Initial Boundary Value Problems for
Scalar and Vector Burgers Equations
By K. T. Joseph and P. L. Sachdev
In this article we study Burgers equation and vector Burgers equation with
initial and boundary conditions. First we consider the Burgers equation in
the quarter plane x > 0, t > 0 with Riemann type of initial and boundary
conditions and use the Hopf–Cole transformation to linearize the problems
and explicitly solve them. We study two limits, the small viscosity limit and the
large time behavior of solutions. Next, we study the vector Burgers equation
and solve the initial value problem for it when the initial data are gradient
of a scalar function. We investigate the asymptotic behavior of this solution
as time tends to infinity and generalize a result of Hopf to the vector case.
Then we construct the exact N-wave solution as an asymptote of solution of
an initial value problem extending the previous work of Sachdev et al. (1994).
We also study the limit as viscosity parameter goes to 0. Finally, we get an
explicit solution for a boundary value problem in a cylinder.
1. Introduction
The nonlinear parabolic partial differential equation
1
ν
ut + u2 x = uxx
2
2
(1)
Address for correspondence: Professor P. L. Sachdev, Department of Mathematics, Indian Institute of
Science, Bangalore 560012, India.
STUDIES IN APPLIED MATHEMATICS 106:481–505
481
© 2001 by the Massachusetts Institute of Technology
Published by Blackwell Publishers, 350 Main Street, Malden, MA 02148, USA, and 108 Cowley Road,
Oxford, OX4 1JF, UK.
482
K. T. Joseph and P. L. Sachdev
was first introduced by J. M. Burgers [1] as the simplest model for fluid flow.
This equation describes the interplay between nonlinearity and diffusion, see
Sachdev [2] and the references therein for physical interpretation and important solutions. A remarkable feature of this equation is that its solution with
initial conditions of the form
ux 0 = u0 x
(2)
can be explicitly written down. In fact Hopf [3] and Cole [4] independently
showed that the equation (1) can be linearized through the transformation
u = −ν
vx
v
(3)
More precisely, assume that the initial data u0 x are integrable in every
finite interval and that
x
u0 ydy = ox2 0
then Hopf [3] showed that if vx t satisfies the linear heat equation
with initial condition
vt =
ν
v
2 xx
1
vx 0 = exp −
ν
(4)
y
0
u0 zdz
(5)
then ux t defined by (3) solves (1) and (2) and conversely, if ux t is a
solution of (1) and (2), then vx t defined by (3) is a solution of (4) and (5)
up to a time-dependent multiplicative factor that is irrelevent in (3). Solving
for v from (4) and (5) and substituting it into (3), he obtained explicit formula
for the solution of (1) and (2); namely,
y
x−y
2
u zdz
t exp − x−y
− 0 0ν
dy
R
2νt
uν x t =
(6)
y
x−y2
0 u0 zdz
exp
−
−
dy
R
2νt
ν
and studied the asymptotic behavior of uν x t as t → ∞. He also constructed explicit weak entropy solution of the inviscid Burgers equation.
Lighthill [5] discovered the N-wave solution of the Burgers equation (1);
namely,
1
∞
U x t =
1
t2 1+
x/t 2
1
t2
c0
2 exp x
2tν
(7)
Boundary Value Problems for Burgers Equations
483
Sachdev et al. [6] showed that this solution can be obtained as time asymptotic of a pure initial value problem and found explicitly the constant c0 in
terms of one lobe area of the initial data.
In a more recent paper, [7] proposed the vector Burgers equation
Ut + U∇U =
ν
U
2
(8)
They observed that if we seek special solutions of the form
U = ∇x φ
(9)
where φx t is a scalar function on Rn × 0 ∞, equation (8) leads to
1
ν
∇ φt + ∇φ2 − φ = 0
(10)
2
2
They used this observation to find special solutions of Burgers equation in
cylindrical coordinates with axisymmetry.
Although there are many results for pure initial value problem, boundary value problem has been studied less. Explicit solutions of the Burgers
equation (1) in the quarter plane with integrable initial data and piecewise
constant boundary data were constructed by [8] using Hopf–Cole transformation. He obtained a formula for its weak limit as viscosity parameter goes
to 0. Using maximum principle, this formula for weak limit was extended to
general boundary data. When the initial and boundary data are two different
constants, the problem was considered in some detail by [9] for ν small. Five
distinct cases arose depending on the relative magnitudes of the constants
appearing in the initial and boundary conditions. The solutions required the
introduction of corner layers, shock layers, boundary layers, and transition
layers. In the present article, we write an explicit form of the solution and
recover various cases in the limit ν → 0. We also consider large time behavior of the solutions. Again, different cases arise depending on the relative
magnitudes of the constants that appear in the initial and boundary conditions. It would be desirable to have solution for nonconstant initial boundary problem, but as [9] point out, that leads to an integral equation that
seems difficult to solve. Using a generalized Hopf–Cole transformation, [10]
and [11], in a series of papers, studied (1) with more general boundary conditions that include the flux condition at the origin. The asymptotic profile
at infinity of the solution of (1) with flux condition was obtained by [12]. A
table of solutions of the Burgers equation is contained in [13].
The aim of the present article is to study Burgers equation (1) and the vector Burgers equation (8) with initial and boundary conditions and investigate
the asymptotic behavior as t tends to ∞ and as ν tends to 0. In Section 2, we
treat the Burgers equation (1) in the quarter plane with Reimann type initial boundary data, solve it exactly, and study the asymptotic limits as t tends
484
K. T. Joseph and P. L. Sachdev
to ∞ or as the viscosity coefficient ν tends to 0. In Section 3 we study the vector Burgers equation, solve exactly the initial value problem for it when the
initial data can be written as gradients of a scalar function and carry out the
study of the limits as t tends to ∞ or ν tends to 0. We also study a boundary
value problem for the vector Burgers equation in the same section.
2. Initial boundary value problem for Burgers equation and
its asymptotics
In this section, we consider the Burgers equation
1
ν
ut + u2 x = uxx
2
2
(11)
in x > 0, t > 0, with initial condition
ux 0 = uI
(12)
u0 t = uB (13)
and the boundary condition
where uI and uB are constants. Using standard Hopf–Cole transformation,
we reduce this problem to a linear one and solve it explicitly. Before the
statement of the result we introduce some notations:
∞
uI y
x − y2
ν
A− x t =
−
dy
exp −
2tν
ν
0
∞
x + y2
uI y
ν
exp −
−
dy
A+ x t =
2tν
ν
0
∞
uB y
x + y2
ν
+
dy
(14)
exp −
B x t =
2tν
ν
0
With these notations we prove the following result.
A solution of the initial boundary value problem (11)–(13) is given by
uν xt = uB
+
uI −uB Aν− xt−Aν+ xt
2
x
Aν− xt+Aν+ xt+2uB t exp− 2tν
+2uB /νuB t −xBν xt
(15a)
when uI + uB = 0 and
uν x t = uB +
when uI + uB = 0.
uI − uB Aν− x t − Aν+ x t
Aν− x t +
uI −uB ν
A x t
uI +uB +
+
2uB
Bν x t
uI +uB (15b)
Boundary Value Problems for Burgers Equations
485
To prove this result, we use Hopf–Cole transformation. First, we set
∞
wx t = −
uy t − uI dy
(16)
x
From (11)–(13), we get
wx 2
ν
+ uI wx = wxx 2
2
wx 0 = 0 wx 0 t = uB − uI wt +
(17)
Now setting
w
v = exp −
ν
(18)
we get from (17)
ν
v 2 xx
νvx 0 t + uB − uI v0 t = 0
vt + uI vx =
vx 0 = 1
(19)
Making the final transformation
uI x uI 2 t
p = exp −
+
v
ν
2ν
(20)
we get from (19) the following problem for p:
ν
p
2 xx
u x
px 0 = exp − I
ν
νpx 0 t + uB p0 t = 0
pt =
This problem has the explicit solution (see [8])
∞ 1
x + y2
x − y2
px t =
+ exp −
exp −
2tν
2tν
2πtν1/2 0
u y
2uB /ν
· exp − I dy +
ν
2πtν1/2
∞ ∞
uI y
uB y − z x + z2
−
−
dzdy
·
exp −
ν
2tν
ν
0
y
(21)
486
K. T. Joseph and P. L. Sachdev
Now from (16), (18), and (20), we have
uν x t = wx + uI = −νlog vx + uI
= −νlog px + uI /ν + uI = −νlog px In other words
uν = −ν
px
p
(22)
To write the formula in a convenient form we note that
∞
u z x + z2
∂x
exp B −
dz
ν
2tν
y
∞
u z x + z2 dz
exp B ∂z exp −
=
ν
2tν
y
∞
uB z x + z2
x + y2
uB
uB y
exp
−
−
−
dz = − exp
ν
2tν
ν
ν
2tν
y
(23)
Here, the first equality is obtained by just writing the exponential of a sum
as a product and using the fact that
x + z2
x + z2
∂x exp −
= ∂z exp −
2νt
2νt
and the second equality is obtained by the use of integration by parts.
Similarly,
∞
x + y2
−uI y
∂x
−
dy
exp
ν
2tν
0
x2 u ∞
−uI y
x + y2
exp
+ I
−
dy (24)
= − exp −
2tν
ν
ν
2tν
0
and
−uI y
x − y2
−
dy
ν
2tν
0
x2 u ∞
x − y2
−uI y
− I
−
dy exp
= exp −
2tν
ν
ν
2tν
0
∂x
∞
exp
Now, it follows from (21), (23), (24), and (25) that
1
px x t =
uI Aν− x t − uI Aν+ x t
1
−ν2πνt 2
+ 2uB Aν+ x t
u
+ 2uB B C ν x t
ν
(25)
Boundary Value Problems for Burgers Equations
487
and
1
px t =
1
2πνt 2
Aν− x t + Aν+ x t +
2uB ν
C x t ν
where Aν− , Aν+ and Bν are given by (14), and
C ν x t =
∞
0
∞
y
u y − z x + y2
uy
exp − B
−
− I dzdy
ν
2tν
ν
Substituting these formulas for p and px in (22), we get
uν x t =
=
uI Aν− x t − uI Aν+ x t + 2uB Aν+ x t + uB 2uνB C ν x t
Aν− x t + Aν+ x t + 2uνB C ν x t
uI − uB Aν− x t − Aν+ x t + uB Aν− x t + Aν+ x t +
Aν− x t + Aν+ x t +
uI − uB Aν− x t − Aν+ x t
=
+ uB Aν− x t + Aν+ x t + 2uνB C ν x t
2uB ν
C x t
ν
2uB ν
C x t
ν
(26)
Now we express C ν in terms of Aν+ and Bν . There are two cases to consider.
The first one is when uI + uB = 0 . In this case,
x + z2
uB y − z uI y
−
−
dzdy
C x t =
exp −
ν
ν
2tν
0
y
∞ ∞
u z x + z2
dzdy
1
exp B −
=
ν
2tν
0
y
∞
x + y2
uB y
−
dy
y exp
=
ν
2tν
0
∞
x + y2
uB y
−
dy
∂y exp
= −tν
ν
2tν
0
∞
x + y2
uB y
−
dy
exp
+ uB t − x
ν
2tν
0
∞
−x2 x + y2
uB y
+ uB t − x
−
dy
exp
= tν exp
2tν
ν
2tν
0
x2 + uB t − xBν x t
(27)
= tν exp −
2tν
ν
∞
∞
488
K. T. Joseph and P. L. Sachdev
When uI + uB = 0,
∞
C ν x t =
x + z2
u y − z uI y
−
−
dzdy
exp − B
ν
ν
2tν
0
y
∞
∞
uI + uB y
uB z x + z2
−
dzdy
exp −
exp
=
ν
ν
2tν
0
y
∞
uI + uB y
ν
∂y exp −
=−
uI + uB 0
ν
∞
uB z x + z2
−
dzdy
exp
·
ν
2tν
y
∞
x + y2
uB y
ν
−
dy
exp
=
uI + uB 0
ν
2tν
∞
x + y2
uI y
−
dy
exp −
−
ν
2tν
0
ν
=
Bν x t − Aν+ x t
(28)
uI + uB ∞
Using (27) for the case uI + uB = 0 and (28) for the case uI + uB = 0 in (26),
the formula (25) for uν follows.
2.1. Study of the limit as ν → 0
Here, we compute the pointwise limit of uν x t as the viscosity parameter
ν → 0. Let Hx be the usual Heaviside function and s = uI + uB /2,
then for x ≥ 0, t ≥ 0, we have the following formula for the pointwise limit
function ux t = limν→0 uν x t.
Case 1. uB > uI and uI + uB > 0
ux t = uI Hx − st + uB Hst − x
Case 2. uI > 0, uB > 0 and uI > uB
ux t = uB HuB t − x + x/tHuI t − xHx − uB t + uI Hx − uI t
Case 3. uI < 0 and uI + uB ≤ 0
ux t = uI
Case 4. uI = 0 and uB < 0 or uI > 0 and uB ≤ 0 and uI + uB = 0 or
uI > 0 and uI + uB = 0,
ux t = x/tHuI t − x + uI Hx − uI t
Boundary Value Problems for Burgers Equations
489
To study the limν→0 uν x t, we write (15) for uν in terms of the standard
“erfc” and use its asymptotic form. For this, we denote
∞
exp−y 2 dy
(29)
erfcy =
y
∞
ux
u2I t
y − x + uI t2
− I
dy
exp −
2ν
ν
2tν
0
∞
2
ux
ut
exp−y 2 dy
= tν1/2 exp I − I
−x+uI t
2ν
ν
2tν1/2
2
uI x
−x + uI t
uI t
1/2
−
erfc
exp
= tν
2ν
ν
2νt1/2
Aν− x t = exp
Similarly,
ux
x + uI t
u2I t
+ I erfc
2ν
ν
2νt1/2
(31)
u x
x − uB t
u2B t
− B erfc
2ν
ν
2νt1/2
(32)
Aν+ x t = tν1/2 exp
and
Bν x t = tν1/2 exp
(30)
To study the asymptotic behavior of uν as ν → ∞ we study the behavior of
Aν− , Aν+ and Bν using the asymptotic expansions of the erfc; namely,
1
1
1
− 3 +o 3
(33)
exp−y 2 y → ∞
erfcy =
2y
4y
y
and
erfc−y = π
1/2
1
1
1
− 3 +o 3
−
exp−y 2 y → ∞
2y
4y
y
From (30)–(34), we have the following as ν → 0. If −x + uI t > 0
x2
tν
Aν− x t ≈
exp −
−x + uI t
2νt
and for −x + uI t < 0
Aν− xt ≈ 2πtν1/2 exp
tν
x2
u2I t uI x
−
−
exp −
2ν
ν
−x+uI t
2νt
For x + uI t > 0
Aν+ x t
tν
x2
≈
exp −
x + uI t
2νt
(34)
(35a)
(35b)
(35c)
490
K. T. Joseph and P. L. Sachdev
and for x + uI t < 0
Aν+ xt ≈ 2πtν1/2 exp
For x − uB t > 0
tν
x2
u2I t uI x
+
−
exp −
2ν
ν
x+uI t
2νt
x2
tν
exp −
B x t ≈
x − uB t
2νt
ν
and for x − uB t < 0
Bν xt ≈ 2πtν1/2 exp
(35d)
(35e)
tν
x2
u2B t uB x
−
−
exp −
2ν
ν
x−uB t
2νt
(35f)
There are several cases to consider. First, we consider
Case 1. uB > uI , uI + uB > 0.
First we take up the subcase uB > 0, uI < 0, and uI + uB > 0. The line
x = uI + uB /2t divides the quarter plane x > 0, t > 0 into two regions.
In the region x < uI + uB /2t, x − uB t ≤ uI − uB /2t < 0. Using the
asymptotic formulae (35b)–(35e) in (15b) we get, as ν → 0,
uν x t ≈ uB +
uI − uB π 1/2 − O1 exp uνI x π 1/2 + O1 exp uνI x +
2uB
π 1/2
uI +uB −uI exp uB2ν
−x +
uI +uB t
2
Because uI < 0 and uB − uI > 0, we get, for x < uI + uB /2t,
lim uν x t = uB ν→0
By a similar argument, we have, for x > uI + uB /2t and x < uB t,
ν
u x t ≈ uB +
uI − uB π 1/2 − O1 exp uνI x π 1/2 + O1 exp uνI x +
2uB
π 1/2
uI +uB −uI exp uB2ν
−x +
uI +uB t
2
so that
lim uν x t = uB + uI − uB = uI ν→0
If x > uI + uB t/2 and x > uB t, as before, from (15b) and (35b)–(35e),
we get
uν x t ≈ uB +
uI − uB π 1/2 − O1 exp uνI x π 1/2 + O1 exp uνI x + Oν 1/2 so that
lim uν x t = uB + uI − uB = uI ν→0
The other subcases uB > 0, uI > 0, uI < uB , and uB > 0, uI = 0 are similar.
Boundary Value Problems for Burgers Equations
491
Case 2. uI > 0 uB > 0 uB < uI .
In the region x − uB t < 0, we have −x + uI t > 0, and so from (15b) and
(35a), (35c), and (35f) we get
uν x t ≈ uB +
uI − uB 2tν1/2
2−x+uI t
+
2tν1/2
2−x+uI t
uI −uB 2tν1/2
uI +uB 2x+uI t
+
2tν1/2
2x+uI t
−
2uB π 1/2
uI +uB
B t
exp x−u
2tν
2
Therefore, for x < uB t, we have
lim uν x t = uB ν→0
On the other hand, in the region x > uI t, we have x > uB t so that
uν x t ≈ uB
2tν1/2
x−uI t2
uI − uB π 1/2 − 2x+u
exp
−
2tν
I t
+
2
2tν1/2 uI −uB x−uI t2
2tν1/2 2uB
1/2
I t
π − 2x+u t u +u exp − 2tν
+ x−u t u +u exp − x−u
2tν
I
I
B
B
I
B
and we get
lim uν x t = uI ν→0
In the region uB t < x < uI t, we have
uν x t ≈ uB
2tν1/2
x2
x2
exp− 2tν
− 2x+u
exp− 2tν
t
I
+ 2tν1/2
x2 x2 uI −uB 2tν1/2
2tν1/2
2uB
x2
exp− 2tν + u +u 2x+u t exp − 2tν + 2x−u
exp − 2tν
2−x+u t
t u +u
uI − uB I
2tν1/2
2−x+uI t
I
B
I
B
I
B
and we find that
ν
lim u x t = uB +
ν→0
uI − uB 1
−x+uI t
+
uI −uB
uI +uB
1
−x+uI t
·
1
x+uI t
On simplification, we have, for uB t < x < uI t,
lim uν x t =
ν→0
This completes case 2.
x
t
+
1
x+uI t
2uB
1
· x−u
uI +uB
Bt
−
492
K. T. Joseph and P. L. Sachdev
Case 3. uI < 0, uI + uB ≤ 0.
First we consider the subcase uI < 0, uB < 0, uI − uB < 0. We get from
(15b) and (35b)–(35e),
1/2
2uI x u
−
u
π
−
O1
exp
I
B
ν
uν x t = uB +
2
uI −uB
2uI x
2uB
1/2
I t
π + O1 u +u exp ν + O1 u +u exp − x−u
2νt
I
B
I
B
so that, for x > 0,
lim uν x t = uB + uI − uB = uI ν→0
For the other subcases the analysis is similar and so we omit the details.
Case 4. uI = 0 and uB < 0 or uI > 0 and uB ≤ 0 and uI + uB non-zero
or uI > 0 and uI + uB = 0.
For the first subcase, for x > 0, we get
1/2
x2
exp−
π 1/2 + 2tν
2tν
2x
uν x t ≈ uB − uB
x2 x2 2tν1/2
2tν1/2
1/2
π − 2x exp − 2tν + 2x−u t exp − 2tν
B
so that
lim uν x t = uB − uB = 0
ν→0
For the subcase uI > 0, uB < 0, uI + uB not zero, following the previous
analysis, we have in the region x > uI t,
uν x t
2tν1/2
x−uI t2
uI − uB π 1/2 − 2x+u
exp
−
2tν
I t
= uB +
1/2
2
x−uI t
2tν1/2
x−uI t2
2uB
B 2tν
π 1/2 + uuI −u
exp
−
exp
−
+
+u 2x+u t
2tν
u +u 2x+u t
2tν
I
B
I
I
B
−
1
x+uI t
B
So we get for x > uI t,
lim uν x t = uI ν→0
In the region x < uI t, we have
lim uν x t = uB +
ν→0
uI − uB 1
−x+uI t
+
1
−x+uI t
uI −uB
1
uI +uB x+uI t
+
2uB
1
uI +uB x−uB t
=
x
t
The other subcases can be treated similarly. This completes the proof of the
result.
Boundary Value Problems for Burgers Equations
493
2.2. Large time behavior
Next, we study the limit of uν x t as t → ∞ where ux t is the solution of
(1)–(3) given by 15j j = 1 2 with a fixed coefficient of viscosity ν > 0. We
prove the following result.
1. If uI < 0, uB < 0 and uI > uB , then
u x
1
uB − uI
I
+ c c = log
lim ux t = −uI coth
t→∞
ν
2
uI + u B
(36)
2. If uI < 0, uI < uB and uB + uI < 0, then
lim ux t = −uI tanh
t→∞
u x
1
u − uB
I
+ c c = log I
ν
2
uI + u B
(37)
3. If uI = 0 and uB < 0, we have
lim ux t =
t→∞
uB
1 − uB xν
(38)
4. For all other cases,
lim ux t = uB t→∞
(39)
To prove the result first, we get the asymptotic behavior of Aν− , Aν+ , and
B as t tends to infinity. Using the asymptotic expansions (33) and (34) in
(30)–(32) we have, as t → ∞,
ν
x2
tν
exp −
uI > 0
≈
−x + uI t
2νt
2
uI x
uI t
ν
1/2
−
exp
A− x t ≈ 2πtν
2ν
ν
2
x
tν
uI < 0
exp −
−
−x + uI t
2νt
x2
tν
ν
exp −
uI > 0
A+ x t ≈
x + uI t
2νt
2
uI x
uI t
ν
1/2
+
exp
A+ x t ≈ 2πtν
2ν
ν
2
x
tν
exp −
uI < 0
−
x + uI t
2νt
tν
x2
ν
exp −
uB < 0
B x t ≈
x − uB t
2νt
Aν− x t
(40a)
(40b)
(40c)
(40d)
(40e)
494
K. T. Joseph and P. L. Sachdev
and
u2B t
uB x
B x t ≈ 2πtν
exp
−
2ν
ν
2
x
tν
exp −
uB > 0
−
x − uB t
2νt
ν
1/2
(40f)
Aν− x t ≈
x
π 1/2
−
uI = 0
2
2tν1/2
(40g)
Aν− x t ≈
x
π 1/2
+
uI = 0
2
2tν1/2
(40h)
Bν x t ≈
x
π 1/2
+
uB = 0
2
2tν1/2
(40i)
First, we consider the case uI < 0, uB < 0. Using (40b), (40d), and (40e) in
(15b), we get
ux t ≈ uB
u2 t
u2 t
uI − uB π 1/2 exp 2νI − uνI x − π 1/2 exp 2νI + uνI x
+
u2 t
u2I t
uI x
2uB 2νt1/2
x2
B
+
π 1/2 exp 2νI − uνI x + π 1/2 uuI −u
exp
+
exp
−
+u
2ν
ν
u +u x−u t
2νt
I
B
I
B
B
So, in this case, we get
uI − uB exp− uνI x − exp uνI x lim ux t = uB +
B
t→∞
exp uνI x exp− uνI x + uuI −u
+u
I
B
On simplification we get
lim ux t = uI
t→∞
exp− uνI x −
exp− uνI x +
uI −uB
uI +uB
uI −uB
uI +uB
exp uνI x exp uνI x (41)
Now, if uI > uB , by rewriting (41) we get (36), the first part of the result. If
uI < uB , we get (37) of the part two.
Now consider uI < 0 and uB = 0. Using (40b), (40d), and (40i) in (15b),
we get
u x
exp− uνI x − exp uνI x I
lim ux t = uI
uI x
uI x = −uI tanh
t→∞
exp− ν + exp ν ν
Now, consider the case uI < 0, uB > 0 and uI + uB = 0. As before, we have
from (15b) and (40b), (40d), and (40f) the limit
lim ux t = uB
t→∞
+
uI − uB exp− uνI x − exp uνI x exp− uνI x +
uI −uB
uI +uB
exp uνI x +
2uB
uI +uB
exp− uBν x exp
u2B −u2I t
2ν
(42)
Boundary Value Problems for Burgers Equations
495
Therefore,
lim ux t = uI
exp− uνI x −
t→∞
exp− uνI x +
uI −uB
exp uνI x uI +uB
uI −uB
exp uνI x uI +uB
provided uB + uI < 0, because in this case, u2B − u2I < 0. On simplification,
we get (37). When uI + uB > 0, we have u2B − u2I > 0, and so (42) gives
lim uν x t = uB t→∞
Now, take up the case uI < 0, uB > 0 and uI +uB = 0. From (15a) and (40b),
(40d), and (40f) (after dividing numerator and dinominator of the resulting
u2 t
expression by π 1/2 2πtν1/2 exp νI we get
uν xt ≈ uB
+
uI −uB exp− uνI x −exp uνI x 2uB t
x
exp− uνI x +exp uνI x + π 1/2 2πtν
1/2 exp− 2νt −
2
u2I t
+ 2uB uνB t−x exp− uBν x 2ν
so that
lim uν x t = uB t→∞
When uI = 0 and uB < 0, using (40f), (40g), and (40h) in (15b), we get
uν x t ≈ uB −
x
uB 22πtν1/2 2νt
1/2
x
22πtν1/2 2νt
1/2 +
2π 1/2 tν
x−uB t
(43)
From (43), we get for uI = 0 and uB < 0,
lim uν x t = uB 1 −
t→∞
x
x − uν
=
B
uB
1 − uBν x
Now, we consider the case uI > 0. First let uB < 0 and uI + uB = 0. We have
from (15b) and (40a), (40c), and (40e) after cancelling out common terms,
ν
ν
uI − uB u −x/t
− u +x/t
I
I
lim uν x t = uB +
uI −uB
t→∞
ν
ν
uI − uB u −x/t + u +u u +x/t + u 2u+uB
I
I
B
I
I
B
tν
x−uB t
= uB The case uI + uB = 0 is similar; here we use (15a) instead of (15b).
Now, consider the case uI > 0, uB > 0. As before, using the asymptotics
496
K. T. Joseph and P. L. Sachdev
(40a), (40c), and (40f) in (15b) and dividing both numerator and dinominator
by π 1/2 tν exp−x2 /2νt in the resulting expression, we get
lim uν x t
t→∞
= uB +
uI − uB uI − uB 1
uI t−x
+
1
uI t−x
uI −uB
1
uI +uB uI t+x
−
1
uI t+x
2uB
2π1/2
uI +uB
+
2
= uB B t
exp x−u
2νt
The case uI > 0 and uB = 0 is similar. This completes the proof of the result.
Consider the stationary problem for 0 < x < ∞
νqxx = q2 x q0 = uB q∞ = uI It can be easily checked that this problem has solution if uI < 0 and uB < uI
or uI < 0 and uI + uB < 0 or uI = 0 and uB < 0. Solving the problem
explicitly, we get the functions on the right-hand side of (36)–(38). Here,
we have shown that they are time asymptotes of boundary value problems
(11)–(13).
3. Higher dimensional extensions
Consider vector equivalent of Burgers equation studied by [7]; namely,
Ut + U∇U =
ν
U
2
(44)
[7] observed that if we seek a solution U of (44), which is gradient of some
scalar function φ,
U = ∇x φ
(45)
then equation (44) becomes
ν
∇φ2
∇x φt +
− φ = 0
2
2
This leads to
φt +
∇φ2
ν
− φ = f t
2
2
(46)
where f t is an arbitrary function of t. Because we are interested in the
space derivative ∇x φ, and this is independent of f t, we let f t = 0. If we
are given initial data for U that are gradients of some scalar function φ0 of
the form,
Ux 0 = ∇x φ0 x
(47)
Boundary Value Problems for Burgers Equations
497
it is enough to find a solution φ of
φt +
∇φ2
ν
− φ = 0
2
2
(48)
with initial condition
φx 0 = φ0 x
(49)
We may then use (45) to get the solution U of (44) and (47). To solve (48)
and (49) we use the Hopf–Cole transformation
φ
θx t = exp −
ν
(50)
Using (50) in (48) and (49), we get the linear problem
ν
θ
2
φ x
θx 0 = exp − 0
ν
θt =
(51)
(52)
Solving (51) and (52), we have
1
θ x t =
n
2πνt 2
ν
x − y2
φ0 y
exp −
−
dy
2νt
ν
Rn
(53)
From (45), (50), and (52), we see that the solution to (44) and (47) is given
by
U ν = −ν
∇θν
θν
(54)
From (53) and (54), it follows that
φ0 y
x−y2
x−y
exp
−
−
dy
R
t
2νt
ν
U ν x t =
2
exp − x−y
− φ0νy dy
Rn
2νt
n
First, we study the asymptotic behavior of this solution as t → ∞.
(55)
498
K. T. Joseph and P. L. Sachdev
3.1. Asymptotic behavior as t → ∞, a general result
Here, we assume initial data of the form (47)
U ν x 0 = ∇x φ0 x
where φ0 satisfies the estimate
n
φ0 x =
1
φi0 xi + o1 x → ∞
(56)
where φi0 x i = 1 2 n are differentiable functions from R1 to R1 .
Assume further that both the limits
+
i
lim φi0 xi = φ−
i lim φ0 xi = φi
xi →−∞
xi →∞
(57)
exist. With the notations
ξ=
x
ki =
1
νt 2
−
φ+
i − φi ν
(58)
and
zi2
dzi
exp −
2
−∞
2
∞
z
ki
exp − i dzi + exp
2
2
ξi
k
gi ξi = exp − i
2
ξi
(59)
for i = 1 2 n, we prove the following result.
Let U ν x t be the solution of (44) and (47) given by the formula (53)
and (54) with φ0 satisfying the conditions (56) and (57), then we have the
following limit as t → ∞ uniformly in ξ in bounded subsets of Rn :
21 g1 ξ1 g2 ξ2 gn ξn t
1
2
lim
U νt ξ t = −
t→∞ ν
g1 ξ1 g2 ξ2 gn ξn (60)
To prove this result, we follow [3]. Consider θν given by (53), where φ0 satisfies the conditions (56) and (57). After a change of variable, the expression
for θν becomes
1
z2
1 1
ν
2
θ x t =
exp −
− φ0 x − tν z dz
n
2
ν
2π 2 Rn
1
z2
1
1
2
− φ0 tν ξ − z dz
exp −
=
n
2
ν
2π 2 Rn
Boundary Value Problems for Burgers Equations
499
Now, using (56) we get
n
1
1
z2
1
i
2 ξ − z −
exp
−
φ
dzi
tν
n
i
i
2
ν 1 0
2π 2 Rn
2
∞
n
1 i
zi
1
1
2
=
exp − − φ0 tν ξi − zi dzi
1
2
ν
i 2π 2 −∞
θν x t ≈
(61)
as t → ∞ uniformly on bounded sets of ξ in Rn . Now take the i-th term in
this product. Following the argument of Hopf [3], we get
2
∞
1 i
zi
φ+
1
i
2
exp − − φ0 tν ξi − zi dzi ≈ exp −
2
ν
ν
−∞
2
2
ξi
−
∞
z
φ
z
exp − i dzi + exp − i
exp − i dzi (62)
×
2
ν
2
ξi
−∞
Thus we have from (61) and (62),
2
ξi
1 +
z
2π lim θ ξνt t =
exp − φi
exp − i dzi
t→∞
ν
2
−∞
i
2
∞
1
z
+ exp − φ−
exp − i dzi ν i
2
ξi
n
2
ν
1
2
n
(63)
Similarly, we get
1
n
1
2π 2 lim νt 2 θxν l ξνt 2 t
t→∞
2
ξi
1 +
zi
1 −
=
dzi + exp − φi
exp − φi
exp −
ν
2
ν
−∞
i=l
2
2
∞
zi
φ−
ξ
φ+
l
l
exp −
dzi exp −
− exp −
exp − l ×
2
ν
ν
2
ξi
1
(64)
1
Because t/ν 2 U ν = −νt 2 ∇x θν /θν and θν is bounded away from 0, we have,
from (63) and (64),
ν
θxν n
θx1 θxν 2
1
1
1
ν
ν
lim t/ν 2 U νt 2 ξ t = − lim νt 2
t→∞
t→∞
θν θν
θ
gn ξn g1 ξ1 =−
g1 ξ1 gn ξn This completes the proof.
Here, we observe that for Burgers equation; that is, when n = 1, the
parameter k1 can be computed in terms of the mass of the initial data,
∞
−∞
∞
+
−
u0 ydy −
u0 ydy =
u0 ydy
νk1 = φ1 − φ1 =
x0
x0
−∞
500
K. T. Joseph and P. L. Sachdev
and then formula (60) with n = 1 is exactly Hopf’s [3] result. Also, we note
that the limit function obtained in the result written in the x t variable;
namely,
U x t = −ν log v1 x1 t
x1
−ν log v2 x2 t
x2
−ν log vn xn t
xn
where for i = 1 2 n,
xi /νt 21
2
ki
z
vi xi t = exp −
exp − i dzi
2
2
−∞
∞
2
k
z
+ exp i
exp − i dzi
1
2
2
xi /νt 2
is an exact solution of (44).
3.2. N-wave solution of vector Burgers equation
We start with the solution
θx t =
x2
1
n
+ t − 2 exp −
c0
2tν
of the heat equation (51) where c0 is a constant. Its space gradient is
1 x −n
x2
2
t exp −
∇x θx t = −
νt
2tν
Consider the function U ∞ = −ν∇θ/θ. By earlier discussion, this is an exact
solution of the vector Burgers equation (44). On simplification, we get
U ∞ x t =
t 1/2 1 +
x
t 1/2
n
t2
c0
2 exp x
2tν
(65)
For n = 1, this exact solution of the Burgers equation was discovered by [5].
Sachdev et al. [6] showed that it can be obtained as time asymptotic of a pure
initial value problem. This solution is called the N-wave solution. Here, we
generalize it for the vector equation (44).
We choose special initial data for (44) whose mass is zero and that is antisymmetric with respect to the origin. To solve the problem explicitly, we need
these data to be written as gradients. To construct such initial data, we consider the ball in Rn of radius l0 with center 0; namely, B0 l0 = x x ≤ l0 .
We take the initial condition as
Ux 0 = xχx≤l0 x
(66)
Boundary Value Problems for Burgers Equations
501
where χx≤l0 x is the characteristic function of the set x x ≤ l0 . Note
that this initial data can be written as the gradient
∇
l2
x2
χx≤l0 x + 0 1 − χx≤l0 x 2
2
So the earlier analysis holds, and we get the following formula for the solution
of (44) and (66):
Ux t = −ν
∇Q
Q
(67)
where Q is given by
Qx t =
1
x − y2
y2 +
dy
exp −
2ν
t
y≤l0 2 l0
x − y2
1
dy
exp −
exp −
+
2ν
2νt
2πνtn/2
y>l0 1
2πνtn/2
(68)
We shall prove the following result.
Let Ux t be the solution of (44) and (66) as given by (67) and (68), then
we have
lim Ux t = U ∞ x t
t→∞
1
uniformly in the variable ξ = x/2tν 2 belonging to a bounded subset of Rn ,
where U ∞ is given by (65) with
2
l
2
exp 2ν0 z2
l0
exp −
(69)
dz − exp −
B0 l0 c0 =
n
2ν
2ν
2πν 2
y≤l0 Here, B0 l0 denotes the volume, if space dimension n ≥ 3, area if n = 2,
and length if n = 1, of B0 l0 .
To prove this result, first we note that Qx t can be written as
2
l
(70)
Qx t = I1 + exp − 0 I2
2ν
where
1
x − y2
y2 +
dy
exp −
2ν
t
y≤l0 1
x − y2
dy
exp −
I2 =
n
2νt
2πνt 2 y>l0 I1 =
1
n
2πνt 2
502
K. T. Joseph and P. L. Sachdev
It can be easily checked by making use of error function and its asymptotics
that, as t → ∞,
2
2
2
exp − x
exp − x
2νt
z
2νt
dz I2 ≈ 1 −
exp −
B0 l0 (71)
I1 ≈
n
n
2ν
2πνt 2
2πνt 2
z≤l0 Substituting these asymptotics in (70), we get
2
2
exp − x
l0
2νt
+
Qx t ≈ exp −
n
2ν
2πνt 2
2
z2
l0
×
exp −
dz − exp −
B0 l0 2ν
2ν
z≤l0 (72)
Similarly,
x2 1 x 1 exp − 2νt
∇x Qx t ≈ − 1 1
ν t 2 t 2 2πνt 21
2
l0
z2
dz − exp −
B0 l0 ×
exp −
2ν
2ν
z≤l0 (73)
Using (72) and (73) in (67), we get
Ux t ≈
x
t
1
2
·
1
1
t2
x2 exp − 2νt
z2 l2 exp −
dz − exp − 0 B0 l0 2ν
2ν
2πνt
z≤l0 ×
2
l2 exp − x 2 2
l
z
2νt
dz − exp − 0 B0 l0 exp − 0 +
exp −
n
2ν
2ν
2ν
2πνt 2
z≤l0 n
2
On rearranging the terms, we get
1
Ux t ≈
1
t2 1+
x/t 2
n
t2
c0
2
exp x
2tν
where c0 is given by (69). This completes the proof.
3.3. Study of the limit ν → 0
First, we remark that the following analysis gives an explicit formula for φν ,
the solution of
ν
1
φt ∇φ2 = φ
2
2
φx 0 = φ0 x
(74)
Boundary Value Problems for Burgers Equations
503
namely,
φ x t = −ν log
φ0 y
x − y2
−
dy exp −
2νt
ν
Rn
1
ν
1
2πνt 2
(75)
Following the analysis of Hopf [3] and Lax [14], see also Joseph [8], we
get an explicit formula for the solution of the initial value problem for the
Hamilton–Jacobi equation
1
φt + ∇φ2 = 0
2
φx 0 = φ0 x
(76)
(77)
namely,
1
φx t = lim φν x t = min φ0 y + x − y2 y
ν→0
2t
(78)
Note that this is the explicit formula derived for the viscosity solution of
(77) derived by other methods, see [15]. Furthermore, it was shown by [15]
that φx t is Lipschitz continuous when φ0 x is and for almost every x t
there is a unique minimizer for (78), which we call y0 x t. Now, to study
the limit limν→0 U ν x t we note that (54) can be written as
U ν x t =
Rn
x−y
dµνxt y
t
(79)
where, for each x t and ν, dµνxt y is a probability measure given explicitly by
2
exp − x−y
− φ0νy dy
2νt
dµνxt y = φ0 y
x−y2
−
dy
n exp −
R
2νt
ν
(80)
Following the argument of Hopf [3] and Lax [14], it can easily be seen that
this measure tends to the δ-measure concentrated at y0 x t, the minimizer
of (78), which is unique for almost every x t. So, for almost all x t, we
get from (79) and (80)
lim U ν x t =
ν→0
x − y0 x t
t
where y0 x t is as before a minimizer of (78).
(81)
504
K. T. Joseph and P. L. Sachdev
3.4. A boundary value problem
The method described before can be used to solve some boundary value
problems as well. Let - be a bounded connected open set with a smooth
boundary. Consider the cylindrical domain D = - × 0 ∞. Consider the
problem
ν
U
2
Ux 0 = ∇φ0 x x ∈ Ut + U · ∇U =
nx t · Ux t∂-×0∞ = α
(82)
Here, φ0 is a smooth function from - to R1 , α is real constant, and nx t is
the unit outward normal of the boundary points of D. We note that nx t ·
Ux t is the normal component of Ux t at the boundary point x t and
when α = 0, (82) says that Ux t is tangential to the boundary point x t.
Here again, we seek a solution of the form Ux t = ∇φx t and, as before,
we get
Ux t = −ν
∇θ
θ
(83)
where θ satisfy the linear problem
ν
θ
2
θt =
φ x
θx 0 = exp − 0
ν
ν∂n θ + αθ∂- × 0 ∞ = 0
(84)
The explicit solution of (84) is
θx t =
∞
cn exp−λn t · φn x
(85)
0
where
φ x
φn xdx
exp − 0
ν
-
cn =
and λn and φn are eigenvalues and normalized eigenfunctions of the eigenvalue problem in -:
ν
− φ = λφ
2
ν∂n φ + αφ∂- = 0
Boundary Value Problems for Burgers Equations
505
Using (85) in (83), we get
∞
c exp−λn t∇φn x
Ux t = −ν 1 ∞ n
1 cn exp−λn tφn x
Letting t → ∞ in (86) we get
lim Ux t = −ν
t→∞
(86)
∇φ1
φ1
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TIFR Mumbai
Indian Institute of Science
(Received April 3, 2000)