Results 1 to 9 of 9

Thread: Fredholm Intergral Equation

  1. #1
    Junior Member
    Joined
    Aug 2007
    Posts
    34
    Awards
    1

    Fredholm Intergral Equation

    I need to find a general solution,

    $\displaystyle f$ ∈ $\displaystyle E [0, \infty)$

    for the Fredholm intergral equation:

    $\displaystyle f(x) = \gamma cos (x) + \delta sin (x) + \lambda (Mf)(x) $

    where $\displaystyle M$ is the intergral operator defined by

    $\displaystyle (Mf)(x):= \int_0^{\infty} x^{a-1} e^{-bt} f(t)dt$

    Where $\displaystyle a$ and $\displaystyle b$ are postive constants and $\displaystyle \gamma, \delta$ and $\displaystyle \lambda $∈ ℝ constants.

    I think, that I have to rewrite it:

    $\displaystyle f(x)= \gamma cos (x) + \delta sin (x) + \lambda \int_0^{\infty} x^{a-1} e^{-bt} f(t)dt$

    And then assuming that λ is a fixed unknown scalar, and so can be incorporated into the equation as follows:

    $\displaystyle f(x)= \gamma cos (x) + \delta sin (x) +\int_0^{\infty} x^{a-1} e^{-bt} f(t)dt$


    I'm not sure if this is right, and I don't really understand what to do next, can someone help please?

    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Apr 2008
    From
    moscow
    Posts
    4
    IMHO
    The next step after this:
    $\displaystyle
    f(x)= \gamma cos x + \delta sin x + \lambda \int_0^{\infty} x^{a-1} e^{-bt} f(t)dt
    $
    is to say: if equation invloved has a solution, its solution can be represented in the following way:
    $\displaystyle f(x)=\gamma cos x + \delta sin x+\lambda x^{a-1} C_1$,
    where $\displaystyle C_1=\int_0^{\infty} e^{-bt}f(t)dt=\int_0^{\infty} e^{-bt}(\gamma cost + \delta sin t + \lambda t^{a-1} C_1)dt$
    Knowing $\displaystyle C_1$ you will find general solution by substituting it into the first equation in this message.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Aug 2007
    Posts
    34
    Awards
    1

    okay

    Thanks, for that, but if I substitute $\displaystyle C_1$ into the first equation, then won't I just be back where I started?

    I.e.



    Just with $\displaystyle x$ replaced with $\displaystyle t$'s
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Apr 2008
    From
    moscow
    Posts
    4
    Did you check this?
    Note that you have equation for $\displaystyle C_1$ where this stuff takes part in both sides of the equation.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Aug 2007
    Posts
    34
    Awards
    1
    Well basically,

    if I let $\displaystyle C_1=\int_0^{\infty} e^{-bt}f(t)dt$

    and subsitute it into

    $\displaystyle f(x)= \gamma cos x + \delta sin x + \lambda \int_0^{\infty} x^{a-1} e^{-bt} f(t)dt$

    Then all I get is

    $\displaystyle
    f(x)=\gamma cos x + \delta sin x+\lambda x^{a-1} C_1
    $

    And keep going round in circles.

    Maybe at this point I need to solve the integral, but I am not sure how to do this?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Apr 2008
    From
    moscow
    Posts
    4
    No. Look at this:
    You have:
    $\displaystyle f(x)= \gamma cos x + \delta sin x + \lambda \int_0^{\infty} x^{a-1} e^{-bt} f(t)dt $
    This equation has a solution:
    $\displaystyle f(x)=\gamma cos x + \delta sin x+\lambda x^{a-1} C_1 $
    where $\displaystyle C_1=\int_0^{\infty} e^{-bt}f(t)dt$
    Lets find $\displaystyle C_1$ in terms of $\displaystyle \lambda$:
    $\displaystyle C_1=\int_0^{\infty} e^{-bt}f(t)dt=\int_0^{\infty} e^{-bt}(\gamma cost + \delta sin t + \lambda t^{a-1} C_1)dt = I_1 + I_2+ I_3$
    $\displaystyle I_1=\gamma \int_0^{\infty} e^{-bt} cost dt = \frac{b \gamma}{b^2+1}$
    $\displaystyle I_2=\delta \int_0^{\infty} e^{-bt} sint dt= \frac{\delta}{b^2+1}$
    $\displaystyle I_3=-\lambda C_1 K$.
    So:
    $\displaystyle C_1=\frac{b \gamma + \delta}{b^2+1}-\lambda C_1 K$
    $\displaystyle C_1=\frac{b \gamma + \delta}{(b^2+1)(1+\lambda K)}$
    if $\displaystyle \lambda$ doesn't equal $\displaystyle -\frac{1}{K}$.
    Now you should substitute $\displaystyle C_1$ into
    $\displaystyle f(x)=\gamma cos x + \delta sin x+\lambda x^{a-1} C_1 $
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Aug 2007
    Posts
    34
    Awards
    1

    Thanks!

    Thank you very much that was really helpful!!!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Aug 2007
    Posts
    34
    Awards
    1

    Fredholm Integral Equation

    I need to find a general solution,

    $\displaystyle f$ ∈ $\displaystyle E [0, \infty)$

    for the Fredholm intergral equation:

    $\displaystyle f(x) = \gamma cos (x) + \delta sin (x) + \lambda (Mf)(x) $

    where $\displaystyle M$ is the intergral operator defined by

    $\displaystyle (Mf)(x):= \int_0^{\infty} x^{a-1} e^{-bt} f(t)dt$

    Where $\displaystyle a$ and $\displaystyle b$ are postive constants and $\displaystyle \gamma, \delta$ and $\displaystyle \lambda $∈ ℝ constants.

    Note that the algebra can be reduced using the Gamma function and standard Laplace Transforms.

    I have started a solution to this, which is attached, can anyone tell me if this is all I have to do to get the general solution or is there more, and if so can anyone advise what I do next please?
    Attached Files Attached Files
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Aug 2007
    Posts
    34
    Awards
    1

    My Solution??? Is it correct?

    I have the following as the general solution and someone please help me, and advise me if this is correct please?

    $\displaystyle
    f(x) = \gamma cos (x) + \delta sin (x) + \frac{\lambda x^{a-1}(b \gamma + \delta)}{(b^2+1)(1+\lambda K)}
    $

    I would be really greatful if someone could help so that I could continue on the next part of the question please...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. intergral equation consiquence..
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Sep 18th 2011, 12:51 AM
  2. A Fredholm integral equation
    Posted in the Advanced Applied Math Forum
    Replies: 7
    Last Post: Apr 19th 2011, 09:49 AM
  3. A second Fredholm integral equation
    Posted in the Advanced Applied Math Forum
    Replies: 0
    Last Post: Apr 19th 2011, 01:30 AM
  4. Homogeneous Fredholm equation of the second kind
    Posted in the Advanced Applied Math Forum
    Replies: 4
    Last Post: Jul 13th 2010, 08:15 AM
  5. Fredholm integral equation
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: Aug 7th 2009, 06:07 AM

Search Tags


/mathhelpforum @mathhelpforum