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B Euler's method for second order DE

  1. Oct 4, 2016 #1
    Can the Euler approximation method be used to solve higher order DE?
    I have ##\ddot x=\omega^2 x## which i rewrite as ##y''=\omega^2y##. initial conditions y(0)=0, y'(0)=1.
    The Euler method: ##y_{n+1}=y_n+h\cdot y'_n##. i use this to make:
    $$y''_{n+1}=y'_n+h\cdot y''_n~~\rightarrow~~\omega^2y_{n+1}=y'_n+h\omega^2 y_n$$
    $$\omega^2(y_n+hy'_n)=y'_n+h\omega^2 y_n$$
    $$\rightarrow~y'_n=\left[ \frac{\omega^2(1-h)}{1-\omega^2h} \right]y_n$$
    But this contradicts the initial condition y'(0)=1, after i substitute y(0)=0 in the formula i found.
     
  2. jcsd
  3. Oct 4, 2016 #2

    BvU

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    Wouldn't the Euler method dictate $$y'_{n+1} = y'_n + h\omega^2y_n$$ ?
     
  4. Oct 6, 2016 #3

    Krylov

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    Yes, every explicit first order method can be used to solve higher-order DE by writing the DE in vector form as a system of first-order DE, also see below.
    I do not quite understand the purpose of introducing ##y##, but it does no harm. Also, this equation looks like a free harmonic oscillator. Are you sure there should not be a minus in front of ##\omega^2##?
    I think that the OP and @BvU both have one half of the method for this problem :wink:.

    As I alluded to at the start, you can solve higher order DE by converting them into a system of first order DE. Although not strictly necessary, this works best by introducing new symbols for the derivatives. For this example, set ##v := y'##, then the DE ##y''(t) = \omega^2y(t)## is equivalent to
    $$
    \left\{
    \begin{align*}
    y'(t) &= v(t)\\
    v'(t) &= \omega^2y(t)
    \end{align*}
    \right.
    $$
    with the initial conditions ##y(0) = 0##, ##v(0) = 1##. Then Euler forward with stepsize ##h > 0## just reads
    $$
    \begin{align*}
    \begin{bmatrix}
    y_{n+1}\\
    v_{n+1}
    \end{bmatrix}
    &\approx
    \begin{bmatrix}
    y_n\\
    v_n
    \end{bmatrix}
    +
    h
    \begin{bmatrix}
    y'(nh)\\
    v'(nh)
    \end{bmatrix}\\
    &=
    \begin{bmatrix}
    y_n\\
    v_n
    \end{bmatrix}
    +
    h
    \begin{bmatrix}
    v_n\\
    \omega^2y_n
    \end{bmatrix}
    \end{align*}
    $$
    where ##y_n := y(nh)## and ##v_n := v(nh)## for ##n = 0, 1,\ldots##. The first component is what the OP called the Euler method, while the second component appears in post #2.
     
  5. Oct 7, 2016 #4
    It is taken from a mathematics book, teaching DE, not a physics book.
    I thank you very much BvU and Krylov.
    How do you quote exactly also the equations in the posts, since when i use the "+Quote" function that is given in this site, or i mark a part of a post and select "+Quote" from the context menu that appears, and then insert these quotes in the reply box the equations come out messed, while the text is good. i manually fix these equations, almost completely write them anew.
    And also, how can you copy a poster's name, like Krylov did in the last post:
    Snap1.jpg
     
  6. Oct 7, 2016 #5

    Krylov

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    Ok. Whenever I see an ##\omega^2## somewhere, I start to think of ##\tfrac{k}{m}##.
    You are welcome!
    I recognize this and it is a bit uncomfortable. Probably there are better ways, but what usually do, is put the cursor where I want the quote (with formulas) to appear and then press "reply". This quotes the entire message with formulas correctly. Then I remove the excess text that I do not want quoted.

    You can also right-click on an equation, then "Show Math As", then "TeX Commands". This opens a pop-up containing the ##\LaTeX## source. That context menu has some other options, too, it appears.
    You write the member name with an "@" in front, like @Karol. When the member has enabled this, it will show him a message that he has been "mentioned" in a certain thread.
     
  7. Oct 7, 2016 #6
    Thank you very much BvU and Krylov
     
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