If the initial solution is the unstable background $(2),$ the corresponding fundamental solution of the Lax pair is \begin{align*}\Psi_0(\lambda)&=\frac{1}{\sqrt{2\mu(\mu+\lambda)}}\,e^{it\sigma_3}\left[\begin{matrix}e^{\Theta(\lambda)} &-(\mu+\lambda)e^{-\Theta(\lambda)}\\(\mu+\lambda)e^{\Theta(\lambda)}&e^{-\Theta(\lambda)}\end{matrix}\right],\qquad (36)\\ \Theta(\lambda)&\equiv i\mu(x+2\lambda t), \end{align*} where $\sigma_3=\operatorname{diag}(1, -1)$ is the Pauli matrix and $\mu, \lambda$ are complex parameters satisfying the constraint $$\mu^2=1+\lambda^2.\qquad (37)$$ EsRW 05. Verify that $(36)$ is the fundamental solution of $(4)$ corresponding to $(2),$ and satisfying $\det(\Psi_0)=1.$
I'm verifying the Nonlinear Schrödinger equation using the Zakharov-Shabat (ZS) Lax pair condition. I've almost finished but I can't solve this last thing.
The last equation has three matrices with the same determinant, so I thought to diagonalize them in order to find the solution, which is $\mu^2=1+\lambda^2.$
Could you help me to find that please?
Ok, I'm going to post the entire exercise, I hope I am clear this time! Thank for your help!
Best Answer
You have a lot of moving pieces, here.
MatrixExp
function that will do this for us. We get $$e^{it\sigma_3}=\left[\begin{matrix}e^{it} &0 \\ 0 &e^{-it}\end{matrix}\right]. $$ Hence, \begin{align*} \Psi_0&=\frac{1}{\sqrt{2\mu(\mu+\lambda)}}\left[\begin{matrix}e^{it} &0 \\ 0 &e^{-it}\end{matrix}\right]\left[\begin{matrix}e^{i\mu(x+2\lambda t)} &-(\mu+\lambda)e^{-i\mu(x+2\lambda t)}\\(\mu+\lambda)e^{i\mu(x+2\lambda t)}&e^{-i\mu(x+2\lambda t)}\end{matrix}\right] \\ &=\frac{1}{\sqrt{2\mu(\mu+\lambda)}}\left[\begin{matrix}e^{i[\mu(x+2\lambda t)+t]} &-(\mu+\lambda)e^{-i[\mu(x+2\lambda t)-t]}\\(\mu+\lambda)e^{i[\mu(x+2\lambda t)-t]}&e^{-i[\mu(x+2\lambda t)+t]}\end{matrix}\right]. \end{align*}