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The Wronskian and linear independence of a ODE solution set

  1. Aug 30, 2010 #1
    1. The problem statement, all variables and given/known data

    Hi

    I seem to remember that if you have a homogenous ODE

    y'' + p(t)y' + q(t)y = 0 which have the solutions y1 and y2. Where we are told that

    y1(t) [tex]\neq 0[/tex]

    then y1 and y2 are linear independent.

    I found the simular claim on sosmath.com but are they simply saying as long as y1 is not a multiplum of y2 then y1 and y2 will always be linear independent?

    THis is very confusing because I learned in linear algebra that in order for vectors to be linear independent then the weight c1=c2 = 0 and if for instance c1 = 1 and c2 = 2 then they are linear dependent.If you people get my point?
     
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  3. Aug 30, 2010 #2

    Mark44

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    Linear independence, whether of vectors or functions, is very similar.

    Two vectors are linearly independent if the only solution to the equation c1v1 + c2v2 = 0 is c1 = c2 = 0. Note that c1 = c2 = 0 is a solution to this equation if the vectors are linearly dependent, but this is not the only solution.


    Similarly, two functions are linearly independent if the only solution to the equation c1f1(t) + c2f2(t) = 0, for all t in the common domain of these functions.

    For your question about the ODE, if you have two functions y1(x) and y2(x), and neither one is the zero function, the two functions will be linearly independent if neither one is a multiple of the other. For example, if the functions are cos(x) and 2cos(x), neither is the zero function, but each one is some multiple of the other, so they are linearly dependent. Going back to the definition of linear independence, the equation c1cos(x) + c22cos(x) = 0, has many solutions where c1 and c2 are not both zero -- namely, c1 = -2, c2 = 1.

    On the other hand, if the two solutions are cos(x) and sin(x), neither function is the zero function, and neither function is a multiple of the other, so it turns out that these functions are linearly independent.
     
  4. Aug 30, 2010 #3
    Hi and thanks for your answer,

    So no need to use the properties of Wronskian to show linear independence of y1 and y2?

    Final question. I need to show that W(y1,y2) = 1 by my question here is the following. Since p and q are unknown do I then just discard them and write up the general solution

    y = c1 exp(r1*t) + c2 exp(r2*t) and take take the wronskian on solution? In order to show that w(y1,y2) = 1 ??
     
  5. Aug 30, 2010 #4

    vela

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    If you have just two functions, you can usually tell by inspection if they are multiples of each other, so it's relatively easy to see if two functions are not independent. But that's not the same as proving linear independence. Using the Wronskian is a quick way to do that if it works.
    No, you can't do that because the solutions will depend on what p(x) and q(x) are. Can you give us the complete problem you're trying to do?
     
  6. Aug 31, 2010 #5
    Its a three part question vela,

    First part is having that ode mentioned in orginally post.

    y'' + p(x)y' + q(x)y = 0

    where x belongs to the interval I and [tex]x_0 \in I[/tex] assume that [tex]y_1(x) \neq 0[/tex]


    Show that the second solution of the ode can be expressed

    [tex]y_2(x) = y_1(x) \int_{x_0}^{x} \frac{1}{y_1(t)^2}e^{-\int_{x_0}^{t} p(u) du}dt[/tex]

    Assuming that [tex]y_1(x)[/tex] is a solution of the original equation. Then
    [tex]y_2(x) = y_1(x) \cdot v(x)[/tex] is the second solution of the ode where v(x) is an unknown funct.

    by taking the derivative of [tex]y_2(x)[/tex] (since this a solution) I plug the respective first order and second order derivative of y2 into the original equation and obtain

    [tex]\frac{v''(t)}{v'(t)} + 2\cdot \frac{y_1'(t)}{y_1(t)} + p(u) = 0[/tex]

    This eqn is seperable and thus by v'' = w

    then I obtain

    [tex]\frac{w'(t)}{w(t)} + 2\cdot \frac{y_1'(t)}{y_1(t)} = - p(u) du[/tex]

    and by integrating on both sides with the their respective variables

    I get

    [tex]ln|w(t) \cdot y_1^2(t)| = - \int p(u) du[/tex]

    I take exp on both sides of the equality and obtain

    [tex]w(t) \cdot y_1^2(t) = e^{- \int p(u) du}[/tex]

    which by integrating again yields

    [tex]v = c_1\cdot \int e^{- \int p(u) du}dt + c_2[/tex] and by choosing c1= 1 and c2 = 0 then I obtain

    [tex]y_2(x) = y_1(x) \int_{x_0}^{x} \frac{1}{y_1(t)^2}e^{-\int_{x_0}^{t} p(u) du}dt[/tex]

    which is the second solution for the ode. where the solution set is defined on the interval [tex]x_0 \leq t \leq x[/tex] and integrant p on the subinterval [tex]x_0 \leq u \leq t[/tex]

    What I know about p and q that they are functions defined on I.

    That leads to the second question, vela.

    2) Show that (I,y1) and (I,y2) are linear independent.

    Well I can't check the wronskian for the ode since p and q aren't givin exact? Do I choose a couple of functions to represent p and q and calculate the solution set of y1 and y2 and if this yields a wronskian different from zero then by magic their solution funct y1(x) and y2(x) are linear independent?
     
    Last edited: Aug 31, 2010
  7. Aug 31, 2010 #6

    vela

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    You can calculate the Wronskian

    [tex]W=y_1'y_2-y_1y_2'[/tex]

    using the product rule and the fundamental theorem of calculus to find the derivative of y2. It should simplify down to

    [tex]W=e^{-\int p(u) du}[/tex]

    You can only get W=1 if you can assume p(x)=0.
     
  8. Aug 31, 2010 #7
    thank you vela ;)

    What about my answer of question one. Have I assumed it correctly?
     
  9. Aug 31, 2010 #8

    vela

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    p(u) should be p(t) in the differential equation. You change the variable when you integrate.

    [tex]\int_a^t \left(\frac{v''(u)}{v'(u)} + 2\cdot \frac{y_1'(u)}{y_1(u)}\right)\,du = -\int_a^t p(u)\,du[/tex]

    which gives you

    [tex]\ln\left|\frac{w(t) y_1^2(t)}{w(a)y_1^2(a)}\right| = - \int_a^t p(u) du[/tex]

    The stuff that depends on the lower limit a are just constants; they correspond to what you called c1. As you noted, you can always rescale y1 to make it equal to 1.
     
  10. Aug 31, 2010 #9
    Okay but besides from the this it looks okay? :)

    But am I correct to assume that what I do do obtain y2 is to define p(u) over a subinterval of (x0,x)??
     
  11. Aug 31, 2010 #10

    vela

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    It looks fine to me. I'm not sure what you're asking about defining p(x). The function has to be defined on I.
     
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