Fredholm Intergral Equation

**ubhik** · April 22nd 2008, 01:45 AM

I need to find a general solution,

$f$ ∈ $E [0, \infty)$

for the Fredholm intergral equation:

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

where $M$ is the intergral operator defined by

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

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

I think, that I have to rewrite it:

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

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

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

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

**kolobok** · April 22nd 2008, 02:53 AM

IMHO
The next step after this:
$ 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:
$f(x)=\gamma cos x + \delta sin x+\lambda x^{a-1} C_1$ ,
where $C_1=\int_0^{\infty} e^{-bt}f(t)dt=\int_0^{\infty} e^{-bt}(\gamma cost + \delta sin t + \lambda t^{a-1} C_1)dt$
Knowing $C_1$ you will find general solution by substituting it into the first equation in this message.

**ubhik** · April 23rd 2008, 01:10 AM

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

I.e.

Just with $x$ replaced with $t$ 's

**kolobok** · April 23rd 2008, 02:32 AM

Did you check this?

Note that you have equation for $C_1$ where this stuff takes part in both sides of the equation.

**ubhik** · April 23rd 2008, 02:43 AM

Well basically,

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

and subsitute it into

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

Then all I get is

$ 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?

**kolobok** · April 23rd 2008, 08:25 AM

No. Look at this:
You have:
$f(x)= \gamma cos x + \delta sin x + \lambda \int_0^{\infty} x^{a-1} e^{-bt} f(t)dt$
This equation has a solution:
$f(x)=\gamma cos x + \delta sin x+\lambda x^{a-1} C_1$
where $C_1=\int_0^{\infty} e^{-bt}f(t)dt$
Lets find $C_1$ in terms of $\lambda$ :
$C_1=\int_0^{\infty} e^{-bt}f(t)dt=\int_0^{\infty} e^{-bt}(\gamma cost + \delta sin t + \lambda t^{a-1} C_1)dt = I_1 + I_2+ I_3$
$I_1=\gamma \int_0^{\infty} e^{-bt} cost dt = \frac{b \gamma}{b^2+1}$
$I_2=\delta \int_0^{\infty} e^{-bt} sint dt= \frac{\delta}{b^2+1}$
$I_3=-\lambda C_1 K$ .
So:
$C_1=\frac{b \gamma + \delta}{b^2+1}-\lambda C_1 K$
$C_1=\frac{b \gamma + \delta}{(b^2+1)(1+\lambda K)}$
if $\lambda$ doesn't equal $-\frac{1}{K}$ .
Now you should substitute $C_1$ into
$f(x)=\gamma cos x + \delta sin x+\lambda x^{a-1} C_1$

**ubhik** · April 24th 2008, 04:39 AM

Thank you very much that was really helpful!!!

**ubhik** · April 30th 2008, 05:15 AM

I need to find a general solution,

$f$ ∈ $E [0, \infty)$

for the Fredholm intergral equation:

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

where $M$ is the intergral operator defined by

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

Where $a$ and $b$ are postive constants and $\gamma, \delta$ and $\lambda$ ∈ 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?

**ubhik** · May 6th 2008, 11:32 AM

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

$ 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...

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