# Existence and Uniqueness

 P: 212 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.
 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 $\tilde{x} = q(x)$ 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 $$\tilde{x}[/itex], [tex] ||q^{-1}(\tilde{x}_1) - q^{-1}(\tilde{x}_2)|| \leq M||\tilde{x}_1 - \tilde{x}_2||$$, then $$||f(x_1)-f(x_2)||\leq LM ||\tilde{x}_1 - \tilde{x}_2||$$. 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 $$\tilde{x}[/itex]. Does that look correct?  P: 341 Keep in mind that [tex]\dot{q}(x) = \frac{\partial q}{\partial x}\dot{x}= f(x)$$ 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 $$\dot{x} = \hat{f}(x)$$
 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 $$\dot{q}(x) = f(x)$$ if possible.
 P: 341 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.