substitution in a non linear differential equation and to get a nicer form well I had this equation at the begining
$$
i \frac{\partial u}{\partial{z}} + \frac{1}{2 k_0} \frac{\partial^2 u}{\partial x^2} +\frac{1}{2}k_0 n_1 F(z) x^2 u-\frac{i[g(z) -\alpha(z)]}{2}u + k_0 n_2|u|^2 u = 0,
$$
If I substitute $X=x/w_0$, $Z=z/L_D$, $G=(g(z)-\alpha(z))L_D$, $U=u\sqrt{k_o n_2 L_D}$, $L_D=k_0w_0^2$ $w_0=(k_0^2n_1)^{-1/4}$
I am not getting this equation
$$
i\frac{\partial U}{\partial Z} + \frac{1}{2} \frac{\partial^2 U}{\partial X^2} + F(Z) \frac{X^2}{2} U - \frac{i}{2} G(Z) U + |U|^2 U = 0
$$
but getting 2nd term multiplied with $w_0$, could any one tell me am I right or wrong? I must say I have calculated it 3 times. please help. well replace $f(z)$ by $F(Z)$ and $g(z)$ by $G(Z)$
 A: Lets see. Taking the change of variables $X = \frac{x}{w_0}$, $Z = \frac{z}{L_D}$ and $U = \frac{u}{C}$, we have
\begin{align}
\frac{\partial}{\partial x} &= \frac{\partial X}{\partial x}\frac{\partial}{\partial X} = \frac{1}{w_0}\frac{\partial}{\partial X},\\
\frac{\partial^2}{\partial x^2} &= \frac{1}{w_0^2}\frac{\partial}{\partial X},\\
\frac{\partial}{\partial z} &= \frac{\partial Z}{\partial z}\frac{\partial}{\partial Z} = \frac{1}{L_D}\frac{\partial}{\partial Z}.
\end{align}
Substituting into the equation, 
\begin{multline}
\frac{i}{C L_D} \frac{\partial U}{\partial Z} + \frac{1}{2 C k_0 w_0^2} \frac{\partial^2 U}{\partial x^2} + \frac{w_0^2 k_0 n_1}{2 C} f(L_D Z) X^2 U \\ - \frac{i \big[g(L_D Z) - \alpha(L_D Z)\big]}{2 C} U + \frac{k_0 n_2}{|C|^2 C} |U|^2 U = 0.
\end{multline}
Multiplying by $C L_D$ the hole equation,
\begin{multline}
i \frac{\partial U}{\partial Z} + \frac{L_D}{2 k_0 w_0^2} \frac{\partial^2 U}{\partial x^2} + \frac{w_0^2 k_0 n_1 L_D}{2} f(L_D Z) X^2 U \\ - \frac{i L_D\big[g(L_D Z) - \alpha(L_D Z)\big]}{2} U + \frac{k_0 n_2 L_D}{|C|^2} |U|^2 U = 0.
\end{multline}
Taking $L_D = k_0 w_0^2$,
\begin{multline}
i \frac{\partial U}{\partial Z} + \frac{1}{2} \frac{\partial^2 U}{\partial x^2} + \frac{w_0^4 k_0^2 n_1}{2} f(k_0 w_0^2 Z) X^2 U \\ - \frac{i k_0 w_0^2\big[g(k_0 w_0^2 Z) - \alpha(k_0 w_0^2 Z)\big]}{2} U + \frac{k_0^2 w_0^2 n_2}{|C|^2} |U|^2 U = 0.
\end{multline}
Let $w_0 = (k_0^2 n_1)^{-1/4}$, then $L_D = \frac{1}{\sqrt{n_1}}$ and
\begin{multline}
i \frac{\partial U}{\partial Z} + \frac{1}{2} \frac{\partial^2 U}{\partial x^2} + \frac{1}{2} f(n_1^{-1/2} Z) X^2 U \\ - \frac{i \big[g(n_1^{-1/2} Z) - \alpha(n_1^{-1/2} Z)\big]}{2 \sqrt{n_1}} U + \frac{k_0 n_2}{\sqrt{n_1} |C|^2} |U|^2 U = 0.
\end{multline}
Taking $C = \frac{\sqrt{k_0 n_2}}{n_1^{1/4}} = \sqrt{k_0 n_2 L_D}$,
\begin{multline}
i \frac{\partial U}{\partial Z} + \frac{1}{2} \frac{\partial^2 U}{\partial x^2} + \frac{1}{2} f(n_1^{-1/2} Z) X^2 U \\ - \frac{i \big[g(n_1^{-1/2} Z) - \alpha(n_1^{-1/2} Z)\big]}{2 \sqrt{n_1}} U + |U|^2 U = 0.
\end{multline}
Denoting
\begin{align}
F(Z) &= f\left(\tfrac{Z}{\sqrt{n_1}}\right),\\
G(Z) &= \frac{1}{\sqrt{n_1}}\left[g\left(\tfrac{Z}{\sqrt{n_1}}\right) - \alpha\left(\tfrac{Z}{\sqrt{n_1}}\right)\right]
\end{align}
we have
$$
i \frac{\partial U}{\partial Z} + \frac{1}{2} \frac{\partial^2 U}{\partial x^2} + \frac{1}{2} F(Z) X^2 U - \frac{i}{2} G(Z) U + |U|^2 U = 0.
$$
What we have done here is written the original equation in adimensional form. Judging by the form of the equation, $L_D$ has [time] dimensions, $w_0$ has [space] dimesions, and $u$ has [space/time²] dimensions.
Please doublecheck the math.
