Quasilinear Partial Differential Equations
of First Order
Basic concepts
Notation: Points in R𝑛 are denoted x = (𝑥1 , 𝑥2 , . . . , 𝑥𝑛 ) or y = (𝑦1 , 𝑦2 , . . . , 𝑦𝑛 ),
but for points in R2 or R3 we often write (𝑥, 𝑦) instead of (𝑥1 , 𝑥2 ) and (𝑥, 𝑦, 𝑧)
insteade of (𝑥1 , 𝑥2 , 𝑥3 ). Partial derivatives have alternative notations:
∂𝑢
= 𝑢′𝑥𝑗 = 𝑢′𝑗 ,
∂𝑥𝑗
∂2𝑢
= 𝑢′′𝑥𝑗 𝑥𝑘 = 𝑢′′𝑗𝑘 ,
∂𝑥𝑘 ∂𝑥𝑗
etc.
Definition 0.1. A linear differential equation of first order in a domain 𝐷 in
R𝑛 is an equation of the form
𝑛
∑
𝑗=1
𝑎𝑗 (x)
∂𝑢
+ 𝑏(x)𝑢(x) + 𝑐(x) = 0,
∂𝑥𝑗
where the coefficient functions 𝑎1 , 𝑎2 , . . . , 𝑎𝑛 , 𝑏, 𝑐 are continuous functions in the
domain 𝐷. A quasilinear differential equation of first order is an equation of
the form
(1)
𝑛
∑
𝑎𝑗 (x, 𝑢(x))
𝑗=1
∂𝑢
+ 𝑏(x, 𝑢(x)) = 0.
∂𝑥𝑗
A solution to the equation is a continuously differentiable function 𝑢, defined in
Ω, which makes the equality hold true for all x ∈ 𝐷.
Remark: One might find it unnecessary to demand the coefficient functions to
be continuous, but we do it anyway, for convenience. Furthermore, one might
question the provision that the solution be continuously differentiable. In the
equation 𝑢′𝑥1 (𝑥1 , 𝑥2 ) = 𝑐(𝑥1 , 𝑥2 ), for example, no derivative with respect to 𝑥2
occurs, and the equation would have a meaning also for functions that are not
differentiable (or not even continuous) with respect to 𝑥2 . Still we demand (to
begin with) that the solution be continuously differentiable in (𝑥1 , 𝑥2 ). One reason for this is that we might want to make a change of coordinates, which would
otherwise be problematic. However, we will have reason to return to this discussion and to modify this standpoint in connection with the concept weak solution.
Introductory examples
We will consider some elementary examples on partial differential equations for
functions (𝑥, 𝑦) 7→ 𝑢(𝑥, 𝑦) in R2 , but first we consider the following example of
an ordinary differential equation, the simplest example imaginable:
𝑢′ (𝑥) = 0.
What are the solutions to this equation? One might be tempted to immediately
answer ”the constant functions”. However, if we consider the problem more critically, we must admit that the answer is as incomplete and unprecise as the question is. Nothing is said about the solution’s domain of definition. It is true that
1
a differentiable function of one variable, defined on an open subset of R, whose
derivative vanishes identically, is constant on each connected component of its
domain of definition, but the constants may be different on different components.
Thus, if a function 𝑢, defined on the set {𝑥 ∈ R; 0 < 𝑥 < 1 or 2 < 𝑥 < 3},
satisfies the differential equation 𝑢′ (𝑥) = 0, we can only conclude that it is constant on (0, 1) and constant on (2, 3). This does not mean that it is a constant
function.
Next, we take the simplest possible example of a partial differential equation
for a function (𝑥, 𝑦) 7→ 𝑢(𝑥, 𝑦) of two variables:
∂𝑢
= 0.
∂𝑥
All functions which depend only on 𝑦, i.e. all functions of the form 𝑢(𝑥, 𝑦) =
𝜙(𝑦), with a continuously differentiable function 𝜙 of one variable, are solutions
to this equation. Whether there are more solutions or not, depends on the
domain where the equation is considered.
Exercise 0.2. Construct a solution to the equation ∂𝑢
∂𝑥 = 0 in the domain
{(𝑥, 𝑦) ∈ R2 ; 1 < 𝑥2 + 𝑦 2 < 2} which is not of the form 𝑢(𝑥, 𝑦) = 𝜙(𝑦).
Exercise 0.3. Formulate a condition on the domain 𝐷 that guarantees that all
solutions in 𝐷 to the equation ∂𝑢
∂𝑥 = 0 are of the form 𝑢(𝑥, 𝑦) = 𝜙(𝑦).
To avoid the phenomenon indicated above, we will now confine ourself to
solutions defined in the whole plane. The solutions to the equation 𝑢′𝑥 = 0
are the functions of the form 𝑢(𝑥, 𝑦) = 𝜙(𝑦), with a continuously differentiable
function 𝜙 of one variable. Thus there are many solutions. To single out a
unique solution one could impose some condition on the solution, for example
by prescribing its values on some curve. A differential equation together with
such a condition constitutes a problem. We will encounter problems of various
types: Cauchy problem, Dirichlet problem, Neumann problem, etc.
Example: Consider the equation 𝑢′𝑥 (𝑥, 𝑦) = 0 and look for a solutions (in the
whole plane) with prescribed values on some curve 𝐶. For which curves 𝐶 is
this problem well-posed?
Solution: Since the general solution is 𝑢(𝑥, 𝑦) = 𝜙(𝑦), i.e. any 𝐶 1 -function that
depends only on 𝑦, the problem will be well-posed (for any prescribed function
on 𝐶) if and only if the curve 𝐶, for any 𝜂 ∈ R, contains exactly one point with
𝑦-coordinate 𝜂. If it contains two or more points with a certain 𝑦-coordinate, we
cannot prescribe the values on 𝐶 arbitrarily, because 𝑢 would have to assume
the same value at all these points. On the other hand, if there is some number 𝜂
that does not occur as 𝑦-coordinate for any point on 𝐶, then the constant value
of 𝑢 on the line 𝑦 = 𝜂 would not be determined by the prescription of values on
𝐶. The lines 𝑦 = 𝜂 (i.e. the horizontal lines) are called characteristics for the
equation.
Example: Same task as in the previous example, but for the equation
3𝑢′𝑥 − 4𝑢′𝑦 = 0.
2
Solution: The solution of the problem depends on the observation that the
equation is essentially the same as the previous one. The left-hand member is
the scalar procuct of the the gradient of 𝑢 and the vector (3, −4). If we divide
by 5, in order to normalize this directional vector, we get
3 ′
4
𝑢 − 𝑢′ = 0.
5 𝑥 5 𝑦
Now the equation says that the directional derivative of 𝑢 in the direction (3, −4)
is zero, i.e. 𝑢 is constant on each line with this direction. The solutions are the
functions that are constant on each of the lines 4𝑥 + 3𝑦 = 𝑐. To get a well-posed
problem, one can prescribe the values of 𝑢 on some curve that intersects each
of these lines exactly once, for example on some line which does not have the
direction (3, −4). The lines 4𝑥 + 3𝑦 = 𝑐 are characteristic for the equation.
We now consider a linear equation with variable coefficient:
𝑎(𝑥, 𝑦)
∂𝑢
∂𝑢
+ 𝑏(𝑥, 𝑦)
= 0.
∂𝑥
∂𝑦
We can interpret the left member of the equation as a multiple of the directional derivative of 𝑢 in the direction (𝑎(𝑥, 𝑦), 𝑏(𝑥, 𝑦)). The only difference from
the previous examples is that this direction now varies from point to point.
The characteristics will no longer be straight lines, but curves in the plane.
Let 𝛾 be a curve, which at each of its points (𝑥, 𝑦) has the tangential direction (𝑎(𝑥, 𝑦), 𝑏(𝑥, 𝑦)). Such a curve is called a characteristic curve and can be
parametrized
{
𝑥(𝑡)
˙
= 𝑎(𝑥(𝑡), 𝑦(𝑡))
𝑡 → (𝑥(𝑡), 𝑦(𝑡)), where
𝑦(𝑡)
˙ = 𝑏(𝑥(𝑡), 𝑦(𝑡))
For the values of 𝑢 along such a curve we have
𝑑
′
𝑢(𝑥(𝑡), 𝑦(𝑡)) = 𝑥(𝑡)
˙
= 𝑢′𝑥 (𝑥(𝑡), 𝑦(𝑡)) + 𝑦(𝑡)𝑢
˙
𝑦 (𝑥(𝑡), 𝑦(𝑡))
𝑑𝑡
= 𝑎(𝑥(𝑡), 𝑦(𝑡))𝑢′𝑥 (𝑥(𝑡), 𝑦(𝑡)) + 𝑏(𝑥(𝑡), 𝑦(𝑡))𝑢′𝑦 (𝑥(𝑡), 𝑦(𝑡)) = 0.
Thus 𝑢 is constant along each such curve. If the functions 𝑎 and 𝑏 are Lipschitz continuous (for example continuously differentiable), there is exactly one
characteristic curve through each point, and we get a well-posed problem if we
prescribe the values of 𝑢 on any curve that intersects each characteristic curve
exactly once. In particular this curve should itself not be characteristic.
Example: Solve the Cauchy problem
(𝑥2 + 1)
∂𝑢
∂𝑢
+ 2𝑥
= 0, 𝑢(0, 𝑦) = 𝜙(𝑦),
∂𝑥
∂𝑦
where 𝜙 is a given function, continuously differentiable on R.
Solution: We start by finding the characteristic curves from the system of
ordinary differential equations
{
𝑥˙ = 𝑥2 + 1
𝑦˙ = 2𝑥
3
The first equation yields
1=
𝑥˙
𝑑
= (arctan 𝑥), arctan 𝑥 = 𝑡 + 𝐶.
𝑥2 + 1
𝑑𝑡
There is no loss of generality in putting 𝐶 = 0 here (this amounts only to a
normalization of the parametrization), so we get
𝑥 = tan 𝑡, −
𝜋
𝜋
<𝑡< .
2
2
Now the second equation gives
0 = 𝑦˙ − 2𝑦 tan 𝑡 = cos2 𝑡
(
2𝑦 sin 𝑡 )
𝑑[ 𝑦 ]
𝑦˙
,
−
= cos2 𝑡
2
3
cos 𝑡
cos 𝑡
𝑑𝑡 cos2 𝑡
so we get
𝑥 = tan 𝑡, 𝑦 = 𝐶 cos2 𝑡 =
𝐶
.
1 + 𝑥2
𝐶
The characteristic curves are the curves 𝑦 = 1+𝑥
2 , and we easily see that there is
exactly one characteristic curve through each point of the plane. The equation
says that the function 𝑢 is constant on each characteristic curve. The general
solution is any 𝐶 1 - function with this property. Now we have the Cauchy data
𝑢(0, 𝑦) = 𝜙(𝑦). Fix a point (¯
𝑥, 𝑦¯) in the plane. It lies on exactly one character𝐶
istic curve 𝑦 = 1+𝑥
¯(1 + 𝑥
¯2 ). This curve passes
2 , namely, the one with 𝐶 = 𝑦
(
)
through (0, 𝐶), so the value at (¯
𝑥, 𝑦¯) (should be )𝜙(𝐶) = 𝜙 𝑦¯(1 + 𝑥
¯2 ) .
Answer: The solution is 𝑢(𝑥, 𝑦) = 𝜙 𝑦(1 + 𝑥2 ) .
Exercise 0.4. Verify the result of the preceding example.
Systematic considerations
For a homogeneous linear equation
𝑛
∑
𝑗=1
𝑎𝑗 (x)
∂𝑢
= 0,
∂𝑥𝑗
the characteristic curves are the curves in R𝑛 that are everywhere tangent to
the vector (𝑎1 (x), 𝑎2 (x), . . . , 𝑎(x𝑛 )). Solutions to the equation are the functions
that are constant along each characteristic curve. This means that the equation
is essentially solved when tha characteristic curves are found. This amounts to
solving the system of ordinary differential equations
𝑥˙ 𝑗 = 𝑎𝑗 (𝑥1 (𝑡), 𝑥2 (𝑡), . . . , 𝑥𝑛 (𝑡)), 𝑗 = 1, 2, . . . , 𝑛.
We now consider the quasilinear equation (1), and we look for solutions, implicitely given by some equation
(2)
Φ(𝑥1 , 𝑥2 , . . . , 𝑥𝑛 , 𝑢(x)) = 0.
The partial differential equation is essentially solved when the function Φ (defined in R𝑛+1 ) is found. We differentiate both members of the equation (2) with
respect to each variable 𝑥𝑗 :
Φ′𝑗 (x, 𝑢(x)) + Φ′𝑛+1 (x, 𝑢(x))𝑢′𝑗 (x) = 0
4
whence
0=
𝑛
∑
(
)
𝑎𝑗 (x, 𝑢(x)) Φ′𝑗 (x, 𝑢(x)) + Φ′𝑛+1 (x, 𝑢(x))𝑢′𝑗 (x)
𝑗=1
=
𝑛
∑
𝑎𝑗 (x, 𝑢(x))Φ′𝑗 (x, 𝑢(x)) + Φ′𝑛+1 (x, 𝑢(x))
𝑗=1
𝑛
∑
𝑎𝑗 (x, 𝑢(x))𝑢′𝑗 (x).
𝑗=1
Thus we see that 𝑢 satisfies (1) if and only if
𝑛
∑
𝑎𝑗 (x, 𝑢(x))Φ′𝑗 (x, 𝑢(x)) − 𝑏(𝑥, 𝑢(x))Φ′𝑛+1 (x, 𝑢(x)) = 0.
𝑗=1
This is the case if the fuction Φ satisfies the homogeneous linear equation
(3)
𝑛
∑
𝑎𝑗 (z)Φ′𝑗 (z) − 𝑏(z)Φ′𝑛+1 (z) = 0
𝑗=1
in R𝑛+1 .
We see that if a 𝐶 1 -function Φ of (𝑛 + 1) variables satisfies (3), then any
function 𝑢, defined implicitly by (2), is a solution to (1). On the other hand,
assume that Φ is a 𝐶 1 -function in a subdomain of R𝑛+1 and, at some point
(0)
(0)
(0)
z(0) = (𝑥1 , 𝑥2 , . . . , 𝑥𝑛 , 𝑦 (0) ) we have Φ′𝑛+1 (z(0) ) = 0. Then the equation (2)
(0)
(0)
(0)
defines a 𝐶 1 -function 𝑢 in a neighbourhood of x(0) = (𝑥1 , 𝑥2 , . . . , 𝑥𝑛 ), and
this function satisfies
𝑢′𝑗 (x) = −
Φ′𝑗 (x, 𝑢(x))
, 𝑢(x(0) ) = 𝑦 (0) .
Φ′𝑛+1 (x, 𝑢(x))
The conclusion is that if any function 𝑢, defined by (2), satisfies (1), then Φ
satisfies (3). We have thus reduced the problem of solving the quasilinear equation (1) to that of solving the homogeneous linear equation (3). The characteristic curves of (3) (which are curves in R𝑛+1 ) are often called characteristic
curves for the equation (1). In other words, there is a certain ambiguity in the
literature concerning the concept characteristic curve. Now we take the concept
characteristic curve in this modified sense, i.e. the characteristic curves for (1)
are curves in R𝑛+1 . Then the following holds (assuming that the coefficients 𝑎𝑗
and 𝑏 are continuously differentiable in some open domain Ω in R𝑛+1 ):
1. Through each point of Ω there is exactly one characteristic curve.
2. If 𝑆 ⊂ Ω is an integral surface (i.e. graph of a solution to (1)), and if 𝑆
has some point in common with a certain characteristic curve, then this
characteristic curve lies completely in 𝑆.
3. If two integral surfaces have a point in common, then they have the whole
characteristic curve through this point in common (follows from the previous statement).
4. If the graph of a 𝐶 1 -function 𝑢 is generated by characteristic curves, then
𝑢 is a solution to (1).
5
The Cauchy problem
Consider the problem
(𝐸)
(𝐶)
∑𝑛
∂𝑢
𝑎𝑗 (x, 𝑢(x)) ∂𝑥
+ 𝑏(x, 𝑢(x)) = 0, x ∈ 𝐷
𝑗
𝑢(x) = 𝜙(x),
x∈𝑆
𝑗=1
where 𝐷 is a domain in R𝑛 , 𝑆 a hypersurface in 𝐷 and 𝜙 a given function on
𝐷. The Cauchy condition (𝐶) means that the (𝑛 − 1)-dimensional surface in
R𝑛+1 ,
Σ := {(x, 𝑦); x ∈ 𝑆, 𝑦 = 𝜙(x)}
shall be a part of the graph of the solution 𝑢. This graph is thus generated by
the characteristics passing through Σ. The problem is well-posed if and only
if this geometrical problem does not degenerate, i.e. if no characteristic is tangent to Σ. To get a mental picture of the situation, imagine a family of curves
(characteristics) in R3 such that, for each point in space, exactly one curve from
the family passes through this point. Now imagine another curve Σ, that is not
tangent to any of the caharacteristics. Then those characteristics that intersect
Σ generate a surface. If, however, some characteristics are tangent to Σ (or
worse, if Σ itself is characteristic), then the construction breaks down.
Example: Solve the Cauchy problem
(𝑥2 + 1)
2𝑥𝑦 ∂𝑢
∂𝑢
+
= 2𝑥𝑢, 𝑦 > 0,
∂𝑥 𝑥2 + 1 ∂𝑦
𝑢(0, 𝑦) = ln 𝑦.
Solution: We are looking for an integral surface in R3 , consisting of characteristic curves and containing the curve 𝑥 = 0, 𝑦 = 𝑠, 𝑧 = ln 𝑠, 𝑠 > 0. this leads
to
⎧
˙
= 𝑥2 + 1,
𝑥(0) = 0
⎨ 𝑥(𝑡)
𝑦(𝑡)
˙ = 𝑥2 + 1,
𝑦(0) = 𝑠
⎩
𝑧(𝑡)
˙ = 2𝑥𝑧,
𝑧(0) = ln 𝑠,
1
ln 𝑠
i.e. 𝑥 = tan 𝑡, 𝑦 = 𝑠𝑒 2 (1−cos 2𝑡) , 𝑧 = cos
2 𝑡 This is a parametrization (with
parameters 𝑠 and 𝑡) of the integral surface we are looking for. We eliminate 𝑠
and 𝑡:
𝑥2
1
1
1 − 𝑥2
𝑥2
− 1+𝑥
− 12 (1−cos 2𝑡)
2
(1 − cos 2𝑡) = (1 −
)
=
,
𝑠
=
𝑦𝑒
=
𝑦𝑒
,
2
2
1 + 𝑥2
1 + 𝑥2
(
𝑧 = ln 𝑠 cos2 𝑡 = ln 𝑦 −
𝑥2 )
(1 + 𝑥2 ) = (1 + 𝑥2 ) ln 𝑦 − 𝑥2 .
1 + 𝑥2
Svar: 𝑢(𝑥, 𝑦) = (1 + 𝑥2 ) ln 𝑦 − 𝑥2 .
Exercise 0.5. Solve the following Cauchy problem in R2 :
𝑥𝑢′𝑥 + 𝑦𝑢′𝑦 = 2𝑢 ln 𝑢, 𝑢(𝑥, 1) = 𝑒𝑥
2
−1
.
Exercise 0.6. Solve the following Cauchy problem in the half-space 𝑥 > 0 of
R3 :
∂𝑢 ( ∂𝑢 ∂𝑢 )
2
+
+
tan 𝑢 = 0, 𝑢(1, 𝑦, 𝑧) = arctan(𝑦 + 𝑧).
∂𝑥
∂𝑦
∂𝑧
6