From: rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.math.research
Subject: Re: Any ideas on skipping from x'=-g(x) right to \int h(x_t)dt ><infty?
Date: 17 Sep 1995 17:52:47 GMT

In article <43dm3e$cce@senator-bedfellow.mit.edu>,
Lones A Smith <lasmith@athena.mit.edu> wrote:
>Suppose I know that x'=-g(x) where g(0)=0 and g is increasing for x>0.
>Initially, say x_0>0 is known. I wish to know whether 
>
>\int_{t=0}^infty  h(x_t) dt >infty     or    <infty
>
>where h(0)=0 and h is increasing for x>0. 
>BUT I do not have any functional forms for g and h, and thus cannot 
>simply solve the DE and then plug its solution into the integral.
>
>Is there any way of using known relationships between g and h near 0
>to definitively resolve this problem? 

No, but the behaviour of  g  and  h  across (0, oo) will do. Your
integral may be written
	\int_{t=0}^\infty [h(x(t))/g(x(t))].g(x(t))dt,
which is useful since g(x(t))=x'(t) by assumption. Thus the change-of-variables
formula gives this integral as
	\int_{x=x_0}^\lastx  [h(x)/g(x)] dx
where  \lastx  is the upper bound on the image  x(0,oo). (warning: this need 
not be \infty, as the example  g(x)=x^2+1 shows.)

Thus what you need to get a finite integral is precisely that  |h/g| be
integrable across this range. The relative behaviours of  h  and  g  near  0
will only determine whether or not  h/g  is locally integrable there; you
definitely need to know the long-term behaviour of this quotient to
resolve the problem as given.

dave

==============================================================================

From: rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.math.research
Subject: Re: Any ideas on skipping from x'=-g(x) right to \int h(x_t)dt ><infty?
Date: 18 Sep 1995 08:18:05 GMT

In article <43hn9f$qa@watson.math.niu.edu>, I wrote:

>In article <43dm3e$cce@senator-bedfellow.mit.edu>,
>Lones A Smith <lasmith@athena.mit.edu> wrote:
>>Suppose I know that x'=-g(x) where g(0)=0 and g is increasing for x>0.
                         ^
One teensy little minus sign escaped my attention.

So you have  dx/dt = -g(x). Separating variables,  G(x) = t,  where  G
is some antiderivative of  -1/g. Since  g  is positive and increasing, 
G'=-1/g is negative and increasing, meaning  G  is decreasing and concave
up. These adjectives describe the graph of  G,  which shows  t  as a
function of  x; the graph of  x  as a function of  t  is then a reflected
version of this, which is also decreasing and concave up. The fact that
-1/g  has a pole at zero does not mean  G  does (e.g. if  g(x) = x^(1/2) and
x(t)=(1-t)^2.) Thus the inverse function describing  x  as a function of  t
may not have a domain including all of  [0, \infty)  (at least, not a
domain on which  dx/dt  continues to be a negative function of  x.)

Since in the sequel you seem to want to let t--> \infty, you must be
assuming that  g  has a high enough order of vanishing at  0  that it is
not locally integrable there. Then we see  lim_{t->\infty} x(t) must be zero.
(OK, that's not completely obvious but it does follow from the assumption
that the given properties of  g  hold on all of  [0, x_0].)

>>Initially, say x_0>0 is known. I wish to know whether 
>>
>>\int_{t=0}^infty  h(x_t) dt >infty     or    <infty
>>
>>where h(0)=0 and h is increasing for x>0. 
...
>Your integral may be written
>	\int_{t=0}^\infty [h(x(t))/g(x(t))].g(x(t))dt,
>which is useful since g(x(t))=x'(t) by assumption. Thus the change-of-variables
>formula gives this integral as
>	\int_{x=x_0}^\lastx  [h(x)/g(x)] dx
>where  \lastx  is the upper bound on the image  x(0,oo). (warning: this need 
>not be \infty, as the example  g(x)=x^2+1 shows.)

I have no idea what that example was supposed to show. (Oh, yes I do, but
I'll pretend I don't; it's less embarassing.) Anyway, in view of what we said
about the graph of  x(t), we get the integral to be more properly
	\int_0^{x=x_0} [h(x)/g(x)] dx.

>Thus what you need to get a finite integral is precisely that  |h/g| be
>integrable across this range. The relative behaviours of  h  and  g  near  0
>will only determine whether or not  h/g  is locally integrable there; you
>definitely need to know the long-term behaviour of this quotient to
>resolve the problem as given.

For "long-term" read "on [0, x_0]", but since you assumed  g  and  h  both
to be increasing, this is equivalent to assuming  h/g  is locally integrable
at both  0  and  x_0. Since you have assumed  g  vanishes
reasonably quickly as  x -> 0, h  will have to vanish a little faster there.
Near  x_0, it is sufficient to assume that  h  stays bounded, or increases
to infinity fairly slowly, although if  g  increases to infinity at  x_0
then  h  can increase a little faster. 

Just in case you're tempted to jump to a conclusion:  h/g  need not be
monotonic near either endpoint; there can be wild swings in the magnitude
of  h/g  which of course can enable or disallow integrability.

dave

==============================================================================

Date: Mon, 18 Sep 95 15:29:18 CDT
From: rusin (Dave Rusin)
To: lasmith@MIT.EDU
Subject: Re: Any ideas on skipping from x'=-g(x) right to \int h(x_t)dt ><infty?

>     Thanks for the reply. I realized ex post that the problem would be much
>easier in the "autonomous" case than in the other case I really cared about,
>namely 
>
>\int t h(x_t)dt ><infty
>
>Do you know of any general approach to a (modified) problem like this, or had
>you just noticed an ad hoc approach that you do not think extends?

No problemo. Let  K(t)  be an antiderivative of  h(x(t)). Then the integrateion
by parts formula writes your integral as  t K(t) - \int K(t) dt.
Now,  K we find as in the previous post:
  K =\int h(x(t)) dt = \int (h/g)(x) dx,
that is,  K(t) = H(x(t))  where  K(x)  is an antiderivative of  h/g. So
that last term in integration by parts is  \int H(x(t)) dt  which looks
just like yesterday's integral but capitalized! It comes out to
\int H(x)/g(x)  dx.
	To recap: we find functions of  x  with  H'(x) = h(x)/g(x) and
L'(x) = H(x) / g(x); then your integral is
	t H(x(t)) - L(x(t))
and is finite as long as  H  and  L  are finite on [0, x_0], with now
the extra condition that   H(x)  decreases to zero quickly enough as
z --> 0  (so that  lim_{t --> oo} t*H(x(t))  stays finite). Oh, I suppose
there are other cases in which the integral can be finite (with the
growth in  H  and  L  cancelling out) but your interest seems to be
in gaining the desired conclusions by asserting enough conditions
about the behaviours of  h  and  g  near 0.

By the way, the assumption that  h  and  g  are defined and increasing
_all_ the way from 0  to  x_0  removes one of the problems I left you with
in the last post: in that case  h/g  will be bounded (and hence integrable)
near  x_0, so that only the behaviour at  x=0  matters.

I will also comment (with regard to your other post) that for any
decreasing function  f  on [0,oo)  we have that  \int f dt < oo  iff
Sum(f(n), n=1, ...., oo)  is finite; applied to  f(t) = h(x(t)), this
shows that the integral mimics the discrete problem correctly. On
the other hand, the condition  x'=-g(x)  does not give quite the same
assumptions as  x_n+1 - x_n = -g(x_n), so I don't quite know what to
say about that part (that's why I didn't followup).

dave

==============================================================================

Date: Tue, 19 Sep 95 16:40:24 -0400
From: [Permission pending]
To: rusin@math.niu.edu (Dave Rusin)
Subject: Re: Any ideas on skipping from x'=-g(x) right to \int h(x_t)dt ><infty?

Hello!

>>     Thanks for the reply. I realized ex post that the problem would be much
>>easier in the "autonomous" case than in the other case I really cared about:
>>
>>\int t h(x_t)dt ><infty
>>
>>Do you know of any general approach to a (modified) problem like this, or
>>had you just noticed an ad hoc approach that you do not think extends?


>	To recap: we find functions of  x  with  H'(x) = h(x)/g(x) and
>L'(x) = H(x) / g(x); then your integral is
>	t H(x(t)) - L(x(t))
>and is finite as long as  H  and  L  are finite on [0, x_0], with now
>the extra condition that   H(x)  decreases to zero quickly enough as
>z --> 0  (so that  lim_{t --> oo} t*H(x(t))  stays finite).

The problem with this approach is that it is not autonomous.
You cannot deduce this condition without knowing the solution
x(t) of the DE.

>I will also comment (with regard to your other post) that for any
>decreasing function  f  on [0,oo)  we have that  \int f dt < oo  iff
>Sum(f(n), n=1, ...., oo)  is finite; applied to  f(t) = h(x(t)), this
>shows that the integral mimics the discrete problem correctly. On
>the other hand, the condition  x'=-g(x)  does not give quite the same
>assumptions as  x_n+1 - x_n = -g(x_n), so I don't quite know what to
>say about that part (that's why I didn't followup).

Basically, I am thinking of the Euler constant (I think that is what it is 

called), namely .577.... which is the difference between the curve
1/x between x=1 and x=infty and the staircase f(x)=1/n for n<=x<n+1.
That is, both are infinite because they differ by a finite number.

If this is true in general, for monotonic functions g and h as I described,
then one integral is finite whenever the other is. 


So, is the diference between the functions always finite?

[sig deleted]

PS I think I shall send you the greater context for this, since you
do seem to be interested, and I am eager for any insights you might 

have.