1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Linear homoheneous ODE question

  1. Jun 19, 2010 #1
    1. The problem statement, all variables and given/known data

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

    p(x), q(x) continuous for x = (-1,1)


    2. Relevant equations

    I. prove that if y1 is an answer to the ODE and y1(0) = y'1(0) = 0.
    then: y1 = 0 for all x = (-1,1).

    II. if y1 is an answer to the ODE, how would you find the second non-dependent answer to the ODE y2?

    III. prove that the two solutions you got, y1 and y2 are independent.

    3. The attempt at a solution

    Hello!

    I just cant wrap my head around this question.
    i know i can answer II if i knew what p(x) and q(x) were.

    general answer:
    y(x)=A*y1+B*y2
    y'(x)=A*y'1+B*y'2
    y''(x)=A*y''1+B*y''2

    A,B=const.

    if I insert the above in the ODE i get:
    y2'' + p(0)y2' + q(0)y2 = 0

    also i can prove that they are linearly independent by the Wronskian(y1,y2)(x)

    can someone please help me with the idea behind answering I. ? :)
     
    Last edited: Jun 19, 2010
  2. jcsd
  3. Jun 19, 2010 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Since p and q are continuous on (-1, 1), this equation satifies the conditions for the "existance and uniqueness" theorem- there exist only one function satifying this equation with taking specfic values of y(0) and y'(0).

    It is clear that y(x)= 0 for all x in (-1, 1) satisfies the the equation and, certainly y(0)= 0, y'(0)= 0. Since that solution is unique, y1(x)= y(x)= 0.

     
  4. Jun 19, 2010 #3
    thank you for your reply.

    the part i don't understand is how you extrapolate from knowing the answer at a single point (x=0) to knowing that y1=0 for all x (-1,1) ?

    of course it's known that in a linear equation always exists a trivial answer y(x)=0, but its only one answer, how can it help me to determine that y1=0 for all x=(-1,1)?

    also how can i find p(0) and q(0), if i can ?
     
  5. Jun 19, 2010 #4

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Did you read the first paragraph of HallsofIvy's post?

    In your problem, not only do you know y=0 is a solution that satisfies the initial conditions, you know it is the only solution that does.
     
  6. Jun 20, 2010 #5
    again, I don't understand how knowing that if y=0 if a single solution to the ODE, can i know that y1=0 for all x (-1,1)?

    if its so then my answer y(x) to the ODE consists only of one solution C*y2(x) because the other linearly independent solution y1(x)=0, but it couldn't be true because i need to have 2 linearly independent solutions to answer an linear ODE.

    also if you see what's my problem is can you please explain it to me, clearly i don't understand what's going on, so don't assume i know the answer :)
     
  7. Jun 20, 2010 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Do you know "unique" means? Do you know what the "existence and uniqueness theorem" is?

    And no one said that "the other linearly independent solution y1(x)=0". y(x)= 0 is always a solution to a linear homogeneous differential equation. The two "linearly independent solutions" must be non-zero solutions since any set of vectors containing the 0 vector is NOT linearly independent.
     
  8. Jun 20, 2010 #7
    This was the question:

    I. prove that if y1 is an answer to the ODE and y1(0) = y'1(0) = 0.
    then: y1 = 0 for all x = (-1,1).

    can someone non-cryptically explain why it is so? :)

    HallsofIvy thank you for your response but its a semi-theoretical question that I'm trying to understand, it's not homework. also I'm done with my ODE course

    and if anyone is kind enough i would love to have a full explanation.
     
  9. Jun 20, 2010 #8

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You're confusing the solutions to the differential equation and the solution to the initial value problem. There is only one solution that satisfies the differential equation and the initial conditions.
     
  10. Jun 20, 2010 #9

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    So what you are really asking for is a proof of the "existence and uniqueness" theorem! You would start by proving:

    If f(x,y) is continuous in both variables and "Lipschitz" in y (differentiable with respect to y is sufficient but not necessary) in some neighborhood of [itex](x_0, y_0)[/itex], then there exist a unique function y(x) satisfying dy/dx= f(x,y) with [itex]y(x_0)= y_0[/itex].

    That's a fairly complicated theorem in its own right but any good differential equations text book should give it. Look for "existence" or "uniqueness" in the index.

    To extend that to second order equations, you do this:

    Let u= y' so that u'= y" and the equation y"+ p(x)y'+ q(x)y= 0 becomes u'+ p(x)u+ q(x)y= 0 or u'= -p(x)u- q(x)y. Together with y'= u that gives you two linked first order differential equations. Now Let
    [tex]Y(x)= \begin{pmatrix}u(x) \\ y(x)\end{pmatrix}[/tex]
    so we can write
    [tex]\begin{pmatrix}u'(x)\\ y'(x)\end{pmatrix}= Y'= \begin{pmatrix}-p(x) & -q(x) \\1 & 0\end{pmatrix}\begin{pmatrix}u \\ y\end{pmatrix}= AY[/tex]
    so the second order differential equation becomes a first derivative equation in the vector function Y(x).

    As long as p(x) and q(x) are differentiable, this equation satisfies the existence and uniqueness theorem above (of course you have to prove that you can extend it to vector functions- that's not hard). The condition that y(0)= 0, y'(0)= u(0)= 0 becomes
    [tex]Y(0)= \begin{pmatrix}u(0) \\ y(0)\end{pmatrix}= \begin{pmatrix}0 \\ 0 \end{pmatrix}[/tex].

    Since the function y(x)= 0 for all x satisfies y''+ p(x)y'+ q(x)y= 0 as well as y(0)= 0, y'(0)= 0, it follows from the "uniquness" that it is the only function that satisfies that equation with those conditions.

    Of more interest, in my opinion, is the fact that there exist a unique function, say [itex]y_1(x)[/itex], satisfying that equation and the conditions [itex]y_1(0)= 1[/itex], [itex]y_1'(0)= 0[/itex] and there exist a unique function, [itex]y_2(x)[/itex], satisfying the equation and [itex]y_2(0)= 0[/itex], [itex]y_2'(0)= 1[/itex].

    Suppose y is any equation satisfying the differential equation. Then y(0) is some number, say A, and y'(0) is some number, say B. It follows that [itex]y(x)= Ay_1(x)+ By_2(x)[/itex]! Why? Because [itex]y'(x)= Ay_1'(x)+ By_2'(x)[/itex] and [itex]y"(x)= Ay_1"(x)+ By_2"(x)[/itex] so that [itex]y"+ p(x)y'+ q(x)y= (Ay_1"+ By_2")+ p(x)(Ay_1'+ By_2')+ q(x)(Ay_1+ By_2)[/itex][itex]= A(y_1"+ p(x)y_1'+ q(x)y_1)+ B(y_2"+ p(x)y_1'+ q(x)y_2)= A(0)+ B(0)= 0[/itex] while [itex]Ay_1(0)+ By_2(0)= A(1)+ B(0)= A[/itex] and [itex]Ay_1'(0)+ By_2'(0)= A(0)+ B(1)= B[/itex].

    That is, y(x) and [itex]Ay_1(x)+ By_2(x)[/itex] satisfy the same differential equation as well as the same "initial conditions" and so, by the uniqueness property, are equal.

    Further, it is easy to show that [itex]y_1(x)[/itex] and [itex]y_2(x)[/itex] are "independent": if [itex]Ay_1(x)+ By_2(x)= 0[/itex] (meaning equal to 0 for all x), then, taking x= 0, A(1)+ B(0)= 0 so A= 0. Differentiating both sides, [itex]Ay_1'(x)+ By_2'(x)= 0[/itex] and again taking x= 0, A(0)+ B(1)= 0 so B= 0.

    That tells us that [itex]y_1(x)[/itex] and [itex]y_2(x)[/itex] for a basis for the set of all solutions to that equation and so the set of all solutions is a vector space of dimension 2.

    That can be extended to show that the set of all solutions to an [itex]n^{th}[/itex] order, linear, homogeneous differential equation is a vector space of dimension n.
     
    Last edited: Jun 20, 2010
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Linear homoheneous ODE question
Loading...