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Let u(x,t)=v(x-sig t), (here sig = sigma)
Then ut=-sig v', uxx=v''
and we obtain the equation:
v''+sig v'+f(v)=0, where f(v)=-2v3+3v2-v
we want v(s) to go to 1 as s tends to infinity, to go to zero as s tends to -infinity,
and we want that v' tends to zero as s tends to + or – infinity.
If we write the ODE as a system:
w'=-f(v) – sig w.
Then we want an equilibrium (also known as stationary) point at P=(1,0) which is the case since f(1)=0 and an equilibrium points at O=(0,0), which is the case since f(0) =0 and for the solution (v(s),w(s)) to go to 0 as s tends to -infinity and to go to P as s tends to +infinity.
Linear stability analysis shows that both P and O are saddle points for the linearization, which is the same for both and of the form [0 1; 1 -sig]: the determinant is negative, and this means one positive eigenvalue and one negative eigenvalue for each....