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From: slawek on 25 Apr 2010 10:35

Uzytkownik "A N Niel" <anniel(a)nym.alias.net.invalid> napisal w wiadomosci
grup dyskusyjnych:250420100551037516%anniel(a)nym.alias.net.invalid...
> In article <4bd40edd$0$19181$65785112(a)news.neostrada.pl>, slawek
> <slawek(a)host.pl> wrote:
>
>> I am looking for solutions of this equation:
>>
>> y'[t] = y[t] - a y[t] y[t-1] y[t-2]
>>
>> It would be assumed that y[t] =0 for t < 0 .
>>
>> TIA
>> slawek
>>
>
> then y'[t]=y[t] for 0 <= t < 2 and since y[0]=0 we get y[t]=0
> for 0 <= t <= 2 . Continue in this way to get y[t]=0 for all t.

y[t] = 0 for t < 0, but y[0] > 0, therefore y[t] = y[0] Exp[t] for 0 <= t
<= 2

Your "ladder method" is still suitable, but not so easy as seems to be.

slawek


From: slawek on 25 Apr 2010 10:38

Uzytkownik "William Elliot" <marsh(a)rdrop.remove.com> napisal w wiadomosci
grup dyskusyjnych:20100425035622.G21838(a)agora.rdrop.com...
> On Sun, 25 Apr 2010, A N Niel wrote:
>> <slawek(a)host.pl> wrote:
>>
>>> I am looking for solutions of this equation:
>>>
>>> y'[t] = y[t] - a y[t] y[t-1] y[t-2]
>>>
>>> It would be assumed that y[t] =0 for t < 0 .
>>
>> then y'[t]=y[t] for 0 <= t < 2 and since y[0]=0 we get y[t]=0
>> for 0 <= t <= 2 . Continue in this way to get y[t]=0 for all t.
>>
> How do you get that y(0) = 0?

By an assumption. There is an arbitrary asumption that y[0] > 0 and that
y[0] is a "small value".


From: slawek on 25 Apr 2010 10:43

Uzytkownik "A N Niel" <anniel(a)nym.alias.net.invalid> napisal w wiadomosci
grup dyskusyjnych:250420100801543817%anniel(a)nym.alias.net.invalid...
> In article <20100425035622.G21838(a)agora.rdrop.com>, William Elliot
> <marsh(a)rdrop.remove.com> wrote:
>
>> On Sun, 25 Apr 2010, A N Niel wrote:
>> > <slawek(a)host.pl> wrote:
>> >
>> >> I am looking for solutions of this equation:
>> >>
>> >> y'[t] = y[t] - a y[t] y[t-1] y[t-2]
>> >>
>> >> It would be assumed that y[t] =0 for t < 0 .
>> >
>> > then y'[t]=y[t] for 0 <= t < 2 and since y[0]=0 we get y[t]=0
>> > for 0 <= t <= 2 . Continue in this way to get y[t]=0 for all t.
>> >
>> How do you get that y(0) = 0?
>
> y differentiable => y continuous

Not so simple. The equation "works" for t > 0. The time t = 0 is a "big
bang" time - not covered by the model. A kind of an fluctuation, some small
perturbance which start a grow up of y function.

slawek


From: William Elliot on 26 Apr 2010 05:29
On Sun, 25 Apr 2010, slawek wrote:
> Uzytkownik "A N Niel"
>> <marsh(a)rdrop.remove.com> wrote:
>>
>>> On Sun, 25 Apr 2010, A N Niel wrote:
>>> > <slawek(a)host.pl> wrote:
>>> >
>>> >> I am looking for solutions of this equation:
>>> >>
>>> >> y'[t] = y[t] - a y[t] y[t-1] y[t-2]
>>> >>
>>> >> It would be assumed that y[t] =0 for t < 0 .
>>> >
>>> > then y'[t]=y[t] for 0 <= t < 2 and since y[0]=0 we get y[t]=0
>>> > for 0 <= t <= 2 . Continue in this way to get y[t]=0 for all t.
>>> >
>>> How do you get that y(0) = 0?
>>
>> y differentiable => y continuous
>
> Not so simple. The equation "works" for t > 0. The time t = 0 is a "big bang"
> time - not covered by the model. A kind of an fluctuation, some small
> perturbance which start a grow up of y function.

Ok, so 0 is not in the domain of y, that is, y(0) is undefined.
Otherwise, you've only the trivial solution y(t) = 0, for all t,
which is the big fizzle.

See my other post to you about non-trivial solutions
when 0 is excluded from the domain of y.



>
> slawek
>
>
>
From: slawek on 26 Apr 2010 13:10
The equation y'[t] = y[t] - y[t] y[t-1] y[t-2] has the solutions:

plots

http://drop.io/8cnw583

or (in TEX)

\left\{x\to \text{Function}\left[\{t\},\begin{cases}
e^t & t<0 \\
e^{t+\frac{\alpha }{2 e^3}-\frac{1}{2} e^{-3+2 t} \alpha } & 0\leq t<1 \\
e^{-e^2+e^{2+\frac{\alpha }{2 e^3}-\frac{1}{2} e^{-5+2 t}
\alpha }+t+\frac{\alpha }{2 e^3}-\frac{\alpha }{2 e}} & 1\leq t<2 \\
e^{-e^2+e^{2-e^2+e^{2+\frac{\alpha }{2 e^3}-\frac{1}{2} e^{-7+2 t}
\alpha }+\frac{\alpha }{2 e^3}-\frac{\alpha }{2 e}}+t+\frac{\alpha }{2
e^3}-\frac{\alpha }{2 e}} & 2\leq t<3 \\
e^{3-e^2+e^{2-e^2+e^{2+\frac{\alpha }{2 e^3}-\frac{\alpha }{2
e}}+\frac{\alpha }{2 e^3}-\frac{\alpha }{2 e}}+\frac{\alpha }{2
e^3}-\frac{\alpha }{2 e}-\int_1^3 e^{-3-2 e^2-\frac{\alpha }{e}}
\left(e^{3+2 e^2+\frac{\alpha }{e}}-e^{e^{2-e^2+e^{2+\frac{\alpha }{2
e^3}-\frac{1}{2} e^{-9+2 K[1]} \alpha }+\frac{\alpha }{2
e^3}-\frac{\alpha }{2 e}}+e^{2+\frac{\alpha }{2 e^3}-\frac{1}{2} e^{-9+2
K[1]} \alpha }+\frac{\alpha }{e^3}+2 K[1]} \alpha \right) \, dK[1]+\int_1^t
e^{-3-2 e^2-\frac{\alpha }{e}} \left(e^{3+2
e^2+\frac{\alpha }{e}}-e^{e^{2-e^2+e^{2+\frac{\alpha }{2 e^3}-\frac{1}{2}
e^{-9+2 K[1]} \alpha }+\frac{\alpha }{2 e^3}-\frac{\alpha }{2
e}}+e^{2+\frac{\alpha }{2 e^3}-\frac{1}{2} e^{-9+2 K[1]}
\alpha }+\frac{\alpha }{e^3}+2 K[1]} \alpha \right) \, dK[1]} & 3\leq t\leq
4 \\
\text{Indeterminate} & \text{True}
\end{cases}
\right]\right\}


slawek


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