
#1
Feb1508, 08:49 AM

P: 212

I am familiar with the existence and uniqueness of solutions to the system
[tex] \dot{x} = f(x) [/tex] requiring [tex]f(x)[/tex] to be Lipschitz continuous, but I am wondering what the conditions are for the system [tex] \dot{q}(x) = f(x) [/tex]. It seems like I could make the same argument for there existing a unique [tex]q(x)[/tex] provided [tex]f(x)[/tex] is Lipschitz with respect to [tex]q(x)[/tex]. Then if [tex]q(x)[/tex] is invertible or onetoone or whatever the proper math term is, then I can get a unique [tex]x[/tex]. 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. 



#2
Feb1808, 10:44 AM

P: 212

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



#3
Feb1808, 12:03 PM

P: 341

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



#4
Feb1808, 03:12 PM

P: 212

Existence and Uniqueness
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
[tex] \dot{q}(x) = f(x) [/tex] if possible. 



#5
Feb1908, 11:41 AM

P: 341

Yes, but proving if the [tex] \hat{f}(x) [/tex] is Lipschitz, is much more easier. Then you can say, OK now we multiply the differential equation from the left with some nonvanishing function [tex]h(x)[/tex] and then take
[tex]h(x)=\frac{\partial g}{\partial x}[/tex] 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 [tex] E \dot{x} = Ax[/tex] where E is not invertible. People usually dive into the problem by saying that the pencil [tex]\lambda E  A[/tex] 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. 


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