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GERF
Symbols
GERFSolve
Taggar`GERF`
G
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▪
The following options can be given:
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▪
"OutputMode" takes the following values:
"
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s
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)
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▪
The balancing constant for wave transformation is calculated automatically for the given equation. However, if and when the package fails to calculate one, "BalanceConstant" be used as described above.
▪
G
E
R
F
S
o
l
v
e
accepts single nonlinear partial differential equations in integral or fractional order.
▪
For fractional order equations,
F
r
a
c
t
i
o
n
a
l
D
may be used and the package regards it as the conformal derivative.
▪
T
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η
=
x
1
x
1
+
x
2
x
2
+
.
.
.
+
x
n
x
n
+
t
t
.
C
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a
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a
b
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v
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.
▪
The coefficients in ansatz are taken to be
i
's for appropriate ranges of
i
.
▪
For systems of equations, the
m
th coefficient in the ansatz for
k
is taken as
k
m
.
▪
The number of unique equations and the number of unique dependent variables must be equal to each other.
Examples
(
1
1
)
Basic Examples
(
2
)
This is the Burgers' equation in
(
1
+
1
)
-dimensions:
I
n
[
1
]
:
=
b
u
r
g
e
r
s
=
∂
t
u
[
x
,
t
]
+
u
[
x
,
t
]
∂
x
u
[
x
,
t
]
-
ν
∂
x
,
x
u
[
x
,
t
]
0
O
u
t
[
1
]
=
(
0
,
1
)
u
[
x
,
t
]
+
u
[
x
,
t
]
(
1
,
0
)
u
[
x
,
t
]
-
ν
(
2
,
0
)
u
[
x
,
t
]
0
Solve it using GERF method:
I
n
[
2
]
:
=
G
E
R
F
S
o
l
v
e
[
b
u
r
g
e
r
s
,
u
[
x
,
t
]
]
O
u
t
[
2
]
=
u
[
x
,
t
]
0
+
2
ν
x
1
+
x
x
+
t
-
0
x
-
ν
2
x
Plot the solution:
I
n
[
3
]
:
=
P
l
o
t
3
D
0
+
1
1
+
x
x
+
1
2
t
-
2
0
x
-
1
x
/
.
{
0
2
,
1
-
1
,
x
-
1
}
,
{
x
,
-
5
,
5
}
,
{
t
,
-
5
,
5
}
O
u
t
[
3
]
=
Obtain the solution sets as well:
I
n
[
4
]
:
=
s
o
l
s
=
G
E
R
F
S
o
l
v
e
[
b
u
r
g
e
r
s
,
u
[
x
,
t
]
,
"
O
u
t
p
u
t
M
o
d
e
"
A
l
l
]
O
u
t
[
4
]
=
-
1
0
,
1
2
ν
x
,
t
-
0
x
-
ν
2
x
,
u
[
x
,
t
]
0
+
2
ν
x
1
+
x
x
+
t
-
0
x
-
ν
2
x
Verify that the solutions are valid:
I
n
[
5
]
:
=
p
i
c
k
=
s
o
l
s
〚
1
〛
;
b
u
r
g
e
r
s
/
.
u
F
u
n
c
t
i
o
n
[
{
x
,
t
}
,
p
i
c
k
〚
1
,
2
〛
]
O
u
t
[
5
]
=
T
r
u
e
Another example, Calogero–Bogoyavlenskii–Schiff equation in
(
2
+
1
)
-dimensions:
I
n
[
1
]
:
=
c
b
s
=
∂
x
,
t
u
[
x
,
y
,
t
]
+
4
∂
x
u
[
x
,
y
,
t
]
∂
x
,
y
u
[
x
,
y
,
t
]
+
2
∂
x
,
x
u
[
x
,
y
,
t
]
∂
y
u
[
x
,
y
,
t
]
+
∂
x
,
x
,
x
,
y
u
[
x
,
y
,
t
]
0
;
Solve it:
I
n
[
2
]
:
=
s
o
l
s
=
G
E
R
F
S
o
l
v
e
[
c
b
s
,
u
[
x
,
y
,
t
]
]
O
u
t
[
2
]
=
u
[
x
,
y
,
t
]
0
-
2
x
1
+
x
x
+
y
y
-
t
2
x
y
,
u
[
x
,
y
,
t
]
-
1
1
+
x
x
+
(
1
+
x
x
)
-
1
+
0
Verification:
I
n
[
3
]
:
=
T
a
b
l
e
[
c
b
s
/
.
u
F
u
n
c
t
i
o
n
[
{
x
,
y
,
t
}
,
p
i
c
k
〚
i
,
2
〛
]
,
{
i
,
L
e
n
g
t
h
@
s
o
l
s
}
]
O
u
t
[
3
]
=
{
T
r
u
e
,
T
r
u
e
}
S
c
o
p
e
(
4
)
O
p
t
i
o
n
s
(
3
)
A
p
p
l
i
c
a
t
i
o
n
s
(
2
)
"
"