exam 1 Partial Differential Equations
November 2024 - Review - Semester 5, 2024/2025

Exercise 3 (Solution)

(6) Let us solve the following PDE using the method of characteristics
 
ut + ux − xuy = 0, u (x, y , 0) = x 2 + 2y e −x .

The characteristic equations are

t ′ (s ) = 1, x ′ (s ) = 1, y ′ (s ) = −x (s ), z ′ (s ) = 0.

The solution is:

t = s + a, x = s + b, y = −s 2 /2 − bs + c, z = d.

From the initial condition t (0) = 0 =⇒ a = 0. Then,

t = s, x − t = b, y + t 2 /2 + (x − t )t = c, z = d.

The solution is then

u (x, y , t ) = f (x − t, y + t 2 /2 + (x − t )t ).

From the initial condition

f (x, y ) = (x 2 + 2y )e −x .
Consequently,

u (x, y , t ) = ((x − t )2 + 2y + t 2 + 2(x − t )t )e t−x ,

= (x 2 + 2y )e t−x .

(7) Let us solve the following PDE

ux + uy + u = e x +2y , u (x, 0) = 0.

The characteristic equations are

x ′ (s ) = 1, y ′ (s ) = 1, z ′ (s ) = e x (s )+2y (s ) − z (s ).

The solution is:

x (s ) = s + a, y (s ) = s + b.
From the initial condition (y (0) = 0)

y = s, x − y = a.

Then,
1
z ′ (s ) = e 3s +a − z (s ), i.e., z (s ) = ce −s + e 3s +a .
4
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 NHSM Midterm Exam 1 Partial Differential Equations
November 2024 - Review - Semester 5, 2024/2025

Then  
1
e y
z − e 2y +x = c.
4
The solution is  
1
e y
u (x, y ) − e 2y +x = f (x − y ).
4
From the initial condition f (x ) = − 14 e x . Thus

1  2y +x  1
u (x, y ) = e − e x−2y = e x sinh (2y ).
4 2

Exercise 4 (Solution)

(i) The quantity of vehicles between a and b can be expressed as:

Z b
Q(t ) = ρ(x, t )dx.
a

(ii) The difference between the traffic inflow (at x = a) and outflow (x = b) is:

D (t ) = ρ(a)u (a, t ) − ρ(b )u (b, t ),

∆q
because, roughly speaking, ρ = ∆x and u (x, t ) = ∆t ,
∆x
where q is the traffic quantity
of vehicles. Thus,
∆q ∆x ∆q
ρ(x, t )u (x, t ) = · = .
∆x ∆t ∆t
The quantity ∆q∆t (x, t ) represents the variation of q over time. Consequently,
∆q
∆t (b, t )
is the inflow and ∆q
∆t (b, t ) is the outflow.

Combining the two previous relations, we get Q′ (t ) = D (t ), i.e.,

Z b
ρt (x, t )dx = ρ(a)u (a, t ) − ρ(b )u (b, t ),
a
Z b
= − (ρu )x (x, t )dx.
a

Consequently,
Z b
[ρt (x, t ) + (ρu )x (x, t )] dx = 0.
a

As this equality holds for any interval [a, b ], we deduce the relation

ρt + (uρ)x = 0.

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 NHSM Midterm Exam 1 Partial Differential Equations
November 2024 - Review - Semester 5, 2024/2025

(iii) Assume that u = 1 − ρ. This assumption is appropriate because when the density is
high, the velocity is reduced. u is inversely proportional to ρ. In this case, we get:

ρt + ((1 − ρ)ρ)x = 0.

Consequently, we have:
ρt + (1 − 2ρ)ρx = 0.

(iv) Assume that u = c − ϵ(ρx /ρ). The velocity u is composed of a constant component c
(the maximum speed they can travel when traffic density is low or when there is little
to no congestion ρx = 0) and a variable component −ϵ(ρx /ρ). This choice captures
the realistic behavior of drivers adjusting their speed based on local traffic conditions,
balancing free-flow conditions with the need to slow down as congestion increases.
– When traffic density is constant (ρ′x = 0), the speed u equals the free-flow speed c.
– When traffic density is increasing (ρ′x > 0), meaning there is more congestion
ahead, the speed decreases as drivers slow down in response to the denser traffic.
– When traffic density is decreasing (ρ′x < 0), meaning the road is clearing ahead,
vehicles may speed up, as indicated by a decrease in u.
In this case, we get:  
ρx
ρt + ( c − ϵ ) ρ = 0.
ρ x
We have,
 
ρx
ρt + (c − ϵ )ρ = ρt + (cρ − ϵρ′x )x ,
ρ x
= ρt + cρx − ϵρxx .

We deduce the convection-diffusion equation

ρt = ϵ(ρ)xx − cρx .

(v) After calculation, we get ρt + ρx [f (ρ) + ρf ′ (ρ)] = 0. This is a quasilinear first order
PDE.