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B5b Applied Partial Differential Equations 4–
4 First-order nonlinear equations
4.1 Introduction
Now we consider general first-order nonlinear scalar PDEs, that is ones that are not necessarily
quasi-linear. The general form of such an equation is
F (p, q, u, x, y) = 0, (4.1)
where we use
∂u
∂x
= p,
∂u
∂y
= q (4.2)
as shorthand, so that
∂p
∂y
∂q
∂x
The case of quasilinear equations corresponds to F being a linear function of p and q, i.e.
F (p, q, u, x, y, ) = a(x, y, u)p + b(x, y, u)q − c(x, y, u). (4.4)
4.2 Charpit’s equations
If we differentiate (4.1) with respect to x and y, we obtain
∂F
∂p
∂p
∂x
∂F
∂q
∂q
∂x
∂F
∂x
− p
∂F
∂u
, (4.5a)
∂F
∂p
∂p
∂y
∂F
∂q
∂q
∂y
∂F
∂y
− q
∂F
∂u
, (4.5b)
or, using (4.3),
∂F
∂p
∂p
∂x
∂F
∂q
∂p
∂y
∂F
∂x
− p
∂F
∂u
, (4.6a)
∂F
∂p
∂q
∂x
∂F
∂q
∂q
∂y
∂F
∂y
− q
∂F
∂u
. (4.6b)
So, if we define characteristics or rays as curves
x(τ ), y(τ )
satisfying
dx
dτ
∂F
∂p
dy
dτ
∂F
∂q
4–2 OCIAM Mathematical Institute University of Oxford
then, along these curves,
dp
dτ
∂F
∂x
− p
∂F
∂u
dq
dτ
∂F
∂y
− q
∂F
∂u
We therefore have a system of four ODEs for x, y, p and q satisfied along the rays. Recall,
though, that in general F depends on u also, so to close the system we also need an ODE for
u along the rays, namely
du
dτ
∂u
∂x
dx
dτ
∂u
∂y
dy
dτ
= p
∂F
∂p
+ q
∂F
∂q
In summary, we have the following system of ODEs for x, y, p, q and u, known as Charpit’s
equations:
dx
dτ
∂F
∂p
, (4.10a)
dy
dτ
∂F
∂q
, (4.10b)
dp
dτ
∂F
∂x
− p
∂F
∂u
, (4.10c)
dq
dτ
∂F
∂y
− q
∂F
∂u
, (4.10d)
du
dτ
= p
∂F
∂p
+ q
∂F
∂q
. (4.10e)
It is easily verified that these reduce to the usual characteristic equations
dx
dτ
= a,
dy
dτ
= b,
du
dτ
= c, (4.11)
for quasi-linear equations where F takes the form (4.4).
4.3 Boundary data
As for quasilinear scalar equations, Cauchy data specifies u along some curve Γ in the
(x, y)-plane:
x = x 0 (s), y = y 0 (s), u = u 0 (s), (4.12)
for s in some (possibly infinite) interval. We also require initial conditions for p and q, say
p = p 0 (s), q = q 0 (s), which are obtained by differentiating u 0 with respect to s and using the
PDE (4.1):
du 0
ds
= p 0
dx 0
ds
+ q 0
dy 0
ds
, F (p 0 , q 0 , u 0 , x 0 , y 0 ) = 0. (4.13)
By the implicit function theorem, the two equations (4.13) may be solved uniquely (in prin-
ciple, if not explicitly) for p 0 and q 0 provided the condition
dx 0
ds
∂F
∂q 0
dy 0
ds
∂F
∂p 0
4–4 OCIAM Mathematical Institute University of Oxford
0
1
2
-0.
0
1 0
1
0
1
2
0
x
u
y
-2 -1 1 2
-0.
1
y
x
(a) (b)
Figure 1: (a) Rays for a sugar heap on an elliptical spoon with a = 2 and b = 1; the bold line
marks the ridge. (b) The corresponding pile height u(x, y).
0
2
4
0
2
0
1
2
0
2
4 0
5
u
x
y
Figure 2: The height u(x, y) of a sandpile outside an elliptical hole with a = 2 and b = 1.
B5b Applied Partial Differential Equations 4–
for some constants a and b, and the solution is given parametrically by
x = a cos(s) −
bτ cos(s) √ a^2 sin^2 (s) + b^2 cos^2 (s)
, y = b sin(s) −
aτ sin(s) √ a^2 sin^2 (s) + b^2 cos^2 (s)
, u = τ. (4.24)
The rays and solution surface are shown in Figure 1. Notice that a ridge line, across which p and q are discontinuous, forms along the x-axis, between the x = −(a^2 − b^2 )/a and x = +(a^2 − b^2 )/a. In this figure a = 2 and b = 1; if a < b then the ridge forms at the corresponding position along the y-axis. If the other root is taken for p and q, then the rays propagate out of the ellipse as τ is increased from zero, and the parametric solution is now
x = a cos(s) +
bτ cos(s) √ a^2 sin^2 (s) + b^2 cos^2 (s)
, y = b sin(s) +
aτ sin(s) √ a^2 sin^2 (s) + b^2 cos^2 (s)
, u = τ. (4.25)
This corresponds to a sandpile on a table with an elliptical hole, as shown in Figure 2.
4.4 Proof that Charpit’s method works
If we differentiate F along a ray, we find that
dF
dτ
∂F
∂p
dp
dτ
∂F
∂q
dq
dτ
∂F
∂u
du
dτ
∂F
∂x
dx
dτ
∂F
∂y
dy
dτ
Since the boundary condition (4.13) sets F to zero on the initial curve Γ, it must therefore
be zero everywhere along the rays passing through Γ. Hence p, q and u satisfy the equation
F (p, q, u, x, y) = 0 everywhere in the domain of definition where there are rays emanating
from Γ.
This is not quite sufficient to prove that u is a solution of the original nonlinear PDE. We
still have to show that the functions p and q that result from solving Charpit’s equations are
equal to ∂u/∂x and ∂u/∂y respectively. To do this, we first prove that φ ≡ 0, where
∂u
∂s
− p
∂x
∂s
− q
∂y
∂s
By differentiating φ with respect to τ and rearranging, we obtain
∂F
∂s
∂F
∂u
∂s
∂u
− p
∂F
∂p
− q
∂F
∂q
The final term is identically zero by (4.10), and we have already shown that F ≡ 0 in the
domain of definition, which implies that ∂F/∂s ≡ 0. Hence φ satisfies
∂F
∂u
with, by (4.13), φ = 0 at τ = 0. Provided ∂F/∂u is bounded, it follows from Picard’s theorem
that φ ≡ 0 in the domain of definition.
From this fact and Charpit’s equation for u, we obtain two simultaneous equations for
∂u/∂x and ∂u/∂y:
∂x
p +
∂y
q =
∂u
∂x
∂u
∂x
∂y
∂u
∂y
, (4.30a)
∂x
∂s
p +
∂y
∂s
q =
∂u
∂s
∂x
∂s
∂u
∂x
∂y
∂s
∂u
∂y
. (4.30b)
B5b Applied Partial Differential Equations 4–
-0.4 -0.2 0.2 0.
-1.
-0.
1
2
xy = 1/
xy = 1/
-0.4 -0.2 0.2 0.
-1.
-0.
1
2
x x
(a) y (b) y
Figure 3: Rays for Example 4.2. (a) u 0 (s) = s^2 ; (b) u 0 (s) = s^3.
and their envelope is found parametrically by differentiating with respect to s:
x =
2 u′′ 0 (s)
, y = s − u′ 0 (s) u′′ 0 (s)
For example, if u 0 (s) = s^2 , then the solution in explicit form is
u =
y^2 1 − 4 x
The rays in this case are given by y = s(1 − 4 x) so, as illustrated in Figure 3(a), they all pass through the point (1/ 4 , 0), and the solution is defined in x < 1 / 4. If u 0 (s) = s^3 , then the solution is
u = s^3 (1 − 9 xs), where s =
1 − 24 xy 12 x
The rays are given by y = s − 6 s^2 x, and their envelope is the curve 24 xy = 1, as illustrated in Figure 3(b). The solution is therefore defined in the region 24 xy < 1 (which is where s is real).
4.6 Geometrical optics
The propagation of sound or light waves in two spatial dimensions is governed by the wave
equation
∂^2 ψ
∂x^2
∂^2 ψ
∂y^2
c^2
∂^2 ψ
∂t^2
where ψ is some state variable such as pressure or electric field and c is the wave speed. We
look for time-periodic (or “monochromatic”) solutions with constant frequency ω by setting
ψ(x, y, t) = φ(x, y)e−iωt. (4.43)
Then φ satisfies the Helmholtz equation
∇^2 φ + k^2 φ =
∂^2 φ
∂x^2
∂^2 φ
∂y^2
+ k^2 φ = 0, (4.44)
where k = ω/c is the wavenumber (i.e. 2 π divided by the wavelength).
4–8 OCIAM Mathematical Institute University of Oxford
e
e
e incident
reflected
Figure 4: Illustration of Snell’s law.
The theory of geometrical optics arises from the limit k → ∞, which is valid over length-
scales much longer than the wavelength. To analyse the behaviour of (4.44) in this limit, we
use the so-called WKBJ method, which involves writing φ in the form
φ(x, y) = A(x, y)eiku(x,y), (4.45)
where A and u represent the amplitude and phase respectively of the solution. Then (4.44)
becomes
∇^2 A + ik
A∇^2 u + 2∇A · ∇u
+ k^2 A
1 − |∇u|^2
We seek solutions in which A is an asymptotic expansion of the form
A ∼ A 0 +
A 1
k
A 2
k^2
At leading order, (4.46) implies that u satisfies the Eikonal equation
|∇u|^2 = 1. (4.48)
Then the successive terms in the amplitude expansion satisfy the transport equations
2 ∇u · ∇A 0 + A 0 ∇^2 u = 0, (4.49a)
2 ∇u · ∇An + An∇^2 u = i∇^2 An− 1 , n ≥ 1. (4.49b)
The Eikonal equation (4.48) may be solved exactly as in Example 4.1. The rays correspond
to light rays and all the familiar properties of geometrical optics, for example that light travels
in straight lines, follow from the solution of Charpit’s equations.
Example 4.3 Reflecting plane waves
One obvious solution of (4.48) is u = x, which corresponds to a plane wave moving in the x-direction. Now we examine what happens if such a wave impinges on a reflecting wall given by a curve Γ in the (x, y)-plane. We decompose the state variable φ into an incident wave φI , namely a plane wave with constant amplitude a, and a reflected wave φR:
φ = φI + φR, φI = aeikx. (4.50)
Now we apply the WKBJ ansatz to φR:
φR = Aeiku(x,y). (4.51)
4–10 OCIAM Mathematical Institute University of Oxford
which may be solved to give τ = 12 cos(s). (4.58)
Thus J = 0 on the curve given parametrically by
x = cos(s)
1 − 12 cos(2s)
, y = sin(s) − 12 cos(s) sin(2s) = sin^3 (s), (4.59)
and such a curve is called a caustic. A single-valued solution may be obtained by truncating the rays when they reach the caustic, as shown in Figure 5(b). A caustic similar to this may often be observed if a bright light is shone on a cup of coffee. An alternative way to find the caustic is to note that the rays are given by
F (x, y; s) = x sin(2s) − y cos(2s) − sin(s) = 0. (4.60)
Their envelope may therefore be found by solving
F =
∂F
∂λ
that is x sin(2s) − y cos(2s) − sin(s) = 2x cos(2s) + 2y sin(2s) − cos(s) = 0, (4.62)
whose simultaneous solution reproduces (4.59).
Example 4.4 illustrates that a single-valued ray solution may be obtained by truncating
rays at any caustic where the Jacobian is zero. It may be shown that the asymptotic ansatz
(4.45) breaks down, with A → ∞ as the caustic is approached. The method of matched
asymptotic expansions yields the appropriate correction in the neighbourhood of a caustic
and allows the behaviour in the dark zone beyond the caustic (corresponding to complex
rays) to be found.
