Laplace Transform

  1. anv

    anv 4

    I have a proplem to analitic calculate this :

    (k/p^2)*LaplaceTransform[Sqrt[z^2+2p*z]/(z+p)],z,k]+
    (1/p^2)*LaplaceTransform[-Sqrt[z^2+2p*z]/(z+p)^2],z,k]

    Im(k)=0, k>0
    Im(p)=0, p>0

    The Mathematica 5 doesn't calculate this.

    Very glad to help.
     
  2. jcsd
  3. dextercioby

    dextercioby 12,324
    Science Advisor
    Homework Helper

    Let me make sure i've got it straight.I'll think about it later.So u wanna compute
    [tex] I=\frac{k}{p^{2}}L(\frac{\sqrt{z^{2}+2pz}}{z+p})+\frac{1}{p^{2}}L[-\frac{\sqrt{z^{2}+2pz}}{(z+p)^{2}}] [/tex]
    ??That's an ugly "animal"... :tongue2:

    Did u try other version of Mathematica ?? :tongue2:

    Daniel.

    PS.I'll work on it...I smell some Bessel functions...
     
  4. anv

    anv 4

    You undestand me right!

    I didn't use other version, but I found the new release 5.1 at http://www.wolfram.com
     
  5. dextercioby

    dextercioby 12,324
    Science Advisor
    Homework Helper

    [tex] I=L(\frac{\sqrt{z^{2}+2pz}}{z+p})=\int_{0}^{+\infty} \frac{\sqrt{z^{2}+2pz}}{z+p}e^{-sz} dz [/tex](1)

    [tex]J=L[-\frac{\sqrt{z^{2}+2pz}}{(z+p)^{2}}]=-\int_{0}^{+\infty} \frac{\sqrt{z^{2}+2pz}}{(z+p)^{2}}e^{-sz} dz [/tex](2)

    The first integral,i complete the square under the sq.root and make a substitution:
    [tex] I=\int_{0}^{+\infty} \frac{\sqrt{(\frac{z+p}{p})^{2}-1}}{\frac{z+p}{p}} e^{-sz} dz [/tex](3)

    And now i make the substitution
    [tex] \frac{z+p}{p}\rightarrow \cosh t [/tex](4)
    [tex]dz=p\sinh t dt [/tex](5)
    The limits of integration are the same ([itex] \arg\cosh 1 =0 [/itex]).
    The exponential becomes:
    [tex] e^{-sz}=e^{sp}e^{-sp\cosh t} [/tex](6)

    Therefore
    [tex] I=p e^{sp}\int_{0}^{+\infty} e^{-sp\cosh t}\frac{\sinh^{2} t}{\cosh t} dt[/tex] (7)

    Can u convince 'Mathematica' to evaluate this integral??Maybe numerically...

    Using the same kind of substitution,for evaluating the second integral (L.transform),u get
    [tex] J=e^{sp} \int_{0}^{+\infty} e^{-sp\cosh t}\frac{\sinh^{2} t}{\cosh^{2} t} dt [/tex] (8)

    Again,i don't know what to do to it.

    Daniel.
     
    Last edited: Jan 5, 2005
  6. anv

    anv 4

    Thx for help, but I don't simplify your answer.

    I'm going use the answer (if it possible) in math model.
    Numerical answer is suit, but if it in the form:
    f(k,p)*numerical_answer

    I try to apply the teory residues for a solve.
    What do you think about?

    P.S. I should note that the root of initial problem is


    \int {0} {\inf} { exp(-rk)/r*(1/r+k) }dt

    r = sqrt ( p^2 + (vt)^2 )
     
  7. anv

    anv 4

    without ambiguity:

    \int {0} {\inf} { (exp(-rk)/r)*(1/r+k) }dt

    r = sqrt ( p^2 + (vt)^2 )


    p.s. my calculation is big, that is why I don't insert its.
     
  8. dextercioby

    dextercioby 12,324
    Science Advisor
    Homework Helper


    I'm sorry,but it doesn't work with residues,because the functions doesn't have poles."Im(p)=0, p>0",and the integration is not on entire R,but only on its positive semiaxis.,where the denominator is never zero.

    Daniel.
     
Know someone interested in this topic? Share this thead via email, Google+, Twitter, or Facebook

Have something to add?