- #1
sarrah1
- 66
- 0
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
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