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I swear I have seen this type of ODE before, but I can't remember how to attack it. In general, I would like to know how to solve $$\left(f'(z)\right)^m = c\,G(z)^n$$ where $m,\;n \in \mathbb{N}$ and $G(z)$ is just a polynomial in $f(z)$.

This sounds hard, so I would be happy with, $$\left(f'(z)\right)^2 = c\,G(z)^n,$$ though this latter equation may be too difficult too. For my homework, though, I need to know, $$\left(f'(z)\right)^2 = \left( c\,f^3 + f^2 \right).$$ It was also suggested in the homework question to utilize $$g^2 = 3\,c-f.$$ If I do this without thinking I get an equation $$\left(f'(z)\right)^2 = c\,f(z)^2,$$ which seems much easier, though I am still a little rattled by the plus-minus.

In case it matters, this is a related to a method for solving the Korteweg-deVries equation, $$u_t + u\,u_x + u_{xxx} = 0.$$ I have seen some solutions (but did not understand them) where the polynomial was "factored" into 3 roots... something like that. I just don't want to know the answer, but how to get it. Please keep in mind that this is my first class in PDEs.

Thanks for any help!

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Yes, I was there too.... I was thinking for some reason there may be some complications due to the sign of the square root. Anyway, thanks! –  nate Feb 1 '12 at 0:30
    
"I have seen some solutions (but did not understand them) where the polynomial was "factored" into 3 roots" - and that is precisely the Weierstrass approach. –  J. M. Feb 8 '12 at 18:17
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3 Answers

up vote 2 down vote accepted

The equation is not too hard to solve.

$$\begin{align*} \left(\frac{dy}{dz}\right)^2 &= cy^3+y^2\\ \frac{dy}{y \sqrt{cy+1}} &= dz\\ cy+1 &= u\\ 2 \frac{du}{u^2-1} &= dz\\ - 2 \tanh^{-1} u + C&= z\\ u &= \tanh \frac{-z+C}{2}\\ \sqrt{cy+1} &= -\tanh \frac{z-C}{2}\\ cy+1 &= \tanh^2 \frac{z-C}{2}\\ y &= \frac{\tanh^2 \frac{z-C}{2}-1}{c}\\ y &= -\left({c\times \cosh^2 \frac{z-C}{2}}\right)^{-1} \end{align*}$$

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Aaahhh, I think that the $g^2 = 3\,c-f$$ must have been a hint towards a substitution in the integration. Turns out it was just confusing the way it was suggested in the problem. Now I see it as a straight-forward problem. Thanks all! –  nate Feb 1 '12 at 1:16
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There is this famous differential equation ... $$ [\wp'(z)]^2 = 4[\wp(z)]^3 - g_2\wp(z) - g_3 $$ Weierstrass Elliptic Functions

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Ahh. Thank you - I was also wondering how I was going to solve this pde if I wasn't able to get rid of 2 BC's - would lead to a third-order polynomial in p(z) just like you mentioned. Thanks! –  nate Feb 1 '12 at 23:31
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Here's how to complete GEdgar's Weierstrass solution. Start with the differential equation

$$(f^\prime (z))^2 = c\,f(z)^3 + f(z)^2$$

and introduce a new function $g(z)$ satisfying the relation

$$f(z)=\frac4{c}g(z)-\frac1{3c}$$

(This is essentially equivalent to "depressing" a cubic equation plus a rescaling.)

We can then derive a differential equation for $g(z)$ through this substitution:

$$(g^\prime (z))^2=4g(z)^3-\frac1{12}g(z)+\frac1{216}$$

and comparing this with the Weierstrass differential equation, we find that

$$g(z)=\wp\left(z;\frac1{12},-\frac1{216}\right)$$

We then compute the discriminant

$$\Delta=\left(\frac1{12}\right)^3-27\left(-\frac1{216}\right)^2=0$$

and find that the underlying cubic is degenerate. Abramowitz and Stegun give appropriate formulae for the case of $\Delta=0$, $g_2 > 0$, and $g_3 < 0$ (see formula 18.12.3); applying the formula listed there yields

$$\wp\left(z;\frac1{12},-\frac1{216}\right)=\frac1{12}+\frac14\mathrm{csch}^2\left(\frac{z}{2}\right)$$

and thus

$$f(z)=\frac4{c}\left(\frac1{12}+\frac14\mathrm{csch}^2\left(\frac{z}{2}\right)\right)-\frac1{3c}=\frac1{c}\mathrm{csch}^2\left(\frac{z}{2}\right)$$

Note that I only derived a particular solution; the modifications necessary for a general solution (i.e., taking the constant(s) of integration into account) is left as an exercise to the reader.

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