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
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
The DE is...
$\displaystyle u^{''} = a + b u^{3} - u = f(u)$ (1)
... so that it contains only $\displaystyle u^{''}$ and $\displaystyle u$. The approach for solving such type of equation is the substitution...
$\displaystyle u^{''} = \frac{d u^{'}}{dx}= \frac{d u^{'}}{du} \frac{du}{dx} = u^{'} \frac{du{'}}{du} $ (2)
... so that (1) becomes...
$\displaystyle u^{'} du^{'} = (a + b u^{3} - u) du $ (3)
The (3) is a separated variables first order ODE the solution of which is...
$\displaystyle 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...
$\displaystyle 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
$\displaystyle \chi$ $\displaystyle \sigma$
Integrals of the type...
$\displaystyle \int f[x,\sqrt{p(x)}]\cdot dx$
... where $\displaystyle f(*)$ is a rational function and $\displaystyle p(*)$ 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
$\displaystyle \chi$ $\displaystyle \sigma$
Hey guys. That's really an interesting problem in transformation theory and involves the solution of the general elliptic equation $\displaystyle y''=A+By+Cy^2+Dy^3$. The following is adapted from "Introduction to Nonlinear Differential and Integral Equations" by H. Davis.
We wish to get the DE into the form:
$\displaystyle \left(\frac{dz}{dx}\right)^2=(1-z^2)(1-k^2 z^2)=\Delta^2(z)$
which has the solution $\displaystyle z=\text{sn}(x,k)$ where $\displaystyle \text{sn}$ is the Jacobi elliptic function. We can do that by writing $\displaystyle y''=A-y+by^3$ in it's elliptic form $\displaystyle \left(y'\right)^2=h^2(y-\alpha)(y-\beta)(y-\gamma)(y-\delta)=h^2\Delta_2^2(y)$ (multiply the general elliptic equation by y' and integrate) and then using the transformations:
$\displaystyle z^2=\frac{\beta-\delta}{\alpha-\delta}\frac{y-\alpha}{y-\beta}$
$\displaystyle k^2=\frac{(\beta-\gamma)(\alpha-\delta)}{(\alpha-\gamma)(\beta-\delta)}$
$\displaystyle M^2=\frac{(\beta-\delta)(\alpha-\gamma)}{4}$
and here is the challenging part. Show:
$\displaystyle \frac{1}{\Delta(z)}\frac{dz}{dx}=\frac{M}{\Delta_2 (y)}\frac{dy}{dx}$
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:
$\displaystyle z=\text{sn}(hMx,k)$
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.
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