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B5b Applied Partial Differential Equations 1–
1 First-order quasi-linear equations
1.1 Introduction
Definitions
In this section we consider partial differential equations (PDEs) of the following form
a(x, y, u)
∂u
∂x
+ b(x, y, u)
∂u
∂y
= c(x, y, u). (1.1)
Here x and y are independent variables, a, b and c are given smooth (i.e. continuously
differentiable) functions, and u(x, y) is a scalar function for which we would like to solve.
Equation (1.1) is known as a first-order quasilinear partial differential equation: first-order
since there are no second or higher derivatives and quasilinear because it is linear in its highest
derivatives (there are no nonlinear terms like (∂u/∂x)
). Equations of this type arise in many
areas of mathematical modelling, including fluid mechanics and traffic flow. They also provide
a relatively straightforward introduction to some important concepts, such as Cauchy data,
characteristics and weak solutions, that will be applied to more complicated equations later
in the course.
Two special cases of (1.1) are worth mentioning. First, if a and b are independent of u,
then (1.1) becomes the semilinear equation
a(x, y)
∂u
∂x
+ b(x, y)
∂u
∂y
= c(x, y, u). (1.2)
If it also happens that c is a linear function of u, then we have
a(x, y)
∂u
∂x
+ b(x, y)
∂u
∂y
= α(x, y)u + β(x, y), (1.3)
which is a linear equation. It is generally the case that linear equations are significantly
better-behaved and easier to solve than nonlinear ones.
Equations like (1.1) often describe the evolution of a quantity u (representing e.g. traffic
density or fluid velocity) in space and time. In such cases, to emphasise the fact that one
independent variable represents time, we will use x and t as independent variables instead of
x and y, writing (1.1) as
a(x, t, u)
∂u
∂t
+ b(x, t, u)
∂u
∂x
= c(x, t, u). (1.4)
1–2 OCIAM Mathematical Institute University of Oxford
a,b,c )
T
(
Characteristics
Characteristic
Initial
curve
projections
o
z = u
y
K
s
x
Figure 1: Schematic showing the characteristics, parameterised by τ and pointing in the di-
rection (a, b, c)
T
, emerging from the initial curve, which is parameterised by s. The projection
of the initial curve onto the (x, y) plane is Γ and the projection of the characteristics onto the
(x, y) plane are the characteristic projections.
1.2 Characteristics
Geometric definition
We can think of the solution u(x, y) we are seeking as defining a surface z = u(x, y) in
three-dimensional space. The normal to this surface is in the direction
n ∝ ∇
(
u(x, y) − z
)
=
∂u
∂x
∂u
∂y
− 1
(1.5)
and the PDE (1.1) can therefore be written as
a
b
c
· n = 0. (1.6)
It follows that the vector (a, b, c)
T
is everywhere tangent to the solution surface.
We can construct curves that are everywhere tangent to (a, b, c)
T
by solving the simulta-
neous ODEs
dx
dτ
= a(x, y, u),
dy
dτ
= b(x, y, u),
du
dτ
= c(x, y, u). (1.7)
Such curves are called characteristics of the PDE (1.1). Their projections onto the (x, y)
plane, i.e. the plane curves
(
x(τ ), y(τ )
)
are called characteristic projections.
1–4 OCIAM Mathematical Institute University of Oxford
In Example 1.1, the PDE (1.10) is semilinear. In such cases, the characteristic projections
satisfy the ODEs
dx
dτ
= a(x, y),
dy
dτ
= b(x, y), (1.15)
which are independent of the solution u. The standard theory of phase planes may be applied
to the ODEs (1.15); for example, there is in general a unique characteristic projection through
each point in the (x, y) plane except at critical points where a and b are both zero. Once
(1.15) have been solved to find the characteristic projections in the (x, y) plane, we find that
u satisfies the decoupled ODE
du
dτ
= c
(
x(τ ), y(τ ), u
)
(1.16)
along each characteristic projection.
For general quasilinear equations, the characteristic projections depend on the solution;
the three ODEs (1.7) are coupled and must be solved simultaneously.
Example 1.2 Solve the PDE
∂u
∂t
∂u
∂x
= 1, (1.17)
for u(x, t) in t > 0 , subject to the initial condition u = x on t = 0.
The characteristics are given by
dt
dτ
= 1,
dx
dτ
= u,
du
dτ
= 1, (1.18)
and the initial data may be parametrised by
t = 0, x = s, u = s at τ = 0. (1.19)
Solving for t first, we see that t ≡ τ and thus we may replace τ by t henceforth. The initial-value
problem for u has the solution
u = s + t, (1.20)
so that the problem for x becomes
dx
dt
= s + t, x = s when t = 0, (1.21)
whose solution is
x = s + st +
t
. (1.22)
Now we can solve (1.22) for s and substitute it into (1.20) to obtain the solution in explicit form:
u =
x + t +
t
1 + t
. (1.23)
Alternative method of solution
The characteristic equations (1.7) may be rearranged to give
dx
a(x, y, u)
=
dy
b(x, y, u)
=
du
c(x, y, u)
. (1.24)
Suppose we can spot two linearly independent first integrals of these ODEs, of the form
f (x, y, u) = const and g(x, y, u) = const. Then the general solution of the PDE (1.1) may be
written in the implicit form
f (x, y, u) = F
(
g(x, y, u)
)
, (1.25)
where F is an arbitrary function.
B5b Applied Partial Differential Equations 1–
Example 1.3 Return to the problem considered in Example 1.2. The characteristic equations may
be written as
dt
1
=
dx
u
=
du
1
(1.26)
and then rearranged to two ODEs:
du
dt
= 1,
dx
dt
= u. (1.27)
These may be integrated to give
u = t + C
, x =
t
t + C
, (1.28)
where C
and C
are constants. Our two first integrals are, therefore, C
= f (x, t, u) = u − t and
C
= g(x, t, u) = x −
t
− C
t = x −
t
− (u − t)t. The general solution is found by setting f = F (g),
which leads to
u = t + F
(
x +
t
− ut
)
, (1.29)
where F is an arbitrary function. It may readily be verified that any u(x, t) satisfying the implicit
equation (1.29) is a solution of (1.17).
The function F is found by fitting the initial data: u = x when t = 0 leads to F (x) ≡ x, that is
u = t + x +
t
− ut, which reproduces the solution (1.23).
This procedure works because the equation f (x, y, u) = const defines a one-parameter
family of surfaces, as does g(x, y, u) = const, and characteristics are lines of intersection be-
tween one member from each of these two families. Now, any surface defined by an equation
of the form f = F (g) has the property that f is constant whenever g is constant. It follows
that such a surface is composed of a family of characteristics, as indicated in Figure 1, and is
thus a solution surface for the PDE (1.1).
Example 1.4 For the PDE
yu
∂u
∂x
− xu
∂u
∂y
= x − y, (1.30)
the characteristic equations
dx
dτ
= yu,
dy
dτ
= −xu,
du
dτ
= x − y, (1.31)
may be rearranged to give [Exercise]
d
dτ
(
x
)
=
d
dτ
(
u
)
= 0. (1.32)
It follows that the general solution is
u
= − 2 x − 2 y + F (x
), (1.33)
where F is an arbitrary function.
1.3 Cauchy data
Geometric interpretation
The term Cauchy data refers to the boundary data that, when applied to a PDE, in principle
determine the solution, at least locally. For the first-order quasilinear PDE (1.1), Cauchy
B5b Applied Partial Differential Equations 1–
Cauchy–Kowalevski theorem
A necessary condition for a unique solution u to exist in a neighbourhood of Γ is for the first
derivatives of u to be determined on Γ. Differentiation of u
and use of the chain rule leads
to
du
ds
=
∂u
∂x
dx
ds
∂u
∂y
dy
ds
. (1.36)
The partial differential equation (1.1) and (1.36) form a pair of simultaneous equations for
∂u/∂x and ∂u/∂y on the curve Γ. We can therefore solve uniquely for these first derivatives
so long as the determinant of the system is nonzero, i.e.
∣
∣
∣
∣
∣
a b
dx
ds
dy
ds
∣
∣
∣
∣
∣
= a
dy
ds
− b
dx
ds
6 = 0. (1.37)
If this condition is satisfied, then both u and its first derivatives are uniquely determined on
the curve Γ, which is clearly the first step in extending the solution away from Γ. Notice
that the criterion (1.37) is equivalent to requiring Γ not to be tangent to a characteristic
projection, as argued above via geometrical reasoning.
When the determinant in (1.37) is zero, there is either no solution for ∂u/∂x and ∂u/∂y
or an infinite number of solutions (this is an instance of the Fredholm Alternative). By
eliminating between (1.1) and (1.36) we find that
1
a
dx
ds
=
1
b
dy
ds
6
1
c
du
ds
⇒ no solution, (1.38a)
1
a
dx
ds
=
1
b
dy
ds
=
1
c
du
ds
⇒ many solutions. (1.38b)
The latter equality is the exceptional case, seen in Example 1.1, in which the variation of
u along Γ just happens to agree with the differential equation (1.16) satisfied along the
characteristic projection.
The process outlined above can be continued to obtain higher derivatives of u. If ∂u/∂x
is known, for example, then further differentiation with respect to s gives
d
ds
(
∂u
∂x
)
=
∂
u
∂x
dx
ds
∂
u
∂x∂y
dy
ds
, (1.39)
while differentiation of (1.1) with respect to x yields
a
∂
u
∂x
∂
u
∂x∂y
∂a
∂x
∂u
∂x
∂b
∂x
∂u
∂y
∂a
∂u
(
∂u
∂x
)
∂b
∂u
∂u
∂x
∂u
∂y
=
∂c
∂x
∂c
∂u
∂u
∂x
. (1.40)
Now we have a pair of simultaneous equations for ∂
u/∂x
and ∂
u/∂x∂y. The condition for
this system to have a unique solution is identical to (1.37).
So long as a, b and c are analytic, so that this differentiation may be continued indefinitely,
we can continue this argument to show that the condition (1.37) allows the derivatives of u
to all orders to be defined uniquely at Γ. Thus a Taylor series for u(x, y) may be constructed
about the initial data curve Γ, and it can be shown that this series has a nonzero radius
of convergence. This is the starting point for the proof of the Cauchy–Kowalevski theorem,
which states that (1.1) has a unique analytic solution in some neighbourhood of Γ, provided
a, b and c are analytic and satisfy the condition (1.37).
1–8 OCIAM Mathematical Institute University of Oxford
0.0 0.5 1.0 1.5 2.
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t
x
Figure 2: The characteristic projections given by equation (1.43).
1.4 Domain of definition
Bounded initial curve
In Example 1.2, we are given u = x along the whole x-axis. In general, however, the initial
data may only be given on a finite or semi-infinite initial curve Γ. In such cases, the solution
is only defined in the region penetrated by characteristic projections that intersect Γ. This
region, which is bounded by the characteristic projections that pass through the end points
of Γ, is called the domain of definition.
Example 1.6 Solve the partial differential equation
∂u
∂t
∂u
∂x
= u (1.41)
for u(x, t) in t > 0 , subject to the initial condition u = x when t = 0, 0 < x < 1.
The characteristics are given by
dt
dτ
= 1,
dx
dτ
= xu,
du
dτ
= u, (1.42)
and the initial data may be parameterised by t = 0, x = s, u = s, 0 < s < 1. The solution in
parametric form is
x = s exp
(
s(e
t
− 1)
)
, u = se
t
, 0 < s < 1 , (1.43)
and the characteristic projections are shown in Figure 2. The domain of definition is the region
0 < x < exp (e
t
− 1).
Note that s may be eliminated from (1.43) to obtain the solution in the implicit form
x = u exp
(
u − t − ue
−t
)
, (1.44)
but there is no explicit formula for u(x, t) in terms of elementary functions.
1–10 OCIAM Mathematical Institute University of Oxford
0.0 0.5 1.0 1.5 2.
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Γ
Figure 4: The characteristic projections for Example 1.8.
Nonuniqueness
The domain in which the solution is properly defined may also be limited by u ceasing to be
a unique function of x and y. Provided the coefficients a, b and c are well-behaved and u
does not blow up, the method of characteristics outlined in section 1.2 always allows us to
determine in principle the solution in parametric form:
(
x(s, τ ), y(s, τ ), u(s, τ )
)
. Then u
may in principle be found as a function of x and y so long as there is a unique transformation
from (s, τ ) to (x, y). By the Inverse Function Theorem, a sufficient condition is that the
Jacobian of the transformation be finite and nonzero:
J =
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
∂x
∂τ
∂x
∂s
∂y
∂τ
∂y
∂s
∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
= a
∂y
∂s
− b
∂x
∂s
6 = 0, ∞. (1.50)
Note that this reproduces the condition (1.37) for u to be determined in the neighbourhood
of Γ.
A unique correspondence between (s, τ ) and (x, y) implies that a unique characteristic
projection passes through each point in the (x, y) plane. Where J becomes zero, this typically
signals that the characteristic projections start to cross each other. For semilinear equations,
this can only happen at critical points of the phase-plane problem (1.15).
Example 1.8 Solve the PDE problem
x
∂u
∂x
∂u
∂y
= 0, u = y on x = 1, 0 < y < 1 , (1.51)
and determine the region in which the solution is defined by the boundary data.
The characteristic equations
dx
dτ
= x,
dy
dτ
= y,
du
dτ
= 0, (1.52)
and initial data x = 1, y = s, u = s, at τ = 0, 0 < s < 1 , lead to
x = e
, y = se
, u = s, 0 < s < 1. (1.53)
B5b Applied Partial Differential Equations 1–
0 1 2 3 4 5 6 0 1 2 3 4 5 6
0 0
1 1
2 2
3 3
4 4
5 5
Domain of
definition
J = 0
(a) (b)
t t
x x
Figure 5: (a) The characteristic projections for Example 1.9. (b) The domain of definition,
bounded by the curve on which J = 0.
We can eliminate s and τ to obtain the explicit solution u = y/x in 0 < y/x < 1. This solution is
evidently not uniquely defined at the origin where, as shown in Figure 4, the characteristic projections
all cross and where J becomes zero. We cannot continue the solution beyond this point, so the domain
of definition is 0 < y/x < 1 , x > 0.
For more general quasilinear equations, the characteristic projections depend on the so-
lution u, so the restriction that they may only cross at critical points no longer holds. The
generic situation is that J = 0 along curves in the (x, y) plane. On these curves, the solution
surface starts to fold over itself such that u ceases to be a single-valued function of x and
y. Since u usually represents a physical quantity (such as pressure, temperature or asset
price), it cannot be multivalued. Moreover when the solution surface develops a fold, the
first derivatives of u become unbounded, so the PDE (1.1) ceases to make sense. For these
reasons, we have to cut off the domain of definition along any curves on which J is zero.
Example 1.9 Solve the PDE problem
∂u
∂t
∂u
∂x
= 0, t > 0 , (1.54)
u = sin(x), 0 ≤ x ≤ π, t = 0, (1.55)
and find the region in the (x, t) plane where the solution is uniquely defined.
The solution is u = sin s, t = τ , x = s + τ sin s in parametric form, or u = sin(x − tu) in
implicit form. The characteristic projections (found by fixing s and varying τ ) are straight lines and
are illustrated in Figure 5(a). We can see that they start to cross a finite distance from the initial data
t = 0. The Jacobian is ∂(x, y)/∂(s, τ ) = 1 + t cos s. The curve on which J = 0 is, therefore, given
parametrically by x = s − tan(s), t = − sec(s) and is illustrated in Figure 5(b). The solution is defined
in the region bounded by this curve, the characteristic projections x = 0 and x = 2π, and t = 0.
In Figure 6, we visualise the solution by plotting snapshots of u versus x at different times t. The
initial u = sin(x) steepens as t increases from zero, becoming multi-valued for t > 1. When t = 1, the
Jacobian first reaches zero at x = π, where |∂u/∂x| becomes unbounded.
B5b Applied Partial Differential Equations 1–
Although the first derivatives of u are discontinuous across C, u itself is assumed to be
continuous, so u
= u
. It follows that du
/dξ = du
/dξ and therefore
dx
dξ
[
∂u
∂x
]
dy
dξ
[
∂u
∂y
]
= 0, (1.60)
where [f ]
= f
− f
represents the jump across C.
Since u
are both classical solutions of the partial differential equation (1.1), we have
a
∂u
∂x
∂u
∂y
= c
. (1.61)
Recall that u is continuous across C and, therefore, so are a, b and c: a
= a
= a and so
forth. By subtracting the equations on either side of c, we thus find
a
[
∂u
∂x
]
[
∂u
∂y
]
= 0. (1.62)
In (1.60) and (1.62), we have a homogeneous linear system for [∂u/∂x]
and [∂u/∂y]
,
which must therefore be zero unless the determinant of the system is zero. In other words,
the first derivatives must be continuous unless
b
dx
dξ
− a
dy
dξ
= 0. (1.63)
This is identical to the equation for a characteristic projection. Thus, the first derivatives of
u may only be discontinuous across a characteristic projection. Indeed, this may be used as
an alternative definition of what a characteristic projection is: it is a curve across which the
first derivatives of u may be discontinuous.
Example 1.10 Consider the partial differential equation
∂u
∂x
∂u
∂y
= 1, (1.64)
subject to the boundary condition
u(x, 0) =
{
0 x < 0 ,
x x > 0.
(1.65)
The characteristic equations
dx = dy = du (1.66)
give the general solution
u = x + f (x − y). (1.67)
The boundary condition gives
f (s) =
{
−s s < 0 ,
0 s > 0 ,
(1.68)
and the solution is therefore
u =
{
y x < y,
x x > y.
(1.69)
Notice that the first derivatives of u are discontinuous across the characteristic y = x that passes
through the origin, but u itself is continuous.
1–14 OCIAM Mathematical Institute University of Oxford
1.6 Weak solutions
In the previous section we showed that classical solutions may be patched together in such a
way that the first derivatives of u are discontinuous across a characteristic projection. Now
we attempt to do the same for solutions in which u itself has a jump across some curve
in the (x, y) plane. Selecting a unique solution is inherently more problematic when u is
discontinuous, as the following example illustrates.
Example 1.11 Consider the partial differential equation
∂u
∂x
∂u
∂y
= 0, (1.70)
subject to
u(0, y) =
{
0 y < 0 ,
1 y ≥ 0.
(1.71)
Now we try to find a curve y = f (x) such that the solution is
u =
{
0 y < f (x),
1 y ≥ f (x).
(1.72)
This clearly satisfies the partial differential equation, everywhere except on y = f (x), and the initial
condition so long as f (0) = 0. Otherwise there does not appear to be any unique way of choosing f.
Note, though, that u is constant along each of the characteristic projections which, for this linear
partial differential equation, are given by y = x + const. We therefore have u = 0 on the characteristic
projections leaving x = 0, y < 0 , and u = 1 on those that come from x = 0, y ≥ 0. This implies that
u = 0 in y < x and u = 1 in y ≥ x, i.e. that the correct choice is f = x.
Example 1.11 illustrates a plausible way of selecting a unique solution for semilinear equa-
tions, for which the characteristic projections are determined independently of the solution.
Suppose the initial data have a discontinuity at some point s = s
along Γ. On either side
of s = s
, a unique solution is determined on each characteristic projection leaving Γ. This
suggests that the discontinuity in u simply propagates along the characteristic projection
through s
(e.g. the line y = x in Example 1.11).
Unfortunately this approach does not work for quasilinear equations. The problem is that
the characteristic projections cannot be found in advance: they depend on the solution. If u
has a discontinuity at s = s
, then so does the slope of the characteristic projections leaving
Γ either side of s
. The two possible outcomes are illustrated schematically in Figure 7. In
diagram (a), the slopes of the characteristic projections leaving either side of s = s
are such
that they diverge. There is therefore a region between the limiting characteristic projections
in which we do not know the solution. In Figure 7(b), the characteristic projections from
either side of s = s
cross, so there is a region in which they overlap and in which there are
therefore two possible solutions for u.
To resolve these difficulties, we now reformulate the problem in such a way that it makes
sense even if u is discontinuous. The idea is to turn the partial differential equation (1.1)
into an integral equation since, although a discontinuous function does not have well-defined
derivatives, it may readily be integrated. The first step is to write the partial differential
equation in so-called conservation form
∂P
∂x
∂Q
∂y
= R, (1.73)
where P , Q and R are functions of x, y and u.
1–16 OCIAM Mathematical Institute University of Oxford
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C
D
D
C
C
Figure 9: Schematic showing the shock C dividing D into two regions D
and D
. The
integration paths on either side of C are denoted C
and C
.
differentiable test function ψ, assumed to vanish on γ, to obtain
∂P
∂x
ψ +
∂Q
∂y
ψ = Rψ, (1.77)
which may be rewritten in the form
∂
∂x
(P ψ) +
∂
∂y
(Qψ) = P
∂ψ
∂x
∂ψ
∂y
Now we integrate both sides of this equation over the region D:
∫ ∫
D
∂
∂x
(P ψ) +
∂
∂y
(Qψ) dxdy =
∫ ∫
D
P
∂ψ
∂x
∂ψ
∂y
We apply Green’s theorem to the left-hand side and use the fact that ψ is assumed to vanish
on γ:
∫
ψ (P dy − Q dx) =
∫ ∫
D
P
∂ψ
∂x
∂ψ
∂y
A so-called weak solution of the partial differential equation (1.73) is a function u that
satisfies (1.80) for any suitably differentiable test function ψ. If u is continuously differen-
tiable, then the steps that led from (1.73) to (1.80) may be reversed. Thus, any continuously
differentiable u satisfying (1.80) is a classical solution of (1.73). However, (1.80) makes sense
when u is non-differentiable or even discontinuous, while the original partial differential equa-
tion (1.73) does not. This is because, by using Green’s theorem, we have removed the need
to differentiate u: only the test function is differentiated.
1.7 Shocks
Now we show how the weak formulation (1.80) allows us to make sense of solutions in which
u is discontinuous across some curve C in the (x, y) plane. Such a curve is called a shock ; this
name arises from the occurrence of such solutions in gas dynamics. If the shock is initiated
on Γ, then it will divide our integration domain D into two regions D
and D
, as shown in
B5b Applied Partial Differential Equations 1–
Figure 9. Thus the area integral on the right-hand side of (1.80) can be written as
∫ ∫
D
P
∂ψ
∂x
∂ψ
∂y
∫ ∫
D
P
∂ψ
∂x
∂ψ
∂y
∫ ∫
D
P
∂ψ
∂x
∂ψ
∂y
Now, inside each of D
and D
, the solution is supposed to be continuously differentiable,
so we can write
∫ ∫
D
i
P
∂ψ
∂x
∂ψ
∂y
=
∫ ∫
D
i
∂
∂x
(P ψ) +
∂
∂y
(Qψ) + ψ
(
R −
∂P
∂x
−
∂Q
∂y
)
dxdy, (1.82)
where i = 1 or 2. The term in brackets is identically zero, because of (1.73), and Green’s
theorem may be applied to the remainder to give
∫ ∫
D
i
P
∂ψ
∂x
∂ψ
∂y
∮
∂D
i
ψ(P dy − Q dx). (1.83)
As indicated in Figure 9, the integration curves ∂D
and ∂D
comprise sections of Γ and γ
joined to curves C
and C
adjacent to the shock on either side. When the two integrals are
summed, the result is
∮
∂D
+∂D
ψ(P dy − Q dx) =
∮
Γ+C
−C
ψ(P dy − Q dx), (1.84)
since ψ is zero on γ. Notice the difference in sign because C
and C
are traversed in different
directions. The right-hand side of (1.80) may therefore be written in the form
∫ ∫
D
P
∂ψ
∂x
∂ψ
∂y
∮
Γ+C
−C
ψ(P dy − Q dx). (1.85)
The integral along Γ cancels with the left-hand side of (1.80), and we are left with
∫
C
ψ(P dy − Q dx) −
∫
C
ψ(P dy − Q dx) = 0, (1.86)
or ∫
C
ψ([P ]
dy − [Q]
dx) = 0, (1.87)
where [F ]
denotes the jump in F across the shock C.
Since this relation must hold for any (suitably smooth) test function ψ, the term in
brackets must be identically zero and we therefore obtain
dy
dx
=
[Q]
[P ]
. (1.88)
This so-called Rankine–Hugoniot condition determines the slope of the shock in terms of the
discontinuities in P and Q across it.
B5b Applied Partial Differential Equations 1–
0 1 2 3 4 5 6
0
1
2
3
4
5
t
Shock
x
Figure 11: Characteristic projections and shock for Example 1.15.
The great advantage of allowing weak solutions in which u is discontinuous is that we
can eliminate multiple-valued solutions like that shown in Figure 6. As soon as the Jacobian
becomes zero, we insert a shock that prevents the characteristic projections from crossing. In
principle, the Rankine–Hugoniot condition (1.88) determines the position of the shock, and
we can find the solution on either side by the usual characteristic methods.
Example 1.15 We return to the problem from Example 1.9, whose solution, in parametric form, is
x = s + t sin(s), u = sin(s). Recall that the Jacobian first reaches zero at x = π, t = 1, and for t > 1
the solution is multivalued. We can prevent this from occurring by, instead, allowing the solution to
be discontinuous, i.e. by initiating a shock at x = π, t = 1.
Suppose the shock is at x = X(t), and that the characteristic projections entering the shock from
either side have parameters s
(t), so that
X = s
) = s
). (1.95)
We also have the Rankine–Hugoniot condition
dX
dt
=
u
2
=
sin(s
) + sin(s
)
2
(1.96)
and X(1) = π. The system (1.95, 1.96) may be solved to give
t =
π − s
(t)
sin[s
(t)]
=
π − s
(t)
sin[s
(t)]
, X(t) ≡ π. (1.97)
The characteristic projections are shown in Figure 11, along with the shock at x = π, t > 1. Compare
with Figure 5(a) and notice that, by introducing a shock, we prevent the characteristic projections from
crossing and thus keep the solution single-valued.
Snapshots of the solution u(x, t) are plotted versus x in Figure 12 for t = 1. 0 , 1. 1 , 1. 2 , 1. 3. Notice
the growing discontinuity in u when t > 1.
1–20 OCIAM Mathematical Institute University of Oxford
t = 1.0 t = 1.
t = 1.2 t = 1.
x
x x
u u
u u
x
Figure 12: The solution u(x, t) from Example 1.15, plotted versus x for t = 1. 0 , 1. 1 , 1. 2 , 1 .3.
1.8 Nonuniqueness of weak solutions
We have shown how one may construct weak solutions by patching together classical solutions
and applying the Rankine–Hugoniot condition (1.88) across any shocks where u is discontin-
uous. This allows us to avoid the unphysical possibility of u becoming a multiple-valued
function. Unfortunately (1.88) is in general not enough to determine the solution uniquely.
There are two further problems to be addressed.
Problem 1
The Rankine–Hugoniot condition (1.88) depends on the functions P and Q that appear in
the conservation form (1.73). However, there may be many different ways of expressing
the same PDE in conservation form, and a different conservation form leads to a different
Rankine–Hugoniot condition.
Example 1.16 The PDE
∂u
∂x
∂u
∂y
= 0 (1.98)
may be written in many different conservation forms, including
∂u
∂x
∂
∂y
(
u
)
= 0 or
∂
∂x
(
u
)
∂
∂y
(
u
)
= 0. (1.99)
These lead to the two different Rankine–Hugoniot conditions
dy
dx
=
[
u
]
[u]
=
u
2
or
dy
dx
=
[
u
]
[
u
]
=
2
(
u
u
)
3 (u
)
, (1.100)
respectively.
