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Laplace Transform Problem - Peacewise Functions

  1. Apr 21, 2009 #1
    Okay so this problem has been giving me nightmares ever since I laid eyes on it.


    The problem: y'' - y = f(t), y(0) = 1, y'(0) = 0

    f(t) = 1, 0 (less than equal to) t (less than) pi/2
    f(t) = sin(t), t (greater than equal to) pi/2



    Any help would be appreciated putting me in the right direction.

    Thank you in advance.


    --Julie
     
  2. jcsd
  3. Apr 22, 2009 #2
    I think the best way forward would be:

    1-1.jpg
     
  4. Apr 22, 2009 #3
    except f(t) isn't 1, read the post again. for piecewise functions use the unit step (heaviside) function.
     
  5. Apr 23, 2009 #4

    djeitnstine

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    You can rewrite f(t) in terms of the heaviside function as matticus said, then perform the transform.

    [tex]f(t)=1+H(t-\frac{\pi}{2})+\left(sin(t)-1\right)H(t-\frac{\pi}{2})[/tex]

    Edit since latex isn't working....

    f(t)=1+H(t-[pi/2])+[sin(t)-1]*H(t-[pi/2])

    where H(t-[pi/2]) denotes the unit step function.
     
  6. Apr 25, 2009 #5
    Hi I was thinking about the same approach as xfunctionx.

    Please advise to see whether his approach is correct or not.

    Thanks
     
  7. Apr 25, 2009 #6

    djeitnstine

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    maths.tuition welcome to PF.

    If you are speaking of doing the same thing that xfunctionx did for bathtub's differential equation that would be wrong

    f(t) = 1, 0 (less than equal to) t (less than) pi/2
    f(t) = sin(t), t (greater than equal to) pi/2

    does not simply equal 1 for all t, please note I already stated the proper form of the function(in terms of the unit step function) to get the the laplace transform for f(t)

    "f(t)=1+H(t-[pi/2])+[sin(t)-1]*H(t-[pi/2])"
     
    Last edited: Apr 25, 2009
  8. Apr 25, 2009 #7
    I think this last expression is wrong. Musn't this be:

    [tex]L\left{H(t)-H\left(t-\frac{\pi}{2}\right)+sin(t)H\left(t-\frac{\pi}{2}\right)\right}= L\left{H(t)-H\left(t-\frac{\pi}{2}\right)+cos\left(t-\frac{\pi}{2}\right)H\left(t-\frac{\pi}{2}\right)\right}[/tex]

    There seems to be something definitely wrong with the latex generation, it keeps posting things from older replies of other posts of me :grumpy:

    coomast
     
  9. Apr 25, 2009 #8

    djeitnstine

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    You are right the correct function is

    f(t)=1+H(t-[pi/2])+[sin(t)-2]*H(t-[pi/2])

    one had to subtract 2 from sin and not 1 or else you would be left with 2-1+sin{t} = 1+sin(t)

    thanks for catching that mistake
     
  10. Apr 29, 2009 #9
    OK, next attempt for a good latex formula. I think it is still not OK, I got:

    [tex]L\left[H(t)-H\left(t-\frac{\pi}{2}\right)+sin(t)H\left(t-\frac{\pi}{2}\right)\right]= L\left[H(t)-H\left(t-\frac{\pi}{2}\right)+cos\left(t-\frac{\pi}{2}\right)H\left(t-\frac{\pi}{2}\right)\right][/tex]

    coomast
     
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