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Laplace Initial Value Problem

  1. May 20, 2005 #1

    I have to solve this equation using laplace transform

    ty''+(t-1)y'+y=t^2 y(0)=0 y'(0)=0

    I tried it using matlab and get to this point


    I think ive gone wrong somewhere

    Any Help would be appreciated.

    Last edited: May 20, 2005
  2. jcsd
  3. May 20, 2005 #2


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    Hello sssr,

    Suppose I've waited long enough for someone better than me to weigh in and they still may, but I don't think you can do this:

    [tex]\mathcal{L}\{ty^{'}\}=\int_0^{\infty} te^{-st}y^{'}dt[/tex]

    Let alone the second derivative! Might be a convolution for it that I'm not aware of however. Please someone say so if so.

    And in fact, I don't see how the power series method can work either.

    Ergo: Unless someone else says otherwise, my only recourse if I were solving it would be old faithful: numerically. :smile:
  4. May 20, 2005 #3
    It seems to me if you apply parts enough times, the only hitch is in evaluating

    [tex]\int_0^{\infty} te^{-st}ydt[/tex]

    But both ty'' and ty' reduce to that and known transforms, if I'm not mistaken.

    Then again, small comfort, since that's not so easy to evaluate anyway.

  5. May 20, 2005 #4


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    I don't think that integral can be evaluated Justin.

    Also Sssr, The ODE is singular at t=0 thus the initial value problem you proposed is, as I see it anyway, ill-poised (no offense). However, good for t>0 you know.

    So what do you think guys?
  6. May 20, 2005 #5
    Hi thanks everyone for the replies
    I actually made a mistake in my working and now
    have come to this point


    So can anyone help in where do i go from here

  7. May 20, 2005 #6
    Not sure if this is helpful or not.

    if we have

    [tex]F(s) = \int_0^\infty e^{-st} f(t) \ dt = \mathcal{L}\{f(t) \}[/tex]


    [tex]F'(s) = \mathcal\{ -t f(t)\}[/tex]

    and in general

    [tex]F^{(n)}(s) = \mathcal{L} \{ (-t)^n f(t)} \}[/tex]
  8. May 21, 2005 #7
    Given that

    [tex]ty''+(t-1)y'+y = t^2 \qquad y(0) = 0, y'(0) = 0[/tex]

    [tex]-\mathcal{L} \{- ty''\} + -\mathcal{L} \{-ty' \} - \mathcal{L}\{ y'\} + \mathcal{L}\{y\} = \frac {2}{s^3}[/tex]

    I force a negative sign in there so I can use those two formulas above.

    [tex]- \frac {d}{ds}\bigg[ s^2Y(s) - sy(0) - y'(0) \bigg] - \frac {d}{ds} \bigg [ sY(s) - y(0) \bigg] - \bigg[ sY(s) - y(0)\bigg] + Y(s) = \frac {2}{s^3}[/tex]

    Where [tex]\mathcal{L}\{ y\} = Y(s)[/tex]

    You will have to compute those derivatives using the product rule.
  9. May 21, 2005 #8
    Thanks Corneo
    I've done that part and get

    Y'(s)*(-s-s^2) + Y(s)*(-3s) = -2/s^3

    and then simplify

    Y'(s) + Y(s)*(3/(s^2+s) = -2/(s^5+s^4)

    Then solving that i end up with


    and this is where i am stuck - how do i solve for C1

  10. May 21, 2005 #9


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    Very nice guys. I see I should have waited a bit longer and have learned a lot. I'll try and solve it too now. That's just a typo right Corneo. Should be:

    [tex]F'(s) = \mathcal{L}\{ -t f(t)\}[/tex]

    Edit: Ok Corneo, I think you did that tutorial for my benefit as Sssr already knew that. Thanks a bunch. :smile:
    Last edited: May 21, 2005
  11. May 21, 2005 #10


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    Alright, even though I'm hesitant to say anything now since I was off from the get-go, I feel I have it now. I think you have an extra minus sign up there. With regards to the constant of integration, just invert the Laplace transform with the 'c' in place and see what you get.

    I'm still unclear about what I think is a singularity of the ODE at t=0. However, the solution I obtained (and verified by back-substitution) does not have one.

    Edit: Oh yea, thanks guys. Interesting problem and I learned a lot! :smile:
    Last edited: May 21, 2005
  12. May 21, 2005 #11
    I haven't had much luck with this problem. I don't know how to solve for Y(s). I thought it would be seperable, but I don't think so.
  13. May 21, 2005 #12


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    Hello guys. Sssr, allow me to summarize what I've done with Corneo's kind help:

    So we have:

    [tex]ty''+(t-1)y'+y = t^2 \qquad y(0) = 0, y'(0) = 0[/tex]

    And we let [tex]\mathcal{L}\{ y\} = Y(s)[/tex]

    So I make the transforms as Corneo suggest, do the differentiation and integration and come up with:


    Note I kept the c. However I wish to point out that in Kreyszig, which goes over this type of problem for Laguerre's ODE (learning this only after Corneo's post), they set it equal to zero but I don't understand why.

    The inverse Laplace transform of the above is:


    That is:


    This satisfies the ODE (and the initial conditions) for any value of c. However, I suspect it needs to be 0 but don't know why.

    Also, I thought I would arrive at a "specific" solution without arbitrary constants. Can anyone explain this to me?
  14. May 23, 2005 #13
    saltydog & Corneo just wanna say thanks for all your help with this

    Thanks heaps
  15. May 31, 2005 #14
    In the brief intermission from me being overworked that I'm stealing from myself, did we figure out what the story with the c was? I noted that the numerical solution Mathematica found set c to zero, as saltydog said, but I've had little time to look into it more than that.

  16. Jun 2, 2005 #15


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    Hello Justin. I've been unavailable for a while and am only now catching up. You know, I hate it when I don't understand something in math and I don't understand the bit about the constant of integration up there. I'm a purist with things like that and will no doubt eventually trek 100 miles to the nearest university math department and just start asking questions. They're usually very nice about things like that you know. I'll try and research it a bit via the library first though. It's a pleasant ride.
  17. Jun 6, 2005 #16


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    Some follow-up

    I've been looking into the problem above concerning the constant of integration and the lack of uniqueness of solutions. Today I spoke with a professor I know concerning this equation and he suggested I look into uniqueness requirements for the following equation:


    His suggestion is that uniqueness will not be guaranteed at the singular points of this equation. That is, for the equation:

    [tex]ty^{''}+(t-1)y^{'}+y=t^2;\quad y(0)=0,\quad y^{'}(0)=0[/tex]

    The solution obtained via Laplace transform:


    May in fact be the non-unique solution for this equation at the singular point t=0.

    I might add Mathematica's NDSolve will return a solution only if initial conditions at t[itex]\neq[/itex]0 are supplied.

    This is only a proposal and needs to be verified. Think I'll work on it.
    Last edited: Jun 6, 2005
  18. Jun 7, 2005 #17
    *sighs* If only I didn't have to write an essay and do a few more finals, move out of my room, then drive for 30-some-odd hours, I could look into it more. But alas, I've been screwed by finals week. :frown:

    So don't bust your butt on my account, salty, because I won't be able to do anything for a week, at least.

  19. Jun 7, 2005 #18


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    Hello Justin. I have a suggestion for you that works for me: that 30 hour trip, stop frequently and eat meals. It makes it a much more pleasant trip. Oh yea, the math up there, that's dessert for me, you know, like cheese cake with strawberries :smile: and I'm finding it a new challenge. Keep in mind that resolution of the problem is not my primary objective but rather the journey and I stop often. :wink:
    Last edited: Jun 7, 2005
  20. Jun 7, 2005 #19
    Unfortunately, it's a 30 hour drive that must be completed in one weekend. Sometimes I wonder about the things I get myself into... But there's a bubble bath waiting for me at the end of it, so it'll all be worth it.

    And while I'm driving for two days straight, you get to wander around in the playground. Oh well. At least I'm done with the essay now.

  21. Jun 7, 2005 #20


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    You sure that's right?

    When I assume the solution is of the form:

    y(t) := e^{-t} p(t) + q(t)

    Where p and q are arbitrary polynomials, and plug into the equation, I get this as the only solultion:

    y(t) = A t^2 e^{-t} + \frac{1}{3} t^2

    And plugging this back into the original equation works.

    Oh, I guess this is your solution -- you just didn't collapse the (1/6)(-2+3C) into a single constant A. Silly me!

    By the way, where is the other family of solutions? I should have two arbitrary constants available to me!!! It's probably a rational function times an exponential... or maybe a different exponential?

    Anyways, back to the problem at hand -- uniqueness of solution.

    It helps to think geometrically, I think. A differential equation is nonsingular when you can solve for the value of the second derivative, if you're given the value and first derivative at that point.

    So, if you have a solution curve that starts at your favorite point and with your favorite slope, there's a unique solution for its curvature near that point.

    (This sort of thing is much easier to picture with a first order equation! Anyways...)

    It's a powerful theorem that there exists at least one solution, but once you have that, it's intuitively clear that it must be locally unique, by looking at an appropriate "infinitessimal" neighborhood of the point.

    Now, when you have a singular point, you cannot solve for the curvature. Either no solution is possible, or all values work. Since these solutions extend into areas where it is locally unique, you can look at it in the other direction... solutions "pinch" together as they approach this point, making you unable to distinguish them near the singular point.
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