LECTURE 8: METHOD OF CHARACTERISTIC AND SEPARATION OF VARIABLES
MINGFENG ZHAO
July 17, 2015
Method of Characteristic
For a general first order PDE F (x, y, u, ux , uy ) = 0, the method of characteristics discovers curves (called characteristic
curves or just characteristics) along which the PDE becomes an ordinary differential equation (ODE). Once the ODE
is found, it can be solved along the characteristic curves and transformed into a solution for the original PDE.
Consider the PDE
(1)
a(x, y, u)ux + b(x, y, u)uy = f (x, y, u).
Since the normal vector to the surface graph z = u(x, y) in R3 is given by (ux (x, y), uy (x, y), −1), then the
equation a(x, y, u)ux + b(x, y, u)uy = f (x, y, u) is equivalent to the geometric statement that the vector field
(a(x, y, u), b(x, y, u), f (x, y, u)) is tangent to the surface z = u(x, y) at every point.
In other words, the graph of the solution must be a union of integral curves of this vector field. These integral curves
are called the characteristic curves of the original PDE. Hence, if a particular parametrization s of the curves is fixed,
then we have



x0 (s) = a(x, y, z)



y 0 (s) = b(x, y, z)




 z 0 (s) = f (x, y, z).
(2)
If (x(s), y(s), z(s)) is a solution to (2), then
f (x(s), y(s), u(x(s), y(s))
= z 0 (s)
=
d
u(x(s), y(s)) = ux (x(s), y(s))x0 (s)ds + uy (x(s), y(s))y 0 (s)ds
ds
= ux (x(s), y(s))dx(s) + uy (x(s), y(s))dy(s).
That is, the PDE (1) holds along the curve (x(s), y(s), u(x(s), y(s))).
1
2
MINGFENG ZHAO
Example 1. Consider the non-homogeneous transport equation:
(3)



ut + cux = f (x, t),

 IC:
u(x, 0) = g(x),
x ∈ R,
t > 0,
x ∈ R.
For each x0 ∈ R and fixed, since u(x0 , 0) = g(x0 ), then we need to find the characteristic curve through the point
(0, x0 , g(x0 )), that is, we should solve:



t0 (s) = 1





 x0 (s) = c
 u0 (s) = f (x(s), t(s)),






 t(0) = 0, x(0) = x ,
0
u(0) = g(x0 ).
So we get
Z
t(s) = s,
x(s) = x0 + cs,
s
and u(s) =
f (x + cτ, τ ) dτ + g(x0 ).
0
Notice that u(s) = u(x(s), t(s)), then
Z
u(x0 + ct, t) =
t
f (x0 + cτ, τ ) dτ + g(x0 ).
0
Since x0 ∈ R is arbitrary, then the solution to (3) is:
Z
t
f (x − ct + cτ, τ ) dτ + g(x − ct) for x ∈ R and t ≥ 0.
u(x, t) =
0
Solving the heat equation
For a heat conduction process with source on a uniform rod with length L, that is, we have the heat equation
ut = c2 uxx for 0 < x < L and t > 0, in this case, we need to know not only the situation of the temperature at the
endpoints x = 0 and x = L, but also the initial temperature u(x, 0) on the bar. Hence we need to solve the problem:



ut = c2 uxx + f (x, t), 0 < x < L, t > 0,

 IC:
u(x, 0) = f (x),
0 ≤ x ≤ L.
For the boundary condition, there are three common types:
• Dirichlet boundary condition, where the end is held at a prescribed temperature:
BC:
u(0, t) = g(t),
u(L, t) = h(t),
t > 0.
LECTURE 8: METHOD OF CHARACTERISTIC AND SEPARATION OF VARIABLES
3
• Neumann boundary condition, where the end is given the heat flux:
BC:
ux (0, t) = g(t),
uy (L, t) = h(t),
t > 0.
• Robin boundary condition, where the heat exchange at endpoints results from the end of the bard being
placed in a heat thermal reservoir at a fixed temperature:
BC:
a(t)u(0, t) + b(t)ux (0, t) = g(t),
c(t)u(L, t) + d(t)ux (L, t) = h(t),
t > 0.
Solving the heat equation without source with homogeneous Dirichlet boundary condition
Let’s consider the heat equation with the homogeneous Dirichlet boundary condition:







(4)






ut = c2 uxx ,
0 < x < L,
BC:
u(0, t) = 0,
u(L, t) = 0,
IC:
u(x, 0) = f (x),
t > 0,
t > 0,
0 ≤ x ≤ L.
Let’s first look at the heat equation and the homogeneous boundary condition:



ut = c2 uxx , 0 < x < L, t > 0,
(5)

 BC: u(0, t) = 0, u(L, t) = 0,
t > 0.
In order to solve (4), first let’s find all ”fundamental” solutions which ”could” be a sequence of solutions
∞
{φn (x, t)}∞
n=1 , secondly, suppose that we can write f (x) as an ”infinite” linear combination of {φn (x, 0)}n=1 , that
∞
X
is, f (x) =
an φn (x, 0) for some sequence {an }∞
n=1 . It’s easy to see that any linear combination of any two solutions
n=1
to (5) is still a solution to (5), then
∞
X
an φn (x, t) is a “solution” to (5).
n=1
A separated solution to (5) is of the form u(x, t) = X(x)T (t), then
ut (x, t)
= X(x)T 0 (t)
ux (x, t)
= X 0 (x)T (t)
uxx (x, t)
= X 00 (x)T (t)
0
= ut − c2 uxx
= X(x)T 0 (t) − c2 X 00 (x)T (t).
4
MINGFENG ZHAO
So we get
−
(6)
T 0 (t)
X 00 (x)
=−
:= λ.
2
c T (t)
X(x)
X 00 (x)
, then λt (x, t) = 0. By (6), then λx (x, t) = 0. So λ(x, t) is a constant, which is called the
X(x)
separation constant. Since u(0, t) = u(L, t) = 0, then X(x) satisfies
Let λ(x, t) = −


 X 00 + λX = 0,
(7)

 X(0) = X(L) = 0.
We are expected to find non-zero solutions to (7). Let X = erx be a solution to X 00 + λX = 0, then X 0 = rerx ,
X 00 = r2 erx , and
0
=
X 00 + λX = r2 erx + λerx
=
erx (r2 + λ).
Then r2 + λ = 0. We have three cases:
a. If λ = 0, that is, r2 = 0. So r = 0, which implies that the general solution to X 00 + λX = X 00 = 0 is
X(x) = C1 + C2 x. Since X(0) = X(L) = 0, then
C1 = C1 + C2 L = 0.
Then C1 = C2 = 0.
b. If λ > 0, that is, λ = µ2 for some µ > 0. So r2 + µ2 = 0, that is, r = ±µi, which implies that the general
solution to X 00 + λX = X 00 + µ2 X = 0 is X(x) = C1 cos(µx) + C2 sin(µx). Since X(0) = X(L) = 0, then
C1 = C1 cos(Lµ) + C2 sin(Lµ) = 0.
Then
C1 = 0,
and C2 sin(Lµ) = 0.
In order to make X to be a non-zero solution, then we need sin(Lµ) = 0, that is, Lµ = nπ for some n ∈ N,
nπ 2
nπ 2
. For λ =
, then all solution to (7) are
that is, λ = µ2 =
L
L
nπ Xn (x) = C sin
x .
L
LECTURE 8: METHOD OF CHARACTERISTIC AND SEPARATION OF VARIABLES
5
c. If λ < 0, that is, λ = −µ2 for some µ > 0. So r2 − µ2 = 0, that is, r = ±µ, which implies that the general
solution to X 00 + λX = X 00 − µ2 X = 0 is X(x) = C1 eµx + C2 e−µx . Since X(0) = X(L) = 0, then
C1 + C2 = C1 eLµ + C2 e−Lµ = 0.
Since µ > 0, then C1 = C2 = 0.
nπ 2
In summary, (7) has non-zero solution if and only if λ =
nπ non-zero solutions to (7) are X(x) = C sin
x .
L
nπ 2
When λ =
, by (6), then
L
0
(8)
for some n ∈ N. When λ =
L
nπ 2
L
in (7), all
= T 0 (t) + c2 λT (t)
nπc 2
T (t).
= T 0 (t) +
L
It’s easy to see that all solutions to (8) are given by:
T (t) = Ce−
n2 π 2 c2
L2
t
.
In summary, for any n ∈ N, we can find a non-zero solution to (5):
un (x, t) = e−
n 2 π 2 c2
L2
t
sin
nπ x .
L
n
nπ o∞
nπ x . So if f (x) is an “infinite” linear combination of un (x, 0) = sin
x
, that
Notice that un (x, 0) = sin
L
L
n=1
∞
∞
∞
nπ nπ X
X
X
n 2 π 2 c2
is, f (x) =
an sin
x , then informally we have u(x, t) =
an un (x, t) =
an e− L2 t sin
x is a solution
L
L
n=1
n=1
n=1
to (4).
Example 2. Consider the problem:













ut = uxx ,
0 < x < 1,
u(1, t) = 0,
t > 0,
BC:
u(0, t) = 0,
t > 0,
IC:
u(x, 0) = 2 sin(πx) + 4 sin(3πx),
0 ≤ x ≤ 1.
By the previous discussion, then a solution is given by:
2
2
u(x, t) = 2e−π t sin(πx) + 4e−9π t sin(3πx).
6
MINGFENG ZHAO
Remark 1. Now, for solving (4), our task is to find the “infinite” linear combination of
∞
nπ X
that is, f (x) =
x . There are three natural questions about this:
an sin
L
n=1
n
nπ o∞
un (x, 0) = sin
x
,
L
n=1
I. For what conditions do we need for a function f (x) on [0, L] such that f (x) can be an “infinite” linear combination
n
nπ o∞
of un (x, 0) = sin
x
?
L
n=1
∞
n
nπ o∞
nπ X
II. If f (x) is an “infinite” linear combination of un (x, 0) = sin
x
x , how
, that is, f (x) =
an sin
L
L
n=1
n=1
to compute the coefficients an ’s?
∞
n
nπ o∞
nπ X
III. If f (x) is an “infinite” linear combination of un (x, 0) = sin
x
x , is
, that is, f (x) =
an sin
L
L
n=1
n=1
∞
∞
nπ X
X
n 2 π 2 c2
the function u(x, t) =
x really a solution to (4)?
an un (x, t) =
an e− L2 t sin
L
n=1
n=1
Solving the heat equation without source with homogeneous Neumann boundary condition
Let’s look the following heat equation with the homogeneous Neumann boundary condition:
(9)













ut = c2 uxx ,
0 < x < L,
BC:
ux (0, t) = 0,
IC:
u(x, 0) = f (x),
t > 0,
ux (L, t) = 0,
t > 0,
0 ≤ x ≤ L.
Let’s first look at the heat equation and the homogeneous boundary condition:
(10)



ut = c2 uxx ,

 BC:
ux (0, t) = 0,
0 < x < L,
t > 0,
ux (L, t) = 0,
t > 0.
A separated solution to (10) is of the form u(x, t) = X(x)T (t), then
ut (x, t)
= X(x)T 0 (t)
ux (x, t)
= X 0 (x)T (t)
uxx (x, t)
= X 00 (x)T (t)
0
= ut − c2 uxx
= X(x)T 0 (t) − c2 X 00 (x)T (t).
LECTURE 8: METHOD OF CHARACTERISTIC AND SEPARATION OF VARIABLES
7
So we get
−
(11)
T 0 (t)
X 00 (x)
=
−
:= λ.
c2 T (t)
X(x)
X 00 (x)
, then λt (x, t) = 0. By (11), then λx (x, t) = 0. So λ(x, t) is a constant. Since ux (0, t) =
X(x)
ux (L, t) = 0, then X(x) satisfies


 X 00 + λX = 0,
(12)

 X 0 (0) = X 0 (L) = 0.
Let λ(x, t) = −
We are expected to find non-zero solutions to (12). Let X = erx be a solution to X 00 + λX = 0, then
X0
= rerx
X 00
= r2 erx
0
= X 00 + λX
= r2 erx + λerx
= erx (r2 + λ).
a. If λ = 0, that is, r2 = 0. So r = 0, which implies that the general solution to X 00 + λX = X 00 = 0 is
X(x) = C1 + C2 x. Since X 0 (x) = C2 , and X 0 (0) = X 0 (L) = 0, then C2 = 0, which implies that all solutions to
(12) are
X(x) = C.
b. If λ < 0, that is, λ = µ2 for some µ > 0. So r2 + µ2 = 0, that is, r = ±µi, which implies that the general solution
to X 00 + λX = X 00 + µ2 X = 0 is X(x) = C1 cos(µx) + C2 sin(µx). Since X 0 (x) = −C1 µ sin(µx) + C2 µ cos(µx)
and X 0 (0) = X 0 (L) = 0, then
C2 µ = −C1 µ sin(µL) + C2 µ cos(µL) = 0.
Since µ > 0, then
C1 sin(µL) = 0,
and C2 = 0.
In order to make X to be a non-zero solution, then we need sin(Lµ) = 0, that is, Lµ = nπ for some n ∈ N,
nπ 2
nπ 2
that is, λ = µ2 =
. For λ =
, then all solution to (12) are
L
L
nπ Xn (x) = C cos
x .
L
8
MINGFENG ZHAO
c. If λ > 0, that is, λ = −µ2 for some µ > 0. So r2 − µ2 = 0, that is, r = ±µ, which implies that the general
solution to X 00 + λX = X 00 − µ2 X = 0 is X(x) = C1 eµx + C2 e−µx . Since X 0 (x) = C1 µeµx − C2 µe−µx and
X 0 (0) = X 0 (L) = 0, then
C1 − C2 = C1 µeµL − C2 µe−µL = 0.
Since µ > 0, then C1 = C2 = 0.
In summary, (12) has non-zero solution if and only if λ =
nπ all non-zero solutions to (12) are X(x) = C cos
x .
L
nπ 2
, by (10), then
When λ =
L
0
(13)
nπ 2
L
nπ 2
S
for some n ∈ N {0}. When λ =
in (12),
L
= T 0 (t) + c2 λT (t)
nπc 2
= T 0 (t) +
T (t).
L
It’s easy to see that all solutions to (13) are given by:
T (t) = Ce−
In summary, for any n ∈ N
S
n2 π 2 c2
L2
t
.
{0}, we can find a non-zero solution to (10):
un (x, t) = e−
n2 π 2 c2
L2
t
cos
nπ x .
L
nπ n
nπ o∞
Notice that un (x, 0) = cos
x . So if f (x) is an “infinite” linear combination of un (x, 0) = cos
x
, that
L
L
n=0
∞
∞
∞
nπ nπ X
X
X
n2 π 2 c2
is, f (x) =
an cos
x , then informally we have u(x, t) =
an un (x, t) =
an e− L2 t cos
x is a solution
L
L
n=0
n=0
n=0
to (9).
Example 3. Consider the problem:













ut = uxx ,
0 < x < 1,
t > 0,
BC:
ux (0, t) = 0,
ux (1, t) = 0,
t > 0,
IC:
u(x, 0) = 2 cos(πx) + 4 cos(3πx),
0 ≤ x ≤ 1.
By the previous discussion, then a solution is given by:
2
2
u(x, t) = 2e−π t cos(πx) + 4e−9π t cos(3πx).
LECTURE 8: METHOD OF CHARACTERISTIC AND SEPARATION OF VARIABLES
Remark 2. Now, for solving (9), our task is to find the “infinite” linear combination of
∞
nπ X
that is, f (x) =
x . There are three natural questions about this:
an cos
L
n=1
9
n
nπ o∞
un (x, 0) = cos
x
,
L
n=1
I. For what conditions do we need for a function f (x) on [0, L] such that f (x) can be an “infinite” linear combination
n
nπ o∞
of un (x, 0) = cos
x
?
L
n=1
∞
nπ n
nπ o∞
X
II. If f (x) is an “infinite” linear combination of un (x, 0) = cos
x
x ,
, that is, f (x) =
an cos
L
L
n=1
n=1
how to compute the coefficients an ’s?
∞
nπ n
nπ o∞
X
x
, that is, f (x) =
an cos
x , is
III. If f (x) is an “infinite” linear combination of un (x, 0) = cos
L
L
n=1
n=1
∞
∞
nπ X
X
n 2 π 2 c2
the function u(x, t) =
an un (x, t) =
an e− L2 t cos
x really a solution to (9)?
L
n=1
n=1
Solving the heat equation without source on a circular ring (with a periodic boundary
condition)
Consider a thin circular wire in which there is no radial temperature dependence, the heat distribution in the ring is
determined by the following initial







(14)
BC:






IC:
value problem with periodic boundary conditions:
ut = c2 uxx ,
−L < x < L,
t > 0,
u(−L, t) = u(L, t), ux (−L, t) = ux (L, t),
u(x, 0) = f (x),
t > 0,
−L ≤ x ≤ L.
Let’s first look at the heat equation and the homogeneous boundary condition:



ut = c2 uxx , −L < x < L, t > 0,
(15)

 BC: u(−L, t) = u(L, t), u (−L, t) = u (L, t),
t > 0.
x
x
A separated solution to (15) is of the form u(x, t) = X(x)T (t), then
ut (x, t)
=
X(x)T 0 (t)
ux (x, t)
=
X 0 (x)T (t)
uxx (x, t)
=
X 00 (x)T (t)
0
=
ut − c2 uxx
=
X(x)T 0 (t) − c2 X 00 (x)T (t).
10
MINGFENG ZHAO
So we get
−
(16)
T 0 (t)
X 00 (x)
=
−
:= λ.
c2 T (t)
X(x)
X 00 (x)
, then λt (x, t) = 0. By (16), then λx (x, t) = 0. So λ(x, t) is a constant. Since u(−L, t) = u(L, t)
X(x)
and ux (−L, t) = ux (L, t), then X(x) satisfies


 X 00 + λX = 0,
(17)

 X(−L) = X(L), X 0 (−L) = X(L).
Let λ(x, t) = −
We are expected to find non-zero solutions to (17). Let X = erx be a solution to X 00 + λX = 0, then X 0 = rerx ,
X 00 = r2 erx , and
0
= X 00 + λX = r2 erx + λerx
= erx (r2 + λ).
a. If λ = 0, that is, r2 = 0. So r = 0, which implies that the general solution to X 00 + λX = X 00 = 0 is
X(x) = C1 + C2 x. Since X(−L) = X(L), that is, C1 − C2 L = C1 + LC2 , then C2 = 0. Since X 0 (x) = C2 , then
X 0 (−L) = X 0 (L), which implies that all solutions to (17) are
X(x) = C.
b. If λ > 0, that is, λ = µ2 for some µ > 0. So r2 + µ2 = 0, that is, r = ±µi, which implies that the general
solution to X 00 + λX = X 00 + µ2 X = 0 is X(x) = C1 cos(µx) + C2 sin(µx). Since X(−L) = X(L), that is,
C1 cos(µL) − C2 sin(Lµ) = C1 cos(µL) + C2 sin(Lµ), then
C2 sin(µL) = 0.
Since X 0 (x) = −C1 µ sin(µx) + C2 µ cos(µx) and X 0 (−L) = X 0 (L), that is, C1 µ sin(µL) + C2 µ cos(µL) =
−C1 µ sin(µL) + C2 µ cos(µL), since µ > 0.
C1 µ sin(µL) = 0.
In order to make X to be a non-zero solution, then we need sin(Lµ) = 0, that is, Lµ = nπ for some n ∈ N,
nπ 2
nπ 2
that is, λ = µ2 =
. For λ =
, then all solution to (17) are
L
L
nπ nπ Xn (x) = C1 cos
x + C2 sin
x .
L
L
LECTURE 8: METHOD OF CHARACTERISTIC AND SEPARATION OF VARIABLES
11
c. If λ < 0, that is, λ = −µ2 for some µ > 0. So r2 − µ2 = 0, that is, r = ±µ, which implies that the
general solution to X 00 + λX = X 00 − µ2 X = 0 is X(x) = C1 eµx + C2 e−µx . Since X(−L) = X(L), that is,
C1 e−µL + C2 eµL = C1 eµL + C2 e−µL , then
C1 = C2 .
Since X 0 (x) = C1 µeµx − C2 µe−µx , and X 0 (−L) = X 0 (L), that is, C1 µe−µL − C2 µeµL = C1 µeµL − C2 µe−µL ,
since µ > 0, then
C1 = −C2
Then C1 = C2 = 0.
nπ 2
nπ 2
S
In summary, (17) has non-zero solution if and only if λ =
for some n ∈ N {0}. When λ =
in (17),
L
L
nπ nπ
all non-zero solutions to (17) are X(x) = C1 cos
x + C2 sin
x .
L
L
nπ 2
When λ =
, by (15), then
L
0
(18)
= T 0 (t) + c2 λT (t)
nπc 2
T (t).
= T 0 (t) +
L
It’s easy to see that all solutions to (18) are given by:
T (t) = Ce−
In summary, for any n ∈ N
S
n2 π 2 c2
L2
t
.
{0}, we can find two non-zero solution to (15):
un (x, t) = e−
n 2 π 2 c2
L2
t
cos
nπ x ,
L
vn (x, t) = e−
n2 π 2 c2
L2
t
sin
nπ x .
L
nπ nπ x and vn (x, 0) = sin
x . So if f (x) is an “infinite” linear combination of
Notice that un (x, 0) = cos
L
L
∞ h
n
nπ nπ o∞
nπ nπ i
X
, that is, f (x) =
un (x, 0) = cos
x , vn (x, 0) = sin
x
an cos
x + bn sin
x , then informally
L
L
L
L
n=0
n=0
∞
∞
h
nπ nπ i
X
X
n2 π 2 c2
we have u(x, t) =
[an un (x, t) + bn vn (x, t)] =
e− L2 t an cos
x + bn sin
x is a solution to (14).
L
L
n=0
n=0
Example 4. Consider the problem:













ut = uxx ,
−1 < x < 1,
t > 0,
BC:
u(−1, t) = u(1, t), ux (−1, t) = ux (1, t),
IC:
u(x, 0) = 2 sin(πx) + 4 cos(3πx),
t > 0,
0 ≤ x ≤ 1.
12
MINGFENG ZHAO
By the previous discussion, then a solution is given by:
2
2
u(x, t) = 2e−π t sin(πx) + 4e−9π t cos(3πx).
Remark 3. Now, for solving (14), our task is to find the “infinite” linear combination of
∞ h
n
nπ nπ o∞
nπ nπ i
X
un (x, 0) = cos
x , vn (x, 0) = sin
x
, that is, f (x) =
an cos
x + bn sin
x . There
L
L
L
L
n=1
n=1
are three natural questions about this:
I. For what conditions do we need for a function f (x) on [0, L] such that f (x) can be an “infinite” linear combination
n
nπ nπ o∞
of un (x, 0) = cos
x , vn (x, 0) = sin
x
?
L
L
n=1
n
nπ nπ o∞
II. If f (x) is an “infinite” linear combination of un (x, 0) = cos
x , vn (x, 0) = sin
x
, that is, f (x) =
L
L
n=1
∞ h
nπ i
nπ X
x + bn sin
, how to compute the coefficients an ’s or bn ’s?
an cos
L
L
n=1
nπ o∞
n
nπ x , vn (x, 0) = sin
x
, that
III. If f (x) is an “infinite” linear combination of un (x, 0) = cos
L
L
n=1
∞
∞
i
h
X
X
nπ
nπ
x + bn sin
x , is the function u(x, t) =
[an un (x, t) + bn v(x, t)] =
is, f (x) =
an cos
L
L
n=1
n=1
∞
h
nπ nπ i
X
n2 π 2 c2
e− L2 t an cos
x + bn sin
x really a solution to (14)?
L
L
n=1
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca