Solution 1 to problem l2o35
Expressions |
Parameters |
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Expressions
The solution is given through the following expressions:
b17=0
a2*b5
b16=-------
6*a1
2
- a2 *b5
b15=-----------
6*a1*a3
b14=0
- 2*a2*b5
b13=------------
a3
- 2*a2*b5
b12=------------
a3
- a2*b5
b11=----------
a3
b10=0
- 12*a1*b5
b9=-------------
5*a3
a2*b5
b8=-------
3*a1
a2*b5
b7=-------
6*a1
b6=b5
2
a2 *b5
b4=---------
2*a1*a3
2*a2*b5
b3=---------
a3
a2*b5
b2=-------
a3
3*a1*b5
b1=---------
5*a3
a6= - a2
a5=0
a4= - 2*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,b5
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
3 2
u =(---*u *a1 *b5 + u *a1*a2*b5*u + 2*u *u *a1*a2*b5
s 5 5x 3x 2x x
1 1 2 2
+ ---*u *f(v,v)*a2*a3*b5 + ---*u *a2 *b5*u + f(v ,v )*a1*a3*b5
6 x 2 x x 2x
1
+ f(v,v )*a1*a3*b5 + ---*f(v,v )*a2*a3*b5*u)/(a1*a3)
3x 3 x
v = - 2*v *a1 - v *a2*u
t 3x x
12 2
v =( - u *v *a1*a2*b5 - 2*u *v *a1*a2*b5 - ----*v *a1 *b5
s 2x x x 2x 5 5x
1 1 2 2
- 2*v *a1*a2*b5*u + ---*v *f(v,v)*a2*a3*b5 - ---*v *a2 *b5*u )/(a1*a3)
3x 6 x 6 x