Results 1 to 2 of 2

Math Help - Quick-ish DE question

  1. #1
    Newbie
    Joined
    Aug 2009
    Posts
    20

    Quick-ish DE question

    Hey guys,

    I'm doing some numerical lin.alg work that requires me to work by hand. I'm having trouble with the following question:-

    Consider the IVP dy/dx = U
    x, y(0) = 1
    *U is some constant

    We wish to nd the solution at a given, xed value of x > 0 using a forward-Euler method. To this end we divide the interval [0; x] into n equally spaced intervals of length h = x=n and use the forward-Euler scheme

    {Y}_{j+1} = {Y}_{j} + U*h*{Y}_{j+1}

    The question I'm struggling with is the following
    :

    Show that the forward Euler scheme implies that {Y}_{j} = (1+Uj)^n and hence,

    log{Y}_{n} = nlog(1+Uj)


    Anyone got any tips, a start or something so that I can get through this question? I'm just completely lost with it.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,347
    Thanks
    30
    Quote Originally Posted by exphate View Post
    Hey guys,

    I'm doing some numerical lin.alg work that requires me to work by hand. I'm having trouble with the following question:-

    Consider the IVP dy/dx = U x, y(0) = 1
    *U is some constant

    We wish to nd the solution at a given, xed value of x > 0 using a forward-Euler method. To this end we divide the interval [0; x] into n equally spaced intervals of length h = x=n and use the forward-Euler scheme

    {Y}_{j+1} = {Y}_{j} + U*h*{Y}_{j+1}

    The question I'm struggling with is the following:

    Show that the forward Euler scheme implies that {Y}_{j} = (1+Uj)^n and hence,

    log{Y}_{n} = nlog(1+Uj)


    Anyone got any tips, a start or something so that I can get through this question? I'm just completely lost with it.
    Are your sure you mean

    {Y}_{j+1} = {Y}_{j} + U*h*{Y}_{j+1}

    or

    {Y}_{j+1} = {Y}_{j} + U*h*{Y}_{j}

    For the latter, it's a difference equation

     <br />
Y_{j+1} = (1 + h U) Y_j<br />

    We look for a solution of the form Y_j = c\rho^j. When you substitute you get

    c \rho^{j+1} = c (1 + hU)\rho^j giving \rho = 1 + hU

    Thus, Y_j = c (1 + hU)^j. The initial condition (when j = 0) gives c = 1.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. One more quick question
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 11th 2010, 11:06 PM
  2. [SOLVED] Quick question
    Posted in the Statistics Forum
    Replies: 0
    Last Post: October 15th 2010, 07:07 PM
  3. quick question
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: November 22nd 2009, 04:45 AM
  4. Quick LOG question!
    Posted in the Algebra Forum
    Replies: 5
    Last Post: September 27th 2007, 07:43 PM
  5. Quick's quick question
    Posted in the Number Theory Forum
    Replies: 22
    Last Post: July 9th 2006, 04:38 PM

Search Tags


/mathhelpforum @mathhelpforum