# Existence and Uniqueness of Solution to an ODE

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.

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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.

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