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.
 

FAQ: Find domain where function is Lipschitz

What does it mean for a function to be Lipschitz?

A function \( f \) is said to be Lipschitz continuous if there exists a constant \( L \geq 0 \) such that for all pairs of points \( x \) and \( y \) in its domain, the inequality \( |f(x) - f(y)| \leq L|x - y| \) holds. The smallest such \( L \) is called the Lipschitz constant.

How do you determine if a function is Lipschitz on a given domain?

To determine if a function is Lipschitz on a given domain, you need to find a Lipschitz constant \( L \) such that the inequality \( |f(x) - f(y)| \leq L|x - y| \) holds for all \( x \) and \( y \) in that domain. This often involves calculating or estimating the derivative of the function and ensuring that it is bounded.

What is the relationship between differentiability and Lipschitz continuity?

If a function \( f \) is differentiable on a closed interval and its derivative \( f' \) is bounded on that interval, then \( f \) is Lipschitz continuous on that interval. Specifically, if \( |f'(x)| \leq M \) for all \( x \) in the interval, then \( f \) is Lipschitz with Lipschitz constant \( L = M \).

Can a function be Lipschitz but not differentiable?

Yes, a function can be Lipschitz continuous but not differentiable. A classic example is the absolute value function \( f(x) = |x| \), which is Lipschitz continuous with Lipschitz constant \( L = 1 \) but is not differentiable at \( x = 0 \).

How is the domain of Lipschitz continuity related to the function's behavior at the boundaries?

The domain of Lipschitz continuity is often influenced by the behavior of the function at the boundaries. If the function or its derivative becomes unbounded near the boundaries, the function may not be Lipschitz continuous on the entire domain. In such cases, the domain can be restricted to exclude problematic points or intervals where the Lipschitz condition fails.

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