ordinary differential equation: $(f'(z))^2 = c\,f(z)^3 + f(z)^2$

Question

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!

Answer 1

The equation is not too hard to solve.

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

Answer 2

There is this famous differential equation ... [\wp'(z)]^2 = 4[\wp(z)]^3 - g_2\wp(z) - g_3 Weierstrass Elliptic Functions

Answer 3

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.