Solution 3 to problem l2o35
Expressions |
Parameters |
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Back to problem l2o35
Expressions
The solution is given through the following expressions:
b17=2*b16
3*a2*b16
b15=----------
a3
6*a2*b16
b14=----------
a3
6*a1*b16
b13=----------
a3
12*a1*b16
b12=-----------
a3
12*a1*b16
b11=-----------
a3
6*a1*b16
b10=----------
a3
18 2
----*a1 *b16
5
b9=--------------
a2*a3
b8=6*b16
b7=3*b16
6*a1*b16
b6=----------
a2
12*a1*b16
b5=-----------
a2
3*a2*b16
b4=----------
a3
12*a1*b16
b3=-----------
a3
6*a1*b16
b2=----------
a3
18 2
----*a1 *b16
5
b1=--------------
a2*a3
a6=a2
a5=a2
a4=a1
Parameters
Apart from the condition that they must not vanish to give a
non-trivial solution, the following parameters are free:
a1,a2,a3,b16
Relevance for the application:
The solution given above tells us that the system {u_s, v_s}
is a 5'th order symmetry for the 3'rd order system {u_t,v_t}
where u=u(t,x) is a scalar function, v=v(t,x) is a vector
function of arbitrary dimension and f(..,..) is the scalar
product between two such vectors:
u =u *a1 + u *a2*u + f(v,v )*a3
t 3x x x
18 2
u =(----*u *a1 *b16 + 6*u *a1*a2*b16*u + 12*u *u *a1*a2*b16
s 5 5x 3x 2x x
2 2
+ 3*u *f(v,v)*a2*a3*b16 + 3*u *a2 *b16*u + 12*f(v ,v )*a1*a3*b16
x x x 2x
+ 6*f(v,v )*a1*a3*b16 + 6*f(v,v )*a2*a3*b16*u)/(a2*a3)
3x x
v =u *a2*v + v *a1 + v *a2*u
t x 3x x
v =(6*u *a1*a2*b16*v + 12*u *v *a1*a2*b16 + 12*u *v *a1*a2*b16
s 3x 2x x x 2x
2 18 2
+ 6*u *a2 *b16*u*v + ----*v *a1 *b16 + 6*v *a1*a2*b16*u
x 5 5x 3x
2 2
+ v *f(v,v)*a2*a3*b16 + 3*v *a2 *b16*u + 2*f(v,v )*a2*a3*b16*v)/(a2*a3)
x x x