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Does any iterative equation of n-th order which does not inclute derivatives of order higher than 1 have exactly n independent solutions?

Let's designate n-th iterate of a function $y(x)$ as $y^{[n]}(x)$

Is it true that the equation

$$F(y^{[n]},y^{[n-1]},...,y,x,y')=0$$

has exactly n independent solutions?

A method of solving some classes of iterative equations was proposed in this paper: http://faculty.kfupm.edu.sa/math/akca/papers/cheng.pdf

It gives exactly n indepentent solutions for each solved iterative equation of n-th order.

For example, the equation

$$y^{[2]}-y'=0$$

has exactly two solutions:

$y_1(x) = e^{\frac{\pi}{3} (-1)^{1/6}} x^{\frac{1}{2}+\frac{i \sqrt{3}}{2}}$

$y_2(x) = e^{\frac{\pi}{3} (-1)^{11/6}} x^{\frac{1}{2}+\frac{i \sqrt{3}}{2}}$

The question here is it the property of any iterative equation, not just of those which suitable for this method.

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Perhaps you should say what $F$ is allowed? –  Gerald Edgar Oct 17 '10 at 20:53
    
For the simplicity count it as a polynomial. –  Anixx Oct 17 '10 at 22:02
7  
What do you mean by an "independent solution"? The solutions don't seem to form a vector space, so I don't know what you mean. –  Dylan Thurston Oct 17 '10 at 22:11

1 Answer 1

Do you have to use $y'$? An involution is a solution to $y^{[2]}=x$. Pick a point $P=(k,k)$, draw a curve descending monotonically to $P$ and then reflect in the line $y=x$ to get the rest of the function. $y=x$, $y=-x+b$ (for arbitrary $b$) and $y=\frac{b}{x}$ are three examples. $y=\frac{ax+b}{cx-a}$ is another.

later As another example of higher degree consider $y^{[4]}=sin^{[4]}(x)$ then there are solutions such as having y be sin(x) for $x \ge -11$ but $\sin(x)+3\pi$ for $x<=-11$

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Ok, but what about higher orders? –  Anixx Oct 18 '10 at 7:29
    
And, by the way this equation has multiple solutions only because of special character of x function. If we take $y^{[2]}=\exp(x)$ there one should expect no more than only one or two solutions (it is really interesting, one or two though). –  Anixx Oct 18 '10 at 8:19
    
Well, $y^{[2k]}=x$ would have at least these solutions if not more. $y=x$ certainly is a special function, but what do you see as especially pertinent about it (Meaning... I have to think about it). –  Aaron Meyerowitz Oct 18 '10 at 12:10
    
It is special because $y\circ x$= y and $x \circ y = y$, the same role as with 0 in addotion and 1 in multiplication. This function is sometimes designated id. –  Anixx Oct 18 '10 at 12:27

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