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Title: Uniqueness for Solutions
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Series: Ordinary Differential Equations
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YouTube-Title: Ordinary Differential Equations 10 | Uniqueness for Solutions
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Forum: Ask a question in Mattermost
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Quiz: Test your knowledge
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Subtitle on GitHub: ode10_sub_eng.srt missing
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Timestamps
00:00 Introduction
00:58 Initial value problem
01:42 Uniqueness of solutions under local Lipschitz condition
02:55 Proof of Theorem
04:22 Put x0 where solutions diverge
05:13 Define difference vector
08:15 Correction: see details about Lipschitz constant in the PDF version
09:53 Make epsilon small enough
11:06 Distinct solutions for the initial value problem don’t exist
12:02 Credits
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Subtitle in English (n/a)
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Quiz Content
Q1: Let $v: \mathbb{R}^n \rightarrow \mathbb{R}^n$ be a locally Lipschitz continuous function. What can we say about the initial value problem
$$ \dot{x} = v(x) \,, ~ x(0) = x_0$$A1: It has at most one solution.
A2: Two different solutions can exist, but then the orbits don’t cross.
A3: There exists at least one solution, but uniqueness is not given in general.
Q2: Consider the initial value problem
$$ \dot{x} = v(x) \,, ~ x(0) = x_0\,.$$What does it mean that we two distinct solution $\alpha_1$ and $\alpha_2$?
A1: There is $\varepsilon > 0$ such that $\alpha_1(\varepsilon) \neq \alpha_2(\varepsilon)$.
A2: We have $\alpha_1(0) \neq \alpha_2(0)$.
A3: There is $\varepsilon > 0$ such that $\alpha_1(\varepsilon) =\alpha_2(\varepsilon)$.
A4: We have $\alpha_1(0) \neq \alpha_2(\varepsilon)$ for all $\varepsilon > 0$.
Q3: What is correct if $v: \mathbb{R}^n \rightarrow \mathbb{R}^n$ is locally Lipschitz continuous at $x_0$?
A1: There is $\delta > 0$ and $L>0$ such that for all $y,z \in B_{\delta}(x_0)$: $\| v(y) - v(z) \| \leq L \| y - z \|$.
A2: There is $\delta > 0$ such that for all $L>0$ there are $y,z \in B_{\delta}(x_0)$: $\| v(y) - v(z) \| \leq L \| y - z \|$.
A3: For all $\delta > 0$ there is $L>0$ such that for all $y,z \in B_{\delta}(x_0)$: $\| v(y) - v(z) \| \leq L \| y - z \|$.
A4: For all $\delta > 0$ there is $L>0$ such that for all $y,z \in B_{\delta}(x_0)$: $\| v(y) - v(z) \| > L \| y - z \|$.
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Last update: 2026-07
