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Homework Help: Second order ODE proof

  1. Oct 23, 2012 #1

    Zondrina

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    1. The problem statement, all variables and given/known data

    I'm pretty sure this is a typo?

    http://gyazo.com/802746486cc68852e5384d5a12aed596

    2. Relevant equations

    See the image ^.

    3. The attempt at a solution

    I believe the theorem they're talking about, is that you can write the general solution of a second order ODE :

    [itex]L[y] = y'' + p(t)y' + q(t)y = 0[/itex] Where L is the differential operator on a solution y.

    In the form : c1y1 + c2y2 ⇔ y1, y2 were both linearly independent solutions to the homogeneous system. That is :

    [itex]W(y_1, y_2) ≠ 0[/itex] where W is the wronskian of y1, y2.

    So a bit of analysis here, when x > 0, y1 = y2 = x3, so that W(y1, y2) = 0. Also when x < 0, y1 = x3 and y2 = -y1. So W(y1, y2) = 0 as well in this case.

    Now, in part a) I showed that a linear combination of these two was indeed a solution by taking the required derivatives and blah blah, the boring stuff.

    Now it's mind boggling me that this can't be a typo because this is a counter-example to this theorem apparently, so I must be missing something crucial here.

    Any pointers on this?
     
  2. jcsd
  3. Oct 23, 2012 #2

    Dick

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    It is true that if the wronskian is nonzero then the functions must be linearly independent. But is it true that if the wronskian is zero then they must linearly dependent?? Does your theorem really say that?
     
  4. Oct 23, 2012 #3

    Zondrina

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    Indeed, the theorem stated that if the Wronskian was non-zero for two solutions y1 and y2, then the solution y = c1y1 + c2y2 creates a fundamental solution set over the interval a < t < b.
     
  5. Oct 23, 2012 #4

    Zondrina

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    I read this quickly on wiki : http://en.wikipedia.org/wiki/Wronskian

    The Wronskian and linear independence :

    If the functions fi are linearly dependent, then so are the columns of the Wronskian as differentiation is a linear operation, so the Wronskian vanishes. So the Wronskian can be used to show that a set of differentiable functions is linearly independent on an interval by showing that it does not vanish identically.

    A common misconception is that W = 0 everywhere implies linear dependence, but Peano (1889) pointed out that the functions x2 and |x|x have continuous derivatives and their Wronskian vanishes everywhere, yet they are not linearly dependent in any neighborhood of 0. There are several extra conditions which ensure that the vanishing of the Wronskian in an interval implies linear dependence. Peano (1889) observed that if the functions are analytic, then the vanishing of the Wronskian in an interval implies that they are linearly dependent. Bochner (1901) gave several other conditions for the vanishing of the Wronskian to imply linear dependence; for example, if the Wronskian of n functions is identically zero and the n Wronskians of n–1 of them do not all vanish at any point then the functions are linearly dependent. Wolsson (1989a) gave a more general condition that together with the vanishing of the Wronskian implies linear dependence.

    The part that caught my eye is highlighted.

    EDIT : Is it because we have a saddle point for |x| in the neighborhood of zero because they're not differentiable at zero?
     
  6. Oct 23, 2012 #5

    Zondrina

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    So for x > 0, the slope of |x| is positive ( Is 1 really ). For x < 0, the slope of |x| is negative ( -1 to be exact ). Thus we have a saddle point at x = 0 and |x| is not differentiable at 0.

    Thus [itex]|x^3| = |x^2||x| = x^2|x|[/itex] is also not differentiable at the origin.
     
  7. Oct 23, 2012 #6

    Dick

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    You might also note your equation doesn't start with y'', it starts with x^2*y''. It's singular at x=0. y=c*x is another linearly independent solution. You might want to check your theorems assumption about p and q.
     
  8. Oct 23, 2012 #7

    Dick

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    |x^3| is differentiable at the origin. It's second derivative is also continuous.
     
  9. Oct 23, 2012 #8

    Zondrina

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    Ahhhhh so take this from before :[itex]L[y] = y'' + p(t)y' + q(t)y = 0[/itex]

    If p and q are continuous, then I will be able to find a general solution. Now though, we have a co-efficient in front of y''. That is r(x) = x2. So we would get :

    [itex]L[y] = x^2y'' + 3xy' + 3y = 0[/itex]
    [itex]L[y] = y'' + \frac{3}{x}y' + \frac{3}{x^2}y = 0[/itex]

    If x > 0, y1 and y2 create my linear combination y which is a solution to this system.

    If x < 0, we observe a similar case.

    If x = 0, then our co-efficients p and q are discontinuous and we are not able to find a solution.
     
  10. Oct 23, 2012 #9

    Dick

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    If your p and q aren't continuous that just means your theorem doesn't apply. Doesn't mean you can't find a solution. I think we've found three linearly independent ones already.
     
  11. Oct 23, 2012 #10

    Zondrina

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    I'm not seeing what you're getting at. Is this problem about showing there are other linearly independent solutions to this problem?
     
  12. Oct 23, 2012 #11

    Dick

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    Yes! y=x^3, y=|x^3| and y=x are all linearly independent solutions. You have too many! You just have to say why your theorem doesn't apply. It says you should have only two.
     
    Last edited: Oct 23, 2012
  13. Oct 23, 2012 #12

    Zondrina

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    Shortened version of the theorem :

    If the Wronskian is non-zero, then the solutions are linearly independent.
    If the Wronskian IS zero, then the solutions are linearly dependent.

    So because there are other linearly independent solutions who's Wronskian's are non-zero say y1 = x and y2 = c/x then we know the theorem is not contradicted?

    This seems weak to me?
     
  14. Oct 24, 2012 #13

    Dick

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    That is FALSE. Isn't that the point here with x^3 and |x^3|?

    What theorem are you talking about? Can you quote it in full?
     
  15. Oct 24, 2012 #14

    Zondrina

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    Literally the theorem I stated in my last post, the shortened version is the one he gave in class apparently.
     
  16. Oct 24, 2012 #15

    Dick

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    If you mean what you said in post 12, that's not generally true. Probably because it's 'shortened'.
     
  17. Oct 24, 2012 #16

    Zondrina

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    Waiiiiiiit a second here. I thought for awhile about this. Here's the theorem I think he's talking about from the book :

    http://gyazo.com/fac8f81914efdedde890eeaef24d734a

    So, it's easy to show the Wronskian of y1 and y2 is equal to zero for x > 0 and x < 0.

    NOW, I didn't actually consider what happens on an interval CONTAINING x = 0.

    Suppose that there exist real scalars c1 and c2 such that : c1y1 + c2y2 = 0

    Then for all x in our interval containing 0 we have :

    c1x3 + c2|x|3 = 0

    Now for x > 0 , c1x3 + c2x3 = 0 which tells us x3(c1 + c2) = 0 which yields c1 + c2 = 0

    Now a similar case for x < 0 yields c2 - c2 = 0.

    Solving both of these equations yields c1 = c2 = 0 and hence y1 and y2 are linearly independent because there exists a trivial relation of linear dependance on our interval containing x = 0. Hence there is a point on our interval where W ≠ 0 and thus the theorem is not contradicted.
     
  18. Oct 24, 2012 #17

    Dick

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    Yes, you have shown that x^3 and |x^3| are linearly independent over an interval containing positive and negative numbers. But their wronskian IS zero everywhere even at zero. Now why does that not contradict your theorem??
     
  19. Oct 24, 2012 #18

    Zondrina

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    I think that's because my DE is not in the form y'' + py' + qy = 0. I would have to divide through by x^2 which would create discontinuities at zero so that my p and q are not continuous?
     
  20. Oct 24, 2012 #19

    Dick

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    Yeah, that's one thing. But the theorem as quoted didn't give any conditions on p and q. Does it give them elsewhere? The other out would be to notice c1*x^3+c2*|x^3| doesn't span all possible solutions. Why not?
     
  21. Oct 24, 2012 #20

    Zondrina

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    Ahhh there's a theorem wayyyy back stating that p and q have to be continuous functions on the interval. So because p and q are not continuous everywhere on the interval in this case, it follows that y1 and y2 are not solutions over the entire interval?
     
  22. Oct 24, 2012 #21

    Dick

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    Don't go back to saying that again. It's wrong. y1 and y2 are solutions over all the reals in the original equation. I thought you checked that. p and q becoming discontinous after you divide by x^2 doesn't change that. It may change whether you can rely on theorems about the solutions of y''+py'+qy=0 with p and q continuous.
     
  23. Oct 24, 2012 #22

    Zondrina

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    Okay I'm fairly confident now that these two theorems :

    http://gyazo.com/6212935f0c5e359748a623a88b309b2e
    http://gyazo.com/428dbfb9ce40f8633cbc588e5c629953

    Are the ones being called into question and are the only theorems relevant to this question at all. Every other theorem involves initial values which we do not have here.

    Now, Theorem 3.2.4 states that, if and only if the Wronskian of y1 and y2 is not zero everywhere, then the linear combination c1y1 + c2y2 contains all solutions of (2).

    So, is it that the linear combination here does not contain all the possible solutions since the Wronskian is zero everywhere?
     
  24. Oct 24, 2012 #23

    Dick

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    Yes! The Wronskian is zero everywhere. But c1y1 + c2y2 does not contain all solutions. Why? That would save the theorem from contradiction.
     
  25. Oct 24, 2012 #24

    Zondrina

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    This is confusing me for some reason. I took the required derivatives and did all the substituting to show that y1 and y2 were both solutions to the DE and then I showed their combination y = y1 + y2 was also a solution.

    So from part (a) I deduced that y1 and y2 were linearly independent solutions to my DE.

    Then when I went to part (b), I found the Wronskian for both cases when x ≥ 0 and x < 0. Turns out W is identically zero just as desired which WOULD mean that y1 and y2 were linearly dependent.

    I think what saves the theorem from contradiction is that one solution is a multiple of the other? I'm a bit lost about this really.
     
  26. Oct 24, 2012 #25

    Dick

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    YOU have proved that the solutions are linearly independent. That would rule out them being multiples of each other, right? YOU also showed that the wronskian is zero everywhere. The only way out is to show that c1*x^3+c2*|x^3| does not describe ALL solutions. I gave you a HUGE hint a number of posts back. Where is it?
     
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