How to form a second order ODE ..

[Narasimham]

eliminating arbitrary constants p,q from sin(p*x)/p+ sin(q*x)?/q ?

[Narasimham]

How to form a second order ODE eliminating arbitrary constants p,q from
y = sin(p*x)/p+ sin(q*x)/q ?

[Gregory Toomey]

Narasimham wrote:

> How to form a second order ODE eliminating arbitrary constants p,q from
> y = sin(p*x)/p+ sin(q*x)/q ?

Your sentence contains a predicate but no subject. I suggest you go to school.
http://www.uottawa.ca/academic/arts/writcent/hypergrammar/subjpred.html

gtoomey

[Narasimham]

"How to form ice from water ?".. Is that sentence correct ? I assume it
to be correct. Replace ice with " a second order ODE" , and water with
the equation [ "y = sin(p*x)/p+ sin(q*x)/q" ] that can be grammatically
treated as an object or as single entity serving as noun. Eliminating
(a gerund or verb-noun) was short for " by eliminating" an obvious
implication in that adjective phrase.

After you read
http://www.uottawa.ca/academic/arts/writcent/hypergrammar/phrfunc.html#prepositional%20phrase

we shall discuss the grammar further in another more appropriate
newsgroup.

[Gregory Toomey]

Narasimham wrote:

> "How to form ice from water ?".. Is that sentence correct ?

No. Its not a sentence, its a predicate without a subject.

gtoomey

[William Elliot]

On Fri, 4 Nov 2005, Narasimham wrote:

> How to form a second order ODE eliminating arbitrary constants p,q from
> y = sin(p*x)/p+ sin(q*x)/q ?
>

How nice, compositionally correct. That and an equation without every
space removed just to make it hard to read. Thus you deserve the courtesy
of an answer, in spite of the grammatical correct verses idiomatic use of
language debate your post has interestingly been afflicted with by another.
(Note how I cleverly avoided ending a sentence with a proposition.)

y = (sin px)/p + (sin qx)/q (1)

y' = cos px + cos qx (2)

y" = -p.sin px - q.sin qx (3)

q = (1/x)arccos(y' - cos px) (4) via (2)

Substitute (4) into (1) and solve for
p = p(x,y,y") (5)
in terms of x, y, y".

Substitute (4) and (5) into (3) to get equation in x, y, y, y"
and no p's and q's. That's how to mind your p's and q's. ;-)

[Robert Low]

William Elliot wrote:
> How nice, compositionally correct. That and an equation without every
> space removed just to make it hard to read. Thus you deserve the courtesy
> of an answer, in spite of the grammatical correct verses

versus

If you're going to be so picky, be correct.

[john_r...@sagitta-ps.com]

William Elliot wrote:
>
> On Fri, 4 Nov 2005, Narasimham wrote:
>
> > How to form a second order ODE eliminating arbitrary constants p,q
> > from y = sin(p*x)/p+ sin(q*x)/q ?
>
> How nice, compositionally correct. That and an equation without every
> space removed just to make it hard to read. Thus you deserve the courtesy
> of an answer, in spite of the grammatical correct verses idiomatic use of
> language debate your post has interestingly been afflicted with by another.
> (Note how I cleverly avoided ending a sentence with a proposition.)
>
> y = (sin px)/p + (sin qx)/q (1)
>
> y' = cos px + cos qx (2)
>
> y" = -p.sin px - q.sin qx (3)
>
> q = (1/x)arccos(y' - cos px) (4) via (2)
>
> Substitute (4) into (1) and solve for
> p = p(x,y,y") (5)
> in terms of x, y, y".

So far so good, but how do you solve for p at this point?

> Substitute (4) and (5) into (3) to get equation in x, y, y, y"
> and no p's and q's. That's how to mind your p's and q's. ;-)

There might not _be_ a 2nd order ODE with (1) as its general
solution.

If we assume the supposed ODE is f(y, y', y'') = 0 then

0 = @f/@p = (@f/@y.@y/@p) + (@f/@y'.@y'/@p) + (@f/@y''.@y''/@p)

and similarly for q, so that either all three of @y/@p : @y/@q,
@y'/@p : @y'/@q, and @y''/@p : @y''/@q are equal, or else:

| @y'/@p @y''/@p | | @y''/@p @y/@p |
| | . @f/@y' = | | . @f/@y
| @y'/@q @y''/@q | | @y''/@q @y/@q |

and since @f/@y and @f/@y' are independent of p and q, the latter
requires the ratios of the two determinants to be independent of
p and q.

I haven't checked; but it seems unlikely that either condition
is satisfied.

The simplest approachs seems to be to find first and higher
derivatives, solve pairs of linear equations in sin(px) and
cos(px) and equate the sum of the resulting squares to 1.

For example, I think (based on a back of an envelope calculation,
which may be wrong!) subtracting the result of two such square
sums gives for p != q:

(p.q)^2.y^2 + y''^2 + (p^2 + q^2).y'^2 + 2.y'.y''' = 0

So if you can find another relation involving only the sum and
product of p^2 and q^2 then you can solve for each separately
and the required ODE can be expressed as the sum of squares
of the derivative of each equated to zero.

But I suppose it's possible, although rather perplexing, that
there's _no_ ODE, of however high an order and/or degree, for
which your expression is a solution (if p:q is irrational).

[fernando revilla]

Independently of the linguistic shades, the problem is clearly enunciated.

Fernando.

[William Elliot]

From: john_r...@sagitta-ps.com
Newsgroups: sci.math
Subject: Re: How to form a second order ODE ..

William Elliot wrote:
>
> On Fri, 4 Nov 2005, Narasimham wrote:
>
> > > How to form a second order ODE
> > > eliminating arbitrary constants p,q
> > > from y = sin(p*x)/p+ sin(q*x)/q ?
>

> > y = (sin px)/p + (sin qx)/q (1)
> > y' = cos px + cos qx (2)
> > y" = -p.sin px - q.sin qx (3)
>
> > q = (1/x)arccos(y' - cos px) (4) via (2)
>
> > Substitute (4) into (1) and solve for
> > p = p(x,y,y') (5)
> > in terms of x, y, y'.

> So far so good, but how do you solve for p at this point?

What me worry? Graphic calculators can do anything. ;-)

> > Substitute (4) and (5) into (3) to get equation in x, y, y, y"
> > and no p's and q's. That's how to mind your p's and q's. ;-)

> There might not _be_ a 2nd order ODE with
> (1) as its general solution.

I'd be unsurprised.

> If we assume the supposed ODE is f(y, y', y'') = 0 then

Oh, not g(x,y, y', y") = 0 ?

> 0 = @f/@p = (@f/@y.@y/@p) + (@f/@y'.@y'/@p) + (@f/@y''.@y''/@p)
> and similarly for q,

y_p y'_p y"_p * f_y = 0 f_y /= 0
y_q y'_q y"_q f_y' 0 f_y' 0
f_y" f_y" 0

> so that either all three of @y/@p : @y/@q,
> @y'/@p : @y'/@q, and @y''/@p : @y''/@q are equal, or else:

> | @y'/@p @y''/@p | | @y''/@p @y/@p |
> | | . @f/@y' = | | . @f/@y
> | @y'/@q @y''/@q | | @y''/@q @y/@q |

> and since @f/@y and @f/@y' are independent of p and q, the latter
> requires the ratios of the two determinants to be independent of
> p and q.

Don't get it.

--
0 y_p y'_p y"_p * g_x = 0 g_x /= 0
0 y_q y'_q y"_q g_y 0 g_y 0
g_y' g_y' 0
g_y" g_y" 0

----

[john_r...@sagitta-ps.com]

William Elliot wrote:
>
> [...]

>
> > If we assume the supposed ODE is f(y, y', y'') = 0 then
>
> Oh, not g(x,y, y', y") = 0 ?

Yes I should have expressed it like that; but it doesn't affect
the following logic because presumably @x/@p = @x/@q = 0.

> > 0 = @f/@p = (@f/@y.@y/@p) + (@f/@y'.@y'/@p) + (@f/@y''.@y''/@p)
> > and similarly for q,
>

> [...]

>
> y_p y'_p y"_p * f_y = 0 f_y /= 0
> y_q y'_q y"_q f_y' 0 f_y' 0
> f_y" f_y" 0

I don't understand this. Have some characters have been stripped out?

I can see how an ODE independent of p and q can be constructed
in principle, starting from:

(p.q)^2.y^2 - y''^2 + (p^2 + q^2).y'^2 + 2.y'.y''' = 0
[*]

This gives (p + q)^2 = polynomial in (pq) with coefficients which are
rational functions in y, y', y'', ..

I think identities similar to [*] constructed from higher derivatives
have
coefficients which are products of (p.q)^r and p^s + p^(s-1).q + ..
q^s.
The latter can be expressed as a polynomial in p+q with coefficients
which are powers of p.q, so factoring p+q out of odd powers of p+q
in every such polynomial we obtain an identity of the form:

p+q = "function of y, y', .. and p.q and (p+q)^2"

Squaring both sides of this, and plugging into the result the value
of (p+q)^2 obtained above we obtain after rearrangement another
identity (although with higher derivatives of y) which is polynomial
in p.q and with coefficients rational in y, y', ..

Every such identity can be divided by the original such identity,
thus keeping its degree in p.q below that of the original. Thus
eventually we obtain more identities than the degree of the
original, each distinct as they involve higher and higher
derivatives of y, and the condition for p.q to be non-zero
can be expressed by a determinant in the coefficients of
these identities, which are independent of p and q.

Needless to say, the resulting ODE will almost certainly be
hideously long and complicated!

[Robert Israel]

In article <1131152745.0...@g44g2000cwa.googlegroups.com>,

Narasimham <math...@hotmail.com> wrote:
>How to form a second order ODE eliminating arbitrary constants p,q from
>y = sin(p*x)/p+ sin(q*x)/q ?

Maybe not second order, but...

Write sin(p*x), sin(q*x), cos(p*x), cos(q*x) as sp, sq, cp, cq
respectively. These are of course related by the equations
sp^2 + cp^2 = 1, sq^2 + cq^2 = 1.
We want to eliminate the six variables sp, cp, sq, cq, p, q
from these equations and
y = sp/p + sq/q
y' = cp + cq
etc.
Now "generically" you should be able to eliminate six variables
from seven equations, ending up with a polynomial equation involving
y, y', y'', y''', y'''', i.e. a fourth-order ODE.
Singular comes up with
[use fixed-width font]

2 3 2 3 3
y'''' y - 4 y'''' y''' y' y + y'''' y'' y

2 2 2 2 2 4
- 3 y'''' y'' y' y + 2 y'''' y'' y + 3 y'''' y'' y' y

2 6 4
- 4 y'''' y'' y' y - y'''' y' + 4 y'''' y'

2 2 3 2 2 2 2 4
- y''' y'' y + 2 y''' y'' y' y - y''' y' y

2 2 3 2 2 3
+ 4 y''' y' y + 2 y''' y'' y' y - 4 y''' y'' y' y

2 5 3
+ 4 y''' y'' y' y + 2 y''' y'' y' - 8 y''' y'' y'

5 2 4 2 4 3 4 3 2
- y'' y + 2 y'' y' y - 3 y'' y - y'' y' + 4 y'' y' = 0

Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

[William Elliot]

On Sun, 6 Nov 2005 john_r...@sagitta-ps.com wrote:
> William Elliot wrote:
> >
> > > If we assume the supposed ODE is f(y, y', y'') = 0 then
> > Oh, not g(x,y, y', y") = 0 ?
>
> Yes I should have expressed it like that; but it doesn't affect
> the following logic because presumably @x/@p = @x/@q = 0.
>

Yes, I came to that conclusion, that your insights remained intact with
the inclusion of the above note or foot note.

> > > 0 = @f/@p = (@f/@y.@y/@p) + (@f/@y'.@y'/@p) + (@f/@y''.@y''/@p)
> > > and similarly for q,
> >
> > [...]
> >
> > y_p y'_p y"_p * f_y = 0 f_y /= 0
> > y_q y'_q y"_q f_y' 0 f_y' 0
> > f_y" f_y" 0
>
> I don't understand this. Have some characters have been stripped out?
>

Common alternative subscript notation f_x(x,y,z) = @f/@x
The matrix bars were not included, but I should have.

Problem is with square matrices and determinate notation.
One ascii resolution is to make function det:R^2n -> R
and let the | bars be used for matrices only.

<Snip>

> Needless to say, the resulting ODE will almost certainly be
> hideously long and complicated!
>

Oh boy, just what I need. Nope better not get into this much, I've yet to
reply to quasi regarding his counter example to an open question about
communative rings all of zero divisors execpt 1.

[Narasimham]

Thanks Robert. However, can we not successfully handle 6 unknown
variables p, q, sp, sq, cp, cq with 5 equations ( y = sp/p + sq/q,
y' = cp + cq , y'' = -p sp - q sq, y''' = -p^2 cp - q^2 cq, sp^2 +
cp^2 = 1, sq^2 + cq^2 = 1) in order to obtain a third order ODE ?

I ought to have at first asked about forming a simpler first order ODE
eliminating p in y = sp / p. Even this presents an implicit function
problem like y/x = sqrt(1 - y' ^2)/ arc cos( y' ) :) TIA