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Nth order linear ode, why do we have n general solutions?

  1. Jun 5, 2015 #1
    hi, I looked up the existence and uniqueness of nth order linear ode and I grasped the idea of them, but still kind of confused why we get n numbers of general solutions.
     
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  3. Jun 5, 2015 #2

    SteamKing

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  4. Jun 7, 2015 #3

    HallsofIvy

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    A basic property of linear homogeneous equations is that the set of solutions forms a vector space. That is, any linear combination of solutions, [itex]a_1y_1+ a_2y_2[/itex] is again a solution. One can show that, for an nth order homogeneous differential equation, this vector space has dimension n. That is, there exist n independent solutions such that any solution can be written in terms of those n solutions.

    To prove that, consider the differential equation [itex]a_n(x)y^{(n)}(x)+ a_{n-1}y^{(n-1)}(x)+ \cdot\cdot\cdot+ a_2y''(x)+ a_1y'(x)+ a_0y(x)= 0[/itex].
    with the following initial values:

    I) [itex]y(0)= 1[/itex], [itex]y'(0)= y''(0)= \cdot\cdot\cdot= y^{(n-1)}(0)= 0[/itex]
    By the fundamental "existence and uniqueness" theorem for initial value problems, there exist a unique function, [itex]y_0(x)[/itex], satisfying the differential equation and those initial conditions.

    II) [itex]y(0)= 0[/itex], [itex]y'(0)= 1[/itex], [itex]y''(0)= y'''(0)= \cdot\cdot\cdot= y^{(n-1)}(0)= 0[/itex]
    Again, there exist a unique solution, [itex]y_1(x)[/itex], satisfying the differential equation and those initial conditions.

    III) [itex]y(0)= y'(0)= 0[/itex], [itex]y''(0)= 1[/itex], [itex]y'''(0)= \cdot\cdot\cdot= y^{(n-1)}(0)= 0[/itex]
    Again, there exist a unique solution, [itex]y_2(x)[/itex], satisfying the differential equation and those initial conditions.

    Continue that, shifting the "= 1" through the derivatives until we get to
    X) [itex]y(0)= y'(0)= \cdot\cdot\cdot= y^{(n-1)}(0)= 0[/itex], [itex]y^{(n-1)}(0)= 1[/itex].

    First, any solution to the differential equation can written as a linear combination of those n solutions:
    Suppose y(x) is a solution to the differential equation. Let [itex]A_0= y(0)[/itex], [itex]A_1= y'(0)[/itex], etc. until [itex]A_{n-1}= y^{(n-1)}(0)[/itex].
    Then [itex]y(x)= A_0y_0(x)+ A_1y_1(x)+ \cdot\cdot\cdot+ A_{n-1}y_{n-1}(x)[/itex]. That can be shown by evaluating both sides, and their derivatives, at x= 0.

    Further, that set of n solutions are independent. To see that suppose that, for some numbers, [itex]A_0, A_1, \cdot\cdot\cdot, A_{n-1}[/itex], [itex]A_0y_0(x)+ A_1y_1(x)+ \cdot\cdot\cdot+ A_{n-1}y_{n-1}(x)= 0[/itex]- that is, is equal to 0 for all x. Taking x= 0 we must have [itex]A_0(1)+ A_1(0)+ \cdot\cdot\cdot+ A_{n-1}(0)= A_0= 0[/itex]. Since that linear combination is 0 for all x, it is a constant and its derivative is 0 for all x. That is [itex]A_0y_0'(x)+ A_1y_1'(0)+ \cdot\cdot\cdot+ A_{n-1}y'(0)= 0[/itex] for all x. Set x= 0 to see that [itex]A_1= 0[/tex]. Continue taking derivatives and setting x= 0 to see that each [itex]A_i[/itex], in turn, is equal to 0.
     
  5. Jun 8, 2015 #4


    Thanks! it helped alot!
     
  6. Jun 17, 2015 #5

    diff14.png (5)
    diff27.png (17)

    - using picard's iteration in vector form, to prove nth order linear ODE's existence & uniqueness.

    ex
    diff31.png (21)

    diff32.png (22)

    (http://ghebook.blogspot.ca/2011/10/differential-equation.html)


    Hi, I actually did picard's iteration and found that without n initial conditions, nth order linear ODE will have n number of constants as we assume initial conditions are some arbitrary constants.

    since function spaces are vector spaces, solutions span n dimensional vector space (not very sure of this)

    If we do picard's iteration,

    y(n-1)=y0(n-1)+∫y(n)dx
    y(n-2)=y0(n-2)+∫y(n-1)dx
    =y(n-2)=y0(n-2)+∫y0(n-1)+∫y(n)dx
    .
    .
    .
    iteration goes on and on until the error is sufficiently decreased.


    if we assume each initial conditions are some constants, we will eventually sort out the solution function y w.r.t constants after sufficient number of iteration is done. Then it will look like,

    y (c0 c1 c2 - - - - - - )[y1 y2 y3 y4 y5 - - - - - - ] <-- (should be vertical)


    which is in the form of y = c1y1+c2y2+c3y3 . . .

    and so on
    is this correct?
    If not, how can we show that solution space has n number of basis?


    ---------------------------------------------------------------------------------------------------------------------------

    Also, I have two questions about Strum Liouville 2nd order ODE.


    1. if we look at the Strum-Liouville 2nd order ODEs, there is an eigenvalue term within the equation. it seems like we are adding one more constant in the equation, which imposes a restriction to find a solution (n+1 constants with n conditions).

    [m(x)y']+[λr(x)-q(x)]y
    =m(x)[y''+P(x)y'+Q(x)y]
    =m(x)[y''+P(x)y'+(λr(x)-q(x))]
    =0

    ∴ [y''+P(x)y'+(λr(x)-q(x))]=0

    if we do Picard's iteration, then we have one more constant λ along with n constants..


    2. I don't understand how eigenvalue directly influence the solutions of 2nd order ODE


    Thanks always, Your answers help me a lot!
     
    Last edited: Jun 17, 2015
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