Let $\mathbf{v}:(a,b)\to\mathbb{R}^2$ be a given continuous function and $t_0\in (a,b)$ a fixed point. Is it true that the following problem has a unique continuous solution $\mathbf{r}:(a,b)\to\mathbb{R}^2$:

$\left\{\begin{array}{ll}\mathbf{v}(t)=\mathbf{r}(t)+\displaystyle\lim_{s\searrow t} \frac{\mathbf{r}(s)-\mathbf{r}(t)}{||\mathbf{r}(s)-\mathbf{r}(t)||}\\ \mathbf{r}(t_0)=\mathbf{v}(t_0)+(\cos\theta_0,\sin\theta_0)\end{array}\right.$ ?


closed as off-topic by Ricardo Andrade, Alexandre Eremenko, Yemon Choi, S. Carnahan Nov 11 '14 at 23:06

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  • 1
    $\begingroup$ It's far from that one. Anyway, that question has not a complete answer... $\endgroup$ – Bogdan Nov 11 '14 at 14:10
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    $\begingroup$ To make sense of the equation classically, $r$ has to be pointwise differentiable and by the continuity assumptions $C^1$. Your equation is a linear ODE, which can be solved explicitly (componentwise if you wish) and uniqueness is guaranteed by standard theory. $\endgroup$ – Joonas Ilmavirta Nov 11 '14 at 15:15
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    $\begingroup$ To justify Joonas' statement that $\mathbf{r}$ is $C^1$: see en.wikipedia.org/wiki/Semi-differentiability#Application $\endgroup$ – Ricardo Andrade Nov 11 '14 at 15:29
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    $\begingroup$ Why do you need $||$ in the denominator if $s>t$? If written in ordinary notation what you are asking is to solve the differential equation $r'+r=v$ with an initial condition. As the equation is linear, a unique solution always exists. $\endgroup$ – Alexandre Eremenko Nov 11 '14 at 21:52
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    $\begingroup$ @Bogdan: the denominator $\|\mathbf{s}-\mathbf{t}\|$ in your equation seems to refer to a vector quantity (norm and boldface); however s and t are real variables. Isn't that it was a typo for $\|\mathbf{r}(s)-\mathbf{r}(t)\|$? $\endgroup$ – Pietro Majer Nov 11 '14 at 23:14