finding general solution (difficult differential equation)

**polo** · May 9th 2010, 11:24 PM

Hi, all

Can someone tell me what the general solution is for this differential equation, if any?

u(x)''= A + b*u(x)^3 - u(x)

To me there seems to be no general solution for it!

Thanks in advance,

Polo

**chisigma** · May 10th 2010, 12:29 AM

The DE is...

$u^{''} = a + b u^{3} - u = f(u)$ (1)

... so that it contains only $u^{''}$ and $u$ . The approach for solving such type of equation is the substitution...

$u^{''} = \frac{d u^{'}}{dx}= \frac{d u^{'}}{du} \frac{du}{dx} = u^{'} \frac{du{'}}{du}$ (2)

... so that (1) becomes...

$u^{'} du^{'} = (a + b u^{3} - u) du$ (3)

The (3) is a separated variables first order ODE the solution of which is...

$u^{'2} = 2 (a u + \frac{b}{4} u^{4} - \frac{u^{2}}{2}) + c_{1} \rightarrow u^{'} = \pm \sqrt {\frac{b}{2} u^{4} - u^{2} + 2 a u + c_{1}}$ (4)

The (4) is a separable variables first order ODE again the solution of which is...

$x = \pm \int \frac{du}{\sqrt {\frac{b}{2} u^{4} - u^{2} + 2 a u + c_{1}}} + c_{2}$ (5)

Now the solution of the integral in (5) requires a little more effort

...

Kind regards

$\chi$ $\sigma$

**polo** · May 10th 2010, 01:03 AM

Hi,

It’s exactly with this part I have problems with.
How do you solve such a beast (equation 5)?

Is there any solution for it?

Thanks in advance

Polo

**chisigma** · May 10th 2010, 01:39 AM

Integrals of the type...

$\int f[x,\sqrt{p(x)}]\cdot dx$

... where $f(*)$ is a rational function and $p(*)$ a polynomial od degree 3 or 4 are called elliptic integrals and their solution is even today not very confortable. See here...

Elliptic Integral -- from Wolfram MathWorld

Kind regards

$\chi$ $\sigma$

**shawsend** · May 10th 2010, 05:45 AM

Hey guys. That's really an interesting problem in transformation theory and involves the solution of the general elliptic equation $y''=A+By+Cy^2+Dy^3$ . The following is adapted from "Introduction to Nonlinear Differential and Integral Equations" by H. Davis.

We wish to get the DE into the form:

$\left(\frac{dz}{dx}\right)^2=(1-z^2)(1-k^2 z^2)=\Delta^2(z)$

which has the solution $z=\text{sn}(x,k)$ where $\text{sn}$ is the Jacobi elliptic function. We can do that by writing $y''=A-y+by^3$ in it's elliptic form $\left(y'\right)^2=h^2(y-\alpha)(y-\beta)(y-\gamma)(y-\delta)=h^2\Delta_2^2(y)$ (multiply the general elliptic equation by y' and integrate) and then using the transformations:

$z^2=\frac{\beta-\delta}{\alpha-\delta}\frac{y-\alpha}{y-\beta}$

$k^2=\frac{(\beta-\gamma)(\alpha-\delta)}{(\alpha-\gamma)(\beta-\delta)}$

$M^2=\frac{(\beta-\delta)(\alpha-\gamma)}{4}$

and here is the challenging part. Show:

$\frac{1}{\Delta(z)}\frac{dz}{dx}=\frac{M}{\Delta_2 (y)}\frac{dy}{dx}$

That expression is a little easier to show if the factored elliptic equation is only a cubic and involving simpler transformation rules and is the version I went through. I assume the quartic is just more messy. Anyway, it then follows:

$z=\text{sn}(hMx,k)$

and by inverting the expression for z above, the expression for y follows. Seems to me though the roots in general are complex which then means the solution is complex-analytic. I say give a good 45 minute presentation on this Polo and you don't have to take the final exam. Some nice plots too.

**polo** · May 10th 2010, 10:22 PM

Dear Shawsend,

Although that the factored elliptic equation in cubic form is for the moment more than I can handle. However, it seems doable.

But I have to admit that higher factored elliptical equation is just for the moment above my head.

Thank you very much for your help, I appreciate very much.

N.B.: the thank you bottom does not seem to function

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