Existance of unique solutions for differential equation

[fortune]

Hi guys,

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 continuous 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

 

[Nino MMP]

!</code>Yes, your solution is correct. The conditions are sufficient for the existence of a unique solution to the differential equation. However, there may be other conditions that can be used to make the solution more narrow. For instance, you could add a condition that requires A(t) and B(x,t) to be differentiable, or that the matrix A(t) must be nonsingular for all t. These additional conditions will depend on the specific problem you are trying to solve.