Autonomous ODE


by williamrand1
Tags: autonomous
williamrand1
williamrand1 is offline
#1
Jan23-13, 02:35 PM
P: 21
Hi everyone,

Im looking for an autonomous first order ode that has the following properties.

For dependent variable x:

x(t=∞)=0

x(t=-∞)=0

and the function x(t) has one maximum.

Any help would be great.

Rgds...
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pasmith
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#2
Jan23-13, 04:30 PM
HW Helper
P: 774
This is impossible.

Suppose the maximum is at [itex]t = t_0[/itex]. Then there exist [itex]t_1 < t_0 < t_2[/itex] such that [itex]x(t_1) = x(t_2)[/itex], but [itex]\dot x(t_1) = -\dot x(t_2)[/itex]. There is no way to express that requirement in an autonomous first order ODE.

You are going to need a second-order autonomous ODE, as should be obvious from the fact that you want to satisfy two boundary conditions.
JJacquelin
JJacquelin is offline
#3
Jan24-13, 01:06 AM
P: 744
What do you think of y' = - y^(3/2) ?

williamrand1
williamrand1 is offline
#4
Jan24-13, 11:16 AM
P: 21

Autonomous ODE


Quote Quote by pasmith View Post
This is impossible.

Suppose the maximum is at [itex]t = t_0[/itex]. Then there exist [itex]t_1 < t_0 < t_2[/itex] such that [itex]x(t_1) = x(t_2)[/itex], but [itex]\dot x(t_1) = -\dot x(t_2)[/itex]. There is no way to express that requirement in an autonomous first order ODE.

You are going to need a second-order autonomous ODE, as should be obvious from the fact that you want to satisfy two boundary conditions.
Thanks pasmith

Could you explain why it is not possible?
williamrand1
williamrand1 is offline
#5
Jan24-13, 11:18 AM
P: 21
Quote Quote by JJacquelin View Post
What do you think of y' = - y^(3/2) ?
Thanks JJ

Is there an exact solution to this?
JJacquelin
JJacquelin is offline
#6
Jan24-13, 02:54 PM
P: 744
dy/dx = -y^(3/2)
dx = - dy/y^(3/2)
x = (2 / y^(1/2)) +C
y^(1/2) = 2/(x-C)
y = 4/(x-C)
Mute
Mute is offline
#7
Jan24-13, 10:40 PM
HW Helper
P: 1,391
Quote Quote by JJacquelin View Post
dy/dx = -y^(3/2)
dx = - dy/y^(3/2)
x = (2 / y^(1/2)) +C
y^(1/2) = 2/(x-C)
y = 4/(x-C)
That has a divergence, not a maximum, though! I'm not sure that's what williamrand1 is looking for.

williamrand1, what about trying to take a function that you know has the properties you desire, differentiate it, and then see if you can rewrite the derivative in terms of x(t), with no explicit time dependence?
JJacquelin
JJacquelin is offline
#8
Jan25-13, 12:38 AM
P: 744
Hi williamrand1 !

Then, what about this one :
y' = -2y*sqrt(ln(1/y))
which solution is : y = exp(-(x+c))


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