finding general solution (difficult differential equation)

Feb 2010
8
0
Bucuresti
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

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Mar 2009
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near Piacenza (Italy)
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 (Thinking)...

Kind regards

\(\displaystyle \chi\) \(\displaystyle \sigma\)
 
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Feb 2010
8
0
Bucuresti
Hi,


It’s exactly with this part I have problems with.
[FONT=&quot]How do you solve [/FONT][FONT=&quot]such a beast ([/FONT][FONT=&quot]equation 5)?

Is there any solution for it?


Thanks in advance

Polo

[/FONT]
 

chisigma

MHF Hall of Honor
Mar 2009
2,162
994
near Piacenza (Italy)
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

\(\displaystyle \chi\) \(\displaystyle \sigma\)
 
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shawsend

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Aug 2008
903
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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. :)
 
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Feb 2010
8
0
Bucuresti
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