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Laplace Transform of

  1. Jun 29, 2012 #1
    L[y2] = ?
     
  2. jcsd
  3. Jun 29, 2012 #2

    chiro

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    Hey maistral and welcome to the forums.

    Because of the square you are going to have use integration by parts. Let's assume for the moment you have a known y and don't want to use it for say solving DE's. Then you will get using integration by parts:

    = [-1/s*e^(-st)*2*y'(t)y(t)](t=0,t=infinity) + 2/s (t=0,t=infinity)[Integral](e^(-st)y'(t)y(t))dt
    = [-1/s*e^(-st)*2*y'(t)y(t)](t=0,t=infinity) + 2/s L[y'(t)y(t)]

    Know you can evaluate the LHS part by knowing initial conditions, but the RHS can only be evaluated if you know y(t), y'(t) or some special kind of information that can be used to get the laplace transform.

    As you can see, for ODE's, this is problematic in the general case unless you have specialized information or other means to get F(s) which is the whole point of using Laplace transforms.

    One possibility though is if you can find a way to get a common function inside the L[] terms for each one (not necessarily y(t) but any expression involving y including derivatives), then what you could do is get your laplace transform in terms of that and find the expression in terms of the combination of terms inside the operator.

    This is not going to be easy if it can be done, but it might give you some food for thought.
     
  4. Jul 1, 2012 #3
    Oh my gosh, my nose just bled. Please keep it simple for at least an engineering student to understand :(

    I was trying to figure out how to solve this equation with Laplace transformation-

    y' = 1.25EXP(-0.25x) - 0.5y^2

    I don't know how to get the transform of y^2.
     
  5. Jul 1, 2012 #4
    Hi !
    Since the ODE isn't linear, I am not sure that Laplace transform method is the best one.
    In fact, the ODE is of the Riccati kind and can be rather easily linearized thanks to the change :
    y=2F'/F
    This leads to a linear second order ODE which unknown function F can be expressed in terms of Bessel functions.
     
  6. Jul 1, 2012 #5
    I'm slowly starting to get it.

    I read this article in wikipedia: http://en.wikipedia.org/wiki/Riccati_equation

    are the q0, q1, and q2 constants or functions of x? because the S = q2q0 looks like a constant multiplication of sorts. Or i have to multiply the functions?

    also, v' = v^2 + R(x)v + S(x) -> what is this lol

    I so hate wikipedia notations =.=
     
  7. Jul 1, 2012 #6

    chiro

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    These are functions of the independent variable x [i.e. R(x) and S(x)], where v is a function of x.
     
  8. Jul 2, 2012 #7
    Hi !

    there is no need to do again the proof of the change of function y(x) -> u(x) which is shown by Wikipedia.
    Just apply it : y(x)=-u'/(q2*u) : see attachment.
     

    Attached Files:

  9. Jul 2, 2012 #8
    A while ago I was slowly able to understand how it was derived (except for a few reservations which aren't clear enough), and I was able to end up with JJac's final equation (i used decimals).

    About the derivation,
    v' = -(u'/u)
    v' = -(u"/u) + (u'/u)^2
    v' = -u"/u + v^2 <<< this v^2, did he just disregard the negative sign because it's raised to two?


    And about the final equation (I'm setting L(u) = U):
    My IVs are: x = 0 ; y = 0.1; y' = 1.245 (is this correct? I substituted x and y on the original dy/dx ODE to get 1.245).

    u" - 0.625EXP(-0.25x) u = 0
    sU - su(0) - u'(0) - 0.625 L[EXP(-0.25x)*u] = 0

    Now all sorts of headaches appear :S
    1) How am I to translate y(0) to u(0)? Will it be easier to just set it as a constant A then carry it through the differential equation's solution? Or is there another way?
    2) I know the s-differentiation property which could allow me to take the Laplace of L[x*u] (It's the L[-tf(t)] right?). I have no idea how to deal with the EXP beside the u. Or should I take the Taylor series of EXP? o.o
     
  10. Jul 2, 2012 #9
    A while ago I was slowly able to understand how it was derived (except for a few reservations which aren't clear enough), and I was able to end up with JJac's final equation (i used decimals).

    About the derivation,
    v' = -(u'/u)
    v' = -(u"/u) + (u'/u)^2
    v' = -u"/u + v^2 <<< this v^2, did he just disregard the negative sign because it's raised to two?


    And about the final equation (I'm setting L(u) = U):
    My IVs are: x = 0 ; y = 0.1; y' = 1.245 (is this correct? I substituted x and y on the original dy/dx ODE to get 1.245).

    u" - 0.625EXP(-0.25x) u = 0
    sU - su(0) - u'(0) - 0.625 L[EXP(-0.25x)*u] = 0

    Now all sorts of headaches appear :S
    1) How am I to translate y(0) to u(0)? Will it be easier to just set it as a constant A then carry it through the differential equation's solution? Or is there another way?
    2) I know the s-differentiation property which could allow me to take the Laplace of L[x*u] (It's the L[-tf(t)] right?). I have no idea how to deal with the EXP beside the u. Or should I take the Taylor series of EXP? o.o!

    If also possible, can you guys show me at least a working example that I can build on? I didn't expect doing this equation that I gave the analytical way is too much. I just wanted to see the beauty of closed-form solutions instead of running Runge-Kutta all the time :(

    Thank you very much for the time guys.
     
  11. Jul 2, 2012 #10
    Obviously, a change of variable X=exp(-x/4) allows to simplify the ODE (see attachment).
    This leads to a new ODE of the Bessel kind, but not on a standard form.
    Another change of variable allows to reduce the ODE to a standard Bessel equation.
    This leads to the solutions u(x). See attachment.
    Then, you will have to express the derivative u'(x) and the final solution will be obtained :
    y(x) = 2 u' / u
     

    Attached Files:

  12. Jul 4, 2012 #11
    The solution y(x) of the ODE, in case of IVs : x = 0 ; y = 0.1 is given in attachment :
     

    Attached Files:

  13. Jul 4, 2012 #12
    Good Lord, my head exploded.

    Even if the answer was given already, I ain't really.. happy with it. I didn't understand fully how to deal with these Riccati equations. If possible (and you have the luxury of time, ofc), can you please give me a very very basic example of a Riccati equation that I can practice and work on? I really want to learn it by heart, and the examples found here in the internet make my head implode in horror.

    Thank you very much! :D
     
  14. Jul 4, 2012 #13
    Hello maistrail !

    Don't be obsessed by "Riccati". In fact, it is the easier part of the job.
    No need for Wolfram which might confuse you with the entire demo.
    Simply, you have to do a change of function :
    Starting from the Riccati ODE :
    dy/dx = q0(x)+q1(x)*y(x)+q2(x)*y2
    Let y(x) = -(1/q2)*u’(x) / u(x)
    I suppose that you know what is a change of function and how to do it : this is a basic and general knowledge that everybody dealing with ODEs is aware of.
    So, you do this change of function and you obtain a linear ODE which unknown is u(x).
    That’s all : No more “Riccati”. Now, you have to solve a linear ODE and you can chose any method that you know for solving linear ODE’s.

    For example, in the case of dy/dx = 1.25 exp(-0.25 x) – 0.5 y2
    q2 = -0.5
    Thanks to the change of function y(x)=-2 u'(x)/u(x) you obtain :
    u’’(x) - (5/8)exp(-0.25 x) u(x) = 0
    This is no longer a Riccati ODE : This is a second order linear ODE.
    So the difficulty of your problem was not because the first equation was of the Riccati kind.
    The difficulty was in solving the second order linear ODE, which is of the Bessel kind, but not on a standard form. The knowledge of solving not simple Bessel ODE was then required.
    Of course, depending on the coefficients of the initial Riccati ODE, the resulting second order linear ODE is not allways of the Bessel kind. It can be any sort of linear ODE, sometimes easy to solve, sometimes hard.

    One last thing : do not expect a simpler result than the function y(x) given at end of my preceeding post. Any analytical method for solving the ODE will leads to the same result. If you try the Laplace method for solving the second order linear ODE, probably, this would be rather arduous and the final result, with the Bessel functions, will be exactly the same.
     
    Last edited: Jul 4, 2012
  15. Jul 8, 2012 #14
    Argh, my net broke.

    Anyway, I still need a little help. I need to build it myself :( To ease myself of all those qn(x)'s, I use this notation:

    y' = A + By + Cy2

    where A, B, and C are functions of x.

    So to solve this analytically, I do a change of variables. I let y = -(1/C)*(u'/u). What would be my y'?

    is y' = -1/C * [(u*u" - (u')2)/u2] + [u'/u * -c'/c2]?

    Is there an easier way for this? surely this looks ugly, but there might be other ways that I'm not aware of.. Also, whatever happened to the (u')2? Ofc unless my answer is wrong, and someone has to review differentiation e.e

    Sorry for the barrage of questions. I'm such a noob :cry:
     
  16. Jul 8, 2012 #15
    As of this moment I'm using Wiki's strategy of attacking the equation, but with a bit of form memorization. it goes like this:

    y' = A + By + Cy2.

    Since I understood Wiki's derivation, I think I could allow myself to memorize these:

    Equation becomes u" - Ru' + Su = 0; where R = B + C'/C and S = AC.

    Is it safe to follow this notation? e.e

    Anyway, so if say, for example, my equation is y' = 1.25x - 0.5y2, my initial condition is y(0) = 1. Since it's a Riccati, I'll use my convention: R = 0, S = -0.625; my differential equation is now

    u" - 0.625u = 0. My subject requires the use of Laplace Transforms, that's where the new problem starts. What would be my u(0) and u'(0)? Will it be related to the y(0) = 1 and y'(0) = 0.5 (i just substituted x = 0, y = 1 on the original DE to get this, I don't even know if this is correct)? Thanks in advance
     
    Last edited: Jul 8, 2012
  17. Jul 9, 2012 #16
    Not u*u'' but u''/u :
    y' = -1/C * [((u"/u) - (u')2)/u2] + [u'/u * -C'/C2]
    In the cqase of your ODE C is constant :
    y' = -1/C * [((u"/u) - (u')2)/u2]
    When you bing back y and y' into the first EDO and simplify, the term (u')2)/u2] disappears.
    Why do you refuse to bring back y = -(1/C)*(u'/u) and y' = -1/C * [((u"/u) - (u')2)/u2] into y' = A + By + Cy2 ?
    If you did it, you will see that it is very easy and that it is the simplest way to obtain the linear ODE.
    If C is function of x, it is a little more complicated, but a linear ODE is obtained anyway.
     
  18. Jul 11, 2012 #17
    Uh, so for example. The equation generated from the conversion is u" - 0.625u = 0.

    How do I solve it using the Laplace transform? My course requires the usage of Laplace transformation :(
     
  19. Jul 11, 2012 #18
    If the Laplace transform of u(x) is U(s), what is the Laplace transform of u''(x) ?
     
  20. Jul 11, 2012 #19
    Oh, the question wasn't clear. My bad.

    The problem is what constants should I use. Say for example, y(0) = 1.

    the equation is in terms of u, u" - 0.625u = 0

    The laplace transform is s2U - su(0) - u'(0) - 0.625U = 0.

    What would be my u(0) and u'(0)? Thanks. :D
     
  21. Jul 11, 2012 #20
    If u(0) and u'(0) are not specified in the wording of the problem, then consider them as paramaters. For example, let u(0)=C1 and u'(0)=C2. They will appear into the general solution of the ODE. Since it is a second order ODE, the general solution must include two arbitrary constants anyway.
     
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