Laplace Transform


by anv
Tags: laplace, transform
anv
anv is offline
#1
Jan4-05, 06:48 PM
P: 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.
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dextercioby
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#2
Jan4-05, 08:24 PM
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Quote Quote by anv
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.
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"...

Did u try other version of Mathematica ??

Daniel.

PS.I'll work on it...I smell some Bessel functions...
anv
anv is offline
#3
Jan5-05, 03:52 AM
P: 4
You undestand me right!

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

dextercioby
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#4
Jan5-05, 07:54 AM
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Laplace Transform


[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.
anv
anv is offline
#5
Jan5-05, 12:50 PM
P: 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 )
anv
anv is offline
#6
Jan5-05, 12:54 PM
P: 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.
dextercioby
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#7
Jan5-05, 01:07 PM
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P: 11,863
Quote Quote by anv
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.

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.


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