Solving a Limiting Situation with Fredholm & Volterra Integral Equations

[sarrah1]

Hi

I have a recurrence relation

$u(n+1)(x)=\lambda\int_{a}^{b} \,u(n)(s)ds+\mu\int_{a}^{x} \,u(n)(s)ds$ , $u(0)(x)=1$ ,
$\lambda$,$\mu$ are + constants

beware that I failed in writing $u$ subscript $n$ so what you see as $u(n)(x)$ is $u(x)$ in which $u$ is subscript of $n$ or $n+1$

so for $n=1$ we get $u(1)(x)=\lambda(b-a)+\mu(x-a)$

and I wish to study the behaviour of $u(n)(x)$ as $n$ tends to infinity

Now if the 1st term on the R.H.S is nonexistent. I get that

$u(n)(x)=\mu^n\frac{(x-a)^n}{n!}$ so I get for a bounded $x$ that $u(n)(x)$ tends to zero as $n$ tends to infinity

So what if the first terms exist . I cannot find inequalities of this sort. I have the feeling that the same result occurs provided that $\lambda (b-a)<1$ i.e. irrespective of \mu

this problem is related to fredholm and volterra integral equations

grateful Sarrah

 

[jvicens]

Hello Sarrah,

Thank you for sharing your recurrence relation with us. I am a scientist and I would be happy to assist you in understanding the behavior of $u(n)(x)$ as $n$ tends to infinity.

Firstly, I would like to clarify that your recurrence relation is a form of Volterra integral equation. This type of equation is commonly used to model various physical and biological systems. In your case, it seems to represent a population growth or decay problem.

Now, let's consider the case when the first term on the right-hand side is non-existent. As you correctly pointed out, $u(n)(x)$ tends to zero as $n$ tends to infinity. This is because the term $\mu^n$ gets smaller and smaller as $n$ increases, and eventually becomes negligible. This result is independent of the value of $\mu$, as long as it is positive.

Next, let's consider the case when the first term is present. In this case, the behavior of $u(n)(x)$ depends on the values of $\lambda$ and $\mu$. As you mentioned, if $\lambda(b-a)<1$, then $u(n)(x)$ will still tend to zero as $n$ tends to infinity. This is because the term $\lambda(b-a)^n$ will decrease faster than $\mu^n$, making the overall expression approach zero.

However, if $\lambda(b-a)>1$, then the behavior of $u(n)(x)$ will depend on the relative magnitudes of $\lambda$ and $\mu$. If $\lambda>\mu$, then $u(n)(x)$ will tend to infinity as $n$ tends to infinity. This is because the term $\lambda(b-a)^n$ will dominate and make the overall expression grow exponentially. On the other hand, if $\lambda<\mu$, then $u(n)(x)$ will tend to zero as $n$ tends to infinity. This is because the term $\mu^n$ will dominate and make the overall expression approach zero.

In conclusion, the behavior of $u(n)(x)$ as $n$ tends to infinity depends on the values of $\lambda$ and $\mu$, and whether or not the first term on the right-hand side is present. I hope this helps in understanding your recurrence relation better. Good luck with your study of Fredholm and Volterra integral equations.