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Newton's first differential equation
Highlights:
Historical:
"But this will appear plainer by an Example or two. ..."
(Newton
1671) --- After outlining his general method for finding solutions
of differential equations. The equation in this module is his first
significant example.
Mathematical: Newton obtained the solution
of a differential equation satisfying a given initial
condition in terms of infinite series. At each stage of
his series solution, he inserted the series into his
differential equation and integrated the resulting polynomial.
Prerequisites:
Linear ODEs
Equation:
y˙ ⁄ x˙ = 1 - 3x + y + x x + x y
Quantities:
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| x | : "the Correlate Quantity" |
| y | : "the Relate Quantity" |
| x˙ | : "the Fluxion of x" |
| y˙ | : "the Fluxion of y" |
Discussion:
Newton's book,
ANALYSIS Per Quantitatum, SERIES, FLUXIONES,
AC DIFFERENTIAS: cum Enumeratione Linearum
TERTII ORDINIS,
consists of one dozen problems.
His second problem
"PROB. II An Equation is being proposed, including the Fluxions
of Quantities, to find the Relations of those Quantities to one another"
is devoted to a general method of finding
the solution of an initial-value problem for a scalar differential
equation in terms of infinite series.
The equation above is his first significant example in this section.
Newton thought of Mathematical quantities as being generated by a continuous
motion. He called such a flowing quantity a fluent (variable),
and referred to its rate of change as the fluxion of fluent of the quantity
and denoted it by a dot over the quantity.
He denoted the change of Relate Quantity (dependent variable)
with respect to the Correlate Quantity (independent variable) with the
ratio of their fluxions.
If we reinterpret Newton in our current calculus jargon as
x˙ = dx/dt, y˙ = dy/dt, and y˙ ⁄ x˙ =
dy/dx, then Newton's differential equation becomes
dy/dx = 1 - 3x + y + x2 + x y.
Newton's Solution:
Now, we will paraphrase Newton's steps and obtain several terms of his power series
solution y(x) of his differential equation satisfying the initial
condition y(0) = 0. Start with the first term
y = 0 + …
and insert it into the differential equation to obtain
y′ = 1 + … .
Now, integrate this with respect to x,
y = x + …
to obtain the next term in the series. Inserting this series for y into
the differential equation, yields
y′ = 1 - 2x + …
integration of which gives
y = x - x2 + … .
The next iteration of this process gives
y′ = 1 - 2x + x2 + …
and
y = x - x2 + (1/3)x3 + … .
Newton continues several more iterations and arrives at the solution
y = x - x2 + (1/3)x3 - (1/6)x4 +
(1/30)x5 - (1/45)x6 + … .
Newton's Demonstration:
It is prudent to verify that a proposed solution of a differential
equation indeed satisfies the differential equation. Here is how
Newton demonstrates the validity of his solution:
DEMONSTRATION
"56. And thus we have solved the Problem, but the
demonstration is still behind. And in so great a variety of
matters, that we may not
derive it synthetically, and with too great perplexity, from
its genuine foundations, it may be sufficient to point it out
thus in
short, by way of Analysis. That is, when any Equation is
propos'd, after you have finish'd the work, you may try whether
from the derived Equation you can return back to the Equation
propos'd ... And thus from
y˙ = 1 - 3x + y + x x + x y
is derived
y = x - x 2 + (1/3)x 3 - (1/6)x 4 +
(1/30)x 5 - (1/45)x 6 , &c.
And thence by Prob. I.
y˙ = 1 - 2x + x 2 - (2/3)x 3 + (1/6)x 4 -
(2/15)x 5 , &c.
Which two values of y˙ agree with each other, as appears by
substituting
x - xx + (1/3)x 3 - (1/6)x 4 +
(1/30)x 5 , &c. instead of y in the first value."
Dynamics:
A series solution of an initial-value problem, in principle,
should yield better approximations to the solution
as more terms of the series are included.
In Figure 1., third through sixth-order approximations of the Newton's series
solution are plotted.
Phaser simulation:
Figure 1. Third through sixth order series approximations
of the solution of Newton's differential equation.
Click on the picture to load it into your local Phaser.
In Figure 2., the actual solution of Newton ODE satisfying
the initial condition y(0) = 0 is plotted in blue.
The additional solution in yellow satisfies the initial
condition y(0) = 1; Newton's series of this solution is
given in the Suggested Explorations below.
Phaser simulation:
Figure 2. Two solutions of Newton's differential equation.
Click on the picture to load it into your local Phaser.
It was indicated above that one can expect better approximations
as more terms of the series are included.
However, this expectation holds only locally near
the initial condition, but not globally. Indeed, the fourth-order
approximation appears to be more faithful to the actual solution
than the fifth-order approximation.
Mathematical results:
Newton ODE is a scalar linear differential equation for
which there exists a formula for the solutions.
Full details of the use of this formula
for Newton ODE is available in a PDF file.
You will notice in this file, however, that
one of the requisite integrations involves the error function
which cannot be expressed in terms of elementary functions.
Suggested Explorations:
Newton solved his equation for the initial value y(0) = 1 as well.
His answer, in this case, is
y = 1 + 2x + x3 + (1/4)x4 +
(1/4)x5 + … .
Demonstrate the validity of Newton's solution a la Newton.
This solution is plotted in yellow in Figure 2 above.
It is very interesting to observe that Newton calculates
up to sixth-order (even) terms for the blue solution
while he stops at the fifth-order (odd) terms
for the yellow solution. Series solutions become
more accurate with additional terms; this is true
locally but not necessarily globally. Why do you think
Newton stopped at the fifth-order terms for the yellow solution
while continued to the sixth-order terms for the blue solution?
Load the Phaser Project in Figure 2.
Click the left mouse button at several locations along the
vertical axis to mark
additional initial conditions. Now click the Go button to see
the additional solutions.
"I said before, that these Solutions may be performed by
an infinite variety of ways. This may be done if you assume at
pleasure not only the initial quantity of the upper series, but
any other given quantity for the first Term of the Quote, and
then you may proceed as before. ...Or if you make use of any
Symbol, say a, to represent the first Term indefinitely, by the
same method of Operation (which I shall here set down)
y = a + x + ax - xx + axx + (1/3)x3 +
(2/3)ax3 + …
which being found, you may substitute 1, 2, 0, (1/2), or any
other number, and thereby obtain the Relation between x and y an
infinite variety of ways."
In other words, Newton computed the solution of his equation
for the initial condition y(0) = a. Verify his answer.
Find the solution satisfying the
general initial condition y(x 0) = y 0. Hint: Find the power
series expansion in powers of (x - x 0).
Newton also studied differential equations whose
right-hand-sides are more complicated than polynomials in x and
y. In this case, he first expanded the differential equation
into a power series and proceeded as before. Here is such an
example.
"32. And after the same manner the Equation
y˙ ⁄ x˙ = 3y - 2x + x/y - 2y/(xx) being
proposed; if, by reason of the Terms x/y and 2y/(xx),
I write 1 - y for y, 1-x for x, there will arise
y˙ ⁄ x˙ = 1 - 3y + 2x + (1-x)/(1-y) + (2y
-2)/(1 - 2x + x2).
But the Term (1 - x)/(1 - y) by infinite Division gives
1 - x + y - xy + y2 - xy2 + y3
- xy3, &c. and the Term (2y - 2)/(1 -2x + xx)
by a like Division gives
2y - 2 + 4xy - 4x + 6x2y - 6x2 +
8x3y - 8x3 + 10x4y
-10x4, &c. Therefore
y˙ ⁄ x˙ = -3x + 3xy + y2 - xy2
+ y3 - xy3, &c. + 6x2y - 6x2
+ 8x3y - 8x3 + 10x4y - 10x4, &c.
Perform the "infinite Divisions" and verify Newton's
calculations.
Related modules:
None.
References:
HAIRER, E., NORSETT, S., and WANNER [1987]. Solving Ordinary
Differential Equations.
Springer-Verlag, New York.
NEWTON, I. [1736].
The Method of Fluxions and Infinite Series;
with its Applications to the Geometry of Curve-lines
by the Inventor Sir Isaac Newton, Kt.,
Late President of the Royal Society. /
Translated from the Author's Latin Original not yet made
publick. /
To which is subjoin'd, A perpetual Comment upon the whole Work,
Consisting of Annotations, Illustrations, and Supplements,
In Order to make this Treatise A Compleat Institution for the
use of Learners. /
By John Colson.
London: Printed by Henry Woodfall; And Sold by John Nourse,
at the Lamb without Temple-Bar. M.DCCXXVI.
O'CONNOR, J.J. and ROBERTSON, E.F. Sir Isaac Newton:
http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Newton.html
Printable Document:
An expanded and printable version of this module in PDF format is available at:
http://www.phaser.com/modules/historic/newton/isaac_newton_ode.pdf
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