Non-linear waves and solitons: verifying solution for Nonlinear Schrödinger Equation

inverse-problemsmatrix equationssoliton-theory

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!

This is the first page of the exercise
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Best Answer

You have a lot of moving pieces, here.

  1. Construct $U$ and $V$ with the given expressions, and $u_0(x,t)=e^{2it}.$ Note that $u_x=0$ and $\overline{u}_x=0.$ \begin{align*} U&=\left[\begin{matrix}0 &u\\ \overline{u} &0\end{matrix}\right]=\left[\begin{matrix}0 &e^{2it}\\ e^{-2it} &0\end{matrix}\right] \\ V&=\left[\begin{matrix}|u|^2 &iu_x \\ -i\,\overline{u}_x &-|u|^2\end{matrix}\right] =\left[\begin{matrix}1 &0 \\ 0 &-1\end{matrix}\right]. \end{align*}
  2. Next we build the AKNS operators $X$ and $T$ as follows: \begin{align*} X&=-i\lambda\sigma_3+iU=-i\lambda\left[\begin{matrix}1 &0\\0&-1\end{matrix}\right]+i\left[\begin{matrix}0 &e^{2it}\\ e^{-2it} &0\end{matrix}\right]=\left[\begin{matrix}-i\lambda &ie^{2it}\\ ie^{-2it} &i\lambda\end{matrix}\right] \\ T&=2\lambda X+iV =2\lambda \left[\begin{matrix}-i\lambda &ie^{2it}\\ ie^{-2it} &i\lambda\end{matrix}\right]+i\left[\begin{matrix}1 &0 \\ 0 &-1\end{matrix}\right] =\left[\begin{matrix}i-2\lambda^2i &2\lambda i e^{2it} \\ 2\lambda i e^{-2it} &-i+2\lambda^2 i\end{matrix}\right]. \end{align*}
  3. Moving on, we calculate $\Psi_0;$ but it has an annoying term $e^{it\sigma_3}$ in it. That is, we have to exponentiate a matrix. Fortunately, we do not have to go to the trouble of finding the eigenvalues and eigenvectors and exponentiating the result. Mathematica has the 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*}
  4. The only missing pieces to $(4)$ are $\Psi_x$ and $\Psi_t,$ which for our case, of course, are really $\dfrac{\partial}{\partial x}\Psi_0$ and $\dfrac{\partial}{\partial t}\Psi_0,$ respectively. I get \begin{align*} \dfrac{\partial}{\partial x}\Psi_0&= \frac{1}{\sqrt{2\mu(\mu+\lambda)}}\left[ \begin{matrix} i\mu e^{i[\mu(x+2\lambda t)+t]} &-(\mu+\lambda)(-i\mu)e^{-i[\mu(x+2\lambda t)-t]}\\ (\mu+\lambda)(i\mu)e^{i[\mu(x+2\lambda t)-t]} &-i\mu e^{-i[\mu(x+2\lambda t)+t]} \end{matrix}\right] \quad\text{and} \\ \dfrac{\partial}{\partial t}\Psi_0&= \frac{1}{\sqrt{2\mu(\mu+\lambda)}}\left[ \begin{matrix} (i(2\mu\lambda+1))e^{i[\mu(x+2\lambda t)+t]} &(\mu+\lambda)(i(2\mu\lambda-1))e^{-i[\mu(x+2\lambda t)-t]}\\ (\mu+\lambda)(i(2\mu\lambda-1))e^{i[\mu(x+2\lambda t)-t]} & -i(2\mu\lambda+1)e^{-i[\mu(x+2\lambda t)+t]}\end{matrix}\right]. \end{align*} To see that $\Psi_0$ satisfies \begin{align*} \Psi_x&=X\Psi \\ \Psi_t&=T\Psi \end{align*} should now be a matter of algebra.
  5. Speaking of algebra, we can note that if $\Psi\mapsto\alpha\Psi,$ whether $\Psi$ is a solution or not remains unchanged, because $\partial_x(\alpha\Psi)=\alpha\,\partial_x\Psi,$ and $X(\alpha\Psi)=\alpha X\Psi$, and similarly for the $t$ equation. Because M.SE has a rather narrow window for typing up equations, this has a practical value in that we can ignore constants out in front of $\Psi$. That is, we need to check that \begin{align*} &\frac{1}{\sqrt{2\mu(\mu+\lambda)}}\left[\begin{matrix} i\mu e^{i[\mu(x+2\lambda t)+t]} &-(\mu+\lambda)(-i\mu)e^{-i[\mu(x+2\lambda t)-t]}\\ (\mu+\lambda)(i\mu)e^{i[\mu(x+2\lambda t)-t]} &-i\mu e^{-i[\mu(x+2\lambda t)+t]}\end{matrix}\right] \\ =&\left[\begin{matrix}-i\lambda &ie^{2it}\\ ie^{-2it} &i\lambda\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*} or \begin{align*} &\left[ \begin{matrix} i\mu e^{i[\mu(x+2\lambda t)+t]} &-(\mu+\lambda)(-i\mu)e^{-i[\mu(x+2\lambda t)-t]}\\ (\mu+\lambda)(i\mu)e^{i[\mu(x+2\lambda t)-t]} &-i\mu e^{-i[\mu(x+2\lambda t)+t]} \end{matrix} \right] \\ =&\left[ \begin{matrix} -i\lambda &ie^{2it}\\ ie^{-2it} &i\lambda \end{matrix} \right] \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*} Performing the matrix multiplication on the RHS yields \begin{align*}&X\Psi_0= \\ &i\left[ \begin{matrix} -\lambda e^{i[\mu(x+2\lambda t)+t]}+(\mu+\lambda)e^{2it}e^{i[\mu(x+2\lambda t)-t]} &\lambda(\mu+\lambda)e^{-i[\mu(x+2\lambda t)-t]}+e^{2it}e^{-i[\mu(x+2\lambda t)+t]} \\ e^{-2it}e^{i[\mu(x+2\lambda t)+t]}+\lambda(\mu+\lambda)e^{i[\mu(x+2\lambda t)-t]} &-(\mu+\lambda)e^{-2it}e^{-i[\mu(x+2\lambda t)-t]}+\lambda e^{-i[\mu(x+2\lambda t)+t]} \end{matrix} \right] \end{align*} The $e^{2it}$ and $e^{-2it}$ have the effect of enabling the exponentials to be like terms, with the following spectacular simplification: \begin{align*}&X\Psi_0 \\ &=i\left[ \begin{matrix} -\lambda e^{i[\mu(x+2\lambda t)+t]}+(\mu+\lambda)e^{i[\mu(x+2\lambda t)+t]} &\lambda(\mu+\lambda)e^{-i[\mu(x+2\lambda t)-t]}+e^{-i[\mu(x+2\lambda t)-t]} \\ e^{i[\mu(x+2\lambda t)-t]}+\lambda(\mu+\lambda)e^{i[\mu(x+2\lambda t)-t]} &-(\mu+\lambda)e^{-i[\mu(x+2\lambda t)+t]}+\lambda e^{-i[\mu(x+2\lambda t)+t]} \end{matrix} \right] \\ &=i\left[ \begin{matrix} \mu e^{i[\mu(x+2\lambda t)+t]} &[\lambda(\mu+\lambda)+1]e^{-i[\mu(x+2\lambda t)-t]} \\ [\lambda(\mu+\lambda)+1]e^{i[\mu(x+2\lambda t)-t]} &-\mu e^{-i[\mu(x+2\lambda t)+t]} \end{matrix} \right] \\ &=i\left[ \begin{matrix} \mu e^{i[\mu(x+2\lambda t)+t]} &[\lambda\mu+\mu^2]e^{-i[\mu(x+2\lambda t)-t]} \\ [\lambda\mu+\mu^2]e^{i[\mu(x+2\lambda t)-t]} &-\mu e^{-i[\mu(x+2\lambda t)+t]} \end{matrix} \right] \\ &=i\mu\left[ \begin{matrix} e^{i[\mu(x+2\lambda t)+t]} &(\lambda+\mu)e^{-i[\mu(x+2\lambda t)-t]} \\ (\lambda+\mu)e^{i[\mu(x+2\lambda t)-t]} &-e^{-i[\mu(x+2\lambda t)+t]} \end{matrix} \right] \\ \end{align*} At this point, a bit of simplification of $\partial_x\Psi_0$ is in order. At last count, we had \begin{align*} &\phantom{=}\partial_x\Psi_0 \\ &=\left[ \begin{matrix} i\mu e^{i[\mu(x+2\lambda t)+t]} &-(\mu+\lambda)(-i\mu)e^{-i[\mu(x+2\lambda t)-t]}\\ (\mu+\lambda)(i\mu)e^{i[\mu(x+2\lambda t)-t]} &-i\mu e^{-i[\mu(x+2\lambda t)+t]} \end{matrix} \right] \\ &=\left[ \begin{matrix} i\mu e^{i[\mu(x+2\lambda t)+t]} &i\mu(\mu+\lambda)e^{-i[\mu(x+2\lambda t)-t]}\\ i\mu(\mu+\lambda)e^{i[\mu(x+2\lambda t)-t]} &-i\mu e^{-i[\mu(x+2\lambda t)+t]} \end{matrix} \right] \\ &=i\mu\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*} Now you can see that these are, indeed, equal. The $\partial_t \Psi_0=T\Psi_0$ case should work out similarly.
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