THE PERTURBED BURGERS EQUATION
Lecture notes – Yair Zarmi
Solutions of the unperturbed equation
Consider the unperturbed Burgers equation
wt  2 ww x  Dw xx
Subscripts denote derivatives. (D is the diffusion coefficient,  the viscosity in fluid dynamics.)
With the Cole-Hopf transformation:
w
D fx
 f
one obtains:
f  ft  D f xx x  f x  ft  D f xx   0
The unperturbed Burgers equation is solved by f that obeys a linear diffusion equation:
ft  D fxx  0 (*)
This is what saying that the Burgers equation is “linearisable” means.
To get an idea about the necessary boundary conditions that need to be imposed, let us study the
solutions of the equation (*). We are interested in wave solutions:
f x,t   Fx  V t 
where V is some fixed velocity. Denoting
  x  Vt
The equation (*) becomes
V f  D f  0
This equation is solved by
V
V 2 
V 
f  a  b exp    a  b exp  x 
t 
D 
D
D 
-2-
This yields for the unperturbed Burgers equation the solution
V
bV
V 2 
exp  x 
t 

D
D 
wx,t  
V
V 2 
a  b exp  x 
t
D 
D
This is a "shock wave" or "front" solution. For V>0, one has at a fixed time
x  
0
wx,t   
V 
w x x,t   0
x  
x  
(**)
The width of the transition region between the two asymptotic values of w(x,t) is (D/V). (The
analysis in Burgers' book is an expansion for small D (small viscosity)). The front propagates to
the left at a speed V. In the paper, they discuss solutions of this type.
Solution of the integrable perturbed equation
For simplicity, I now choose =D=1. The following perturbed Burgers equation is integrable
wt  2 ww x  wxx   3w 2 wx  3w wxx  3w x 2  wxxx 
because it is linearised by the same Cole-Hopf transformation, which now yields:
f  ft  f xx    f xxx x  f x  ft  fxx   fxxx   0
The "linearisable" perturbed Burgers equation is solved by f that obeys the linear equation:
ft  f xx    f xxx  0
If we look for solutions of the type
f x,t   Fx  V t 
,
  x  Vt
then we obtain for f the equation
V f  f    f   0
Looking for front solutions, we search for f of the form
f  exp 
-3-
and obtain for 
V    2    3  0
The roots are
0



   1  1  4   V


 1  1  4  V


 2    V 1    V  2   V

 2    V  1 V   1    V  2   V
2

 O   V 

2
3

3 
 O   V  



The solution for f is:
 



a  b exp V  1    V   2   V  2  O   V  3 

f  
c exp V  1  V   1    V   2   V  2  O   V  3

 

The term with (1/) in the exponent generates a front the width of which is O(). It may cause
troubles if used in higher-order perturbation expansion. The analysis of the boundary conditions at
x ±∞ follows the one presented earlier.
Analysis of the general perturbed equation
Consider now the general perturbed Burgers equation
ut  2uux  uxx   31 u 2 ux  32 uuxx  33 ux 2  4 uxxx 
The O() terms are of the types that appear in the expansion of the fluid dynamical equations that
lead to the Burgers equation in lowest order.
The perturbed equation may not be "linearisable." We write a near-identity transformation (NIT):
u  w     2  
We assume no explicit dependence of  on x or on t:
   v,w,wx ,wxx ,
x
where
v   wdx
x0
-4-
The lower limit of integration, x0, is arbitrary. If x0 is not specified, the integral is defined up to a
constant in x, which may depend on time. Let us choose x0=∞, so as to be able to impose
boundary condition (**).
The normal form notation is the equation for the time dependence of the zero-order term::
wt  U0  U1   2 U2 
Inserting the NIT and the normal form in the original equation, we find in lowest order:
U0  2w wx  w xx
Some useful identities:
2
U0, x  2w x  2w wxx  wxxx
x
U
dx  wx,t   w x x,t   wx0 ,t   wx x 0 ,t 
2
2
0
x0
Choosing x0=∞, and zero-boundary condition at x0=∞, the contribution of the lower limit to the
integral vanishes.
Other useful identities:
x
vt   wt dx 

x
 U
0
 U1   2U2 
dx

 t   w wt   wx wtx   v vt 
 w 2 ww x  wxx   U1 

 wx 2wx 2  2 wwxx  wxxx  U1, x 
x
 2
 v w  wx   U1 




 dx 

(Note: in the last line of this identity the coefficient of the v term had a contribution of the form
 wx 0   wx x 0 
2
which I have already canceled, due to the choice x0=∞. This will be consistently done in the
following.)
-5-
 x   w wx   wx wxx  v w
 xx   ww w x 2   wx wx w xx 2   vv w 2 
2 wwx wx wxx  2 wv w wx  2w x v w wxx 
 w wxx  w x wxxx   v wx
Using all these relations, one obtains the following relation in O() :



 

U1  w 2 w wx  wxx  w x 2 wx 2  2w wxx  w xxx   v w 2  w x 
 w 2w wx   wx 2 ww xx   v 2 w 2  2 wx 
2
2
2
 ww w x   wx wx w xx   vv w 
2wwx w x wxx  2 wv ww x  2 wx v w wxx 
 w wxx   wx wxxx   v wx 
3 1 w 2 w x  3 2 ww xx  3 3 w x 2   4 w xxx
The underlined terms cancel out. What remains is:

2

 
2  w w 3 
U1  w 2  v   vv  wx 2  ww  2 wx  33  w xx 2  wx wx 
w wx 2 wv  ww xx
wx v

 3 2  wx w xx
2
ww x
x
1
wx 2   wxxx 4 
We treat this as a polynomial in w, wxx, v, ... We want U1 to be a “resonant” term, namely, a
symmetry of the unperturbed equation. The definition of a resonant term is through the vanishing
of its Lie-bracket with the unperturbed operator:
2ww
x

 wxx ,U1  0
This gives for U1 the form:
U1   4 3w 2 wx  3w wxx  3w x 2  wxxx 
This form guarantees that the normal form equation through first order is linearisable by the ColeHopf transformation, hence the normal form is integrable.
The remainder has to satisfy several identities. The first three obvious required identities are:
Coefficient of wxx2=0:  wx w x  0
-6-
 wwx  0
Coefficient of wxwxx=0:
Coefficient of wwxx=34:
 wx v  23  4  2 
The first identity implies
  Aw,v w x  Bw,v
The second identity implies
A  Av 
The third identity implies
Av 
3
2
 4  2  A  32  4  2 v  
where  is a free constant (which cannot be determined in this order). Hence,
  wx 
3
2
4  2 w x v  Bw,v
Inserting this result in the expression for U1, we get
U1  w 2 Bv  Bvv  wx 2 Bww  3 3 
w wx 2 Bwv  wx 2 Bw,v
w 2 wx 31 
3
2
4   2  w wxx 3 4 wxxx  4
We have to kill the term
w
2
Bv  Bvv 
[No terms of the form (w2 multiplied by functions of v) are desired in U1.] Hence, we must have
Bv  Bvv  0  B  Cwexp v  Dw
This yields
U1  wx 2 Cww exp v   Dww  33 
w wx 2Cw exp  v wx 2C exp v   2 D 
w 2 wx 31 
3
2
4   2  w wxx 3 4 wxxx  4
-7-
To get rid of the term proportional to
w x exp v 
(there are no v dependent terms in U1) we must have
wCw  C  0

C  w
Here  is a constant. This also eliminates the term
w x 2 Cww exp v
Hence, we obtain
U1  wx 2 Dww  33  wx 2 D 
w 2 wx 31 
3
2
4   2 w wxx 3 4 wxxx  4
The next identity is:
D
Coefficient of wx2=34:
3
2
4  3 w 2  k w  l
where k and l are constants. To avoid a term linear in wx, we must have k=l=0, yielding:
  wx  w exp v  
U1  w 2 w x 31 
3
2
3
2
4   2wx v  32 4  3 w 2
 4  2   3 4  3  4 3wx 2  3w wxx  w xxx 
We now have to take care of the remaining term:
w 2 wx 31  32 4  2   3 4  3 
For its coefficient to be 34, the relation
21   2  2 3   4  0
(***)
has to be satisfied. Otherwise, U1 cannot have the "linearisable" form. One approach is to add the
term, which cannot be eliminated in U1, which then obtains the form:
U1   4 3w wx  3w wxx  3w x  wxxx   31  33  32 2  32 4 w wx
2
2
 4 F3  w 
2
Z w 
-8-
However, as now U1 does not have the "linearisable" form, the normal form equation is not
integrable, hence the notion of “obstacle” to integrability. Obstacles to integrability appear in
higher orders as well. Another approach is to insist on the original idea of the normal form being
comprised of resonant terms only, and to include the “obstacle” term in the equation for the firstorder correction, . One conclusion is obvious: The assumption that the normal form expansion
can be carried out under the assumption that all entities in the NIT and the normal form are
differential polynomials in the zero-order term, w, does not work.
Comment
Look at the final form of :
  wx  w exp v  
3
2
4   2wx v  32 4  3 w 2
There are two arbitrary terms in 
 wx
and
 wexp v 
The equations are transparent to them. The first one may be modified by varying the lower limit
of integration in the definition of v, through the term:
3
2
4  2 wx v
The contribution of the lower limit of integration in the definition of v has the general form
 32  4  2 vx 0,t w x
which is linear in wx. Thus, the choice of the boundary point can affect . For x0=∞ one has
v(x0,t)=0, thus, to de-coupling of  from the contribution of the lower limit.
Effect of obstacle on first-order term
If one solves the problem with a zero-order term which is a wave front with a single velocity (of
the type discussed at the beginning), then one CAN solve for the first-order term, , because then
the obstacle does not appear. One then finds:
  1  2 2  3  2  4 wx v 
1
2
21   2  3  2 4 u 2
Now going to the general problem (where the zero-order solution is NOT a single-velocity wave),
we assume that  has a term, the form of which is like that obtained for a single-velocity wave
solution + a correction, which we’ll denote by :
-9-
  1  2 2  3  2  4 wx v 
1
2
21   2  3  2  4 u 2  
The equation that  obeys is:
t  2 wx   2 w  x  xx    21   2  2 3  4 w 2 wx  w wxx  wx 2 
Notice that the coefficient of the obstacle is just the quantity that needs to vanish, for there NOT to
be an obstacle. However, the obstacle may vanish even if that coefficient does not. This happens
in the case if w is a single-velocity wave. If w is not, then the obstacle may exist.
Let’s look at the specific case of a two-velocity wave:
w
  
   k˜  k   

1  Bexp k x  k˜ t  C expk x  k˜ t 
k1 Bexp k1 x  k˜1 t  k2 Cexp k2 x  k˜2 t
i
1
1
2
i
4
ki 2

i  1,2
2
Then the obstacle does not vanish. However, a close study of the obstacle, , shows the following
behaviour. For fixed x, when t is negative,  vanishes exponentially in time, as t  ∞, while for
t  +∞,   k1 k2 wx.
Thus, asymptotically, for large negative t, the obstacle vanishes, hence the correction  vanishes
exponentially in time. Asymptotically, for large positive t,  may be replaced by wx, converting the
equation for  into
t  2 wx   2 w  x  xx  2 1   2  2 3   4 k1 k2 wx
This is obviously solved by

1
2
2 1  2  2  3   4k1 k2   21 21  2  23  4 k1 k2
Thus, asymptotically for large positive t,  goes to a constant. The transition between the two
constant asymptotic values is a function of t and x.