The Transport Equation
Consider a fluid, flowing with velocity, V, in a thin straight tube whose cross section will be
denoted by A. Suppose the fluid contains a contaminant whose concentration at position x
at time t will be denoted by u(x,t). Then at time t, the amount of contaminant in a section of
the tube between positions, x 1 and x 2 is given by the expression
x
X x 2 uÝx, tÞ A dx = amount of contaminant in Ýx 1 , x 2 Þ at time t
1
Similarly, we can write an expression for the amount of contaminant that flows through a
plane located at position, x, during the time interval from t 1 to t 2
t
X t 2 uÝx, tÞ A V dt = amount of contaminant flowing through a plane at position, x,
1
during the interval Ýt 1 , t 2 Þ
Then an equation expressing a material balance for the contaminant can be written as
follows,
x
x
t
t
X x 2 uÝx, t 2 Þ A dx = X x 2 uÝx, t 1 Þ A dx + X t 2 uÝx 1 , tÞ A V dt ? X t 2 uÝx 2 , tÞ A V dt
1
1
1
1
i.e., the amount of contaminant in the section Ýx 1 , x 2 Þ at time t 2 equals the amount of
contaminant in the section Ýx 1 , x 2 Þ at time t 1 , plus the amount of contaminant that flowed
through the plane at position x 1 during the time interval Ýt 1 , t 2 Þ minus the amount of
contaminant that flowed through the plane at position x 2 during the time interval Ýt 1 , t 2 Þ. Of
course this equation is based on the assumption that there are no other sources of
contaminant in the tube and there is no loss of contaminant through the walls of the tube.
Now the fundamental theorem of calculus implies that
x
x
x
t
X x 2 uÝx, t 2 Þ A dx ? X x 2 uÝx, t 1 Þ A dx = X x 2 X t 2 / t uÝx, tÞ A dt dx
1
1
1
t2
t2
1
1
1
t
x
X t uÝx 1 , tÞ A V dt ? X t uÝx 2 , tÞ A V dt = ? X t 2 X x 2 / x uÝx, tÞ A V dx dt
1
1
and, combining these results with the balance equation leads to
x
t
t
x
X x 2 X t 2 / t ß uÝx, tÞ Aà dt dx + X t 2 X x 2 / x ßuÝx, tÞ A Và dx dt = 0.
1
1
1
1
If we assume that this equality holds for every segment Ýx 1 , x 2 Þ in the tube and for each time
interval Ýt 1 , t 2 Þ, and if the function uÝx, tÞ and its partial derivatives of order one are
continuous functions of x and t, then it follows from an elementary property of continuous
functions that
/ t ß uÝx, tÞ Aà + / x ßuÝx, tÞ A Và = 0
for all Ýx, tÞ.
If the fluid velocity V and the cross section of the tube, A, are constants, then this equation
reduces to
/ t uÝx, tÞ + V / x uÝx, tÞ = 0
for all Ýx, tÞ.
This is the so called ”transport equation” in one dimension.
Since this equation contains partial derivatives of order at most equal to one, it is called a
first order partial differential equation. It is, moreover, a linear partial differential
equation. This terminology relates to the fact that if we define an operator, L, as follows
LßuÝx, tÞà = / t uÝx, tÞ + V / x uÝx, tÞ
1
then it is a simple matter to verify that
LßC 1 u 1 Ýx, tÞ + C 2 u 2 Ýx, tÞà = C 1 Lßu 1 Ýx, tÞà + C 2 Lßu 2 Ýx, tÞà.
Any operator with this property is called a linear operator (any function of one variable fÝxÞ
with the property that fÝc 1 x 1 + c 2 x 2 Þ = c 1 fÝx 1 Þ + c 2 fÝx 2 Þ is a function whose graph is a straight
line), and any partial differential equation (PDE) expressing an equality for a linear partial
differential operator is called a linear equation.
Problem 1 Show that any PDE that contains terms involving products of the unknown
function with its derivatives is not a linear PDE. What is the form of the most general linear
first order PDE for a function of 2 variables, uÝx, tÞ? What is it for a function of 4 variables,
uÝx, y, z, tÞ?
Problem 2 Show that for any smooth function of one variable, FÝxÞ, we have that
uÝx, tÞ = FÝx ? VtÞ
solves
/ t uÝx, tÞ + V / x uÝx, tÞ = 0
Devise a similar such solution for the equation,
/ t uÝx, y, z, tÞ + V 1 / x uÝx, y, z, tÞ + +V 2 / y uÝx, y, z, tÞ + +V 3 / z uÝx, y, z, tÞ = 0.
Consider now the more general situation involving the flow of a fluid, flowing with velocity, 3
v,
n
in a region U in R . Suppose the fluid contains a contaminant whose concentration at
position 3
x at time t will be denoted by uÝx3, tÞ. Then at time t, the amount of contaminant in an
arbitrary ball B in U is given by the expression
X uÝx3, tÞ dx = amount of contaminant in B at time t
B
Similarly, the outflow through the boundary of the ball during the time interval Ýt 1 , t 2 Þ is given
by
t
X t 2 X/B uÝx3, tÞ 3v 6 3n dS dt = outflow through the boundary of the ball during
1
the time interval Ýt 1 , t 2 Þ
where 3
n denotes the outward unit normal to /B, the boundary the surface of the ball. Now
the material balance equation becomes
t
XB uÝx3, t 2 Þ dx = XB uÝx3, t 1 Þ dx ? X t 2 X/B uÝx3, tÞ 3v 6 3n dS dt
1
expressing the fact that the amount of contaminant in the ball at time t 2 equals the amount
of contaminant in the ball at time t 1 , less that amount that has flowed out of the ball through
the boundary. As before, we can use the fundamental theorem to write
t
XB uÝx3, t 2 Þ dx ? XB uÝx3, t 1 Þ dx = X t 2 XB / t uÝx3, tÞ dxdt
1
Recall now that the divergence theorem asserts that for an arbitrary smooth vector field,
3 Ýx3Þ,
G
3 Ýx3Þ 6 3
3 dx.
X G
n dS = X div G
/B
B
3 = uÝx3, tÞ 3
Then for G
v we get
t
t
X t 2 XB / t uÝx3, tÞ dxdt + X t 2 XB divßuÝx3, tÞ 3và dx dt = 0.
1
1
2
Problem 3 Show that for any smooth scalar function, uÝx3, tÞ , and any constant vector
3
v, divßuÝx3, tÞ 3
và = 3
v 6 grad uÝx3, tÞ
It follows from the result of the problem that since B is an arbitrary ball in U, and Ýt 1 , t 2 Þ is
similarly arbitrary, then if u and its derivatives of order one are all continuous in U,
v 6 grad uÝx3, tÞ = 0,
in U for all t.
/ t uÝx3, tÞ + 3
This is the transport equation in n-dimensions.
Solutions for First Order Equations
Consider first the problem of finding the general solution for the equation
/ t uÝx, tÞ + V / x uÝx, tÞ = 0
for all Ýx, tÞ.
By a solution to the equation, we mean a function, uÝx, tÞ , that is continuous and has
continuous first derivatives at all points (x,t), and in addition is such that the continuous
function / t uÝx, tÞ + V / x uÝx, tÞ is equal to zero at all points. If uÝx, tÞ has all these
properties, we say that uÝx, tÞ is a classical solution for the PDE.
Suppose that
x = xÝsÞ
t = tÝsÞ
?K < s < K
is the parametric description of some curve C in the x-t plane. Then for u = uÝx, tÞ a smooth
function of x and t, it is clear that du/ds = / x uÝx, tÞ x v ÝsÞ + / t uÝx, tÞ t v ÝsÞ expresses the
derivative of u along the curve C. In particular, if the curve C is such that
x v ÝsÞ = V
Then
and
t v ÝsÞ = 1, i.e.,
du/ds = / x uÝx, tÞ V + / t uÝx, tÞ = 0,
xÝsÞ = Vs + x 0 , and tÝsÞ = s + t 0
asserts that u is constant along C
from which it is apparent that u = uÝx, tÞ solves the PDE if and only if u is constant along
C. It is also clear that if V is a constant, then C is a straight line. The parametric
representation for this straight line is then
xÝsÞ = Vs + x 0
tÝsÞ = s + t 0
?K < s < K
Eliminating the parameter, s, leads to the implicit equation x ? Vt = x 0 ? Vt 0 =: C 0 , for the
straight line passing through the point Ýx 0 , t 0 Þ in the x-t plane. Clearly, for FÝxÞ any smooth
function of one variable, uÝx, tÞ = FÝx ? VtÞ is a smooth function of x and t which is constant
along C. Then by our previous observation, u = uÝx, tÞ solves the PDE if and only if
uÝx, tÞ = FÝx ? VtÞ for FÝxÞ any smooth function of one variable. Note that the general
solution to a linear first order partial differential equation contains an arbitrary function, in
contrast to the general solution for a linear first order ordinary differential equation which
contains an arbitrary constant.
Now consider the more general problem of finding the most general solution for the
equation
v 6 4 uÝx3, tÞ = 0
/ t uÝx3, tÞ + 3
for all Ýx3, tÞ.
3
3
x=3
xÝsÞ = Ýx 1 ÝsÞ, ..., x N ÝsÞÞ
If
t = tÝsÞ
?K < s < K
is the parametric description of some curve C in R N+1 then for u = uÝx, tÞ a smooth function
of x and t, it is clear that
du/ds = / 1 uÝx, tÞ x v1 ÝsÞ + ... + / N uÝx, tÞ x vN ÝsÞ + / t uÝx, tÞ t v ÝsÞ
expresses the derivative of u along the curve C. If the curve C is such that
3
3
vs + x 0 , and tÝsÞ = s + t 0
v and t v ÝsÞ = 1, i.e.,
xÝsÞ = 3
x v ÝsÞ = 3
then
du/ds = 3
v 6 4u + / t uÝx, tÞ = 0,
asserts that u is constant along C, (obviously C is a straight line in R N+1 Þ. By the previous
argument, the following statements are equivalent:
v 6 4 uÝx3, tÞ = 0
1Þ u = uÝx3, tÞ is a solution of / t uÝx3, tÞ + 3
vs + x 0 , and tÝsÞ = s + t 0
2Þ u = uÝx3, tÞ is constant along the straight line 3
xÝsÞ = 3
3Þ uÝx3, tÞ = fÝx3 ? 3
vtÞ for fÝx3Þ any smooth scalar valued function of N variables
Problem 4 Show that the three statements are, in fact, equivalent and that C is a straight
line if 3
v is a constant.
Statement 3 here asserts that the general solution for the equation given in statement 1 is
any function of the form uÝx3, tÞ = fÝx3 ? 3
vtÞ where f is any smooth function from R N into R 1 .
Evidently the solution to this equation is a very long way from being unique. On the other
hand, consider the problem of finding a function uÝx3, tÞ which satisfies the conditions
v 6 4 uÝx3, tÞ = FÝx3, tÞ
for all 3x, and all t > 0,
a) / t uÝx3, tÞ + 3
b)
uÝx3, 0Þ = gÝx3Þ,
for all 3x
The condition a) is an inhomogeneous version of the equation in statement 1). The term
inhomogeneous is used to express the fact that the right side of the equation is not zero
and, in fact, contains the forcing function, FÝx3, tÞ. The equation in statement 1) is said to
be a homogeneous equation, since the right side of the equation is zero. The condition b)
is referred to as an initial condition since it specifies the state variable, uÝx3, tÞ, at the initial
time t = 0. The functions FÝx3, tÞ and gÝx3Þ are called the data for the problem and are
assumed to be given. We will show that for each pair of data functions, FÝx3, tÞ and gÝx3Þ, we
can find a function uÝx3, tÞ which satisfies both of the conditions a) and b). Then we say,
uÝx3, tÞ is a solution to the initial value problem. Note that this construction does not
gaurantee that the solution is unique (although we may be able to show later that the
solution is unique).
We will describe the construction in the simplest case that N = 1. Then the initial value
problem (IVP) assumes the form
/ t uÝx, tÞ + V / x uÝx, tÞ = FÝx, tÞ,
uÝx, 0Þ = gÝxÞ,
? K < x < K,
t > 0,
?K < x < K
First, let u 1 Ýx, tÞ denote the solution of this IVP in the case that FÝx, tÞ = 0. The general
4
solution of the homogeneous PDE is any function of the form, u 1 Ýx, tÞ = fÝx ? VtÞ, where f
denotes a smooth function of one variable. In order to satisfy the initial condition we must
have, u 1 Ýx, 0Þ = fÝxÞ = gÝxÞ, from which it follows that u 1 Ýx, tÞ = gÝx ? VtÞ. Here we see that
in order for the homogeneous initial value problem to have a solution, the given initial
function must be continuously differentiable.
Now let u 2 Ýx, tÞ denote the solution of the IVP in the case that gÝxÞ = 0 but FÝx, tÞ is not
zero. Then
/ t u 2 Ýx, tÞ + V / x u 2 Ýx, tÞ = FÝx, tÞ,
u 2 Ýx, 0Þ = 0,
? K < x < K,
t > 0,
?K < x < K
The inhomogeneous equation is equivalent to
d/ds áu 2 ÝxÝsÞ, tÝsÞÞâ = FÝxÝsÞ, tÝsÞÞ,
where
x v ÝsÞ = V
t v ÝsÞ = 1; i.e.,
and
xÝsÞ = Vs + x 0 , and tÝsÞ = s + t 0
Then for arbitrary parameter values s 2 > s 1 ,
s2
u 2 ÝxÝs 2 Þ, tÝs 2 ÞÞ = u 2 ÝxÝs 1 Þ, tÝs 1 ÞÞ + X d/ds áu 2 ÝxÝsÞ, tÝsÞÞâds
s1
s2
= u 2 ÝxÝs 1 Þ, tÝs 1 ÞÞ + X FÝxÝsÞ, tÝsÞÞ ds
s1
If we choose s 2 = 0 > s 1 = ?t, then x 0 = xÝs 2 Þ, t 0 = tÝs 2 Þ. That is,
and
xÝs 2 Þ = V0 + x,
xÝs 1 Þ = VÝ?tÞ + x, and
tÝs 2 Þ = 0 + t
tÝs 1 Þ = ?t + t
and
0
t
?t
0
u 2 Ýx, tÞ = u 2 Ýx ? Vt, 0Þ + X FÝx + Vs, t + sÞ ds = 0 + X FÝx + VÝS ? tÞ, SÞ dS
where we made the change of variable, S = t + s in the integral. Now it is easy to show that
t
uÝx, tÞ = u 1 Ýx, tÞ + u 2 Ýx, tÞ = gÝx ? VtÞ + X FÝx ? VÝt ? SÞ, SÞ dS
0
satisfies both of the conditions a) and b); i.e., this is a solution of the initial value problem.
Problem 5 Verify that the solution constructed here does, in fact, satisfy both conditions in
the IVP. What are the conditions on F(x,t) in order for this solution to be valid? Verify that in
the case of general N, the solution of the IVP is given by
3 tÞ + X t FÝx3 ? V
3 Ýt ? SÞ, SÞ dS
uÝx3, tÞ = gÝx3 ? V
0
Suppose N = 1, F = 0 and that the initial data function, gÝxÞ, is such that
gÝxÞ =
1
if x < 1
2
if x > 1
Then
5
uÝx, tÞ =
1
if x ? Vt < 1
2
if x ? Vt > 1
formally satisfies the IVP but since the first derivatives of uÝx, tÞ fail to exist at points on the
line x ? Vt = 1, it cannot be said to be a solution in the classical sense. In the coming weeks
we will try to determine whether there is some weaker sense in which this can be said to be
a solution of the IVP.
Method of Characteristics
The method of finding a solution for the transport equation in the previous examples is a
special case of the so called ”method of characteristics”. If we consider a more general first
order equation in 2 variables,
AÝx, tÞ / t uÝx, tÞ + BÝx, tÞ / x uÝx, tÞ + CÝx, tÞ uÝx, tÞ = FÝx, tÞ ,
then a curve C in the x-t plane is said to be a characteristic curve for this PDE if C is a
solution curve for the following system of ordinary differential equations,
t v ÝsÞ = AÝxÝsÞ, tÝsÞÞ,
x v ÝsÞ = BÝxÝsÞ, tÝsÞÞ.
Note that along any characteristic curve C, the PDE reduces to an ordinary differential
equation
d/dsá uÝxÝsÞ, tÝsÞÞâ + CÝxÝsÞ, tÝsÞÞ uÝxÝsÞ, tÝsÞÞ = FÝxÝsÞ, tÝsÞÞ.
Note that although the linear PDE reduces to a linear ODE along characteristic curves, the
ordinary differential equations which produce the characteristics may be nonlinear. In the
examples we have considered, the coefficients A, B and C were constants and
correspondingly, the characteristics turned out to be straight lines. In general, when the
coefficients in the PDE are not constant, then the characteristics turn out to be a family of
curves. We will return to the method of characteristics later when considering conservation
laws.
6