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I have to solve the following differential equation: $$\frac{d^2\phi(x)}{dx^2}=\phi(x)^2$$ The solution is given by:

$$\phi(x)=6\wp(x+c_1,0,c_2)$$ where $c_1,c_2$ are constants. I have the following boundary conditions: $$\phi(-L)=\phi(L)=\phi_0$$ How can I solve the previous ODE taking into account the conditions at the boundaries? Thank you in advance.

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up vote 1 down vote accepted

$$\phi''= \phi^2 $$ $$\phi'\phi''= \phi'\phi^2 $$

$$\int \phi'\phi'' dx=\int \phi^2 \phi'dx$$

$$\frac{\phi'^{2}}{2} = \frac{\phi^3}{3} +k $$

$$ \phi'^{2} = \frac{2 \phi^3}{3} +2k = \frac{2 \phi^3}{3} +c_1 $$

$$\phi' = \sqrt{\frac{2 \phi^3}{3} +c_1} $$

$$\frac{\phi'}{\sqrt{\frac{2 \phi^3}{3} +c_1}} =1 $$ $$\int \frac{\phi'}{\sqrt{\frac{2 \phi^3}{3} +c_1}} dx=\int dx$$

$$\int \frac{1}{\sqrt{\frac{2 \phi^3}{3} +c_1}} d\phi=x+c_2$$

I asked the integral to the wolfram and it includes elliptic integral of the first kind . The solution is here from wolfram integrator. Sorry it seems that there is no easy expression of the solution for that ODE.

You will need to take the solution from wolfram and to put the boundary conditions into that complex equation that wolfram offered and then you need to find $c_1,c_2$. If there is a simplier expression of the solution I would like to learn it.

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