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    Existence and Uniqueness of Solution to an ODE

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    See the attached file.
    Express the 2nd order ODE
    d_t^2 u=(d^2 u)/(dt)^2 =sin?(u)+cos?(Ï?t) Ï? Z/{0}
    u(0)=a
    d_t u(0)=b
    as a system of 1st order ODEs and verify that there exists a global solution by invoking the global existence and uniqueness Theorem.

    Useful information:
    Global existence and uniqueness Theorem:
    The ordinary differential equation
    d_t â-u=â-f(t,â-u (t))
    â-u (0)=â-u_0
    has a unique solution if â-fâ??C^0 (I)Ã-Lipschitz(L_â?? (R)), f is continuous with respect to 1st variable and Lipschitz with respect to 2nd variable.

    Lipschitz Continuity: A function g:Iâ?'R is Lipschitz continuous if â??Î?>0 such that
    â?-g(â-x)-g(â-y)â?-â?¤Î?â?-â-x-â-yâ?-â??â-x,â-yâ??I.

    NB: â- means vector value.

    © BrainMass Inc. brainmass.com October 5, 2022, 2:02 pm ad1c9bdddf
    https://brainmass.com/math/ordinary-differential-equations/existence-uniqueness-solution-ode-429916

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    SOLUTION This solution is FREE courtesy of BrainMass!

    To express the 2nd order ODE as a system of 1st order ODEs, we let v = u_t. Then we have

    u_t = v
    v_t = sin u + cos(omega t)

    with the initial conditions

    u(0) = a
    v(0) = b

    These ODEs may be written in vector form as

    (1) d_t w = f(t,w(t))

    where w(t) = (u(t),v(t)) and f(t,w(t)) = (v, sin u + cos(omega t)).

    To verify that there exists a global solution to (1), we apply the Global Existence and Uniqueness Theorem. According to this theorem, the ODE given by (1) has a unique solution if f is continuous with respect to t and if
    there exists a constant L > 0 such that for all vectors w1=(u1,v1) and w2=(u2,v2) in R^2 and for all t in R, we have

    (2) ||f(t, w2) - f(t, w1)|| <= L ||w2 - w1|| = L sqrt((u2 - u1)^2 + (v2 - v1)^2).

    Now

    ||f(t, w2) - f(t,w1)|| = ||(v2, sin u2 + cos(omega t)) - (v1, sin u1 + cos(omega t)||
    = sqrt((v2 - v1)^2 + (sin u2 - sin u1)^2)
    <= sqrt((v2 - v1)^2 + (u2 - u1)^2)

    so (2) holds with L = 1. Therefore, (1) has a unique solution, as well as our original 2nd order ODE.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com October 5, 2022, 2:02 pm ad1c9bdddf>
    https://brainmass.com/math/ordinary-differential-equations/existence-uniqueness-solution-ode-429916

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