Poisson equation with one boundary condition at infinity Given the one-dimensional Poisson equation:
$${d^2\over dx^2}V(x)=\exp(-\beta V(x))$$
it's possible to find a solution given the initial conditions. But, if we have these boundary conditions: 
$$V(0)=V_0$$
$$\lim_{x\to\infty}V(x)=0$$
can someone help me to find an analytical solution if any?
For the sake of simplicity, you can consider the easier ODE:
$${d^2\over dx^2}V(x)=\exp(-V(x))$$
 A: $\dfrac{d^2V}{dx^2}=e^{-\beta V}$
$2\dfrac{dV}{dx}\dfrac{d^2V}{dx^2}=2e^{-\beta V}\dfrac{dV}{dx}$
$\int2\dfrac{dV}{dx}\dfrac{d^2V}{dx^2}dx=\int2e^{-\beta V}\dfrac{dV}{dx}dx$
$\int2\dfrac{dV}{dx}d\left(\dfrac{dV}{dx}\right)=\int2e^{-\beta V}~dV$
$\left(\dfrac{dV}{dx}\right)^2=\dfrac{2C_1^2}{\beta}-\dfrac{2e^{-\beta V}}{\beta}$
$\dfrac{dV}{dx}=\pm\sqrt{\dfrac{2}{\beta}}\sqrt{C_1^2-e^{-\beta V}}$
$\dfrac{dV}{\sqrt{C_1^2-e^{-\beta V}}}=\pm\sqrt{\dfrac{2}{\beta}}dx$
$\int\dfrac{dV}{\sqrt{C_1^2-e^{-\beta V}}}=\int\pm\sqrt{\dfrac{2}{\beta}}dx$
For $\int\dfrac{dV}{\sqrt{C_1^2-e^{-\beta V}}}$ ,
Let $u=e^{-\beta V}$ ,
Then $V=\ln u$
$dV=\dfrac{du}{u}$
$\therefore\int\dfrac{dV}{\sqrt{C_1^2-e^{-\beta V}}}=\int\dfrac{du}{u\sqrt{C_1^2-u}}=C-\dfrac{2}{C_1}\tanh^{-1}\dfrac{\sqrt{C_1^2-u}}{C_1}=C-\dfrac{2}{C_1}\tanh^{-1}\dfrac{\sqrt{C_1^2-e^{-\beta V}}}{C_1}$
Hence $-\dfrac{2}{C_1}\tanh^{-1}\dfrac{\sqrt{C_1^2-e^{-\beta V}}}{C_1}=\pm\sqrt{\dfrac{2}{\beta}}x+c$
$\tanh^{-1}\dfrac{\sqrt{C_1^2-e^{-\beta V}}}{C_1}=\mp\left(\dfrac{C_1x}{\sqrt{2\beta}}+C_2\right)$
$\dfrac{\sqrt{C_1^2-e^{-\beta V}}}{C_1}=\mp\tanh\left(\dfrac{C_1x}{\sqrt{2\beta}}+C_2\right)$
$\sqrt{C_1^2-e^{-\beta V}}=\mp~C_1\tanh\left(\dfrac{C_1x}{\sqrt{2\beta}}+C_2\right)$
$C_1^2-e^{-\beta V}=C_1^2\tanh^2\left(\dfrac{C_1x}{\sqrt{2\beta}}+C_2\right)$
$e^{-\beta V}=C_1^2\text{sech}^2\left(\dfrac{C_1x}{\sqrt{2\beta}}+C_2\right)$
$V=-\dfrac{1}{\beta}\ln\left(C_1^2\text{sech}^2\left(\dfrac{C_1x}{\sqrt{2\beta}}+C_2\right)\right)$
$V(0)=V_0$ :
$-\dfrac{1}{\beta}\ln\left(C_1^2\text{sech}^2C_2\right)=V_0$
$C_1^2\text{sech}^2C_2=e^{-\beta V_0}$
$C_1^2=e^{-\beta V_0}\cosh^2C_2$
$C_1=\pm~e^{-\frac{\beta V_0}{2}}\cosh C_2$
$\therefore V=-\dfrac{1}{\beta}\ln\left(e^{-\beta V_0}\cosh^2C_2~\text{sech}^2\left(\dfrac{\pm~xe^{-\frac{\beta V_0}{2}}\cosh C_2}{\sqrt{2\beta}}+C_2\right)\right)$
difficult to find $C_2$ when $V(\infty)=0$
