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Existence and Uniqueness

by LeBrad
Tags: existence, uniqueness
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LeBrad
#1
Feb15-08, 08:49 AM
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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 one-to-one 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.
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LeBrad
#2
Feb18-08, 10:44 AM
LeBrad's Avatar
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?
trambolin
#3
Feb18-08, 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]

LeBrad
#4
Feb18-08, 03:12 PM
LeBrad's Avatar
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.
trambolin
#5
Feb19-08, 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 non-vanishing 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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