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Laplace Transform problem

  1. Oct 30, 2005 #1
    I have the problem,
    [tex]ty(t)= \int_{0}^{t}\tau^{\alpha-1}y(t-\tau)d\tau[/tex]
    subject to the constraint that [tex]\int_{0}^{\infty}y(t)dt=1[/tex].
    In need to get the answer in the form of, Y(p)=something (where Y(p) is the Laplace transform of y(t)).
    I can see that the right hand side is [tex]Y(p)\frac{\Gamma(\alpha)}{p^a}[/tex], but how could I evaluate the left hand side?
     
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  3. Oct 30, 2005 #2

    Tide

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    That wasn't very clear but I think you are saying you want to solve the integral equation for y(t) and you are attempting to do so using the Laplace transform?

    For the left side you just need to note that [itex]t e^{-st} = - \frac {d}{ds}e^{-st}[/itex], reverse the order of differentiation and integration and arrive at the desired result.
     
  4. Oct 30, 2005 #3
    I'm actually just looking at a past exam question. It only asks me to find Y(p). I don't need to invert it afterwards.

    OK, so we use [tex]\int_{0}^{\infty}ty(t)e^{-pt}dt[/tex]
    [tex]=-\int_{0}^{\infty}\frac{d}{dp}(e^{-pt}y(t))dt[/tex]
    [tex]=-\frac{d}{dp}\int_{0}^{\infty}e^{-pt}y(t)dt[/tex]
    [tex]=-\frac{d}{dp}Y(p)[/tex]

    and so, [tex]Y(p)\frac{\Gamma(\alpha)}{p^a}+\frac{d}{dp}Y(p)=0[/tex]

    It concerns me, though, that I haven't actually used [tex]\int_{0}^{\infty}y(t)dt=1[/tex] in answering the question.
     
  5. Oct 30, 2005 #4

    Tide

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    You should have used that condition when you evaluated the transform of the right side of the original equation.
     
  6. Oct 31, 2005 #5
    I just used the property that [tex]L[\int_{0}^{t}g(\tau)y(t-\tau)d\tau;t\rightarrow p]=G(p)Y(p)[/tex] to slove for the right hand side. I can't actually see how I'd use [tex]\int_{0}^{\infty}y(t)dt=1[/tex] to do that.
     
  7. Oct 31, 2005 #6

    Tide

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    Good point - that's the convolution theorem. Offhand, then, I can't see where the integral constraint comes in.
     
  8. Oct 31, 2005 #7

    CarlB

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    I'm going to guess that you're suppose to take the Laplace transform of [tex]t y(t)[/tex] and get [tex]-dY/dt[/tex], and then solve the differential equation:

    [tex]-dY/dp = Y(p)\frac{\Gamma(\alpha)}{p^a}[/tex]

    The definite integral he gave you will be necessary to pick out the particular solution of the DE that you need. Isn't it something like [tex]Y(0)[/tex]?

    Carl
     
    Last edited: Oct 31, 2005
  9. Oct 31, 2005 #8

    saltydog

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    If I may summarize to help me an others understand this:

    Letting:

    [tex]\mathcal{L}\left\{y(x)\right\}=\int_0^\infty e^{-pt}y(t)dt=Y(p)[/tex]
    we take the Laplace Transform of both sides of the eqution:
    [tex]\mathcal{L}\left\{ty(t)\right\}=\mathcal{L}\left\{ \int_{0}^{t}\tau^{\alpha-1}y(t-\tau)d\tau\right\}[/tex]

    Noting that the expression in brackets on the RHS is the inverse transform for the product:

    [tex]Y(p)H(p)[/tex]

    where:

    [tex]H(p)=\mathcal{L}\left\{t^{\alpha-1}\right\}[/tex]

    we arrive at the IVP:

    [tex]-Y^{'}(p)=Y(p)\frac{\Gamma(\alpha)}{p^\alpha};\quad Y(0)=\int_{0}^{\infty}e^{-0t}y(t)dt=1[/tex]

    Hoplite, you can solve for Y(p) right? Anyway, I get a form of Y(p) in which I am unable to invert. Would like to know the answer though.

    Edit: Alright, I can invert it if alpha is 1/2 (I ain't proud). So what's the range for alpha?
     
    Last edited: Nov 1, 2005
  10. Nov 1, 2005 #9

    saltydog

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    Alright, I have a question. Do I need to take this elsewhere? You guys mind?

    So I solve that equation in Y(p) above and using the initial condition given, I get, well, what do you get Hoplite? Anyway, so I let alpha=1/2 and arrive at a solution:

    [tex]y(t)=\frac{e^{-\pi/t}}{t^{3/2}}[/tex]

    So I wish to back-substitute this solution into the integral equation:

    [tex]\frac{e^{-\pi/t}}{\sqrt{t}}=\int_0^t \frac{e^{-\pi/(t-\tau)}}{(t-\tau)^{3/2}\sqrt{\tau}}d\tau[/tex]

    So, how do I show that the integral is equal to the LHS?
     
    Last edited: Nov 1, 2005
  11. Nov 1, 2005 #10

    CarlB

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    The DE can be solved by bringing Y stuff to the LHS and p stuff to the RHS for an arbitrary value of alpha. Then [tex]\int Y'/Y dp = \ln(Y)+c[/tex] works on the LHS and the RHS looks like [tex]\int p^{-\alpha} dp = p^{1-\alpha}/(1-\alpha) + c[/tex].

    Carl
     
  12. Nov 1, 2005 #11

    saltydog

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    I got question #2. I just have such high hopes for integral and integro-differential equations in artificial intelligence. You know those guys at Blue Column are to use . . . but I digress.

    Anyway, looking at the integral expression (for alpha=1/2) when t=1:

    [tex]\int_0^1 \frac{e^{-\pi/(1-\tau)}}{(1-\tau)^{3/2}\sqrt{\tau}}d\tau[/tex]

    Note that this has a vertical asymptote at tau=0. How do I know that the integral converges?

    Hoplite, where you at with this anyway? I'm expecting plots you know. :smile:
     
  13. Nov 1, 2005 #12

    CarlB

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    The factor of [tex]e^{-\pi/(1-\tau)}[/tex] should go to zero a lot faster than the denominator.

    Carl
     
  14. Nov 1, 2005 #13

    saltydog

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    Hey Carl. That's right for the upper limit. That is:

    [tex]\mathop\lim\limits_{x\to 1^-} \frac{e^{-\pi/(1-x)}}{(1-x)^{3/2}\sqrt{x}}=0[/tex]

    However for the lower limit, the numerator just goes to [itex]e^{-\pi}[/itex] and we're left with an expression:

    [tex]\frac{e^{-\pi}}{0}[/tex]

    However, I believe I have it using the p-test and limit tests:

    P-test:

    [tex]\int_0^1 \frac{1}{x^p}dx[/tex]

    is convergent iff p<1.


    Limit test:
    If:

    [tex]\mathop\lim\limits_{t\to 0}\frac{f(x)}{g(x)}=1[/tex]

    then:

    [tex]\int_0^1 f(x)dx[/tex]

    is convergent iff:

    [tex]\int_0^1 g(x)dx[/tex]

    is convergent.

    So let:

    [tex]f(x)=\frac{e^{-\pi/(1-x)}}{(1-x)^{3/2}\sqrt{x}}[/tex]

    and:

    [tex]g(x)=\frac{e^{-\pi}}{\sqrt{x}}[/tex]

    then:

    [tex]\mathop\lim\limits_{x\to 0}\frac{f(x)}{g(x)}=1[/tex]

    and becaues of the p-test:

    [tex]\int_0^1 g(x)dx[/tex]

    converges and therefore:

    [tex]\int_0^1 f(x)dx[/tex]

    does so as well.

    QED
     
    Last edited: Nov 1, 2005
  15. Nov 1, 2005 #14
    What the question says about alpha is, "Indicate any restrictions on alpha that may be needed for your solution to be valid. You are not required to invert the transform." As far as I can see, he wants the answer, alpha is not a negative integer because the presence of of the gamma function.
    But I've got it now. It was this advice from Saltydog that I needed: [tex] Y(0)=\int_{0}^{\infty}e^{-0t}y(t)dt[/tex]

    So, using,
    [tex]-\int\frac{dY(p)}{dp}=\Gamma(\alpha)\int p^{-\alpha}dp[/tex]
    So, for alpha is not equal to 1: (How do you do the inequality sign in LaTex?)
    [tex]Y(p)=k.exp(-\frac{\Gamma(\alpha)}{1-\alpha}p^{1-\alpha})[/tex]
    But Y(0)=1 => k=1.
    [tex]Y(p)=exp(-\frac{\Gamma(\alpha)}{1-\alpha}p^{1-\alpha})[/tex]
    Or, if alpha=1:
    [tex]Y(p)=\frac{k}{p}[/tex]
    But this blows up at p=0, so alpha cannot equal one.
    (I'd love to know how many students actually got this one right.)

    You mean residue plots? Not gonna happen, I think. :smile:
     
    Last edited: Nov 1, 2005
  16. Nov 1, 2005 #15

    saltydog

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    Not invert the transform? That's like giving you a bicycle but saying you can't make the wheels go round and round.:yuck:

    Thanks Hoplite. It was a very interesting problem for me.:smile:
     
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