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let $y(x)$ be a continuous function $0\le y(x) \le1$ over the support $x_\min\le x \le x_\max$, for $x_\min < x_\max$ strictly positive values.

let $y'(x)$ be the first order derivative of $y(x)$.

let $y(x)^k$ be the $k$-th power of the function $y(x)$.

Solve the Cauchy problem:

$$a(x)y'(x) = b(x)y(x)^{-3} - cy(x)^{-2} - y(x)^{-1}$$

$$y(x_0) = y_0$$

where $a(x)=a_1 + a_2x$ with $a_1>0, a_2<0$ and $b(x)=b_1 + b_2x$ with $b_1<0, b_2>0$

and $x_0$, $y_0$, $c$ are positive constant real values.

I am interested in a solution such that $y'(x)<0$. I proved existence and uniqueness, but I am looking for the analytic expression of $y(x)$.

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I guess this won't help, but what do you mean by "particular"? –  Maesumi Jul 11 '13 at 13:46
    
the particular solution, so out of the general solution, the only trajectory that satisfies the boundary condition –  user85884 Jul 11 '13 at 14:02
    
I can prove the existence and uniqueness of a solution y(x) such that y'(x)>0. But I want to find the analytical expression for the solution (if any) –  user85884 Jul 11 '13 at 14:04
    
Nonlinear equation with two generic functions involved... I would not expect any explicit solutions, unless $a$ and $b$ are known functions, and pretty special at that. –  40 votes Jul 11 '13 at 23:41

2 Answers 2

up vote 0 down vote accepted

$(a_1+a_2x)y'(x)=(b_1+b_2x)y(x)^{-3}-cy(x)^{-2}-y(x)^{-1}$

$(a_1+a_2x)\dfrac{dy}{dx}=\dfrac{b_1+b_2x}{y^3}-\dfrac{c}{y^2}-\dfrac{1}{y}$

$\left(\dfrac{b_2x}{y^3}-\dfrac{1}{y}-\dfrac{c}{y^2}+\dfrac{b_1}{y^3}\right)\dfrac{dx}{dy}=a_2x+a_1$

Let $u=x-\dfrac{y^2}{b_2}-\dfrac{cy}{b_2}+\dfrac{b_1}{b_2}$ ,

Then $x=u+\dfrac{y^2}{b_2}+\dfrac{cy}{b_2}-\dfrac{b_1}{b_2}$

$\dfrac{dx}{dy}=\dfrac{du}{dy}+\dfrac{2y}{b_2}+\dfrac{c}{b_2}$

$\therefore\dfrac{b_2u}{y^3}\left(\dfrac{du}{dy}+\dfrac{2y}{b_2}+\dfrac{c}{b_2}\right)=a_2\left(u+\dfrac{y^2}{b_2}+\dfrac{cy}{b_2}-\dfrac{b_1}{b_2}\right)+a_1$

$\dfrac{b_2u}{y^3}\dfrac{du}{dy}+\dfrac{2}{y^2}+\dfrac{c}{y^3}=a_2u+\dfrac{a_2y^2}{b_2}+\dfrac{a_2cy}{b_2}+a_1-\dfrac{a_2b_1}{b_2}$

$\dfrac{b_2u}{y^3}\dfrac{du}{dy}=a_2u+\dfrac{a_2y^2}{b_2}+\dfrac{a_2cy}{b_2}+\dfrac{a_1b_2-a_2b_1}{b_2}-\dfrac{2}{y^2}-\dfrac{c}{y^3}$

$\dfrac{du}{dy}=\dfrac{a_2y^3u}{b_2}+\dfrac{a_2y^5}{b_2^2}+\dfrac{a_2cy^4}{b_2^2}+\dfrac{(a_1b_2-a_2b_1)y^3}{b_2^2}-\dfrac{2y}{b_2}-\dfrac{c}{b_2}$

This belongs to an Abel equation of the second kind.

In fact all Abel equation of the second kind can be transformed into Abel equation of the first kind.

Let $u=\dfrac{1}{v}$ ,

Then $\dfrac{du}{dy}=-\dfrac{1}{v^2}\dfrac{dv}{dy}$

$\therefore-\dfrac{1}{v^3}\dfrac{dv}{dy}=\dfrac{a_2y^3}{b_2v}+\dfrac{a_2y^5}{b_2^2}+\dfrac{a_2cy^4}{b_2^2}+\dfrac{(a_1b_2-a_2b_1)y^3}{b_2^2}-\dfrac{2y}{b_2}-\dfrac{c}{b_2}$

$\dfrac{dv}{dy}=-\left(\dfrac{a_2y^5}{b_2^2}+\dfrac{a_2cy^4}{b_2^2}+\dfrac{(a_1b_2-a_2b_1)y^3}{b_2^2}-\dfrac{2y}{b_2}-\dfrac{c}{b_2}\right)v^3-\dfrac{a_2y^3v^2}{b_2}$

Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2

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My academic interest are PDEs and ODEs. Personally, I'd like to say that there would be explicit solutions for such nonlinear ODE in the first place.

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This does not answer the question. It is more appropriate as a comment. –  Daryl Jul 13 '13 at 10:21
    
thanks, this does not answer the question tough and I am still looking for –  user85884 Jul 13 '13 at 11:40

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