Analytical Solutions for the Beam Trajectory

For the nondimensionalized electron beam equations. we want to solve:

$$\begin{array}{c}\frac{d^2}{d{\xi }^2}\left(\begin{array}{c}{\Lambda }_x\\ {\Lambda }_z\end{array}\right)=\left(\begin{array}{cc}0& -\sin {\Lambda }_z\\ \sin {\Lambda }_z& 0\end{array}\right)\left(\begin{array}{c}{\dot{\Lambda }}_x\\ {\dot{\Lambda }}_z\end{array}\right)\end{array}$$

Assume that $|{\dot{\Lambda }}_x|\gg |{\dot{\Lambda }}_z|$ and ${\dot{\Lambda }}_z={\Omega }_{0z}$, where ${\Omega }_{0z}$ is a constant, such that the equations reduce to:

$$\begin{array}{c}\frac{d^2}{d{\xi }^2}\left(\begin{array}{c}{\Lambda }_x\\ {\Lambda }_z\end{array}\right)=\left(\begin{array}{cc}0& -\sin {\Lambda }_z\\ \sin {\Lambda }_z& 0\end{array}\right)\left(\begin{array}{c}0\\ {\Omega }_{0z}\end{array}\right)\end{array}$$

The initial conditions are:

$$\begin{array}{c}{\Lambda }_x(0)={\Lambda }_z(0)=0,{\dot{\Lambda }}_x(0)={\Omega }_{0z},{\dot{\Lambda }}_z(0)={\Omega }_{0z}\end{array}$$ $$\begin{array}{c}\frac{d^2}{d{\xi }^2}\left(\begin{array}{c}{\Lambda }_x\\ {\Lambda }_z\end{array}\right)=\left(\begin{array}{cc}0& -\sin {\Lambda }_z\\ \sin {\Lambda }_z& 0\end{array}\right)\left(\begin{array}{c}0\\ {\Omega }_{0z}\end{array}\right)=\left(\begin{array}{c}-{\Omega }_{0z}\sin {\Lambda }_z\\ 0\end{array}\right)\end{array}$$ $$\begin{array}{rl}& {\ddot{\Lambda }}_z=0\\ & {\Lambda }_z=A\xi +B\end{array}$$

where $A$ and $B$ are arbitrary constants

$$\begin{array}{c}{\Lambda }_z(0)=B=0\end{array}$$ $$\begin{array}{c}{\dot{\Lambda }}_z(0)=A={\Omega }_{0z}\end{array}$$

Therefore, $\boxed{{\Lambda }_z(\xi )={\Omega }_{0z}\xi }$

$$\begin{array}{c}{\ddot{\Lambda }}_x=-{\Omega }_{0z}\sin {\Lambda }_z=-{\Omega }_{0z}\sin ({\Omega }_{0z}\xi )\end{array}$$ $$\begin{array}{c}{\dot{\Lambda }}_x=\cos ({\Omega }_{0z}\xi )+A\end{array}$$ \begin{equation*} \Lambda_x = \frac{1}{\Omega_{0z}} \sin(\Omega_{0z} \xi) + A \xi + B \end{equation*} $$ $$ \begin{equation*} \dot \Lambda_x (0) = 1 + A = \Omega_{0z} \to A= \Omega_{0z}-1 \end{equation*} $$\begin{array}{c}{\Lambda }_x(0)=B=0\end{array}$$

Therefore, $\boxed{{\Lambda }_x(\xi )=\frac{1}{{\Omega }_{0z}}\sin ({\Omega }_{0z}\xi )+({\Omega }_{0z}-1)\xi }$