Results 1 to 7 of 7

Math Help - Perturbation question

  1. #1
    Newbie
    Joined
    Aug 2009
    Posts
    21

    Perturbation question

    I'm trying to solve this problem but having some difficulty, could somebody please help me out?

    y" + (1+ey)y = 0. Subject to y(0)=1, y'(0)=0

    I've used the expansion y(x)=y0(x) + ey1(x) + ...

    when i differentiate this twice and substitute into my eqn and collect powers of e i get two equations:

    y0" + y0 = 0
    y1" + y1 = 0

    I think this is wrong and not really sure what to do.

    Thanks for any help (and sorry about the poor layout of the problem)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    The DE has the form...

    y^{''} = f(y) (1)

    ... and its solution is...

     x= \int \frac{dy}{\sqrt {2 \phi(y) + c_{1}}} + c_{2} (2)

    ... where...

     \phi(y) = \int f(y)\cdot dy (3)

    Here we have...

     f(y)= - y\cdot (1+e^{y}) (4)

    ... so that is...

     x= \int \frac{dy}{\sqrt{- y^{2} - 2y e^{y} +2 e^{y} + c_{1}}} + c_{2} (5)

    The integral in (5) has no elementary approach ...

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,387
    Thanks
    52
    Quote Originally Posted by bubblygirl90 View Post
    I'm trying to solve this problem but having some difficulty, could somebody please help me out?

    y" + (1+ey)y = 0. Subject to y(0)=1, y'(0)=0

    I've used the expansion y(x)=y0(x) + ey1(x) + ...

    when i differentiate this twice and substitute into my eqn and collect powers of e i get two equations:

    y0" + y0 = 0
    y1" + y1 = 0

    I think this is wrong and not really sure what to do.

    Thanks for any help (and sorry about the poor layout of the problem)
    If y'' + y + \varepsilon y^2 = 0, then substituting y = y_0 + \varepsilon y_1 + O\left(\varepsilon^2\right) gives

     <br />
y''_0 + \varepsilon y''_1 + y_0 + \varepsilon y_1 + \varepsilon \left(y_0 + \varepsilon y_1 \right)^2 + O\left(\varepsilon^2\right) = 0<br />

    or

     <br />
O(1)\;\;\;y''_0 + y_0 = 0<br />
     <br />
O(\varepsilon)\;\;\;y''_1 + y_1 + y^2_0 = 0<br />

    with ICs y_0(0) = 1, y'_0(0) = 0, y_1(0) = 0, y'_1(0) = 0, .

    The first one is easy - soln y_0 = \cos x. Use this in the second and solve for y_1.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Aug 2009
    Posts
    21
    Thank you Danny, that really helped me out a lot!
    I'm just having some trouble with this last bit now.

    If i have: y1" + y1 = -(cosx)^2

    then would the homogeneous solution be cos x? or A + B(e^x)?
    and also I'm completely unsure how to find the particular solution
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Aug 2009
    Posts
    21
    Quote Originally Posted by Danny View Post
    If y'' + y + \varepsilon y^2 = 0, then substituting y = y_0 + \varepsilon y_1 + O\left(\varepsilon^2\right) gives

     <br />
y''_0 + \varepsilon y''_1 + y_0 + \varepsilon y_1 + \varepsilon \left(y_0 + \varepsilon y_1 \right)^2 + O\left(\varepsilon^2\right) = 0<br />

    or

     <br />
O(1)\;\;\;y''_0 + y_0 = 0<br />
     <br />
O(\varepsilon)\;\;\;y''_1 + y_1 + y^2_0 = 0<br />

    with ICs y_0(0) = 1, y'_0(0) = 0, y_1(0) = 0, y'_1(0) = 0, .

    The first one is easy - soln y_0 = \cos x. Use this in the second and solve for y_1.
    Thank you so much ive got an answer which when i check it it works
    y = cos(x) + e( cos(x) - 1/2 - 1/6 cos(2x) )

    The only thing is i havent used the initial conditions anywhere other than working out the solution of y0 = cos(x).. do i need them for working out the value of epsilon?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,387
    Thanks
    52
    Quote Originally Posted by bubblygirl90 View Post
    Thank you so much ive got an answer which when i check it it works
    y = cos(x) + e( cos(x) - 1/2 - 1/6 cos(2x) )

    The only thing is i havent used the initial conditions anywhere other than working out the solution of y0 = cos(x).. do i need them for working out the value of epsilon?
    Your second solution should be

    y_1 = c_1 \cos x + c_2 \ sin x - \frac{1}{2} - \frac{1}{6} \cos 2x

    Here you'd use the IC's.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Sep 2009
    From
    Athens, Greece
    Posts
    20

    Question Another perturbation question

    I was going to open a new thread, but I'll post my question here! Does anybody know what singular perturbation methods we could apply and what will get for the bvp problem below?

    \varepsilon y''+(t-\frac{1}{2})y=0
    0<t<1
    0<\varepsilon<<1
    with boundary values
    y(0)=1
    y(1)=2
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Perturbation Problem
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: February 21st 2011, 07:29 AM
  2. Replies: 0
    Last Post: February 1st 2011, 01:54 PM
  3. Taylor Perturbation
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: January 24th 2011, 06:13 AM
  4. Perturbation Theory help
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: January 18th 2010, 05:28 AM
  5. Help for a Perturbation problem
    Posted in the Advanced Applied Math Forum
    Replies: 5
    Last Post: February 14th 2009, 04:22 PM

Search Tags


/mathhelpforum @mathhelpforum