Find domain where function is Lipschitz

  • #1
psie
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Homework Statement
Reduce the ODEs ##x'''+x^2=1, x''=x^{-1/2}## and ##x''=\sqrt{1+(x')^2}## to a system of first order and find a naturally defined region ##\Omega## where the right hand side satisfies a Lipschitz condition.
Relevant Equations
##f## satisfies a Lipschitz condition in the ##x##-variable in a set ##\Omega## if ##\lVert f(t,x)-f(t,y)\rVert\leq L\lVert x-y\rVert##, for some positive constant ##L##.
The reduction is simple in all cases. For the first one, put ##x_1=x, x_2=x'## and ##x_3=x''##. Let ##\pmb{x}=(x_1,x_2,x_3)##. Then we get $$\pmb{x}'= \begin{pmatrix}x_1' \\ x_2' \\ x_3' \end{pmatrix}=\begin{pmatrix}x_2 \\ x_3 \\ 1-x_1^2 \end{pmatrix}=\pmb{f}(\pmb{x}),$$ where ##\pmb{f}(\pmb{x})=(f_1(\pmb{x}),f_2(\pmb{x}),f_3(\pmb{x}))=(x_2,x_3,1-x_1^2)##.

Similarly, ##\pmb{g}(\pmb{x})=(x_2,x_1^{-1/2})## and ##\pmb{h}(\pmb{x})=(x_2,\sqrt{1+(x_2)^2})## for the other two ODEs.

In the first case, I'm interested in finding a subset of ##\mathbb R^3## such that I can bound ##\lVert \pmb{f}(\pmb{x})-\pmb{f}(\pmb{x})\rVert##. I'm unsure how to approach this in all cases, whether to use the definition of some norm directly or the mean value theorem. In the latter case, I'm unsure how the mean value theorem applies to vector-valued functions of a vector. Anyway, grateful for any help.
 
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  • #2
Perhaps start by calculating [itex]f(x) - f(y)[/itex], and see if you can write [tex]
\|f(x) - f(y)\|^2 = A(x,y)|x_1 - y_1|^2 + B(x,y)|x_2 - y_2|^2 + C(x,y)|x_3 - y_3|^2[/tex] for positive functions [itex]A[/itex], [itex]B[/itex] and [itex]C[/itex]. How can you then guarantee that [tex]
\|f(x) - f(y)\|^2 \leq L^2\|x - y\|^2[/tex] for some [itex]L > 0[/itex]?
 
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  • #3
Good idea, however, when it comes to ##\pmb{h}(\pmb{x})=(x_2,\sqrt{1+(x_2)^2})##, there is no ##x_1## coordinate included in the components. Maybe this is not a problem. Using the Euclidean norm: $$\lVert \pmb{h}(\pmb{x})-\pmb{h}(\pmb{y})\rVert^2=(x_2-y_2)^2+\left(\sqrt{1+x_2^2}-\sqrt{1+y_2^2}\right)^2 $$ How can I handle the second term, i.e. ##\left(\sqrt{1+x_2^2}-\sqrt{1+y_2^2}\right)^2##, so that it potentially doesn't cause any trouble?

As an alternative approach, I think a vector-valued function is Lipschitz iff all of its components are. Therefor we can apply the mean value theorem to the components, which is fairly simple in this case.
 

1. What is the definition of Lipschitz continuity?

Lipschitz continuity is a mathematical property of a function where the rate of change of the function is bounded by a constant. In other words, the function cannot have a slope that is too steep or too flat between any two points on its graph.

2. How is Lipschitz continuity different from other types of continuity?

Lipschitz continuity is a stronger form of continuity compared to other types, such as uniform continuity. It requires the function to have a bounded rate of change, whereas other types only require the function to have a small enough change in its output when the input is close enough.

3. Why is it important to find the domain where a function is Lipschitz?

Finding the domain where a function is Lipschitz can help determine the stability and convergence of numerical methods used to solve equations involving the function. It can also provide insight into the behavior of the function and its graph.

4. How can I determine if a function is Lipschitz on a given domain?

To determine if a function is Lipschitz on a given domain, you can use the definition of Lipschitz continuity and check if the function satisfies the condition of having a bounded rate of change. You can also use the Mean Value Theorem to check for Lipschitz continuity.

5. Can a function be Lipschitz on one domain but not on another?

Yes, a function can be Lipschitz on one domain but not on another. Lipschitz continuity is a local property, meaning it only needs to hold on a small interval of the function's domain. Therefore, a function can be Lipschitz on one domain but not on another if the rate of change of the function is bounded on one domain but not on the other.

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