The derivation in the paper: $$ V=-\frac{\left( kT/e \right) \left( x+ct \right)}{ct},\ E=\frac{kT/\left( ec \right)}{t} $$
$$ \left\{ \begin{array}{l} \varepsilon _0\dfrac{\mathrm{d}^2V}{\mathrm{d}x^2}=n_0e\left[ \exp \left( \dfrac{eV}{kT} \right) -1 \right]&\text{for}\,\,x<0\ \ \ \left( Eq.1 \right)\\ \varepsilon _0\dfrac{\mathrm{d}^2V}{\mathrm{d}x^2}=n_0e\exp \left( \dfrac{eV}{kT} \right) &\text{for}\,\,x>0 \ \ \ \left( Eq.2 \right)\\ \end{array} \right. \ $$ Integration then gives: $$ \left\{ \begin{array}{l} \dfrac{1}{2}\varepsilon _0E^2=n_0kT\left[ \exp \left( \dfrac{eV}{kT} \right) -1-\dfrac{eV}{kT} \right] &\text{for}\,\,x<0\ \left( Eq.3 \right)\\ \dfrac{1}{2}\varepsilon _0E^2=n_0kT\exp \left( \dfrac{eV}{kT} \right) &\text{for}\,\,x>0\ \left( Eq.4 \right)\\ \end{array} \right. \ $$ where $E=-\dfrac{\mathrm{d}V}{\mathrm{d}x}$. The boundary conditions employed here are $\dfrac{\mathrm{d}V}{\mathrm{d}x}=0$, $V=0$ at $x=-\infty$ and $\dfrac{\mathrm{d}V}{\mathrm{d}x}=0$, $V=-\infty$ at $x=+\infty$.
The above content is the introduction and derivation in the paper. My aim is to derive Eq.3 and Eq.4 from Eq.1 and Eq.2.
But the result I derived is different from the paper, and I don't know where I am wrong. Taking Eq.2 to derive Eq.4 as an example, my derivation is as follows: $$ \varepsilon _0\frac{d^2V}{dx^2}=n_0e\exp \left( \frac{eV}{kT} \right) \Rightarrow \varepsilon _0E^2=n_0kT\exp \left( \frac{eV}{kT} \right) $$ It is obvious that the left side of the equation is missing by 1/2.
I am not sure how you did the derivation but here are the steps $$ \varepsilon _0\frac{d^2V}{dx^2}=n_0e\left[ \exp \left( \frac{eV}{kT} \right) -1 \right] \,\,\text{for}\,\,x<0 $$ we have $$ \varepsilon _0\frac{d^2V}{dx^2} \frac{dV}{dx} = n_0e\left[ \exp \left( \frac{eV}{kT} \right) -1 \right]\dfrac{dV}{dx} $$ where we multiply $V'$ so we can integrate once or $$ \varepsilon _0\frac{1}{2}\frac{d}{dx}V'^2 = n_0e\frac{d}{dx}\left[\frac{kT}{e}\exp \left( \frac{eV}{kT} \right) - V\right] $$ this leads to $$ \varepsilon _0\frac{1}{2}V'^2 = n_0e\left[\frac{kT}{e}\exp \left( \frac{eV}{kT} \right) - V\right] + C $$ where $C$ is the integration constant we can re-write as $$ \varepsilon _0\frac{1}{2}E^2 = n_0e\left[\frac{kT}{e}\exp \left( \frac{eV}{kT} \right) - V\right] + C $$ using the boundary conditions $E =0, V=0$ this means $$ 0 = n_0e\left[\frac{kT}{e}\cdot 1 - 0\right] + C $$ or $$ C = -n_0e\frac{kT}{e} $$ this leads to $$ \varepsilon _0\frac{1}{2}E^2 = n_0e\left[\frac{kT}{e}\exp \left( \frac{eV}{kT} \right) - V\right] -n_0e\frac{kT}{e} = n_0e\left[\frac{kT}{e}\exp \left( \frac{eV}{kT} \right) - \frac{kT}{e} - V\right] $$ we finally arrive at $$ \varepsilon _0\frac{1}{2}E^2= n_0kT\left[\exp \left( \frac{eV}{kT} \right) - 1 - \frac{e}{kT}V\right] $$ e.g. Eq. 3