Wronskians
Keltus
I have a question about wronskians. If y_1 and y_2 are functions, we use the
Wronskian to determine if they are linearly independent or linearly
dependent.
My text says
a) "If y_1 and y_2 are linearly dependent, then their wronskian = 0 on I"
b) "If y_1 and y_2 are linearly independent, then their wronskian <> 0 at
each point of I"
I'm guessing the different usage of grammar "on I" and "at each point of I"
means something different. Now what I want to know is, if the wronskian is
2x or something, and the interval is (-1,1) such that the wronskian is 0 at
some points, and not 0 at some points, is it linearly independent of
dependent? I'm guessing independent but I'm not sure just by reading the
text. Can anyone clear this up for me?
Cheers,
~Jeff
David C. Ullrich
wrote:
>Hi,
>
>I have a question about wronskians. If y_1 and y_2 are functions, we use the
>Wronskian to determine if they are linearly independent or linearly
>dependent.
Can't be done! Here's an example:
Let f(x) = x^2, and let g(x) = 2x^2 for x > 0, x^2 for x <= 0. Then
f and g are _independent_, although the Wronskian equals 0
at _every_ point.
>My text says
>a) "If y_1 and y_2 are linearly dependent, then their wronskian = 0 on I"
>b) "If y_1 and y_2 are linearly independent, then their wronskian <> 0 at
>each point of I"
I hope it doesn't say that for any old functions y1 and y2. It's
talking about two solutions to a second-order differential
homogeneous linear differential equation, right?
>I'm guessing the different usage of grammar "on I" and "at each point of I"
>means something different. Now what I want to know is, if the wronskian is
>2x or something, and the interval is (-1,1) such that the wronskian is 0 at
>some points, and not 0 at some points, is it linearly independent of
>dependent?
_If_ we're talking about solutions to a (second-order linear
homogeneous) differential equation on an interval then that
can't happen - either the Wronskian is 0 at every point or it
is 0 at _no_ point.
(Um, if the coefficient of y'' is 1 and the other
coefficients are differentiable, anyway...)
>I'm guessing independent but I'm not sure just by reading the
>text. Can anyone clear this up for me?
>
>Cheers,
> ~Jeff
>
******************
David C. Ullrich
Robert Israel
Keltus <purple...@yahoo.com> wrote:
>I have a question about wronskians. If y_1 and y_2 are functions, we use the
>Wronskian to determine if they are linearly independent or linearly
>dependent.
This has bothered me for some time: DE texts such as Boyce and DiPrima
tend to promote this, but it's really rather silly to do so IMHO. To check
whether y_1 and y_2 are linearly dependent on some interval, all you have
to be able to do is tell whether their ratio is constant. Using the
Wronskian, you would check whether y_1(s) y_2'(s) - y_1'(s) y_2(s) = 0 for
s in the interval. Frankly, I don't see the advantage in the latter
approach.
>My text says
>a) "If y_1 and y_2 are linearly dependent, then their wronskian = 0 on I"
>b) "If y_1 and y_2 are linearly independent, then their wronskian <> 0 at
>each point of I"
>I'm guessing the different usage of grammar "on I" and "at each point of I"
>means something different.
Not really. The subtlety is that the negation of "the wronskian = 0 on I"
is not "the wronskian <> 0 on I" but "the wronskian <> 0 at some point of
I". Note that (b) is true for solutions of a homogeneous second-order
linear differential equation, but not for arbitrary differentiable functions.
> Now what I want to know is, if the wronskian is
>2x or something, and the interval is (-1,1) such that the wronskian is 0 at
>some points, and not 0 at some points, is it linearly independent of
>dependent? I'm guessing independent but I'm not sure just by reading the
>text. Can anyone clear this up for me?
They are linearly independent. But they aren't both solutions of a
homogeneous second-order linear differential equation.
Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
Justin Davis
"David C. Ullrich" <ull...@math.okstate.edu> wrote in message
news:contbvgvuannm7ia3...@4ax.com...
> On Sun, 11 May 2003 23:58:18 GMT, "Keltus" <purple...@yahoo.com>
> wrote:
>
> >Hi,
> >
> >I have a question about wronskians. If y_1 and y_2 are functions, we use
the
> >Wronskian to determine if they are linearly independent or linearly
> >dependent.
>
> Can't be done! Here's an example:
>
> Let f(x) = x^2, and let g(x) = 2x^2 for x > 0, x^2 for x <= 0. Then
> f and g are _independent_, although the Wronskian equals 0
> at _every_ point.
>
I'm a bit confused by this. Are we talking linear dependence?
Virgil
A quick Google search for Wronskian turned up over 7000
hits. Several of the first few seemed appropriate to your
question.
For example:
http://mathworld.wolfram.com/Wronskian.html
http://www.sosmath.com/diffeq/second/linearind/linearind.html
http://www.math.pku.edu.cn/stu/eresource/wsxy/sxrjjc/wk/Encyc
lopedia/math/w/w160.htm
Dr. Richard L. Hall
Robert Israel
Sorry, I should have said they aren't both solutions of a differential
equation y" + p(x) y' + q(x) y = 0 where p and q are continous on
(-1,1).
For example, x and x^2 are solutions of x^2 y" - 2 x y' + y = 0.
Keltus
"Robert Israel" <isr...@math.ubc.ca> wrote in message
news:b9mn6s$m65$1...@nntp.itservices.ubc.ca...
okay that clears things up. I wasn't sure with the usage of "at each point
of I". I guess it's the same as "at some point of I"
thanks
Virgil
isr...@math.ubc.ca (Robert Israel) wrote:
> This has bothered me for some time: DE texts such as Boyce and DiPrima
> tend to promote this, but it's really rather silly to do so IMHO. To check
> whether y_1 and y_2 are linearly dependent on some interval, all you have
> to be able to do is tell whether their ratio is constant. Using the
> Wronskian, you would check whether y_1(s) y_2'(s) - y_1'(s) y_2(s) = 0 for
> s in the interval. Frankly, I don't see the advantage in the latter
> approach.
Using the quotient rule for the derivative of ratio
y_2/y_1, the numerator of the derivative is exactly the
Wronskian, which will be zero on an inteval iff the ratio
y_2/y_1 is constant on that interval.
As the tests are exactly equivalent in result, one should
use in any situation the one easiest to apply in that
situation.
Travis Willse
> This has bothered me for some time: DE texts such as Boyce and DiPrima
> tend to promote this, but it's really rather silly to do so IMHO. To check
> whether y_1 and y_2 are linearly dependent on some interval, all you have
> to be able to do is tell whether their ratio is constant. Using the
> Wronskian, you would check whether y_1(s) y_2'(s) - y_1'(s) y_2(s) = 0 for
> s in the interval. Frankly, I don't see the advantage in the latter
> approach.
It really is silly to use the Wronskian for two functions in an application,
but in a text it illustrates the validity of the method whose real beauty
lies in testing the linear independence of three or more equations, no two
of which are linear multiples of each other. Say, {sinh(x), cosh(x), e^x}.
Regards,
Travis Willse
G. Marete
As can be verified from any basic book for ODEs:
If y_1 and y_2 are two solutions of a linear ordianry differential
equation on an interval I, then their Wronskian is either 0 on (all) I
or <> 0 on (all) I. In particluar, if the Wronskian is 0 at any point,
then it is 0 on (all) I.
As a consequence, to determine linear dependence/independence, one
only needs to check at a single point.
In fact, in the case of constant coefficents, one has the formula:
W(t)= exp(-a1*(t-x))*W(x), where x is any fixed point and a1 is the
coeffeicient for the first derivative.
What I have said above then follows from this formula.
Again, the proofs can be found in an basic textbook of Ordinary Diff.
Eq. Persoanlly, I think that Coddington's book is one of the best.
Regards,
Marete.
"Keltus" <purple...@yahoo.com> wrote in message news:<3ebed5b2$1...@news03.toast.net>...
Keith Ramsay
| Wronskian to determine if they are linearly independent or linearly
| dependent.
|
| My text says
| a) "If y_1 and y_2 are linearly dependent, then their wronskian = 0 on I"
Assuming both functions are differentiable on the interval, so that the
Wronskian is defined. In general, if a finite collection of functions has
a linear dependency, and they are differentiable enough times, then the
determinants satisfy the same dependency, and hence the rows of the matrix
whose determinant is being taken are linearly dependent, which makes the
determinant zero wherever it's defined.
| b) "If y_1 and y_2 are linearly independent, then their wronskian <> 0 at
| each point of I"
I think there's an additional assumption here, such as that y_1 and y_2
are solutions to a certain kind of differential equation. Otherwise,
consider y_1 = x^2 and y_2 = x^3. They are linearly independent even though
the Wronskian determinant is zero when x=0. But they aren't both solutions
to the same differential equation of the kind you're considering.
Keith Ramsay
David C. Ullrich
>Hello:
>
>As can be verified from any basic book for ODEs:
>
>If y_1 and y_2 are two solutions of a linear ordianry differential
>equation on an interval I, then their Wronskian is either 0 on (all) I
>or <> 0 on (all) I. In particluar, if the Wronskian is 0 at any point,
>then it is 0 on (all) I.
Not _quite_. Of course the equation has to be homogeneous,
and a more subtle point is that it has to have the form
y'' + b(x) y' + c(x) y = 0;
if it's
a(x) y'' + b(x) y' + c(x) y = 0
where a(x) vanishes at some point then what you said is
not true.
******************
David C. Ullrich
David C. Ullrich
wrote:
Yes. Which part are you confused by, the claim that the functions
are independent or the claim that the Wronskian is identically 0?
******************
David C. Ullrich
David C. Ullrich
It's not hard to give examples of independent functions whose
Wroonskian vanishes _identically_, in fact.
>But they aren't both solutions
>to the same differential equation of the kind you're considering.
>
>Keith Ramsay
******************
David C. Ullrich
Bill Dubuque
>Keltus <purple...@yahoo.com> wrote:
>
>> can we use the Wronskian to determine linear dependence? [...]
>
> Can't be done! Here's an example:
>
> f,g are _independent_, but Wronskian = 0 at _every_ point [...]
Actually it can be done in a local manner, as in the following
THEOREM If f1,..,fn are n-1 times differentiable on interval I
and they have Wronskian W(f1,..,fn) vanishing at all points in I
then f1,..,fn are linearly dependent on some *subinterval* of I
Proof: We employ the following easily proved Wronskian identity:
W(g f1,..,g fn) = g^n W(f1,..,fn) which immediately implies
W(f1,..., fn) = f1^n W((f2/f1)',..,(fn/f1)') where f1 != 0
Proceed by induction on n. Theorem is clearly true if n = 1.
Suppose now n > 1 and W(f1,..,fn) = 0 at every point in I.
If f1 = 0 throughout I then f1,..,fn are dependent on I.
Else f1 is nonzero at some point of I so also throughout some
subinterval J < I since f1 is continuous (being differentiable
by hypothesis). By above identity W((f2/f1)',..,(fn/f1)') = 0
throughout J so by induction there exists a subinterval K of J
where the Wronskian's arguments are linearly dependent, i.e.
on K: c2(f2/f1)' +...+ cn(fn/f1)' = 0, all ci'=0; some cj!=0
=> ((c2 f2 +...+ cn fn)/f1)' = 0 via ( )' linear
=> c2 f2 +...+ cn fn = c1 f1 for some c1, c1' = 0
Therefore f1,..,fn are linearly dependent on K < I. QED
This theorem has as immediate corollaries the well-known results
that the vanishing of the Wronskian throughout an interval I is
a necessary and sufficient condition for linear dependence of
(1) functions analytic on I
(2) functions satisfying a monic homogeneous linear differential
equation whose coefficients are continuous throughout I.
-Bill Dubuque
David C. Ullrich
wrote:
>David C. Ullrich <ull...@math.okstate.edu> wrote:
>>Keltus <purple...@yahoo.com> wrote:
>>
>>> can we use the Wronskian to determine linear dependence? [...]
>>
>> Can't be done! Here's an example:
>>
>> Let f(x) = x^2, g(x) = 2x^2 for x > 0, x^2 for x <= 0. Then
>> f,g are _independent_, but Wronskian = 0 at _every_ point [...]
>
>Actually it can be done in a local manner, as in the following
>
>THEOREM If f1,..,fn are n-1 times differentiable on interval I
>and they have Wronskian W(f1,..,fn) vanishing at all points in I
>then f1,..,fn are linearly dependent on some *subinterval* of I
Huh, thanks. Never really thought about it, but I'd always
assumed that the example could be jacked up to one where
they were independent on no subinterval.
>Proof: We employ the following easily proved Wronskian identity:
>
> W(g f1,..,g fn) = g^n W(f1,..,fn) which immediately implies
>
> W(f1,..., fn) = f1^n W((f2/f1)',..,(fn/f1)') where f1 != 0
>
>Proceed by induction on n. Theorem is clearly true if n = 1.
>Suppose now n > 1 and W(f1,..,fn) = 0 at every point in I.
>If f1 = 0 throughout I then f1,..,fn are dependent on I.
>Else f1 is nonzero at some point of I so also throughout some
>subinterval J < I since f1 is continuous (being differentiable
>by hypothesis). By above identity W((f2/f1)',..,(fn/f1)') = 0
>throughout J so by induction there exists a subinterval K of J
>where the Wronskian's arguments are linearly dependent, i.e.
>
>on K: c2(f2/f1)' +...+ cn(fn/f1)' = 0, all ci'=0; some cj!=0
>
> => ((c2 f2 +...+ cn fn)/f1)' = 0 via ( )' linear
>
> => c2 f2 +...+ cn fn = c1 f1 for some c1, c1' = 0
>
>Therefore f1,..,fn are linearly dependent on K < I. QED
>
>This theorem has as immediate corollaries the well-known results
>that the vanishing of the Wronskian throughout an interval I is
>a necessary and sufficient condition for linear dependence of
>(1) functions analytic on I
>(2) functions satisfying a monic homogeneous linear differential
>equation whose coefficients are continuous throughout I.
>
>-Bill Dubuque
******************
David C. Ullrich
Jason Pawloski
>In article <b9mn6s$m65$1...@nntp.itservices.ubc.ca>,
>Robert Israel <isr...@math.ubc.ca> wrote:
>>In article <3ebed5b2$1...@news03.toast.net>,
>>Keltus <purple...@yahoo.com> wrote:
>
>>> Now what I want to know is, if the wronskian is
>>>2x or something, and the interval is (-1,1) such that the wronskian is 0 at
>>>some points, and not 0 at some points, is it linearly independent of
>>>dependent? I'm guessing independent but I'm not sure just by reading the
>>>text. Can anyone clear this up for me?
>
>>They are linearly independent. But they aren't both solutions of a
>>homogeneous second-order linear differential equation.
>
>Sorry, I should have said they aren't both solutions of a differential
>equation y" + p(x) y' + q(x) y = 0 where p and q are continous on
>(-1,1).
>
>For example, x and x^2 are solutions of x^2 y" - 2 x y' + y = 0.
In some cases, wouldn't it be more efficient to use a Wronskian if you have
large rational equations which don't easily divide, or have many solutions
which need to be teached for linear dependence?
Yes, in my DE course we learned to use the Wronskian to check for linear
dependence. You're right, it was a waste of time, because simple inspection
and a high-school level knowledge of numbers was usually more than enough to
determine linear (in)dependence.
Jason Pawloski