Solution 2 to problem l2o35
Expressions |
Parameters |
Relevance |
Back to problem l2o35
Expressions
The solution is given through the following expressions:
b17=2*b16
a2*b16
b15=--------
a3
2*a2*b16
b14=----------
a3
b13=0
b12=0
2*a1*b16
b11=----------
a3
2*a1*b16
b10=----------
a3
b9=0
b8=6*b16
b7=3*b16
6*a1*b16
b6=----------
a2
18*a1*b16
b5=-----------
a2
5*a2*b16
b4=----------
a3
20*a1*b16
b3=-----------
a3
10*a1*b16
b2=-----------
a3
2
6*a1 *b16
b1=-----------
a2*a3
1
a6=---*a2
3
1
a5=---*a2
3
a4=0
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
2
u =(6*u *a1 *b16 + 10*u *a1*a2*b16*u + 20*u *u *a1*a2*b16
s 5x 3x 2x x
2 2
+ 3*u *f(v,v)*a2*a3*b16 + 5*u *a2 *b16*u + 18*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
1 1
v =---*u *a2*v + ---*v *a2*u
t 3 x 3 x
v =(2*u *a1*b16*v + 2*u *v *a1*b16 + 2*u *a2*b16*u*v + v *f(v,v)*a3*b16
s 3x 2x x x x
2
+ v *a2*b16*u + 2*f(v,v )*a3*b16*v)/a3
x x