Q-conditional symmetry as a source of exact solutions of generalized Burgers equation
In this paper the conditional symmetry to a hyperbolic generalization of Burgers equation is studied. Employment of the generalized symmetry enabled to obtain new exact solutions, describing the evolution of various wave patterns.
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Інститут математики НАН України
2006
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Transactions of Institute of Mathematics of NAS of Ukraine| _version_ | 1872552954629718016 |
|---|---|
| author | Kutafina, E. Vladimirov , V. Кутафіна , Є. Владіміров, В. |
| author_facet | Kutafina, E. Vladimirov , V. Кутафіна , Є. Владіміров, В. |
| author_institution_txt_mv | [
{
"author": "E. Kutafina",
"institution": null
},
{
"author": "V. Vladimirov ",
"institution": null
}
] |
| author_sort | Kutafina, E. |
| baseUrl_str | https://trim.imath.kiev.ua/index.php/trim/oai |
| collection | OJS |
| datestamp_date | 2020-08-13T10:42:30Z |
| description | In this paper the conditional symmetry to a hyperbolic generalization of Burgers equation is studied. Employment of the generalized symmetry enabled to obtain new exact solutions, describing the evolution of various wave patterns. |
| first_indexed | 2026-08-04T01:07:38Z |
| format | Article |
| fulltext |
Збiрник праць Iнституту математики НАН України 2006, т.3, N 2, 170–181
УДК 517.912:512.816
Q-conditional symmetry as a source
of exact solutions of generalized
Burgers equation
E.V. KUTAFINA, V.A. VLADIMIROV
AGH University of Science and Technology, Kraków, Poland
E-mail: kutafina@mat.agh.edu.pl, vladimir@mat.agh.edu.pl
В роботi проаналiзовано умовну симетрiю гiперболiчного узагальнен-
ня рiвняння Бюргерса. Використання узагальненої симетрiї дозволи-
ло одержати новi точнi розв’язки, що описують рiзноманiтнi хвильовi
структури.
In this paper the conditional symmetry to a hyperbolic generalization of
Burgers equation is studied. Employment of the generalized symmetry
enabled to obtain new exact solutions, describing the evolution of various
wave patterns.
1. Introduction. In last few years we dealt with different methods of
obtaining analytical solutions of nonlinear PDE’s that are not completely
integrable [1–3], paying special attention to the following generalization
of Burgers equation (GBE) [4]:
τutt − κuxx +Auux +But +Hux = f(u).
Here and henceforth lower indices mean partial derivatives with respect
to corresponding variables. The classical symmetry methods [5, 6] are
very popular in obtaining exact solutions to nonlinear PDEs, but for non-
zero constants classical symmetries of GBE are reduced to the generators
of translations ∂t and ∂x, giving rise to travelling-wave solutions. So in
this study we proceed further on and look for solutions which cannot be
described in terms of travelling waves. To do this we employ so-called
Q-conditional symmetry methods [7–11].
Let us consider equation
F (x, t, u, ux, ut, . . . ) = 0. (1)
Q-conditional symmetry as a source of solutions to GBE 171
It is of common knowledge, that within the classical Lie algorithm [5,6]
we look for the operator
Q = ξ1∂x + ξ2∂t + φ∂u, (2)
such that
Q̂F |F (x,t,u,ux,ut,... )=0 = 0, (3)
where Q̂ denotes the proper prolongation of Q.
Looking for the Q-conditional symmetry, we pose the additional con-
dition:
Q(u(x, t)− u) = 0 = ξ1 ux + ξ2 ut − φ (4)
and solve the equations:
Q̂F |F (x,t,u,ux,ut,... )=0, Q=0, Q1=0, Q2=0,... = 0, (5)
where Q = 0, Q1u = 0, Q2u = 0, . . . denote equation (4) and its
differential consequences of the corresponding orders. The additional
condition allows finding much wider classes of reductions to GBE.
2. Brief overview of the cases. We deal with the equation:
τutt − κuxx +Auux +But +Hux =
= f(u) = λ0 + λ1u+ λ2u
2 + λ3u
3. (6)
To examine the conditional symmetry of (6), we consider it together
with the equation (4). Here ξ1, ξ2, φ depend on the variables x, t, u.
We assume that τ , κ, A, λ3 are non-zero and examine symmetries of the
system (4), (6). Let us notice, that whenever ξ1 (or ξ2) is non-zero, it
can be scaled to 1.
Case I: ξ2 = 1, ξ1 6= 0. Using (6), (4) and its differential consequen-
ces we can eliminate ut and all the second derivatives of u(t, x)1. After
computing the prolongation of Q and performing the splitting procedure
we obtain four determining equations:
e1 = 3f(u)(κ− τξ2
1)ξ1u + τφ2(2τξ1ξ1u
2 + κξ1uu − τξ2
1ξ1uu)−
1Let us notice that in case when κ − τξ21 = 0 the above procedure fails. This
situation is thoroughly examined in section 3.
172 E.V. Kutafina, V.A. Vladimirov
− 2κτξ1uφt + 2τ2ξ2
1ξ1uφt +Bκξ1t − 2Hτξ1ξ1t − 2Auτξ1(ξ1t +
+Bτξ2
1ξ1t + 2κτφuξ1t + 2τ2ξ2
1φuξ1t + 2τ2ξ1ξ1t
2 + 2κτξ1φtu −
− 2τ2ξ3
1φtu + φ(−2τ2ξ3
1φuu + 2τξ1(κφuu − ξ1u(H +Au−
− 2τξ1t))− κ(A+ 2Bξ1u− 2τξ1tu) + τξ2
1(A+ 4(B + τφu)ξ1u−
− 2τξ1tu)) + κτξ1tt− τ2ξ2
1ξ1tt−Hκξ1x −Auκξ1x + 2Bκξ1ξ1x−
−Hτξ2
1ξ1x −Auτξ2
1ξ1x + 4κτξ1φuξ1x − 2κτξ1ξ
2
1x + 2κ2φxu −
− 2κτξ2
1φxu − κ2ξ1xx + κτξ2
1ξ1xx = 0,
e2 = −τφ2(2ξ1(B + τφ1u)ξ1u + κφ1uu − τξ2
1φ1uu)−Bκφt+
+Bτξ2
1φt − 2τ2ξ1φt(ξ1t − κτ(φtt + τ2ξ2
1(φtt −Hκφx −
−Auκφx +Hτξ2
1φx +Auτξ2
1φx + 2κτ(ξ1tφx − 2κτφtξ1x +
+ 2κτξ1φxξ1x + f(u)((−κ+ τξ2
1)(φu + 2(τξ1(ξ1t + κξ1x)) +
+ φ((κ− τξ2
1)f ′(u)− 2(−(τf(u)ξ1ξ1u) + τξ1(τξ1uφt + (B +
+ τφuξ1t − τ2ξ2
1φxu + κ(τφxu − τξ1uφx + (B + τφuξ1x))) +
+ κ2φxx − κτξ2
1φxx = 0,
e3 = −4τ2φξ2
1ξ
2
1u − 2ξ1u((H +Au)κ+ τξ3
1(B + τφu −
− τξ2
1(H +Au− 2τξ1t − κξ1(B + τ(φu − 2τξ1x)) + (κ−
− τξ2
1)((κ− τξ2
1)φuu − 2(τφξ1ξ1uu + τξ1(ξ1xt + κξ1xt)) = 0,
e4 = 2τξ1ξ
2
1u + κξ1uu − τξ2
1ξ1uu = 0.
Since the first three equations are very complicated, we start our
analysis from the last and the simplest one.
Case Ii: ξ1u 6= 0. Introducing the new function ξ1u = Ψ(ξ1) and
consequently ξ1uu = Ψ′(ξ1)Ψ(ξ1), we obtain the integrable equation:
2τξ1Ψ(ξ1)
2
+ κΨ′(ξ1)Ψ(ξ1)− τξ2
1Ψ′(ξ1)Ψ(ξ1) = 0. (7)
Equation (7) is satisfied by the following function:
ξ1 =
√
κ
τ
tanh
(√
κτc1(x, t)(u+ c2(x, t)
)
. (8)
Function φ can be calculated from e3=0. Unfortunately e1=0 gives us
either τ = 0 or κ = 0 or λ3 = λ2 = A = 0. So all the possibilities are in
contradiction with our assumptions.
Q-conditional symmetry as a source of solutions to GBE 173
Case Iii: ξ1u = 0. To solve e4 = 0 we can also put ξ1u = 0. Then the
third determining equation (e3 = 0) takes the form: (κ − τξ2
1)φuu = 0.
Since (κ− τξ2
1) 6= 0, then φuu = 0. In other words, φ = a(x, t)u+ b(x, t)
and the remaining determining equations are as follows:
e1 = −(Aκ(ua+ b)) +Aτ(ua+ b)ξ2
1 + 2κτξ1at − 2τ2ξ3
1at +
+Bκξ1t + 2κτaξ1t − 2Hτξ1(ξ1t − 2Auτξ1ξ1t +Bτξ2
1ξ1t +
+ 2τ2aξ2
1ξ1t + 2τ2ξ1ξ
2
1t + κτξ1tt − τ2ξ2
1ξ1tt + 2κ2ax −
− 2κτξ2
1ax −Hκξ1x −Auκξ1x + 2Bκξ1ξ1x + 4κτaξ1ξ1x −
−Hτξ2
1ξ1x −Auτξ2
1ξ1x − 2κτξ1ξ
2
1x − κ2ξ1xx + κτξ2
1ξ1xx = 0,
e2 = −κaf(u) + τaf(u)ξ2
1 + κ(ua+ b)f ′(u)− τ(ua+ b)ξ2
1f
′(u)−
− 2κτ(ua+ b)at + 2τ2(ua+ b)ξ1
2at −Bκ(uat + bt) +
+Bτξ2
1(uat + bt)− 2Bτ(ua+ b)ξ1ξ1t − 2τ2a(ua+ b)ξ1ξ1t +
+ 2τf(u)ξ1ξ1t − 2τ2ξ1(uat + bt)ξ1t − κτ(uatt + btt) +
+ τ2ξ2
1(uatt + btt)−Hκ(uax + bx)−Auκ(uax + bx) +
+Hτξ2
1(uax + bx) +Auτξ2
1(uax + bx) + 2κτ(ξ1t(uax + bx)−
− 2Bκ(ua+ b)ξ1x − 2κτa(ua+ b)ξ1x + 2κf(u)ξ1x −
− 2κτ(uat + bt)ξ1x + 2κτξ1(uax + bx)ξ1x + κ2(uaxx + bxx)−
− κτξ2
1(uaxx + bxx).
Taking into account that f(u) = λ3u
3 + λ2u
2 + λ1u + λ0, we collect
the terms at different powers of u. Nullifying them, we obtain (among
others) the following equations:
a(x, t)(κ− τξ1)
2
+ 2τξ1ξ1t + κξ1x + τξ2
1ξ1x = 0,
a(x, t)(κ− τξ12) + τξ1ξ1t + κξ1x = 0.
The above system can be presented in the following equivalent form:
a(x, t) = −ξ1x, ξ1ξ1x = −ξ1t.
The second equation is a model equation of gas dynamics. Its general
solution is as follows [12]:
x = tξ1 + Φ(ξ1),
where Φ(·) is an arbitrary function.
174 E.V. Kutafina, V.A. Vladimirov
For Φ(ξ1) = 0
ξ1(x, t) = x/t, a(x, t) = −1/t
and remaining equations of this group are satisfied providing the follo-
wing conditions are fulfilled:
b(x, t) =
−λ2
3tλ3
, B = 0, λ0 =
λ3
2
27λ2
3
,
A =
−3Hλ3
λ2
, λ1 =
λ2
2
3λ3
.
Using the ansatz
u(x, t) =
−λ2x+ 3λ3R( tx )
3λ3x
,
we get in this case the reduced equation
λ2(κξ2
1 − τ)R′′(ξ1) + 3Hλ3ξ1R
′(ξ1)R(ξ1) + λ3λ2R(ξ1)3 +
+ 4κλ2ξ1R
′(ξ1) + 3Hλ3R(ξ1)2 + 2κλ2R(ξ1) = 0. (9)
Equation (9) is a nonlinear non-autonomous ODE which is, generally
speaking, nonintegrable. A particular solution can be obtained using
the ansatz-based method [1], slightly modified for the purposes of non-
autonomous case. Thus, we represent the solution in the form
R(ξ1) =
p
1 + q(ξ1) exp(α(ξ1))
. (10)
The numerator of equation (9) contains then different powers of
Exp (α(ξ1)). Splitting the equation we can calculate the constants:
p = ±
√
2κ/λ3, H = ∓2/3
√
2κ/λ3λ2,
and the function q(ξ1). The simplified solutions are as follows
R(ξ1) = ±
√
2κF (ξ1)
√
λ3(F (ξ1) +
√
κξ2
1 − τ)
,
R(ξ1) = ±
√
2κ
√
λ3(1 + F (ξ1)
√
κξ2
1 − τ)
,
where F (ξ1) = arctanh (
√
κ/τξ1).
Q-conditional symmetry as a source of solutions to GBE 175
Case II: ξ2 = 1, ξ1 = 0. With these assumptions determining equati-
ons are as follows:
e1 = φuu = 0,
e2 = f(u)φu +Bφt − φ(f ′(u)− 2τφtu) + τφtt +
+Hφx +Auφx − κφxx = 0,
e3 = Aφ− 2κφxu = 0.
From the firs equation we get the representation for u:
φ(x, t, u) = a(x, t)u+ b(x, t).
On virtue of this, the last equation becomes as follows:
A[b(x, t) + a(x, t)u]− 2κax(x, t) = 0.
Since a(x, t) and b(x, t) are independent of u and A 6= 0 we must put
a(x, t) = 0 and as a consequence b(x, t) = 0. The condition (4) gives
ut(x, t) = 0 and then u = u(x).
Case III: ξ2 = 0, ξ1 = 1. We get in this case three determining
equations:
e1 = φuu = 0,
e2 = φut = 0,
e3 = Aφ2 + f(u)φu +Bφt + τφtt +Hφx +Auφx −
− φ(f ′(u) + 2κφxu)− κφxx = 0.
From the first two equations we obtain that
φ(x, t, u) = a(x)u+ b(x, t).
The last equation then takes the form:
af(u) +A(b+ au)2 − 2κax(b+ au) +Bbt +H(axu+ bx) + (11)
+Au(axu+ bx)− (b+ au)f ′(u) + τbtt − κ(axxu+ bxx) = 0.
Expression (11) can be splitted by powers of u. The coefficient of u3 is
−2λ3a(x) while that of u2 is −λ2a+Aa2−3λ3b+Aa
′. If we put a(x) = 0,
then this implies b(x, t) = 0 and we again encounter the situation where
176 E.V. Kutafina, V.A. Vladimirov
u is function of one independent variable. So we rather put λ3 = 0,
λ2 6= 0. Then a(x) can be easily calculated:
a(x) =
λ2e
xλ2
A
Ae
xλ2
A − eλ2c1
. (12)
Nullifying the coefficient of u in (11), we obtain a first order linear equa-
tion, which gives us the form of b(x, t):
b(x, t) =
e
xλ2
A (λ2e
λ2c1(−AH + κλ2) +A2e
xλ2
A c2(t))
A2(eλ2c1 −Ae
xλ2
A )2
. (13)
Substituting (12), (13) into (11) and equating to zero the coefficient
of u0 (the free term), we obtain a very complicated expression which
determines an unknown function c2(t). To our luck, this expression can
be splitted as well. This time we equate to zero coefficients of e3c1λ2 ,
e3xλ2/A, e2c1λ2+xλ2/A and e(c1+2x/A)λ2 (equations obtained this way we
denote as (eqn1)–(eqn4)).
Solving (eqn1) we get
λ0 =
1
A4
(AH − κλ2)(A2λ1 −AHλ2 + κλ2
2). (14)
Adding (eqn2) to (eqn3), we obtain the expression
A [Hλ2 − c2(t)] [λ2 (AH − κλ2)−Ac2(t)] = 0.
Thus, c2(t) must be constant. Under this assumption we finally solve
equations (eqn3)–(eqn4) and conclude that
c2 = λ2(H − κλ2/A).
Now equation (4) takes on the form
ux =
λ2e
xλ2
A
Ae
xλ2
A − eλ2c1
u+
e
xλ2
A λ2(−AH + κλ2)
A2(eλ2c1 −Ae
xλ2
A )
. (15)
Solving (15), we obtain:
u =
1
A2
[
−AH + kλ2 +A2
(
ec1λ2 −Ae
xλ2
A
)
c3(t)
]
. (16)
Q-conditional symmetry as a source of solutions to GBE 177
Fig. 1. The example solution of the form (16), A = λ2 = c = τ = κ = 1,
H = −1, c4 = c5 = 0.
Finally, inserting (16) into (6), we obtain determining equation for un-
known function c3(t):
A2(τc′′3(t) +Bc′3(t))−A2cλ2c3(t)2 −
− (A2λ1 − 2AHλ2 + 2κλ2
2)c3(t) = 0, (17)
where c = ec1λ2 . Some special solutions to this equation can be easily
found. Let us introduce the new variable g(c3) = c′3(t). Then c′′3(t) = g′g
and (17) becomes as follows:
A2(τg′(c3)g(c3) +Bg(c3))−A2cλ2c
2
3 −
− (A2λ1 − 2AHλ2 + 2κλ2
2)c3 = 0.
For B = 0 it can be solved just by two integrations:∫
τg′(c3)g(c3)dg −A2cλ2c
3
3 − (A2λ1 − 2AHλ2 + 2κλ2
2)c3dc3 =
= 1
2g(c3)2 − 1
3A
2cλ2c
2
3 − 1
2 (A2λ1 − 2AHλ2 + 2κλ2
2)c23 = c4. (18)
Returning back to the variable c3(t) we obtain:
c′3(t) = ±
√
a3c33 + a2c23 + a0, (19)
where a3 = 2cλ2/(3τ), a2 = (A2λ1 − 2AHλ2 + 2κλ2
2))/(A2τ), a0 =
2c4/(A
2τ). After integration the function c3(t) can be written in terms
178 E.V. Kutafina, V.A. Vladimirov
of elliptic functions. In case when c4 = 0 we get:
t+ c5 =
−2
√
a2
arctanh
√
a2 + a3c3(t)
a2
.
Under additional conditions A = λ2 = c = τ = κ = 1, H = −1, c5 = 0
the solution of the GBE takes the form:
u(x, t) = 2 + (exp(x)− 1)
(
15
2 sech2
(√
5
2 t
))
, (20)
and it is illustrated in Fig. 1.
3. Solutions of the case ξ2 = 1, κ− τξ2
1 = 0. Let us consider
system (6), (4) in the case then ξ2 = 1, ξ1 = ε
√
κ/τ , ε = ±1. After using
the condition (4), together with the proper differential consequences,
equation (6) becomes as follows:
[H −Bε
√
κ/τ +Au− 2ε
√
κτφu]ux +
+ (B + τφu)φ+ τ − ε
√
κτφx − f(u) = 0. (21)
Next we denote the coefficient of ux by k1 and the rest of (21) by k2.
Solving the system k1 = k2 = 0 we get:
φ = φ(u) =
1
2ε
√
κτ
(A/2
√
τu2 + u(H
√
τ − εB
√
κ)) + c
with an extra conditions
λ0 = 1/2c(B + εH
√
κ/τ), λ1 =
−B2κ+H2τ + 2Acε
√
κτ3/2
4κτ
,
λ2 =
A(3H − εB
√
κ/τ)
8κ
, λ3 =
A2
8κ
.
To calculate the solution u(x, t), let us return to equation (4):
ut +
√
κ/τux = φ = a0 + a1u+ a2u
2,
where a0 = c, a1 = (H
√
τ − εB
√
κ)/(2ε
√
κτ), a2 = A/(4ε
√
κτ). Writing
it in the characteristic form
dx√
κ/τ
=
dt
1
=
du
a0 + a1u+ a2u2
,
Q-conditional symmetry as a source of solutions to GBE 179
-10 -5 5 10
-2
-1
1
2
3
4
Fig. 2. Temporal evolution of solution described by formula (25) in case when
Γ(ω) = sin[2.25ω].
-5 5 10 15 20 25 30
-2
-1
1
2
3
4
Fig. 3. Temporal evolution of solution described by formula (25) in case when
Γ(ω) = −ω2.
2 3 4 5
5
6
7
8
9
10
11
Fig. 4. A blow-up regime described by the formula (26) in case when Γ(ω) =
−3.75ω2 + 5.
180 E.V. Kutafina, V.A. Vladimirov
we easily get the general solution
x+ Ψ(
√
κ/τt− x) =
√
κ
τ
∫
du
a0 + a1u+ a2u2
, (22)
where Ψ(·) is and arbitrary function. Below we present some special
solutions of (22).
For A = τ = κ = ε = B = 1, H = 0 equation (4) takes the form
ut + ux =
1
4
(
u2 − 2u+ 4c
)
. (23)
General solution of this equation depends on whether ∆ = 1 − 4c is
positive or not. For c < 1/4 solution of (23) is as follows:
u(t, x) =
u2G(ω)ex
√
∆
2 − u1
G(ω)ex
√
∆
2 − 1
, (24)
where u1 = 1 +
√
∆, u2 = 1 −
√
∆, G(·) is an arbitrary function of
ω = x− t. Putting c = −2 and G(ω) = −eΓ(ω) we obtain the formula
u(x, t) = 2
2− exp[3x/2 + Γ(ω)]
1 + exp[3x/2 + Γ(ω)]
. (25)
For c = −2 and G(ω) = eΓ(ω) we get the solution
u(x, t) = 2
exp[3x/2 + Γ(ω)] + 2
1− exp[3x/2 + Γ(ω)]
. (26)
If c > 1/4 then solution to (23) is as follows:
u(x, t) = 1 + βarctg
[
βx
4
+G(ω)
]
, (27)
where β =
√
|1− 4c|. This solution is always singular.
If c = 1/4 then solution to (23) is as follows:
u(x, t) = 1 +
1
G(ω)− x
4
. (28)
This solution is also singular.
Q-conditional symmetry as a source of solutions to GBE 181
Let us give examples of solutions corresponding to formulae (25)
and (26). Thus, inserting Γ(ω) = sin[2.25ω] into equation (26), we obtain
an oscillating kink-like solution, shown in Fig. 2. For Γ(ω) = −ω2 this
solution produces a “dark” soliton with growing “support” (Fig. 3).
In contrast to (25), solution (26) is always singular. For Γ(ω) =
−3.75ω2 + 5 its evolution is shown in Fig. 4, in which we see how an
initial localized wave pack grows in amplitude and in a finite time gives
rise to a blow-up regime.
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| id | oai:trim.imath.kiev.ua:article-454 |
| institution | Transactions of Institute of Mathematics of NAS of Ukraine |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-08-04T01:07:38Z |
| publishDate | 2006 |
| publisher | Інститут математики НАН України |
| record_format | ojs |
| resource_txt_mv | trimimathkievua/b3/443787e9d2a02ec851a0a7709b2782b3.pdf |
| spelling | oai:trim.imath.kiev.ua:article-4542020-08-13T10:42:30Z Q-conditional symmetry as a source of exact solutions of generalized Burgers equation Q-умовна симетрія як джерело точних розвя’зків рівняння Бюргерса Kutafina, E. Vladimirov , V. Кутафіна , Є. Владіміров, В. In this paper the conditional symmetry to a hyperbolic generalization of Burgers equation is studied. Employment of the generalized symmetry enabled to obtain new exact solutions, describing the evolution of various wave patterns. В роботі проаналізовано умовну симетрію гіперболічного узагальнення рівняння Бюргерса. Використання узагальненої симетрії дозволило одержати нові точні розв'язки, що описують різноманітні хвильові структури. Інститут математики НАН України 2006-11-14 Article Article application/pdf https://trim.imath.kiev.ua/index.php/trim/article/view/454 Transactions of Institute of Mathematics, the NAS of Ukraine; Vol. 3 No. 2 (2006): Symmetry and Integrability of Equations of Mathematical Physics (Dedicated to the 70-th Anniversary of Professor W.I. Fushchych); 170-181 Сборник Трудов Института математики НАН Украины; Том 3 № 2 (2006): Симетрія та інтегровність рівнянь математичної фізики (До 70-річчя від дня народження Вільгельма Ілліча Фущича); 170-181 Збірник Праць Інституту математики НАН України; Том 3 № 2 (2006): Симетрія та інтегровність рівнянь математичної фізики (До 70-річчя від дня народження Вільгельма Ілліча Фущича); 170-181 3083-7529 1815-2910 en https://trim.imath.kiev.ua/index.php/trim/article/view/454/451 Авторське право (c) 2020 Є. Кутафіна , В. Владіміров http://creativecommons.org/licenses/by/4.0 |
| spellingShingle | Kutafina, E. Vladimirov , V. Кутафіна , Є. Владіміров, В. Q-conditional symmetry as a source of exact solutions of generalized Burgers equation |
| title | Q-conditional symmetry as a source of exact solutions of generalized Burgers equation |
| title_alt | Q-умовна симетрія як джерело точних розвя’зків рівняння Бюргерса |
| title_full | Q-conditional symmetry as a source of exact solutions of generalized Burgers equation |
| title_fullStr | Q-conditional symmetry as a source of exact solutions of generalized Burgers equation |
| title_full_unstemmed | Q-conditional symmetry as a source of exact solutions of generalized Burgers equation |
| title_short | Q-conditional symmetry as a source of exact solutions of generalized Burgers equation |
| title_sort | q-conditional symmetry as a source of exact solutions of generalized burgers equation |
| url | https://trim.imath.kiev.ua/index.php/trim/article/view/454 |
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