What Are the Conditions for Uniqueness in Nonlinear Differential Systems?

[LeBrad]

I am familiar with the existence and uniqueness of solutions to the system

\dot{x} = f(x)

requiring f(x) to be Lipschitz continuous, but I am wondering what the conditions are for the system

\dot{q}(x) = f(x).

It seems like I could make the same argument for there existing a unique q(x) provided f(x) is Lipschitz with respect to q(x). Then if q(x) is invertible or one-to-one or whatever the proper math term is, then I can get a unique x. Is that correct? If that's the case it looks like I'm just doing a nonlinear change of coordinates, showing uniqueness in that coordinate system, and then having a unique map back to the original coordinate system.

 

[LeBrad]

Ok, since nobody complained I'm going to assume what I said is correct. In that case, I want to show Lipschitzness of f(x) with respect to q(x). If I define \tilde{x} = q(x) and assume f is Lipschitz with respect to x, then

||f(x_1)-f(x_2)||\leq L ||x_1 - x_2|| = L ||q^{-1}(\tilde{x}_1) - q^{-1}(\tilde{x}_2)||.

So if q^{-1}(\tilde{x}) is Lipschitz with respect to \tilde{x}[/itex],<br /> <br /> ||q^{-1}(\tilde{x}_1) - q^{-1}(\tilde{x}_2)|| \leq M||\tilde{x}_1 - \tilde{x}_2||,<br /> <br /> then <br /> <br /> ||f(x_1)-f(x_2)||\leq LM ||\tilde{x}_1 - \tilde{x}_2||.<br /> <br /> So it seems it is sufficient to show that f(x) is Lipschitz with respect to x and that q^{-1}(\tilde{x}) is Lipschitz with respect to \tilde{x}[/itex]. Does that look correct?

 

[trambolin]

Keep in mind that

\dot{q}(x) = \frac{\partial q}{\partial x}\dot{x}= f(x)

If nonzero or invertible in general (otherwise you have what is called a singular or descriptor system), \frac{\partial q}{\partial x} is also a function of x might be carried to the other side and you have another \dot{x} = \hat{f}(x)

 

[LeBrad]

Yeah, I know I can do that, but I was trying to keep that as a last resort. I have reason to keep it in the form
\dot{q}(x) = f(x)
if possible.

 

[trambolin]

Yes, but proving if the \hat{f}(x) is Lipschitz, is much more easier. Then you can say, OK now we multiply the differential equation from the left with some non-vanishing function h(x) and then take
h(x)=\frac{\partial g}{\partial x}

What I am trying to say is you have a point there, but it does not bring much difference into the problem nature. But, if you can prove that without inverting the function, then you have a nice result. Such as analyzing the properties of the linear singular system

E \dot{x} = Ax

where E is not invertible. People usually dive into the problem by saying that the pencil \lambda E - A is regular, does not have impulsive modes etc. You will definitely need some more assumptions to handle that issue when it becomes a general nonlinear differential system.