Solution 4 to problem l2o35
Expressions |
Parameters |
Relevance |
Back to problem l2o35
Expressions
The solution is given through the following expressions:
1
---*a5*b6
2
b17=-----------
a4
1
---*a5*b6
2
b16=-----------
a4
2
2*a5 *b6
b15=----------
a3*a4
2
2*a5 *b6
b14=----------
a3*a4
3*a5*b6
b13=---------
a3
9
---*a5*b6
2
b12=-----------
a3
7
---*a5*b6
2
b11=-----------
a3
a5*b6
b10=-------
a3
9
----*a4*b6
10
b9=------------
a3
2*a5*b6
b8=---------
a4
a5*b6
b7=-------
a4
3
b5=---*b6
2
2
- 2*a5 *b6
b4=-------------
a3*a4
5
- ---*a5*b6
2
b3=--------------
a3
- a5*b6
b2=----------
a3
1
- ----*a4*b6
10
b1=---------------
a3
a6=2*a5
a2=0
a1=0
Parameters
Apart from the condition that they must not vanish to give a
non-trivial solution, the following parameters are free:
a3,a4,a5,b6
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 =f(v,v )*a3
t x
1 2 5
u =( - ----*u *a4 *b6 - u *a4*a5*b6*u - ---*u *u *a4*a5*b6
s 10 5x 3x 2 2x x
2 2 3
+ u *f(v,v)*a3*a5*b6 - 2*u *a5 *b6*u + ---*f(v ,v )*a3*a4*b6
x x 2 x 2x
+ f(v,v )*a3*a4*b6 + 2*f(v,v )*a3*a5*b6*u)/(a3*a4)
3x x
v =u *a5*v + v *a4 + 2*v *a5*u
t x 3x x
7 9
v =(u *a4*a5*b6*v + ---*u *v *a4*a5*b6 + ---*u *v *a4*a5*b6
s 3x 2 2x x 2 x 2x
2 9 2
+ 2*u *a5 *b6*u*v + ----*v *a4 *b6 + 3*v *a4*a5*b6*u
x 10 5x 3x
1 2 2 1
+ ---*v *f(v,v)*a3*a5*b6 + 2*v *a5 *b6*u + ---*f(v,v )*a3*a5*b6*v)/(a3*a4)
2 x x 2 x