# Finding out the stability of a complicated matricial differential equation about its fixed points

Suppose you have an equation ${{d\bf{A}}\over{dx}}= A B^\dagger A - B$,
where A,B are $n\times n$ unitary matrices ;
B is defined by $B_{ij}= -\frac{1}{2}\alpha_i\delta_{ij}A_{ij}$.

The expression isn't particularly important but note that the RHS is pretty complicated - B being a projection of A of sorts.

The fixed points of this equation are determined by equating the RHS to a zero matrix. I need to -

1. find out the fixed points to this system; for the $3\times 3$ case all of these may be guessed, except for one.
2. calculate the stabilities around these fixed points.
EDIT: by "calculating stabilities about a fixed point" i mean the following -

(i) Obtain the fixed point(s) $A^*$ such that ${{d\bf{A}}\over{dx}}\vert_{A=A^*} = f(A^*)= 0$

(ii) if $A^*\mapsto A^*+\epsilon A_1$, the RHS changes, upto the first order in $\epsilon$ to a slightly larger function easily got at by the product rule of differentiation (note that the differential of $B$ is still the projection of $A_1$). Now one has to diagonalise this expression, to obtain choices of $A_1$ that satisfy the form $${{d\epsilon}\over{dx}} = \mu \epsilon \vert_{\mbox{along A_1}}$$ Note that any unitary transformation on A renders the concise relation between A and B moot.

Please suggest analytical approaches to tackle this problem.

EDIT: The fixed points of the equation ${{d\bf{A}}\over{dx}}= A B^\dagger A - A$ with the $\alpha_i$s set to one - are -
(i) The identity element of GL3 - $$\left(\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right)$$ (ii-iv) Cyclic Permutations of - $$\left(\begin{array}{rrr} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right)$$ (v-vi) And the two cases - $\left(\begin{array}{rrr} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right)$ and $\left(\begin{array}{rrr} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array}\right)$

(vii) Define $a\equiv \frac{1}{\sum_i \alpha_i^{-1}}$. The fixed point is - $$A_{ij} = -\frac{a}{\alpha_i}\delta_{ij} + \sqrt{( 1 - \frac{a}{\alpha_i})( 1 - \frac{a}{\alpha_j})}(1-\delta_{ij})$$ Hope this helps! Code for the matrices are here.

If you think you can help with a numerical method for this, visit here.

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This doesn't quite make sense to me: if $A$ starts unitary, it won't remain unitary unless $\frac{d}{dx} (A^\dagger A) = 0$, and it doesn't look to me like that is the case here. – Robert Israel Apr 12 '12 at 8:08
@RobertIsrael thats completely right. thus the product of any valid $A^*$ & $A_1$ has to be anti-hermitian. The fact that $U^\dagger U = 1$ is supposed to be a constraint on the evolution of the matrix $A$ with x, thus. – Debanjan Basu Apr 12 '12 at 11:03
Another problem: unless I am mistaken, the unique fixed point is the diagonal matrix with entries $1/\alpha_i$ on the diagonal but your post alludes to multiple fixed points (that is, unless your $\delta$s are not Dirac deltas...). – Did Apr 12 '12 at 11:38
No! of course not!! I will explain that part to edit the question as well. The html formatting isnt working down here. – Debanjan Basu Apr 12 '12 at 16:17
$$\left(\begin{array}{rrr} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \end{array}\right)$$ is not a unitary matrix. – Robert Israel Apr 12 '12 at 19:51