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Question about independance of functions and wronskian

  1. Oct 5, 2011 #1
    Hi,

    I just want to clarify something written in my textbook - a contradiction of sorts.

    My book says, if i have two functions, Y1 Y2, and their wronskian is 0 at any point on the interval I, the functions are dependant functions.

    However, while doing a problem, I found the wronskian to be 2t^3 for two functions Y1 Y2. Clearly, this wronskian is 0 when t=0. ACcording to the original theoram this would imply the functions are dependant on the interval (-infinity, infinity). However that (Seems like) is the wrong answer.

    Can someone clarify? thanks
     
  2. jcsd
  3. Oct 5, 2011 #2
    to elaborate on the question further:

    consider the linear equation t^2(y``)-3t(y`)+3y=0 for -inf to inf

    Y1 = t and Y2= t^3

    Wronskian of Y1 and Y2 equals 2t^3 implying independance.
    ...

    actually while writing this i think i answered my own question.The DE must be in staqndard format, i.e. y```+py`+qy=0. when i put the equation in that format, t=0 is not defined. therefore tcannot=0 reight/>
     
  4. Oct 5, 2011 #3

    S_Happens

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    Yes, you're supposed to put it in the standard format, but that doesn't address your original question.

    The (limited) test that we do in ODE for linear independence using the Wronskian is to determine whether the Wronskian is equal to (this is where my book says "identically equal to") zero or not over a range, not simply that it can take on zero value at a single point in that range. W = 2t^3 implies linear independence. W = 0 would imply linear dependance.

    There is more to it, but that is beyond my knowledge and really beyond what is required for a basic ODE course.

    Edit- I'm not sure if I was clear so I'll add this. To test for linear independance we use the Wronskian. When you use it, you don't look at W = 2t^3 and say "I can sub in t = 0 and make the Wronskian = 0." W = 2t^3 means the Wronskian is NOT equal to zero across the ENTIRE range we are using. The Wronskian has to be equal to zero without subsituting in any values of the dependant variable.

    If it doesn't make sense work backwards from two solutions to get a DE and then calculate the Wronskian. Ex: y1 = e^x and y2 = e^(2x) (you should know these are linearly indpendant)

    I can't come up with two solutions that would make the Wronskian zero that you can't immediately tell that they differ by a constant. Ex: y1 = x^2 and y2 = 7x^2
     
    Last edited: Oct 5, 2011
  5. Oct 5, 2011 #4

    LCKurtz

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    The situation is different for the Wronskian of two solutions of a second order linear DE than for the Wronskian of any two functions. For two solutions, you have the theorem that the Wronskian is never zero or identically zero, and consequently the Wronskian can be used as a test for linear independence. But this is not true for any two functions as, for example, t2 and t3. For these, the Wronskian is 0 at a t=0 yet the functions are linearly independent. One thing this tells you is that t2 and t3 are not solutions of a 2nd order linear differential equation.

    To summarize: If the Wronskian is nonzero at any point, that settles that the functions are linearly independent for any two functions. But if W = 0 at some point, that doesn't imply linear dependence, except for solutions of the DE.
     
  6. Oct 5, 2011 #5

    S_Happens

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    I'm really wondering if I'm there's some miscommunication here as I don't agree with what I think you're saying.

    First thing, since the OP is using t (and you as well) I'm just going to make sure we are saying that y = y(t). Although it doesn't matter, I usually prefer y = y(x) specifically for this reason.

    I think it's pretty obvious that we're talking about solutions to a 2nd order DE. I am lost when you say that

    They can certainly be solutions to a 2nd order differential. In fact, they are they general solution to [itex]t^{2}y^{''}-4ty^{'}+6y = 0[/itex] are they not? What am I missing here?
     
  7. Oct 5, 2011 #6

    LCKurtz

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    Sorry, I should have included the statement that the leading coefficient is nonzero, which is the usual condition for the Wronskian stuff. I'm talking about solutions to

    y'' + p(t)y' + q(t)y = 0.
     
  8. Oct 5, 2011 #7

    S_Happens

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    Ok, the only reason I put my equation in that form is that I haven't used Latex much and didn't want to have to mess with it. Divide by t2 to put it in that form. The general solution is still

    y = C1t2 + C2t3

    I'm still wondering why you're saying t to a power can't be a solution to a DE. Being able to substitute zero in for t to make the Wronskian zero has nothing to do with anything as far as I know, although I'm taking ordinary differential equations right now. As I can see the OP is simply asking about using the Wronskian of two solutions of a DE to determine linear (in)dependence, doing that (t = 0) is an incorrect application.
     
  9. Oct 5, 2011 #8

    LCKurtz

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    Dividing your example by t2 makes p(t) and q(t) discontinuous, which again violates the usual assumptions. And it is true that t2 and t3 cannot be solutions of y''+p(t)y'+q(t)y = 0 with p and q continuous. Here's a proof:

    If t2 and t3 were solutions of that equation, their Wronskian would be non-zero for all t or identically zero. But it is neither. Therefore they aren't solutions.

    [Edit] Also note that if you consider only an interval excluding t = 0, the Wronskian test does work.
     
  10. Oct 6, 2011 #9

    S_Happens

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    Right. I lost my mind for a while.

    In these cases we've (my class) only been considering intervals not containing 0. It was briefly mentioned up front and then immediately moved to application. Although I always go back to try and pick up the details missed/glossed over in favor of pure problem solving repetition, there's always something lost when it's done this way.
     
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