How to transform a coupled differential equation into a system with diagonal linear part

195 Views Asked by At

Consider the system given by $$iu_t +u_{xx}+2|u|^2u = -v+iu$$ $$iv_t +u_{xx}+2|v|^2v = -u-iu$$

I am trying to transform the system into a system with diagonal linear part. I can solve a problem like this if the system has only one derivative for example $$x_t = f(x,y)$$ $$y_t = g(x,y)$$ Normally I would would find the linearization matrix about about $(0,0)$ and proceed as in this question: Bring the linear part of a ODE system into a diagonal form

However what is confusing me with the system at hand is that the system has derivatives with respect to $x$ and $t$ and I have to deal with strange derivatives like of $|u|^2$.

Could someone provide some help. Thanks.

1

There are 1 best solutions below

0
On

First, I assume that $u$ and $v$ are complex valued, otherwise you just have $|u|^2 = u^2$. For complex valued $u$ and $v$, you can write $u = a + b i$ and $v = c + d i$. In that case, there is no problem in taking derivatives of terms like $u |u|^2$.

Then, for the rest of the linearisation part, it might be insightful to write the system as \begin{equation} i\, \partial_t I \mathbf{w} + \partial_{xx} I \mathbf{w} + \begin{pmatrix} 2 u |u|^2 \\ 2 v |v|^2 \end{pmatrix} = \begin{pmatrix} i & -1 \\ -1 & -i \end{pmatrix} \mathbf{w}, \end{equation} where $\mathbf{w} = \begin{pmatrix} u \\ v \end{pmatrix}$ and $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ is the identity matrix.