Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How do i prove this theorem?

  1. Mar 30, 2008 #1
    Theorem: Existence and Uniqueness

    Let p(t), q(t), and g(t) be continuous on an interval I, then the differential equation

    [tex] y'' + p(t)y' + q(t)y = g(t) \ \ \ \ , y(t_0) = b_0 \ \ \ , y'(t_0) = b_1 [/tex]

    has a unique solution defined for all t in I.

    I have no idea where to start??
  2. jcsd
  3. Mar 30, 2008 #2


    User Avatar
    Science Advisor

    How you would do that depends upon where you are allowed to start.

    Typically, one first proves the basic "existence and uniqueness" theorem for first order equations: If, f(t,y) is continuous in both variables and Lipschitz in y in some neighborhood of (t0,y0), then the differential equation dy/dt= f(t,y) with initial condition y(t0)= y0 has a unique solution in that neighborhood. That can be proven using the Banach fixed point principle. That's the hard part!

    In fact, the Banach fixed point principle is true in Rn, not just R, so the existence and uniqueness theorem is true for y(x) vector valued as well as numeric.

    For your problem, let x= y'(t). Then y"= x' so your differential equation becomes x'+ p(t)x+ q(t)y= g(t) or x'= -p(t)x- q(t)y- g(t). Your second order differential equation is now a system of two first order equations: x'= -p(t)x- q(t)y+ g(t) and y'= x.

    Now let
    [tex]Y(t)= \left(\begin{array}{c}x(t) \\ y(t)\end{array}\right)[/tex]
    and you can state the system of equations as a single vector equation:
    [tex]\frac{dY}{dt}= \left(\begin{array}{c}dx/dt \\ dy/dt\end{array}\right)= \left(\begin{array}{cc}-p(t) & -q(t) \\ 1 & 0\end{array}\right)Y+ \left(\begin{array}{c} q(t) \\ 0\end{array}\right)[/tex]

    [tex]P(t)= \left(\begin{array}{cc}-p(t) & -q(t) \\ 1 & 0\end{array}\right)[/tex]
    [tex]Q(t)= \left(\begin{array}{c} q(t) \\ 0\end{array}\right)[/tex]
    that can be written dY/dt= P(t)Y+ Q(t). Since that is trivially continuous in both t and Y and differentiable (and so Lipschitz) with respect to y for all t and y, it has a unique solution for any initial value
    [tex]Y(t_0)= Y_0= \left(\begin{array}{c}b_1 \\ b_0\end{array}\right)[/tex]

    Since that is a linear differential equation, it is easy to show that the set of all solutions (to the homogeneous equation, dropping Q(t)) forms a vector space. Then we can show that the two solutions satisfying
    [tex]Y_0= \left(\begin{array}{c}1 \\ 0\end{array}\right)[/tex]
    [tex]Y_0= \left(\begin{array}{c}0 \\ 1\end{array}\right)[/tex]
    both span the solution space and are independent- so the set of all solutions to a second order linear differential equation is a two dimensional vector space.
    Last edited by a moderator: Mar 30, 2008
  4. Mar 30, 2008 #3
    thanks Halls. It looks a lill bit out of my domain for the moment, but i think i'll be fine,i'll try to understand it.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook