General Mathematical Theorem?

Bert

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nHello,\n\nIn Schrodinger\'s Statistical Thermodynamics on pg.12 he states->\n\nTo give a more direct proof:\n\nF+U*mu=G\n\nThen, from a general mathematical theorem, the ratio of two integrating factors\n\n1/T and mu is a function of G.\n\n\nWhat general mathematical theorem is he talking about?\n\nIf anyone is familiar with this work could you let me know?\n\nCheers,\n\nBert\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Hello,

In Schrodinger's Statistical Thermodynamics on pg.12 he states->

To give a more direct proof:

[tex]F+U*\mu=G[/tex]

Then, from a general mathematical theorem, the ratio of two integrating factors

1/T and \mu is a function of G.

What general mathematical theorem is he talking about?

If anyone is familiar with this work could you let me know?

Cheers,

Bert

Igor Khavkine

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nOn Mon, 30 Aug 2004 04:57:37 -0400, Bert wrote:\n\n>\n>\n> Hello,\n>\n> In Schrodinger\'s Statistical Thermodynamics on pg.12 he states->\n>\n> To give a more direct proof:\n>\n> F+U*mu=G\n[...]\n\nNot everyone has a copy of this book. Some context and an explanation\nof the notation you are using would be helpful in answering your question.\n\nIgor\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On Mon, 30 Aug 2004 04:57:37 [itex]-0400,[/itex] Bert wrote:

>
>
> Hello,
>
> In Schrodinger's Statistical Thermodynamics on pg.12 he states->
>
> To give a more direct proof:
>
> [itex]F+U*\mu=G[/itex]
[...]

Not everyone has a copy of this book. Some context and an explanation
of the notation you are using would be helpful in answering your question.

Igor

diamonis

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nAs I understand it, having the book, the problem referred to is:\n(for the 2 dimensional case but really asked for any dim)\n\ngiven an integrating factor mu(x,y) for the expression Pdx + Qdy = 0,\nwhere P=P(x,y), Q=Q(x,y) are functions of (x,y) so that\nmu * { Pdx+Qdy } = dz for some function z(x,y) (ie z(x,y)=constant\nsolves the diff eqn Pdx+Qdy=0), then for any other integrating factor\nmu1(x,y) then\nmu1 / mu = F(z), that is, mu1/mu must be a function of z(x,y) only.\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>As I understand it, having the book, the problem referred to is:
(for the 2 dimensional case but really asked for any dim)

given an integrating factor [itex]\mu(x,y)[/itex] for the expression Pdx + Qdy [itex]= 0,[/itex]
where [itex]P=P(x,y), Q=Q(x,y)[/itex] are functions of (x,y) so that
[itex]\mu * { Pdx+Qdy } = dz[/itex] for some function z(x,y) (ie z(x,y)=constant
solves the diff eqn [itex]Pdx+Qdy=0),[/itex] then for any other integrating factor
mu1(x,y) then
mu1 [itex]/ \mu = F(z),[/itex] that is, [itex]mu1/\mu[/itex] must be a function of z(x,y) only.

Stephen Parrott

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\ndiamonis wrote:\n> As I understand it, having the book, the problem referred to is:\n>\n> (for the 2 dimensional case but really asked for any dim)\n>\n>\n>\n> given an integrating factor mu(x,y) for the expression Pdx + Qdy = 0,\n>\n> where P=P(x,y), Q=Q(x,y) are functions of (x,y) so that\n>\n> mu * { Pdx+Qdy } = dz for some function z(x,y) (ie z(x,y)=constant\n>\n> solves the diff eqn Pdx+Qdy=0), then for any other integrating factor\n>\n> mu1(x,y) then\n>\n> mu1 / mu = F(z), that is, mu1/mu must be a function of z(x,y) only.\n>\n\nYour conclusion is correct (modulo technical hypotheses),\nbut I don\'t know of any elementary text that explains clearly the reason.\nI also don\'t know of any advanced text that treats the issue,\nthough I assume there must be some, since physics texts often\nseem to regard it as obvious.\n\nThe problem lies a bit outside the mainstream of the kind of\nmathematics relevant to its proof.\nMy guess is that the most likely place to find a proof\nwould be some old-fashioned text on mathematical physics.\n\nI think of your conclusion as\nan application of standard facts of advanced calculus,\nin particular that a mapping on Euclidean space (the plane in this case)\nis locally invertible at a point if its Jacobian is nonzero at that\npoint. I don\'t know if it is a theorem with a generally recognized name.\n\nThe basic idea is that given a function v = v(x,y) on the plane,\nwe can usually introduce new local coordinates (u,v) so that when\nexpressed in\nnew coordinates, v is the second coordinate function.\nHere "usually" means "generically",\nin the absence of certain degeneracies\nwhich I\'ll initially ignore for expositional simplicity.\n\nFor example, if v(x,y) := sqrt(x^2 + y^2),\nthe curves of constant v are circles centered at the origin,\nand polar coordinates give one way to do this.\n\nIndeed, we can usually do this in an even simpler way\nin which the new first coordinate u coincides with the old first\ncoordinate: u = x.\nIn the previous example v(x,y) := sqrt(x^2 + y^2), this corresponds to\nusing x and the usual radial polar coordinate as new coordinates---this\ncan be done locally at all points off the x-axis.\n\nThe moral is that just about any function can be viewed locally as\na coordinate function with respect to appropriately chosen coordinates,\nso anything that is true about a coordinate function is quite likely\ntrue locally\nfor general functions (perhaps under additional hypotheses to rule out\ndegenerate cases).\n\nIt\'s much easier to think about coordinate functions than arbitrary\nfunctions,\nso this insight is quite a powerful heuristic.\nHere is how it can be used to solve the problem at hand.\n\nConsider a differential form\n\n(1) P(x,y) dx + Q(x,y) dy ,\n\nand an integrating factor m = m(x,y). That is, m is a nonzero function such\nthat\n\n(2) m(x,y) P(x,y) dx + m(x,y) Q(x,y) dy = dz\n\nfor some function z. Suppose, in the spirit of the above observation,\nthat z happens to be a coordinate function, say z(x,y) = y for all x,y.\nThen dz = dy, and (2) becomes\n\n(3) m(x,y) P(x,y) dx + m(x,y) Q(x,y) dy = dy .\n\nEquating coefficients of dx and dy in (3) shows that this can happen\nonly if P(x,y)\nvanishes identically, and also\n\n(4) m(x,y) = 1/Q(x,y) , equivalently Q(x,y) = 1/m(x,y) .\n\nNOTATION: In the following, if f = f(x,y) is a function of two variables,\ndf/dx denotes the partial derivative of f with respect to x,\nand similarly for df/dy.\n\nNow consider a second integrating factor m1 = m1(x,y) for (1).\nSince we now know that P=0, this means that\n\n(5) m1(x,y) Q(x,y) dy = df(x,y) := df/dx dx + df/dy dy\n\nfor some function f = f(x,y).\nAgain equating coeficients of dx on both sides,\nwe see that df/dx = 0. This implies that\nf(x,y) is actually independent of x,\nso we can write f(x,y) = f1(y) for some function f1 of just one variable.\n\nUsing (4), we rewrite (5) as:\n\nm1(x,y)/m(x,y) dy = df1/dy dy ,\n\nso\n\nm1(x,y)/m(x,y) = df1/dy .\n\nThe right side is a function of y alone: call it F(y) := df1/dy.\nSo, we\'ve shown that\n\nm1(x,y)/m(x,y) = F(y) = F(z(x,y)) ,\n\nwhich is the desired conclusion for our special case z(x,y) := y.\n\nNow let\'s examine the general case in which the function z = z(x,y)\nin (2) is not assumed to be of this special form.\nThe plan is to change coordinates so that it *is* of the special form,\nand then check\nthat the coordinate change doesn\'t affect the conclusion.\n\nConsider the mapping on the plane defined by:\n\n(6) (x,y) ---> (x, z(x,y)) .\n\nSuppose that this map is invertible in a neighborhood of a given point\n(x0,y0)\n(a hypothesis which we\'ll examine later).\nUse this map to introduce new coordinates (x,z) as follows:\n\n(7) A point with old coordinates (x,y) is assigned new coordinates (x,\nz(x,y)).\n\nA differential form whose expression in old coordinates was\nP(x,y) dx + Q(x,y) dy will have an expression\nP\'(x,z) dx + Q\'(x,z) dz in new coordinates,\nwhere P\' and Q\' are some functions (of x and z):\n\n(8) P dx + Q dy = P\' dx + Q\' dz .\n\n[From the laws of transformations of differential forms,\none easily obtains expressions for P\' and Q\',\nbut we do not need these.\nFor example, P\' = P + Q dy/dx,\nwhere y is considered as a function of new coordinates x,z.]\n\nThis is somewhat abbreviated notation; to forestall any ambiguity or\nconfusion,\nlet\'s also present (8) more explicitly as:\n\n(8)\' P(x,y) dx + Q(x,y) dy = P\'(x, z(x,y)) dx + Q\'(x, z(x,y)) dz\nfor all x,y.\n\nLet m\' = m\'(x,z) denote the previous integrating factor expressed as\na function of x and z. That is,\n\nm\'(x,z(x,y)) = m(x,y) for all x,y .\n\nRecall from (2) that z = z(x,y) was originally introduced as a function\nsatisfying\n\n(9) dz = m(P dx + Q dy) .\n\nIn terms of the new coordinates x,z, this reads:\n\n(10) dz = m\'(x,z) P\'(x,z) dx + m\'(x,z) Q\'(x,z) dz .\n\nIncidentally, this implies that P\' vanishes identically and\nm\' = 1/Q\', but we don\'t need this.\nIf we were giving the proof from scratch,\nwithout having first worked out\nthe special case in which z = z(x,y) = y is a coordinate function,\nwe would follow that proof from here.\n\nBut since we have already proved the special case,\nwe can simply invoke that result, and we\'re done.\n\nThere is just one subtle point which might bother a careful reader:\nAre the "dz" in (9) and the "dz" in (10) really the same?\nThat is, the definition of the "dz" in (9) is dz := dz/dx dx + dz/dy dy,\nwhile the "dz" in (10) is just "dz" in new coordinates, i.e., "d" of the\nnew z coordinate.\nIs it possible that these two "dz"s are somehow different?\n\nTo settle this, we need a precise definition of "differential form".\nI\'ll leave as an exercise the application of the reader\'s favorite\ndefinition\n(there are several, all essentially equivalent) to check that\nthe two "dz"s really are the same.\n\nAlternatively, just apply the proof of the special case\nstarting with (10) instead of with (3).\nThis alternative completion is less abstract and perhaps more reassuring.\nWe could have given the proof in this way\nwithout doing the special case first,\nbut that would have obscured\nthe generally useful insight which led to the proof,\nnamely that nearly any function\ncan serve locally as a coordinate function.\n\nLOCAL INVERTIBILITY OF THE MAP (6): (x,y) --> (x, z(x,y))\n\nThe above proof assumed the local invertibility of the map (6).\nNow let\'s discuss when that map will be invertible.\n\nA standard result in advanced calculus implies that it will be invertible\nin a neighborhood of a point (x0,y0) provided that its Jacobian does not\nvanish\nat that point.\nCalculation of the Jacobian reveals that this is equivalent\nto the nonvanishing of dz/dy (x0,y0).\n\nSo, the above proof is valid in a neighborhood of (x0,y0)\nso long as dz/dy(x0,y0) is nonzero.\nIf it happens that dz/dy(x0,y0) = 0, we can carry out the same argument\nwith the roles of x and y interchanged, obtaining the desired conclusion\nexcept possibly in "degenerate" cases in which\n\ndz/dx (x0,y0) = 0 = dz/dy (x0,y0) .\n\nSince\n\ndz = dz/dx dx + dz/dy dy\n= m(x,y)P(x,y) dx + m(x,y)Q(x,y) dy,\n\nand since "integrating factors" m are (I assume) nonzero by definition,\nwe are done if we can assume that P and Q cannot vanish simultaneously.\nThis seems a reasonable hypothesis for the desired conclusion that\nm/m1 = F(z) for some function F.\n\nThat some such hypothesis on P and Q is necessary can be seen from\nthe observation that the assertion is false when P and Q vanish identically.\nIn that case, ANY nonvanishing function m is an integrating factor for P\ndx + Q dy,\nso ANY nonvanishing function g(x,y) can be\nthe quotient of two integrating functions.\n\nSimilarly, the assertion can fail\nif P and Q vanish identically on an open set.\nI think the above argument might extend to prove the assertion\nunder the hypothesis that\nP and Q do not vanish identically on any open set,\nbut the simpler hypothesis that\nP and Q do not vanish simultaneously at any point\nis probably sufficient for most physics applications.\n\n\n\n\n\n\n\n\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>diamonis wrote:
> As I understand it, having the book, the problem referred to is:
>
> (for the 2 dimensional case but really asked for any dim)
>
>
>
> given an integrating factor [itex]\mu(x,y)[/itex] for the expression Pdx + Qdy [itex]= 0,[/itex]
>
> where [itex]P=P(x,y), Q=Q(x,y)[/itex] are functions of (x,y) so that
>
> [itex]\mu * { Pdx+Qdy } = dz[/itex] for some function z(x,y) (ie z(x,y)=constant
>
> solves the diff eqn [itex]Pdx+Qdy=0),[/itex] then for any other integrating factor
>
> mu1(x,y) then
>
> mu1 [itex]/ \mu = F(z),[/itex] that is, [itex]mu1/\mu[/itex] must be a function of z(x,y) only.
>

Your conclusion is correct (modulo technical hypotheses),
but I don't know of any elementary text that explains clearly the reason.
I also don't know of any advanced text that treats the issue,
though I assume there must be some, since physics texts often
seem to regard it as obvious.

The problem lies a bit outside the mainstream of the kind of
mathematics relevant to its proof.
My guess is that the most likely place to find a proof
would be some old-fashioned text on mathematical physics.

I think of your conclusion as
an application of standard facts of advanced calculus,
in particular that a mapping on Euclidean space (the plane in this case)
is locally invertible at a point if its Jacobian is nonzero at that
point. I don't know if it is a theorem with a generally recognized name.

The basic idea is that given a function [itex]v = v(x,y)[/itex] on the plane,
we can usually introduce new local coordinates (u,v) so that when
expressed in
new coordinates, v is the second coordinate function.
Here "usually" means "generically",
in the absence of certain degeneracies
which I'll initially ignore for expositional simplicity.

For example, if v(x,y) [itex]:= \sqrt(x^2 + y^2),[/itex]
the curves of constant v are circles centered at the origin,
and polar coordinates give one way to do this.

Indeed, we can usually do this in an even simpler way
in which the new first coordinate u coincides with the old first
coordinate: [itex]u = x[/itex].
In the previous example v(x,y) [itex]:= \sqrt(x^2 + y^2),[/itex] this corresponds to
using x and the usual radial polar coordinate as new coordinates---this
can be done locally at all points off the x-axis.

The moral is that just about any function can be viewed locally as
a coordinate function with respect to appropriately chosen coordinates,
so anything that is true about a coordinate function is quite likely
true locally
for general functions (perhaps under additional hypotheses to rule out
degenerate cases).

It's much easier to think about coordinate functions than arbitrary
functions,
so this insight is quite a powerful heuristic.
Here is how it can be used to solve the problem at hand.

Consider a differential form

(1) P(x,y) [itex]dx + Q(x,y) dy ,[/itex]

and an integrating factor [itex]m = m(x,y)[/itex]. That is, m is a nonzero function such
that

(2) m(x,y) P(x,y) [itex]dx + m(x,y) Q(x,y) dy = dz[/itex]

for some function z. Suppose, in the spirit of the above observation,
that z happens to be a coordinate function, say z(x,y) = y for all x,y.
Then [itex]dz =[/itex] dy, and (2) becomes

(3) m(x,y) P(x,y) [itex]dx + m(x,y) Q(x,y) dy = dy .[/itex]

Equating coefficients of dx and dy in (3) shows that this can happen
only if P(x,y)
vanishes identically, and also

(4) m(x,y) [itex]= 1/Q(x,y) ,[/itex] equivalently Q(x,y) [itex]= 1/m(x,y)[/itex] .

NOTATION: In the following, if [itex]f = f(x,y)[/itex] is a function of two variables,
[itex]df/dx[/itex] denotes the partial derivative of f with respect to x,
and similarly for [itex]df/dy[/itex].

Now consider a second integrating factor [itex]m1 = m1(x,y)[/itex] for (1).
Since we now know that [itex]P=0,[/itex] this means that

(5) m1(x,y) Q(x,y) [itex]dy = df(x,y) := df/dx dx + df/dy dy[/itex]

for some function [itex]f = f(x,y)[/itex].
Again equating coeficients of dx on both sides,
we see that [itex]df/dx = .[/itex] This implies that
f(x,y) is actually independent of x,
so we can write f(x,y) [itex]= f1(y)[/itex] for some function f1 of just one variable.

Using (4), we rewrite (5) as:

[tex]m1(x,y)/m(x,y) dy = df1/dy dy[/itex] ,

so

[itex]m1(x,y)/m(x,y) = df1/dy[/itex] .

The right side is a function of y alone: call it F(y) [itex]:= df1/dy[/itex].
So, we've shown that

[itex]m1(x,y)/m(x,y) = F(y) = F(z(x,y)) ,[/tex]

which is the desired conclusion for our special case z(x,y) [itex]:= y[/itex].

Now let's examine the general case in which the function [itex]z = z(x,y)[/itex]
in (2) is not assumed to be of this special form.
The plan is to change coordinates so that [itex]it *is*[/itex] of the special form,
and then check
that the coordinate change doesn't affect the conclusion.

Consider the mapping on the plane defined by:

(6) (x,y) ---> (x, z(x,y)) .

Suppose that this map is invertible in a neighborhood of a given point
(x0,y0)
(a hypothesis which we'll examine later).
Use this map to introduce new coordinates (x,z) as follows:

(7) A point with old coordinates (x,y) is assigned new coordinates (x,
z(x,y)).

A differential form whose expression in old coordinates was
P(x,y) [itex]dx + Q(x,y) dy[/itex] will have an expression
P'(x,z) [itex]dx + Q'(x,z) dz[/itex] in new coordinates,
where P' and Q' are some functions (of x and z):

(8) P [itex]dx + Q dy =[/itex] P' [itex]dx + Q' dz[/itex] .

[From the laws of transformations of differential forms,
one easily obtains expressions for P' and Q',
but we do not need these.
For example, [itex]P' = P + Q dy/dx,[/itex]
where y is considered as a function of new coordinates x,z.]

This is somewhat abbreviated notation; to forestall any ambiguity or
confusion,
let's also present (8) more explicitly as:

(8)' P(x,y) [itex]dx + Q(x,y) dy = P'(x, z(x,y)) dx + Q'(x, z(x,y)) dz[/itex]
for all x,y.

Let [itex]m' = m'(x,z)[/itex] denote the previous integrating factor expressed as
a function of x and z. That is,

m'(x,z(x,y)) [itex]= m(x,y)[/itex] for all x,y .

Recall from (2) that [itex]z = z(x,y)[/itex] was originally introduced as a function
satisfying

(9) [itex]dz = m(P dx + Q dy) .[/itex]

In terms of the new coordinates x,z, this reads:

(10) [itex]dz = m'(x,z) P'(x,z) dx + m'(x,z) Q'(x,z) dz[/itex] .

Incidentally, this implies that P' vanishes identically and
[itex]m' = 1/Q',[/itex] but we don't need this.
If we were giving the proof from scratch,
without having first worked out
the special case in which [itex]z = z(x,y) = y[/itex] is a coordinate function,
we would follow that proof from here.

But since we have already proved the special case,
we can simply invoke that result, and we're done.

There is just one subtle point which might bother a careful reader:
Are the "dz" in (9) and the "dz" in (10) really the same?
That is, the definition of the "dz" in (9) is [itex]dz := dz/dx dx + dz/dy[/itex] dy,
while the "dz" in (10) is just "dz" in new coordinates, i.e., "d" of the
new z coordinate.
Is it possible that these two "dz"s are somehow different?

To settle this, we need a precise definition of "differential form".
I'll leave as an exercise the application of the reader's favorite
definition
(there are several, all essentially equivalent) to check that
the two "dz"s really are the same.

Alternatively, just apply the proof of the special case
starting with (10) instead of with (3).
This alternative completion is less abstract and perhaps more reassuring.
We could have given the proof in this way
without doing the special case first,
but that would have obscured
the generally useful insight which led to the proof,
namely that nearly any function
can serve locally as a coordinate function.

LOCAL INVERTIBILITY OF THE MAP (6): (x,y) --> (x, z(x,y))

The above proof assumed the local invertibility of the map (6).
Now let's discuss when that map will be invertible.

A standard result in advanced calculus implies that it will be invertible
in a neighborhood of a point (x0,y0) provided that its Jacobian does not
vanish
at that point.
Calculation of the Jacobian reveals that this is equivalent
to the nonvanishing of [itex]dz/dy (x0,y0)[/itex].

So, the above proof is valid in a neighborhood of (x0,y0)
so long as [itex]dz/dy(x0,y0)[/itex] is nonzero.
If it happens that [itex]dz/dy(x0,y0) = 0, we[/itex] can carry out the same argument
with the roles of x and y interchanged, obtaining the desired conclusion
except possibly in "degenerate" cases in which

[itex]dz/dx (x0,y0) == dz/dy (x0,y0)[/itex] .

Since

[itex]dz = dz/dx dx + dz/dy dy= m(x,y)P(x,y) dx + m(x,y)Q(x,y)[/itex] dy,

and since "integrating factors" m are (I assume) nonzero by definition,
we are done if we can assume that P and Q cannot vanish simultaneously.
This seems a reasonable hypothesis for the desired conclusion that
[itex]m/m1 = F(z)[/itex] for some function F.

That some such hypothesis on P and Q is necessary can be seen from
the observation that the assertion is false when P and Q vanish identically.
In that case, ANY nonvanishing function m is an integrating factor for P
[itex]dx + Q[/itex] dy,
so ANY nonvanishing function g(x,y) can be
the quotient of two integrating functions.

Similarly, the assertion can fail
if P and Q vanish identically on an open set.
I think the above argument might extend to prove the assertion
under the hypothesis that
P and Q do not vanish identically on any open set,
but the simpler hypothesis that
P and Q do not vanish simultaneously at any point
is probably sufficient for most physics applications.

Igor Khavkine

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\ndiamonis@hotmail.com (diamonis) wrote in message news:<608c8f11.0409132357.1ea176e3@posting.google.com>...\n> As I understand it, having the book, the problem referred to is:\n> (for the 2 dimensional case but really asked for any dim)\n>\n> given an integrating factor mu(x,y) for the expression Pdx + Qdy = 0,\n> where P=P(x,y), Q=Q(x,y) are functions of (x,y) so that\n> mu * { Pdx+Qdy } = dz for some function z(x,y) (ie z(x,y)=constant\n> solves the diff eqn Pdx+Qdy=0), then for any other integrating factor\n> mu1(x,y) then\n> mu1 / mu = F(z), that is, mu1/mu must be a function of z(x,y) only.\n\nIn addition to Stephen Parrott\'s detailed explanation, I can offer a\nmore pedestrian and rather more direct approach. What I will prove is\nthat mu1(x,y)/mu(x,y) is constant on integral curves of Pdx + Qdy, which\nare also constant contours of z(x,y). The last statement, for most intents\nand purposes, is equivalent to mu1/mu = F(z), but is well posed even when\nthe would-be function F(z) is not single valued.\n\nThe following requires some familiarity with differential forms but\nthe same idea can be used if you\'re just working with partial derivatives\nof mu, mu1, P, and Q while keeping in mind the definition of an\nintegrating factor.\n\nLet me introduce the two differential forms w = P dx + Q dy\nand v = mu dmu1 - mu1 dmu, and a vector field t defined such that t(x,y)\nis a tangent vector to an integral curve of w passing through (x,y).\nIn other words, the integral curves of w and t are identical and we\nhave the identity w(t) = 0.\n\nConsider the differential d(mu1/mu) = (mu dmu1 - mu1 dmu)/mu^2 = v/mu^2.\nFor mu1/mu to be constant on integral curves of w, the quantity\nd(mu1/mu)(t) = v(t)/mu^2 must be identically zero, which is equivalent too\nthe identity v(t) = 0.\n\nSince both mu1 and mu are integrating factors for the equation w = 0,\nby Poincare\'s lemma d(mu w) = dmu /\\ w + mu dw = 0 and\nd(mu1 w) = dmu1 /\\ w + mu dw = 0. Multiplying the last two expressions by\nmu1 and mu respectively and then subtracting gives the identity\n(mu dmu1 - mu1 dmu) /\\ w = v /\\ w = 0.\n\nPlugging t and any other vector field s into the differential form v /\\ w\ngives\n\n(v /\\ w)(t,s) = v(t) w(s) - v(s) w(t) = 0.\n\nThe second term vanishes since w(t) = 0 leaving us with v(t) w(s) = 0,\nfor all vector fields s. But since w is not identically zero, this last\nequation implies that v(t) = 0. Which proves that the function mu1/mu\nis constant on integral curves of the original equation\nv = P dx + Q dy = 0, and hence also on constant contours of either z(x,y)\nor z1(x,y).\n\nHope this helps.\n\nIgor\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>diamonis@hotmail.com (diamonis) wrote in message news:<608c8f11.0409132357.1ea176e3@p...google.com>...
> As I understand it, having the book, the problem referred to is:
> (for the 2 dimensional case but really asked for any dim)
>
> given an integrating factor [itex]\mu(x,y)[/itex] for the expression Pdx + Qdy [itex]= 0,[/itex]
> where [itex]P=P(x,y), Q=Q(x,y)[/itex] are functions of (x,y) so that
> [itex]\mu * { Pdx+Qdy } = dz[/itex] for some function z(x,y) (ie z(x,y)=constant
> solves the diff eqn [itex]Pdx+Qdy=0),[/itex] then for any other integrating factor
> mu1(x,y) then
> mu1 [itex]/ \mu = F(z),[/itex] that is, [itex]mu1/\mu[/itex] must be a function of z(x,y) only.

In addition to Stephen Parrott's detailed explanation, I can offer a
more pedestrian and rather more direct approach. What I will prove is
that [itex]mu1(x,y)/\mu(x,y)[/itex] is constant on integral curves of Pdx + Qdy, which
are also constant contours of z(x,y). The last statement, for most intents
and purposes, is equivalent to [itex]mu1/\mu = F(z),[/itex] but is well posed even when
the would-be function F(z) is not single valued.

The following requires some familiarity with differential forms but
the same idea can be used if you're just working with partial derivatives
of [itex]\mu,[/itex] mu1, P, and Q while keeping in mind the definition of an
integrating factor.

Let me introduce the two differential forms [itex]w = P dx + Q dy[/itex]
and [itex]v = \mu[/itex] dmu1 - mu1 dmu, and a vector field t defined such that t(x,y)
is a tangent vector to an integral curve of w passing through (x,y).
In other words, the integral curves of w and t are identical and we
have the identity [itex]w(t) =[/itex] .

Consider the differential [itex]d(mu1/\mu) = (\mu[/itex] dmu1 - mu1 [itex]dmu)/\mu^2 = v/\mu^2[/itex].
For [itex]mu1/\mu[/itex] to be constant on integral curves of w, the quantity
[itex]d(mu1/\mu)(t) = v(t)/\mu^2[/itex] must be identically zero, which is equivalent too
the identity v(t) = .

Since both mu1 and \mu are integrating factors for the equation [itex]w = 0,[/itex]
by Poincare's lemma [itex]d(\mu w) =[/itex] dmu [itex]/\ w + \mu dw =[/itex] and
d(mu1 [itex]w) =[/itex] dmu1 [itex]/\ w + \mu dw =[/itex] . Multiplying the last two expressions by
mu1 and \mu respectively and then subtracting gives the identity
[itex](\mu[/itex] dmu1 - mu1 dmu) [itex]/\ w = v /\ w = .[/itex]

Plugging t and any other vector field s into the differential form [itex]v /\ w[/itex]
gives

[itex](v /\ w)(t,s) = v(t) w(s) - v(s) w(t) =[/itex] .

The second term vanishes since w(t) = leaving us with v(t) w(s) [itex]= 0,[/itex]
for all vector fields s. But since w is not identically zero, this last
equation implies that v(t) = . Which proves that the function [itex]mu1/\mu[/itex]
is constant on integral curves of the original equation
[itex]v = P dx + Q dy = 0,[/itex] and hence also on constant contours of either z(x,y)
or z1(x,y).

Hope this helps.

Igor

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