number of indep. soloutions to diffyqs?

[Jeremy Price]

I've taken some classes in solving differential equations, and I understand
all the different techniques, and everything, but am confused as to how we
can be assured that our techniques yield *all* possible solutions.
Especially when we do things like, if x1 is a solution, put x2(x) =
x1(x)*f(x) into the differential equation and obtain another solution. Is
there a general way to tell how many linearly independent solutions a
differential equation has, (especially regardless of homogeneity and
linearity)? I understand why these methods work, of course, but can't see
exactly how we can be sure we are obtaining all the solutions. I can see
intuitively a little why it should be the case (especially when using
eigenvalues of a matrix in the matrix form of a diffyq x'=Ax, but would like
something more rigorous).

Thanks,
Jeremy

[G. A. Edgar]

In article <c5lcbq$1ac2$1...@nntp1.u.washington.edu>, Jeremy Price
<cfg...@u.PANTS.washington.edu> wrote:

Good textbooks (the ones with proofs) include "existence and
uniqueness" theorems. That is what you want here. It is something
separate from "techniques" for finding solutions in closed form.

For example...
Coddington, An Introduction to Ordinary Differential Equations.
Dover reprint, around $10.

--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/

[Proginoskes]

"Jeremy Price" <cfg...@u.PANTS.washington.edu> wrote in message news:<c5lcbq$1ac2$1...@nntp1.u.washington.edu>...

Check out the Wronskian.
Nonlinear differential equations are harder to deal with and may have
a variety of solutions.
-- Christopher Heckman

[Kevin O'Neill]

I believe your question is about completeness.
The references cited by the other respondents are very good, but
you might have to consult some more advanced texts as well.

______________________________________________________

"Jeremy Price" <cfg...@u.PANTS.washington.edu> wrote in message
news:c5lcbq$1ac2$1...@nntp1.u.washington.edu...

[Bill Dubuque]

Jeremy Price <cfg...@u.PANTS.washington.edu> wrote:
>
> I've taken some classes in solving differential equations, and I understand
> all the different techniques, and everything, but am confused as to how we

> can be assured that our techniques yield all possible solutions.

> Especially when we do things like, if x1 is a solution, put x2(x) =
> x1(x)*f(x) into the differential equation and obtain another solution. Is
> there a general way to tell how many linearly independent solutions a
> differential equation has, (especially regardless of homogeneity and
> linearity)? I understand why these methods work, of course, but can't see
> exactly how we can be sure we are obtaining all the solutions. I can see
> intuitively a little why it should be the case (especially when using
> eigenvalues of a matrix in the matrix form of a diffyq x'=Ax, but would like
> something more rigorous).

It's very easy to establish unqueness of solutions by way of the
Wronskian and Variation of Parameters. I append my prior post <4>:

Fernando Gonzalez del Cueto <fcu...@yahoo.com> wrote:
:
: how can I prove that the ONLY solutions of the ODE
:
: y'' = -y are the ones given by A cos(x) + B sin(x) ?

Easy: Wronskian W(cos,sin) = cos^2 + sin^2 = 1 != 0, so
Variation of Parameters <2> specialized to _homogeneous_
case shows that any other solution is a linear combination
of cos, sin with _constant_ coef's. Below are easy proofs,
which generalizes to higher-order LODEs (and recurrences).
The proof is slicker in matrix form, e.g. Lemma 4.1 in <3>.

THEOREM If f,g,h are solutions on an interval I of the LODE

y'' = p y' + q y, p,q continuous on I

and gh'-g'h != 0 for all x in I, then there exist constants c,d

such that f = c g + d h on I

PROOF: The below equations [0],[1] have a unique solution (c,d)
since det = W(g,h) = gh'- g'h != 0 on I.

[0] f = c g + d h

[1] f' = c g' + d h'

[2] qf+pf' = f" = c g" + d h" via q[0]+p[1], g"=qg+pg', h"=qh+ph'

[3] 0 = c'g + d'h via [0]'-[1]

[4] 0 = c'g' + d'h' via [1]'-[2] (1b)

The above equations [3],[4] have unique solution (c',d') = (0,0)
since det = gh'-g'h = W(g,h) != 0 on I. Thus c,d are constants.
-------
THEOREM If f,g,h are solutions of the recurrence

y'' = p y' + q y, where y'(n) := y(n+1)

with Wronskian W = gh'-g'h != 0 then there exist constants c,d

such that f = c g + d h

PROOF: [0],[1] below have unique solution (c,d) via det = W != 0

[0] f = c g + d h

[1] f' = c g' + d h' Now q[0] + p[1] yields:

[2] qf+pf' = f" = c g" + d h" via qg+pg' = g", qh+ph' = h"

[3] 0 = (c'-c)g' + (d'-d)h' via [0]'-[1]

[4] 0 = (c'-c)g" + (d'-d)h" via [1]'-[2]

The above equations [3],[4] have solution (c'-c,d'-d) = (0,0),
unique via det = W' = g'h"-g"h' != 0. So c,d are constants,
since c' = c means c(n+1) = c(n).

-Bill Dubuque

<1> L. E. Pursell: A simple uniqueness theory for ordinary linear
homogeneous differential equations, Amer. Math. Monthly, 74, 1967, 47-50
http://links.jstor.org/sici?sici=0002-9890(196701)74:1%3C47%3E

<2> Variation of Parameters
http://planetmath.org/encyclopedia/VariationOfParameters.html
http://ltcconline.net/greenl/courses/204/appsHigherOrder/variationHigher.htm

<3> Marius van der Put: Symbolic analysis of differential equations
http://msri.org/activities/programs/9899/focm/soggy/MSRIintro/van_der_Put2.ps
d:\Symbolic analysis of differential equations -- Put.ps.gz

<4> This post: 2003-11-12, There are no other solutions... how to prove it?
http://google.com/groups?selm=y8zn0b0ix4q.fsf%40nestle.ai.mit.edu

[Bill Dubuque]

By the way, in my prior post I should have used the
standard term "Casoratian" for the difference-theoretic
analog of the Wronskian for differential equations.

I chose the unified terminology in order to more clearly
emphasize the analogy between differential equations
and difference equations (or recurrences if you prefer).

-Bill Dubuque