# solution to non-linear difference equation

Did anyone ever come across the global solution for a non-linear difference equation that looks like this:

$y(t+1)=y(t)+a+b \sqrt{c y(t)+d}$.

The initial condition is $y(0)=y_0$, and a,b,c and d are real numbers.

Any help is more than welcome!

Thanks a lot in advance.

PS: Mathematica is able to find a solution this that looks like $\left\{\left\{y[t]\to a t^2+\text{y0}-\frac{2 t \sqrt{a (d+c \text{y0})}}{\sqrt{c}}\right\},\left\{y[t]\to a t^2+\text{y0}+\frac{2 t \sqrt{a (d+c \text{y0})}}{\sqrt{c}}\right\}\right\}.$ I could use this of course but I'd like to know whether it is possible to solve this analytically by hand. Or, basically, how to get Mathematica's solution. There should be some way of transforming the problem by substitution. But - how?

PPS: Harald Hanche-Olsen just noticed that Mathematica's solution only holds under a special case for $t=1$.

-
I rather suspect that there is no closed form solution. If I am right, you will need to work with the difference equation itself to find out whatever you need to know, such as asymptotic behaviour and other properties of the sequence. –  Harald Hanche-Olsen Oct 8 '12 at 11:06
Well, I can obviously linearize it, characterize the limit etc. But I am really searching for a way to get the global solution that Mathematica spits out. Which basically means that there exists this global solution and thus there should potentially exist a way to find it... –  I_S Oct 8 '12 at 11:10
Is it just me, or does the Mathematica solution only look right for $t=1$ when $2\sqrt{a/c}=b$? –  Harald Hanche-Olsen Oct 8 '12 at 11:37
That's odd but you are correct. The only assumptions I added in Mathematica were non-negativity ones to allow Mathematica to simplify the results. Hm - I am not really used to Mathematica giving `wrong' results. –  I_S Oct 8 '12 at 11:56
Stranger things have happened. All software has bugs. Mathematica is software. –  Harald Hanche-Olsen Oct 8 '12 at 12:27

## 1 Answer

You can try a solution on the form $y(t)=At^2+Bt+C$. Substituting that in the recurrence relation, you get $2At+A+B-a=b\sqrt{cAt^2+cBt+cC+d}$. Square this, collect similar powers of $t$, and compare coefficients to get three equations to determine $A$, $B$, $C$.

But then you need a fourth equation to satisfy the initial condition, which indicates that you cannot find a solution on this form in general. For special choices of the data it will be possible, however. Trying higher degree polynomials seems to lead nowhere.

-
Sorry, but could you be a bit more precise about "compare coefficients to get three equations to determine A, B, C"? Thanks! –  I_S Oct 9 '12 at 9:22
All I mean by this is applying the principle that, if $a_2t^2+a_1t+a_0=b_2t^2+b_1t+b_0$ for all $t$, then $a_k=b_k$ for $k=0$, $1$, $2$. –  Harald Hanche-Olsen Oct 9 '12 at 13:22
Thanks a lot for your help! kind regards. –  I_S Oct 16 '12 at 12:51