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x'=A(t)x(t)+B(x,t)u(t) (1)

x, t belongs to R^n and R+ respectively. u(t) is a function of t.

I am required to find the sufficient conditions for A(t), f(x), B(x,t) and u(t)

for the (1) to have a existence of a unique solution.

In the lecture notes, the prof lists 4 conditions to be sufficient conditions for the

existence of a unique solution for x'=f(x,t):

1. for a fixed t, f(x) is continous.

2. There exists a set S contained in R+ containing at most a finite

number of points per unit interval. S will depend on f(x,t) and will denote

possible discontinuity points. With this set, for x is fixed, f(t) will be discontinuous only when t belongs to S.

3. If x(t) is continous, then f(x,t) is piecewise continous in t

4.satisifies a global Lipschitz condition.

Thus my solution is as follows:

1. for a fixed t: B(x,t) are continuous.

2. for a fixed x, A(t), u(t), B(x,t) are discontinuous only in S

3. A(t)x(t)+B(x,t)u(t) is picewise continuous in t.

4. I derive the global Lipschitz condition:

||A(x1-x2)+U(B(x1)-B(x2))||<=|A|*||(x1-x2)||+|u|*||B(x1)-B(x2)||

so if ||B(x1)-B(x2)||<=K(t)||x1-x2|| then ||A(x1-x2)+U(B(x1)-B(x2))||<=K1||x1-x2||

with k1=max(|A|, |k*u|).

So the fourth condition will be B(x,t) satisfies the Lipschitz condition.

I am not sure if the conditions can be narrower? can you please verify my solution.

Thanks