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[SOLVED] Verify sol exist discrete

dwsmith

Well-known member
Feb 1, 2012
1,673
Verify that an exact solution exist for the logistic difference equation
$$
u_{t+1}=ru_t(1-u_t),\quad r>0
$$
in the form $u_t=A\sin^2(\alpha^t)$ by determining values of r, A and alpha. Is the solution periodic? Oscillatory?

I have yet to encounter a problem that says verify a solution exist. What do they want me to do?
 

chisigma

Well-known member
Feb 13, 2012
1,704
First step is writing the recursive relation as...

$\displaystyle \Delta_{n}= u_{n+1}-u_{n}= (r-1)\ u_{n} -r\ u^{2}_{n}=f(u_{n})\ ,\ r>0$ (1)

Second step is investigating the [qualitative...] behavior of the solution supposing r>1. The function $\displaystyle f(x)=(r-1)\ x- r\ x^{2}$ has one 'repulsive fixed point' [a point $x_{0}$ where is $f(x_{0})=0$ and $f^{'}(x_{0})>0$...] in $x_{-}=0$ and one 'attractive fixed point' [a point $x_{0}$ where is $f(x_{0})=0$ and $f^{'}(x_{0})<0$...] in $x_{+}=1-\frac{1}{r}$. An interesting property of f(*) is that for r>1, no matter what is r, is $f(0)=0$ and $f(1)=-1$. The reader can understand better what follows observing the annexed figure, where f(x) for r=2, r=3 and r=4 are reported. Now we examine different situations…

a) $1<r\le 2$. In this case the sequence will converge monotonically increasing [without oscillations...} at $x_{+}=1-\frac{1}{r}$. As example the case $r=2\ ,\ x_{+}= \frac{1}{2}\ ,\ u_{0}=.1$ is reported...

http://www.wolframalpha.com/input/?i=g(0)=.1+,+g(n+1)=2+g(n)-2+g(n)^2

b) $2<r\le 3$. In this case the sequence will converge at $x_{+}=1-\frac{1}{r}$ 'with oscillations'. As example the case $r=3\ ,\ x_{+}= \frac{2}{3}\ ,\ u_{0}=.1$ is reported...

http://www.wolframalpha.com/input/?i=g(0)=.1+,+g(n+1)=3+g(n)-3+g(n)^2

c) $3<r\le 4$. In this case the 'attractive fixed point' in general cannot be 'achieved' and, with very particular exceptions, the sequence, even if bounded, will diverge. As example the case $r=4\ ,\ x_{+}= \frac{3}{4}\ ,\ u_{0}=.1$ is reported...

http://www.wolframalpha.com/input/?i=g(0)=.1+,+g(n+1)=4+g(n)-4+g(n)^2

d) $r>4$. In this case, with very particular exceptions, any $u_{0}$ will produce a sequence diverging to $- \infty$...

In next posts I will try to investigate about the possibility to achieve some explicit expressions of the solutions of (1)...

Kind regards

$\chi$ $\sigma$

MHB03.PNG
 
Last edited:

chisigma

Well-known member
Feb 13, 2012
1,704
The 'logistic recursive relation'...

$\displaystyle u_{n+1}=r\ u_{n}\ (1-u_{n})$ (1)

... has closed form solutions only for a limited set of values of r. Some years ago 'Monster Wolfram' postulated that an exact solution should have the form...

$\displaystyle u_{n}= \frac{1}{2}\ \{1-f[r^{n}\ f^{-1} (1-2\ u_{0})]\}$ (2)

... and after some time M.Trott and R.Germundsson demonstrated that such a solution exists only for r=-2, r=2 and r=4. For r=4 is $f(x)=\cos x$ so that is...

$\displaystyle u_{n}= \frac{1}{2}\ \{1-\cos [2^{n}\ \cos^{-1} (1-2\ u_{0})]\}= \sin^{2} [2^{n-1}\ \cos^{-1} (1-2\ u_{0})]$ (3)

Kind regards

$\chi$ $\sigma$
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Ok so I can get all the ranges for r's at each respective steady state. That is just solving the logistic equation as is. How is $A\sin^2\alpha^t$ used in this problem. I don't understand what to do with it.
 

chisigma

Well-known member
Feb 13, 2012
1,704
Ok so I can get all the ranges for r's at each respective steady state. That is just solving the logistic equation as is. How is $A\sin^2\alpha^t$ used in this problem. I don't understand what to do with it.
If it is required a solution in the form...

$\displaystyle u_{n}= A\ \sin^{2} \alpha^{n}$ (1)

... with A, r , $\alpha$ and $u_{0}$ constants I'm afraid that the only possibility is $u_{0}=0$ , $A=0$ , with $\alpha$ and $r$ arbitrary. Are You sure that a solution in the form (1) is required?...

Kind regards

$\chi$ $\sigma$
 

dwsmith

Well-known member
Feb 1, 2012
1,673
If it is required a solution in the form...

$\displaystyle u_{n}= A\ \sin^{2} \alpha^{n}$ (1)

... with A, r , $\alpha$ and $u_{0}$ constants I'm afraid that the only possibility is $u_{0}=0$ , $A=0$ , with $\alpha$ and $r$ arbitrary. Are You sure that a solution in the form (1) is required?...

Kind regards

$\chi$ $\sigma$
Yup that is what it says verbatim
 

chisigma

Well-known member
Feb 13, 2012
1,704

dwsmith

Well-known member
Feb 1, 2012
1,673
Can you supply the 'original source', please?...

Kind regards

$\chi$ $\sigma$
J.D. Murray Into to Math Bio 1 ch. 2 question 3

If you google the book, there is a pdf file of it in the first page of google choices
 

chisigma

Well-known member
Feb 13, 2012
1,704
I found the J.D. Murray's book and at the page 76 is written...

3. Verify that an exact solution exists for the logistic difference equation...

$\displaystyle u_{n+1}=r\ u_{n}\ (1-u_{n})\ ,\ r>0$ (1)

... in the form $u_{n}=A\ \sin^{2} \alpha^{n}$ by determining the values for r, A and $\alpha$. Is the solution (i) periodic?(ii) oscillatory? Describe it! If $r>4$ discuss possible solution implications.

In my opinion the author did commit a slip and he intended to write '... in the form $u_{n}=A\ \sin^{2} (a\ \alpha^{n})$' and in this case the solution is...

$\displaystyle u_{n}= \sin^{2} [2^{n-1}\ \cos^{-1} (1-2 u_{0})] \implies r=4\ ,\ A=1\ ,\ \alpha=2\ ,\ a=\frac{\cos^{-1} (1-2 u_{0})}{2}$ (1)

In order to verify that see...

http://mathworld.wolfram.com/LogisticMap.html

It is remarkable the fact that in general the (1) is non periodic and, because is $0<u_{n}<1$, the (1) is an excellent random number generator with 'key' $u_{0}$...

Kind regards

$\chi$ $\sigma$