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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:

$${d^2\over dx^2}V(x)=\exp(-V(x))$$

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Something's wrong --- If $V(x)\to0$ then the equation implies $V''(x)\to1$ as $x\to\infty$. But this isn't possible. –  Bob Pego Feb 10 '12 at 21:42
    
There would exist $a>0$ such that $V''(x)>1/2$ for all $x>a$. Now $V'(x)-V'(a)=\int_a^xV''(t)\,dt>(x-a)/2$, so $V'(x)>1$ for $x>a+2(1+V'(a))$. After one more integration you find $V(x)\to\infty$ as $x\to\infty$, not 0. –  Bob Pego Feb 10 '12 at 22:14
    
@Bob Pego: I see. So the boundary conditions are not compatible with the equation given. –  Riccardo.Alestra Feb 11 '12 at 9:09

1 Answer 1

up vote 1 down vote accepted

$\dfrac{d^2V}{dx^2}=e^{-\beta V}$

$2\dfrac{dV}{dx}\dfrac{d^2V}{dx^2}=2e^{-\beta V}\dfrac{dV}{dx}$

$\int2\dfrac{dV}{dx}\dfrac{d^2V}{dx^2}dx=\int2e^{-\beta V}\dfrac{dV}{dx}dx$

$\int2\dfrac{dV}{dx}d\left(\dfrac{dV}{dx}\right)=\int2e^{-\beta V}~dV$

$\left(\dfrac{dV}{dx}\right)^2=\dfrac{2C_1^2}{\beta}-\dfrac{2e^{-\beta V}}{\beta}$

$\dfrac{dV}{dx}=\pm\sqrt{\dfrac{2}{\beta}}\sqrt{C_1^2-e^{-\beta V}}$

$\dfrac{dV}{\sqrt{C_1^2-e^{-\beta V}}}=\pm\sqrt{\dfrac{2}{\beta}}dx$

$\int\dfrac{dV}{\sqrt{C_1^2-e^{-\beta V}}}=\int\pm\sqrt{\dfrac{2}{\beta}}dx$

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 $V(\infty)=0$

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