How to Show the Wronskian of Three Distinct Solutions Equals Zero?

Click For Summary

Homework Help Overview

The discussion revolves around the Wronskian of three distinct solutions to a second-order linear homogeneous differential equation. Participants are exploring how to demonstrate that the Wronskian of these solutions equals zero, given that the solutions are distinct.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to clarify the implications of having three distinct solutions and how that relates to the Wronskian. Some express confusion about the nature of the Wronskian for three functions and question the linear dependence of the solutions.

Discussion Status

The discussion is active, with participants sharing their thoughts on the relationship between the Wronskian and linear dependence. Some have suggested that the Wronskian must be zero due to the properties of the differential equation, while others are exploring the implications of linear dependence and the conditions under which the Wronskian can be evaluated.

Contextual Notes

Participants note that the second-order differential equation can only have two linearly independent solutions, which raises questions about the linear dependence of three solutions. There is also mention of the difficulty in calculating the Wronskian directly and the implications of Cramer's rule in this context.

BrettJimison
Messages
81
Reaction score
5

Homework Statement


Hello,

I was just looking for a quick tip:
If I have three distinct solutions to a second order linear homogeneous d.e, how would I show that the wronskian of (y1,y2,y3)(x)=0?

I know how to show the wronskian is not zero for a linearly independent set, but I'm confused since I'm given three distinct solutions to a second order.

Homework Equations


I know y=c1y1+c2y2+c3y3 by the principle of super position but I'm not sure how I would go about showing a wronskian evaluated at these is zero...any tips? just need a point in the right direction

The Attempt at a Solution

 
Physics news on Phys.org
BrettJimison said:

Homework Statement


Hello,

I was just looking for a quick tip:
If I have three distinct solutions to a second order linear homogeneous d.e, how would I show that the wronskian of (y1,y2,y3)(x)=0?

I know how to show the wronskian is not zero for a linearly independent set, but I'm confused since I'm given three distinct solutions to a second order.

Homework Equations


I know y=c1y1+c2y2+c3y3 by the principle of super position but I'm not sure how I would go about showing a wronskian evaluated at these is zero...any tips? just need a point in the right direction

The Attempt at a Solution

The Wronskian in the case of a 2nd order DE is a 2 by 2 determinant. What does the Wronskian of three solutions mean?
[Edit] Or maybe you mean the 3 by 3 Wronskian anyway? In which case you know they are linearly dependent from the DE so...
 
Last edited:
LCKurtz said:
The Wronskian in the case of a 2nd order DE is a 2 by 2 determinant. What does the Wronskian of three solutions mean?
[Edit] Or maybe you mean the 3 by 3 Wronskian anyway? In which case you know they are linearly dependent from the DE so...
I'm trying to find a 3x3 wronskian with y1 y2 y3 that is equal to zero...not sure how to do it symbolically. It's a lot easier when I know the functions...
 
What theorems do you have about Wronskians and linear dependence/independence?
 
LCKurtz said:
What theorems do you have about Wronskians and linear dependence/independence?
Well I know the set will be linearly dependet since the wronskian will equal zero. I also know that the set has to be L.D since it has 3 elements and the D.E is only 2nd order.

If
L2[ y1]=0
L2[y2]=0
l2[y3]=0

Then
y1=c1y2+c2y3
y2=c1y1+c2y3
y3=c1y1+c2y2

I then have three functions that I can plug into the wronskian...I tried this and got a soltion that doesn't seem to simplify to zero.
 
BrettJimison said:
Well I know the set will be linearly dependet since the wronskian will equal zero. I also know that the set has to be L.D since it has 3 elements and the D.E is only 2nd order.

If
L2[ y1]=0
L2[y2]=0
l2[y3]=0

Then
y1=c1y2+c2y3
y2=c1y1+c2y3
y3=c1y1+c2y2

I then have three functions that I can plug into the wronskian...I tried this and got a soltion that doesn't seem to simplify to zero.

Also letting a=y1, b=y2, c=y3

My determinatnt simplfies to: a"(bc'-b'c)+a'(b"c-c"c)+a(b'c"-b"c')
 
BrettJimison said:
Well I know the set will be linearly dependet since the wronskian will equal zero.

No, you don't know that. You are trying to show the Wronskian is zero.

I also know that the set has to be L.D since it has 3 elements and the D.E is only 2nd order.

Yes. You do know the 3 functions are linearly dependent and that is why.

Now think about what being linearly dependent tells you about the system of equations
##c_1y_1 + c_2y_2 + c_3y_3 = 0##
##c_1y_1' + c_2y_2' + c_3y_3' = 0##
##c_1y_1'' + c_2y_2'' + c_3y_3'' = 0##
and what that implies about the Wronskian.
 
LCKurtz said:
No, you don't know that. You are trying to show the Wronskian is zero.
Yes. You do know the 3 functions are linearly dependent and that is why.

Now think about what being linearly dependent tells you about the system of equations
##c_1y_1 + c_2y_2 + c_3y_3 = 0##
##c_1y_1' + c_2y_2' + c_3y_3' = 0##
##c_1y_1'' + c_2y_2'' + c_3y_3'' = 0##
and what that implies about the Wronskian.

Well...I know If I put that into matrix form its determintant will be the wronskian. I am taking its determinant now but its not looking like its going to cancell out...

edit: I know that those equations are L.D which means they can be wriiten as linear combos of each other..
 
Last edited:
BrettJimison said:
Well...I know If I put that into matrix form its determintant will be the wronskian. I am taking its determinant now but its not looking like its going to cancell out...
I actually get exactly what I posted previously...hmmm

The thing is: I know the set of equations is L.D and I know the wronskian will be zero. I have no idea how to show its zero though. When I take the determinant of the matrix, I don't get something that cancells

[edit] I understand why its not zero: I am basically evaluating a wronskian of a set of solutions for L3[y] so I won't equal zero...I need to find a way to get the three equations into a different form?
 
Last edited:
  • #10
You aren't expected to actually calculate the determinant. You have ##n## equations in the ##n## unknown ##c_i##'s. What would you know about the ##c_i##'s if the determinant of coefficients (the Wronskian) wasn't zero? What would Cramer's rule tell you?
 
  • #11
LCKurtz said:
You aren't expected to actually calculate the determinant. You have ##n## equations in the ##n## unknown ##c_i##'s. What would you know about the ##c_i##'s if the determinant of coefficients (the Wronskian) wasn't zero? What would Cramer's rule tell you?
If the determinant wasnt zero (wronskian) then that means that there is non trivial solutions to the system which means linear dependence but I don't know how to show that since I can't compute the wronskian to show that it isn't zero...
 
  • #12
BrettJimison said:
If the determinant wasnt zero (wronskian) then that means that there is non trivial solutions to the system which means linear dependence but I don't know how to show that since I can't compute the wronskian to show that it isn't zero...
Im pretty sure cramers rule would just tell us we could find non trivial solutions...
 
  • #13
BrettJimison said:
If the determinant wasnt zero (wronskian) then that means that there is non trivial solutions to the system which means linear dependence but I don't know how to show that since I can't compute the wronskian to show that it isn't zero...

No. Think about what you are saying. Try a little 2 by 2 system:
##x + y = 0##
##2x - y = 0##
Here the determinant of coefficients isn't zero. Solve it with Cramer's rule.
 
  • #14
The SPECIFIC question states:

Suppose y1,y2,y3 are three distinct solutions of L2[y] =0 on I = (a,b)
Show that W(y1,y2,y3)(x)=0 on I.
 
  • #15
LCKurtz said:
No. Think about what you are saying. Try a little 2 by 2 system:
##x + y = 0##
##2x - y = 0##
Here the determinant of coefficients isn't zero. Solve it with Cramer's rule.
Sorry meant to say if the determinant WAS zero...typo, this is what I have been trying to prove the whole time.
 
  • #16
But what if the determinant isn't zero. What does that tell you? What does that say about your system of equations?

I have to run for now. Back in a couple of hours.
 
  • #17
LCKurtz said:
But what if the determinant isn't zero. What does that tell you? What does that say about your system of equations?

I have to run for now. Back in a couple of hours.
Ok Thanks for your time!

If the Det wasnt zero, then equations would be linearly independent and ONLY the trivial solution would work.
I can explain my reasoning in words, but I am having a hard time putting it into math...

Thanks!
 
  • #18
BrettJimison said:
Ok Thanks for your time!

If the Det wasnt zero, then equations would be linearly independent and ONLY the trivial solution would work.
I can explain my reasoning in words, but I am having a hard time putting it into math...

Thanks!
There isn't much math to write, but you need to be careful about your notation. You don't talk about "equations being linearly independent" in this context. You talk about the functions ##y_1,...y_n## being independent. What you have stated above that is correct and relevant is that if the determinant is zero, you can get only trivial solutions for the ##c_i##'s. But you know the solutions are linearly dependent so there is a non-trivial solution for the ##c_i##'s. So, what do you conclude?
 
  • #19
There isn't much math to write, but you need to be careful about your notation. You don't talk about "equations being linearly independent" in this context. You talk about the functions ##y_1,...y_n## being independent. What you have stated above that is correct and relevant is that if the determinant is nonzero, you can get only trivial solutions for the ##c_i##'s. But you know the solutions are linearly dependent so there is a non-trivial solution for the ##c_i##'s. So, what do you conclude?
 
Last edited:
  • Like
Likes   Reactions: BrettJimison
  • #20
LCKurtz said:
There isn't much math to write, but you need to be careful about your notation. You don't talk about "equations being linearly independent" in this context. You talk about the functions ##y_1,...y_n## being independent. What you have stated above that is correct and relevant is that if the determinant is nonzero, you can get only trivial solutions for the ##c_i##'s. But you know the solutions are linearly dependent so there is a non-trivial solution for the ##c_i##'s. So, what do you conclude?
W(y1,y2,y3)(x)=0 Thanks for your time and help!
 

Similar threads

Linearly independent functions with identically zero Wronskian
  • · Replies 1 ·
Replies
1
Views
2K
Understanding Wronskian of solutions to homoeneous 2nd order linear DE
  • · Replies 2 ·
Replies
2
Views
2K
Ordinary Differential equations question
  • · Replies 4 ·
Replies
4
Views
2K
Solve (t^2-1)y'' +4ty'+2y=6t, given two particular solutions
  • · Replies 2 ·
Replies
2
Views
1K
Wronskian Property: Linearly Dependent Functions
  • · Replies 7 ·
Replies
7
Views
2K
The Wronskian and the Derivative of the Wronskian
  • · Replies 2 ·
Replies
2
Views
3K
Construct a second order ODE given the solutions?
  • · Replies 11 ·
Replies
11
Views
3K
Solutions of a linear, second-order, homogeneous differential equation
  • · Replies 1 ·
Replies
1
Views
3K
Discussion problem, wronskian matrix, linear independence of solution
  • · Replies 3 ·
Replies
3
Views
2K
How to Determine Linear Dependence Using Wronskian: A Quick Guide
  • · Replies 2 ·
Replies
2
Views
2K