First Order Partial Differential Equations, Part - 1:
Single Linear and Quasilinear First Order Equations
PHOOLAN PRASAD
DEPARTMENT OF MATHEMATICS
INDIAN INSTITUTE OF SCIENCE, BANGALORE
Definition
First order PDE in two independent variables is a relation
F (x, y; u; ux , uy ) = 0
F a known real function from D3 ⊂ R5 → R
(1)
Examples: Linear, semilinear, quasilinear, nonlinear equations ux + uy = 0
ux + uy = ku, c and k are constant
ux + uy = u2
uux + uy = 0
u2x − u2y = 0
u2x + u2y + 1 = 0
q
ux + 1 − u2y = 0, defined for |uy | ≤ 1
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Symbols for various domains used
In this lecture we denote
by D a domain in R2 where a solution is defined,
by D1 a domain in R2 where the coefficients of a
linear equation are defined and
by D2 is a domain in (x, y, u)-space i.e., R3
finally by D3 a domain in R5 where the function F
of five independent variables is defined.
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Meaning of a classical (genuine) solution u(x, y)
u is defined on a domain D ⊂ R2
u ∈ C 1 (D)
(x, y, u(x, y), ux (x, y), uy (x, y)) ∈ D3 when
(x, y) ∈ D
F (x, y; u(x, y); ux (x, y), uy (x, y)) = 0 ∀(x, y) ∈ D.
We call a classical solution simply a solution.
Otherwise, generalized or weak solution.
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Classification
Linear equation:
a(x, y)ux + b(x, y)uy = c1 (x, y)u + c2 (x, y)
(3)
Non linear equation: All other equations with subclasses
Semilinear equation:
a(x, y)ux + b(x, y)uy = c(x, y, u)
(4)
Quasilinear equation:
a(x, y, u)ux + b(x, y, u)uy = c(x, y, u)
(5)
Nonlinear equation: F (x, y; u; ux , uy ) = 0 where F is not linear in
ux , uy .
Properties of solutions of all 4 classes of equations are quite
different.
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Example 2
ux = 0
General solution in D = R2
u = f (y), f is an arbitrary C 1 function.
Function f is uniquely determined if u is prescribed on any curve no
where parallel to x-axis. On a line parallel to x-axis, we can not
prescribe u arbitrarily.
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Example 3
uy = 0
(6)
General solution in D = R2
u = f (x), f C 1(R).
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Directional derivative
ux = 0 means rate of change of u in direction (1, 0) parallel to x−axis
is zero
i.e. (1, 0).(ux , uy ) = 0
We say ux = 0 is a directional derivative in the direction (1, 0).
Consider a curve with parametric representation x = x(σ), y = y(σ)
given by ODE
dy
dx
= a(x, y),
= b(x, y)
(8)
dσ
dσ
Tangent direction of the curve at (x, y):
(a(x, y), b(x, y))
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Directional derivative contd..
Rate of change of u(x, y) with σ as we move along this
curve is
dx
dy
du
= ux
+ uy
(10)
dσ
dσ
dσ
= a(x, y)ux + b(x, y)uy
which is a directional derivative in the direction (a, b)
at (x, y)
If u satisfies PDE aux + buy = c(x, y, u) then
du
= c(x, y, u)
dσ
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Characteristic equation and compatibility
condition
For the PDE
a(x, y)ux + b(x, y)uy = c(x, y)
Characteristic equations
dy
dx
= a(x, y),
= b(x, y)
dσ
dσ
and compatibility condition
du
= c(x, y)
dσ
(14) and (15) with a(x, y) 6= 0 characteristic equations give
dy
=
dx
compatibility condition becomes
du
=
dx
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(12)
(13)
(14)
b(x, y)
a(x, y)
(15)
c(x, y)
a(x, y)
(16)
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Example 4
yux − xuy = 0
(17)
Characteristic equations are
dy
dx
= y,
= −x
(18)
dσ
dσ
or
dy
x
= − ⇒ y dy+x dx = 0 ⇒ x2+y 2 = constant.
dx
y
(19)
The characteristic curves are circles with centre
at (0,0).
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Example 4 contd..
Compatibility conditions along these curves are
du
= 0 ⇒ u = constant.
dσ
(20)
Hence value of u at (x, y) = value of u at (−x, −y).
u is an even function of x and also of y.
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Example 4 contd..
Will this even function be of the form
u = f (x2 + y 4 )?
The information
u = constant on the circles x2 + y 2 = constant
⇒ u = f (x2 + y 2 )
where f ∈ C 1 (R) is arbitrary.
Every solution is of this form.
u is an even function of x and y but of a special
form.
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Linear first order PDE
a(x, y)ux + b(x, y)uy = c1 (x, y)u + c2 (x, y)
(21)
Let w(x, y) be any solution of the nonhomogeneous
equation (21). Set u = v + w(x, y)
⇒ v satisfies the homogeneous equation
a(x, y)vx + b(x, y)vy = c1 (x, y)v
(22)
Let f (x, y) be a general solution of (22)
⇒ u = f (x, y) + w(x, y)
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Example 5: Equation with constant coefficients
aux + buy = c, a, b, c are constants
(24)
For the homogeneous equation, c = 0, characteristic
equation (with a 6= 0)
dy
b
= ⇒ ay − bx = constant
dx a
(25)
du
= 0 ⇒ u = constant
dx
(26)
Along these
Hence u = f (ay − bx) is general solution of the
homogeneous equation.
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Example 5: Equation with constant coefficients
contd..
For the nonhomogeneous equation, the compatibility condition
du
c
c
= ⇒ u = const + x
dx
a
a
(27)
The constant here is constant along the characteristics
ay − bx = const.
Hence general solution
c
u = f (ay − bx) + x.
a
(28)
Alternatively u = ac x is a particular solution. Hence the result.
Solution of a PDE contains arbitrary elements. For a first order
PDE, it is an arbitrary function.
In applications - additional condition ⇒ Cauchy problem.
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The Cauchy Problem for F (x, y; u; ux , uy ) = 0
Piecewise smooth curve
γ : x = x0 (η), y = y0 (η), η ∈ I ⊂ R
a given function u0 (η) on γ
Find a solution u(x, y) in a neighbourhood of γ
The solution takes the prescribed value u0 (η) on γ
i.e
u(x0 (η), y0 (η)) = u0 (η)
(29)
Existence and uniqueness of solution of a Cauchy
problem requires restrictions on γ, the function F and
the Cauchy data u0 (η).
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The Cauchy Problem contd..
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Example 6a
Solve yux − xuy = 0 in R2
with u(x, 0) = x, x ∈ R
The solution must be an even function of x and y.
But the Cauchy data is an odd function.
Solution does not exist.
Example 6b
Solve yux − xuy = 0 in a domain D
u(x, 0) = x, x ∈ R+
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Example 6b conti....
Solution is u(x, y) = (x2 + y 2 )1/2 , verify with partial
derivatives
x
y
ux = 2
, uy = 2
2
1/2
(x + y )
(x + y 2 )1/2
(31)
Solution is determined in R2 \ {(0, 0)}.
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Example 7
Cauchy problem: Solve
2ux + 3uy = 1
with data u|γ = u(αη, βη) = u0 (η) on
γ : βx − αy = 0; α, β = constant
Characteristic curve through an arbitrary point P 0 (x0 , y 0 ) in
(x, y)- plane.
dx
dy
= 2,
=3
dσ
dσ
⇒ x = x0 + 2σ, y = y 0 + 3σ
is a straight line
3x − 2y = 3x0 − 2y 0
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Example 7 contd..
If P 0 lies on γ, x0 = αη, y 0 = βη, the characteristic
curves are
x = αη + 2σ, y = βη + 3σ, η = const, σ varies
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Example 7 contd..
Compatibility condition
du
= 1 ⇒ u = u0 (x0 , y 0 ) + σ
dσ
(32)
u = u(αη, βη) + σ = u0 (η) + σ
(33)
When P 0 lies on γ
Solution of the Cauchy problem
Solve x = αη + 2σ, y = βη + 3σ for σ and η
σ=
βx − αy
2y − 3x
, η=
2β − 3α
2β − 3α
(34)
Substitute in expression for u
βx − αy
u(x, y) =
+ u0
2β − 3α
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2y − 3x
2β − 3α
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Example 7 contd.. Existence and Uniqueness
The solution exists as long as 2β − 3α 6= 0 i.e., the
datum curve is not a characteristic curve
Uniqueness:
Compatibility condition carries information on the
variation of u along a characteristic in unique way.
This leads to uniqueness.
What happens when 2β − 3α = 0?
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Example 8: Characteristic Cauchy problem
2β − 3α = 0 ⇒ datum curve is a characteristic curve
Choose α = 2, β = 3 ⇒ x = 2η, y = 3η
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Example 8: Characteristic Cauchy problem
contd...
The characteristic Cauchy problem: Solve
2ux + 3uy = 1
with data
1
u(2η, 3η) = u0 (η) = u0 ( x)
2
Since
du0 (η)
d
=
u(2η, 3η) = 2ux + 3uy = 1, using PDE
dη
dη
(36)
The Cauchy data u0 cannot be prescribed arbitrarily on γ.
u0 (η) = η = 21 x, ignoring constant of integration.
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Example 8 contd..
u = 21 x is a particular solution satisfying the
Cauchy data and g(3x − 2y) is solution of the
homogeneous equation.
Hence
1
u = x + g(3x − 2y), g ∈ C 1 and g(0) = 0 (37)
2
is a solution of the Cauchy problem.
Since g is any C 1 function with g(0) = 0, solution
of the Characteristic Cauchy problem is not unique.
We verify an important theorem ”in general,
solution of a characteristic Cauchy problem
does not exist and if exists, it is not unique”.
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Quasilinear equation
a(x, y, u)ux + b(x, y, u)uy = c(x, y, u)
(38)
Since a and b depend on u, it is not possible to interpret
∂
∂
+ b(x, y, u)
(39)
a(x, y, u)
∂x
∂y
as a directional derivative in (x, y)- plane.
We substitute a known solution u(x, y) for u in a and b, then at
∂
any point (x, y), it represents directional derivative ∂σ
in the
direction given by
dy
dx
= a(x, y, u(x, y)),
= b(x, y, u(x, y))
(40)
dσ
dσ
Along these characteristic curves, we get compatibility condition
du
= c(x, y, u(x, y))
(41)
dσ
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Quasilinear equation contd..
(41) is true for every solution u(x, y). The Characteristic
equations
dx
dy
= a(x, y, u),
= b(x, y, u)
dσ
dσ
(42)
along with the Compatibility condition
du
= c(x, y, u)
dσ
forms closed system.
Solution of 3 equations form a 2-parameter family of curves in
(x, y, u)-space are called Monge Curves and their
projections on (x, y)-plane are two parameter family of
characteristic curves.
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Method of solution of a Cauchy problem
Solve
a(x, y, u)ux + b(x, y, u)uy = c(x, y, u)
(43)
in a domain D containing
γ : x = x0 (η), y = y0 (η)
with Cauchy data
u(x0 (η), y0 (η)) = u0 (η)
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Method of solution of a Cauchy problem contd..
Solve
dy
du
dx
= a(x, y, u),
= b(x, y, u),
= c(x, y, u)
dσ
dσ
dσ
(45)
(x, y, u)|σ=0 = (x0 (η), y0 (η), u0 (η))
(46)
x = x(σ, x0 (η), y0 (η), u0 (η)) ≡ X(σ, η)
y = y(σ, x0 (η), y0 (η), u0 (η)) ≡ Y (σ, η)
u = u(σ, x0 (η), y0 (η), u0 (η)) ≡ U (σ, η)
(47)
⇒
Solving the first two for σ = σ(x, y), η = η(x, y) we get the solution
u = U (σ(x, y), η(x, y)) ≡ u(x, y)
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Method of solution of a Cauchy problem contd..
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Quasilinear equations conti...
Theorem:
x0 (η), y0 (η), u0 (η) ∈ C 1 (I), say I = [0, 1]
a(x, y, u), b(x, y, u), c(x, y, u) ∈
C 1 (D2 ), where D2 is a domain in (x, y, u) − space
D2 contains curve Γ in (x, y, u)-space
Γ : x = x0 (η), y = y0 (η), u = u0 (η), η ∈ I
dy0
dη a(x0 (η), y0 (η), u0 (η))
−
dx0
dη b(x0 (η), y0 (η), u0 (η))
6= 0, η ∈ I
There exists a unique solution of the Cauchy problem in a domain D
containing I.
Do not worry about the complex statement in the
theorem. As long as the datum curve γ is not
tangential to a characteristic curve and the functions
involved are smooth, the solution exist and is unique
(see previous slide).
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Example 9
Cauchy problem
ux + uy = u
u(x, 0) = 1 ⇒
x0 = η, y0 = 0, u0 = 1
(49)
Step 1. Characteristic curves
dx
=1⇒x=σ+η
dσ
dy
=1⇒y=σ
dσ
Step 2. Therefore
σ = y, η = x − y
Step 3. Compatibility condition
du
= u ⇒ u = u0 (η)eσ = eσ
dσ
Step 4. Solution u = ey
exists on D = R2
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Example 10
Cauchy problem
ux + uy = u2
u(x, 0) = 1 ⇒
x0 = η, y0 = 0, u0 = 1
(51)
Step 1. Characteristic equations give
x = σ + η, y = σ
Step 2. Compatibility condition gives
1
du
= u2 ⇒ u =
dσ
u0 (η) − σ
1
Step 3. Solution u = 1−y
exists on the domain D = y < 1 and u → +∞ as y → 1−.
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Example 11
Cauchy problem
uux + uy = 0
u(x, 0) = x, 0 ≤ x ≤ 1
(52)
⇒ x = η, y = 0, u = η, 0 ≤ η ≤ 1 at σ = 0
Step 1. Characteristic equations and compatibility
condition
dx
dy
du
= u,
= 1,
=0
(53)
dσ
dσ
dσ
Step 2. Quasilinear equations, characteristics depend
on the solution
u=η
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Example 11 contd..
Step 3.
x = η(σ + 1), y = σ
Step 4. From solution of characteristic equations
x
σ = y and η = y+1
x
Step 5. Solution is u = y+1
, but D =?
Step 6. Characteristic curves are straight lines
x
= η, 0 ≤ η ≤ |
y+1
which meet at the point (−1, 0).
Step 7. u is constant on these characteristics (see next
slide).
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Solution is determined in a wedged shaped region
in (x, y)-plane including the lines y = 0 and y = x + 1.
Figure:
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Example 12
Cauchy problem
uux + uy = 0
u(x, 0) = 12 , 0 ≤ x ≤ 1
(54)
Step 1. Parametrization of Cauchy data
1
⇒ x0 = η, y0 = 0, u0 = , 0 ≤ η ≤ 1.
2
Step 2. The compatibility condition gives
1
u = constant = .
2
Step 3. The characteristic curves are
y − 2x = −2η, 0 ≤ η ≤ 1.
on which solution has the same value u = 12 .
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Example 12 contd..
Step 4. The solution u = 12 of the Cauchy problem is determined in an
infinite strip 2x ≤ y ≤ 2x − 2 in (x, y)-plane.
Important: From examples 11 and 12, we notice that the domain,
where solution of a Cauchy problem for a quasilinear equation is
determined, depends on the Cauchy data.
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General solution
General solution contains an arbitrary function.
Theorem : If φ(x, y, u) = C1 and ψ(x, y, u) = C2 be
two independent first integrals of the ODEs
dy
du
dx
=
=
a(x, y, u)
b(x, y, u)
c(x, y, u)
(56)
and φ2u + ψu2 6= 0, the general solution of the PDE
aux + buy = c is given by
h(φ(x, y, u), ψ(x, y, u)) = 0
(57)
where h is an arbitrary function.
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Example 13
uux + uy = 0
(58)
dy
du
dx
=
=
(59)
u
1
0
Note 0 appearing in a denominator to be properly interpreted
⇒ u = C1
x − C1 y = C2
⇒ x − uy = C2 ⇒
(60)
General solution is given by
φ(u, x − uy) = 0
or u = f (x − uy)
(61)
where h and f arbitrary functions.
Note : Solution of this nonlinear equation may be very difficult.
Numerical method is generally used.
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Example 14
Consider the differential equation
1
(y + 2ux)ux − (x + 2uy)uy = (x2 − y 2 )
2
(62)
The characteristic equations and the compatibility condition are
dx
dy
du
=
= 1 2
y + 2ux
−(x + 2uy)
(x − y 2 )
2
(63)
To get one first integral we derive from these
2du
xdx + ydy
= 2
2
2
2u(x − y )
x − y2
(64)
which immediately leads to
ϕ(x, y, u) ≡ x2 + y 2 − 4u2 = C1
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Example 14 contd..
For another independent first integral we derive a
second combination
2du
ydx + xdy
=
y 2 − x2
x2 − y 2
(66)
ψ(x, y, u) ≡ xy + 2u = C2
(67)
which leads to
The general integral of the equation (55)is given by
h(x2 + y 2 − 4u2 , xy + 2u) = 0
x2 + y 2 − 4u2 = f (xy + 2u)
(68)
where h or f are arbitrary functions of their arguments.
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Example 14 contd..
Consider a Cauchy problem u = 0 on x − y = 0
⇒ x = η, y = η, u = 0
From (58) and (60) we get 2η 2 = C1 and η 2 = C 2 which
gives C1 = 2C2 . Therefore, the solution of the Cauchy
problem is obtained, when we take h(ϕ, ψ) = ϕ − 2ψ.
This gives, taking only the suitable one,
o
1 np
2
u=
(x − y) + 1 − 1
(69)
2
We note that the solution of the Cauchy problem is
determined uniquely at all points in the (x, y)-plane.
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Exercise
1. Show that all the characteristic curves of the partial
differential equation
(2x + u)ux + (2y + u)uy = u
through the point (1,1) are given by the same straight
line x − y = 0
2. Discuss the solution of the differential equation
uux + uy = 0, y > 0, −∞ < x < ∞
with Cauchy data
(
α 2 − x2
u(x, 0) =
0
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for |x| ≤ α
for |x| > α.
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Exercise contd..
3. Find the solution of the differential equation
m 1 − u ux − mM uy = 0
r
satisfying
My
u(0, y) =
ρ−y
where m, r, ρ, M are constants, in a neighourhood of
the point x = 0, y = 0.
4. Find the general integral of the equation
(2x − y)y 2 ux + 8(y − 2x)x2 uy = 2(4x2 + y 2 )u
and deduce the solution of the Cauchy problem when
1
the u(x, 0) = 2x
on a portion of the x-axis.
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1
R. Courant and D. Hilbert. Methods of Mathematical Physics,
vol 2: Partial Differential Equations. Interscience Publishers, New
York, 1962.
2
L. C. Evans. Partial Differential Equations. Graduate Studies in
Mathematics, Vol 19, American Mathematical Society, 1999.
3
F. John. Partial Differential Equations. Springer-Verlag, New
York, 1982.
4
P. Prasad. (1997) Nonlinearity, Conservation Law and Shocks,
Part I: Genuine Nonlinearity and Discontinuous Solutions,
RESONANCE- Journal of Science Education by Indian Academy
of Sciences, Bangalore, Vol-2, No.2, 8-18.
P. Prasad. (1997) Nonlinearity, Conservation Law and Shocks,
Part II: Stability Consideration and Examples, RESONANCEJournal of Science Education by Indian Academy of Sciences,
Bangalore, Vol-2, No.7, 8-19.
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1
P. Prasad. Nonlinear Hyperbolic Waves in Multi-dimensions.
Monographs and Surveys in Pure and Applied Mathematics,
Chapman and Hall/CRC, 121, 2001.
2
P. Prasad. A theory of first order PDE through propagation of
discontinuities. Ramanujan Mathematical Society News Letter,
2000, 10, 89-103; see also the webpage:
3
P. Prasad and R. Ravindran. Partial Differential Equations.
Wiley Eastern Ltd, New Delhi, 1985.
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Thank You!
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