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Science Forum Index » Mathematics Forum » Perturbation problem
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| Henk Jansen |
 Posted: Mon Dec 15, 2003 6:13 pm |
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I have the following perturbation problem --- perhaps someone can give
me some help...
Given the nonlinear ODE problem:
dy/dt = u cos(x) - v (1)
y dx/dt = - u sin(x) (2)
with boundary conditions
y(0) = y0 , y(T) = 0
x(0) = x0 , x(T) = 0
where u << v so that e:=u/v<<1 is a small parameter. When e->0, (2)
vanishes and T=p0/v so that T = p0/v + O(e)
The 'simple' question is: which asymptotic expansion will give me an
approximate solution y(t,e), x(t,e) for t in [0,T(e)]?
After several attempts the best I have been able to produce is an
implicit (i.e. nonlinear) algebraic solution in y:
0 = f(t,x0,p0,e,y) (3)
x = g(t,x0,e) (4)
My intuition says that with the correct expansion it should be
possible to obtain an explicit solution to (1) and (2).
My procedure giving (3),(4) proceeds as follows. I first rewrite
(1),(2) in terms of small parameter e:
dy/(v dt) = e cos(x) - 1 (5)
y dx/(v dt) = -e sin(x) (6)
Defining the dimensionless time
s:=vt/y(t) => y(t)s = vt
where s in [0,00] for t in [0,T] (notice that y(T)=0). Hence
dy/dt s + y(t) ds/dt = v =>
ds/dt = ( v - dy/dt s ) / y(t) =>
= ( v - (u cos(x) - v ) s ) / p(s) =>
ds/(v dt) = ( 1 - ( e cos(x) - 1 ) s ) / p(s) (7)
Substituting ds for (v dt) in (5),(6) with (7) and p:=y/y0 gives:
dp/ds = p(s) (e cos(x) -1) / ( 1 - (e cos(x) - 1) s ) (
dx/ds = e sin(x) / ( 1 - (e cos(x) - 1) s ) (9)
p(0) = 1 , p(00) = 0
x(0) = x0 , x(00) = 0
Using the following asymptotic expansions:
p = 1 - s + p1(x0,s) e + p2(x0,s) e^2 + O(e^3) (10) => dp/ds = ...
x = x0 + x1(x0,s) e + x2(x0,s) e^2 + O(e^3) (11) => dx/ds = ...
and substituting these in (, (9) gives ODEs for the functions p1,
p2, x1, x2 that can be solved for. Unfortunately, since, by
definition, s involves y (=y0*p), (10) and (11) are nonlinear in y
(c.q. p), requiring a numerical procedure to solve for. I believe this
could possibly be avoided with a slightly more powerful expansion. The
question is which ...
Thanks,
-- Henk |
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| Robert Israel |
| Posted: Mon Dec 15, 2003 7:11 pm |
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In article <4a5aaa89.0312151513.57129711@posting.google.com>,
Henk Jansen <burley@zonnet.nl> wrote:
Quote: I have the following perturbation problem --- perhaps someone can give
me some help...
Given the nonlinear ODE problem:
dy/dt = u cos(x) - v (1)
y dx/dt = - u sin(x) (2)
with boundary conditions
y(0) = y0 , y(T) = 0
x(0) = x0 , x(T) = 0
where u << v so that e:=u/v<<1 is a small parameter. When e->0, (2)
vanishes and T=p0/v so that T = p0/v + O(e)
Too many boundary conditions! Two of them, say y(0) = y0, x(0) = x0,
will specify the solution. No guarantee that there will be any T
where both x and y are 0. And I don't understand your last statement
there: for u = 0, you should get x = x0 and y = y0 - v t. So T = y0/v
will make y(T) = 0 but not x(T) = 0.
Quote: The 'simple' question is: which asymptotic expansion will give me an
approximate solution y(t,e), x(t,e) for t in [0,T(e)]?
I get
x(t) = x0 + e sin(x0) ln(1 - v/y0 t) + O(e^2)
y(t) = y0 - (v - e v cos(x0)) t + O(e^2)
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2 |
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| Henk Jansen |
| Posted: Sat Dec 20, 2003 4:31 pm |
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israel@math.ubc.ca (Robert Israel) wrote in message news:<brliji$rp3$1@nntp.itservices.ubc.ca>...
Quote: In article <4a5aaa89.0312151513.57129711@posting.google.com>,
Henk Jansen <burley@zonnet.nl> wrote:
I have the following perturbation problem --- perhaps someone can give
me some help...
Given the nonlinear ODE problem:
dy/dt = u cos(x) - v (1)
y dx/dt = - u sin(x) (2)
with boundary conditions
y(0) = y0 , y(T) = 0
x(0) = x0 , x(T) = 0
where u << v so that e:=u/v<<1 is a small parameter. When e->0, (2)
vanishes and T=p0/v so that T = p0/v + O(e)
Too many boundary conditions! Two of them, say y(0) = y0, x(0) = x0,
will specify the solution. No guarantee that there will be any T
where both x and y are 0. And I don't understand your last statement
there: for u = 0, you should get x = x0 and y = y0 - v t. So T = y0/v
will make y(T) = 0 but not x(T) = 0.
The 'simple' question is: which asymptotic expansion will give me an
approximate solution y(t,e), x(t,e) for t in [0,T(e)]?
I get
x(t) = x0 + e sin(x0) ln(1 - v/y0 t) + O(e^2)
y(t) = y0 - (v - e v cos(x0)) t + O(e^2)
Thanks, that's correct; I found the same solution (including
second-order terms) in my first approach which I wrote as
x(t) = x0 + e sin(x0) ln(1 - s) + sin(x0)^2 (ln(1-s) + 1/2) e^2 +
O(e^3) (a)
p(t) = 1 - (1 - e cos(x0)) s - sin(x0)^2 (1-s) (1-ln(1-s)) e^2 +
O(e^3) (b)
where p:=y/y0 and s:=vt/y0. The problem only is that, even when
including second-order terms, the solution is not defined for s > 1
and the solution for x(t) quickly breaks down when s > 0.6 (I have a
Maple worksheet that nicely shows what happens). The solution for p(t)
was good (it's a near perfect linearly decreasing function in time).
Therefore I tried a renormalization (time scaling):
s = r + r1 e + r2 e^2 + O(e^3)
substitution of which in (a) and (b) told me what functions to take
for r1 and r2 to remove the secular terms so that I obtained
s = r ( 1 + cos(x0) e + cos(x0)^2 e^2 ) + O(e^3)
so that the scaled time becomes
r = ( 1 - cos(x0) e ) s + O(e^4)
and with t \in [0,1] the solution would now be valid for scaled time r
Quote: 1. Also, substituting r for s in (a), (b) offered me indeed a much
better solution for x(t) but a poorer solution for p(t). This
conclusion set me off in the direction outlined in my original email
to this list using a normalized time s:=vt/y(t) instead of y0. This
solved the problem: very good solutions were obtained for x(t) as well
as y(t) but, unfortunately, the solution for y(t) was in implicit
form, as I could have guessed from
y( t( y( t( y(..) ) ) ) )
Thus the question still stands: which perturbation gives me an
explicit solution for x(t), y(t) that are valid for t \in [0,T] with
T = y0/v + T1 e + T2 e^2 + O(e^3) ? (end of
question) |
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