A functional-differential equation
PiSigma
satisfying f'(x) = f(f(x)) and f(0) = 1.
I would appreciate any help on this matter.
Graziano
--
Ciao,
Graziano
Graziano
--
Ciao,
Graziano
alainv...@yahoo.fr
relation f '(x) = f( f(x) ) implies f '(f(x) ) =f(f '(x) )=f^[3](x) ,
so f and f ' commute ;
since f(0) = 1 we can' t expect a solution like
f(x) = c*x^a ( c and a constant values to define).
Sorry, no more to tell ya ,
Bon courage, Alain.
Robert Israel
A solution of your functional-differential equation is [an appropriate
branch of]
f(x) = (sqrt(3)+i)/2 exp(pi sqrt(3)/6) x^((1 + i sqrt(3))/2)
for x <> 0, but of course that's not real-valued and doesn't satisfy the
initial condition.
Do you have any reason to believe your problem has a solution?
Global existence is not all that common, even for ordinary
differential equations.
An integral form of your equation is
f(x) = 1 + int_0^x f(f(t)) dt
I don't know if iteration of this with a clever starting point
might converge to a solution. But a couple of attempts didn't
give promising results.
Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
C6...@shaw.ca
A continuuing problem... My Google newsreader show this as f(x) =
f(f(x)), rather than the form f '(x) = f(f(x)) as revealed by later
posts. Somehow, the "prime" gets stripped off. I am trying, with this
post, to solve the stripping problem by leaving a space between the f
and the prime. Hopefully, others will do the same, if it works.
R.G. Vickson
PS: the prime is not stripped off in the "preview post" mode, but it is
omitted in raw reader mode.
dwwdkddb
Robert Israel
Try f(x) = a x^p and see what equations a and p have to satisfy.
PiSigma
"Robert Israel" <isr...@math.ubc.ca> wrote in message
news:dm4vnp$m0t$1...@nntp.itservices.ubc.ca...
> In article <dm3qck$kc9$1...@ulysses.noc.ntua.gr>, PiSigma <psi...@tee.gr> wrote:
> >I want to find a function f : R -> R
> >satisfying f'(x) = f(f(x)) and f(0) = 1.
> >I would appreciate any help on this matter.
>
> A solution of your functional-differential equation is [an appropriate
> branch of]
> f(x) = (sqrt(3)+i)/2 exp(pi sqrt(3)/6) x^((1 + i sqrt(3))/2)
> for x <> 0, but of course that's not real-valued and doesn't satisfy the
> initial condition.
>
> Do you have any reason to believe your problem has a solution?
> Global existence is not all that common, even for ordinary
> differential equations.
I don't know if there exists a solution. A non-global, real-valued
solution would also be nice even without the initial condition (but
not the zero solution).
john_r...@sagitta-ps.com
I find that for some insane reason the Google reply box
uses a variable pitch font, which means line endings
that appear tidy there become higgledy-piggledy when
displayed fixed-pitch in preview and published format!
To avoid this, I have to prepare replies in notepad and
then paste it into the reply box, which is quite tedious
(but also a very good idea for non-trivial-length posts,
as Google has an alarming habit of failing on posts and
when Internet Explorer is used losing the text!!)
If f'(x) = f(f(x)) can't we just differentiate again,
WRT x, and use the chain rule on the RHS, to obtain
f''(x) = df/df . df/dx = f'(x) and thus conclude that
f'(x) = c.e^x, so that f(x) = c.e^x + d and then the
original condition requires x - 1 = c.(e^x - 1) ?
For any given constant, c, this gives a single point
(for real solutions). So the condition is incompatible
with a curve, and therefore arguably, as single points
can't have a derivative, there is no solution.
Unfortunately the solution above also isn't compatible
with Robert Israel's solution, which leads me to wonder
if my assumption is somehow wrong. But a sanity check
with g(x) = x^2 and h(x) = g(g(x)) = (g(x))^2, so that
h(x) = x^4, shows that the chain rule works with this
example: dh/dx = dh/d(g(x)) . dg/dx = 2g(x).2x = 4x^3.
Robert Israel
>
>"Robert Israel" <isr...@math.ubc.ca> wrote in message
>news:dm4vnp$m0t$1...@nntp.itservices.ubc.ca...
>> In article <dm3qck$kc9$1...@ulysses.noc.ntua.gr>, PiSigma <psi...@tee.gr> wrote:
>> >I want to find a function f : R -> R
>> >satisfying f'(x) = f(f(x)) and f(0) = 1.
>> >I would appreciate any help on this matter.
>>
>> A solution of your functional-differential equation is [an appropriate
>> branch of]
>> f(x) = (sqrt(3)+i)/2 exp(pi sqrt(3)/6) x^((1 + i sqrt(3))/2)
>> for x <> 0, but of course that's not real-valued and doesn't satisfy the
>> initial condition.
>>
>> Do you have any reason to believe your problem has a solution?
>> Global existence is not all that common, even for ordinary
>> differential equations.
>
>I don't know if there exists a solution. A non-global, real-valued
>solution would also be nice even without the initial condition (but
>not the zero solution).
The integral equation
f(x) = f(a) + int_a^x f(f(t)) dt
suggests considering iteration of the operator T where
Tf(x) = a + int_a^x f(f(t)) dt
Consider this on continuous functions on the interval J = [a-r,a+r]
where r > 0.
Note that if f maps J into J, then so does Tf if
|a| + r <= 1. For convenience, I'll write M = |a| + r.
Moreover, Tf is differentiable in (a-r,a+r)
with (Tf)'(x) = f(f(x)), so in particular |(Tf)'| <= M.
Now note that
|Tf(x) - Tg(x)| <= int_a^x |f(f(t)) - g(g(t))| dt
<= int_a^x (|f(f(t)) - f(g(t))| + |f(g(t)) - g(g(t))|) dt
Suppose f and g map J into J, with
|f'| <= M and |f - g| <= D on this interval. Then
|Tf(x) - Tg(x)| <= int_a^x (M D + D) dt <= r (M + 1) D
We want T to be a contraction on a suitable set of functions,
so assume r (M + 1) < 1. Thus we consider the iteration
f_1(x) = a
f_{n+1} = T(f_n)
and we get estimates
|f_{n+1}(x) - f_n(x)| <= r (M + 1) |f_n(x) - f_{n-1}(x)|
<= ... <= (r (M+1))^(n-1) a r for x in J.
So if |a| + r < 1 and r (|a| + r + 1) < 1 the sequence f_n converges
uniformly to a function f on the interval J, and T(f) = f. Thus
f'(x) = f(f(x)) and f(a) = a. Moreover, f can be written formally
as a power series around x = a:
f(x) = a + a (x - a) + a^2 (x - a)^2/2 + (a^3 + a^4) (x - a)^3/6 + ...
and I think this should converge for |x - a| < r.
Robert Israel
<john_r...@sagitta-ps.com> wrote:
>
>C6...@shaw.ca wrote:
>>
>> PiSigma wrote:
>> >
>> > I want to find a function f : R -> R
>> > satisfying f'(x) = f(f(x)) and f(0) = 1.
>> > I would appreciate any help on this matter.
>If f'(x) = f(f(x)) can't we just differentiate again,
>WRT x, and use the chain rule on the RHS, to obtain
>f''(x) = df/df . df/dx = f'(x)
No. The chain rule says d/dx (f(f(x)) = f'(f(x)) f'(x).
john_r...@sagitta-ps.com
My first reply to this hasn't appeared, and hopefully
it never will because differentiating by parts to give
"f'' = df/df . df/dx" and assuming "df/df" equals 1
is nonsense as this is actually d(f(f(x))/d(f(x)) !
However, I think one can do better by thinking of x
and f(x) as both being parametrized by some other
value whose inverse is t = g(x) and for which f
is a differential operator.
If we let f(x) = dx/dg + d^2(x)/dg^2 then f' = f o f
gives:
df/dx = (d/dg + d^2/dg^2)(dx/dg + d^2(x)/dg^2)
= d^2(x)/dg^2 + 2.d^3(x)/dg^3 + d^4(x)/dg^4
= dx/dg . d/dx(dx/dg + 2.d^2(x)/dg^2 + d^3(x)/dg^3)
Integrating this by parts WRT x as it stands using the
rule "first times integral of second minus integral of
(diff of first times integral of second)" and noticing
that the "diff of first" WRT x, i.e. d/dx(dx/dg), is
d/dg(dx/dx) which is zero, and therefore the second
integral is a constant we obtain a 2nd order ODE in
G := dx/dg
(f =) G + dG/dg = G.(G + 2G.dG/dg + d^2(G)/dg^2)
In principle, this can be solved as G = G(g) with
a couple of constants of integration and then f(x)
and x both expressed parametrically in terms of g.
(Expressing x = x(g) requires another integration,
and hence _another_ constant of integration - where
the hell are they all coming from ?!)
Trying the same trick with just f(x) = dx/dg doesn't
work, and nor does incorporating functions of x as
coefficients of the operator, as this prevents dx/dg
being factored in the expression for df/dx (except
possibly by imposing elaborate conditions on these
functions, which seems fairly pointless as there is
a simple alternative of assuming they are constant
which we have done).
But one could adjust the constant coefficients to
try and end up with an ODE in G which can be more
readily solved, e.g. whose terms can be collected
entirely into exact derivatives.
Cheers
John R Ramsden (jhnr...@yahoo.com.uk)
* Remove m from com to reply
* From address is defunct
john_r...@sagitta-ps.com
john_r...@sagitta-ps.com wrote:
>
> [..] noticing that the "diff of first" WRT x, i.e. d/dx(dx/dg),
> is d/dg(dx/dx) which is zero,
No this isn't right either, because x and g are not independent.
In fact d/dx(dx/dg) = d/dx(1 / dg/dx) = - d^2(g)/dx^2 / (dg/dx)^2
However, one can make a lot more progress by assuming:
f(x) = (ax + b).dx/dg
and I'll post the details to tomorrow (unless someone else
gets time to nip in first ..)
Robert Israel
More generally, suppose f maps the disk U = {z: |z - a| <= r} in the
complex
plane into itself, and satisfies f(b) = c where b and c are in U. This
leads to
the operator
Tf(z) = c + int_b^z f(f(t)) dt
and then Tf also maps U into itself if |c - a + a (a - b)| <= (1 - |a|
- |a-b| - r) r.
If f is the image of such a function under T, |f'(z)| <= |a| + r on U,
and then
T is a strict contraction on such functions if (r + |b-a|) | (|a| + r +
1) < 1.
If so, the sequence f_n with f_1 = c and f_{n+1} = T f_n converges
uniformly
on U to a function f that is analytic in the interior of U, satisfying
your equation
with f(b) = c. The region of analyticity could turn out to be larger.
For example, you could take a = b = 0, c = 1/4, r = 1/2. The Taylor
series of
the function f, which by the above analysis is guaranteed to converge
for
|z| < 1/2, is approximately
1/4+ .3388226478 x + .06310286283 x^2 + .01058266737 x^3 +.001876688787
x^4 +.0003615747008 x^5 + .00007425636819 x^6 + .00001596727294 x^7
+ .000003555019431 x^8 + .0000008135397704 x^9 + ...
This approximation maps the disk of radius approximately 3.5 centred at
0 into itself,
and the series appears to converge in that disk.
Robert Israel
Robert Israel wrote:
> Robert Israel wrote:
> > In article <dm6d8f$1tf3$1...@ulysses.noc.ntua.gr>, PiSigma <psi...@tee.gr> wrote:
> > >
> > >"Robert Israel" <isr...@math.ubc.ca> wrote in message
> > >news:dm4vnp$m0t$1...@nntp.itservices.ubc.ca...
> > >> In article <dm3qck$kc9$1...@ulysses.noc.ntua.gr>, PiSigma <psi...@tee.gr> wrote:
> > >> >I want to find a function f : R -> R
> > >> >satisfying f'(x) = f(f(x)) and f(0) = 1.
> > >> >I would appreciate any help on this matter.
> > >>
> > >> A solution of your functional-differential equation is [an appropriate
> > >> branch of]
> > >> f(x) = (sqrt(3)+i)/2 exp(pi sqrt(3)/6) x^((1 + i sqrt(3))/2)
> > >> for x <> 0, but of course that's not real-valued and doesn't satisfy the
> > >> initial condition.
> > >>
> > >> Do you have any reason to believe your problem has a solution?
> > >> Global existence is not all that common, even for ordinary
> > >> differential equations.
> > >
> > >I don't know if there exists a solution. A non-global, real-valued
> > >solution would also be nice even without the initial condition (but
> > >not the zero solution).
Still more generally, suppose U is a convex open proper subset of the
complex plane
containing b and c, and let S be the set of analytic functions f: U ->
closure(U) such
that f(b) = c. Suppose T (as defined above) maps S into itself. Now S
is convex and compact in the topology of uniform convergence on compact
subsets of U, and the Tychonoff fixed point theorem says that T has a
fixed point.
This fixed point will be an analytic function on U that satisfies f'(z)
= f(f(z)) and f(b) = c.
michaeld
Robert Israel has given some local existence results. In fact if I
haven't made an error it can be shown there is a global solution f:
R->R but it is strictly decreasing and has a fixed point in (-1,0).
On the other hand it's easy to show there is no global solution f: R->R
with f(0) > 0. Such a solution would have to have f(x) > 0 for all x >
0 - otherwise let x = inf{x' > 0 | f(x') = 0}. Note that x > 0 and f(x)
= 0 but f'(x) = f(0) > 0, which is impossible.
It quickly follows that f^(n)(x) > 0 for all x > 0 so that f(x)/x^n ->
infinity as x->infinity, for all n. In particular, we can find a > 1
with f(a) > a. A straightforward induction using the
functional-differential equation shows then that:
f^(n)(a) >= a^(n(n-1)/2+1)
so that f^(n)(a)/n! -> infinity as n->infinity. This contradicts the
remainder form of Taylor's theorem:
f(a+h) = f(a) + f'(a)h/1! + ... + f^(n)(a)h^n/n! +
f^(n+1)(a+th)t^(n+1)/(n+1)!
>= f(a) + f'(a)h/1! + ... + f^(n)(a)h^n/n!
(as f^(n+1) > 0)
because the left hand side is a fixed number and the right hand side
tends to infinity as n->infinity.
Michael
Robert Israel
michaeld <mich...@cantab.net> wrote:
>PiSigma wrote:
>> I want to find a function f : R -> R
>> satisfying f'(x) = f(f(x)) and f(0) = 1.
>> I would appreciate any help on this matter.
>Robert Israel has given some local existence results. In fact if I
>haven't made an error it can be shown there is a global solution f:
>R->R but it is strictly decreasing and has a fixed point in (-1,0).
I'd be interested in seeing your proof of this.
michaeld
> In article <1133208437....@g47g2000cwa.googlegroups.com>,
> michaeld <mich...@cantab.net> wrote:
> >PiSigma wrote:
> >> I want to find a function f : R -> R
> >> satisfying f'(x) = f(f(x)) and f(0) = 1.
> >> I would appreciate any help on this matter.
>
> >Robert Israel has given some local existence results. In fact if I
> >haven't made an error it can be shown there is a global solution f:
> >R->R but it is strictly decreasing and has a fixed point in (-1,0).
>
> I'd be interested in seeing your proof of this.
OK, well in brief outline:
The formal substitution y = f(x) gives dy/dx = f(y), which suggests the
equation:
integral[-1/2 ... f(x)] dy/f(y) = x + 1/2
Now start off with a local (strictly decreasing) solution f: (-1,0) ->
(-1,0) with a fixed point at -1/2 say - this can be constructed as you
outlined using fixed point theorems. The integrand on the left hand
side of the above makes sense for -1/2 < y < 0 (since f(y) has already
been defined) so this can be used to define the f(x) (in the integral's
limit) upto a point a < -1 such that f(a) = 0. So we now have a
solution f: [a,0) -> (-1,0] with f(a) = 0.
Doing something similar but integrating from -1/2 down to a gives us an
extension to f: [a,infinity) -> (a,0) with f(x) -> a as x->infinity.
Finally integrating from -1/2 to infinity extends f to R. I think then
a simple check shows that f'(x) = f(f(x)). I haven't really thought
about whether this is likely to be analytic - though very naively you'd
probably expect so, because f is extended by inverting integrals of
functions that are (locally) analytic.
Maybe I'll be able to give a better explanation later.
Michael
michaeld
[...]
OK, I'll try and write this out a bit better.
Start with the analytic solution f: [-1,0] -> (-1,0) to f' = f o f with
f(-1/2) = -1/2. This is the solution given by the Taylor series you
wrote down when the fixed point is set to -1/2. You can show using
fixed point theorems that the radius of convergence is > 1/2 so that
the power series converges on [-1,0].
This function is strictly decreasing. Let a = f(-1) so that -1 < a < 0.
Define now:
F(t) = -1 + integral[a ... t] dy/f(y)
This makes sense say for a <= t <= 0 because f is defined and
continuous on [a,0].
Then, defined on [a,0], we have that F is continuous, strictly
decreasing and F(a) = -1. Let b = F(0). Then F: [a,0] -> [b,-1] is a
bijection; in fact it is differentiable with negative derivative, F'(t)
= 1/f(t) (where left/right differentiability is understood at the
endpoints).
Therefore F has an inverse, which is a bijection [b,-1] -> [a,0] that
is differentiable with negative derivative (again understanding
left/right differentiability at the endpoints). Define this inverse to
be f, so that F(f(x)) = x for x in [b,-1]. This is OK because before
now we only defined f on [-1,0] so we haven't defined anything twice
(except for at -1 but both definitions give us the value a).
We now have f: [b,0] -> [-1,0]. To check it is differentiable at x =
-1... well it's clearly continuous here. The right derivative exists
and equals f(f(-1)) = f(a) (because of the original definition of f on
[-1,0].) The left derivative exists and equals 1/F'(a) which is also
f(a). Hence we are differentiable at x = -1. We're clearly
differenentiable everywhere else on [b,0] (again with left/right
derivatives being understood as appropriate).
Therefore f: [b,0] -> [-1,0] is differentiable, with negative
derivative, and with f(b) = 0. And we have that:
x = -1 + integral[a ... f(x)] dy/f(y)
for x in [b,-1].
Since f is differentiable we can differentiate the above equation and
use FTC to deduce that 1 = f'(x)/f(f(x)) so that f'(x) = f(f(x)) for x
in [b,-1]. Therefore this holds on [b,0].
So we have a strictly decreasing f: [b,0] -> [-1,0] satisfying f'(x) =
f(f(x)) and f(b) = 0.
Next look at:
G(t) = b - integral[t ... 0] dy/f(y)
for b < t <= 0. This defines a differentiable function with G(0) = b
and G(t) -> infinity as t->b (since f(t) tends to 0 linearly as t -> b
so its integral diverges). By inverting this we can extend f to
[b,infinity] and this gives us a differentiable f: [b,infinity] ->
[b,0] satisfying f' = f o f.
Finally the extension to R is obtained by looking at:
H(t) = b + integral[0 ... t] dy/f(y)
for 0 <= t < infinity. H maps [0,infinity) to (-infinity,b]
surjectively (as f(t)->b as t->infinity) so we invert this to get a
function from (-infinity,b] to [0,infinity) and we can check again that
f' = f o f everywhere.
So we get a function f: R->R that is differentiable with f' = f o f,
f(-1/2) = -1/2, f(b) = 0, f(x) -> b as x->infinity and f(x) ~ -bx as
x->-infinity for some b < -1.
Does that look right now?
Btw, I changed my mind about the function being analytic. It's probably
locally analytic or something but I personally very much doubt it
extends to an analytic function f: C->C.
Michael
Robert Israel
michaeld <mich...@cantab.net> wrote:
>Btw, I changed my mind about the function being analytic. It's probably
>locally analytic or something but I personally very much doubt it
>extends to an analytic function f: C->C.
So f should be analytic on a neighbourhood of R, but probably not entire.
Indeed, from calculating the first 20 Taylor coefficients of the
series of f around -1/2, it looks to me like the radius of convergence
should be somewhere near 4.
Narasimham
If y =f (x) , y' = f (y) .So if f is given we can solve for y = f (x).
E.g., if f (u) = u, y = c e^x and if f(u) = sqrt(1+u^2), then y =
sinh(x + c) etc.
But if the functional dependence f is not explicitly given, your
problem is indeterminate or underspecified...However, I like your
question, nee a trap.
PiSigma
"Narasimham" <math...@hotmail.com> wrote in message
news:1133382786.2...@g47g2000cwa.googlegroups.com...
> > Find a function satisfying f '(x) = f(f(x))
>
> If y =f (x) , y' = f (y) .So if f is given we can solve for y = f (x).
> E.g., if f (u) = u, y = c e^x and if f(u) = sqrt(1+u^2), then y =
> sinh(x + c) etc.
If f is given there is no unknown function...
>
> But if the functional dependence f is not explicitly given, your
> problem is indeterminate or underspecified...However, I like your
Robert Israel and michaeld have given local solutions, so I don't
see in what sense the problem is indeterminate or underspecified.
> question, nee a trap.
>
??
michaeld
The solution f: R->R I gave is global on R.... it's just that
(according to various heuristic arguments and Robert's numerical
evidence) it does not extend to the complex plane C.
It also violates the initial condition f(0) = 1 you gave, but as I
noted earlier there is no global f: R->R satisfying f' = f o f and f(0)
> 0.
As for questions of uniqueness.... I'm not completely sure whether the
solution f: R->R is the unique solution with f(-1/2) = -1/2. I _think_
I can show that it is, but I have to check a few details. On the other
hand it's very easy to see that the local solutions with positive fixed
points are (locally) unique. For example, there is a unique solution
f:(0,1) -> (0,1) with f(1/2) = 1/2.
Michael
Narasimham
Shall be reverting to it later, doubt if it is adequately well-posed.
For time being took liberty to post your problem into Mathematica
group, hope it is in order...
Among those discussed, Andrzej Kozlowski 's numerical solution for a
"chained" function EQ = { f ' [x] == f [ Cos[x] ], f [0] == 1} works
out alright.
Regards
Narasimham
Dave Rusin
by now. The question was about solutions to the equation
f' ( x ) = f ( f ( x ) )
where, during the thread, the initial condition f ( -1/2 ) = -1/2 was
added. (Equivalently, f(x) = h(x+1/2) - 1/2 where h now satisfies
h'= h o h - 1/2 and h(0)=0.)
In article <dmjjfg$gkj$1...@nntp.itservices.ubc.ca>,
Robert Israel <isr...@math.ubc.ca> wrote:
>So f should be analytic on a neighbourhood of R, but probably not entire.
>Indeed, from calculating the first 20 Taylor coefficients of the
>series of f around -1/2, it looks to me like the radius of convergence
>should be somewhere near 4.
I extended the calculation to the first 50+ coefficients. (The
coefficient of x^n is an integer divided by n! 2^(-1 + n(n-1) ).
The numerators grow to several hundred digits by n=50.)
As Robert suggests, the numerators are of comparable magnitude
after multiplication by 4^n -- but not quite! In the range he
checked, all but the first few of these products are smaller (in
magnitude) than 0.5; but the 23rd is around 0.57; the 31st is 1.06; the
41st is nearly 2.0, the 44th is nearly 4.0, and the 53rd is nearly 8.0.
So I don't have quite so much confidence that the series
converges on a disc that large!
I also checked for divergence of the series on the circle of radius 4.
It seems that the values of the partial sums are growing fastest
when x is near 4 exp(+-2i) , which is interesting (but it's hard
to say much from limited data).
So I am led to ask how one might obtain better information about
the function. Here are a couple of things I tried, which prompt
more questions.
1. What's a good way to solve such an equation numerically?
Maple balked at the request for a numeric solution. I tried to
think of some clever way to expand the function in terms of, say,
a basis of orthogonal polynomials, but there didn't seem to be
a good way to use the defining equation to rewrite the integrals
in any useful way. Some good numerical approximations might give
support to the conjecture that the function is analytic along
the circle, except at those two points. To discover that the
critical points are (numerically) exactly at exp(2i), or exp(2/3 pi)
or whatever might give some clues to the behaviour of the function.
2. Since the defining equation requires computing composites, I
thought of extending the problem in the following way. Suppose
we look for a commuting family of operators H_t on (a neighborhood
of the origin in) the complex plane. In the present case I mean
by that a set of functions H_t(x) such that
H_s( H_t(x) ) = H_t( H_s(x) )
The thinking is that both sides should represent H_{st} (x),
so that in particular H_1(x) will be the identity map. We can
compute the power series necessary to guarantee as well that
H_{-1/2} = h; more generally H_t(x) is the unique function in
the family for which (H_t)'(x) = t. Here are the first few terms
of the expansion:
H_t(x) = t*x + 1/6*t*(t-1)*x^2 + 1/36*(t-1)*t*(3*t+1)*x^3
+ 1/5184*(155*t^2-61*t-85)*t*(t-1)*x^4 + ...
The point of introducing the H's is that it is no longer necessary
to refer to composites; the composite of two H's is another H.
Thus the ODE which mentions composites is equivalent to
dH/dx (x,-1/2) = H (x, 1/4) - 1/2
So if, say, someone knows how to solve delay-differential equations,
e.g. dH/dx(x,t) = H(x,t+3/4) - 1/2, then that person could solve
the original equation (using whatever interpretation of "solve" one
prefers). Not being a differential equations guy myself, I didn't
know how to go any further. (I was hoping to see some nice patterns
involving dH/dx and dH/dt after finding the power series above,
but nothing obvous leaps out.)
One of the reasons I thought to pursue this a bit more was that
in Oct 2000 we had a thread about a differential equation
(x^2 y')' = -x^2 y^2,
which, to my surprise, had an _essential_ singularity some distance away
from the point where we were computing Taylor series. I wonder whether
this equations has the same feature? The old thread had message-ID
<8t3fsc$ms1$1...@news1.math.niu.edu>.
dave