Results 1 to 11 of 11

Math Help - Differential Equations, etc. - Know em?

  1. #1
    Newbie
    Joined
    Dec 2006
    Posts
    9

    Differential Equations, etc. - Know em?

    Two problems...

    P1. Logistic differential equation:

    dx/dt = 5x - 0.1(x^2)

    a) Describe dynamics as t goes to infinity?

    b) Write the finite difference approximation to the equation used in the Euler method?

    c) For what step size values does the diff. approximation approach a fixed point as n goes to infinity? What is the fixed point?

    d) For what step size values does the diff. approximation exhibit oscillations as n goes to infinity?

    tyvm
    Last edited by 12_bladez; May 8th 2007 at 04:03 PM. Reason: ...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by 12_bladez View Post
    Two problems...

    P1. Logistic differential equation:

    dx/dt = 5x - 0.1(x^2)

    a) Describe dynamics as t goes to infinity?
    If x(0)<0 x(t) -> -infty as t -> infty

    If x(0)>0 x(t) is increasing if x(t)<50, and decreasing if x(t)>50, and if fact
    x(t) -> 50 as t -> infty, though a bit more work is needed to prove this.

    b) Write the finite difference approximation to the equation used in the Euler method?
    approximate dx/dt by [x(t+h)-x(t)]/h, so our finite difference approximation
    is:

    x(t+h) = x(t) + h*[5*x(t) - 0.1(x^2(t))]


    c) For what step size values does the diff. approximation approach a fixed point as n goes to infinity? What is the fixed point?

    d) For what step size values does the diff. approximation exhibit oscillations as n goes to infinity?

    P2. Derive the following two formulas for approximating the third derivative and find their error terms; which of the two is more accurate?

    a) f'''(x) = [1/(h^3)][f(x+3h) - 3f(x+2h) + 3f(x+h) - f(x)]

    b) f'''(x) = [1/(2h^3)][f(x+2h) - 2f(x+h) + 2f(x-h) - f(x-2h)]

    tyvm
    Last edited by CaptainBlack; May 8th 2007 at 11:10 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Dec 2006
    Posts
    9
    ty CaptainBlack

    Would you could you elaborate a bit more on 1a? Don't get 1c or 1d either huh? Thanks again.

    P.S. Isn't 1b suppose to be:

    x(t+h) = x(t) + h(5x(t) - 0.1((x(t))^2))

    ???
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by 12_bladez View Post

    P.S. Isn't 1b suppose to be:

    x(t+h) = x(t) + h(5x(t) - 0.1((x(t))^2))

    ???
    Yes, clumsy editing on my part!

    RonL
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by 12_bladez View Post
    ty CaptainBlack

    Would you could you elaborate a bit more on 1a? Don't get 1c or 1d either huh? Thanks again.
    ???
    Quote Originally Posted by 12_bladez View Post
    Two problems...

    P1. Logistic differential equation:

    dx/dt = 5x - 0.1(x^2)

    a) Describe dynamics as t goes to infinity?
    If x(0)<0 x(t) -> -infty as t -> infty

    If x(0)>0 x(t) is increasing if x(t)<50, and decreasing if x(t)>50, and if fact
    x(t) -> 50 as t -> infty, though a bit more work is needed to prove this.
    Clearly if x is ever negative the derivative is negative, and gets more
    negative the more negative x becomes, so there is run away growth
    in the absolute value of x, and the growth is in the direction of increasingly
    more negative x. So as time goes to infty x goes to -infty.

    If x is ever positive but <50 dx/dt is positive so x is growing, but it must
    level off where dx/dt=0, which is at x=50. So if x starts with a value in
    the interval (0,50) it grows closer and closer to 50, so as t->infty
    x->infty.

    If x is ever positive but >50 dx/dt is negative so x is declining, but it must
    level off where dx/dt=0, which is at x=50. So if x starts with a value in
    the >50 it grows closer and closer to 50, so as t->infty
    x->infty.

    If x(0)=50, then dx/dt=0, so x remains at a value of 50. However any small
    disturbance away from x=50 will decay back towards x=50, so x=50 is a
    stable equilibrium.

    If x(0)=0, then dx/dt=0, so x remains at a value of 0. However any small
    disturbance away from x=0 will grow away from x=0, so x=0 is an
    unstable equilibrium.

    RonL
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Dec 2006
    Posts
    9
    Quote Originally Posted by CaptainBlack View Post
    Clearly if x is ever negative the derivative is negative, and gets more
    negative the more negative x becomes, so there is run away growth
    in the absolute value of x, and the growth is in the direction of increasingly
    more negative x. So as time goes to infty x goes to -infty.

    If x is ever positive but <50 dx/dt is positive so x is growing, but it must
    level off where dx/dt=0, which is at x=50. So if x starts with a value in
    the interval (0,50) it grows closer and closer to 50, so as t->infty
    x->infty.

    If x is ever positive but >50 dx/dt is negative so x is declining, but it must
    level off where dx/dt=0, which is at x=50. So if x starts with a value in
    the >50 it grows closer and closer to 50, so as t->infty
    x->infty.

    If x(0)=50, then dx/dt=0, so x remains at a value of 50. However any small
    disturbance away from x=50 will decay back towards x=50, so x=50 is a
    stable equilibrium.

    If x(0)=0, then dx/dt=0, so x remains at a value of 0. However any small
    disturbance away from x=0 will grow away from x=0, so x=0 is an
    unstable equilibrium.

    RonL
    Whenever you say "x" , do you mean x(0), x(t) or just plain x?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    approximate dx/dt by [x(t+h)-x(t)]/h, so our finite difference approximation
    is:

    x(t+h) = x(t) + h*[5*x(t) - 0.1(x^2(t))]


    c) For what step size values does the diff. approximation approach a fixed point as n goes to infinity? What is the fixed point?
    If this approaches a fixed point as t->infty, then [5*x(t) - 0.1(x^2(t))] ->0
    so the fixed point is either 0, or 50.

    Consider the case where x is close to 0.

    x(t+h) = x(t) (1+5h) - 0.1x^2(t)

    but as x is small:

    x(t+h) ~= x(t) (1+5h),

    so:

    |x(t+h)| ~= |x(t)| (1+5h)

    but 1+5h>1, so 0 is not a fixed point of the finite difference equation.

    Now let x(t)=50+epsilon, then

    x(t+epsilon) = (50 + epsilon) + h[5*(50 + epsilon) - 0.1*(50 + epsilon)^2]

    ................ = [- epsilon*h*(epsilon + 50)/10] + epsilon + 50

    ................ ~= epsilon*(1-5 h) +50.

    so this goes to 50 if |1-5h|<1, or h<0.4 (assuming h>0).

    RonL
    Last edited by CaptainBlack; May 8th 2007 at 08:30 PM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Dec 2006
    Posts
    9
    Did u get 1d too by any chance CaptainBlack? Thanks again for a, b and c.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by 12_bladez View Post
    Two problems...

    P1. Logistic differential equation:

    dx/dt = 5x - 0.1(x^2)

    a) Describe dynamics as t goes to infinity?

    b) Write the finite difference approximation to the equation used in the Euler method?

    c) For what step size values does the diff. approximation approach a fixed point as n goes to infinity? What is the fixed point?

    d) For what step size values does the diff. approximation exhibit oscillations as n goes to infinity?

    tyvm
    Quote Originally Posted by CaptainBlack View Post
    Now let x(t)=50+epsilon, then

    x(t+h) = (50 + epsilon) + h[5*(50 + epsilon) - 0.1*(50 + epsilon)^2]

    ................ = [- epsilon*h*(epsilon + 50)/10] + epsilon + 50

    ................ ~= epsilon*(1-5 h) +50.

    so this goes to 50 if |1-5h|<1, or h<0.4 (assuming h>0).

    RonL
    With x=50+epsilon with epsilon small we have:

    x(t+h) ~= epsilon*(1-5 h) +50.

    So x(t+h) ~= 50 - epsilon, when h=0.4, so the solution is oscillitary.

    For larger h the sequence of x's alternates in sign but grown is magintute.

    (note correction to the last line of the previous post)


    RonL
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Dec 2006
    Posts
    9
    So that's not the answer to 1d then? You're just correcting your 1c?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by 12_bladez View Post
    So that's not the answer to 1d then? You're just correcting your 1c?
    No that last post was the answer to 1d.

    I was just pointing out that I had corrected 1c, as if you are like me you
    don't necessarily reread the earlier posts.

    What I did was: changed the smileys back to what they should be
    and corrected the solution of:

    |1-5h|<1

    from h<0.2 to h<0.4, for h>0
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. 'Differential' in differential equations
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: October 5th 2010, 10:20 AM
  2. Replies: 2
    Last Post: May 18th 2009, 03:49 AM
  3. Differential Equations
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 12th 2009, 05:44 PM
  4. Replies: 5
    Last Post: July 16th 2007, 04:55 AM
  5. Replies: 3
    Last Post: July 9th 2007, 05:30 PM

Search Tags


/mathhelpforum @mathhelpforum